Step 1: Understanding the Concept:

This problem is based on Torricelli's Law, which states that the speed of efflux of a fluid through a sharp-edged hole at the bottom of a tank filled to a depth \(h\) is the same as the speed that a body (in this case, a drop of water) would acquire in falling freely from a height \(h\).


Step 2: Key Formula or Approach:

The velocity of efflux \(v\) is given by the formula: \[ v = \sqrt{2gh} \]
where \(g\) is the acceleration due to gravity and \(h\) is the depth of the hole from the free surface of the liquid.


Step 3: Detailed Explanation:

Given values are:
Depth, \(h = 2\) m
Acceleration due to gravity, \(g \approx 10\) m/s\(^2\) (using 10 for standard approximation, or 9.8 for more precision).
Substitute the values into the formula: \[ v = \sqrt{2 \times 10 \times 2} \] \[ v = \sqrt{40} \] \[ v = 2\sqrt{10} m/s \]
If we use \(g = 9.8\) m/s\(^2\): \[ v = \sqrt{2 \times 9.8 \times 2} = \sqrt{39.2} \approx 6.26 m/s \]
Both forms are acceptable depending on the specific exam's conventions, but \(2\sqrt{10}\) is most common in non-calculator exams.


Step 4: Final Answer:

The velocity of water flow is \(2\sqrt{10}\) m/s.
Quick Tip: Always assume \(g = 10 m/s^2\) in competitive exams to quickly check options, unless the answers are very close to each other, in which case use \(9.8 m/s^2\).

Step 1: Understanding the Concept:

When objects are stacked, each object must support the weight of all the objects positioned directly above it. The "force experienced" usually refers to the load it carries on its upper surface.


Step 2: Key Formula or Approach:

The downward force exerted on any stone in the stack is equal to the total weight of the stones resting on top of it. \[ F = n \cdot mg \]
where \(n\) is the number of stones above the stone in question.


Step 3: Detailed Explanation:

The stack consists of 10 stones total.
We are looking at the 6th stone from the bottom.
The stones above the 6th stone are the 7th, 8th, 9th, and 10th stones.
Number of stones above the 6th stone = \(10 - 6 = 4\) stones.
Each stone has a mass '\(m\)'.
The total mass resting on the 6th stone is \(4m\).
Therefore, the downward force (weight) experienced by the 6th stone from the top part of the stack is: \[ F = 4mg \]

Step 4: Final Answer:

The force experienced by the 6th stone from the top load is \(4mg\).
Quick Tip: Draw a quick diagram. Numbering the items from bottom to top helps prevent off-by-one errors when counting how many items are above or below a specific point.

Step 1: Understanding the Concept:

According to Bohr's model of the hydrogen atom, the angular momentum of an electron in a stationary orbit is quantized and is an integral multiple of \(\frac{h}{2\pi}\).


Step 2: Key Formula or Approach:

The angular momentum \(L\) for the \(n^{th}\) orbit is given by Bohr's quantization condition: \[ L_n = \frac{nh}{2\pi} \]
To find the difference, we calculate \(\Delta L = L_5 - L_3\).


Step 3: Detailed Explanation:

Angular momentum of the \(5^{th}\) orbit (\(n = 5\)): \[ L_5 = \frac{5h}{2\pi} \]
Angular momentum of the 3rd orbit (\(n = 3\)): \[ L_3 = \frac{3h}{2\pi} \]
The difference between their angular momenta is: \[ \Delta L = L_5 - L_3 \] \[ \Delta L = \frac{5h}{2\pi} - \frac{3h}{2\pi} \] \[ \Delta L = \frac{(5 - 3)h}{2\pi} \] \[ \Delta L = \frac{2h}{2\pi} \]
Simplify the fraction: \[ \Delta L = \frac{h}{\pi} \]

Step 4: Final Answer:

The difference in angular momentum is \(\frac{h}{\pi}\).
Quick Tip: Remember that angular momentum transitions always occur in discrete steps of \(\frac{h}{2\pi}\). A jump of \(\Delta n\) orbits means a change of \(\Delta n \times \frac{h}{2\pi}\).

Step 1: Understanding the Concept:

The problem involves a single-loop circuit with multiple DC voltage sources (cells) and resistors in series. We need to find the equivalent electromotive force (EMF) and the equivalent total resistance to calculate the current using Ohm's Law.


Step 2: Key Formula or Approach:

For cells connected in series aiding each other, the equivalent EMF is the sum of individual EMFs: \[ E_{eq} = E_1 + E_2 + E_3 \]
The equivalent internal resistance is: \[ r_{eq} = r_1 + r_2 + r_3 \]
The total current \(I\) in the circuit is given by Ohm's Law: \[ I = \frac{E_{eq}}{R_{ext} + r_{eq}} \]


Step 3: Detailed Explanation:

Based on standard circuit diagram conventions and the provided image, all three cells are oriented in the same direction (long line on the left, short thick line on the right), meaning their EMFs are additive (aiding each other).
Total equivalent EMF: \[ E_{eq} = 3V + 4V + 4V = 11V \]
Total internal resistance of the cells: \[ r_{eq} = 0.5\,\Omega + 0.75\,\Omega + 0.75\,\Omega = 2.0\,\Omega \]
The external resistance is: \[ R = 4\,\Omega \]
Total resistance in the circuit: \[ R_{total} = R + r_{eq} = 4\,\Omega + 2.0\,\Omega = 6.0\,\Omega \]
Calculate the current using Ohm's law: \[ I = \frac{E_{eq}}{R_{total}} \] \[ I = \frac{11}{6} A \approx 1.83 A \]


Step 4: Final Answer:

The current through the 4\(\Omega\) resistor is \(\frac{11}{6}\) A.
Quick Tip: Always carefully check the polarity of battery symbols in series circuits. The long line represents the positive terminal. If they are aligned \(+ - \rightarrow + -\), they add up.

Step 1: Understanding the Concept:

The efficiency of a Carnot heat engine (or any ideal reversible heat engine) depends only on the absolute temperatures of the hot reservoir (source) and the cold reservoir (sink).


Step 2: Key Formula or Approach:

The formula for the efficiency \(\eta\) of a heat engine is: \[ \eta = 1 - \frac{T_2}{T_1} \]
where: \(T_1\) = absolute temperature of the source (in Kelvin) \(T_2\) = absolute temperature of the sink (in Kelvin)
Always convert Celsius to Kelvin before using thermodynamic formulas: \(T(K) = T(^\circ C) + 273\).


Step 3: Detailed Explanation:

Given values:
Efficiency, \(\eta = 40% = 0.40\)
Sink temperature, \(T_2 = 27^\circ\text{C}\)
Convert sink temperature to Kelvin: \[ T_2 = 27 + 273 = 300 K \]
Substitute the known values into the efficiency formula to find \(T_1\): \[ 0.40 = 1 - \frac{300}{T_1} \]
Rearrange the equation to solve for the fraction: \[ \frac{300}{T_1} = 1 - 0.40 \] \[ \frac{300}{T_1} = 0.60 \]
Solve for \(T_1\): \[ T_1 = \frac{300}{0.60} \] \[ T_1 = \frac{3000}{6} = 500 K \]
If the options are in Celsius, convert it back: \[ T_1 (^\circC) = 500 - 273 = 227^\circC \]

Step 4: Final Answer:

The temperature of the source is \(500\) K (or \(227^\circ\text{C}\)).
Quick Tip: The most common mistake in thermodynamics problems is forgetting to convert temperatures from Celsius to Kelvin. Always do this first!

Step 1: Understanding the Concept:

When a body at rest explodes, no external forces are acting on the system, so linear momentum is conserved. The two fragments must fly apart with equal and opposite momenta to maintain a total momentum of zero.


Step 2: Key Formula or Approach:

1) Conservation of momentum: \(\vec{p_1} + \vec{p_2} = 0 \implies |p_1| = |p_2| = p\).
2) Kinetic energy in terms of momentum: \(K = \frac{p^2}{2m}\).
The ratio of kinetic energies is inversely proportional to their masses because their momenta are equal.


