Step 1: Understanding the Question:

A chiral carbon atom is a carbon atom that is bonded to four different groups or atoms.

To find the number of chiral carbons, we must first determine the structural formula of 2-chloro-3,4,5-trimethylhexane.


Step 2: Key Formula or Approach:

Draw the structure:

The parent chain is hexane (\(C_{6}\)).

Substituents are at positions 2 (chloro), 3 (methyl), 4 (methyl), and 5 (methyl).

Structure: \( CH_{3} - CH(Cl) - CH(CH_{3}) - CH(CH_{3}) - CH(CH_{3}) - CH_{3} \).


Step 3: Detailed Explanation:

Let's analyze each carbon atom in the chain:

1. \( C_{1} \): Part of a methyl group (\( -CH_{3} \)), has three identical hydrogen atoms. Not chiral.

2. \( C_{2} \): Bonded to H, Cl, \( CH_{3} \), and the rest of the chain \( -CH(CH_{3})CH(CH_{3})CH(CH_{3})CH_{3} \). All four groups are different. Chiral.

3. \( C_{3} \): Bonded to H, \( CH_{3} \), \( -CH(Cl)CH_{3} \), and \( -CH(CH_{3})CH(CH_{3})CH_{3} \). All four groups are different. Chiral.

4. \( C_{4} \): Bonded to H, \( CH_{3} \), \( -CH(CH_{3})CH(Cl)CH_{3} \), and \( -CH(CH_{3})_{2} \). All four groups are different. Chiral.

5. \( C_{5} \): Bonded to H, \( CH_{3} \), \( -CH(CH_{3})CH(CH_{3})CH(Cl)CH_{3} \), and another \( CH_{3} \) (the \( C_{6} \) atom). Since it has two methyl groups, it is not chiral.

6. \( C_{6} \): Part of a methyl group. Not chiral.

Total chiral carbons = 3 (at positions 2, 3, and 4).


Step 4: Final Answer:

There are 3 chiral carbon atoms present in 2-chloro-3,4,5-trimethylhexane.
Quick Tip: Identify chirality by looking for "asymmetric" environments.
Remember that if a carbon is attached to two identical groups (like two methyls at the end of an isopropyl group), it cannot be chiral.

Step 1: Understanding the Question:

We are given the pH of a monoacidic base solution and its concentration. We need to find the percentage dissociation (\( \alpha \times 100 \)).


Step 2: Key Formula or Approach:

1. \( pOH = 14 - pH \)

2. \( [OH^{-}] = 10^{-pOH} \)

3. For a weak base, \( [OH^{-}] = C \alpha \), where \( C \) is concentration and \( \alpha \) is degree of dissociation.

4. Percentage dissociation = \( \alpha \times 100 \)


Step 3: Detailed Explanation:

Given: \( pH = 10 \), \( C = 0.01 M = 10^{-2} M \).

At 298 K:
\[ pOH = 14 - 10 = 4 \]
\[ [OH^{-}] = 10^{-pOH} = 10^{-4} M \]

Using the relation for degree of dissociation:
\[ \alpha = \frac{[OH^{-}]}{C} = \frac{10^{-4}}{10^{-2}} = 10^{-2} = 0.01 \]

Calculating percentage dissociation:
\[ Percentage dissociation = \alpha \times 100 = 0.01 \times 100 = 1% \]


Step 4: Final Answer:

The percentage dissociation of the monoacidic base is 1%.
Quick Tip: Always convert pH to pOH when dealing with basic solutions to find the hydroxide ion concentration.
Remember \( \alpha = \frac{Dissociated concentration}{Initial concentration} \).

Step 1: Understanding the Question:

The order of a reaction with respect to a specific reactant is the exponent of its concentration in the experimentally determined rate law equation.


Step 2: Key Formula or Approach:

General rate law expression: \( R = k[A]^{x}[B]^{y} \).

Here, order w.r.t. A is \( x \) and order w.r.t. B is \( y \).


Step 3: Detailed Explanation:

The given rate law is:
\[ R = k[NO_{2}]^{2} \]

This expression can be written as:
\[ R = k[NO_{2}]^{2}[CO]^{0} \]

Comparing this with the general form, we see that the power to which the concentration of CO is raised is 0.

Therefore, the order of the reaction with respect to CO is 0 (zero).


Step 4: Final Answer:

The order of the reaction with respect to CO is zero.
Quick Tip: If a reactant does not appear in the rate law equation, the reaction is zero-order with respect to that reactant.
The coefficients in the balanced chemical equation do not necessarily determine the order.

Step 1: Understanding the Question:

Interhalogen compounds exist in different physical states (gas, liquid, or solid) depending on their molecular weight and intermolecular forces.


Step 2: Detailed Explanation:

- \( ICl \): It is a ruby red solid at room temperature.

- \( ClF_{3} \): It is a colorless gas.

- \( BrF_{5} \): It is a colorless liquid at room temperature.

- \( IF_{7} \): It is a colorless gas.


Step 3: Final Answer:

Among the given options, \( BrF_{5} \) is the compound that exists in a liquid state at room temperature.
Quick Tip: Memorize the physical states of common interhalogen compounds:
Gases: \( ClF, ClF_{3}, BrF, IF_{7} \).
Liquids: \( BrF_{3}, BrF_{5}, IF_{5} \).
Solids: \( ICl, IBr, ICl_{3} \).

Step 1: Understanding the Question:

We need to determine the relationship between the cell potential (\( E_{cell} \)) and the standard cell potential (\( E^{\circ}_{cell} \)) for the given concentration cell conditions using the Nernst equation.


Step 2: Key Formula or Approach:

Cell reaction: \( Ni_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Ni^{2+}_{(aq)} + 2Ag_{(s)} \)

Number of electrons transferred, \( n = 2 \).

Nernst Equation:
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{n} \log \frac{[Ni^{2+}]}{[Ag^{+}]^{2}} \]


Step 3: Detailed Explanation:

Substitute the given concentrations: \( [Ni^{2+}] = 0.01 M \), \( [Ag^{+}] = 0.01 M \).
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{2} \log \frac{0.01}{(0.01)^{2}} \]
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{2} \log \frac{10^{-2}}{10^{-4}} \]
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{2} \log(10^{2}) \]

Since \( \log(10^{2}) = 2 \):
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{2} \times 2 \]
\[ E_{cell} = E^{\circ}_{cell} - 0.0592 V \]

This means \( E_{cell} \) is less than \( E^{\circ}_{cell} \) by 0.0592 V.


Step 4: Final Answer:

The true statement is that the emf is less than \( E^{\circ}_{cell} \) by 0.0592 V.
Quick Tip: Pay close attention to the stoichiometric coefficients in the Nernst equation log term, as they become powers of the concentrations.
For \( Ag^{+} \), the power is 2, which simplifies the log calculation in this specific case.

Step 1: Understanding the Question:

Effective Atomic Number (EAN) is the total number of electrons around the central metal atom/ion in a complex.


Step 2: Key Formula or Approach:
\[ EAN = Z - X + Y \]

Where:
\( Z \) = Atomic number of central metal atom
\( X \) = Oxidation state of central metal atom
\( Y \) = Total number of electrons donated by ligands


Step 3: Detailed Explanation:

For the complex \( [Co(NH_{3})_{6}]^{3+} \):

- Central metal is Cobalt (Co). Atomic number \( Z = 27 \).

- Let oxidation state be \( x \). \( x + 6(0) = +3 \Rightarrow x = +3 \). So, \( X = 3 \).

- There are 6 ammonia (\( NH_{3} \)) ligands. Each is a monodentate ligand donating 2 electrons.

- Total electrons donated \( Y = 6 \times 2 = 12 \).

Calculation:
\[ EAN = 27 - 3 + 12 \]
\[ EAN = 24 + 12 = 36 \]


Step 4: Final Answer:

The EAN of Co in the given complex is 36.
Quick Tip: Stable complexes often follow the EAN rule where the result equals the atomic number of the next noble gas (for Co, the next noble gas is Krypton, \( Z=36 \)).

Step 1: Understanding the Question:

Amino acids are classified as acidic, basic, or neutral based on the relative number of amino (\( -NH_{2} \)) and carboxyl (\( -COOH \)) groups in their structure.


Step 2: Detailed Explanation:

- Thr (Threonine): Contains one amino and one carboxyl group. It is a neutral amino acid.

- Trp (Tryptophan): Contains one amino and one carboxyl group. It is a neutral amino acid.

- His (Histidine): Contains more amino groups than carboxyl groups. It is a basic amino acid.

- Glu (Glutamic acid): Contains two carboxyl groups and one amino group. Due to the extra carboxyl group, it is an acidic amino acid.


Step 3: Final Answer:

Glu (Glutamic acid) is the acidic amino acid.
Quick Tip: Common acidic amino acids: Aspartic acid (Asp) and Glutamic acid (Glu).
Common basic amino acids: Lysine (Lys), Arginine (Arg), and Histidine (His).

Step 1: Understanding the Question:

The spin-only magnetic moment depends on the number of unpaired electrons in the metal ion.


Step 2: Key Formula or Approach:

1. Write electronic configuration of \( Mn \) and \( Mn^{2+} \).

2. Find number of unpaired electrons (\( n \)).

3. Use formula: \( \mu = \sqrt{n(n+2)} \) Bohr Magnetons (BM).


Step 3: Detailed Explanation:

Atomic number of \( Mn = 25 \).

Ground state electronic configuration: \( [Ar] 3d^{5} 4s^{2} \).

Configuration of \( Mn^{2+} \): \( [Ar] 3d^{5} \).

In the \( 3d \) subshell, there are 5 orbitals. According to Hund's rule, 5 electrons will occupy these singly.

So, number of unpaired electrons \( n = 5 \).

Calculating magnetic moment:
\[ \mu = \sqrt{5(5+2)} \]
\[ \mu = \sqrt{5 \times 7} = \sqrt{35} \approx 5.916 BM \]

Rounding to two decimal places, we get 5.92 BM.


Step 4: Final Answer:

The spin-only magnetic moment for \( Mn^{2+} \) is 5.92 BM.
Quick Tip: A quick shortcut: The magnetic moment value is always "\( n \).something" where \( n \) is the number of unpaired electrons.
For \( n=5 \), the answer must start with 5.

Step 1: Understanding the Question:

For a Body-Centered Cubic (BCC) unit cell, we need to find the edge length (\( a \)) from the atomic radius (\( r \)) and then calculate the volume (\( V = a^{3} \)).


Step 2: Key Formula or Approach:

Relationship between \( a \) and \( r \) in BCC: \( \sqrt{3} a = 4r \Rightarrow a = \frac{4r}{\sqrt{3}} \).

Volume of unit cell: \( V = a^{3} \).


Step 3: Detailed Explanation:

Given: \( r = 1.86 \times 10^{-8} cm \).

Calculate edge length \( a \):
\[ a = \frac{4 \times 1.86 \times 10^{-8}}{1.732} \]
\[ a = \frac{7.44}{1.732} \times 10^{-8} \approx 4.2956 \times 10^{-8} cm \]

Calculate volume \( V \):
\[ V = a^{3} = (4.2956 \times 10^{-8})^{3} \]
\[ V \approx 79.25 \times 10^{-24} cm^{3} \]
\[ V = 7.925 \times 10^{-23} cm^{3} \]

Matching with the closest given option, we get \( 7.951 \times 10^{-23} cm^{3} \).


Step 4: Final Answer:

The volume of the bcc unit cell is \( 7.951 \times 10^{-23} cm^{3} \).
Quick Tip: In BCC, atoms touch along the body diagonal. So \( 4r = Body diagonal = a\sqrt{3} \).
Always ensure units are consistent before calculating power of 3.

Step 1: Understanding the Question:

Cryoscopic constant (\( K_{f} \)) is the proportionality constant in the equation for depression in freezing point (\( \Delta T_{f} \)).


Step 2: Key Formula or Approach:
\[ \Delta T_{f} = K_{f} \cdot m \]

Where \( m \) is the molality of the solution.


Step 3: Detailed Explanation:

Given:
\( \Delta T_{f} = 1.8 K \)
\( m = 0.4 m \) (molal)

Rearranging the formula to find \( K_{f} \):
\[ K_{f} = \frac{\Delta T_{f}}{m} \]
\[ K_{f} = \frac{1.8}{0.4} \]
\[ K_{f} = \frac{18}{4} = 4.5 K kg mol^{-1} \]


Step 4: Final Answer:

The cryoscopic constant of the solvent is \( 4.5 K kg mol^{-1} \).
Quick Tip: Cryoscopic constant is also known as the molal depression constant.
The units are typically \( K kg/mol \).

Step 1: Understanding the Question:

We need to find the enthalpy change for the target reaction: \( N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)} \) using Hess's Law.


Step 2: Key Formula or Approach:

Hess's Law of Constant Heat Summation states that the total enthalpy change for a reaction is the same regardless of whether the reaction occurs in one step or several steps.


Step 3: Detailed Explanation:

Let's add the two given equations:

Equation (i): \( 2H_{2(g)} + N_{2(g)} \rightarrow N_{2}H_{4(g)} \)

Equation (ii): \( N_{2}H_{4(g)} + H_{2(g)} \rightarrow 2NH_{3(g)} \)

Adding (i) and (ii):
\( (2H_{2} + N_{2} + N_{2}H_{4} + H_{2})_{(g)} \rightarrow (N_{2}H_{4} + 2NH_{3})_{(g)} \)

Cancelling \( N_{2}H_{4(g)} \) from both sides and combining \( H_{2} \) terms:
\( N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)} \)

The resulting enthalpy change \( \Delta_{r}H^{\circ} \) will be the sum of individual enthalpies:
\[ \Delta_{r}H^{\circ} = \Delta_{r}H^{\circ}_{1} + \Delta_{r}H^{\circ}_{2} \]
\[ \Delta_{r}H^{\circ} = 95.4 kJ + (-187.6 kJ) \]
\[ \Delta_{r}H^{\circ} = -92.2 kJ \]


Step 4: Final Answer:

The standard enthalpy change for the synthesis of ammonia is -92.2 kJ.
Quick Tip: When adding reactions, simply add their \( \Delta H \) values.
If you have to reverse a reaction, change the sign of its \( \Delta H \).
If you multiply a reaction by a factor, multiply \( \Delta H \) by the same factor.

Step 1: Understanding the Question:

Mesityl oxide is a common name for the IUPAC compound 4-methylpent-3-en-2-one.


Step 2: Detailed Explanation:

Mesityl oxide is formed by the aldol condensation of two acetone molecules followed by dehydration.

Reaction:
\( 2 CH_{3}COCH_{3} \xrightarrow{Base} CH_{3}C(OH)(CH_{3})CH_{2}COCH_{3} \xrightarrow{-H_{2}O} (CH_{3})_{2}C=CH-CO-CH_{3} \).

Analyzing the structure:

- It has a methyl group at carbon 4 of a pent-3-en-2-one chain.

- This corresponds to option (B).


Step 3: Final Answer:

The structural formula of mesityl oxide is \( (CH_{3})_{2}C = CH - CO - CH_{3} \).
Quick Tip: Mesityl oxide is a classic example of an \( \alpha, \beta \)-unsaturated ketone.
Remember its origin from self-condensation of acetone to help identify the skeleton.

Step 1: Understanding the Question:

Solubility of aromatic compounds in water depends on their ability to form intermolecular hydrogen bonds with water molecules.


Step 2: Detailed Explanation:

- Phenol: Relatively soluble due to intermolecular H-bonding with water.

- p-Cresol: Has an additional non-polar methyl group which decreases solubility compared to phenol.

- o-Nitrophenol: Undergoes intramolecular hydrogen bonding between the hydroxy group and the nitro group. This significantly reduces the availability of the \( -OH \) group to form hydrogen bonds with water molecules.

- p-Nitrophenol: Forms intermolecular hydrogen bonds with water molecules and is more soluble than the ortho isomer.

As a result, o-nitrophenol has the lowest solubility among the options.


Step 3: Final Answer:

o-Nitrophenol has the lowest solubility in water.
Quick Tip: Intramolecular hydrogen bonding (chelation) leads to lower boiling points and lower water solubility compared to isomers that form intermolecular hydrogen bonds.

Step 1: Understanding the Question:

The reaction is electrophilic aromatic substitution (chlorination) of chlorobenzene.


Step 2: Key Formula or Approach:

The chlorine atom already present on the benzene ring is ortho and para directing due to the resonance effect (\( +R \)).


Step 3: Detailed Explanation:

When chlorobenzene reacts with chlorine in the presence of anhydrous \( FeCl_{3} \), the incoming electrophile \( Cl^{+} \) attacks the ortho and para positions.

The para-substituted product (p-dichlorobenzene) is the major product due to less steric hindrance at the para position compared to the ortho position.

Looking at the provided images:

- Image A shows 1,2-dichlorobenzene (ortho).

- The bottom left image shows 1,3-dichlorobenzene (meta).

- Image C shows 1,3,5-trichlorobenzene.

- The bottom right image shows 1,4-dichlorobenzene (para).

Therefore, the para product is the major one.


Step 4: Final Answer:

The major product is p-dichlorobenzene.
Quick Tip: Halogens on a benzene ring are deactivating but ortho-para directing.
Para isomers are usually major in such substitutions unless special directing or steric factors favor ortho.

Step 1: Understanding the Question:

Calcium carbide (\( CaC_{2} \)) is an industrial source for a specific simple alkyne upon hydrolysis.


Step 2: Detailed Explanation:

The reaction between calcium carbide and water is a hydrolysis reaction:
\[ CaC_{2(s)} + 2H_{2}O_{(l)} \rightarrow Ca(OH)_{2(aq)} + C_{2}H_{2(g)} \]

The gas produced is \( C_{2}H_{2} \), which is commonly known as acetylene or ethyne.


Step 3: Final Answer:

The alkyne formed is ethyne.
Quick Tip: Calcium carbide contains the carbide ion \( [C \equiv C]^{2-} \).
Adding water simply protonates this ion to form ethyne.

Step 1: Understanding the Question:

Electrical conductivity requires the presence of mobile charged particles, such as free electrons or free ions.


Step 2: Detailed Explanation:

- Solid sodium chloride (NaCl): It is an ionic compound. In solid state, ions are fixed in a crystal lattice and cannot move. Thus, it does not conduct electricity.

- Aqueous potassium chloride (KCl): When dissolved in water, KCl dissociates into \( K^{+} \) and \( Cl^{-} \) ions which are free to move. It is a good conductor.

- Graphite (s): It has delocalized \( \pi \)-electrons due to its layered structure. These electrons can move, making it a conductor.

- Copper metal (s): Like all metals, it has a "sea of mobile electrons" that facilitates electrical conduction.


Step 3: Final Answer:

Solid sodium chloride is a nonconductor of electricity.
Quick Tip: Ionic solids only conduct electricity in molten state or when dissolved in a polar solvent like water.
Always check the physical state (solid vs aqueous/molten) for ionic compounds.

Step 1: Understanding the Question:

We need to convert a given mass of NaOH into millimoles.


Step 2: Key Formula or Approach:

1. \( Moles = \frac{Mass}{Molar Mass} \)

2. \( Millimoles = Moles \times 1000 \)


Step 3: Detailed Explanation:

Molar mass of Sodium Hydroxide (\( NaOH \)):

- Na = 23 g/mol

- O = 16 g/mol

- H = 1 g/mol

Total = \( 23 + 16 + 1 = 40 g/mol \).

Calculate moles:
\[ Moles = \frac{0.160 g}{40 g/mol} = 0.004 mol \]

Calculate millimoles:
\[ Millimoles = 0.004 \times 1000 = 4.00 mmol \]


Step 4: Final Answer:

The number of millimoles is 4.00.
Quick Tip: \( 1 mole = 1000 millimoles \).
Alternatively, \( Millimoles = \frac{Mass in mg}{Molar Mass} \).
\( 0.160 g = 160 mg \). So, \( 160 / 40 = 4 \).

Step 1: Understanding the Question:

The acidity or basicity of a salt solution depends on the strength of the acid and base from which the salt is derived.


Step 2: Detailed Explanation:

- \( NH_{4}NO_{3} \): Derived from weak base (\( NH_{4}OH \)) and strong acid (\( HNO_{3} \)). Solution is acidic.

- \( Na_{2}CO_{3} \): Derived from strong base (\( NaOH \)) and weak acid (\( H_{2}CO_{3} \)). The carbonate ion undergoes hydrolysis to produce \( OH^{-} \). Solution is basic.

- \( NaNO_{3} \): Derived from strong base (\( NaOH \)) and strong acid (\( HNO_{3} \)). Solution is neutral.

- \( CuSO_{4} \): Derived from weak base (\( Cu(OH)_{2} \)) and strong acid (\( H_{2}SO_{4} \)). Solution is acidic.


Step 3: Final Answer:
\( Na_{2}CO_{3} \) forms a basic solution in water.
Quick Tip: Salt of (Strong Base + Weak Acid) \( \rightarrow \) Basic.
Salt of (Weak Base + Strong Acid) \( \rightarrow \) Acidic.
Salt of (Strong Base + Strong Acid) \( \rightarrow \) Neutral.

Step 1: Understanding the Question:

This question asks for the oxidation state changes of nitrogen atoms in the thermal decomposition of ammonium nitrite.


Step 2: Detailed Explanation:

The compound \( NH_{4}NO_{2} \) consists of two different ions: the ammonium ion (\( NH_{4}^{+} \)) and the nitrite ion (\( NO_{2}^{-} \)).

1. Oxidation state of N in \( NH_{4}^{+} \):
\( x + 4(+1) = +1 \Rightarrow x = -3 \).

2. Oxidation state of N in \( NO_{2}^{-} \):
\( y + 2(-2) = -1 \Rightarrow y = +3 \).

In the product \( N_{2} \), the oxidation state of N is 0.

Changes:

- Nitrogen from \( NH_{4}^{+} \) goes from -3 to 0 (loss of electrons, oxidation).

- Nitrogen from \( NO_{2}^{-} \) goes from +3 to 0 (gain of electrons, reduction).

Since one nitrogen atom is oxidized and the other is reduced, nitrogen is both oxidized and reduced in this reaction (Comproportionation).


Step 3: Final Answer:

Nitrogen is oxidised as well as reduced.
Quick Tip: This is a specific type of redox reaction called comproportionation, where two different oxidation states of the same element react to give a single intermediate oxidation state.

Step 1: Understanding the Question:

Electronegativity is the tendency of an atom to attract a shared pair of electrons. It follows specific trends in the periodic table.


Step 2: Detailed Explanation:

All the given elements (Be, Mg, Ca, Sr) belong to Group 2 of the periodic table (Alkaline Earth Metals).

The elements are arranged in the following order down the group:

Beryllium (Be) \( \rightarrow \) Magnesium (Mg) \( \rightarrow \) Calcium (Ca) \( \rightarrow \) Strontium (Sr).

Periodic Trend: Electronegativity decreases as we move down a group because the atomic radius increases and the nuclear attraction for outer electrons weakens.

Therefore, the element at the top of the group (Be) has the highest electronegativity.


Step 3: Final Answer:

Beryllium (Be) has the highest electronegativity among the options.
Quick Tip: Electronegativity increases across a period and decreases down a group.
Fluorine is the most electronegative element overall, but within Group 2, Beryllium leads.

Step 1: Understanding the Question:

Colloids are classified based on the physical state of the dispersed phase and the dispersion medium.


Step 2: Detailed Explanation:

Fog consists of tiny droplets of water (liquid) suspended in the air (gas).

- Dispersed Phase: Liquid

- Dispersion Medium: Gas

A colloidal system where a liquid is dispersed in a gas is specifically called a liquid aerosol.


Step 3: Final Answer:

Fog is an example of an aerosol.
Quick Tip: Remember:
Liquid in Gas = Aerosol (Fog, Mist, Clouds)
Solid in Gas = Aerosol (Smoke, Dust)
Gas in Liquid = Foam (Whipped cream)
Liquid in Liquid = Emulsion (Milk)

Step 1: Understanding the Question:

Aryl halides are generally unreactive towards nucleophilic substitution due to partial double bond character of the \(C-X\) bond. This reactivity can be increased by the presence of electron-withdrawing groups (EWGs).


Step 2: Detailed Explanation:

- The presence of an EWG like the nitro group (\(-NO_{2}\)) at ortho or para positions stabilizes the negative charge in the intermediate Meisenheimer complex through resonance (mesomeric effect).

- This makes the \(C-Cl\) bond easier to break (increases reactivity).

- In m-nitrochlorobenzene, the nitro group cannot stabilize the carbanion intermediate via resonance because the negative charge never resides on the carbon atom bearing the nitro group.

- Therefore, the meta isomer is less reactive than the ortho and para isomers.

- 2,4,6-trinitrochlorobenzene has three such groups and is the most reactive.

- Consequently, m-nitrochlorobenzene has the most difficulty in breaking the \(C-X\) bond among the given substituted chlorobenzenes.


Step 3: Final Answer:

m-Nitrochlorobenzene has the most difficulty in breaking the \(C-X\) bond compared to the other options.
Quick Tip: Reactivity order for nucleophilic aromatic substitution:
2,4,6-trinitro \(>\) 2,4-dinitro \(>\) p-nitro \(\approx\) o-nitro \(>\) m-nitro \(>\) chlorobenzene.
Resonance stabilization by EWGs only occurs from ortho and para positions.

Step 1: Understanding the Question:

Nanoparticles have unique properties that differ from bulk materials, leading to various industrial and medical applications.


Step 2: Detailed Explanation:

Silver nanoparticles (AgNPs) are widely known for their potent antimicrobial and antibacterial properties.

They release silver ions (\(Ag^{+}\)) which interact with bacterial cell walls and metabolic enzymes, leading to the death of the microorganism.

Because of this efficiency, they are commonly used as coatings for water filters, air filters, and medical bandages to act as disinfectants.


Step 3: Final Answer:

Silver nanoparticles are used for coating filter materials as bacterial disinfectants.
Quick Tip: Silver has been used as an antimicrobial agent for centuries. At the nanoscale, its surface area increases dramatically, making it much more effective at lower concentrations.

Step 1: Understanding the Question:

According to Dalton's Law of partial pressures and the ideal gas law, at constant volume and temperature, the partial pressure of a gas is directly proportional to its number of moles (\( p_{i} \propto n_{i} \)).


Step 2: Key Formula or Approach:

Number of moles (\( n \)) = \( \frac{Mass(m)}{Molar Mass(M)} \)

Since all gases have equal mass \( m \), then \( n \propto \frac{1}{M} \).

Partial pressure \( p_{i} \propto \frac{1}{M} \).

Minimum partial pressure corresponds to the maximum molar mass.


Step 3: Detailed Explanation:

Let's find the molar masses (\(M\)) of the given gases:

- \( M(H_{2}) = 2 g/mol \)

- \( M(He) = 4 g/mol \)

- \( M(Ne) = 20 g/mol \)

- \( M(CO_{2}) = 12 + 2(16) = 44 g/mol \)

Comparing the values: \( 44 > 20 > 4 > 2 \).

Since \( CO_{2} \) has the largest molar mass, it will have the lowest number of moles for a fixed mass.

Therefore, \( CO_{2} \) exerts the minimum partial pressure.


Step 4: Final Answer:
\( CO_{2} \) exerts the minimum partial pressure.
Quick Tip: For equal masses of different gases:
Lightest gas (lowest M) \(\rightarrow\) Max Moles \(\rightarrow\) Max Partial Pressure.
Heaviest gas (highest M) \(\rightarrow\) Min Moles \(\rightarrow\) Min Partial Pressure.