Step 3: Detailed Explanation:

Let the two fragments have masses \(m_1 = m\) and \(m_2 = 3m\).
Since they have equal magnitude of momentum \(p\), their kinetic energies are: \[ K_1 = \frac{p^2}{2m_1} = \frac{p^2}{2m} \] \[ K_2 = \frac{p^2}{2m_2} = \frac{p^2}{2(3m)} = \frac{p^2}{6m} \]
The total kinetic energy released is \(E\): \[ E = K_1 + K_2 = \frac{p^2}{2m} + \frac{p^2}{6m} \]
Find a common denominator to add them: \[ E = \frac{3p^2}{6m} + \frac{p^2}{6m} = \frac{4p^2}{6m} = \frac{2p^2}{3m} \]
We need to find the kinetic energy of the fragment of mass \(m\), which is \(K_1\):
From the total energy equation, we can express \(\frac{p^2}{2m}\) in terms of \(E\): \[ E = \frac{4}{3} \left( \frac{p^2}{2m} \right) \] \[ E = \frac{4}{3} K_1 \]
Solving for \(K_1\): \[ K_1 = \frac{3}{4} E \]
Alternatively, use the shortcut ratio method: \[ K_1 : K_2 = \frac{1}{m_1} : \frac{1}{m_2} = \frac{1}{m} : \frac{1}{3m} = 3 : 1 \]
Therefore, \(K_1\)'s share of the total energy \(E\) is: \[ K_1 = \left( \frac{3}{3 + 1} \right) E = \frac{3}{4} E \]

Step 4: Final Answer:

The kinetic energy of the fragment of mass \(m\) is \(\frac{3}{4}E\).
Quick Tip: In an explosion into two parts, the kinetic energy divides in the inverse ratio of the masses. The lighter fragment always carries away the majority of the kinetic energy.

Step 1: Understanding the Concept:

A current-carrying loop generates its own magnetic field and behaves like a magnetic dipole. It has a magnetic dipole moment \(\vec{M}\) which is perpendicular to the plane of the loop.


Step 2: Key Formula or Approach:

When a magnetic dipole \(\vec{M}\) is placed in a uniform external magnetic field \(\vec{B}\), it experiences a torque given by: \[ \vec{\tau} = \vec{M} \times \vec{B} \]
The magnitude of the torque is \(\tau = MB \sin\theta\), where \(\theta\) is the angle between the magnetic moment and the magnetic field.


Step 3: Detailed Explanation:

Because the loop is freely suspended, the torque will cause it to rotate.
The loop will continue to rotate until it reaches a position of stable equilibrium.
Stable equilibrium occurs when the potential energy \(U = -\vec{M} \cdot \vec{B} = -MB \cos\theta\) is minimized.
This minimum occurs when \(\theta = 0^\circ\), meaning the magnetic moment vector \(\vec{M}\) aligns perfectly parallel to the external magnetic field \(\vec{B}\).
Since the magnetic moment vector is perpendicular to the plane of the loop, aligning \(\vec{M}\) parallel to \(\vec{B}\) means the plane of the loop itself becomes perpendicular to the magnetic field direction.


Step 4: Final Answer:

The loop will experience a torque and rotate to align its magnetic moment parallel to the magnetic field, meaning its plane becomes perpendicular to the magnetic field lines.
Quick Tip: Always distinguish between the direction of the "plane of the loop" and the "area vector/magnetic moment". They are perpendicular to each other. Stable equilibrium aligns the area vector with the B-field.

Step 1: Understanding the Concept:

When a magnetic dipole is rotated in a uniform magnetic field, work must be done against the magnetic torque. This work done is stored as the change in magnetic potential energy of the dipole.


Step 2: Key Formula or Approach:

The work done \(W\) in rotating a magnetic dipole of moment \(M\) in a uniform magnetic field \(B\) from an initial angle \(\theta_1\) to a final angle \(\theta_2\) is given by the change in potential energy: \[ W = U_{final} - U_{initial} = (-MB \cos\theta_2) - (-MB \cos\theta_1) \] \[ W = MB (\cos\theta_1 - \cos\theta_2) \]


Step 3: Detailed Explanation:

Given values:
Initial angle, \(\theta_1 = 60^\circ\)
Final angle, \(\theta_2 = 90^\circ\)
Substitute these angles into the work formula: \[ W = MB (\cos 60^\circ - \cos 90^\circ) \]
We know the standard trigonometric values: \(\cos 60^\circ = \frac{1}{2}\) \(\cos 90^\circ = 0\)
Substituting these in gives: \[ W = MB \left( \frac{1}{2} - 0 \right) \] \[ W = \frac{MB}{2} \]

Step 4: Final Answer:

The work done to rotate the dipole is \(\frac{MB}{2}\).
Quick Tip: Remember the potential energy formula \(U = -MB \cos\theta\). Work done by an external agent is \(\Delta U = U_f - U_i\). Be careful with the negative signs to avoid sign errors.

Step 1: Understanding the Concept:

The potential energy \(U\) stored in a capacitor network connected to a constant voltage source \(V\) depends on its equivalent capacitance \(C_{eq}\).
The formula is \(U = \frac{1}{2} C_{eq} V^2\).
To find the configuration with maximum stored energy, we must identify which circuit has the maximum equivalent capacitance.


Step 2: Key Formula or Approach:

For capacitors in series: \(\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots\)
For capacitors in parallel: \(C_{eq} = C_1 + C_2 + \dots\)
Evaluate \(C_{eq}\) for each provided configuration.


Step 3: Detailed Explanation:

Let's analyze the equivalent capacitance for each given circuit:

Circuit (a):
The top branch has two capacitors \(C\) in series. Their equivalent capacitance is \(C_{top} = \frac{C \times C}{C + C} = \frac{C}{2}\).
This top branch is in parallel with the bottom capacitor \(C\).
Total equivalent capacitance: \(C_a = \frac{C}{2} + C = 1.5 C\).

Circuit (b):
All three capacitors are connected in series.
Total equivalent capacitance: \(\frac{1}{C_b} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C} \implies C_b = \frac{C}{3} \approx 0.33 C\).

Circuit (c):
All three capacitors are connected in parallel.
Total equivalent capacitance: \(C_c = C + C + C = 3 C\).

Comparing the equivalent capacitances: \(C_c (3C) > C_a (1.5C) > C_b (0.33C)\).
Since energy \(U\) is directly proportional to \(C_{eq}\), circuit (c) stores the maximum potential energy.


Step 4: Final Answer:

The maximum potential energy is stored in configuration c).
Quick Tip: To maximize capacitance (and thus stored energy for a given voltage), connect components in parallel. To minimize it, connect them in series.

Step 1: Understanding the Concept:

This problem relates to an astronomical telescope. Unless stated otherwise, we assume it is in "normal adjustment" (image formed at infinity) because this is the standard operating mode where equations are simplest. Note: The unit 'cm' on magnifying power in the question text is a typo, as magnification is dimensionless. We treat the value as \(M = 75\).


Step 2: Key Formula or Approach:

For a telescope in normal adjustment:
1) Magnifying power magnitude: \(M = \frac{f_o}{f_e}\)
2) Tube length (distance between objective and eyepiece): \(L = f_o + f_e\)
where \(f_o\) is the focal length of the objective lens and \(f_e\) is the focal length of the eyepiece.


Step 3: Detailed Explanation:

Given values:
Tube length, \(L = 76\) cm
Magnifying power, \(M = 75\)
Using the magnification formula: \[ \frac{f_o}{f_e} = 75 \implies f_o = 75 f_e \]
Substitute this expression for \(f_o\) into the tube length formula: \[ L = f_o + f_e \] \[ 76 = (75 f_e) + f_e \] \[ 76 = 76 f_e \]
Solving for \(f_e\): \[ f_e = \frac{76}{76} = 1 cm \]
Now, substitute \(f_e\) back into the equation for \(f_o\): \[ f_o = 75 \times 1 cm \] \[ f_o = 75 cm \]

Step 4: Final Answer:

The focal length of the objective is \(75\) cm and the eyepiece is \(1\) cm.
Quick Tip: In an astronomical telescope, the objective lens always has a much larger focal length compared to the eyepiece (\(f_o \gg f_e\)) to achieve high magnification and light-gathering power.

Step 1: Understanding the Concept:

Displacement is a vector quantity that represents the shortest straight-line distance from the initial position to the final position. Since the person moves in two perpendicular directions, we can use the Pythagorean theorem to find the magnitude of the final displacement.


Step 2: Key Formula or Approach:

1) Distance = Speed \(\times\) Time
2) Vector displacement magnitude: \(|\vec{D}| = \sqrt{d_1^2 + d_2^2}\) for two perpendicular path segments \(d_1\) and \(d_2\).


Step 3: Detailed Explanation:

Calculate the distance for the first leg of the journey (from A to B):
Speed \(v_1 = 20\) km/h
Time \(t_1 = 2\) hours
Distance \(d_1 = v_1 \times t_1 = 20 \times 2 = 40\) km.
Let's align this movement along the x-axis.

Calculate the distance for the second leg of the journey (perpendicular to AB):
Speed \(v_2 = 20\) km/h
Time \(t_2 = 30\) minutes = \(0.5\) hours (Time must be in consistent units)
Distance \(d_2 = v_2 \times t_2 = 20 \times 0.5 = 10\) km.
Since it is perpendicular, let's align this movement along the y-axis.

The person forms a right-angled triangle with their path. The net displacement is the hypotenuse.
Magnitude of displacement: \[ |\vec{D}| = \sqrt{d_1^2 + d_2^2} \] \[ |\vec{D}| = \sqrt{40^2 + 10^2} \] \[ |\vec{D}| = \sqrt{1600 + 100} \] \[ |\vec{D}| = \sqrt{1700} \]
Simplify the radical: \[ |\vec{D}| = \sqrt{100 \times 17} = 10\sqrt{17} km \]

Step 4: Final Answer:

The displacement of the man is \(10\sqrt{17}\) km.
Quick Tip: Always ensure time units match speed units before calculating distance (e.g., convert 30 minutes to 0.5 hours). Drawing a quick vector diagram prevents confusion.

Step 1: Understanding the Concept:

We need to find the magnification of an image formed by a concave mirror. We can calculate this using the mirror formula to first find the image distance, and then the magnification formula, or use a direct relation between focal length, object distance, and magnification.


Step 2: Key Formula or Approach:

1) Cartesian Sign Convention:
Focal length of concave mirror, \(f\) is negative.
Object distance, \(u\) is negative (placed in front of the mirror).
2) Mirror formula: \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)
3) Magnification formula: \(m = -\frac{v}{u}\)
Alternatively, combine them for a faster direct formula: \(m = \frac{f}{f - u}\).


Step 3: Detailed Explanation:

Given values with sign convention:
Focal length, \(f = -80\) m
Object distance, \(u = -100\) m

Using the direct formula for magnification: \[ m = \frac{f}{f - u} \]
Substitute the values: \[ m = \frac{-80}{-80 - (-100)} \] \[ m = \frac{-80}{-80 + 100} \] \[ m = \frac{-80}{20} \] \[ m = -4 \]
A magnification of \(-4\) indicates the image is real (negative sign), inverted, and 4 times larger than the object.

Optional verification via mirror formula:
\(\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-80} - \frac{1}{-100} = -\frac{5}{400} + \frac{4}{400} = -\frac{1}{400} \implies v = -400\) m. \(m = -\frac{v}{u} = -\frac{-400}{-100} = -4\). Both methods yield the same result.


Step 4: Final Answer:

The magnification of the image produced is \(-4\).
Quick Tip: Memorize the direct magnification formulas \(m = \frac{f}{f - u}\) and \(m = \frac{f - v}{f}\). They save precious time by avoiding the intermediate step of calculating \(v\) or \(u\).

Step 1: Understanding the Concept:

The question asks to identify a physical quantity that shares the same dimensional formula as the Boltzmann constant (\(k_B\)). We must first find the dimensions of \(k_B\) and then match it with known quantities.


Step 2: Key Formula or Approach:

The Boltzmann constant relates kinetic energy to temperature. A fundamental relation is the average kinetic energy of a gas molecule: \(E = \frac{3}{2} k_B T\)
From this, the dimensional formula for \(k_B\) is \([k_B] = \frac{[Energy]}{[Temperature]}\).


Step 3: Detailed Explanation:

Let's find the dimensional formula for \(k_B\):
Dimensions of Energy \([E] = [M L^2 T^{-2}]\)
Dimensions of Temperature \([T] = [K]\) \[ [k_B] = \frac{[M L^2 T^{-2}]}{[K]} = [M L^2 T^{-2} K^{-1}] \]
Now let's check common physical quantities that involve energy per unit temperature:
1) Entropy (\(S\)): Defined thermodynamically as \(dS = \frac{dQ}{T}\), where \(dQ\) is heat energy.
Dimensions of Entropy = \(\frac{[Energy]}{[Temperature]} = [M L^2 T^{-2} K^{-1}]\).
2) Heat Capacity (\(C\)): Defined as \(C = \frac{\Delta Q}{\Delta T}\), the heat energy required to change temperature.
Dimensions of Heat Capacity = \(\frac{[Energy]}{[Temperature]} = [M L^2 T^{-2} K^{-1}]\).
Both Entropy and Heat Capacity have the exact same dimensional formula as the Boltzmann constant.


Step 4: Final Answer:

The dimension of \(k_B\) is the same as Entropy or Heat Capacity.
Quick Tip: Grouping physical quantities by their dimensional formulas is a powerful shortcut for exam questions. \(k_B\), Entropy, and Heat Capacity all represent "Energy per unit Kelvin".

Step 1: Understanding the Concept:

When the platform accelerates, the block tends to slip backward due to inertia. In the non-inertial reference frame of the platform, the block experiences a backward pseudo force. Static friction must provide the forward force to accelerate the block with the platform and prevent relative slipping.


Step 2: Key Formula or Approach:

1) Pseudo force on the block: \(F_p = ma\)
2) Maximum static friction: \(f_{s,max} = \mu_s N = \mu_s mg\)
To prevent slipping, the required frictional force must be less than or equal to the maximum available static friction: \(F_p \le f_{s,max}\).


Step 3: Detailed Explanation:

Given values:
Mass of block, \(m = 5\) kg
Acceleration of platform, \(a = 3\) m/s\(^2\)
The required force to keep the block moving with the platform is \(F = ma\). This force is provided purely by static friction \(f_s\).
So, \(f_s = m \cdot a\).
The condition to prevent slipping is that this required friction does not exceed the maximum possible static friction: \[ f_s \le \mu_s \cdot N \]
Since the surface is horizontal, the normal force \(N = mg\). \[ ma \le \mu_s \cdot mg \]
Notice that the mass \(m\) cancels out from both sides: \[ a \le \mu_s \cdot g \]
Solving for the minimum coefficient of static friction \(\mu_s\): \[ \mu_s \ge \frac{a}{g} \]
Substitute the values (using \(g = 10\) m/s\(^2\) for typical competitive exam simplicity): \[ \mu_{s,minimum} = \frac{3}{10} = 0.3 \]
If we strictly use \(g = 9.8\) m/s\(^2\): \(\mu_{s,minimum} = \frac{3}{9.8} \approx 0.306\). Both are conceptually correct.


Step 4: Final Answer:

The minimum coefficient of static friction required is \(0.3\).
Quick Tip: Notice that the mass of the block is irrelevant in finding the minimum coefficient of friction for horizontal acceleration. The condition simplifies directly to \(\mu_s \ge \frac{a}{g}\).

Step 1: Understanding the Concept:

The question text appears slightly garbled from memory. It most likely asks for the change in gravitational potential energy (\(\Delta U\)) of a mass \(m\) when raised to a height \(h\) from the surface of the Earth. The approximation \(mgh\) is invalid here because the height \(h\) is comparable to the radius of the Earth \(R\).


Step 2: Key Formula or Approach:

The general formula for the change in gravitational potential energy when a body is raised from the Earth's surface to a height \(h\) is: \[ \Delta U = \frac{mgh}{1 + \frac{h}{R}} \]
where \(m\) is the mass, \(g\) is gravity at surface, \(h\) is height, and \(R\) is Earth's radius.


Step 3: Detailed Explanation:

Given values:
Height, \(h = 3.6 \times 10^6\) m
Radius of Earth, \(R = 6.4 \times 10^6\) m
First, let's find the ratio of \(h\) to \(R\) to simplify the calculation: \[ \frac{h}{R} = \frac{3.6 \times 10^6}{6.4 \times 10^6} = \frac{3.6}{6.4} = \frac{36}{64} \]
Simplify the fraction by dividing numerator and denominator by 4: \[ \frac{h}{R} = \frac{9}{16} \]
Now, substitute this ratio into the \(\Delta U\) formula: \[ \Delta U = \frac{mg(h)}{1 + \frac{h}{R}} \]
Substitute \(h = \frac{9}{16} R\) into the numerator: \[ \Delta U = \frac{mg\left(\frac{9}{16}R\right)}{1 + \frac{9}{16}} \]
Calculate the denominator: \[ 1 + \frac{9}{16} = \frac{16}{16} + \frac{9}{16} = \frac{25}{16} \]
Now put it back together: \[ \Delta U = \frac{\frac{9}{16} mgR}{\frac{25}{16}} \]
The \(16\) in the denominators cancels out: \[ \Delta U = \frac{9}{25} mgR \]

Step 4: Final Answer:

The change in gravitational potential energy is \(\frac{9}{25} mgR\).
Quick Tip: Always use the exact formula \(\Delta U = \frac{mgh}{1 + h/R}\) whenever the height is given as a large value (e.g., in thousands of km) comparable to the Earth's radius.

Step 1: Understanding the Concept:

Simple Harmonic Motion (SHM) is symmetric and periodic. The particle oscillates between a negative extreme, a mean (equilibrium) position, and a positive extreme.


Step 2: Key Formula or Approach:

The total time for one complete cycle (e.g., mean \(\rightarrow\) positive extreme \(\rightarrow\) mean \(\rightarrow\) negative extreme \(\rightarrow\) mean) is defined as the Time Period, \(T\).
By symmetry, the cycle can be divided into four identical time intervals.