Step 1: Understanding the Question:

The type of overlap depends on the hybridization of the atoms involved in the bond.


Step 2: Detailed Explanation:

Acetylene (ethyne) has the formula \( H - C \equiv C - H \).

Each carbon atom in acetylene is triple-bonded to another carbon and single-bonded to a hydrogen atom.

- The carbon atoms are \( sp \) hybridized to accommodate the linear geometry.

- The hydrogen atom has its valence electron in a \( 1s \) orbital.

- The \( C-H \) sigma (\( \sigma \)) bond is formed by the axial overlap of one \( sp \) hybrid orbital from carbon and the \( 1s \) orbital of hydrogen.

Thus, the overlap is \( sp - s \).


Step 3: Final Answer:

The overlap involved in \( C-H \) bonds in acetylene is \( sp-s \).
Quick Tip: Hybridization of carbon:
Alkane (\(C-C\)): \( sp^3 \)
Alkene (\(C=C\)): \( sp^2 \)
Alkyne (\(C\equiv C\)): \( sp \)
In hydrocarbons, the \(H\) atom always uses its \(s\) orbital.

Step 1: Understanding the Question:

Interhalogen compounds of the type \( XY_{3} \) have the central halogen \( X \) in a +3 oxidation state.


Step 2: Key Formula or Approach:

Use VSEPR theory to determine the geometry:

1. Total valence electrons of central halogen (Group 17) = 7.

2. Number of electrons used for 3 single bonds with \( Y \) = 3.

3. Remaining electrons = 4, which equals 2 lone pairs.

4. Steric number = 3 bond pairs + 2 lone pairs = 5.


Step 3: Detailed Explanation:

For a steric number of 5, the electron domain geometry is trigonal bipyramidal (\( sp^3d \)).

According to VSEPR theory, the two lone pairs will occupy equatorial positions to minimize repulsion.

The resulting molecular geometry (shape) formed by the three bond pairs is T-shaped.

Due to lone pair-bond pair repulsions, the bond angles are slightly less than 90\(^{\circ}\), making it a "Bent T-shape".


Step 4: Final Answer:

The shape of the interhalogen compound where the central atom is in the +3 oxidation state is Bent 'T' shape.
Quick Tip: Standard interhalogen shapes:
\( XY \): Linear (+1 state)
\( XY_{3} \): T-shaped (+3 state)
\( XY_{5} \): Square pyramidal (+5 state)
\( XY_{7} \): Pentagonal bipyramidal (+7 state)

Step 1: Understanding the Question:

A homologous series is a series of organic compounds with the same functional group and similar chemical properties in which successive members differ by a fixed unit.


Step 2: Key Formula or Approach:

Successive members in a homologous series differ by a \( -CH_{2} \) (methylene) group.

Difference in molar mass = Atomic mass of C + 2 \( \times \) Atomic mass of H.


Step 3: Detailed Explanation:

For the alkane series (\( C_{n}H_{2n+2} \)):

- 3rd homologue is Propane (\( C_{3}H_{8} \)).

- 4th homologue is Butane (\( C_{4}H_{10} \)).

Difference = \( (C_{4}H_{10}) - (C_{3}H_{8}) = CH_{2} \).

Molar mass of \( CH_{2} = 12 + 2(1) = 14 g/mol \).


Step 4: Final Answer:

The difference in molar masses is 14 g mol\(^{-1}\).
Quick Tip: In any homologous series (alkanes, alkenes, alcohols, etc.), the difference between any two \textbf{adjacent} members is always \( 14 g/mol \).

Step 1: Understanding the Question:

Saponification is the alkaline hydrolysis of fats or oils (triglycerides) to produce glycerol and soap.


Step 2: Key Formula or Approach:

Reaction:

1 Triglyceride + 3 NaOH \( \rightarrow \) 1 Glycerol + 3 Soap (Sodium salts of fatty acids).


Step 3: Detailed Explanation:

From the balanced chemical equation, the stoichiometric ratio between the triglyceride and glycerol is \( 1:1 \).

This means:

1 mole of triglyceride produces 1 mole of glycerol.

Therefore, \( n \) moles of triglyceride will produce \( n \) moles of glycerol.


Step 4: Final Answer:

The number of moles of glycerol produced is \( n \).
Quick Tip: A triglyceride is a tri-ester of glycerol. Hydrolyzing one molecule of triglyceride "releases" exactly one molecule of the glycerol backbone.

Step 1: Understanding the Question:

The field strength of a ligand is determined by its position in the spectrochemical series, which ranks ligands based on the crystal field splitting energy (\( \Delta_{o} \)) they produce.


Step 2: Detailed Explanation:

The partial spectrochemical series is:
\( I^{-} < Br^{-} < S^{2-} < SCN^{-} < Cl^{-} < F^{-} < OH^{-} < ox^{2-} < H_{2}O < NCS^{-} < edta^{4-} < NH_{3} < en < CN^{-} < CO \)

Comparing the given options:

- \( I^{-} \): At the very beginning of the series (Weakest).

- \( S^{2-} \): Weak field, but stronger than \( I^{-} \).

- \( en \) (ethylenediamine): Strong field ligand.

- \( CO \) (carbonyl): Strongest known field ligand.

Thus, iodide (\( I^{-} \)) has the lowest field strength.


Step 3: Final Answer:

The lowest field strength ligand is \( I^{-} \).
Quick Tip: Halide ions are typically weak field ligands, and their strength increases with decreasing size: \( I^{-} < Br^{-} < Cl^{-} < F^{-} \).

Step 1: Understanding the Question:

Polyesters are polymers containing the ester functional group in their main chain.


Step 2: Detailed Explanation:

- PHBV (Poly-hydroxybutyrate-co-hydroxyvalerate): It is a copolymer of 3-hydroxybutanoic acid and 3-hydroxypentanoic acid. These monomers are joined by ester linkages, making it a biodegradable polyester.

- Urea formaldehyde: This is a thermosetting resin, not a polyester.

- Thermocol: This is expanded polystyrene, an addition polymer.

- Polyacrylonitrile (PAN): Also known as Orlon or Acrilan, it is an addition polymer of acrylonitrile.


Step 3: Final Answer:

PHBV is a polyester fibre.
Quick Tip: The most famous polyester is Terylene (Dacron). PHBV is important to remember as a biological, biodegradable polyester.

Step 1: Understanding the Question:

The question asks to identify the weakest ligand among the choices based on the spectrochemical series.


Step 2: Detailed Explanation:

Based on the experimental determination of crystal field splitting energy, ligands are arranged as:
\( I^{-} < Br^{-} < S^{2-} < Cl^{-} < \dots < NH_{3} < en < CN^{-} < CO \).

Iodide ion (\( I^{-} \)) is at the start of the series, meaning it causes the least splitting and is the weakest (lowest field strength) ligand.


Step 3: Final Answer:

The lowest field strength ligand is \( I^{-} \).
Quick Tip: Ligands with the same coordinating atom usually follow periodic trends. For halides, splitting power increases as Electronegativity increases: \( I < Br < Cl < F \).

Step 1: Understanding the Question:

The basic strength of methyl-substituted amines in aqueous solution is determined by the combined effects of inductive effect (\( +I \)), solvation effect (hydrogen bonding with water), and steric hindrance.


Step 2: Detailed Explanation:

1. Inductive effect: Basic strength should increase with the number of methyl groups: \( 3^{\circ} > 2^{\circ} > 1^{\circ} > NH_{3} \).

2. Solvation effect: Basic strength increases with the ability to form H-bonds with water: \( NH_{3} > 1^{\circ} > 2^{\circ} > 3^{\circ} \).

3. Steric hindrance: Higher for tertiary amines, making the lone pair less available for donation.

For methyl groups, the experimental order in water is:

Secondary (\( 2^{\circ} \)) \(>\) Primary (\( 1^{\circ} \)) \(>\) Tertiary (\( 3^{\circ} \)).

Order: \( (CH_{3})_{2}NH > CH_{3}NH_{2} > (CH_{3})_{3}N \).


Step 3: Final Answer:

The correct decreasing order is (CH\(_{3}\))\(_{2}\)NH \(>\) CH\(_{3}\)NH\(_{2}\) \(>\) (CH\(_{3}\))\(_{3}\) N.
Quick Tip: Remember the shortcut codes for aqueous basicity:
Methyl group: 213 (\( 2^{\circ} > 1^{\circ} > 3^{\circ} \))
Ethyl group: 231 (\( 2^{\circ} > 3^{\circ} > 1^{\circ} \))

Step 1: Understanding the Question:

A zero effective magnetic moment (\( \mu_{eff} = 0 \)) occurs when an ion has no unpaired electrons (all electrons are paired).


Step 2: Detailed Explanation:

The electronic configuration of lanthanoids in the +3 oxidation state involves losing 2 electrons from the \( 6s \) orbital and 1 from the \( 5d \) or \( 4f \) orbital.

- \( Ho^{3+} \): Has unpaired electrons in \( 4f \) subshell.

- \( Lu^{3+} \): Lutetium (\( Z = 71 \)) has ground state config \( [Xe] 4f^{14} 5d^{1} 6s^{2} \). In \( Lu^{3+} \), it loses \( 5d^1 \) and \( 6s^2 \), leaving \( [Xe] 4f^{14} \). The \( 4f \) subshell is completely filled, meaning \( n = 0 \).

- \( Pr^{3+} \): Has unpaired electrons.

- \( Er^{3+} \): Has unpaired electrons.

Since \( Lu^{3+} \) has zero unpaired electrons, its magnetic moment is zero.


Step 3: Final Answer:

Lutetium (Lu) exhibits zero effective magnetic moment in the +3 state.
Quick Tip: Only \( La^{3+} (f^0) \) and \( Lu^{3+} (f^{14}) \) are diamagnetic (zero magnetic moment) among the common lanthanoid +3 ions.

Step 1: Understanding the Question:

Solutions that obey Raoult's law over the entire range of concentration are called ideal solutions.


Step 2: Detailed Explanation:

Ideal solutions are formed between components that have similar chemical structures and similar intermolecular forces.

- Phenol and aniline: Show negative deviation due to strong H-bonding between components.

- Chloroform and acetone: Show negative deviation due to H-bonding.

- Ethanol and acetone: Show positive deviation as acetone breaks the H-bonding of ethanol.

- Benzene and toluene: These are both non-polar hydrocarbons with very similar structures and London dispersion forces. They form a nearly ideal solution.


Step 3: Final Answer:

The mixture of Benzene and toluene obeys Raoult's law.
Quick Tip: Look for "structural siblings" (e.g., n-hexane/n-heptane, chlorobenzene/bromobenzene) to find ideal solutions.

Step 1: Understanding the Question:

A reaction is spontaneous if \( \Delta G < 0 \). The Gibbs free energy is defined as \( \Delta G = \Delta H - T\Delta S \).


Step 2: Key Formula or Approach:

At equilibrium, \( \Delta G = 0 \), so \( T_{eq} = \frac{\Delta H}{\Delta S} \).

For a reaction to be spontaneous below this temperature (\( T < T_{eq} \)), we need to examine the signs of \( \Delta H \) and \( \Delta S \).


Step 3: Detailed Explanation:

- If \( \Delta H < 0 \) (exothermic) and \( \Delta S < 0 \) (decrease in entropy):
\[ \Delta G = (-value) - T(-value) = -|\Delta H| + T|\Delta S| \]

At low temperatures (below \( T_{eq} \)), the magnitude of the negative term (\( |\Delta H| \)) is greater than the positive term (\( T|\Delta S| \)). So \( \Delta G < 0 \).

Thus, the reaction is spontaneous only at lower temperatures.

- If \( \Delta H > 0 \) and \( \Delta S > 0 \), the reaction becomes spontaneous above the equilibrium temperature.

- If \( \Delta H < 0 \) and \( \Delta S > 0 \), spontaneous at all temperatures.

- If \( \Delta H > 0 \) and \( \Delta S < 0 \), non-spontaneous at all temperatures.


Step 4: Final Answer:

The correct conditions are \( \Delta H < 0 \) and \( \Delta S < 0 \).
Quick Tip: Exothermic reactions with a decrease in entropy are "Enthalpy driven" and occur spontaneously only at low temperatures.

Step 1: Understanding the Question:

Nodes are regions in space around the nucleus where the probability of finding an electron is zero.


Step 2: Key Formula or Approach:

- Radial nodes = \( n - l - 1 \)

- Angular nodes = \( l \)

- Total nodes = \( n - 1 \)

Where \( n \) is the principal quantum number and \( l \) is the azimuthal quantum number.


Step 3: Detailed Explanation:

For a \( 2s \) orbital:

- \( n = 2 \)

- For an s-orbital, \( l = 0 \)

Calculation:

Total nodes = \( 2 - 1 = 1 \).

(Specifically, this is one radial node).


Step 4: Final Answer:

The number of nodes in the \( 2s \) orbital is 1.
Quick Tip: An orbital with principal quantum number \( n \) will always have \( n - 1 \) total nodes.
1s has 0 nodes, 2s has 1, 3s has 2, and so on.

Step 1: Understanding the Question:

In a homologous series of aldehydes, the boiling point increases with the increase in molar mass.


Step 2: Detailed Explanation:

As the size of the carbon chain increases, the magnitude of Van der Waals forces (dispersion forces) increases.

- Methanal (\( HCHO \)): 1 Carbon atom (smallest).

- Ethanal (\( CH_{3}CHO \)): 2 Carbon atoms.

- Propanal (\( CH_{3}CH_{2}CHO \)): 3 Carbon atoms.

- Pentanal (\( CH_{3}(CH_{2})_{3}CHO \)): 5 Carbon atoms.

Methanal is actually a gas at room temperature, while the others are liquids. It has the lowest molecular weight and the weakest intermolecular attractions.


Step 3: Final Answer:

Methanal has the lowest boiling point.
Quick Tip: Boiling point \( \propto \) Molar Mass (for similar functional groups).
Methanal (Formaldehyde) is the simplest aldehyde and is a gas, whereas larger ones are liquids or solids.

Step 1: Understanding the Question:

A phenol is a compound where the hydroxyl (\( -OH \)) group is directly attached to an aromatic ring.


Step 2: Detailed Explanation:

Let's analyze the structures in the images:

- Option A: Shows 3-nitrophenol. The \( -OH \) is directly on the benzene ring. It is a phenol.

- Option B: Shows 2-naphthol. Although it's a bicyclic system (naphthalene), the \( -OH \) is directly attached to the aromatic ring. It belongs to the phenol family.

- Option C: Shows 4-bromophenol. The \( -OH \) is directly on the benzene ring. It is a phenol.

- Option D: Shows benzyl alcohol (\( C_{6}H_{5}CH_{2}OH \)). In this molecule, the hydroxyl group is attached to a saturated carbon atom (\( sp^3 \) hybridized), which is then attached to the ring. This is an aromatic alcohol, not a phenol.


Step 3: Final Answer:

Option D (Benzyl alcohol) is not a phenol.
Quick Tip: Phenols: Ar-OH (attached to \( sp^2 \) aromatic carbon).
Aromatic Alcohols: Ar-C-OH (attached to \( sp^3 \) side-chain carbon).
Phenols are acidic, while aromatic alcohols are neutral like aliphatic alcohols.

Step 1: Understanding the Question:

The question asks to identify the reactant 'A' which reacts with acetic anhydride in the presence of an acid catalyst (H\(^+\)) to produce Aspirin and Acetic acid. This reaction is a classic example of acetylation.


Step 2: Key Formula or Approach:

The reaction given is the synthesis of Aspirin. We need to know the chemical structure of Aspirin and the reactants involved in its preparation.

Aspirin is chemically known as acetylsalicylic acid. Its synthesis involves the acetylation of salicylic acid.


Step 3: Detailed Explanation:

The reaction is the esterification of the phenolic hydroxyl group (-OH) of salicylic acid using acetic anhydride.

The structure of Salicylic acid is a benzene ring with a carboxylic acid group (-COOH) and a hydroxyl group (-OH) at ortho positions.

The structure of Acetic anhydride is (CH\(_3\)CO)\(_2\)O.

The reaction proceeds as follows:

Salicylic acid (A) reacts with acetic anhydride. The acetyl group (CH\(_3\)CO-) from acetic anhydride replaces the hydrogen atom of the hydroxyl group of salicylic acid.
\[ Salicylic acid + Acetic anhydride \xrightarrow{H^+} Acetylsalicylic acid (Aspirin) + Acetic acid \]
Therefore, the reactant 'A' is Salicylic acid.


Step 4: Final Answer:

Based on the synthesis reaction of Aspirin, the compound 'A' is Salicylic acid.
Quick Tip: Remember the common names and structures of important organic compounds. The synthesis of Aspirin is a very common reaction in organic chemistry, often taught as an example of acetylation. Knowing that Aspirin is acetylsalicylic acid directly points to salicylic acid as the starting material.

Step 1: Understanding the Question:

The question shows a two-step reaction starting from ethanenitrile and asks for the final product 'B'. The reagents suggest a specific named reaction.


Step 2: Key Formula or Approach:

The sequence of reactions is the Stephen's reduction (or Stephen's aldehyde synthesis).

Step 1: Reduction of a nitrile with stannous chloride (SnCl\(_2\)) and hydrochloric acid (HCl) to form an iminium salt (A).

Step 2: Hydrolysis of the iminium salt with water (H\(_3\)O\(^+\)) to form an aldehyde (B).

The general reaction is: R-CN \( \xrightarrow{1. SnCl_2/HCl} \) \( \xrightarrow{2. H_3O^+} \) R-CHO


Step 3: Detailed Explanation:

The starting material is Ethanenitrile, which has the formula CH\(_3\)-C\( \equiv \)N.

Step 1: Ethanenitrile is treated with SnCl\(_2\) and HCl. The nitrile is reduced to an iminium chloride salt (intermediate A).
\[ CH_3C \equiv N + 2[H] \xrightarrow{SnCl_2/HCl} [CH_3CH=NH_2]^+Cl^- \quad (Ethaniminium chloride, A) \]
Step 2: The intermediate iminium salt (A) is then hydrolyzed by warming with water (acidic hydrolysis, H\(_3\)O\(^+\)). The C=N bond is broken, and the carbon atom is converted to a carbonyl group (C=O), forming an aldehyde. The nitrogen atom forms ammonium chloride (NH\(_4\)Cl).
\[ [CH_3CH=NH_2]^+Cl^- \xrightarrow{H_2O, \Delta} CH_3CHO + NH_4Cl \]
The product 'B' is CH\(_3\)CHO, which is known as Ethanal (or acetaldehyde).


Step 4: Final Answer:

The final product 'B' of the Stephen's reduction of ethanenitrile is Ethanal.
Quick Tip: Stephen's reduction is a specific method for preparing aldehydes from nitriles. Remember the reagent combination (SnCl\(_2\)/HCl followed by hydrolysis) and the product type (aldehyde). This will help you quickly identify the outcome of such reactions in exams.

Step 1: Understanding the Question:

The question asks to calculate the work done by an ideal gas during an isothermal expansion against a constant external pressure. This indicates an irreversible process.


Step 2: Key Formula or Approach:

The formula for work done in an irreversible expansion against a constant external pressure is:
\[ W = -P_{ext} \Delta V \]
where \(P_{ext}\) is the constant external pressure and \(\Delta V\) is the change in volume (\(V_{final} - V_{initial}\)). We also need the conversion factor between L-atm (or dm\(^3\)-atm) and Joules. 1 L-atm = 101.3 J (approximately).


Step 3: Detailed Explanation:

Given values:

- Moles of gas, n = 1 mole

- Initial volume, \(V_{initial}\) = 2 dm\(^3\)

- Final volume, \(V_{final}\) = 2.8 dm\(^3\)

- External pressure, \(P_{ext}\) = 1 atm

Note: 1 dm\(^3\) = 1 L.


First, calculate the change in volume, \(\Delta V\):
\[ \Delta V = V_{final} - V_{initial} = 2.8 dm^3 - 2.0 dm^3 = 0.8 dm^3 \]
Now, calculate the work done in dm\(^3\)-atm:
\[ W = -P_{ext} \Delta V = - (1 atm) \times (0.8 dm^3) = -0.8 dm^3-atm \]
Finally, convert the work done from dm\(^3\)-atm to Joules (J). The standard conversion factor is 1 L-atm = 101.325 J. Since 1 dm\(^3\) = 1 L, 1 dm\(^3\)-atm = 101.325 J.
\[ W = -0.8 \times 101.325 J \] \[ W = -81.06 J \]
This value is very close to option (B).


Step 4: Final Answer:

The calculated work done is -81.06 J, which corresponds to option (B) -81.04 J. The small difference is likely due to rounding of the conversion factor.
Quick Tip: Pay close attention to the wording. "Against constant external pressure" signifies an irreversible process, for which the formula is \(W = -P_{ext}\Delta V\). If the expansion were reversible, the formula would be \(W = -nRT \ln(V_2/V_1)\). Always check the units and be ready to convert L-atm to Joules (1 L-atm \(\approx\) 101.3 J).

Step 1: Understanding the Question:

The question asks to calculate the osmotic pressure of an electrolyte solution, given its molarity, temperature, van't Hoff factor, and the universal gas constant.


Step 2: Key Formula or Approach:

The formula for osmotic pressure (\(\Pi\)) of an electrolyte solution is given by the van't Hoff equation:
\[ \Pi = i \times C \times R \times T \]
where:

- \(i\) is the van't Hoff factor

- \(C\) is the molar concentration (Molarity) of the solution

- \(R\) is the universal gas constant

- \(T\) is the absolute temperature in Kelvin


Step 3: Detailed Explanation:

Given values:

- Molarity, \(C\) = 0.2 M (or 0.2 mol/dm\(^3\))

- Temperature, \(T\) = 300 K

- van't Hoff factor, \(i\) = 1.6

- Gas constant, \(R\) = 0.0821 atm dm\(^3\) K\(^{-1}\) mol\(^{-1}\)


Substitute these values into the osmotic pressure formula:
\[ \Pi = (1.6) \times (0.2 mol/dm^3) \times (0.0821 atm dm^3 K^{-1} mol^{-1}) \times (300 K) \]
Now, perform the calculation:
\[ \Pi = 1.6 \times 0.2 \times 0.0821 \times 300 atm \] \[ \Pi = 0.32 \times 0.0821 \times 300 atm \] \[ \Pi = 0.32 \times 24.63 atm \] \[ \Pi = 7.8816 atm \]
Rounding to two decimal places, the osmotic pressure is 7.88 atm.


Step 4: Final Answer:

The calculated osmotic pressure is 7.88 atm, which matches option (B).
Quick Tip: For electrolyte solutions, always remember to include the van't Hoff factor (\(i\)) in colligative property calculations. For non-electrolytes, \(i=1\). Ensure that the units of R, C, T, and P are consistent. Here, R is in atm dm\(^3\), so the pressure \(\Pi\) will be in atm.

Step 1: Understanding the Question:

The question asks to find the total number of unit cells in a given mass of a metal. We are provided with the total mass of the metal and the mass of a single unit cell (indirectly).


Step 2: Key Formula or Approach:

The total number of unit cells in a sample can be calculated by dividing the total mass of the sample by the mass of a single unit cell.
\[ Number of unit cells = \frac{Total mass of metal}{Mass of one unit cell} \]
The mass of one unit cell can be found using its density (\(\rho\)) and volume (\(V_{uc}\)).
\[ Mass of one unit cell = density \times volume of unit cell = \rho \times V_{uc} \]

Step 3: Detailed Explanation:

Given values:

- Total mass of metal = 0.79 g

- Product of density and volume of unit cell (\(\rho \times V_{uc}\)) = 1.58 \( \times \) 10\(^{-22}\) g.


This product directly gives us the mass of one unit cell.

Mass of one unit cell = 1.58 \( \times \) 10\(^{-22}\) g.


Now, we can calculate the number of unit cells:
\[ Number of unit cells = \frac{0.79 g}{1.58 \times 10^{-22} g} \] \[ Number of unit cells = \frac{0.79}{1.58} \times 10^{22} \]
Since 0.79 is exactly half of 1.58 (1.58 / 2 = 0.79):
\[ Number of unit cells = 0.5 \times 10^{22} \]
To express this in standard scientific notation:
\[ Number of unit cells = 5.0 \times 10^{21} \]

Step 4: Final Answer:

The total number of unit cells in 0.79 g of the metal is 5.0 \( \times \) 10\(^{21}\), which corresponds to option (D).
Quick Tip: Recognize that the "product of density and volume of unit cell" is simply the mass of one unit cell. This simplifies the problem significantly. The question becomes a straightforward division of total mass by the mass of a single unit.

Step 1: Understanding the Question:

The question requires identifying a secondary amine from a given list of compounds. Amines are classified based on the number of alkyl or aryl groups attached to the nitrogen atom.

- Primary (1° amine): One alkyl/aryl group attached to nitrogen (R-NH\(_2\)).

- Secondary (2° amine): Two alkyl/aryl groups attached to nitrogen (R\(_2\)NH).

- Tertiary (3° amine): Three alkyl/aryl groups attached to nitrogen (R\(_3\)N).


Step 2: Detailed Explanation:

Let's analyze the structure of each option:

(A) Phenylmethanamine: The structure is C\(_6\)H\(_5\)CH\(_2\)-NH\(_2\). The nitrogen atom is bonded to one carbon group (the benzyl group) and two hydrogen atoms. This is a primary amine.


(B) N-Methylethanamine: The name indicates an ethane chain with an amine group, and a methyl group on the nitrogen atom. The structure is CH\(_3\)CH\(_2\)-NH-CH\(_3\). The nitrogen atom is bonded to two carbon groups (an ethyl group and a methyl group) and one hydrogen atom. This is a secondary amine.


(C) N,N-Dimethylethanamine: The name indicates an ethanamine with two methyl groups on the nitrogen. The structure is CH\(_3\)CH\(_2\)-N(CH\(_3\))\(_2\). The nitrogen atom is bonded to three carbon groups (one ethyl and two methyl groups) and no hydrogen atoms. This is a tertiary amine.


(D) Prop-2-en-1-amine (Allylamine): The structure is CH\(_2\)=CH-CH\(_2\)-NH\(_2\). The nitrogen atom is bonded to one carbon group (the allyl group) and two hydrogen atoms. This is a primary amine.


Step 3: Final Answer:

Based on the analysis, N-Methylethanamine is the only secondary amine in the list.
Quick Tip: To quickly classify an amine, look at the nitrogen atom. If it's bonded to one carbon, it's primary. Two carbons, secondary. Three carbons, tertiary. The nomenclature "N-" prefix is a key indicator of substitution on the nitrogen atom. "N-alkyl" means one substitution (likely secondary), while "N,N-dialkyl" means two substitutions (tertiary).

Step 1: Understanding the Question:

The question asks to identify the biodegradable polymer from the given options. A biodegradable polymer is a polymer that can be decomposed by the action of living organisms, usually microbes, into water, carbon dioxide, and biomass.


Step 2: Detailed Explanation:

Let's analyze each polymer:

(A) Nylon 6: It is a polyamide synthesized from the monomer caprolactam. It is a synthetic polymer and is non-biodegradable.


(B) Nylon 6,6: It is a polyamide synthesized from two monomers: hexamethylenediamine and adipic acid. It is also a synthetic polymer and is non-biodegradable.