Step 3: Detailed Explanation:

Let's trace one full oscillation starting from the mean position:
1) Mean position to positive extreme: Takes time \(t_1\).
2) Positive extreme back to mean position: Takes time \(t_2\).
3) Mean position to negative extreme: Takes time \(t_3\).
4) Negative extreme back to mean position: Takes time \(t_4\).
Because SHM is completely symmetrical around the mean position, the time taken for each of these four distinct segments is exactly the same: \[ t_1 = t_2 = t_3 = t_4 \]
The sum of these four segments equals one full time period \(T\): \[ t_1 + t_2 + t_3 + t_4 = T \] \[ 4 \cdot t_1 = T \] \[ t_1 = \frac{T}{4} \]
Therefore, the time taken to move from the mean position to either extreme position is one-quarter of the total time period.


Step 4: Final Answer:

The time taken is \(\frac{T}{4}\).
Quick Tip: Using the reference circle model for SHM is highly effective here. Mean to extreme corresponds to rotating an angle of \(90^\circ\) (or \(\frac{\pi}{2}\) radians). Since a full \(360^\circ\) takes time \(T\), \(90^\circ\) takes \(\frac{T}{4}\).

Step 1: Understanding the Concept:

In the photoelectric effect, the work function (\(\Phi\)) is the minimum energy required to eject an electron from the surface of a metal. It is directly related to the threshold wavelength (\(\lambda_0\)), which is the maximum wavelength of light capable of causing photoelectric emission.


Step 2: Key Formula or Approach:

The relationship between work function and threshold wavelength is: \[ \Phi = \frac{hc}{\lambda_0} \]
To quickly calculate energy in electron-volts (eV) when wavelength is given in Angstroms (Å), use the convenient constant \(hc \approx 12400 eV\cdotÅ\) (or \(12420\) for slightly more precision, but \(12400\) is standard for quick exam math).


Step 3: Detailed Explanation:

Given value:
Threshold wavelength, \(\lambda_0 = 6000\) Å
Using the shortcut formula: \[ \Phi (in eV) = \frac{12400}{\lambda_0 (in Å)} \]
Substitute the given wavelength: \[ \Phi = \frac{12400}{6000} eV \]
Cancel the zeros: \[ \Phi = \frac{124}{60} eV \]
Divide by 4: \[ \Phi = \frac{31}{15} eV \]
Perform the division: \[ \Phi \approx 2.066 eV \]

Step 4: Final Answer:

The work function is approximately \(2.06\) eV.
Quick Tip: Memorize the value \(hc = 12400 eV\cdotÅ\) (or \(1240 eV\cdotnm\)). It dramatically speeds up all modern physics calculations involving photons and energy conversions.

Step 1: Understanding the Concept:

Kinetic energy is defined by the mass and velocity of a particle. When different particles have the same velocity, their kinetic energies will depend entirely on their respective masses. The particle with the greatest mass will have the highest kinetic energy.


Step 2: Key Formula or Approach:

The formula for kinetic energy is: \[ K.E. = \frac{1}{2}mv^2 \]
Let the mass of a proton be \(m\).
The mass of a deuteron (1 proton + 1 neutron) is approximately \(2m\).
The mass of an alpha particle (2 protons + 2 neutrons) is approximately \(4m\).


Step 3: Detailed Explanation:

Since all particles have the same velocity \(v\), we can write their kinetic energies as:
1) For the proton: \[ K.E._p = \frac{1}{2}mv^2 \]
2) For the deuteron: \[ K.E._d = \frac{1}{2}(2m)v^2 = 2 \left( \frac{1}{2}mv^2 \right) = 2 K.E._p \]
3) For the alpha particle: \[ K.E._\alpha = \frac{1}{2}(4m)v^2 = 4 \left( \frac{1}{2}mv^2 \right) = 4 K.E._p \]
Comparing the kinetic energies: \(K.E._p < K.E._d < K.E._\alpha\).
Therefore, the alpha particle has the highest kinetic energy.


Step 4: Final Answer:

The \(\alpha\) - particle has higher K.E.
Quick Tip: Always memorize the relative masses and charges of common subatomic particles: Proton (\(m, +e\)), Deuteron (\(2m, +e\)), Alpha (\(4m, +2e\)). For same velocity, \(KE \propto m\).

Step 1: Understanding the Concept:

This problem is an application of the conservation of angular momentum. When a swimmer jumps from a height, the net external torque acting on him is zero (ignoring air resistance), so his angular momentum remains constant.


Step 2: Key Formula or Approach:

The principle of conservation of angular momentum states: \[ L = I\omega = constant \]
where \(I\) is the moment of inertia and \(\omega\) is the angular velocity (spin rate).
To increase the number of spins (\(\omega\)), the swimmer must decrease his moment of inertia (\(I\)).


Step 3: Detailed Explanation:

Moment of inertia \(I\) depends on the distribution of mass relative to the axis of rotation (\(I = \sum mr^2\)).
When the swimmer stretches his arms and legs, his mass is distributed further from his axis of rotation, which increases his moment of inertia (\(I\)). This would decrease his spin rate (\(\omega\)).
Conversely, when he curls his body (tucks his knees to his chest), he brings his mass closer to the axis of rotation. This decreases his moment of inertia (\(I\)).
Since \(L = I\omega\) is constant, a decrease in \(I\) must result in a corresponding increase in \(\omega\) to keep the product constant. Therefore, curling his body increases his spin rate.


Step 4: Final Answer:

He must curl his body to decrease his moment of inertia.
Quick Tip: Angular momentum \(L = I\omega\) is always conserved in mid-air dives and jumps. Remember the inverse relationship: decreasing radius (curling up) decreases \(I\) and drastically increases spin speed \(\omega\).

Step 1: Understanding the Concept:

When two ideal gases are mixed without any heat loss to the surroundings, the total internal energy of the system is conserved. Both gases are monoatomic, meaning they have the same molar heat capacity at constant volume.


Step 2: Key Formula or Approach:

Conservation of internal energy: \(U_{total} = U_A + U_B\).
Internal energy of an ideal gas is \(U = nC_vT = \frac{N}{N_A} C_v T\), where \(N\) is the number of molecules and \(N_A\) is Avogadro's number. \[ (n_A + n_B) C_v T_{mix} = n_A C_v T_A + n_B C_v T_B \]
Since \(C_v\) is the same for both monoatomic gases, it cancels out, leaving: \[ T_{mix} = \frac{N_A T_A + N_B T_B}{N_A + N_B} \]


Step 3: Detailed Explanation:

Given values:
Number of molecules of gas A, \(N_A = 10^{24}\)
Number of molecules of gas B, \(N_B = 10^{23}\)
Temperature of gas A, \(T_A = 300\) K
Temperature of gas B, \(T_B = 400\) K

Substitute these into the mixture temperature formula: \[ T_{mix} = \frac{(10^{24} \times 300) + (10^{23} \times 400)}{10^{24} + 10^{23}} \]
Factor out \(10^{23}\) from the numerator and denominator to simplify: \[ T_{mix} = \frac{10^{23}(10 \times 300 + 1 \times 400)}{10^{23}(10 + 1)} \] \[ T_{mix} = \frac{3000 + 400}{11} \] \[ T_{mix} = \frac{3400}{11} \]
Perform the division: \[ T_{mix} \approx 309.09 K \]

Step 4: Final Answer:

The temperature of the mixture is approximately 309 K.
Quick Tip: For mixing identical types of gases (e.g., both monoatomic), the mixture temperature is simply the weighted average of their initial temperatures based on the number of moles (or molecules): \(T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}\).

Step 1: Understanding the Concept:

Silicon (Si) and Germanium (Ge) are Group 14 intrinsic semiconductors. Doping them with elements from different groups changes their electrical properties.
- Doping with Group 13 elements (trivalent) creates "holes" as majority carriers, resulting in a p-type semiconductor.
- Doping with Group 15 elements (pentavalent) provides extra electrons as majority carriers, resulting in an n-type semiconductor.


Step 2: Key Formula or Approach:

Identify the group of each dopant mentioned:
Group 13 (p-type): Boron (B), Aluminum (Al), Gallium (Ga), Indium (In).
Group 15 (n-type): Nitrogen (N), Phosphorus (P), Arsenic (As), Antimony (Sb), Bismuth (Bi).


Step 3: Detailed Explanation:

Let's evaluate each statement:
1) Indium is Group 13. Doping creates p-type. (Statement is False)
2) Phosphorus is Group 15. Doping creates n-type. (Statement is False)
3) Arsenic is Group 15. Doping creates n-type. (Statement is True)
4) Gallium is Group 13. Doping creates p-type. (Statement is False)
5) Antimony is Group 15. Doping creates n-type. (Statement is False)
Only statement 3 correctly identifies the resulting semiconductor type for the given dopant.