(C) Nylon 2-nylon 6: It is a polyamide copolymer. It is synthesized from two different amino acid monomers: glycine (which has 2 carbon atoms) and aminocaproic acid (which has 6 carbon atoms). The presence of amide linkages that are similar to those in proteins makes it susceptible to microbial degradation. Therefore, it is a biodegradable polymer.


(D) HDP (High-Density Polyethylene): It is an addition polymer of ethene. It consists of long hydrocarbon chains and is very resistant to chemical and biological degradation, making it non-biodegradable.


Step 3: Final Answer:

Among the given options, Nylon 2-nylon 6 is the only biodegradable polymer.
Quick Tip: Remember the key examples of biodegradable polymers like PHBV (Poly-\(\beta\)-hydroxybutyrate-co-\(\beta\)-hydroxy valerate) and Nylon 2-nylon 6. These are often contrasted with common non-biodegradable synthetic polymers like polyethene, PVC, nylon 6, and nylon 6,6.

Step 1: Understanding the Question:

The question provides the standard EMF (E°\(_cell\)) of a galvanic cell and asks for the new EMF (E\(_cell\)) when the concentrations of both ions are changed by the same factor. We need to use the Nernst equation to solve this.


Step 2: Key Formula or Approach:

The Nernst equation relates the cell potential (E\(_cell\)) to the standard cell potential (E°\(_cell\)) and the reaction quotient (Q):
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log_{10} Q \quad (at 298 K) \]
First, we need to write the overall cell reaction to determine the number of electrons transferred (n) and the expression for Q.


Step 3: Detailed Explanation:

The cell notation is Cd\(_{(s)}\) | Cd\(^{+2}\) || Cu\(^{+2}\) | Cu\(_{(s)}\).

Oxidation at Anode: Cd\(_{(s)}\) \( \rightarrow \) Cd\(^{+2}_{(aq)}\) + 2e\(^-\)

Reduction at Cathode: Cu\(^{+2}_{(aq)}\) + 2e\(^-\) \( \rightarrow \) Cu\(_{(s)}\)

Overall Cell Reaction: Cd\(_{(s)}\) + Cu\(^{+2}_{(aq)}\) \( \rightarrow \) Cd\(^{+2}_{(aq)}\) + Cu\(_{(s)}\)


From the balanced reaction, the number of electrons transferred, n = 2.

The reaction quotient, Q, is given by:
\[ Q = \frac{[Products]}{[Reactants]} = \frac{[Cd^{+2}]}{[Cu^{+2}]} \]
(Note: The concentrations of pure solids Cd\(_{(s)}\) and Cu\(_{(s)}\) are taken as 1).


Initial Condition (Standard state):

[Cd\(^{+2}\)] = 1 M, [Cu\(^{+2}\)] = 1 M.
\(Q = \frac{1}{1} = 1\).
\(E_{cell} = E^\circ_{cell} - \frac{0.0591}{2} \log(1) = E^\circ_{cell} - 0 = E^\circ_{cell}\).

Given E°\(_cell\) = 0.74 V.


New Condition:

The concentrations of both Cd\(^{+2}\) and Cu\(^{+2}\) decrease by 10 times.

New [Cd\(^{+2}\)] = 1 M / 10 = 0.1 M.

New [Cu\(^{+2}\)] = 1 M / 10 = 0.1 M.


Now, calculate the new reaction quotient, Q':
\[ Q' = \frac{[New Cd^{+2}]}{[New Cu^{+2}]} = \frac{0.1 M}{0.1 M} = 1 \]
Calculate the new cell EMF using the Nernst equation:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{2} \log_{10} Q' \] \[ E_{cell} = 0.74 V - \frac{0.0591}{2} \log_{10}(1) \]
Since log\(_10\)(1) = 0:
\[ E_{cell} = 0.74 V - 0 = 0.74 V \]

Step 4: Final Answer:

The emf of the cell remains unchanged at +0.740 V because the ratio of the ion concentrations (the reaction quotient Q) did not change.
Quick Tip: In a galvanic cell, if the concentrations of all ionic species in the reaction quotient Q are changed by the same factor, the value of Q remains unchanged. Consequently, the cell EMF will also remain unchanged from its initial value.

Step 1: Understanding the Question:

The question asks for the equilibrium concentration of lead ions (Pb\(^{2+}\)) in a solution containing a known concentration of sulfide ions (S\(^{2-}\)). This is a problem involving the solubility product constant (K\(_{sp}\)).


Step 2: Key Formula or Approach:

For a sparingly soluble salt like lead(II) sulfide (PbS), the dissolution equilibrium is:
\[ PbS_{(s)} \rightleftharpoons Pb^{2+}_{(aq)} + S^{2-}_{(aq)} \]
The solubility product expression is:
\[ K_{sp} = [Pb^{2+}] [S^{2-}] \]
We are given K\(_{sp}\) and [S\(^{2-}\)], and we need to find [Pb\(^{2+}\)].


Step 3: Detailed Explanation:

Given values:

- K\(_{sp}\) for PbS = 8.0 \( \times \) 10\(^{-28}\)

- Concentration of sulphide ions, [S\(^{2-}\)] = 1 \( \times \) 10\(^{-11}\) mol/dm\(^3\) (or M)

Let the equilibrium concentration of Pb\(^{2+}\) be [Pb\(^{2+}\)].


Rearrange the K\(_{sp}\) formula to solve for [Pb\(^{2+}\)]:
\[ [Pb^{2+}] = \frac{K_{sp}}{[S^{2-}]} \]
Substitute the given values into the equation:
\[ [Pb^{2+}] = \frac{8.0 \times 10^{-28}}{1 \times 10^{-11}} \] \[ [Pb^{2+}] = 8.0 \times 10^{(-28 - (-11))} \] \[ [Pb^{2+}] = 8.0 \times 10^{-17} \]
The equilibrium concentration of Pb\(^{2+}\) ions is 8.0 \( \times \) 10\(^{-17}\) mol/dm\(^3\). Note that Pb\(^{++}\) is another way to write Pb\(^{2+}\).


Step 4: Final Answer:

The calculated concentration of Pb\(^{2+}\) is 8.0 \( \times \) 10\(^{-17}\) M, which matches option (C).
Quick Tip: This type of problem is a direct application of the K\(_{sp}\) expression. It demonstrates the common ion effect. The presence of sulfide ions from another source suppresses the dissolution of PbS, leading to a very low concentration of Pb\(^{2+}\) ions. Always write down the equilibrium and the K\(_{sp}\) expression first.

Step 1: Understanding the Question:

The question describes a first-order reaction and gives its half-life (the time for 50% completion). It then asks for the total percentage of reactant that has reacted after a period of two half-lives.


Step 2: Key Formula or Approach:

For a first-order reaction, the half-life (t\(_{1/2}\)) is constant. This means that in every half-life period, the amount of reactant decreases by 50% of its value at the beginning of that period.

Given: t\(_{1/2}\) = 16 minutes (time for 50% completion).

We need to find the percentage reacted after 32 minutes.

Notice that 32 minutes = 2 \( \times \) 16 minutes = 2 \( \times \) t\(_{1/2}\).


Step 3: Detailed Explanation:

Let the initial amount of reactant be 100%.

After the first half-life (16 minutes):

- Amount of reactant reacted = 50% of the initial amount.

- Amount of reactant remaining = 100% - 50% = 50% of the initial amount.


After the second half-life (another 16 minutes, for a total of 32 minutes):

- The reaction will proceed by another 50%, but this is 50% of the amount \textit{remaining after the first half-life.

- Amount of reactant remaining after 32 mins = 50% of (50% of initial amount) = 0.5 \( \times \) 50% = 25% of the initial amount.


The question asks for the percentage of reactant reacting in 32 minutes.

- Total percentage reacted = Initial percentage - Final percentage remaining

- Total percentage reacted = 100% - 25% = 75%.


Step 4: Final Answer:

After 32 minutes (two half-lives), 75% of the reactant has reacted. This corresponds to option (D).
Quick Tip: For first-order reactions, you can quickly calculate the amount remaining after 'n' half-lives using the formula: Amount remaining = (Initial Amount) / 2\(^n\). In this case, n = 32 mins / 16 mins = 2. So, amount remaining = 100% / 2\(^2\) = 100% / 4 = 25%. The amount reacted is 100% - 25% = 75%.

Step 1: Understanding the Question:

The question provides the standard cell notation for a galvanic cell and states that its EMF is positive. A positive EMF indicates a spontaneous reaction. We need to derive the balanced overall cell reaction from the notation.


Step 2: Key Formula or Approach:

The cell notation is given in the format: Anode | Anode electrolyte || Cathode electrolyte | Cathode.

- At the anode (left side), oxidation occurs.

- At the cathode (right side), reduction occurs.

We need to write the half-reactions for the anode and cathode and then combine them to get a balanced overall reaction.


Step 3: Detailed Explanation:

The given cell notation is A\(_{(s)}\) | A\(^+\)(1M) || B\(^{+2}\)(1M) | B\(_{(s)}\).

Anode (Oxidation) Half-Reaction:

The left side shows A\(_{(s)}\) being oxidized to A\(^+\).
\[ A_{(s)} \rightarrow A^{+}_{(aq)} + 1e^- \]
Cathode (Reduction) Half-Reaction:

The right side shows B\(^{+2}\) being reduced to B\(_{(s)}\).
\[ B^{+2}_{(aq)} + 2e^- \rightarrow B_{(s)} \]
Balancing the Overall Reaction:

To get the overall reaction, the number of electrons lost in oxidation must equal the number of electrons gained in reduction.

- The anode reaction involves 1 electron.

- The cathode reaction involves 2 electrons.

To balance the electrons, we must multiply the anode half-reaction by 2.
\[ 2 \times (A_{(s)} \rightarrow A^{+}_{(aq)} + 1e^-) \quad \Rightarrow \quad 2A_{(s)} \rightarrow 2A^{+}_{(aq)} + 2e^- \]
Now, add the modified anode half-reaction and the cathode half-reaction:
\[ 2A_{(s)} \rightarrow 2A^{+}_{(aq)} + 2e^- \] \[ B^{+2}_{(aq)} + 2e^- \rightarrow B_{(s)} \]
\hrule \[ 2A_{(s)} + B^{+2}_{(aq)} \rightarrow 2A^{+}_{(aq)} + B_{(s)} \]
The 2e\(^-\) on both sides cancel out. This is the balanced, spontaneous cell reaction since the EMF is positive.


Step 4: Final Answer:

The correct cell reaction is 2A\(_{(s)}\) + B\(^{+2}_{(aq)}\) \( \rightarrow \) 2A\(^{+}_{(aq)}\) + B\(_{(s)}\). This matches option (B) (assuming typos in the question's option rendering are corrected to A\(^+\) and B\(^{2+}\)).
Quick Tip: Remember the mnemonic "An Ox, Red Cat" (Anode-Oxidation, Reduction-Cathode) and "LOAN" (Left-Oxidation-Anode-Negative) for galvanic cells. The cell notation always places the anode on the left and the cathode on the right. Balancing electrons is the key step to finding the correct overall reaction.

Step 1: Understanding the Question:

The question asks for the argument (angle) of a complex number given in a fractional form. To find the argument, we first need to simplify the complex number into the standard form \(z = a + bi\).


Step 2: Key Formula or Approach:

To simplify a fraction of complex numbers, we multiply the numerator and the denominator by the conjugate of the denominator.

If \(z = a + bi\), the argument is given by \(arg(z) = \tan^{-1}\left(\frac{b}{a}\right)\), considering the quadrant in which the point (a, b) lies.


Step 3: Detailed Explanation:

Given complex number \( z = \frac{13-5i}{4-9i} \).

The conjugate of the denominator (4 - 9i) is (4 + 9i).

Multiply the numerator and denominator by (4 + 9i):
\[ z = \frac{(13-5i)(4+9i)}{(4-9i)(4+9i)} \]
Calculate the numerator:
\[ (13-5i)(4+9i) = 13(4) + 13(9i) - 5i(4) - 5i(9i) \] \[ = 52 + 117i - 20i - 45i^2 \]
Since \(i^2 = -1\):
\[ = 52 + 97i - 45(-1) = 52 + 97i + 45 = 97 + 97i \]
Calculate the denominator:
\[ (4-9i)(4+9i) = 4^2 - (9i)^2 = 16 - (81i^2) \] \[ = 16 - 81(-1) = 16 + 81 = 97 \]
Now, substitute the numerator and denominator back into the expression for z:
\[ z = \frac{97 + 97i}{97} = \frac{97(1+i)}{97} = 1 + i \]
The complex number is \(z = 1 + 1i\). This is in the form \(a + bi\), with \(a = 1\) and \(b = 1\).

Since both \(a\) and \(b\) are positive, the complex number lies in the first quadrant.

The argument is:
\[ arg(z) = \tan^{-1}\left(\frac{b}{a}\right) = \tan^{-1}\left(\frac{1}{1}\right) = \tan^{-1}(1) = \frac{\pi}{4} \]

Step 4: Final Answer:

The argument of the complex number z is \( \frac{\pi}{4} \).
Quick Tip: Instead of simplifying, you can use the property \(arg(z_1/z_2) = arg(z_1) - arg(z_2)\). However, simplifying to the form \(a+bi\) is often more straightforward and less prone to errors, especially when the resulting number is simple like \(1+i\).

Step 1: Understanding the Question:

The question provides a relationship between \(\sin\theta\) and a variable \(x\), and asks for the value of an expression involving \(\sin(3\theta)\) and \(x^3\). The key is to find a relationship between \(\frac{1}{2}\left(x^3 + \frac{1}{x^3}\right)\) and \(\sin(3\theta)\) using the given equation.


Step 2: Key Formula or Approach:

We will use the given equation and the cubic expansion formula \((a+b)^3 = a^3 + b^3 + 3ab(a+b)\). We will also use the triple angle identity for sine: \(\sin(3\theta) = 3\sin\theta - 4\sin^3\theta\).


Step 3: Detailed Explanation:

We are given:
\[ \sin\theta = \frac{1}{2}\left(x + \frac{1}{x}\right) \implies 2\sin\theta = x + \frac{1}{x} \]
Let's cube both sides of this equation:
\[ (2\sin\theta)^3 = \left(x + \frac{1}{x}\right)^3 \] \[ 8\sin^3\theta = x^3 + \left(\frac{1}{x}\right)^3 + 3(x)\left(\frac{1}{x}\right)\left(x + \frac{1}{x}\right) \] \[ 8\sin^3\theta = \left(x^3 + \frac{1}{x^3}\right) + 3(1)\left(x + \frac{1}{x}\right) \]
We know that \(x + \frac{1}{x} = 2\sin\theta\). Substitute this back into the equation:
\[ 8\sin^3\theta = \left(x^3 + \frac{1}{x^3}\right) + 3(2\sin\theta) \] \[ 8\sin^3\theta = \left(x^3 + \frac{1}{x^3}\right) + 6\sin\theta \]
Now, isolate the term \(\left(x^3 + \frac{1}{x^3}\right)\):
\[ x^3 + \frac{1}{x^3} = 8\sin^3\theta - 6\sin\theta \]
The expression we need to evaluate is \( \sin 3\theta + \frac{1}{2}\left(x^3 + \frac{1}{x^3}\right) \).

Let's find the value of \( \frac{1}{2}\left(x^3 + \frac{1}{x^3}\right) \):
\[ \frac{1}{2}\left(x^3 + \frac{1}{x^3}\right) = \frac{1}{2}(8\sin^3\theta - 6\sin\theta) = 4\sin^3\theta - 3\sin\theta \]
This expression is related to the triple angle formula for sine. Recall:
\[ \sin(3\theta) = 3\sin\theta - 4\sin^3\theta \]
Therefore,
\[ 4\sin^3\theta - 3\sin\theta = -(3\sin\theta - 4\sin^3\theta) = -\sin(3\theta) \]
So, we have found that:
\[ \frac{1}{2}\left(x^3 + \frac{1}{x^3}\right) = -\sin(3\theta) \]
Now, substitute this into the original expression we need to find:
\[ \sin 3\theta + \frac{1}{2}\left(x^3 + \frac{1}{x^3}\right) = \sin 3\theta + (-\sin 3\theta) = 0 \]

Step 4: Final Answer:

The value of the expression is 0.
Quick Tip: Whenever you see expressions like \(x + 1/x\), \(x^2 + 1/x^2\), or \(x^3 + 1/x^3\), think about algebraic manipulation by squaring or cubing the base expression. The structure often leads to cancellations or connections with trigonometric identities, as seen in this problem.

Step 1: Understanding the Question:

The question asks for the coordinates of the incenter of a triangle with given vertices A, B, and C. The incenter is the point of concurrency of the angle bisectors of a triangle.


Step 2: Key Formula or Approach:

The coordinates of the incenter (I) of a triangle with vertices A(\(x_1, y_1\)), B(\(x_2, y_2\)), and C(\(x_3, y_3\)) are given by the formula:
\[ I = \left( \frac{ax_1 + bx_2 + cx_3}{a+b+c}, \frac{ay_1 + by_2 + cy_3}{a+b+c} \right) \]
where \(a, b, c\) are the lengths of the sides opposite to vertices A, B, and C, respectively.

We need to calculate the side lengths first.


Step 3: Detailed Explanation:

The given vertices are A(0,0), B(3,0), and C(0,-4).

Let's assign (\(x_1, y_1\)) = (0,0), (\(x_2, y_2\)) = (3,0), (\(x_3, y_3\)) = (0,-4).


Calculate side lengths:

- Side \(a\) is the length of BC (opposite to vertex A):
\[ a = \sqrt{(0-3)^2 + (-4-0)^2} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
- Side \(b\) is the length of AC (opposite to vertex B):
\[ b = \sqrt{(0-0)^2 + (-4-0)^2} = \sqrt{0^2 + (-4)^2} = \sqrt{16} = 4 \]
- Side \(c\) is the length of AB (opposite to vertex C):
\[ c = \sqrt{(3-0)^2 + (0-0)^2} = \sqrt{3^2 + 0^2} = \sqrt{9} = 3 \]
The perimeter is \(a+b+c = 5+4+3 = 12\).


Calculate incenter coordinates:

- x-coordinate of incenter:
\[ I_x = \frac{a x_1 + b x_2 + c x_3}{a+b+c} = \frac{5(0) + 4(3) + 3(0)}{12} = \frac{0 + 12 + 0}{12} = \frac{12}{12} = 1 \]
- y-coordinate of incenter:
\[ I_y = \frac{a y_1 + b y_2 + c y_3}{a+b+c} = \frac{5(0) + 4(0) + 3(-4)}{12} = \frac{0 + 0 - 12}{12} = \frac{-12}{12} = -1 \]
So, the coordinates of the incenter are (1, -1).


Step 4: Final Answer:

The calculated coordinates of the incenter are (1, -1). Comparing this result with the given options, none of the options match. There seems to be an error in the question's options. For the given vertices, the correct incenter is (1, -1).
Quick Tip: For a right-angled triangle with vertices at the origin (0,0) and on the axes at (p, 0) and (0, q), the inradius \(r\) is given by \(r = \frac{p+q-h}{2}\), where \(h\) is the hypotenuse. The incenter is at (\(r, r\)) if in the first quadrant, (\(-r, r\)), etc. Here, p=3, q=-4, h=5. \(r = \frac{3+4-5}{2} = 1\). The incenter for vertices (0,0), (3,0), (0,-4) is at (r, -r), which is (1, -1). This is a quick check for right-angled triangles on axes.

Step 1: Understanding the Question:

We need to find the equations of the tangent lines to a given circle. The condition is that these tangents must be perpendicular to a given line.


Step 2: Key Formula or Approach:

1. Find the slope of the given line.

2. Find the slope of the required tangents using the condition of perpendicularity (\(m_1 m_2 = -1\)).

3. Use the equation of a tangent to a circle \(x^2 + y^2 = r^2\) with slope \(m\): \(y = mx \pm r\sqrt{1+m^2}\).

4. Rearrange the equation into the form given in the options.


Step 3: Detailed Explanation:

1. Analyze the circle:

The equation of the circle is \(x^2 + y^2 = 36\).

This is a circle centered at the origin (0,0) with radius \(r^2 = 36\), so \(r=6\).


2. Find the slope of the tangent:

The given line is \(5x + y - 2 = 0\). We can write it in slope-intercept form \(y = -5x + 2\).

The slope of this line is \(m_1 = -5\).

The tangents are perpendicular to this line. Let the slope of the tangent be \(m\).

The condition for perpendicular lines is \(m \times m_1 = -1\).
\[ m \times (-5) = -1 \implies m = \frac{-1}{-5} = \frac{1}{5} \]
So, the slope of the required tangents is \(m = 1/5\).


3. Find the equation of the tangents:

The equation of a tangent with slope \(m\) to the circle \(x^2+y^2=r^2\) is:
\[ y = mx \pm r\sqrt{1+m^2} \]
Substitute \(m = 1/5\) and \(r = 6\):
\[ y = \frac{1}{5}x \pm 6\sqrt{1 + \left(\frac{1}{5}\right)^2} \] \[ y = \frac{1}{5}x \pm 6\sqrt{1 + \frac{1}{25}} \] \[ y = \frac{1}{5}x \pm 6\sqrt{\frac{25+1}{25}} \] \[ y = \frac{1}{5}x \pm 6\sqrt{\frac{26}{25}} \] \[ y = \frac{1}{5}x \pm 6\frac{\sqrt{26}}{5} \]
4. Rearrange the equation:

Multiply the entire equation by 5 to clear the denominator:
\[ 5y = x \pm 6\sqrt{26} \]
Rearrange to match the form in the options:
\[ x - 5y \pm 6\sqrt{26} = 0 \]
This equation represents the two parallel tangents.


Step 4: Final Answer:

The equations of the tangents are \( x - 5y \pm 6\sqrt{26} = 0 \), which matches option (A).
Quick Tip: Remember the standard forms for tangents to a circle. For \(x^2+y^2=r^2\), the tangent equation is \(y = mx \pm r\sqrt{1+m^2}\). For a circle \((x-h)^2+(y-k)^2=r^2\), the equation is \((y-k) = m(x-h) \pm r\sqrt{1+m^2}\). Always start by identifying the circle's center and radius.

Step 1: Understanding the Question:

We are given two piecewise functions, \(f(x)\) and \(g(x)\), and we need to find the range of their composition, \(f(g(x))\). The range of \(f(g(x))\) is the set of all possible output values of \(f\) when the input is taken from the range of \(g\).


Step 2: Key Formula or Approach:

1. Determine the domain of the composite function \(f(g(x))\). This is the set of \(x\) in the domain of \(g\) for which \(g(x)\) is in the domain of \(f\).

2. Determine the range of the inner function, \(g(x)\), over its domain. Let's call this Range(\(g\)).

3. Determine the range of the outer function, \(f(x)\), for the inputs that come from Range(\(g\)).


Step 3: Detailed Explanation:

1. Analyze the functions and their domains:

Domain of \(f\), \(D_f = [-3, 2] \cup (-1, 0) = [-3, 2]\).

Domain of \(g\), \(D_g = [-2, 3] \cup [0, 1] = [-2, 3]\).

The domain of \(f(g(x))\) is the domain of \(g\), which is \(x \in [-2, 3]\).


2. Find the range of g(x):

We need to find the set of all values \(g(x)\) can take for \(x \in [-2, 3]\).

- For \(x \in [-2, 3]\), \(g(x) = -x\). The range for this part is \(g(-2) = 2\) and \(g(3) = -3\). Since it's a line, the range is \([-3, 2]\).

- For \(x \in [0, 1]\), \(g(x) = x\). The range for this part is \([0, 1]\).

The overall range of \(g(x)\) is the union of these ranges: \([-3, 2] \cup [0, 1] = [-3, 2]\).

So, Range(\(g\)) = \([-3, 2]\).


3. Find the range of f(y) for y \(\in\) Range(g):

The range of \(g(x)\) becomes the set of inputs for \(f(x)\). So we need to find the range of \(f(y)\) for \(y \in [-3, 2]\).

Let's look at the definition of \(f(x)\):
\( f(x) = \begin{cases} 3-x, & -1 < x < 0
1+\frac{5x}{3}, & -3 \le x \le 2 \end{cases} \)

The input domain we are interested in is \(y \in [-3, 2]\). This entire interval falls into the second piece of the definition of \(f(x)\).

So, we need to find the range of the function \(h(y) = 1 + \frac{5y}{3}\) on the interval \(y \in [-3, 2]\).

Since \(h(y)\) is a linear function with a positive slope (5/3), it is monotonically increasing. Its minimum and maximum values will occur at the endpoints of the interval.

- At the minimum input \(y = -3\):

\(f(-3) = 1 + \frac{5(-3)}{3} = 1 - 5 = -4\).

- At the maximum input \(y = 2\):

\(f(2) = 1 + \frac{5(2)}{3} = 1 + \frac{10}{3} = \frac{3+10}{3} = \frac{13}{3}\).

Since the function is continuous on this interval, the range of \(f(g(x))\) is the closed interval from the minimum to the maximum value.

Range of \(f(g(x))\) = \([-4, \frac{13}{3}]\).


Step 4: Final Answer:

The range of the composite function (fog)(x) is \( [-4, \frac{13}{3}] \), which matches option (C).
Quick Tip: To find the range of a composite function \(f(g(x))\), first find the range of the inner function \(g(x)\). Then, use this range as the domain for the outer function \(f(x)\) to find the final range. For piecewise functions, be careful to match the range of \(g(x)\) to the correct piece(s) of the definition of \(f(x)\).

Step 1: Understanding the Question:

The question asks for the coordinates of the foci of a given conic section. We first need to identify the type of conic and its standard form.


Step 2: Key Formula or Approach:

1. Rearrange the given equation into the standard form of an ellipse, \(\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1\) (for a vertical ellipse) or \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) (for a horizontal ellipse), by completing the square.

2. Identify the center (h, k), and the values of \(a^2\) and \(b^2\).

3. Calculate the distance from the center to the foci, \(c\), using the formula \(c^2 = a^2 - b^2\).

4. Determine the coordinates of the foci, which are (h, k \( \pm \) c) for a vertical ellipse or (h \( \pm \) c, k) for a horizontal ellipse.


Step 3: Detailed Explanation:

The given equation is \( 25x^2 + 16y^2 - 150x = 175 \).

1. Complete the square:

Group the x-terms and y-terms:
\[ (25x^2 - 150x) + 16y^2 = 175 \]
Factor out the coefficient of the squared term from the x-group:
\[ 25(x^2 - 6x) + 16y^2 = 175 \]
To complete the square for \(x^2 - 6x\), we take half of the coefficient of x (-6), which is -3, and square it to get 9. We add and subtract 9 inside the parenthesis:
\[ 25(x^2 - 6x + 9 - 9) + 16y^2 = 175 \] \[ 25((x - 3)^2 - 9) + 16y^2 = 175 \]
Distribute the 25:
\[ 25(x - 3)^2 - 25(9) + 16y^2 = 175 \] \[ 25(x - 3)^2 - 225 + 16y^2 = 175 \]
Move the constant term to the right side:
\[ 25(x - 3)^2 + 16y^2 = 175 + 225 \] \[ 25(x - 3)^2 + 16y^2 = 400 \]
Divide the entire equation by 400 to get 1 on the right side:
\[ \frac{25(x - 3)^2}{400} + \frac{16y^2}{400} = 1 \] \[ \frac{(x - 3)^2}{16} + \frac{y^2}{25} = 1 \]
2. Identify parameters:

This is the standard form of a vertical ellipse because the denominator under the \(y^2\) term (25) is larger than the denominator under the \(x^2\) term (16).

- Center (h, k) = (3, 0).

- \(a^2 = 25 \implies a = 5\) (major radius).

- \(b^2 = 16 \implies b = 4\) (minor radius).

3. Calculate c:
\[ c^2 = a^2 - b^2 = 25 - 16 = 9 \] \[ c = \sqrt{9} = 3 \]
4. Find the foci:

For a vertical ellipse, the foci are located at (h, k \( \pm \) c).

Foci = (3, 0 \( \pm \) 3).

The coordinates of the two foci are (3, 3) and (3, -3). This can be written as (3, \( \pm \)3).


Step 4: Final Answer:

The foci of the conic are at (3, \( \pm \)3), which matches option (B).
Quick Tip: To quickly determine if an ellipse is horizontal or vertical, look at the coefficients of \(x^2\) and \(y^2\) in the standard form. The larger denominator indicates the direction of the major axis. If it's under \(y^2\), the ellipse is vertical. If it's under \(x^2\), it's horizontal. The foci always lie on the major axis.

Step 1: Understanding the Question:

We are asked to evaluate a limit, A, which is of the indeterminate form \(1^\infty\). After finding the value of A, we need to calculate its natural logarithm (log\(_e\) A).


Step 2: Key Formula or Approach:

For limits of the form \( \lim_{x \to a} [f(x)]^{g(x)} \) which result in \(1^\infty\), we can use the following formula:
\[ \lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} [f(x)-1]g(x)} \]
We will also use the standard trigonometric limit: \( \lim_{u \to 0} \frac{\tan u}{u} = 1 \).


Step 3: Detailed Explanation:

The given limit is \( A = \lim_{x \to 0^+} (1 + \tan^2 \sqrt{x})^{\frac{1}{2x}} \).

Here, \(f(x) = 1 + \tan^2 \sqrt{x}\) and \(g(x) = \frac{1}{2x}\).

As \(x \to 0^+\), \( \sqrt{x} \to 0 \), so \( \tan^2 \sqrt{x} \to 0 \). Thus, \(f(x) \to 1\).

As \(x \to 0^+\), \( g(x) = \frac{1}{2x} \to \infty \).

So, the limit is of the form \(1^\infty\).


Using the formula, we can write A as:
\[ A = e^L \]
where \( L = \lim_{x \to 0^+} [ (1 + \tan^2 \sqrt{x}) - 1 ] \times \frac{1}{2x} \)
\[ L = \lim_{x \to 0^+} [ \tan^2 \sqrt{x} ] \times \frac{1}{2x} \] \[ L = \lim_{x \to 0^+} \frac{\tan^2 \sqrt{x}}{2x} \]
To evaluate this limit, we can manipulate it to use the standard limit \( \lim_{u \to 0} \frac{\tan u}{u} = 1 \).

Let's rewrite the expression for L:
\[ L = \frac{1}{2} \lim_{x \to 0^+} \frac{\tan^2 \sqrt{x}}{x} \] \[ L = \frac{1}{2} \lim_{x \to 0^+} \left( \frac{\tan \sqrt{x}}{\sqrt{x}} \right)^2 \]
Let \(u = \sqrt{x}\). As \(x \to 0^+\), \(u \to 0^+\). The limit becomes:
\[ L = \frac{1}{2} \left( \lim_{u \to 0^+} \frac{\tan u}{u} \right)^2 \]
Since \( \lim_{u \to 0^+} \frac{\tan u}{u} = 1 \):
\[ L = \frac{1}{2} (1)^2 = \frac{1}{2} \]
So, the value of the original limit is \( A = e^L = e^{1/2} \).


The question asks for log\(_e\) A:
\[ \log_e A = \log_e (e^{1/2}) \]
Using the property \( \log_b (b^p) = p \):
\[ \log_e A = \frac{1}{2} \]

Step 4: Final Answer:

The value of log\(_e\) A is \( \frac{1}{2} \), which corresponds to option (C).
Quick Tip: Recognizing the indeterminate form \(1^\infty\) is the first and most crucial step. The formula \(e^{\lim (f(x)-1)g(x)}\) is a powerful tool for these types of limits. Always try to manipulate the expression in the exponent to match standard limit forms like \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) or \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \).

Step 1: Understanding the Question:

We need to determine the continuity of the function \(f(x) = 2x - |x - x^2|\) at \(x=1\). A function is continuous at a point \(c\) if the limit as \(x\) approaches \(c\) exists, the function is defined at \(c\), and the limit equals the function's value at \(c\). That is, \( \lim_{x \to c} f(x) = f(c) \).


Step 2: Key Formula or Approach:

To check continuity at \(x=1\), we must evaluate:

1. The left-hand limit (LHL): \( \lim_{x \to 1^-} f(x) \)

2. The right-hand limit (RHL): \( \lim_{x \to 1^+} f(x) \)

3. The value of the function at the point: \(f(1)\)

If LHL = RHL = \(f(1)\), the function is continuous at \(x=1\).

The absolute value function \(|y|\) is defined as \(y\) if \(y \ge 0\) and \(-y\) if \(y < 0\). We need to analyze the sign of \(x-x^2\) around \(x=1\).


Step 3: Detailed Explanation:

Let's analyze the term inside the absolute value: \(g(x) = x - x^2 = x(1-x)\).

- If \(x\) is slightly less than 1 (e.g., \(x=0.9\)), then \(x>0\) and \(1-x>0\), so \(x(1-x) > 0\).

- If \(x\) is slightly greater than 1 (e.g., \(x=1.1\)), then \(x>0\) and \(1-x<0\), so \(x(1-x) < 0\).


1. Left-Hand Limit (LHL): \(x \to 1^-\)

For \(x < 1\) (and \(x > 0\)), \(x - x^2 > 0\). So, \(|x - x^2| = x - x^2\).
\[ f(x) = 2x - (x - x^2) = 2x - x + x^2 = x + x^2 \] \[ LHL = \lim_{x \to 1^-} (x + x^2) = 1 + 1^2 = 2 \]

2. Right-Hand Limit (RHL): \(x \to 1^+\)

For \(x > 1\), \(x - x^2 < 0\). So, \(|x - x^2| = -(x - x^2) = x^2 - x\).
\[ f(x) = 2x - (x^2 - x) = 2x - x^2 + x = 3x - x^2 \] \[ RHL = \lim_{x \to 1^+} (3x - x^2) = 3(1) - 1^2 = 3 - 1 = 2 \]

3. Function Value at x = 1:
\[ f(1) = 2(1) - |1 - 1^2| = 2 - |0| = 2 \]

Since LHL = RHL = f(1) = 2, the function is continuous at x = 1.


Step 4: Final Answer:

The function \(f(x)\) is continuous at x = 1, which corresponds to option (A).
Quick Tip: When dealing with continuity of functions involving absolute values, the key is to analyze the sign of the expression inside the absolute value around the point of interest. This allows you to rewrite the function as a piecewise function without the absolute value, making it easier to evaluate the left-hand and right-hand limits.

Step 1: Understanding the Question:

The question asks us to determine the truth value (True or False) of three conditional statements (implications) labeled (A), (B), and (C), and then to identify which of the final options correctly describes these truth values.


Step 2: Key Formula or Approach:

A conditional statement "If P, then Q" (written as P \( \rightarrow \) Q) is only false when the hypothesis P is true and the conclusion Q is false. In all other cases, the implication is true.
Truth Table for P \( \rightarrow \) Q:

- T \( \rightarrow \) T is True

- T \( \rightarrow \) F is False

- F \( \rightarrow \) T is True

- F \( \rightarrow \) F is True


Step 3: Detailed Explanation:

Let's analyze each statement's truth value.

Statement (A): If 4+3 = 8, then 5+3=9

- Hypothesis (P): "4+3 = 8" is False.

- Conclusion (Q): "5+3 = 9" is False.

- The implication is F \( \rightarrow \) F, which is True. So, statement (A) is True.


Statement (B): If 6 + 4 = 10, then moon is flat

- Hypothesis (P): "6 + 4 = 10" is True.

- Conclusion (Q): "moon is flat" is False.

- The implication is T \( \rightarrow \) F, which is False. So, statement (B) is False.


Statement (C): If both (A) and (B) are true, then 5 + 6 = 17

- Hypothesis (P): "both (A) and (B) are true". We found that (A) is True and (B) is False. Therefore, the statement "both (A) and (B) are true" is False.

- Conclusion (Q): "5 + 6 = 17" is False.

- The implication is F \( \rightarrow \) F, which is True. So, statement (C) is True.


Summary of Truth Values:

- Statement (A) is True.

- Statement (B) is False.

- Statement (C) is True.


Evaluate the Options:

- Option (A): (A) is true while (B) and (C) are false. (Incorrect, C is true).

- Option (B): (A) and (B) are false, while (C) is true. (Incorrect, A is true).

- Option (C): (A) and (C) are true, while (B) is false. (Correct).

- Option (D): (A) is false, but (B) and (C) are true. (Incorrect, A is true and B is false).


Step 4: Final Answer:

The correct description of the truth values is given in option (C).
Quick Tip: The most important rule to remember for logical implication (If P, then Q) is that it's always true unless you start with a true statement (P) and conclude with a false one (Q). This is often summarized as "a false premise can imply anything".

Step 1: Understanding the Question:

The question asks for the ratio of the determinant of the adjoint of matrix \( B \) to the determinant of matrix \( C \), where \( B \) is the adjoint of \( A \) and \( C \) is \( 5A \).


Step 2: Key Formula or Approach:

For a square matrix \( A \) of order \( n = 3 \):

1. \( |A| = \det(A) \)

2. \( |adj A| = |A|^{n-1} = |A|^2 \)

3. \( adj B = adj(adj A) \implies |adj B| = |A|^{(n-1)^2} = |A|^4 \)

4. \( |kA| = k^n |A| = 5^3 |A| = 125|A| \)


Step 3: Detailed Explanation:

First, calculate the determinant of \( A \):
\[ |A| = 1(0 - (-3)) - (-1)(0 - (-6)) + 1(0 - 4) \] \[ |A| = 1(3) + 1(6) - 4 = 3 + 6 - 4 = 5 \]
Now, find \( |adj B| \):
\[ |adj B| = |adj(adj A)| = |A|^4 = 5^4 = 625 \]
Next, find \( |C| \):
\[ |C| = |5A| = 5^3 \times |A| = 125 \times 5 = 625 \]
Calculate the ratio:
\[ \frac{|adj B|}{|C|} = \frac{625}{625} = 1 \]

Step 4: Final Answer:

The ratio is 1.
Quick Tip: Remember the general property \( |adj(adj A)| = |A|^{(n-1)^2} \). For a \( 3 \times 3 \) matrix, this simplifies to \( |A|^4 \). Also, \( |kA| = k^n|A| \) is a frequently tested property.

Step 1: Understanding the Question:

We need to find the number of real solutions for the given quadratic equation in \( \sin x \) within the domain \( [0, 3\pi] \).


Step 2: Key Formula or Approach:
Let \( t = \sin x \). The equation becomes a standard quadratic equation \( 2t^2 + 5t - 3 = 0 \). Solve for \( t \) and then find the corresponding values of \( x \).


Step 3: Detailed Explanation:

Solving \( 2t^2 + 5t - 3 = 0 \):
\[ 2t^2 + 6t - t - 3 = 0 \implies 2t(t + 3) - 1(t + 3) = 0 \] \[ (2t - 1)(t + 3) = 0 \implies t = \frac{1}{2} or t = -3 \]
Since \( \sin x \) must lie in the range \( [-1, 1] \), \( \sin x = -3 \) is not possible.

Thus, \( \sin x = \frac{1}{2} \).

In the interval \( [0, 2\pi] \), \( \sin x = \frac{1}{2} \) at \( x = \frac{\pi}{6} \) and \( x = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \).

In the interval \( [2\pi, 3\pi] \), \( \sin x = \frac{1}{2} \) at \( x = 2\pi + \frac{\pi}{6} = \frac{13\pi}{6} \) and \( x = 3\pi - \frac{\pi}{6} = \frac{17\pi}{6} \).

All these values lie within \( [0, 3\pi] \).

Total number of solutions = 4.


Step 4: Final Answer:

There are 4 solutions.
Quick Tip: Always check the range of trigonometric functions. \( |\sin x| \le 1 \) immediately eliminates extraneous solutions from quadratic forms. Visualizing the sine curve helps in counting roots across periodic intervals quickly.

Step 1: Understanding the Question:

The general second-degree equation represents a pair of lines if its discriminant \( \Delta = 0 \). We first find \( \lambda \), then the angle between the lines.


Step 2: Key Formula or Approach:

For \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \):

1. \( \Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0 \)

2. \( \tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} \right| \)


Step 3: Detailed Explanation:

Comparing with the general equation: \( a=1, 2h=-3 \implies h=-1.5, b=\lambda, 2g=3 \implies g=1.5, 2f=-5 \implies f=-2.5, c=2 \).

Using \( \Delta = 0 \):
\[ 1(\lambda)(2) + 2(-2.5)(1.5)(-1.5) - 1(-2.5)^2 - \lambda(1.5)^2 - 2(-1.5)^2 = 0 \] \[ 2\lambda + 11.25 - 6.25 - 2.25\lambda - 4.5 = 0 \] \[ -0.25\lambda + 0.5 = 0 \implies \lambda = 2 \]
Now calculate \( \tan \theta \):
\[ \tan \theta = \frac{2\sqrt{(-1.5)^2 - (1)(2)}}{1+2} = \frac{2\sqrt{2.25 - 2}}{3} = \frac{2\sqrt{0.25}}{3} = \frac{2 \times 0.5}{3} = \frac{1}{3} \]
Since \( \tan \theta = \frac{1}{3} \), we have \( cot \theta = 3 \).
\( cosec^2 \theta = 1 + cot^2 \theta = 1 + 9 = 10 \).

The required value is \( \frac{10}{\sqrt{10}} = \sqrt{10} \).


Step 4: Final Answer:

The value is \( \sqrt{10} \).
Quick Tip: The angle between a pair of lines depends only on the homogeneous part (\( ax^2+2hxy+by^2 \)). Once you find the coefficients, using \( cosec^2\theta = \frac{(a+b)^2 + 4(h^2-ab)}{4(h^2-ab)} \) can also directly give the result.

Step 1: Understanding the Question:

We need to find the angle between two lines whose direction cosines \( (l, m, n) \) satisfy two given equations.


Step 2: Key Formula or Approach:

Substitute \( m = -(3l + 5n) \) from the linear equation into the quadratic equation to find ratios of \( l, m, n \). These ratios will give the direction ratios of the two lines. Then use \( \cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}} \).


Step 3: Detailed Explanation:

From \( 3l + m + 5n = 0 \), we get \( m = -3l - 5n \).

Substitute into \( 6mn - 2nl + 5lm = 0 \):
\[ 6n(-3l - 5n) - 2nl + 5l(-3l - 5n) = 0 \] \[ -18nl - 30n^2 - 2nl - 15l^2 - 25ln = 0 \] \[ -15l^2 - 45ln - 30n^2 = 0 \implies l^2 + 3ln + 2n^2 = 0 \] \[ (l + n)(l + 2n) = 0 \implies l = -n or l = -2n \]
Case 1: If \( l = -n \), then \( m = -3(-n) - 5n = -2n \). DRs: \( (-n, -2n, n) \to (1, 2, -1) \).

Case 2: If \( l = -2n \), then \( m = -3(-2n) - 5n = n \). DRs: \( (-2n, n, n) \to (-2, 1, 1) \).

Calculate \( \cos \theta \):
\[ \cos \theta = \frac{|1(-2) + 2(1) + (-1)(1)|}{\sqrt{1+4+1}\sqrt{4+1+1}} = \frac{|-2 + 2 - 1|}{6} = \frac{1}{6} \]
Thus, \( \sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \frac{1}{36}} = \frac{\sqrt{35}}{6} \).


Step 4: Final Answer:

The value is \( \frac{\sqrt{35}}{6} \).
Quick Tip: For equations of the form \( al+bm+cn=0 \) and \( fmn+gnl+hlm=0 \), the lines are perpendicular if \( \frac{f}{a} + \frac{g}{b} + \frac{h}{c} = 0 \). Here, \( \frac{6}{-5} + \frac{-2}{1} + \frac{5}{3} \ne 0 \), so calculate \( \cos \theta \) step by step.

Step 1: Understanding the Question:

This is a Binomial Distribution problem with \( n = 6 \) and \( p = \frac{1}{2} \). We need to calculate the probability for the range defined by \( |X - 2| \le 1 \).


Step 2: Key Formula or Approach:

1. \( |X - 2| \le 1 \implies -1 \le X - 2 \le 1 \implies 1 \le X \le 3 \).

2. \( P(X=k) = \binom{n}{k} p^k q^{n-k} \).

3. Since \( p = q = \frac{1}{2} \), \( P(X=k) = \binom{6}{k} \left(\frac{1}{2}\right)^6 = \frac{\binom{6}{k}}{64} \).


Step 3: Detailed Explanation:

We need \( P(1 \le X \le 3) = P(X=1) + P(X=2) + P(X=3) \).
\[ P(X=1) = \frac{\binom{6}{1}}{64} = \frac{6}{64} \] \[ P(X=2) = \frac{\binom{6}{2}}{64} = \frac{15}{64} \] \[ P(X=3) = \frac{\binom{6}{3}}{64} = \frac{20}{64} \]
Summing them up:
\[ P(1 \le X \le 3) = \frac{6 + 15 + 20}{64} = \frac{41}{64} \]

Step 4: Final Answer:

The probability is \( \frac{41}{64} \).
Quick Tip: When \( p = 0.5 \), the binomial distribution is symmetric. This means \( P(X=k) = P(X=n-k) \). Using the row of Pascal's triangle for \( n=6 \) (\( 1, 6, 15, 20, 15, 6, 1 \)) can save time in calculating combinations.

Step 1: Understanding the Question:

We are given two conditions to find vector \( \bar{p} \). One is a dot product constraint, and the other is a cross product relationship.


Step 2: Key Formula or Approach:

1. \( \bar{p} \times \bar{b} = \bar{c} \times \bar{b} \implies (\bar{p} - \bar{c}) \times \bar{b} = 0 \). This means \( \bar{p} - \bar{c} \) is parallel to \( \bar{b} \).

2. So, \( \bar{p} = \bar{c} + \lambda \bar{b} \) for some scalar \( \lambda \).

3. Use \( \bar{p} \cdot \bar{a} = 0 \) to find \( \lambda \).


Step 3: Detailed Explanation:

Substitute \( \bar{p} = \bar{c} + \lambda \bar{b} \) into \( \bar{p} \cdot \bar{a} = 0 \):
\[ (\bar{c} + \lambda \bar{b}) \cdot \bar{a} = 0 \implies \bar{c} \cdot \bar{a} + \lambda (\bar{b} \cdot \bar{a}) = 0 \]
Calculate the dot products:
\[ \bar{c} \cdot \bar{a} = (3, -1, 1) \cdot (1, 1, 0) = 3 - 1 = 2 \] \[ \bar{b} \cdot \bar{a} = (2, 0, -1) \cdot (1, 1, 0) = 2 + 0 = 2 \]
Now solve for \( \lambda \):
\[ 2 + \lambda(2) = 0 \implies 2\lambda = -2 \implies \lambda = -1 \]
Find \( \bar{p} \):
\[ \bar{p} = \bar{c} - 1\bar{b} = (3\hat{i} - \hat{j} + \hat{k}) - (2\hat{i} - \hat{k}) = \hat{i} - \hat{j} + 2\hat{k} \]

Step 4: Final Answer:
The vector is \( \hat{i} - \hat{j} + 2\hat{k} \).
Quick Tip: The relation \( \bar{u} \times \bar{v} = \bar{w} \times \bar{v} \) always implies \( \bar{u} = \bar{w} + \lambda \bar{v} \). This is a very efficient way to solve vector equation problems without expanding into components immediately.

Step 1: Understanding the Question:

First, find the normalization constant \( k \). Then use the given probability to set up an integral and solve for \( a \).


Step 2: Key Formula or Approach:

1. Total probability: \( \int_{0}^{1} f(x) dx = 1 \).

2. \( P(X > a) = \int_{a}^{1} f(x) dx = \frac{20}{27} \).


Step 3: Detailed Explanation:

Finding \( k \):
\[ \int_{0}^{1} k(x - x^2) dx = k \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = k \left( \frac{1}{2} - \frac{1}{3} \right) = \frac{k}{6} = 1 \implies k = 6 \]
Now use \( P(X > a) = \frac{20}{27} \):
\[ \int_{a}^{1} 6(x - x^2) dx = 6 \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_a^1 = 6 \left[ \left(\frac{1}{6}\right) - \left(\frac{a^2}{2} - \frac{a^3}{3}\right) \right] = 1 - 3a^2 + 2a^3 = \frac{20}{27} \] \[ 2a^3 - 3a^2 + 1 - \frac{20}{27} = 0 \implies 2a^3 - 3a^2 + \frac{7}{27} = 0 \]
Testing \( a = \frac{1}{3} \):
\[ 2\left(\frac{1}{27}\right) - 3\left(\frac{1}{9}\right) + \frac{7}{27} = \frac{2}{27} - \frac{9}{27} + \frac{7}{27} = 0 \]
So, \( a = \frac{1}{3} \).


Step 4: Final Answer:

The value of \( a \) is \( \frac{1}{3} \).
Quick Tip: Checking the options after deriving the final polynomial equation is often faster than solving the cubic equation analytically.

Step 1: Understanding the Question:

We need to find the variance of a discrete random variable given its probability distribution table.


Step 2: Key Formula or Approach:

1. \( E(X) = \mu = \sum x_i P(x_i) \)

2. \( E(X^2) = \sum x_i^2 P(x_i) \)

3. \( Var(X) = E(X^2) - [E(X)]^2 \)


Step 3: Detailed Explanation:

From the table: \( X_i = \{0, 1, 2, 3, 4\} \), \( P_i = \{0.4, 0.3, 0.1, 0.1, 0.1\} \).

Calculate \( E(X) \):
\[ E(X) = 0(0.4) + 1(0.3) + 2(0.1) + 3(0.1) + 4(0.1) = 0 + 0.3 + 0.2 + 0.3 + 0.4 = 1.2 \]
Calculate \( E(X^2) \):
\[ E(X^2) = 0^2(0.4) + 1^2(0.3) + 2^2(0.1) + 3^2(0.1) + 4^2(0.1) = 0 + 0.3 + 0.4 + 0.9 + 1.6 = 3.2 \]
Calculate \( Var(X) \):
\[ Var(X) = 3.2 - (1.2)^2 = 3.2 - 1.44 = 1.76 \]

Step 4: Final Answer:

The variance is 1.76.
Quick Tip: Double-check that the sum of probabilities \( \sum P(x_i) = 1 \). In this case \( 0.4+0.3+0.1+0.1+0.1 = 1 \), confirming the distribution is valid.

Step 1: Understanding the Question:

This is an inequality involving inverse trigonometric functions. The domain of both functions is \( [-1, 1] \).


Step 2: Key Formula or Approach:

1. Factorize the expression: \( (\cos^{-1} x - \sin^{-1} x)(\cos^{-1} x + \sin^{-1} x) > 0 \).

2. Use the identity: \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \).

3. Since \( \frac{\pi}{2} > 0 \), the inequality simplifies to \( \cos^{-1} x - \sin^{-1} x > 0 \implies \cos^{-1} x > \sin^{-1} x \).


Step 3: Detailed Explanation:

We need \( \cos^{-1} x > \sin^{-1} x \) for \( x \in [-1, 1] \).

At \( x = \frac{1}{\sqrt{2}} \), \( \cos^{-1} x = \sin^{-1} x = \frac{\pi}{4} \).

For \( x < \frac{1}{\sqrt{2}} \), \( \cos^{-1} x \) increases and \( \sin^{-1} x \) decreases (graphically).

For example, at \( x = 0 \), \( \cos^{-1}(0) = \frac{\pi}{2} \) and \( \sin^{-1}(0) = 0 \), which satisfies \( \frac{\pi}{2} > 0 \).

At \( x = -1 \), \( \cos^{-1}(-1) = \pi \) and \( \sin^{-1}(-1) = -\frac{\pi}{2} \), which satisfies \( \pi > -\frac{\pi}{2} \).

Thus, the condition holds for \( x \in [-1, \frac{1}{\sqrt{2}}) \).


Step 4: Final Answer:

The solution is \( -1 \le x < \frac{1}{\sqrt{2}} \).
Quick Tip: Using graphs of \( \sin^{-1} x \) and \( \cos^{-1} x \) is the easiest way to solve such inequalities. The point of intersection is always \( x = \frac{1}{\sqrt{2}} \).

Step 1: Understanding the Question:

We need to find vector \( \bar{d} \) first using the perpendicularity and dot product conditions, then calculate the magnitude squared of the cross product.


Step 2: Key Formula or Approach:

1. \( \bar{d} = \lambda (\bar{b} \times \bar{c}) \).

2. \( \bar{a} \cdot \bar{d} = 18 \implies \lambda \bar{a} \cdot (\bar{b} \times \bar{c}) = 18 \).

3. \( |\bar{a} \times \bar{d}|^2 = |\bar{a}|^2 |\bar{d}|^2 - (\bar{a} \cdot \bar{d})^2 \).


Step 3: Detailed Explanation:

Calculate \( \bar{b} \times \bar{c} \):
\[ \bar{b} \times \bar{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & -2 & -2
-1 & 4 & 3 \end{vmatrix} = \hat{i}(-6 + 8) - \hat{j}(3 - 2) + \hat{k}(4 - 2) = 2\hat{i} - \hat{j} + 2\hat{k} \]
Let \( \bar{d} = \lambda (2, -1, 2) \).

Given \( \bar{a} \cdot \bar{d} = 18 \implies (2, 3, 4) \cdot \lambda(2, -1, 2) = 18 \implies \lambda(4 - 3 + 8) = 18 \implies 9\lambda = 18 \implies \lambda = 2 \).

So, \( \bar{d} = (4, -2, 4) \).

Now, calculate magnitudes:
\( |\bar{a}|^2 = 2^2 + 3^2 + 4^2 = 4 + 9 + 16 = 29 \).
\( |\bar{d}|^2 = 4^2 + (-2)^2 + 4^2 = 16 + 4 + 16 = 36 \).

Calculate \( |\bar{a} \times \bar{d}|^2 \):
\[ |\bar{a} \times \bar{d}|^2 = (29)(36) - (18)^2 = 1044 - 324 = 720 \]

Step 4: Final Answer:

The value is 720.
Quick Tip: Using the identity \( |\bar{a} \times \bar{b}|^2 + (\bar{a} \cdot \bar{b})^2 = |\bar{a}|^2 |\bar{b}|^2 \) (Lagrange's Identity) is much faster than performing the actual cross product and then finding its magnitude.

Step 1: Understanding the Question:

This is an application of differential equations to a physical process (growth of a balloon).


Step 2: Key Formula or Approach:

1. Volume \( V = \frac{4}{3} \pi r^3 \), Surface Area \( S = 4\pi r^2 \).

2. Given \( \frac{dV}{dt} = kS \).


Step 3: Detailed Explanation:

Differentiate volume with respect to time: \( \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \).

Substitute into the given relation: \( 4\pi r^2 \frac{dr}{dt} = k(4\pi r^2) \implies \frac{dr}{dt} = k \).

Integrate: \( r = kt + C \).

At \( t = 0 \), \( r = 3 \implies 3 = k(0) + C \implies C = 3 \).