Step 4: Final Answer:

Arsenic is n type.
Quick Tip: A useful mnemonic for dopants: Group 13 makes P-type (remember the 'P' in positive holes), Group 15 makes N-type (remember the 'N' in negative electrons).

Step 1: Understanding the Concept:

Radioactive decay involves the emission of particles that alter the mass number (\(A\)) and atomic number (\(Z\)) of a nucleus.
- An alpha (\(\alpha\)) particle is a helium nucleus (\(^4_2He\)), so emitting one decreases the mass number by 4 and the atomic number by 2.
- A beta (\(\beta^-\)) particle is an electron (\(^0_{-1}e\)), so emitting one leaves the mass number unchanged but increases the atomic number by 1.


Step 2: Key Formula or Approach:

Write the balanced nuclear equation: \[ ^{A_{initial}}_{Z_{initial}}X \rightarrow ^{A_{final}}_{Z_{final}}Y + n_\alpha (^{4}_{2}He) + n_\beta (^{0}_{-1}e) \]
Conserve mass numbers (top values): \(A_{initial} = A_{final} + 4n_\alpha + 0\)
Conserve atomic numbers (bottom values): \(Z_{initial} = Z_{final} + 2n_\alpha - 1n_\beta\)


Step 3: Detailed Explanation:

Given the parent nucleus is \(^{223}_{94}X\). It emits 2 \(\alpha\) particles and 2 \(\beta\) particles.
The nuclear reaction is: \[ ^{223}_{94}X \rightarrow ^{a}_{b}Y + 2 (^{4}_{2}\alpha) + 2 (^{0}_{-1}\beta) \]
Apply conservation of mass number to find \(a\): \[ 223 = a + 2(4) + 2(0) \] \[ 223 = a + 8 \] \[ a = 223 - 8 = 215 \]
Apply conservation of atomic number to find \(b\): \[ 94 = b + 2(2) + 2(-1) \] \[ 94 = b + 4 - 2 \] \[ 94 = b + 2 \] \[ b = 94 - 2 = 92 \]
Therefore, the new nucleus \(Y\) has mass number \(a = 215\) and atomic number \(b = 92\).


Step 4: Final Answer:

The values are a = 215 and b = 92.
Quick Tip: Always assume beta decay refers to \(\beta^-\) (electron emission) unless \(\beta^+\) (positron emission) is explicitly stated. \(\beta^-\) decay increases the atomic number by 1.

Step 1: Understanding the Concept:

The magnetic field \(B\) inside a long, tightly wound solenoid is uniform and depends on the current flowing through it and the number of turns per unit length.


Step 2: Key Formula or Approach:

The formula for the magnetic field inside a solenoid is: \[ B = \mu_0 n I \]
where \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7}\) T\(\cdot\)m/A), \(I\) is the current, and \(n\) is the number of turns per unit length.
Since \(n = \frac{N}{L}\) (where \(N\) is total turns and \(L\) is length), we can rewrite the formula as: \[ B = \mu_0 \left( \frac{N}{L} \right) I \]


Step 3: Detailed Explanation:

Given values:
Magnetic field, \(B = 2\pi\) T
Current, \(I = 2\) A
Length, \(L = 2\) m
Permeability, \(\mu_0 = 4\pi \times 10^{-7}\) T\(\cdot\)m/A

Substitute these values into the magnetic field formula: \[ 2\pi = (4\pi \times 10^{-7}) \times \left( \frac{N}{2} \right) \times 2 \]
Simplify the right side (the \(2\) in the numerator and denominator cancel out): \[ 2\pi = (4\pi \times 10^{-7}) \times N \]
Solve for \(N\): \[ N = \frac{2\pi}{4\pi \times 10^{-7}} \] \[ N = \frac{1}{2 \times 10^{-7}} \] \[ N = 0.5 \times 10^7 \] \[ N = 5 \times 10^6 \]
The total number of turns is \(5 \times 10^6\).


Step 4: Final Answer:

The total number of turns is \(5 \times 10^6\).
Quick Tip: Always distinguish between \(N\) (total turns) and \(n\) (turn density = \(N/L\)). Solenoid formulas commonly use \(n\), so it's easy to forget to multiply by the length \(L\) to find the total turns \(N\).

Step 1: Understanding the Concept:

In Young's Double Slit Experiment (YDSE), the fringe width (the distance between two consecutive bright or dark fringes) depends on the wavelength of the light used.


Step 2: Key Formula or Approach:

The formula for fringe width (\(\beta\)) is: \[ \beta = \frac{\lambda D}{d} \]
where \(\lambda\) is the wavelength of light, \(D\) is the distance to the screen, and \(d\) is the distance between the slits.
Since \(D\) and \(d\) are constant for a given experimental setup, the fringe width is directly proportional to the wavelength: \[ \beta \propto \lambda \]


Step 3: Detailed Explanation:

To arrange the fringe widths in increasing order, we must arrange the corresponding wavelengths of the given colors in increasing order.
Recall the visible light spectrum (VIBGYOR). The wavelength increases from Violet to Red.
Comparing the given colors: Blue, Green, Red.
Their wavelengths order is: \[ \lambda_{Blue} < \lambda_{Green} < \lambda_{Red} \]
Because fringe width is directly proportional to wavelength (\(\beta \propto \lambda\)), the order of their fringe widths will be identical: \[ \beta_{Blue} < \beta_{Green} < \beta_{Red} \]


Step 4: Final Answer:

The increasing order of fringe width is Blue \(<\) Green \(<\) Red.
Quick Tip: Remember VIBGYOR: Wavelength (\(\lambda\)) increases from left to right. Since fringe width \(\beta = \lambda D/d\), red light always produces the widest fringes and violet the narrowest in any standard interference pattern.

Step 1: Understanding the Concept:

According to Faraday's Law of Induction, an electromotive force (EMF) is induced in a circuit whenever the magnetic flux linking that circuit changes. We must calculate the initial and final magnetic flux to find the average induced EMF.


Step 2: Key Formula or Approach:

1) Magnetic flux is given by: \(\Phi = BA \cos\theta\)
where \(\theta\) is the angle between the magnetic field vector \(B\) and the area normal vector \(A\).
2) Faraday's Law for average induced EMF: \(|e| = \left| \frac{\Delta \Phi}{\Delta t} \right| = \left| \frac{\Phi_{final} - \Phi_{initial}}{\Delta t} \right|\)


Step 3: Detailed Explanation:

Given values:
Area, \(A = 0.1 cm^2 = 0.1 \times 10^{-4} m^2 = 10^{-5} m^2\) (must convert to standard SI units)
Magnetic field, \(B = 0.5 T\)
Time interval, \(\Delta t = 0.5 s\)

Initial state: The plane of the coil is perpendicular to the magnetic field. This means the normal to the area is parallel to the magnetic field. Thus, the initial angle \(\theta_1 = 0^\circ\). \[ \Phi_{initial} = BA \cos(0^\circ) = BA(1) = BA \]
Final state: The coil is rotated so its plane becomes parallel to the field. This means the normal to the area is perpendicular to the field. Thus, the final angle \(\theta_2 = 90^\circ\). \[ \Phi_{final} = BA \cos(90^\circ) = BA(0) = 0 \]

Now, calculate the magnitude of the average induced EMF: \[ |e| = \left| \frac{\Phi_{final} - \Phi_{initial}}{\Delta t} \right| \] \[ |e| = \left| \frac{0 - BA}{0.5} \right| = \frac{BA}{0.5} \]
Substitute the values of \(B\) and \(A\): \[ |e| = \frac{(0.5 T) \times (10^{-5} m^2)}{0.5 s} \]
The \(0.5\) in numerator and denominator cancel out: \[ |e| = 10^{-5} V \]

Step 4: Final Answer:

The induced emf is \(10^{-5}\) V.
Quick Tip: Always double-check the angle \(\theta\). It is defined as the angle between the magnetic field lines and the normal (perpendicular line) to the surface area, not the surface itself. "Plane perpendicular to field" means \(\theta = 0^\circ\).

Step 1: Understanding the Concept:

This problem deals with the elastic properties of materials. Young's modulus relates the stress applied to a wire to the strain it experiences.
Step 2: Key Formula or Approach:

Young's Modulus (\(Y\)) is defined as the ratio of tensile stress to tensile strain: \[ Y = \frac{Stress}{Strain} = \frac{F/A}{\Delta L / L} \]
where \(F\) is the applied force, \(A\) is the cross-sectional area, and \(\Delta L / L\) is the fractional elongation (strain).
Rearranging to solve for Area (\(A\)): \[ A = \frac{F}{Y \times Strain} \]


Step 3: Detailed Explanation:

Given values:
Force, \(F = 4.5 \times 10^3 N\)
Young's Modulus, \(Y = 9 \times 10^{11} N/m^2\)
Elongation is \(1%\), which means the strain \(\left( \frac{\Delta L}{L} \right) = \frac{1}{100} = 0.01\).

Substitute these values into the rearranged formula: \[ A = \frac{4.5 \times 10^3}{9 \times 10^{11} \times 0.01} \]
Calculate the denominator: \[ 9 \times 10^{11} \times 0.01 = 9 \times 10^9 \]
Now, divide: \[ A = \frac{4.5 \times 10^3}{9 \times 10^9} \] \[ A = \left( \frac{4.5}{9} \right) \times 10^{3 - 9} \] \[ A = 0.5 \times 10^{-6} m^2 \]
To convert square meters to square millimeters (\(1 m^2 = 10^6 mm^2\)): \[ A = 0.5 \times 10^{-6} \times 10^6 mm^2 = 0.5 mm^2 \]

Step 4: Final Answer:

The cross-sectional area of the wire is 0.5 mm\(^2\).
Quick Tip: Percentage elongation directly translates to strain: \(X%\) elongation means strain \(\epsilon = X/100\). Always ensure \(Y\) and \(F\) are in compatible standard units (N/m\(^2\) and N) before calculating Area.

Step 1: Understanding the Concept:

The moment of inertia of a system of discrete particles about a specific axis is the sum of the products of each particle's mass and the square of its perpendicular distance from that axis.


Step 2: Key Formula or Approach:

The formula for the moment of inertia \(I\) about an axis is: \[ I = \sum m_i r_i^2 = m_1 r_1^2 + m_2 r_2^2 + \dots \]
where \(m_i\) is the mass of the \(i\)-th particle and \(r_i\) is its perpendicular distance from the axis of rotation.
For particles located entirely in the x-y plane, their distance from the origin (\(d = \sqrt{x^2+y^2}\)) is exactly equivalent to their perpendicular distance to the z-axis.


Step 3: Detailed Explanation:

Given values:
Mass of first particle, \(m_1 = 4 kg\)
Distance of first particle from origin, \(r_1 = 2 m\)
Mass of second particle, \(m_2 = 6 kg\)
Distance of second particle from origin, \(r_2 = 4 m\)

Because the particles lie in the x-y plane, the z-axis is perpendicular to this plane passing through the origin. The given distances from the origin are therefore the perpendicular distances \(r\) to the z-axis.
Substitute the values into the moment of inertia formula: \[ I_z = m_1(r_1)^2 + m_2(r_2)^2 \] \[ I_z = (4 kg)(2 m)^2 + (6 kg)(4 m)^2 \] \[ I_z = 4(4) + 6(16) \] \[ I_z = 16 + 96 \] \[ I_z = 112 kg\cdotm^2 \]

Step 4: Final Answer:

The moment of inertia about the z-axis is 112 kg m\(^2\).
Quick Tip: For points in the xy-plane, the perpendicular distance to the z-axis is simply the 2D radial distance from the origin \(r = \sqrt{x^2+y^2}\). The perpendicular distance to the x-axis would be \(|y|\), and to the y-axis would be \(|x|\).

Step 1: Understanding the Concept:

When an electron is accelerated from rest through a potential difference \(V\), it acquires kinetic energy \(K = eV\). This kinetic energy is related to its momentum, which in turn determines its de Broglie wavelength.


Step 2: Key Formula or Approach:

The general formula for the de Broglie wavelength of an accelerated particle is \(\lambda = \frac{h}{\sqrt{2mK}}\).
For an electron specifically, substituting the mass \(m\), charge \(e\), and Planck's constant \(h\) yields a very convenient simplified formula: \[ \lambda (in Å) \approx \frac{12.27}{\sqrt{V (in Volts)}} \]
Recall that \(1 nm = 10 Å\).


Step 3: Detailed Explanation:

Given the de Broglie wavelength \(\lambda = 0.122 nm\).
Convert this to Angstroms (Å): \[ \lambda = 0.122 \times 10 Å = 1.22 Å \]
Use the simplified formula for an electron: \[ 1.22 = \frac{12.27}{\sqrt{V}} \]
Rearrange to solve for \(\sqrt{V}\): \[ \sqrt{V} = \frac{12.27}{1.22} \]
Perform the division: \[ \sqrt{V} \approx 10.057 \]
Square both sides to find the accelerating potential \(V\): \[ V \approx (10.057)^2 \approx 101.1 V \]
In standard textbook problems, a wavelength of exactly \(\approx 0.1227 nm\) is used to yield precisely \(100 V\). Given the options, \(100 V\) is clearly the intended standard answer for this approximation.


Step 4: Final Answer:

The accelerating potential is approximately 100 V.
Quick Tip: Memorize the electron de Broglie shortcut formula: \(\lambda = \frac{1.227}{\sqrt{V}}\) nm or \(\frac{12.27}{\sqrt{V}}\) Å. It drastically reduces calculation time and avoids rounding errors from working with tiny constants like \(h\) and \(m_e\).

Step 1: Understanding the Concept:

An equipotential surface is a locus of points where the electric potential is constant. For two opposite charges, the primary "special" equipotential surface is the \(V=0\) surface. A true planar equipotential perpendicular to the axis only exists if the charges are equal and opposite (a dipole). For unequal charges, the \(V=0\) surface is a sphere that intersects the axis joining them. The question typically asks for the null potential point located directly between the charges.


Step 2: Key Formula or Approach:

The electric potential \(V\) at a point due to a point charge \(q\) is \(V = \frac{kq}{r}\).
The total potential at a point between the charges is the sum of potentials from both charges. We set this sum to zero to find the intersection point: \[ V_{total} = V_A + V_B = 0 \implies \frac{k q_A}{x} + \frac{k q_B}{d - x} = 0 \]
where \(x\) is the distance from charge A, and \(d\) is the total separation distance.


Step 3: Detailed Explanation:

Let the distance between charge A and B be \(d\).
Let the point of zero potential between them be at a distance \(x\) from charge A (\(-1 \mu C\)).
The distance from this point to charge B (\(+5 \mu C\)) will be \((d - x)\).
Set the total potential at this point to zero: \[ \frac{k(-1 \mu C)}{x} + \frac{k(+5 \mu C)}{d - x} = 0 \]
Move the negative term to the other side: \[ \frac{k(5 \mu C)}{d - x} = \frac{k(1 \mu C)}{x} \]
Cancel out the common terms \(k\) and \(\mu C\): \[ \frac{5}{d - x} = \frac{1}{x} \]
Cross-multiply to solve for \(x\): \[ 5x = 1(d - x) \] \[ 5x = d - x \]
Add \(x\) to both sides: \[ 6x = d \] \[ x = \frac{d}{6} \]
Thus, the zero potential point between the charges is at a distance of \(d/6\) from charge A.
Step 4: Final Answer:

The distance from charge A is d/6.
Quick Tip: To quickly find the internal null potential point, use the ratio of distances \(x_1/x_2 = |q_1|/|q_2|\). Here \(x_A/x_B = 1/5\), meaning the total distance \(d\) is split into \(1+5=6\) parts. The distance from A is 1 part, or \(d/6\).

Step 1: Understanding the Concept:

The "rate of flow of electrons" is the definition of electric current (\(I\)).
When conductors are connected in parallel, the potential difference (\(V\)) across each of them is the same.
We need to relate current, voltage, and resistance using Ohm's law to find the ratio of resistances.


Step 2: Key Formula or Approach:

Ohm's Law: \(V = IR\), which can be rearranged as \(R = \frac{V}{I}\).
Since \(V\) is constant for all branches in a parallel circuit, resistance \(R\) is inversely proportional to current \(I\): \[ R \propto \frac{1}{I} \]
Therefore, the ratio of resistances is: \[ R_1 : R_2 : R_3 = \frac{1}{I_1} : \frac{1}{I_2} : \frac{1}{I_3} \]


Step 3: Detailed Explanation:

Given the ratio of currents: \[ I_1 : I_2 : I_3 = 3 : 2 : 1 \]
Using the inverse relationship for parallel circuits, the ratio of their resistances is: \[ R_1 : R_2 : R_3 = \frac{1}{3} : \frac{1}{2} : \frac{1}{1} \]
To convert this ratio of fractions into a ratio of whole numbers, find the Least Common Multiple (LCM) of the denominators (3, 2, and 1).
The LCM of 3, 2, and 1 is 6.
Multiply each fraction in the ratio by 6: \[ R_1 : R_2 : R_3 = \left(\frac{1}{3} \times 6\right) : \left(\frac{1}{2} \times 6\right) : \left(\frac{1}{1} \times 6\right) \] \[ R_1 : R_2 : R_3 = 2 : 3 : 6 \]

Step 4: Final Answer:

The resistance ratio is 2:3:6.
Quick Tip: In parallel circuits, \(V\) is constant so \(I \propto 1/R\). In series circuits, \(I\) is constant so \(V \propto R\). To find the inverse ratio \(A:B:C = \frac{1}{a}:\frac{1}{b}:\frac{1}{c}\), don't just reverse the numbers; actually calculate the reciprocals and multiply by the LCM.