At \( t = 2 \), \( r = 9 \implies 9 = k(2) + 3 \implies 2k = 6 \implies k = 3 \).

The equation for radius is \( r = 3t + 3 \).

Now find \( r \) at \( t = 4 \):
\[ r = 3(4) + 3 = 12 + 3 = 15 cm \]

Step 4: Final Answer:

The radius after 4 minutes is 15 cm.
Quick Tip: In this specific scenario, since \( \frac{dV}{dt} \propto S \), the radius increases linearly with time (\( \frac{dr}{dt} = constant \)). You can use simple linear interpolation: radius increased by 6 cm in 2 mins, so it will increase by another 6 cm in the next 2 mins.

Step 1: Understanding the Question:

Three points on a line are collinear. The crossing point on the \( yz \)-plane has \( x = 0 \).


Step 2: Key Formula or Approach:

Use the condition that vectors formed by these points are parallel (direction ratios are proportional). Let \( P_1(a, 1, 6) \), \( P_2(3, 4, b) \), and \( P_3(0, 8.5, -6.5) \).


Step 3: Detailed Explanation:
Direction ratios of line \( P_2P_3 \):
\( (3-0, 4-8.5, b-(-6.5)) = (3, -4.5, b+6.5) \).

Direction ratios of line \( P_1P_2 \):
\( (a-3, 1-4, 6-b) = (a-3, -3, 6-b) \).

Since they are the same line, the DRs are proportional:
\[ \frac{a-3}{3} = \frac{-3}{-4.5} = \frac{6-b}{b+6.5} \]
From the middle term: \( \frac{-3}{-4.5} = \frac{3}{4.5} = \frac{30}{45} = \frac{2}{3} \).

Solving for \( a \):
\[ \frac{a-3}{3} = \frac{2}{3} \implies a-3 = 2 \implies a = 5 \]
Solving for \( b \):
\[ \frac{6-b}{b+6.5} = \frac{2}{3} \implies 3(6-b) = 2(b+6.5) \] \[ 18 - 3b = 2b + 13 \implies 5b = 5 \implies b = 1 \]
Calculate \( 3a + 4b \):
\[ 3(5) + 4(1) = 15 + 4 = 19 \]

Step 4: Final Answer:

The value is 19.
Quick Tip: For any point on a line segment dividing two points in ratio \( k:1 \), you can use the section formula. Here, the point with \( x=0 \) suggests \( 0 = \frac{k(3) + 1(a)}{k+1} \), which gives \( a = -3k \). Identifying the ratio from the y-coordinates makes it fast.

Step 1: Understanding the Question:

This equation is not directly integrable as a linear differential equation in \( y \). Let's check if it's linear in \( x \).


Step 2: Key Formula or Approach:

1. Rewrite as \( \frac{dx}{dy} = 2y - x \implies \frac{dx}{dy} + x = 2y \).

2. This is a linear differential equation of form \( \frac{dx}{dy} + P(y)x = Q(y) \).

3. Integrating factor \( IF = e^{\int P(y) dy} \).

4. Solution: \( x \cdot IF = \int Q(y) \cdot IF dy + C \).


Step 3: Detailed Explanation:

Here \( P(y) = 1 \) and \( Q(y) = 2y \).
\( IF = e^{\int 1 dy} = e^y \).

The solution is:
\[ x e^y = \int 2y e^y dy + c \]
Using integration by parts \( \int u v dy = u \int v dy - \int (u' \int v dy) dy \):
\[ x e^y = 2 [y e^y - \int e^y dy] + c = 2 [y e^y - e^y] + c = 2e^y(y - 1) + c \]
Divide by \( e^y \):
\[ x = 2(y - 1) + c e^{-y} \]

Step 4: Final Answer:

The solution is \( x = 2(y - 1) + ce^{-y} \).
Quick Tip: If a differential equation is difficult to solve as \( \frac{dy}{dx} \), always try reciprocalizing it to \( \frac{dx}{dy} \). This often converts it into a standard linear form.

Step 1: Understanding the Question:

Expand the derivative term first to identify the standard linear form.


Step 2: Key Formula or Approach:

For \( \frac{dy}{dx} + P(x)y = Q(x) \), the integrating factor is \( IF = e^{\int P(x) dx} \).


Step 3: Detailed Explanation:

Expand \( \frac{d}{dx}(xy) = x \frac{dy}{dx} + y \).

The equation becomes:
\[ y + x \frac{dy}{dx} + y = x(\sin x + \log x) \] \[ x \frac{dy}{dx} + 2y = x(\sin x + \log x) \]
Divide by \( x \):
\[ \frac{dy}{dx} + \frac{2}{x} y = \sin x + \log x \]
Here \( P(x) = \frac{2}{x} \).
\[ IF = e^{\int \frac{2}{x} dx} = e^{2 \log x} = e^{\log x^2} = x^2 \]

Step 4: Final Answer:

The integrating factor is \( x^2 \).
Quick Tip: Look for the term multiplying \( y \) after getting \( \frac{dy}{dx} \) alone. Using \( e^{n \log x} = x^n \) is a common simplification in these problems.

Step 1: Understanding the Question:

The number of arbitrary constants in a general solution indicates the order of the differential equation.


Step 2: Detailed Explanation:

There are two arbitrary constants, A and B. Therefore, the order of the differential equation must be 2.

Let's find the equation:

Differentiate once: \( 2Ax + 2Byy' = 0 \implies Ax + Byy' = 0 \dots (i) \)

Differentiate again: \( A + B(yy'' + (y')^2) = 0 \dots (ii) \)

From (i), \( A = -\frac{Byy'}{x} \). Substitute into (ii):
\[ -\frac{Byy'}{x} + B(yy'' + (y')^2) = 0 \]
Since \( B \ne 0 \), divide by B:
\[ -\frac{yy'}{x} + yy'' + (y')^2 = 0 \implies xyy'' + x(y')^2 - yy' = 0 \]
The highest order derivative is \( y'' \) (order 2) and its power is 1 (degree 1).


Step 4: Final Answer:

The differential equation has order 2 and degree 1.
Quick Tip: The order is always equal to the number of independent arbitrary constants. For degree, check the power of the highest order derivative after making the equation free of radicals/fractions.

Step 1: Understanding the Question:

We need to find the direction ratios of each line and then use the angle formula.


Step 2: Detailed Explanation:

Line 1: \( x = y, z = 0 \). In symmetric form, \( \frac{x}{1} = \frac{y}{1} = \frac{z}{0} \).

DRs of line 1 are \( (1, 1, 0) \).

Line 2: \( y = 0, z = 0 \). This is the x-axis. In symmetric form, \( \frac{x}{1} = \frac{y}{0} = \frac{z}{0} \).

DRs of line 2 are \( (1, 0, 0) \).

Calculate the angle \( \theta \):
\[ \cos \theta = \frac{|(1)(1) + (1)(0) + (0)(0)|}{\sqrt{1^2+1^2+0^2}\sqrt{1^2+0^2+0^2}} = \frac{1}{\sqrt{2} \cdot 1} = \frac{1}{\sqrt{2}} \]
Thus, \( \theta = 45^\circ \).


Step 4: Final Answer:

The angle is \( 45^\circ \).
Quick Tip: Line 1 lies in the xy-plane and bisects the angle between the positive x and y axes. Since the second line is the x-axis itself, the angle is naturally half of \( 90^\circ \), which is \( 45^\circ \).

Step 1: Understanding the Question:

We use the square of the sum of three vectors identity.


Step 2: Key Formula or Approach:
\( |\bar{a} + \bar{b} + \bar{c}|^2 = |\bar{a}|^2 + |\bar{b}|^2 + |\bar{c}|^2 + 2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a}) \).


Step 3: Detailed Explanation:

Given \( \bar{a} + \bar{b} + \bar{c} = \bar{0} \), so \( |\bar{a} + \bar{b} + \bar{c}|^2 = 0 \).
\[ 0 = 3^2 + 4^2 + 5^2 + 2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a}) \] \[ 0 = 9 + 16 + 25 + 2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a}) \] \[ 0 = 50 + 2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a}) \] \[ 2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a}) = -50 \] \[ \bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a} = -25 \]

Step 4: Final Answer:

The value is -25.
Quick Tip: Notice that \( 3^2 + 4^2 = 5^2 \). This implies vectors \( \bar{a}, \bar{b} \) are perpendicular and form a right-angled triangle with \( \bar{c} \). The result is always \( -\frac{1}{2}(|\bar{a}|^2 + |\bar{b}|^2 + |\bar{c}|^2) \) when the sum is zero.

Step 1: Understanding the Question:

The area is bounded by a parabola, a line, and vertical boundaries \( x = 0 \) (\( y \)-axis) and \( x = 3 \).


Step 2: Key Formula or Approach:

Area \( = \int_{a}^{b} [f(x) - g(x)] dx \), where \( f(x) \ge g(x) \).


Step 3: Detailed Explanation:

On the interval \( [0, 3] \), \( x^2 + 3 \) is always greater than \( x \).

Area \( = \int_{0}^{3} [(x^2 + 3) - x] dx \)
\[ = \left[ \frac{x^3}{3} + 3x - \frac{x^2}{2} \right]_0^3 \] \[ = \left( \frac{27}{3} + 3(3) - \frac{9}{2} \right) - 0 = 9 + 9 - 4.5 = 18 - 4.5 = 13.5 = \frac{27}{2} \]

Step 4: Final Answer:

The area is \( \frac{27}{2} \) sq. units.
Quick Tip: Always sketch a rough diagram to determine which curve is "upper" and which is "lower". This ensures the calculated area is positive.

Step 1: Understanding the Question:

Use the properties of integration for odd and even functions over symmetric intervals \( [-a, a] \).


Step 2: Key Formula or Approach:

1. \( \int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx \) if \( f \) is even.

2. \( \int_{-a}^{a} f(x) dx = 0 \) if \( f \) is odd.


Step 3: Detailed Explanation:

Let \( f(x) = x^2 \cos x \) and \( g(x) = \log(\frac{\pi - x}{\pi + x}) \cos x \).
\( f(-x) = (-x)^2 \cos(-x) = x^2 \cos x \to \) Even.
\( g(-x) = \log(\frac{\pi + x}{\pi - x}) \cos(-x) = -\log(\frac{\pi - x}{\pi + x}) \cos x \to \) Odd.

Thus, the integral becomes \( 2 \int_{0}^{\pi/2} x^2 \cos x dx + 0 \).

Use integration by parts twice for \( \int x^2 \cos x dx \):
\[ \int x^2 \cos x dx = x^2 \sin x - \int 2x \sin x dx \] \[ = x^2 \sin x - [2x(-\cos x) - \int 2(-\cos x) dx] \] \[ = x^2 \sin x + 2x \cos x - 2\sin x \]
Evaluate from 0 to \( \pi/2 \):
\[ 2 \left[ (\frac{\pi^2}{4})(1) + 0 - 2(1) \right] - 0 = 2(\frac{\pi^2}{4} - 2) = \frac{\pi^2}{2} - 4 \]

Step 4: Final Answer:

The result is \( \frac{\pi^2}{2} - 4 \).
Quick Tip: Function \( \log(\frac{a-x}{a+x}) \) is a very common odd function in integration problems. Recognizing it immediately can eliminate complex parts of the integral.

Step 1: Understanding the Question:

Apply the property \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx \).


Step 2: Detailed Explanation:

Let \( I = \int_{0}^{1} \log(\frac{1-x}{x}) dx \dots (i) \)

Using the property:
\[ I = \int_{0}^{1} \log(\frac{1-(1-x)}{1-x}) dx = \int_{0}^{1} \log(\frac{x}{1-x}) dx \dots (ii) \]
Add (i) and (ii):
\[ 2I = \int_{0}^{1} [\log(\frac{1-x}{x}) + \log(\frac{x}{1-x})] dx \]
Using log property \( \log A + \log B = \log(AB) \):
\[ 2I = \int_{0}^{1} \log(\frac{1-x}{x} \cdot \frac{x}{1-x}) dx = \int_{0}^{1} \log(1) dx = \int_{0}^{1} 0 dx = 0 \]
So \( I = 0 \).


Step 4: Final Answer:

The integral is 0.
Quick Tip: Whenever you have a log of a fraction within limits \( [0, 1] \) or \( [0, a] \), check if swapping \( x \) for \( a-x \) reciprocates the fraction. If so, the integral is likely zero.

Step 1: Understanding the Question:

The vector from origin to the foot of the perpendicular is the normal to the plane. The plane passes through this point.


Step 2: Key Formula or Approach:

The equation of a plane through \( (x_1, y_1, z_1) \) with normal DRs \( (a, b, c) \) is:
\( a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \).


Step 3: Detailed Explanation:

Normal vector \( \bar{n} = \vec{OP} = (2, -1, 4) \).

Point on plane \( P = (2, -1, 4) \).

Equation:
\[ 2(x - 2) + (-1)(y - (-1)) + 4(z - 4) = 0 \] \[ 2x - 4 - y - 1 + 4z - 16 = 0 \] \[ 2x - y + 4z - 21 = 0 \]

Step 4: Final Answer:

The equation is \( 2x - y + 4z - 21 = 0 \).
Quick Tip: The point \( P(2, -1, 4) \) must satisfy the plane equation. Quickly substitute it into the options: \( 2(2) - (-1) + 4(4) = 4 + 1 + 16 = 21 \). Only option (D) satisfies this.

Step 1: Understanding the Question:

Use substitution to convert the trigonometric integral into a rational algebraic function.


Step 2: Key Formula or Approach:

Let \( u = \sin \theta \), then \( du = \cos \theta d\theta \).


Step 3: Detailed Explanation:

Convert the denominator first:
\[ 5 - \cos^2 \theta - 4 \sin \theta = 5 - (1 - \sin^2 \theta) - 4 \sin \theta = \sin^2 \theta - 4 \sin \theta + 4 = (\sin \theta - 2)^2 \]
Now substitute \( u = \sin \theta \):
\[ \int \frac{5u - 2}{(u - 2)^2} du = \int \frac{5(u - 2) + 8}{(u - 2)^2} du = \int \left( \frac{5}{u-2} + \frac{8}{(u-2)^2} \right) du \] \[ = 5 \log |u - 2| - \frac{8}{u - 2} + c \]
Substitute back \( u = \sin \theta \):
\[ = 5 \log |\sin \theta - 2| + \frac{8}{2 - \sin \theta} + c \]

Step 4: Final Answer:

The integral is \( 5 \log |\sin \theta - 2| + \frac{8}{2 - \sin \theta} + c \).
Quick Tip: Whenever you see \( \cos \theta d\theta \) in the numerator, look for ways to express the entire remaining integrand in terms of \( \sin \theta \). Perfect squares in the denominator often lead to simple partial fraction splits.

Step 1: Understanding the Question:

We need to convert the switching circuit into its logical expression, simplify it, and count the resulting unique switches.


Step 2: Detailed Explanation:

From the diagram:

Top branch: \( p \) (representing \( S_1 \)).

Bottom branch: \( q \) (representing \( S_2 \)) in series with a parallel block of \( \sim q \) (representing \( S_2' \)) and \( \sim r \) (representing \( S_3' \)).

The logical statement is \( L \equiv p \lor [q \land (\sim q \lor \sim r)] \).

Simplify using distribution:
\[ L \equiv p \lor [(q \land \sim q) \lor (q \land \sim r)] \] \[ L \equiv p \lor [F \lor (q \land \sim r)] (where F is Contradiction) \] \[ L \equiv p \lor (q \land \sim r) \]
The simplified circuit has 3 switches: \( S_1 \), \( S_2 \), and \( S_3' \).


Step 4: Final Answer:

The alternative circuit has 3 switches.
Quick Tip: Remember the identity \( p \lor (q \land \sim p) \equiv p \lor q \). Here, the nested structure simplifies significantly because \( q \land \sim q \) is always False.

Step 1: Understanding the Question:

Use trigonometric identities to simplify the expression before substituting the value of \( x \).


Step 2: Key Formula or Approach:

Use \( \cos^{-1} x + \sin^{-1} x = \frac{\pi}{2} \).


Step 3: Detailed Explanation:
\[ \cos(2 \cos^{-1} x + \sin^{-1} x) = \cos(\cos^{-1} x + (\cos^{-1} x + \sin^{-1} x)) \] \[ = \cos(\cos^{-1} x + \frac{\pi}{2}) \]
Using the identity \( \cos(A + \frac{\pi}{2}) = -\sin A \):
\[ = -\sin(\cos^{-1} x) \]
We know \( \sin(\cos^{-1} x) = \sqrt{1 - x^2} \).

At \( x = \frac{1}{5} \):
\[ -\sin(\cos^{-1} \frac{1}{5}) = -\sqrt{1 - (\frac{1}{5})^2} = -\sqrt{1 - \frac{1}{25}} = -\sqrt{\frac{24}{25}} = -\frac{\sqrt{24}}{5} \]

Step 4: Final Answer:

The value is \( -\frac{\sqrt{24}}{5} \).
Quick Tip: Recognizing the sum \( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \) is the most common shortcut for such problems. It reduces a complicated triple function composition to a simple basic identity.

Step 1: Understanding the Question:

Use properties of triangles (projection rule) to simplify the given equation.


Step 2: Key Formula or Approach:

1. Projection Rule: \( a = b \cos C + c \cos B \).

2. Area of triangle \( \Delta = \frac{1}{2} ab \sin C \).


Step 3: Detailed Explanation:

The given equation is \( b \sin C (b \cos C + c \cos B) = 42 \).

By projection rule, \( (b \cos C + c \cos B) = a \).

So, \( b \sin C (a) = 42 \implies ab \sin C = 42 \).

We know Area \( \Delta = \frac{1}{2} ab \sin C \).
\[ \Delta = \frac{1}{2} (42) = 21 sq. units \]

Step 4: Final Answer:

The area is 21 sq. units.
Quick Tip: Always look for groups of terms that match the Projection Rule or Sine Rule in such expressions. The bracketed term \( (b \cos C + c \cos B) \) is a classic representation of side \( a \).

Step 1: Understanding the Question:

The question asks us to evaluate the indefinite integral of a rational function.

The integrand is \( \frac{x}{1+x^4} \).

We observe that the derivative of \( x^2 \) is \( 2x \), which is present in the numerator (after adjusting the constant).


Step 2: Key Formula or Approach:

We use the method of substitution and the standard integral formula:
\[ \int \frac{1}{1+u^2} du = \tan^{-1} u + c \]

Step 3: Detailed Explanation:

Let \( u = x^2 \).

Differentiating both sides with respect to \( x \), we get:
\[ du = 2x dx \implies x dx = \frac{1}{2} du \]
Substituting these into the original integral:
\[ \int \frac{x}{1+x^4} dx = \int \frac{1}{1+(x^2)^2} (x dx) \] \[ = \int \frac{1}{1+u^2} \left( \frac{1}{2} du \right) \] \[ = \frac{1}{2} \int \frac{1}{1+u^2} du \] \[ = \frac{1}{2} \tan^{-1} u + c \]
Substituting \( u = x^2 \) back into the result:
\[ = \frac{1}{2} \tan^{-1} x^2 + c \]

Step 4: Final Answer:

The evaluated integral is \( \frac{1}{2} \tan^{-1} x^2 + c \).
Quick Tip: Whenever you see \( x^{2n} \) in the denominator and \( x^{n-1} \) in the numerator, try substituting \( u = x^n \).
This often simplifies the integral into a standard inverse trigonometric form.

Step 1: Understanding the Question:

We need to find the equation of a plane that passes through a given point and contains a line defined by two other points.

Once the equation is found, we identify the intercepts \( p, q, r \) and calculate their sum.


Step 2: Key Formula or Approach:

A plane containing a line joining points \( A(x_1, y_1, z_1) \) and \( B(x_2, y_2, z_2) \) and passing through point \( P(x_3, y_3, z_3) \) has a normal vector \( \vec{n} = \vec{PA} \times \vec{PB} \).


Step 3: Detailed Explanation:

Let the given points be \( P(2, 1, -1) \), \( A(1, 3, 2) \), and \( B(1, 2, 1) \).

Vector \( \vec{PA} = (1-2)\hat{i} + (3-1)\hat{j} + (2-(-1))\hat{k} = -\hat{i} + 2\hat{j} + 3\hat{k} \).

Vector \( \vec{PB} = (1-2)\hat{i} + (2-1)\hat{j} + (1-(-1))\hat{k} = -\hat{i} + \hat{j} + 2\hat{k} \).

The normal to the plane is \( \vec{n} = \vec{PA} \times \vec{PB} \):
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-1 & 2 & 3
-1 & 1 & 2 \end{vmatrix} \] \[ \vec{n} = \hat{i}(4-3) - \hat{j}(-2-(-3)) + \hat{k}(-1-(-2)) = 1\hat{i} - 1\hat{j} + 1\hat{k} \]
The equation of the plane passing through \( P(2, 1, -1) \) with normal \( (1, -1, 1) \) is:
\[ 1(x-2) - 1(y-1) + 1(z-(-1)) = 0 \] \[ x - 2 - y + 1 + z + 1 = 0 \implies x - y + z = 0 \]
This plane passes through the origin.

For a plane passing through the origin, all three intercepts on the axes are 0.

Thus, \( p = 0, q = 0, r = 0 \).

The sum \( p + q + r = 0 + 0 + 0 = 0 \).


Step 4: Final Answer:

The sum of the intercepts is 0.
Quick Tip: If the constant term in the general equation of a plane \( ax + by + cz + d = 0 \) is zero (\( d = 0 \)), the plane passes through the origin.
In such cases, the intercepts on all coordinate axes are zero.

Step 1: Understanding the Question:

The question asks for the indefinite integral of a square root of a quadratic expression.


Step 2: Key Formula or Approach:

We use the method of completing the square to transform the quadratic into the form \( u^2 - a^2 \).

Then we apply the standard integral formula:
\[ \int \sqrt{u^2 - a^2} du = \frac{u}{2} \sqrt{u^2 - a^2} - \frac{a^2}{2} \log |u + \sqrt{u^2 - a^2}| + c \]

Step 3: Detailed Explanation:

The expression is \( x^2 + 3x \).

Completing the square:
\[ x^2 + 3x = x^2 + 3x + \left(\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2 = \left(x + \frac{3}{2}\right)^2 - \frac{9}{4} \]
Let \( u = x + \frac{3}{2} \) and \( a = \frac{3}{2} \). Then \( du = dx \).

The integral becomes:
\[ \int \sqrt{u^2 - a^2} du = \frac{u}{2} \sqrt{u^2 - a^2} - \frac{a^2}{2} \log |u + \sqrt{u^2 - a^2}| + c \]
Substituting back \( u = x + \frac{3}{2} \), \( a^2 = \frac{9}{4} \), and \( u^2 - a^2 = x^2 + 3x \):
\[ = \frac{x + 3/2}{2} \sqrt{x^2 + 3x} - \frac{9/4}{2} \log |x + \frac{3}{2} + \sqrt{x^2 + 3x}| + c \] \[ = \frac{2x + 3}{4} \sqrt{x^2 + 3x} - \frac{9}{8} \log |x + \frac{3}{2} + \sqrt{x^2 + 3x}| + c \]

Step 4: Final Answer:

The result is \( \frac{2x+3}{4} \sqrt{x^2 + 3x} - \frac{9}{8} \log (x + \frac{3}{2} + \sqrt{x^2 + 3x}) + c \).
Quick Tip: For integrals of the form \( \int \sqrt{ax^2 + bx + c} dx \), always start by completing the square.
Remember the coefficients in the result: the first part is always \( \frac{linear term}{2} \times radical \) and the second part involves \( \frac{a^2}{2} \).

Step 1: Understanding the Question:

A line is a normal to a hyperbola. We need to find the relationship between the coefficients \( a \) and \( b \).


Step 2: Key Formula or Approach:

The slope of the normal is the negative reciprocal of the derivative of the curve.

For a curve \( y = f(x) \), slope of tangent \( m_t = \frac{dy}{dx} \) and slope of normal \( m_n = -\frac{1}{dy/dx} \).


Step 3: Detailed Explanation:

The curve is \( xy = 1 \implies y = \frac{1}{x} \).

Differentiating with respect to \( x \):
\[ \frac{dy}{dx} = -\frac{1}{x^2} \]
The slope of the normal at any point \( (x, y) \) is:
\[ m_n = -\frac{1}{-1/x^2} = x^2 \]
Since \( x^2 \) is always strictly positive for any real \( x \ne 0 \), the slope of the normal to this curve must be positive.

The equation of the given line is \( ax + by + c = 0 \), which can be written as \( y = -\frac{a}{b}x - \frac{c}{b} \).

The slope of this line is \( m = -\frac{a}{b} \).

For this line to be a normal, we must have:
\[ -\frac{a}{b} > 0 \]
This inequality holds if \( a \) and \( b \) have opposite signs.

From the options:

(A) \( a, b \) both positive \( \implies \) slope is negative.

(B) \( a > 0, b < 0 \implies -\frac{(+)}{(-)} = (+) \). This is a positive slope.

(C) \( a < 0, b \ge 0 \). If \( b=0 \), the line is vertical (not defined by slope).

(D) \( a, b \) both negative \( \implies \) slope is negative.

Thus, option (B) is the most suitable general choice.


Step 4: Final Answer:

The correct condition is \( a > 0, b < 0 \).
Quick Tip: The curve \( xy=1 \) represents a rectangular hyperbola in the first and third quadrants.
Its tangents always have a negative slope, so its normals must always have a positive slope.

Step 1: Understanding the Question:

We are given the sum of two numbers and need to minimize the sum of their reciprocals.


Step 2: Key Formula or Approach:

Let the numbers be \( x \) and \( y \). We are given \( x + y = 4 \).

We need to minimize \( S = \frac{1}{x} + \frac{1}{y} \).

We can use the AM-HM inequality: \( \frac{x+y}{2} \ge \frac{2}{1/x + 1/y} \).


Step 3: Detailed Explanation:

From AM-HM inequality:
\[ \frac{x+y}{2} \ge \frac{2}{1/x + 1/y} \] \[ \frac{4}{2} \ge \frac{2}{1/x + 1/y} \] \[ 2 \ge \frac{2}{S} \] \[ S \ge 1 \]
The equality holds when \( x = y \).

Since \( x + y = 4 \), we have \( x = y = 2 \).

The minimum value of the sum of reciprocals is \( \frac{1}{2} + \frac{1}{2} = 1 \).


Step 4: Final Answer:

The minimum value is 1.
Quick Tip: For a fixed sum, the sum of reciprocals of positive numbers is minimized when the numbers are equal.
This is a general result of the symmetry of the function.

Step 1: Understanding the Question:

We need to find the equation of the tangent and the normal separately and then multiply them to get the combined equation.


Step 2: Key Formula or Approach:

1. Find \( \frac{dy}{dx} \) to get the slope of the tangent.

2. Use point-slope form for tangent and normal.

3. Multiply the two linear equations.


Step 3: Detailed Explanation:

Curve: \( xy = 15 \implies y = \frac{15}{x} \).

Differentiating: \( \frac{dy}{dx} = -\frac{15}{x^2} \).

At \( (5, 3) \), slope of tangent \( m_t = -\frac{15}{5^2} = -\frac{15}{25} = -\frac{3}{5} \).

Equation of tangent: \( y - 3 = -\frac{3}{5}(x - 5) \implies 5y - 15 = -3x + 15 \implies 3x + 5y - 30 = 0 \).

Slope of normal \( m_n = \frac{5}{3} \).