Step 1: Understanding the Concept:

The photoelectric effect describes the emission of electrons when light hits a material. According to Einstein's photoelectric equation, the energy of the incident photon is used to overcome the material's work function (the minimum binding energy of the electron), with the remaining energy becoming the maximum kinetic energy of the ejected electron.


Step 2: Key Formula or Approach:

Einstein's photoelectric equation is: \[ E_{photon} = \Phi + K_{max} \]
where: \(E_{photon}\) is the energy of the incident photon. \(\Phi\) is the work function of the metal. \(K_{max}\) is the maximum kinetic energy of the emitted electron.
Rearranging the formula to solve for the work function: \[ \Phi = E_{photon} - K_{max} \]


Step 3: Detailed Explanation:

Given values:
Energy of incident photon, \(E_{photon} = 5.2 eV\)

1) For the first metal, Nickel (Ni):
Kinetic energy of emitted electrons, \(K_{Ni} = 1.2 eV\)
Calculate its work function \(\Phi_{Ni}\): \[ \Phi_{Ni} = E_{photon} - K_{Ni} \] \[ \Phi_{Ni} = 5.2 eV - 1.2 eV = 4.0 eV \]

2) For the second metal, Molybdenum (Mo):
Kinetic energy of emitted electrons, \(K_{Mo} = 0.5 eV\)
Calculate its work function \(\Phi_{Mo}\): \[ \Phi_{Mo} = E_{photon} - K_{Mo} \] \[ \Phi_{Mo} = 5.2 eV - 0.5 eV = 4.7 eV \]

Therefore, the work functions are 4.0 eV for Ni and 4.7 eV for Mo.


Step 4: Final Answer:

The work functions are 4.0 eV and 4.7 eV respectively.
Quick Tip: Einstein's photoelectric equation \(E = \Phi + K_{max}\) represents basic energy conservation. The work function \(\Phi\) is a property of the specific metal, while \(K_{max}\) depends on the incident light energy.

Step 1: Understanding the Concept:

The question asks for the commercial, industrial-scale method for the preparation of benzaldehyde. While all listed options can produce benzaldehyde, commercial methods prioritize cost-effectiveness and scalability.


Step 2: Key Formula or Approach:

Identify the standard industrial route for synthesizing benzaldehyde from its most common precursor, toluene.


Step 3: Detailed Explanation:

(A) Hydrogenation with Pd in BaSO\(_4\) is the Rosenmund reduction, used to convert acid chlorides to aldehydes. It is a laboratory method, not typically the primary commercial route for benzaldehyde due to reagent costs.

(B) Side chain chlorination of toluene produces benzal chloride (\(C_6H_5CHCl_2\)). Subsequent hydrolysis of benzal chloride yields benzaldehyde. This is the primary commercial method because chlorine and toluene are relatively cheap bulk chemicals.

(C) Chromyl chloride in acetic acid (or \(CS_2\)) describes the Etard reaction. This is an excellent laboratory method for partial oxidation of toluene to benzaldehyde but is less suitable for massive commercial scale due to the use of chromium reagents.


Step 4: Final Answer:

The commercial preparation is via side chain chlorination followed by hydrolysis.
Quick Tip: Always distinguish between "laboratory methods" (like Etard or Rosenmund) and "commercial methods" (which usually involve cheaper reagents like \(Cl_2\) and \(H_2O\)) in organic synthesis questions.

Step 1: Understanding the Concept:

The question describes a halogen exchange reaction where an alkyl bromide is converted to an alkyl fluoride using a heavy metal fluoride reagent.


Step 2: Key Formula or Approach:

Recall named reactions involving the synthesis of alkyl fluorides, specifically using reagents like \(AgF\), \(Hg_2F_2\), \(CoF_3\), or \(SbF_3\).


Step 3: Detailed Explanation:

Direct fluorination of alkanes is highly exothermic and explosive, making it impractical. Therefore, alkyl fluorides are typically synthesized indirectly via halogen exchange.
Heating an alkyl chloride or alkyl bromide in the presence of a metallic fluoride such as \(AgF\), \(Hg_2F_2\), \(CoF_3\), or \(SbF_3\) results in the replacement of the chlorine or bromine atom with a fluorine atom.
This specific halogen exchange reaction is known as the Swarts reaction.
Reaction: \(CH_3-Br + AgF \rightarrow CH_3-F + AgBr\downarrow\)


Step 4: Final Answer:

The name of the reaction is the Swarts reaction.
Quick Tip: Associate Swarts reaction exclusively with the synthesis of Alkyl Fluorides using heavy metal fluorides. Contrast this with the Finkelstein reaction, which synthesizes Alkyl Iodides using NaI in dry acetone.

Step 1: Understanding the Concept:

Enthalpy of fusion (\(\Delta H_{fus}\)) is the energy required to change a substance from solid to liquid at its melting point. It depends on the strength of intermolecular forces and how efficiently the molecules pack into a solid crystal lattice.


Step 2: Key Formula or Approach:

Evaluate the intermolecular forces (London dispersion, dipole-dipole) and structural symmetry of each molecule to estimate relative crystal lattice strengths.


Step 3: Detailed Explanation:

1) CCl\(_4\) (Carbon tetrachloride): It is a perfectly symmetrical, non-polar tetrahedral molecule. It only has weak London dispersion forces. Due to its high symmetry, it forms "plastic crystals" with a relatively low degree of order just below its melting point, resulting in a very low enthalpy of fusion (\(\approx 2.5\) kJ/mol).

2) CH\(_3\)COCH\(_3\) (Acetone): It is a polar molecule with a net dipole moment due to the carbonyl group. It experiences both London dispersion and dipole-dipole interactions. Its enthalpy of fusion is higher than that of highly symmetric non-polar small molecules (\(\approx 5.7\) kJ/mol).

3) C\(_6\)H\(_6\) (Benzene): Although non-polar, benzene is a flat, highly symmetric planar ring. Flat planar molecules can stack very efficiently in a crystal lattice (strong \(\pi-\pi\) interactions in the solid state), leading to a remarkably stable solid structure compared to its liquid state. This efficient packing requires significantly more energy to disrupt, giving it the highest enthalpy of fusion among the three (\(\approx 9.9\) kJ/mol).

Therefore, the increasing order is CCl\(_4\) \(<\) CH\(_3\)COCH\(_3\) \(<\) C\(_6\)H\(_6\).


Step 4: Final Answer:

The increasing order of enthalpy of fusion is CCl\(_4\) \(<\) CH\(_3\)COCH\(_3\) \(<\) C\(_6\)H\(_6\).
Quick Tip: While stronger intermolecular forces generally mean higher phase change enthalpies, crystal packing efficiency plays a massive role in melting points and enthalpy of fusion. Planar molecules like benzene pack exceptionally well.

Step 1: Understanding the Concept:

The reaction involves the conversion of a diazonium salt into an aryl halide.


Step 2: Key Formula or Approach:

Differentiate between the reagents used in similar named reactions for this specific transformation. The use of copper powder versus a copper(I) salt is the key differentiator.


Step 3: Detailed Explanation:

Benzene diazonium chloride (\(C_6H_5N_2^+Cl^-\)) can be converted to chlorobenzene using two closely related named reactions:
1) Sandmeyer Reaction: Uses cuprous chloride (\(Cu_2Cl_2\) or \(CuCl\)) dissolved in \(HCl\).
2) Gattermann Reaction: Uses finely divided copper powder (\(Cu\)) in the presence of \(HCl\).
The question explicitly states the use of "HCl and Cu powder". Therefore, this modification of the Sandmeyer reaction is known as the Gattermann reaction.


Step 4: Final Answer:

The name of the reaction is the Gattermann reaction.
Quick Tip: Cu Salt = Sandmeyer. Cu Powder = Gattermann. The Gattermann reaction generally gives a slightly lower yield than the Sandmeyer reaction but is often more convenient to perform.

Step 1: Understanding the Concept:

Ionization energy (IE) is the energy required to remove the most loosely bound electron from an isolated gaseous atom. It generally increases across a period from left to right due to increasing effective nuclear charge.


Step 2: Key Formula or Approach:

Write down the electronic configurations of the given elements and look for exceptions to the general trend caused by fully filled or half-filled orbital stability.