Equation of normal: \( y - 3 = \frac{5}{3}(x - 5) \implies 3y - 9 = 5x - 25 \implies 5x - 3y - 16 = 0 \).

Combined equation:
\[ (3x + 5y - 30)(5x - 3y - 16) = 0 \] \[ 15x^2 - 9xy - 48x + 25xy - 15y^2 - 80y - 150x + 90y + 480 = 0 \] \[ 15x^2 + 16xy - 15y^2 - 198x + 10y + 480 = 0 \]

Step 4: Final Answer:

The combined equation is \( 15x^2 + 16xy - 198x + 10y + 480 - 15y^2 = 0 \).
Quick Tip: Always simplify the individual lines first.
Check if the point (5, 3) satisfies the final combined equation; if it doesn't, there is a calculation error.

Step 1: Understanding the Question:

We are given the angle between a line and a plane. We need to use the formula for this angle to find the unknown parameter \( \lambda \).


Step 2: Key Formula or Approach:

The angle \( \theta \) between a line with direction vector \( \vec{d} \) and a plane with normal vector \( \vec{n} \) is given by:
\[ \sin \theta = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}| |\vec{n}|} \]

Step 3: Detailed Explanation:

Line: \( \frac{x}{1} = \frac{y-1}{2} = \frac{z-3}{\lambda} \implies \vec{d} = (1, 2, \lambda) \).

Plane: \( x + 2y + 3z = 6 \implies \vec{n} = (1, 2, 3) \).

Given \( \theta = \cos^{-1} \sqrt{\frac{5}{14}} \implies \cos \theta = \sqrt{\frac{5}{14}} \).

Then \( \sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \frac{5}{14}} = \sqrt{\frac{9}{14}} = \frac{3}{\sqrt{14}} \).

Using the formula:
\[ \frac{|1(1) + 2(2) + \lambda(3)|}{\sqrt{1^2 + 2^2 + \lambda^2} \sqrt{1^2 + 2^2 + 3^2}} = \frac{3}{\sqrt{14}} \] \[ \frac{|5 + 3\lambda|}{\sqrt{5 + \lambda^2} \sqrt{14}} = \frac{3}{\sqrt{14}} \] \[ |5 + 3\lambda| = 3\sqrt{5 + \lambda^2} \]
Squaring both sides:
\[ 25 + 9\lambda^2 + 30\lambda = 9(5 + \lambda^2) \] \[ 25 + 9\lambda^2 + 30\lambda = 45 + 9\lambda^2 \] \[ 30\lambda = 20 \implies \lambda = \frac{2}{3} \]

Step 4: Final Answer:

The value of \( \lambda \) is \( 2/3 \).
Quick Tip: Be careful with the sine and cosine!
The formula for the angle between a line and a plane uses \(\sin\), while for two lines or two planes it uses \(\cos\).

Step 1: Understanding the Question:

This is a related rates problem. We are given the rates of change of dimensions and need to find the rates of change of perimeter and area.


Step 2: Key Formula or Approach:

Perimeter \( P = 2(x + y) \implies \frac{dP}{dt} = 2 \left( \frac{dx}{dt} + \frac{dy}{dt} \right) \).

Area \( A = xy \implies \frac{dA}{dt} = x \frac{dy}{dt} + y \frac{dx}{dt} \).


Step 3: Detailed Explanation:

Given: \( \frac{dx}{dt} = -5 cm/min \) (decrease), \( \frac{dy}{dt} = 3 cm/min \) (increase).

Dimensions: \( x = 5, y = 2 \).

Rate of change of perimeter:
\[ \frac{dP}{dt} = 2(-5 + 3) = 2(-2) = -4 cm/min \]
Rate of change of area:
\[ \frac{dA}{dt} = 5(3) + 2(-5) = 15 - 10 = 5 cm^2/min \]

Step 4: Final Answer:

The rates are -4 and 5 respectively.
Quick Tip: Remember to assign a negative sign to rates representing a decrease.
The product rule is essential for calculating the rate of change of the area.

Step 1: Understanding the Question:

The coefficients of the polynomial \( f(x) \) depend on its own derivatives evaluated at certain points. We need to find these values.


Step 2: Key Formula or Approach:

Let \( f'(1) = a \), \( f''(2) = b \), and \( f'''(3) = c \).

Then \( f(x) = 3x^3 + 2ax^2 + bx + c \).

Differentiate repeatedly to set up a system of equations for \( a, b, c \).


Step 3: Detailed Explanation:
\( f'(x) = 9x^2 + 4ax + b \)
\( f''(x) = 18x + 4a \)
\( f'''(x) = 18 \).

Thus, \( c = f'''(3) = 18 \).

From \( f''(2) = b \):
\[ b = 18(2) + 4a = 36 + 4a \]
From \( f'(1) = a \):
\[ a = 9(1)^2 + 4a(1) + b = 9 + 4a + (36 + 4a) \] \[ a = 45 + 8a \implies -7a = 45 \implies a = -\frac{45}{7} \]
Then \( b = 36 + 4\left(-\frac{45}{7}\right) = \frac{252 - 180}{7} = \frac{72}{7} \).

The polynomial is:
\[ f(x) = 3x^3 + 2\left(-\frac{45}{7}\right)x^2 + \left(\frac{72}{7}\right)x + 18 \] \[ f(x) = \frac{21x^3 - 90x^2 + 72x + 126}{7} \]

Step 4: Final Answer:

The function is \( \frac{1}{7}(21x^3 - 90x^2 + 72x + 126) \).
Quick Tip: Treat the constant derivative values as variables first.
Once you have the derivatives, substituting the specific points leads to a simple linear system.

Step 1: Understanding the Question:

This is a parametric differentiation problem where we need the second derivative.


Step 2: Key Formula or Approach:

1. \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \).

2. \( \frac{d^2y}{dx^2} = \frac{d}{d\theta} \left( \frac{dy}{dx} \right) \cdot \frac{d\theta}{dx} \).


Step 3: Detailed Explanation:

Given \( x = \sin \theta \) and \( y = \sin^3 \theta \).
\( \frac{dx}{d\theta} = \cos \theta \).
\( \frac{dy}{d\theta} = 3 \sin^2 \theta \cos \theta \).
\( \frac{dy}{dx} = \frac{3 \sin^2 \theta \cos \theta}{\cos \theta} = 3 \sin^2 \theta \).

Now, find the second derivative:
\[ \frac{d^2y}{dx^2} = \frac{d}{d\theta}(3 \sin^2 \theta) \times \frac{1}{\cos \theta} \] \[ = (6 \sin \theta \cos \theta) \times \frac{1}{\cos \theta} = 6 \sin \theta \]
At \( \theta = \frac{\pi}{6} \):
\[ \frac{d^2y}{dx^2} = 6 \sin\left(\frac{\pi}{6}\right) = 6 \times \frac{1}{2} = 3 \]

Step 4: Final Answer:

The value is 3.
Quick Tip: Alternatively, observe that \( y = x^3 \).
Then \( \frac{dy}{dx} = 3x^2 \) and \( \frac{d^2y}{dx^2} = 6x \).
At \( \theta = \pi/6 \), \( x = \sin(\pi/6) = 1/2 \), so \( 6(1/2) = 3 \).

Step 1: Understanding the Question:

We need to find the derivative of an implicitly defined function.


Step 2: Key Formula or Approach:

Take the natural logarithm of both sides to simplify the expression before differentiating.


Step 3: Detailed Explanation:
\( x = e^{\tan^{-1} \left( \frac{y}{x^2} - 1 \right)} \implies \log_e x = \tan^{-1} \left( \frac{y}{x^2} - 1 \right) \).

First, find \( y \) at \( x = 1 \):
\( \log_e 1 = \tan^{-1}(y/1 - 1) \implies 0 = \tan^{-1}(y-1) \implies y - 1 = 0 \implies y = 1 \).

Differentiating with respect to \( x \):
\[ \frac{1}{x} = \frac{1}{1 + \left( \frac{y}{x^2} - 1 \right)^2} \times \frac{d}{dx} \left( \frac{y}{x^2} - 1 \right) \] \[ \frac{1}{x} = \frac{1}{1 + \left( \frac{y}{x^2} - 1 \right)^2} \times \left( \frac{x^2 y' - 2xy}{x^4} \right) \]
At \( (1, 1) \):
\[ \frac{1}{1} = \frac{1}{1 + (1-1)^2} \times \left( \frac{1^2 y' - 2(1)(1)}{1^4} \right) \] \[ 1 = 1 \times (y' - 2) \implies y' = 3 \]

Step 4: Final Answer:

The derivative \( dy/dx \) at \( x=1 \) is 3.
Quick Tip: Evaluating the point \( (x, y) \) before fully expanding the derivative expression saves significant time in implicit differentiation.

Step 1: Understanding the Question:

This is a Linear Programming Problem. We need to find the feasible region, identify corner points, and evaluate the objective function.


Step 2: Key Formula or Approach:
1. Plot lines: \( x + 3y = 60 \), \( x + y = 10 \), \( x = y \).

2. Find intersection points (vertices) of the feasible region.

3. Calculate \( z \) at each vertex.


Step 3: Detailed Explanation:

Vertices are formed by the intersections of the boundary lines:

1. \( x+y=10 \) and \( x-y=0 \implies (5, 5) \). \( z(5,5) = 15+25 = 40 \).

2. \( x+3y=60 \) and \( x-y=0 \implies 4x=60 \implies (15, 15) \). \( z(15,15) = 45+75 = 120 \).

3. \( x+y=10 \) and \( y=0 \implies (10, 0) \). \( z(10,0) = 30 \).

4. \( x+3y=60 \) and \( y=0 \implies (60, 0) \). \( z(60,0) = 180 \).

The region bounded by \( y \ge 0 \) and the other lines gives these vertices.

If we consider the bounded region between the two parallel-ish lines and the bisector \( x=y \):

The minimum value is 40 at (5,5) and the maximum value is 120 at (15,15) if restricted to the segment.

Difference = \( 120 - 40 = 80 \).


Step 4: Final Answer:

The difference is 80.
Quick Tip: Often in these problems, the 'difference' refers to values at intersection points of the non-axis constraints.
Check the vertices \( (15,15) \) and \( (5,5) \) first.

Step 1: Understanding the Question:

We need to evaluate a complex scalar triple product involving vector cross products.


Step 2: Key Formula or Approach:

Use the properties of vector products:

1. \( \vec{u} \cdot (\vec{v} \times \vec{w}) = [\vec{u}, \vec{v}, \vec{w}] \).

2. \( (\vec{u} \times \vec{v}) \times \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u} \).

Alternatively, observe the structure: \( \vec{v} \cdot (\vec{u} \times \vec{w}) = \vec{u} \cdot (\vec{w} \times \vec{v}) \).


Step 3: Detailed Explanation:

The expression is \( (2\bar{a} - \bar{b}) \cdot ((\bar{a} \times \bar{b}) \times (\bar{a} + 2\bar{b})) \).

This is a scalar triple product of vectors \( \vec{A} = 2\bar{a} - \bar{b} \), \( \vec{B} = \bar{a} \times \bar{b} \), and \( \vec{C} = \bar{a} + 2\bar{b} \).

We can rewrite it as \( (\bar{a} \times \bar{b}) \cdot ((2\bar{a} - \bar{b}) \times (\bar{a} + 2\bar{b})) \).

Calculate the cross product:
\[ (2\bar{a} - \bar{b}) \times (\bar{a} + 2\bar{b}) = 2(\bar{a} \times \bar{a}) + 4(\bar{a} \times \bar{b}) - (\bar{b} \times \bar{a}) - 2(\bar{b} \times \bar{b}) \]
Since \( \bar{a} \times \bar{a} = 0 \), \( \bar{b} \times \bar{b} = 0 \), and \( \bar{b} \times \bar{a} = -(\bar{a} \times \bar{b}) \):
\[ = 0 + 4(\bar{a} \times \bar{b}) + (\bar{a} \times \bar{b}) - 0 = 5(\bar{a} \times \bar{b}) \]
The final result is \( (\bar{a} \times \bar{b}) \cdot 5(\bar{a} \times \bar{b}) = 5 |\bar{a} \times \bar{b}|^2 \).

Check \( \bar{a} \) and \( \bar{b} \):
\( |\bar{a}|^2 = \frac{1}{10}(3^2 + 1^2) = 1 \).
\( |\bar{b}|^2 = \frac{1}{49}(2^2 + 3^2 + (-6)^2) = 1 \).
\( \bar{a} \cdot \bar{b} = \frac{1}{7\sqrt{10}}(3(2) + 0(3) + 1(-6)) = 0 \).

Since \( \bar{a} \perp \bar{b} \) and both are unit vectors, \( |\bar{a} \times \bar{b}| = |\bar{a}||\bar{b}|\sin 90^\circ = 1 \).

Result \( = 5(1)^2 = 5 \).


Step 4: Final Answer:

The value is 5.
Quick Tip: Simplify the vector expressions before plugging in the coordinates.
Recognizing unit vectors and perpendicularity saves tedious coordinate-wise calculation.

Step 1: Understanding the Question:

The sides of a triangle are in Arithmetic Progression and the largest angle is \( 120^\circ \). We need to find the side lengths and the area.


Step 2: Key Formula or Approach:

1. Let sides be \( a-d, a, a+d \). The largest side is \( a+d = 7 \).

2. Use Cosine Rule: \( c^2 = a^2 + b^2 - 2ab \cos C \).

3. Area \( = \frac{1}{2} ab \sin C \).


Step 3: Detailed Explanation:

The greatest side \( c = 7 \) is opposite to \( 120^\circ \).

Let the sides be \( 7-2d, 7-d, 7 \)? No, let them be \( a, a+d, a+2d \).

Better: let sides be \( x-d, x, x+d \). Greatest side \( x+d = 7 \implies d = 7 - x \).

Cosine rule:
\[ 7^2 = (x-d)^2 + x^2 - 2x(x-d) \cos 120^\circ \] \[ 49 = x^2 - 2xd + d^2 + x^2 + x(x-d) = 3x^2 - 3xd + d^2 \]
Substitute \( d = 7-x \):
\[ 49 = 3x^2 - 3x(7-x) + (7-x)^2 \] \[ 49 = 3x^2 - 21x + 3x^2 + 49 - 14x + x^2 \] \[ 49 = 7x^2 - 35x + 49 \implies 7x^2 - 35x = 0 \]
Since \( x \ne 0 \), \( x = 5 \).

Then \( d = 7 - 5 = 2 \).

Sides are \( 5-2 = 3 \), \( 5 \), and \( 7 \).

Area \( = \frac{1}{2} \times 3 \times 5 \times \sin 120^\circ = \frac{15}{2} \times \frac{\sqrt{3}}{2} = \frac{15\sqrt{3}}{4} m^2 \).


Step 4: Final Answer:

The area is \( \frac{15\sqrt{3}}{4} m^2 \).
Quick Tip: For sides in A.P. with a \( 120^\circ \) angle, the sides always follow the ratio 3:5:7.
This is a very useful property to remember for competitive exams.

Step 1: Understanding the Question:

We use the identity \( ^nC_r + ^nC_{r-1} = ^{n+1}C_r \) repeatedly to simplify the sum.


Step 2: Key Formula or Approach:

Pascal's identity: \( ^nC_r + ^nC_{r-1} = ^{n+1}C_r \).


Step 3: Detailed Explanation:

Starting with the first two terms:
\( ^{15}C_4 + ^{15}C_5 = ^{16}C_5 \).

Add the next term:
\( ^{16}C_5 + ^{16}C_6 = ^{17}C_6 \).

Add the next term:
\( ^{17}C_6 + ^{17}C_7 = ^{18}C_7 \).

Add the final term:
\( ^{18}C_7 + ^{18}C_8 = ^{19}C_8 \).

We are given \( ^{19}C_r = ^{19}C_8 \).

By symmetry, \( ^nC_r = ^nC_{n-r} \).

So \( r = 8 \) or \( r = 19 - 8 = 11 \).


Step 4: Final Answer:

The values of \( r \) are 8 or 11.
Quick Tip: Apply the identity terms one by one from left to right.
Always remember to check both \( r \) and \( n-r \) for equality in combinations.

Step 1: Understanding the Question:

For independent events, the probability of the intersection is the product of the individual probabilities.


Step 2: Key Formula or Approach:

Let \( P(A) = x \) and \( P(B) = y \).
\( P(A \cap B') = P(A)P(B') = x(1-y) = \frac{3}{25} \).
\( P(A' \cap B) = P(A')P(B) = (1-x)y = \frac{8}{25} \).


Step 3: Detailed Explanation:

1. \( x - xy = \frac{3}{25} \)

2. \( y - xy = \frac{8}{25} \)

Subtract (1) from (2):
\( y - x = \frac{5}{25} = \frac{1}{5} \implies y = x + \frac{1}{5} \).

Substitute in (1):
\( x - x(x + 1/5) = 3/25 \)
\( x - x^2 - x/5 = 3/25 \implies \frac{4}{5}x - x^2 = \frac{3}{25} \)

Multiply by 25:
\( 20x - 25x^2 = 3 \implies 25x^2 - 20x + 3 = 0 \).

Solving the quadratic:
\( x = \frac{20 \pm \sqrt{400 - 300}}{50} = \frac{20 \pm 10}{50} \).
\( x = \frac{30}{50} = \frac{3}{5} \) or \( x = \frac{10}{50} = \frac{1}{5} \).

From the options, \( \frac{1}{5} \) is present.


Step 4: Final Answer:

The value of \( P(A) \) is \( 1/5 \).
Quick Tip: Symmetry in equations often allows you to find the difference between variables first.
Check if the calculated value leads to a valid probability \( 0 \le y \le 1 \).

Step 1: Understanding the Question:

Work done is the scalar product of force and displacement.


Step 2: Key Formula or Approach:
\( W = \vec{F} \cdot \vec{d} \).

Displacement vector \( \vec{d} = \vec{r}_Q - \vec{r}_P \).


Step 3: Detailed Explanation:

Force vector \( \vec{F} = 3\hat{i} + 4\hat{j} + 5\hat{k} \).

Displacement vector \( \vec{d} = (2-3)\hat{i} + (3-4)\hat{j} + (4-5)\hat{k} = -\hat{i} - \hat{j} - \hat{k} \).

Work done:
\[ W = (3\hat{i} + 4\hat{j} + 5\hat{k}) \cdot (-\hat{i} - \hat{j} - \hat{k}) \] \[ W = (3)(-1) + (4)(-1) + (5)(-1) = -3 - 4 - 5 = -12 J \]

Step 4: Final Answer:

The work done is -12 J.
Quick Tip: Always calculate displacement as Final Position minus Initial Position.
Negative work indicates that the force is acting in the opposite direction to the displacement.

Step 1: Understanding the Question:

A moving charge constitutes a current. A current loop has a magnetic dipole moment.


Step 2: Key Formula or Approach:

Magnetic Moment \( M = I \times A \).

Current \( I = \frac{e}{T} \), where \( T \) is the time period.


Step 3: Detailed Explanation:

Time period \( T = \frac{2\pi r}{v} \).

Current \( I = \frac{e}{2\pi r / v} = \frac{ev}{2\pi r} \).

Area of the loop \( A = \pi r^2 \).

Magnetic Moment \( M = \left( \frac{ev}{2\pi r} \right) \times \pi r^2 = \frac{evr}{2} \).


Step 4: Final Answer:

The magnetic moment is \( \frac{evr}{2} \).
Quick Tip: Remember the relationship between angular momentum \( L \) and magnetic moment \( M \): \( \frac{M}{L} = \frac{e}{2m} \).
Since \( L = mvr \), then \( M = \left( \frac{e}{2m} \right) mvr = \frac{evr}{2} \).

Step 1: Understanding the Question:

We need to calculate the propagated error in a derived quantity (pressure) from measured quantities (force and length).


Step 2: Key Formula or Approach:

Pressure \( P = \frac{F}{A} = \frac{F}{L^2} \).

Relative error: \( \frac{\Delta P}{P} = \frac{\Delta F}{F} + 2 \frac{\Delta L}{L} \).


Step 3: Detailed Explanation:

Given percentage errors:
\( \frac{\Delta F}{F} \times 100 = 4% \)
\( \frac{\Delta L}{L} \times 100 = 2% \)

The percentage error in pressure is:
\[ % error in P = \left( \frac{\Delta F}{F} + 2 \frac{\Delta L}{L} \right) \times 100 \] \[ = 4% + 2(2%) = 4% + 4% = 8% \]

Step 4: Final Answer:

The percentage error in measurement of pressure is 8%.
Quick Tip: Relative errors always add up for multiplication and division.
When a quantity is raised to a power \( n \), its relative error is multiplied by \( |n| \).

Step 1: Understanding the Question:

This is a problem of electrostatic equilibrium of a charge under the influence of other charges.


Step 2: Key Formula or Approach:

Coulomb's Law: \( F = \frac{k q_1 q_2}{r^2} \).

Net force is the vector sum of forces from all other charges.


Step 3: Detailed Explanation:

Let the positions be \( x_1 = 0 \), \( x_2 = L/2 \), and \( x_3 = L \).

Charges at these positions are \( +3q \), \( Q \), and \( +q \) respectively.

Force on \( +q \) at \( L \) due to \( +3q \) at \( 0 \):
\( F_1 = \frac{k(3q)(q)}{L^2} = \frac{3kq^2}{L^2} \) (directed away from origin).

Force on \( +q \) at \( L \) due to \( Q \) at \( L/2 \):
\( F_2 = \frac{k(Q)(q)}{(L/2)^2} = \frac{4kQq}{L^2} \).

For net force on \( +q \) to be zero:
\[ F_1 + F_2 = 0 \] \[ \frac{3kq^2}{L^2} + \frac{4kQq}{L^2} = 0 \] \[ 3q + 4Q = 0 \implies Q = -\frac{3}{4}q \]
Comparing with \( Q = xq \), we get \( x = -3/4 \).


Step 4: Final Answer:

The value of \( x \) is \( -3/4 \).
Quick Tip: To cancel a repulsive force from a positive charge, the intermediate charge must be of opposite sign (negative) to provide an attractive force.
This eliminates options (A) and (D) immediately.

Step 1: Understanding the Question:

We need to verify the kinetic theory of gases relations between pressure, volume, temperature, and molecular speed.


Step 2: Detailed Explanation:

Let's check each statement:

(A) and (D): Kinetic energy \( KE = \frac{3}{2} PV \implies PV = \frac{2}{3} KE \). Both (A) and (D) are incorrect.

(B): \( T_1 = -73^\circC = 200 K \). \( T_2 = 527^\circC = 800 K \).

r.m.s speed \( v_{rms} \propto \sqrt{T} \).

Ratio: \( \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{800}{200}} = \sqrt{4} = 2 \). Speed is doubled. This is correct.

(C): \( T_1 = 173 K \), \( T_2 = 900 K \). Ratio \( \sqrt{900/173} \approx 2.28 \ne 4 \). Incorrect.


Step 3: Final Answer:

Statement (B) is the only correct statement.
Quick Tip: Always convert temperatures in Celsius to Kelvin (\( K = {}^\circC + 273 \)) before using gas law or kinetic theory formulas.
Ratios of temperatures often yield perfect squares in exam problems.

Step 1: Understanding the Question:

We need to find the change in power of a lens when it is immersed in a liquid with a different refractive index.


Step 2: Key Formula or Approach:

Lens Maker's Formula: \( P = \frac{1}{f} = (\mu_{rel} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \).


Step 3: Detailed Explanation:

In air: \( P_a = (1.5 - 1) K = 0.5 K = 3 D \), where \( K \) represents the curvature part.

In liquid: \( P_l = \left( \frac{1.5}{2} - 1 \right) K = (0.75 - 1) K = -0.25 K \).

Taking the ratio:
\[ \frac{P_l}{P_a} = \frac{-0.25 K}{0.5 K} = -\frac{1}{2} \] \[ P_l = -\frac{1}{2} \times 3 D = -1.5 D \]
The magnitude of the new power is 1.5 D. (The negative sign indicates it now behaves as a concave lens).


Step 4: Final Answer:

The new power of the lens is 1.5 D.
Quick Tip: When a lens is placed in a medium denser than its own material (\( \mu_m > \mu_g \)), its nature reverses: convex becomes concave and vice versa.
The power always decreases in magnitude when placed in any liquid.

Step 1: Understanding the Question:

Removing a coil from a magnetic field causes a change in flux, which induces an emf and a current, leading to charge flow.


Step 2: Key Formula or Approach:

Induced charge \( q = \frac{\Delta \Phi}{R} \), where \( \Delta \Phi \) is the total change in magnetic flux.


Step 3: Detailed Explanation:

Initial flux through the coil: \( \Phi_1 = n B A \).

Final flux after removal: \( \Phi_2 = 0 \).

Change in flux: \( \Delta \Phi = n B A \).

The total charge that flows is:
\[ q = \frac{\Delta \Phi}{R} = \frac{n B A}{R} \]
Rearranging for magnetic flux density \( B \):
\[ B = \frac{q R}{n A} \]

Step 4: Final Answer:

The magnetic flux density is \( \frac{q R}{n A} \).
Quick Tip: Induced charge \( q \) is independent of the time taken for the flux change.
It only depends on the total change in flux and the resistance of the circuit.

Step 1: Understanding the Question:

A photodiode operates in reverse bias. The reverse saturation current is directly proportional to the incident light intensity.


Step 3: Detailed Explanation:

In the given characteristic graph, the reverse current (measured in \( \muA \) on the negative y-axis) increases in magnitude as the light intensity increases.

Curve \( I_4 \) shows the highest magnitude of current for any given reverse voltage.

Therefore, curve \( I_4 \) corresponds to the highest light intensity.


Step 4: Final Answer:

The maximum intensity is \( I_4 \).
Quick Tip: For a photodiode, more photons (higher intensity) generate more electron-hole pairs, which significantly increases the minority carrier concentration and hence the reverse saturation current.
Look for the curve furthest from the horizontal axis in the 3rd quadrant.

Step 1: Understanding the Question:

We need to express the time of flight in terms of the maximum height achieved by a projectile.


Step 2: Key Formula or Approach:

Max Height \( H = \frac{u^2 \sin^2 \theta}{2g} \).

Time of Flight \( T = \frac{2 u \sin \theta}{g} \).


Step 3: Detailed Explanation:

From the height formula:
\[ u \sin \theta = \sqrt{2gH} \]
Substitute this into the time of flight formula:
\[ T = \frac{2 \sqrt{2gH}}{g} = 2 \sqrt{\frac{2gH}{g^2}} = 2 \sqrt{\frac{2H}{g}} \]

Step 4: Final Answer:

The time of flight is \( 2 \sqrt{\frac{2H}{g}} \).
Quick Tip: Notice that the time of flight is exactly double the time taken to fall freely from height \( H \).
Time to fall from \( H \) is \( \sqrt{2H/g} \), so total time is \( 2\sqrt{2H/g} \).

Step 1: Understanding the Question:

This is a standard Doppler Effect problem where both source and observer are moving.


Step 2: Key Formula or Approach:

Apparent frequency \( f' = f \left( \frac{V \pm v_o}{V \mp v_s} \right) \).


Step 3: Detailed Explanation:

Source velocity \( v_s = \frac{V}{3} \) (towards observer \( \implies \) denominator is \( V - v_s \)).

Observer velocity \( v_o = \frac{V}{4} \) (away from source \( \implies \) numerator is \( V - v_o \)).
\[ f' = F \left( \frac{V - V/4}{V - V/3} \right) \] \[ f' = F \left( \frac{3V/4}{2V/3} \right) = F \left( \frac{3}{4} \times \frac{3}{2} \right) \] \[ f' = \frac{9}{8} F \]

Step 4: Final Answer:

The apparent frequency is \( \frac{9}{8} F \).
Quick Tip: Moving towards increases frequency (smaller denominator), moving away decreases frequency (smaller numerator).
Always visualize the relative motion to get the signs right in the Doppler formula.

Step 1: Understanding the Question:

The spectral series of hydrogen are classified by their final energy level \( n_{final} \).


Step 3: Detailed Explanation:

- Lyman Series (\( n_{final} = 1 \)): Ultraviolet (UV) region.

- Balmer Series (\( n_{final} = 2 \)): Visible region.

- Paschen Series (\( n_{final} = 3 \)): Infrared (IR) region.

Analyzing the options:

(A) Ends at \( n=1 \) (Lyman, UV).

(B) and (C) End at \( n=2 \) (Balmer, Visible).

(D) Ends at \( n=3 \). This is the Paschen series, which is in the Infrared region.


Step 4: Final Answer:

Transition from \( n=5 \) to \( n=3 \) gives infrared radiation.
Quick Tip: The higher the final level \( n \), the lower the energy change and longer the wavelength.
UV (high energy, \( n \to 1 \)) \( > \) Visible (\( n \to 2 \)) \( > \) IR (low energy, \( n \to 3, 4, 5 \)).

Step 1: Understanding the Question:

This is a problem of buoyancy and equilibrium in a layered fluid.


Step 2: Key Formula or Approach:

Upthrust \( U = Weight of fluid displaced \).

In equilibrium, Weight of cylinder \( W = Total Upthrust \).


Step 3: Detailed Explanation:

Let the top liquid have density \( \rho \) and the bottom denser liquid have density \( 3\rho \).

Length of cylinder in top liquid \( = 3l/4 \).

Length of cylinder in bottom liquid \( = l/4 \).

Total upthrust \( U = Upthrust from top liquid + Upthrust from bottom liquid \).
\[ U = (A \times 3l/4) \rho g + (A \times l/4) 3\rho g \] \[ U = \frac{3}{4} A l \rho g + \frac{3}{4} A l \rho g = \frac{6}{4} A l \rho g = \frac{3}{2} A l \rho g \]
Weight of cylinder \( W = (A \times l) d g \).

Equating \( W \) and \( U \):
\[ A l d g = \frac{3}{2} A l \rho g \implies d = \frac{3}{2} \rho \]

Step 4: Final Answer:

The density of the cylinder is \( \frac{3}{2} \rho \).
Quick Tip: Density of floating object is always a weighted average of the densities of the liquids it displaces.
\( d = \frac{\rho_1 h_1 + \rho_2 h_2}{h_1 + h_2} \). Here \( d = \frac{\rho(3l/4) + 3\rho(l/4)}{l} = \frac{3/4\rho + 3/4\rho}{1} = 1.5\rho \).

Step 1: Understanding the Question:

We need to find the final steady-state potential when two charged capacitors are connected in parallel.


Step 2: Key Formula or Approach:

Common Potential \( V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} \).


Step 3: Detailed Explanation:

Given: \( C_1 = 100 \muF, V_1 = 20 V \) and \( C_2 = 50 \muF, V_2 = 40 V \).

Common potential \( V \):
\[ V = \frac{(100 \times 20) + (50 \times 40)}{100 + 50} \] \[ V = \frac{2000 + 2000}{150} = \frac{4000}{150} = \frac{400}{15} \] \[ V = \frac{80}{3} V \]

Step 4: Final Answer:

The common potential is \( \frac{80}{3} \) V.
Quick Tip: Conservation of charge is the underlying principle here.
If polarities were opposite, the formula would be \( V = \frac{|C_1 V_1 - C_2 V_2|}{C_1 + C_2} \).

Step 1: Understanding the Question:

An LED is a special p-n junction diode that emits light when forward-biased. Its I-V characteristics are similar to a normal diode but with a higher knee voltage.


Step 3: Detailed Explanation:

LEDs are operated exclusively in forward bias to produce light through radiative recombination of carriers.

The characteristic graph for an LED is strictly in the first quadrant (forward bias).

Graph (d) shows the typical non-linear exponential increase of current after reaching a specific threshold (knee) voltage in the first quadrant.


Step 4: Final Answer:

Graph (d) correctly represents the I-V characteristics of an LED.
Quick Tip: Remember that LEDs have a higher "turn-on" or knee voltage than typical silicon (0.7V) or germanium (0.3V) diodes, usually between 1.8V to 3.0V depending on the color.
Look for the graph that only shows forward-bias behavior.

Step 1: Understanding the Question:

The linear velocity \(\vec{v}\) of a particle rotating with angular velocity \(\vec{\omega}\) at a position vector \(\vec{r}\) is given by the cross product of the two vectors.


Step 2: Key Formula or Approach:

The relation is given by:
\[ \vec{v} = \vec{\omega} \times \vec{r} \]


Step 3: Detailed Explanation:

Given:
\(\vec{\omega} = 3\hat{i} - 4\hat{j} + \hat{k}\)
\(\vec{r} = 5\hat{i} - 6\hat{j} + 6\hat{k}\)

Using the determinant method for the cross product:
\[ \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & -4 & 1
5 & -6 & 6 \end{vmatrix} \]

Expanding along the first row:
\[ \vec{v} = \hat{i} [(-4)(6) - (1)(-6)] - \hat{j} [(3)(6) - (1)(5)] + \hat{k} [(3)(-6) - (-4)(5)] \]
\[ \vec{v} = \hat{i} [-24 + 6] - \hat{j} [18 - 5] + \hat{k} [-18 + 20] \]
\[ \vec{v} = -18\hat{i} - 13\hat{j} + 2\hat{k} \]


Step 4: Final Answer:

The linear velocity is \((-18\hat{i} - 13\hat{j} + 2\hat{k})\).
Quick Tip: Remember the order of the cross product is crucial: \(\vec{v} = \vec{\omega} \times \vec{r}\). Reversing the order to \(\vec{r} \times \vec{\omega}\) will give the same magnitude but opposite direction.
Always verify signs carefully while calculating the \(\hat{j}\) component in the determinant.

Step 1: Understanding the Question:

Brewster's Law states that for a specific angle of incidence (Brewster's angle), the reflected light is completely polarized. The relationship depends on the refractive indices of the two media.


Step 2: Key Formula or Approach:

Brewster's angle \(\theta_p\) is given by:
\[ \tan \theta_p = \frac{\mu_2}{\mu_1} \]

where \(\mu_2\) is the refractive index of the second medium and \(\mu_1\) is the first.


Step 3: Detailed Explanation:

For propagation from air to glass:

Let the refractive index of glass be \(\mu\). For air, \(\mu_{air} = 1\).
\[ \tan \theta = \frac{\mu}{1} \implies \tan \theta = \mu \]

For propagation from glass to air:

Let the new Brewster's angle be \(\theta'\).
\[ \tan \theta' = \frac{1}{\mu} \]

Substituting \(\mu = \tan \theta\):
\[ \tan \theta' = \frac{1}{\tan \theta} = \cot \theta \]

Using trigonometric identity \(\cot \theta = \tan(\frac{\pi}{2} - \theta)\):
\[ \tan \theta' = \tan\left(\frac{\pi}{2} - \theta\right) \]

Thus, \(\theta' = \frac{\pi}{2} - \theta\).


Step 4: Final Answer:

Statement (A) is correct as it accurately reflects the mathematical relationship between the two Brewster angles.
Quick Tip: If light travels from medium 1 to 2, \(\tan \theta_{1 \rightarrow 2} = n_{21}\).
If light travels from medium 2 to 1, \(\tan \theta_{2 \rightarrow 1} = n_{12} = \frac{1}{n_{21}}\).
Since \(\tan \alpha \cdot \tan \beta = 1\) implies \(\alpha + \beta = 90^\circ\), the two angles must be complementary.

Step 1: Understanding the Question:

The question asks about the effect of temperature on the specific resistance (resistivity) of different materials.


Step 2: Detailed Explanation:

1. Conductors: In conductors (metals), as temperature increases, the thermal vibrations of lattice ions increase. This leads to more frequent collisions between free electrons and ions, thereby increasing the resistivity (\(\rho\)). Thus, specific resistance increases with temperature.

2. Semiconductors: In semiconductors, as temperature increases, more covalent bonds break, leading to a significant increase in the concentration of charge carriers (electrons and holes). This effect dominates over the increased collision frequency, resulting in a net decrease in resistivity. Thus, specific resistance decreases with temperature.


Step 3: Final Answer:

Specific resistance increases for a conductor and decreases for a semiconductor as temperature increases.
Quick Tip: Conductors have a positive temperature coefficient of resistance (\(\alpha > 0\)).
Semiconductors and insulators have a negative temperature coefficient of resistance (\(\alpha < 0\)).

Step 1: Understanding the Question:

The gas undergoes a two-step expansion process: first isothermal, then adiabatic. We need to find the final pressure after both steps.


Step 2: Key Formula or Approach:

1. Isothermal process: \(P_1 V_1 = P_2 V_2\)

2. Adiabatic process: \(P_2 V_2^\gamma = P_3 V_3^\gamma\)

3. For a polyatomic gas at moderate temperatures, the ratio of specific heats \(\gamma = \frac{4}{3}\) (assuming 6 degrees of freedom).


Step 3: Detailed Explanation:

Step I: Isothermal expansion

Initial state: \((P, V)\)

Final state: \((P', 3V)\)
\[ P \cdot V = P' \cdot (3V) \implies P' = \frac{P}{3} \]

Step II: Adiabatic expansion

Initial state: \((P', 3V)\)

Final state: \((P_{final}, 24V)\)

Using \(P' \cdot V_{initial}^\gamma = P_{final} \cdot V_{final}^\gamma\):
\[ \frac{P}{3} \cdot (3V)^{4/3} = P_{final} \cdot (24V)^{4/3} \]
\[ P_{final} = \frac{P}{3} \cdot \left(\frac{3V}{24V}\right)^{4/3} \]
\[ P_{final} = \frac{P}{3} \cdot \left(\frac{1}{8}\right)^{4/3} \]

Since \(8 = 2^3\):
\[ P_{final} = \frac{P}{3} \cdot \left((2^3)^{1/3}\right)^{-4} = \frac{P}{3} \cdot (2)^{-4} \]
\[ P_{final} = \frac{P}{3} \cdot \frac{1}{16} = \frac{P}{48} \]


Step 4: Final Answer:

The final pressure of the gas is \(P/48\).
Quick Tip: For polyatomic gases (non-linear molecules like \(NH_3, CH_4\)), common values are \(f = 6\) and \(\gamma = 1 + \frac{2}{6} = \frac{4}{3}\).
In adiabatic calculations involving powers like \(1/8\), check if the base is a power of 2 to simplify the fractional exponent.

Step 1: Understanding the Question:

Work done in moving a charge in an electric field is equal to the change in electrostatic potential energy of the system. Bringing them "2 cm closer" means the final distance is \(10 - 2 = 8\) cm.


Step 2: Key Formula or Approach:

Electrostatic potential energy \(U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r} = k \frac{q_1 q_2}{r}\)

Work done \(W = \Delta U = U_{final} - U_{initial} = k q_1 q_2 \left(\frac{1}{r_f} - \frac{1}{r_i}\right)\)


Step 3: Detailed Explanation:

Given:
\(q_1 = 10 \times 10^{-6} C\)
\(q_2 = 4 \times 10^{-6} C\)
\(r_i = 10 cm = 0.1 m\)
\(r_f = 10 - 2 = 8 cm = 0.08 m\)
\(k = 9 \times 10^9\)

Substitute the values:
\[ W = 9 \times 10^9 \times (10 \times 10^{-6}) \times (4 \times 10^{-6}) \left(\frac{1}{0.08} - \frac{1}{0.1}\right) \]
\[ W = 9 \times 10^9 \times 40 \times 10^{-12} \left(12.5 - 10\right) \]
\[ W = 360 \times 10^{-3} \times (2.5) \]
\[ W = 0.36 \times 2.5 = 0.9 J \]


Step 4: Final Answer:

The work required is 0.9 J.
Quick Tip: Be careful with the phrase "bring them 2 cm closer". It denotes the change in distance, not the final distance.
Convert all units to SI (cm to m, \(\mu C\) to C) before starting calculations.

Step 1: Understanding the Question:

This is a case of a conical pendulum. The tension \(T\) in the string provides the horizontal centripetal force and balances the weight vertically.


Step 2: Key Formula or Approach:

Let \(T\) be tension, \(\theta\) be the angle with vertical.

1. \(T \cos \theta = mg\)

2. \(T \sin \theta = m \omega^2 r\), where \(r = l \sin \theta\) is the radius of the circular path.


Step 3: Detailed Explanation:

From the second equation:
\[ T \sin \theta = m \omega^2 (l \sin \theta) \]

Dividing both sides by \(\sin \theta\) (assuming \(\theta \neq 0\)):
\[ T = m \omega^2 l \]

Given:

Frequency \(f = \frac{1}{\pi}\) rev/s

Angular velocity \(\omega = 2\pi f = 2\pi \left(\frac{1}{\pi}\right) = 2\) rad/s

Length \(= l\)

Mass \(= m\)

Substitute \(\omega = 2\) into the tension formula:
\[ T = m \cdot (2)^2 \cdot l = 4ml \]


Step 4: Final Answer:

The tension in the string is 4 ml.
Quick Tip: In a conical pendulum, the tension component \(T \sin \theta\) depends on the radius.
A common shortcut: for a conical pendulum, the Tension is simply \(m \omega^2 l\) because the \(\sin \theta\) term cancels out from both the force and geometry expressions.

Step 1: Understanding the Question:

A change in magnetic flux through a coil induces an electromotive force (EMF) based on Faraday's law. This EMF causes a current to flow through the total resistance of the circuit.


Step 2: Key Formula or Approach:

1. Induced EMF for \(n\) turns: \(e = -n \frac{\Delta \phi}{\Delta t} = n \frac{\phi_1 - \phi_2}{t}\) (magnitude)

2. Total resistance \(R_{total} = R + \frac{R}{4}\)

3. Induced current \(I = \frac{e}{R_{total}}\)


Step 3: Detailed Explanation:

The total resistance of the circuit is:
\[ R_{total} = R + \frac{R}{4} = \frac{5R}{4} \]

The magnitude of induced EMF in the coil is:
\[ e = n \frac{|\phi_2 - \phi_1|}{t} = n \frac{(\phi_1 - \phi_2)}{t} \]

Induced current is:
\[ I = \frac{n (\phi_1 - \phi_2) / t}{5R/4} \]
\[ I = \frac{4n (\phi_1 - \phi_2)}{5Rt} \]


Step 4: Final Answer:

The induced current is \(\frac{4n(\phi_1 - \phi_2)}{5Rt}\).
Quick Tip: Remember that current depends on the total resistance of the entire closed loop, including external series resistors.
Don't forget the factor '\(n\)' for a coil with multiple turns.

Step 1: Understanding the Question:

The "gravitational pull" refers to the acceleration due to gravity (\(g\)) on the surface. We need to find the relationship between the radii of the Earth and the Moon given the ratios of their \(g\) values and masses.


Step 2: Key Formula or Approach:

Acceleration due to gravity is:
\[ g = \frac{GM}{R^2} \implies R^2 = \frac{GM}{g} \]


Step 3: Detailed Explanation:

Let \(g_e, M_e, R_e\) be Earth's parameters and \(g_m, M_m, R_m\) be Moon's.

Given: \(g_m = \frac{1}{6} g_e\) and \(M_m = \frac{1}{8} M_e\).

Using the formula for \(g\):
\[ \frac{g_m}{g_e} = \frac{M_m}{M_e} \times \left(\frac{R_e}{R_m}\right)^2 \]
\[ \frac{1/6 g_e}{g_e} = \frac{1/8 M_e}{M_e} \times \left(\frac{R_e}{R_m}\right)^2 \]
\[ \frac{1}{6} = \frac{1}{8} \times \left(\frac{R_e}{R_m}\right)^2 \]
\[ \left(\frac{R_e}{R_m}\right)^2 = \frac{8}{6} = \frac{4}{3} \]

Taking the square root on both sides:
\[ \frac{R_e}{R_m} = \sqrt{\frac{4}{3}} \]
\[ R_e = \sqrt{\frac{4}{3}} R_m \]


Step 4: Final Answer:

The radius of the Earth is \((\sqrt{4/3})^{th}\) of the Moon's radius.
Quick Tip: For ratio problems in gravitation, write the formula \(g \propto M/R^2\) and equate the ratio of variables directly to save time.

Step 1: Understanding the Question:

According to Einstein's photoelectric equation, the stopping potential varies linearly with the frequency of incident light. The slope of this graph is related to Planck's constant \(h\).


Step 2: Key Formula or Approach:

Einstein's photoelectric equation:
\[ eV_s = hv - \phi_0 \implies V_s = \left(\frac{h}{e}\right)v - \frac{\phi_0}{e} \]

This is in the form \(y = mx + c\), where the slope \(m = \frac{h}{e}\).


Step 3: Detailed Explanation:

From the given graph, consider two points \((v_1, V_1)\) and \((v_2, V_2)\).

The slope of the graph is given by:
\[ Slope = \frac{\Delta y}{\Delta x} = \frac{V_2 - V_1}{v_2 - v_1} \]

Since \(Slope = \frac{h}{e}\):
\[ \frac{h}{e} = \frac{V_2 - V_1}{v_2 - v_1} \]
\[ h = \frac{e(V_2 - V_1)}{(v_2 - v_1)} \]


Step 4: Final Answer:

The value of Planck's constant is \(\frac{e(V_2 - V_1)}{(v_2 - v_1)}\).
Quick Tip: The slope of the Stopping Potential (\(V\)) vs Frequency (\(v\)) graph is always \(h/e\), which is a universal constant for all metals.
The x-intercept is the threshold frequency (\(v_0\)).

Step 1: Understanding the Question:

Work done in increasing the surface area of a liquid film is the product of surface tension and the increase in surface area. A soap bubble has two free surfaces (inner and outer).


Step 2: Key Formula or Approach:
\[ W = T \times \Delta A_{total} \]

For a soap bubble: \(\Delta A_{total} = 2 \times (4\pi R_2^2 - 4\pi R_1^2)\)


Step 3: Detailed Explanation:

Given diameters \(d_1\) and \(d_2\), the radii are \(R_1 = d_1/2\) and \(R_2 = d_2/2\).

The total surface area of a soap bubble is \(2 \times (4\pi R^2)\).

Initial Area \(A_1 = 2 \times 4\pi (d_1/2)^2 = 2\pi d_1^2\)

Final Area \(A_2 = 2 \times 4\pi (d_2/2)^2 = 2\pi d_2^2\)

Increase in area \(\Delta A = A_2 - A_1 = 2\pi (d_2^2 - d_1^2)\)

Work done \(W = T \cdot \Delta A = T \cdot 2\pi (d_2^2 - d_1^2)\)


Step 4: Final Answer:

The work done is \(2\pi (d_2^2 - d_1^2) T\).
Quick Tip: Always check if the question mentions a "bubble" or a "drop".
Liquid Drop: 1 surface, \(\Delta A = 4\pi(R_2^2 - R_1^2)\).
Soap Bubble: 2 surfaces, \(\Delta A = 8\pi(R_2^2 - R_1^2) = 2\pi(d_2^2 - d_1^2)\).

Step 1: Understanding the Question:

The mechanical energy (Potential Energy) lost by the object as it falls is converted into heat upon impact, which melts the ice. The object also cools from \(4^\circ C\) to \(0^\circ C\), but the problem states its temperature just before hitting the snow is \(0^\circ C\), implying the thermal energy change during fall is already accounted for or negligible compared to the impact energy. We focus on the kinetic energy just before impact, which equals the initial Potential Energy.


Step 2: Key Formula or Approach:

1. Potential Energy \(PE = mgh\)

2. Heat required to melt ice \(Q = m_{ice} L\)

3. Conservation of energy: \(PE = Q\)


Step 3: Detailed Explanation:

Given:
\(m_{object} = 3.5 kg\)
\(h = 2000 m\)
\(g = 10 m/s^2\)
\(L = 3.5 \times 10^5 J/kg\)

Total potential energy converted to heat:
\[ PE = 3.5 \times 10 \times 2000 = 70,000 J \]

This heat melts mass \(m_{ice}\) of ice:
\[ 70,000 = m_{ice} \times (3.5 \times 10^5) \]
\[ m_{ice} = \frac{70,000}{350,000} = \frac{7}{35} = 0.2 kg \]

Converting to grams:
\[ 0.2 kg = 0.2 \times 1000 = 200 g \]


Step 4: Final Answer:

The quantity of ice that melts is 200 gram.
Quick Tip: In energy conversion problems, ensure all units are in SI. \(1 J = 1 kg \cdot m^2/s^2\).
If the specific heat capacity of the object is not provided, assume only the mechanical energy (PE or KE) contributes to melting.

Step 1: Understanding the Question:

For a particle moving in a horizontal circle inside a smooth cone, the normal reaction \(N\) from the surface provides both the centripetal force and counteracts gravity.


Step 2: Key Formula or Approach:

Let \(\theta\) be the semi-vertical angle of the cone.

Vertical equilibrium: \(N \sin \theta = mg\)

Horizontal (centripetal) force: \(N \cos \theta = \frac{mv^2}{r}\)

Dividing the equations: \(\tan \theta = \frac{rg}{v^2}\)

Also, from geometry, \(\tan \theta = \frac{r}{h}\), where \(h\) is the height from vertex.


Step 3: Detailed Explanation:

Equating the two expressions for \(\tan \theta\):
\[ \frac{r}{h} = \frac{rg}{v^2} \]
\[ v^2 = gh \implies v = \sqrt{gh} \]

Given:
\(h = 10 cm = 0.1 m\)
\(g = 10 m/s^2\)

Substitute the values:
\[ v = \sqrt{10 \times 0.1} = \sqrt{1} = 1 m/s \]


Step 4: Final Answer:

The speed of the particle is 1 m/s.
Quick Tip: For a particle on a smooth cone, the speed required to stay at height \(h\) is independent of the radius \(r\) and the mass \(m\).
The final formula \(v = \sqrt{gh}\) is very similar to the velocity at the lowest point of a simple pendulum, but here it applies to conical horizontal motion.

Step 1: Understanding the Question:

We need to find the ratio 'n' of the speeds of two stones such that their centripetal forces are identical.


Step 2: Key Formula or Approach:

Centripetal force \(F = \frac{mv^2}{r}\)


Step 3: Detailed Explanation:

Let '1' denote the lighter stone and '2' denote the heavier stone.
\(m_1 = m, r_1 = r, v_1 = v_1\)
\(m_2 = 3m, r_2 = r/3, v_2 = v_2\)

Given condition: \(F_1 = F_2\)
\[ \frac{m_1 v_1^2}{r_1} = \frac{m_2 v_2^2}{r_2} \]
\[ \frac{m \cdot v_1^2}{r} = \frac{3m \cdot v_2^2}{r/3} \]
\[ \frac{m v_1^2}{r} = \frac{9 m v_2^2}{r} \]
\[ v_1^2 = 9 v_2^2 \]
\[ v_1 = 3 v_2 \]

The question states \(v_1 = n v_2\). Comparing the two, we get \(n = 3\).


Step 4: Final Answer:

The value of \(n\) is 3.
Quick Tip: When comparing forces, always substitute the given mass and radius ratios carefully into the \(mv^2/r\) formula.
Note how the radius in the denominator for the heavier stone (\(r/3\)) flips and multiplies the numerator.

Step 1: Understanding the Question:

In an ideal LC circuit, energy oscillates between the electric field of the capacitor and the magnetic field of the inductor. Energy is conserved.


Step 2: Key Formula or Approach:

Max energy in Capacitor = Max energy in Inductor
\[ \frac{1}{2} C V^2 = \frac{1}{2} L I_{max}^2 \]


Step 3: Detailed Explanation:

Given:
\(C = 4 \mu F = 4 \times 10^{-6} F\)
\(V = 10 V\)
\(L = 10 mH = 10 \times 10^{-3} H = 10^{-2} H\)

Using the energy conservation equation:
\[ C V^2 = L I_{max}^2 \]
\[ (4 \times 10^{-6}) \times (10)^2 = (10^{-2}) \times I_{max}^2 \]
\[ 4 \times 10^{-4} = 10^{-2} \times I_{max}^2 \]
\[ I_{max}^2 = \frac{4 \times 10^{-4}}{10^{-2}} = 4 \times 10^{-2} = 0.04 \]

Taking square root:
\[ I_{max} = \sqrt{0.04} = 0.2 A \]


Step 4: Final Answer:

The maximum current in the coil is 0.2 A.
Quick Tip: Alternatively, you can use \(I_{max} = Q_{max} \omega\), where \(\omega = \frac{1}{\sqrt{LC}}\) and \(Q_{max} = CV\).
\(I_{max} = (CV) \times \frac{1}{\sqrt{LC}} = V \sqrt{\frac{C}{L}}\). Both methods lead to the same result.

Step 1: Understanding the Question:

The total energy in SHM is the sum of Kinetic Energy (KE) and Potential Energy (PE) at any instant. Total energy is also equal to the maximum potential energy at the extreme position.


Step 2: Key Formula or Approach:

1. Total Energy \(E = KE + PE\)

2. \(E = \frac{1}{2} m \omega^2 A^2\)

3. Angular frequency \(\omega = 2\pi f\)


Step 3: Detailed Explanation:

Given:
\(m = 0.2 kg\)
\(f = \frac{25}{\pi} Hz \implies \omega = 2\pi \cdot \frac{25}{\pi} = 50 rad/s\)

At \(x = 0.04 m\), \(KE = 1 J\) and \(PE = 0.6 J\).

Total Energy \(E = 1 + 0.6 = 1.6 J\)

Using the total energy formula:
\[ 1.6 = \frac{1}{2} \times 0.2 \times (50)^2 \times A^2 \]
\[ 1.6 = 0.1 \times 2500 \times A^2 \]
\[ 1.6 = 250 \times A^2 \]
\[ A^2 = \frac{1.6}{250} = \frac{16}{2500} \]

Taking the square root:
\[ A = \sqrt{\frac{16}{2500}} = \frac{4}{50} = \frac{8}{100} = 0.08 m \]


Step 4: Final Answer:

The amplitude of oscillation is 0.08 m.
Quick Tip: In SHM problems with given energies, always find the Total Energy first. It's constant throughout the motion.
Note: At \(x=0.04\), \(PE = \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} (0.2)(50^2)(0.04^2) = 0.1 \times 2500 \times 0.0016 = 0.4 J\). Since the question gives \(PE = 0.6 J\), it might imply the potential energy zero point is not at equilibrium, but calculating \(A\) using Total \(E\) and frequency is the standard path.

Step 1: Understanding the Question:

The figure shows two identical circular coils perpendicular to each other, with a common center. Each creates a magnetic field at the center directed along its axis.


Step 2: Key Formula or Approach:

1. Magnetic field at the center of a circular coil: \(B = \frac{\mu_0 N I}{2R}\)

2. Since the coils are perpendicular, their fields \(\vec{B_1}\) and \(\vec{B_2}\) are perpendicular.

3. Net magnetic field \(B_{net} = \sqrt{B_1^2 + B_2^2}\)


Step 3: Detailed Explanation:

Both coils are identical and carry the same current.
\(B_1 = \frac{\mu_0 N I}{2R}\)
\(B_2 = \frac{\mu_0 N I}{2R}\)

The vectors are at \(90^\circ\) to each other.
\[ B_{net} = \sqrt{\left(\frac{\mu_0 N I}{2R}\right)^2 + \left(\frac{\mu_0 N I}{2R}\right)^2} \]
\[ B_{net} = \frac{\mu_0 N I}{2R} \sqrt{1^2 + 1^2} \]
\[ B_{net} = \frac{\mu_0 N I}{2R} \sqrt{2} \]
\[ B_{net} = \frac{\mu_0 N I}{\sqrt{2}R} \]


Step 4: Final Answer:

The resultant magnetic field at the center is \(\frac{\mu_0 N I}{\sqrt{2}R}\).
Quick Tip: When two identical fields are perpendicular, the resultant is always \(\sqrt{2}\) times the individual field.
If the coils were carrying currents in the same plane, you would add or subtract the fields depending on the current direction.

Step 1: Understanding the Question:

In a "Death Well" (cylindrical wall), a motorcyclist moves in horizontal circles. To prevent the rider from falling down, the frictional force must balance the gravitational weight.


Step 2: Key Formula or Approach:

The normal reaction \(N\) from the wall provides the centripetal force:
\[ N = \frac{mv^2}{R} \]

The frictional force \(f_s\) must balance the weight \(mg\):
\[ f_s = mg \]

Also, for the rider not to slip, \(f_s \leq \mu_s N\).


Step 3: Detailed Explanation:

Substituting the expressions into the inequality:
\[ mg \leq \mu_s \left(\frac{mv^2}{R}\right) \]

Cancelling mass \(m\) from both sides and rearranging for velocity \(v\):
\[ g \leq \frac{\mu_s v^2}{R} \implies v^2 \geq \frac{Rg}{\mu_s} \]
\[ v \geq \sqrt{\frac{Rg}{\mu_s}} \]

Thus, the minimum speed required is \(v_{min} = \sqrt{\frac{Rg}{\mu_s}}\).


Step 4: Final Answer:

The minimum speed required is \(\sqrt{\frac{Rg}{\mu_s}}\).
Quick Tip: Notice that the minimum speed is independent of the mass of the cyclist and the vehicle.
Always remember that in this vertical wall case, friction is vertical and the normal reaction is horizontal.

Step 1: Understanding the Question:

The number of times a listener hears maximum intensity in one second is equal to the beat frequency, which is the absolute difference between the frequencies of the two sound waves.


Step 2: Key Formula or Approach:

A standard wave equation is \(y = A \sin(kx - \omega t)\), where \(\omega = 2\pi f\).

Beat frequency \(b = |f_1 - f_2|\).


Step 3: Detailed Explanation:

From \(y_1 = a \sin(0.2\pi x - 50\pi t)\):
\(\omega_1 = 50\pi \implies 2\pi f_1 = 50\pi \implies f_1 = 25\) Hz.

From \(y_2 = a \sin(0.15\pi x - 46\pi t)\):
\(\omega_2 = 46\pi \implies 2\pi f_2 = 46\pi \implies f_2 = 23\) Hz.

Beat frequency:
\[ b = |25 - 23| = 2 beats/second \]

This means the maximum intensity is heard 2 times in one second.


Step 4: Final Answer:

The listener hears maximum intensity 2 times in one second.
Quick Tip: Beats occur due to the superposition of two waves with slightly different frequencies.
Always look for the coefficient of \(t\) to find angular frequency \(\omega\) and divide it by \(2\pi\) to get the linear frequency \(f\).

Step 1: Understanding the Question:

When two coils are in series and placed close to each other, mutual inductance (\(M\)) comes into play. The net inductance depends on whether the magnetic fields aid or oppose each other.


Step 2: Key Formula or Approach:

For series opposing connection: \(L_{net} = L_1 + L_2 - 2M\).

For identical coils placed "very close", assume perfect coupling coefficient \(k = 1\), so \(M = \sqrt{L_1 L_2}\).


Step 3: Detailed Explanation:

Given:

1. Coils are identical: \(L_1 = L_2 = L\).

2. Winding direction is exactly opposite: This is a series opposing case.

3. Placed very close: Coupling is nearly perfect, so \(M = \sqrt{L \times L} = L\).

Now, substitute these into the net inductance formula:
\[ L_{net} = L + L - 2L = 2L - 2L = 0 \]

The magnetic fluxes produced by the two coils cancel each other out completely.


Step 4: Final Answer:

The net inductance is zero.
Quick Tip: If the winding direction was the same (aiding), the net inductance would be \(L + L + 2L = 4L\).
"Opposite winding" usually implies a non-inductive coil configuration used in making standard resistors.

Step 1: Understanding the Question:

Breaking a large drop into small droplets increases the total surface area, which requires work (energy). This work is provided by multiplying surface tension with the change in surface area.


Step 2: Key Formula or Approach:

1. Volume conservation: \(\frac{4}{3}\pi R^3 = n \times \frac{4}{3}\pi r^3 \implies R^3 = n r^3\)

2. Energy \(W = T \cdot \Delta A = T(n \cdot 4\pi r^2 - 4\pi R^2)\)


Step 3: Detailed Explanation:

Given \(n = 216\):
\(R = \sqrt[3]{216} \cdot r = 6r \implies r = \frac{R}{6}\).

Now, calculate the increase in surface area:
\[ \Delta A = 216 \times 4\pi r^2 - 4\pi R^2 \]
\[ \Delta A = 216 \times 4\pi \left(\frac{R}{6}\right)^2 - 4\pi R^2 \]
\[ \Delta A = 216 \times 4\pi \left(\frac{R^2}{36}\right) - 4\pi R^2 \]
\[ \Delta A = 6 \times 4\pi R^2 - 4\pi R^2 = 24\pi R^2 - 4\pi R^2 = 20\pi R^2 \]

Work done:
\[ W = T \cdot \Delta A = T(20\pi R^2) = 20\pi TR^2 \]

Comparing with the given form \(x TR^2\), we get \(x = 20\pi\).


Step 4: Final Answer:

The value of x is \(20\pi\).
Quick Tip: For \(n\) droplets, the work done is \(W = 4\pi TR^2 (n^{1/3} - 1)\).
Here, \(W = 4\pi TR^2 (216^{1/3} - 1) = 4\pi TR^2 (6 - 1) = 20\pi TR^2\). This formula saves calculation time.

Step 1: Understanding the Question:

A molecule is polar if it has a non-zero net dipole moment. This happens when the center of positive charge does not coincide with the center of negative charge due to asymmetry.


Step 2: Detailed Explanation:

Looking at the provided diagrams:

(a) Represents a bent molecule (like \(H_2O\)). The dipole moments of the two bonds do not cancel each other, resulting in a net dipole moment. Hence, it is polar.

(b) Represents a homonuclear diatomic molecule (like \(N_2\) or \(O_2\)). It is perfectly symmetric; the centers of charge coincide, so it is non-polar.

(c) Represents a linear symmetric molecule (like \(CO_2\)). The individual bond dipoles are equal and opposite, cancelling out to zero. It is non-polar.

(d) Represents a homonuclear diatomic molecule (like \(H_2\)). Similar to (b), it is non-polar.


Step 3: Final Answer:

The molecule labeled (a) is the polar molecule. According to the options provided, this corresponds to choice (B).
Quick Tip: Check for symmetry. Symmetric molecules (linear \(O=C=O\), tetrahedral \(CH_4\)) are generally non-polar as bond dipoles cancel.
Bent or pyramidal shapes often result in polar molecules.

Step 1: Understanding the Question:

The root mean square (r.m.s.) speed is defined as the square root of the average of the squares of the speeds of all individual molecules.


Step 2: Key Formula or Approach:
\[ v_{rms} = \sqrt{\frac{v_1^2 + v_2^2 + ... + v_n^2}{n}} \]


Step 3: Detailed Explanation:

Given speeds: 2, 5, 3, 6, 3, 5 m/s. Number of molecules \(n = 6\).

Square the speeds: \(4, 25, 9, 36, 9, 25\).

Sum of squares:
\[ \sum v^2 = 4 + 25 + 9 + 36 + 9 + 25 = 108 \]

Mean of squares:
\[ \overline{v^2} = \frac{108}{6} = 18 \]

Root Mean Square:
\[ v_{rms} = \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2} \]

Substituting \(\sqrt{2} \approx 1.414\):
\[ v_{rms} = 3 \times 1.414 = 4.242 m/s \]


Step 4: Final Answer:

The r.m.s. speed is approximately 4.24 m/s.
Quick Tip: \(v_{rms}\) is always slightly greater than the average speed. For these values, average speed is \((2+5+3+6+3+5)/6 = 4\) m/s, so \(v_{rms}\) must be \(> 4\) m/s. This helps eliminate options (A) and (B).

Step 1: Understanding the Question:

The power factor in an AC circuit is given by the cosine of the phase angle, which can be expressed in terms of resistance (\(R\)) and impedance (\(Z\)).


Step 2: Key Formula or Approach:

Power Factor \(\cos \phi = \frac{R}{Z}\).


Step 3: Detailed Explanation:

Let \(R\) be the constant resistance.

For the first case:
\(0.5 = \frac{R}{Z_1} \implies Z_1 = \frac{R}{0.5} = 2R\)

For the second case:
\(0.25 = \frac{R}{Z_2} \implies Z_2 = \frac{R}{0.25} = 4R\)

The problem states \(Z_1 = x Z_2\). Substituting the values:
\[ 2R = x (4R) \]
\[ x = \frac{2R}{4R} = \frac{2}{4} = 0.5 \]


Step 4: Final Answer:

The value of x is 0.5.
Quick Tip: Power factor is inversely proportional to impedance if resistance is kept constant. If the power factor halves (from 0.5 to 0.25), the impedance must double. Hence \(Z_1\) is half of \(Z_2\).

Step 1: Understanding the Question:

The given equation represents the superposition of two perpendicular SHMs of the same frequency, or more simply, a combination of two sine and cosine functions.


Step 2: Key Formula or Approach:

For \(x = a_1 \sin \omega t + a_2 \cos \omega t\), the resultant amplitude is \(R = \sqrt{a_1^2 + a_2^2}\).


Step 3: Detailed Explanation:

Let \(A = R \cos \theta\) and \(B = R \sin \theta\).

The equation becomes:
\[ x = R \cos \theta \sin \omega t + R \sin \theta \cos \omega t \]
\[ x = R \sin(\omega t + \theta) \]

This is the standard equation of SHM with amplitude \(R\).

From the substitution:
\[ A^2 + B^2 = R^2 \cos^2 \theta + R^2 \sin^2 \theta = R^2(\cos^2 \theta + \sin^2 \theta) = R^2 \]
\[ R = \sqrt{A^2 + B^2} = (A^2 + B^2)^{1/2} \]


Step 4: Final Answer:

The motion is simple harmonic with amplitude \((A^2 + B^2)^{1/2}\).
Quick Tip: Superposition of any number of SHMs of the same frequency along the same line results in an SHM.
The vector addition of phasors can also be used to find the resultant amplitude quickly.

Step 1: Understanding the Question:

A wire of length \(L\) is bent into a circular coil of \(N\) turns. We need to find how the magnetic moment \(M\) depends on \(L\).


Step 2: Key Formula or Approach:

1. Magnetic moment \(M = N I A\), where \(A = \pi r^2\).

2. Length of wire \(L = N(2\pi r)\).


Step 3: Detailed Explanation:

From the length equation, radius \(r = \frac{L}{2\pi N}\).

Substitute \(r\) into the area formula:
\[ A = \pi r^2 = \pi \left(\frac{L}{2\pi N}\right)^2 = \frac{L^2}{4\pi N^2} \]

Now, substitute \(A\) into the magnetic moment formula:
\[ M = N I A = N I \left(\frac{L^2}{4\pi N^2}\right) \]
\[ M = \frac{I L^2}{4\pi N} \]

For a fixed number of turns \(N\) and current \(I\):
\[ M \propto L^2 \]


Step 4: Final Answer:

The relation is \(M \propto L^2\).
Quick Tip: Magnetic moment is proportional to the area enclosed. For a fixed total length, a single turn (\(N=1\)) coil encloses the maximum area and thus has the maximum magnetic moment.

Step 1: Understanding the Question:

In YDSE, the resultant intensity at a point depends on the phase difference between the two waves, which is directly proportional to the path difference.


Step 2: Key Formula or Approach:

1. Phase difference \(\phi = \frac{2\pi}{\lambda} \Delta x\).

2. Intensity \(I_{res} = I_{max} \cos^2\left(\frac{\phi}{2}\right)\).


Step 3: Detailed Explanation:

Case 1: Path difference \(\Delta x = \lambda\).
\[ \phi_1 = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi \]

Resultant intensity:
\[ I = I_{max} \cos^2\left(\frac{2\pi}{2}\right) = I_{max} \cos^2(\pi) = I_{max} \cdot (-1)^2 = I_{max} \]

So, the maximum intensity is \(I\).

Case 2: Path difference \(\Delta x' = \frac{\lambda}{6}\).
\[ \phi_2 = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{6} = \frac{\pi}{3} = 60^\circ \]

Resultant intensity \(I'\):
\[ I' = I_{max} \cos^2\left(\frac{\phi_2}{2}\right) = I \cos^2\left(\frac{\pi/3}{2}\right) = I \cos^2\left(\frac{\pi}{6}\right) \]

Given \(\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}\):
\[ I' = I \cdot \left(\frac{\sqrt{3}}{2}\right)^2 = I \cdot \frac{3}{4} = \frac{3I}{4} \]


Step 4: Final Answer:

The intensity is \(\frac{3I}{4}\).
Quick Tip: A path difference of \(\lambda\) corresponds to a bright fringe (max intensity).
Path differences of \(\lambda/4\) and \(\lambda/2\) result in intensities of \(I_{max}/2\) and \(0\) respectively. Always map path difference to phase difference first.

Step 1: Understanding the Question:

We use the First Law of Thermodynamics to relate heat, work, and internal energy for various processes.


Step 2: Key Formula or Approach:

First Law of Thermodynamics: \(Q = \Delta U + W_{by}\), where \(W_{by}\) is work done BY the system.

Equivalently, \(Q = \Delta U - W_{on}\), where \(W_{on}\) is work done ON the system.


Step 3: Detailed Explanation:

Given: \(\Delta U = W_{on}\).

Substituting this into the first law:
\[ Q = \Delta U - \Delta U = 0 \]

A process in which there is no heat exchange (\(Q = 0\)) between the system and its surroundings is called an Adiabatic process.

In such a process, all work done on the system is stored as internal energy.


Step 4: Final Answer:

The system undergoes an adiabatic process.
Quick Tip: Isothermal: \(\Delta U = 0 \implies Q = W_{by}\).
Isochoric: \(W = 0 \implies Q = \Delta U\).
Adiabatic: \(Q = 0 \implies \Delta U = -W_{by} = W_{on}\).
Memorizing these simplified forms of the first law for standard processes is very helpful.

Step 1: Understanding the Question:

The principle of a potentiometer is that the EMF of a cell is proportional to its balancing length. The proportionality constant is the potential gradient \(k\).


Step 2: Key Formula or Approach:
\(E = k l = \left(\frac{V}{L}\right)l\), where \(V\) is the potential difference across the total wire length \(L\).


Step 3: Detailed Explanation:

Initial state:

Length \(= L\). Potential gradient \(k_1 = \frac{V}{L}\).

Balancing length \(l_1 = \frac{L}{4}\).

EMF \(E = k_1 l_1 = \left(\frac{V}{L}\right) \cdot \left(\frac{L}{4}\right) = \frac{V}{4}\).

Final state:

New length \(L' = L + \frac{L}{3} = \frac{4L}{3}\).

New potential gradient \(k_2 = \frac{V}{L'} = \frac{V}{4L/3} = \frac{3V}{4L}\).

Let the new balancing length be \(l_2\). For the same cell:
\[ E = k_2 l_2 \]
\[ \frac{V}{4} = \left(\frac{3V}{4L}\right) \cdot l_2 \]
\[ 1 = \left(\frac{3}{L}\right) \cdot l_2 \implies l_2 = \frac{L}{3} \]


Step 4: Final Answer:

The new null point is obtained at \(L/3\).
Quick Tip: If the driver cell voltage remains constant, the balancing length is directly proportional to the total length of the wire (\(l \propto L_{total}\)).
Ratio: \(\frac{l_1}{l_2} = \frac{L_1}{L_2} \implies \frac{L/4}{l_2} = \frac{L}{4L/3} = \frac{3}{4} \implies l_2 = \frac{4}{3} \cdot \frac{L}{4} = \frac{L}{3}\).

Step 1: Understanding the Question:

We need to compare the harmonics of an open pipe and a closed pipe of the same length \(L\).


Step 2: Key Formula or Approach:

1. Open pipe: \(n^{th}\) overtone is \((n+1)^{th}\) harmonic. Frequency \(f_p = (n+1) \left(\frac{v}{2L}\right)\).

2. Closed pipe: \(n^{th}\) overtone is \((2n+1)^{th}\) harmonic. Frequency \(f'_c = (2n+1) \left(\frac{v}{4L}\right)\).


Step 3: Detailed Explanation:

Let the fundamental frequency of the open pipe be \(f = \frac{v}{2L}\).

Second overtone of open pipe \((n=2)\):
\[ f_{o2} = (2+1) \cdot f = 3f \]

Now, consider the closed pipe of same length. Its fundamental frequency is \(f_c = \frac{v}{4L} = \frac{f}{2}\).

Third overtone of closed pipe \((n=3)\):
\[ f'_{c3} = (2 \times 3 + 1) \cdot f_c = 7 \cdot \left(\frac{f}{2}\right) = 3.5f \]

Given: \(f'_{c3} - f_{o2} = 200\) Hz.
\[ 3.5f - 3f = 200 \]
\[ 0.5f = 200 \implies f = 400 Hz \]


Step 4: Final Answer:

The fundamental frequency of the open pipe is 400 Hz.
Quick Tip: Remember: Open pipe harmonics are \(1f, 2f, 3f...\). Closed pipe harmonics (of the same length) are \(0.5f, 1.5f, 2.5f...\) (only odd multiples of \(v/4L\)).
Mapping overtones to harmonics correctly is crucial.

Step 1: Understanding the Question:

The question asks about the change in current when frequency increases, assuming the applied voltage remains constant. Current \(I\) is inversely proportional to reactance \(X\).


Step 2: Key Formula or Approach:

1. Inductive reactance \(X_L = \omega L = 2\pi f L\).

2. Capacitive reactance \(X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}\).

3. Current \(I = \frac{V}{X}\).


Step 3: Detailed Explanation:

Circuit 1 (Pure Inductor):

As frequency \(f\) increases, \(X_L \propto f\) increases.

Since \(I_1 = \frac{V}{X_L}\), an increase in \(X_L\) leads to a decrease in current.

Circuit 2 (Pure Capacitor):

As frequency \(f\) increases, \(X_C \propto \frac{1}{f}\) decreases.

Since \(I_2 = \frac{V}{X_C}\), a decrease in \(X_C\) leads to an increase in current.


Step 4: Final Answer:

The current will decrease in the first circuit and increase in the second circuit.
Quick Tip: Inductors block high-frequency AC (Low pass filters).
Capacitors pass high-frequency AC (High pass filters).
Thinking of their filter behavior helps arrive at the answer quickly.

Step 1: Understanding the Question:

The potential energy in SHM varies with displacement \(x\), reaching its maximum at the extreme positions (\(x = \pm A\)).


Step 2: Key Formula or Approach:

1. Potential Energy \(PE = \frac{1}{2} m \omega^2 x^2\).

2. Maximum Potential Energy \(PE_{max} = \frac{1}{2} m \omega^2 A^2\).


Step 3: Detailed Explanation:

Given: \(PE = \frac{1}{2} PE_{max}\).
\[ \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} \cdot \left(\frac{1}{2} m \omega^2 A^2\right) \]

Divide both sides by \(\frac{1}{2} m \omega^2\):
\[ x^2 = \frac{1}{2} A^2 \]

Taking the square root on both sides:
\[ x = \pm \frac{A}{\sqrt{2}} \]


Step 4: Final Answer:

The displacement is \(\pm \frac{A}{\sqrt{2}}\).
Quick Tip: At \(x = A/2\), \(PE\) is \(1/4\) of \(PE_{max}\).
At \(x = A/\sqrt{2}\), \(PE = KE = \frac{1}{2} E_{total}\). This is a very common point discussed in SHM studies.

Step 1: Understanding the Question:

The de-Broglie wavelength of a charged particle depends on its kinetic energy, which is determined by the accelerating potential difference \(V\).


Step 2: Key Formula or Approach:
\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2meV}} \]

Thus, \(\lambda \propto \frac{1}{\sqrt{V}}\).


Step 3: Detailed Explanation:

Let \(V_1 = 10\) kV and \(V_2 = 20\) kV.

Wavelength ratio:
\[ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}} \]
\[ \frac{\lambda_2}{\lambda} = \sqrt{\frac{10}{20}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]
\[ \lambda_2 = \frac{\lambda}{\sqrt{2}} \]


Step 4: Final Answer:

The new wavelength is \(\frac{\lambda}{\sqrt{2}}\).
Quick Tip: If the potential is doubled, the wavelength decreases by a factor of \(\sqrt{2}\).
If the potential is quadrupled (multiplied by 4), the wavelength is halved.

Step 1: Understanding the Question:

The fundamental frequency of a vibrating string depends on its length, tension, and its mass per unit length (which relates to diameter).


Step 2: Key Formula or Approach:

Frequency \(n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}\)

Mass per unit length \(\mu = volume per unit length \times density = \pi \left(\frac{D}{2}\right)^2 \cdot 1 \cdot \rho = \frac{\pi D^2 \rho}{4}\).

So, \(n = \frac{1}{2L} \sqrt{\frac{T}{\pi D^2 \rho / 4}} = \frac{1}{LD} \sqrt{\frac{T}{\pi \rho}}\).

This shows \(n \propto \frac{\sqrt{T}}{LD}\).


Step 3: Detailed Explanation:

Initial frequency: \(n \propto \frac{\sqrt{T}}{LD}\).

New parameters:
\(T' = 3T\), \(L' = 3L\), \(D' = 2D\).

New frequency \(n'\):
\[ n' \propto \frac{\sqrt{3T}}{(3L)(2D)} \]
\[ n' \propto \frac{\sqrt{3}}{6} \cdot \frac{\sqrt{T}}{LD} \]

Ratio:
\[ \frac{n'}{n} = \frac{\sqrt{3}}{6} = \frac{\sqrt{3}}{2 \times 3} = \frac{\sqrt{3}}{2 \times \sqrt{3} \times \sqrt{3}} = \frac{1}{2\sqrt{3}} \]
\[ n' = \frac{n}{2\sqrt{3}} \]


Step 4: Final Answer:

The new frequency will be \(\frac{n}{2\sqrt{3}}\).
Quick Tip: When dealing with string diameter changes, remember that frequency is inversely proportional to diameter (\(n \propto 1/D\)).
Always write out the ratio formula for all variables to avoid missing any factor.

Step 1: Understanding the Question:

The question asks to find the atomic number \(Z\) (represented by \(P\)) of the daughter nucleus after a series of radioactive decays. Conservation of mass number \(A\) and charge number \(Z\) is used.


Step 2: Key Formula or Approach:

1. \(\alpha\)-decay: \(A \rightarrow A-4, Z \rightarrow Z-2\).

2. \(\beta^-\) (electron) decay: \(A \rightarrow A, Z \rightarrow Z+1\).

3. \(\beta^+\) (positron) decay: \(A \rightarrow A, Z \rightarrow Z-1\).


Step 3: Detailed Explanation:

Initial nucleus: \(^{242}_{92}X\).

Change in Atomic Number (\(Z\)):

- Two \(\alpha\) particles: \(2 \times (-2) = -4\)

- One electron (\(e^-\)): \(1 \times (+1) = +1\)

- Two positrons (\(e^+\)): \(2 \times (-1) = -2\)

Net change in \(Z\):
\[ \Delta Z = -4 + 1 - 2 = -5 \]

Final atomic number \(P\):
\[ P = 92 - 5 = 87 \]

(Note: Net change in \(A = 2 \times (-4) = -8\). Final \(A = 242 - 8 = 234\), which matches the given daughter nucleus).


Step 4: Final Answer:

The value of P is 87.
Quick Tip: Treat the decay particles as "accounting" for \(Z\) and \(A\). Electrons increase \(Z\), positrons and alphas decrease it.
Always verify if the given final mass number matches your calculation to ensure no decay steps were missed.

Step 1: Understanding the Question:

This problem relates path difference to phase difference and resultant intensity in an interference pattern.


Step 2: Key Formula or Approach:

1. Phase difference \(\phi = \frac{2\pi}{\lambda} \Delta x\).

2. Resultant intensity \(I = I_0 \cos^2\left(\frac{\phi}{2}\right)\).


Step 3: Detailed Explanation:

Given path difference \(\Delta x = \frac{\lambda}{4}\).

Calculate phase difference \(\phi\):
\[ \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2} = 90^\circ \]

Now calculate intensity \(I\):
\[ I = I_0 \cos^2\left(\frac{\pi/2}{2}\right) = I_0 \cos^2\left(\frac{\pi}{4}\right) = I_0 \cos^2(45^\circ) \]

Using \(\cos 45^\circ = \frac{1}{\sqrt{2}}\):
\[ I = I_0 \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_0}{2} \]

The ratio \(\frac{I_0}{I}\) is:
\[ \frac{I_0}{I} = \frac{I_0}{I_0 / 2} = \frac{2}{1} \]


Step 4: Final Answer:

The ratio of \(I_0\) to \(I\) is 2 : 1.
Quick Tip: At path difference \(\lambda/4\), the intensity is exactly half of the maximum intensity.
This corresponds to a point halfway between a maximum and a minimum in terms of energy.

Step 1: Understanding the Question:

Points A and B enclose cell \(E_1\). Points A and C enclose both \(E_1\) and \(E_2\). From the circuit diagram, the cells are in series opposing combination when considering points A and C (negative of \(E_1\) to negative of \(E_2\)).


Step 2: Key Formula or Approach:

Potential balancing principle: \(E \propto l\).


Step 3: Detailed Explanation:

Case 1: Potential between A and B.

Only \(E_1\) is in the circuit branch.
\[ E_1 = k \cdot l_1 = k \cdot (3.60) \quad --- (i) \]

Case 2: Potential between A and C.

Both cells are connected in series. Looking at the polarities (positive terminals at A and C, negative terminals connected together at B), the net EMF is \(E_1 - E_2\).
\[ E_1 - E_2 = k \cdot l_2 = k \cdot (0.90) \quad --- (ii) \]

Divide equation (i) by (ii):
\[ \frac{E_1}{E_1 - E_2} = \frac{3.60}{0.90} = 4 \]

Cross multiply:
\[ E_1 = 4E_1 - 4E_2 \]
\[ 4E_2 = 3E_1 \]
\[ \frac{E_1}{E_2} = \frac{4}{3} \]


Step 4: Final Answer:

The ratio \(E_1/E_2\) is 4 : 3.
Quick Tip: Check polarity carefully! If terminals are \((+ -)(+ -)\), it's addition \((E_1 + E_2)\). If they are \((+ -)(- +)\), it's subtraction \((E_1 - E_2)\).
The shorter balancing length for the combined circuit (0.9 m vs 3.6 m) is a clear indicator that the cells are opposing.