Step 3: Detailed Explanation:

Let's examine the electronic configurations:
Be (Z=4): \(1s^2 2s^2\) (Fully filled \(2s\) subshell, relatively stable)
B (Z=5): \(1s^2 2s^2 2p^1\) (Electron to be removed is in a higher energy \(2p\) orbital)
N (Z=7): \(1s^2 2s^2 2p^3\) (Exactly half-filled \(2p\) subshell, extra stability)
O (Z=8): \(1s^2 2s^2 2p^4\) (Electron pairing in \(2p\) causes repulsion, making it easier to remove one electron to achieve the stable \(2p^3\) configuration)

General expected trend: Be \(<\) B \(<\) N \(<\) O
Actual trend considering stability anomalies:
1) Be vs B: The \(2p\) electron of B is shielded by the \(2s\) electrons and is slightly higher in energy, making it easier to remove than a \(2s\) electron from the full subshell of Be. Thus, IE(B) \(<\) IE(Be).
2) N vs O: Removing an electron from oxygen relieves inter-electronic repulsion in the paired \(2p\) orbital to achieve a stable half-filled state, whereas removing an electron from nitrogen disrupts a stable half-filled \(2p^3\) state. Thus, IE(O) \(<\) IE(N).
Combining these, the order is: B \(<\) Be \(<\) O \(<\) N.
The element with the lowest ionization energy among these is Boron (B).


Step 4: Final Answer:

Boron (B) shows the lowest ionization energy.
Quick Tip: Always check for Group 2 vs Group 13 (\(s^2\) vs \(s^2p^1\)) and Group 15 vs Group 16 (\(p^3\) vs \(p^4\)) anomalies when ordering ionization energies across a period.

Step 1: Understanding the Concept:

Dipole moment is a vector sum of individual bond dipoles and lone pair dipoles within a molecule. It depends on the molecular geometry and the electronegativity differences between the atoms.


Step 2: Key Formula or Approach:

Determine the geometry using VSEPR theory and analyze the direction of bond dipoles relative to the lone pair dipole.


Step 3: Detailed Explanation:

1) H\(_2\)O: Bent geometry with two lone pairs on oxygen. The O-H bond dipoles and the lone pair dipoles all point in roughly the same direction, reinforcing each other, leading to a very high net dipole moment (\(\mu \approx 1.85\) D).

2) NH\(_3\): Trigonal pyramidal geometry with one lone pair. The N-H bond dipoles (pointing from H to N) reinforce the lone pair dipole, resulting in a high net dipole moment (\(\mu \approx 1.47\) D).

3) NF\(_3\): Trigonal pyramidal geometry with one lone pair. The N-F bond dipoles (pointing from N to F, as F is more electronegative) act in the opposite direction to the lone pair dipole. This cancellation leads to a very small net dipole moment (\(\mu \approx 0.24\) D).

4) CHCl\(_3\): Tetrahedral geometry, but asymmetric. The three C-Cl bonds have strong dipoles pulling away from the carbon, while the C-H bond has a small dipole pushing towards it. The net result is an intermediate dipole moment (\(\mu \approx 1.04\) D).
Arranging them in increasing order: NF\(_3\) (\(\sim\)0.24 D) \(<\) CHCl\(_3\) (\(\sim\)1.04 D) \(<\) NH\(_3\) (\(\sim\)1.47 D) \(<\) H\(_2\)O (\(\sim\)1.85 D).


Step 4: Final Answer:

The increasing order of dipole moment is NF\(_3\) \(<\) CHCl\(_3\) \(<\) NH\(_3\) \(<\) H\(_2\)O.
Quick Tip: The comparison between NH\(_3\) and NF\(_3\) is a classic exam question. Always remember that opposing vectors (lone pair vs bond dipoles) in NF\(_3\) drastically reduce its overall dipole moment compared to NH\(_3\).

Step 1: Understanding the Concept:

A mixed oxide is an oxide that contains atoms of a single metallic element in two different oxidation states. It can be formally considered as a compound of two simpler oxides.


Step 2: Key Formula or Approach:

Analyze the average oxidation state of the metal in each formula. If it is a non-integer or represents a combination of stable oxidation states, try to decompose the formula into two simpler oxides.


Step 3: Detailed Explanation:

Let's determine the oxidation state of Mn in each option:
(A) MnO\(_2\): Oxidation state of Mn is +4. It is a simple dioxide.
(B) Mn\(_2\)O\(_3\): Oxidation state of Mn is +3. It is a simple sesquioxide.
(C) Mn\(_3\)O\(_4\): The average oxidation state of Mn here is \(+8/3\), which is fractional. This is a strong indicator of a mixed oxide. Mn\(_3\)O\(_4\) can be written as MnO \(\cdot\) Mn\(_2\)O\(_3\), showing it contains manganese in both +2 and +3 oxidation states. It is structurally similar to magnetite (\(Fe_3O_4\)).
(D) Mn\(_2\)O: Oxidation state of Mn is +1. (Though rare, it would be a simple oxide).
Therefore, Mn\(_3\)O\(_4\) is the mixed oxide.


Step 4: Final Answer:

The mixed oxide is Mn\(_3\)O\(_4\).
Quick Tip: Common examples of mixed oxides to memorize for exams include Fe\(_3\)O\(_4\) (FeO\(\cdot\)Fe\(_2\)O\(_3\)), Pb\(_3\)O\(_4\) (2PbO\(\cdot\)PbO\(_2\)), and Mn\(_3\)O\(_4\) (MnO\(\cdot\)Mn\(_2\)O\(_3\)).

Step 1: Understanding the Concept:

The question requires matching common everyday substances with their approximate pH values.


Step 2: Key Formula or Approach:

Recall the general acidic, basic, or neutral nature of these common substances as taught in standard chemistry curricula.


Step 3: Detailed Explanation:

(1) Milk of Magnesia: This is an aqueous suspension of magnesium hydroxide, \(Mg(OH)_2\). It is used as an antacid to neutralize stomach acid, meaning it is alkaline (basic). Its pH is typically around 10 to 10.5.

(2) Milk: Normal cow's milk is very slightly acidic due to the presence of lactic acid. Its pH is usually around 6.5 to 6.8.

(3) Egg white: Also known as albumen, it is naturally slightly alkaline to protect the yolk from bacteria. Its pH ranges from about 7.6 to 7.9.

(4) Black coffee: Coffee contains various weak organic acids (like chlorogenic acid), making it mildly acidic. Its pH is typically around 5.
Matching these characteristics to the provided numbers:
Milk of Magnesia \(\rightarrow\) 10
Milk \(\rightarrow\) 6.8
Egg white \(\rightarrow\) 7.8
Black coffee \(\rightarrow\) 5


Step 4: Final Answer:

The correct matching is: Milk of Magnesia (10), Milk (6.8), Egg white (7.8), Black coffee (5).
Quick Tip: Familiarity with the pH of common substances (like blood \(\sim\)7.4, gastric juice \(\sim\)1.2, pure water = 7) provides great reference points for estimating the pH of other items in matching questions.

Step 1: Understanding the Concept:

This question requires arranging molecules based on their net dipole moment, which is the vector sum of all individual bond dipoles and lone pair moments based on their 3D geometry.


Step 2: Key Formula or Approach:

Identify the molecular geometry using VSEPR theory to see if bond dipoles cancel out perfectly (non-polar) or result in a net vector (polar).


Step 3: Detailed Explanation:

(1) BF\(_3\): Boron trifluoride has a trigonal planar geometry. It is perfectly symmetric. The three B-F bond dipoles are at 120-degree angles to each other in a plane, so their vector sum is exactly zero. \(\mu = 0\) D.

(2) CHCl\(_3\): Chloroform is tetrahedral but asymmetric. The dense electron cloud is pulled towards the three chlorine atoms. It has a moderate net dipole moment. \(\mu \approx 1.04\) D.

(3) NH\(_3\): Ammonia has a trigonal pyramidal shape. The three N-H bond dipoles add constructively with the lone pair dipole on nitrogen, giving a substantial net dipole. \(\mu \approx 1.47\) D.

(4) H\(_2\)O: Water is a bent molecule. The two O-H bond dipoles strongly reinforce the dipoles from the two lone pairs on oxygen, resulting in the highest dipole moment among this set. \(\mu \approx 1.85\) D.

Increasing order: BF\(_3\) (0) \(<\) CHCl\(_3\) (\(\sim\)1.04) \(<\) NH\(_3\) (\(\sim\)1.47) \(<\) H\(_2\)O (\(\sim\)1.85).


Step 4: Final Answer:

The order is BF\(_3\) \(<\) CHCl\(_3\) \(<\) NH\(_3\) \(<\) H\(_2\)O.
Quick Tip: Always identify perfectly symmetrical molecules first (like \(BF_3, CCl_4, CO_2\)). Their dipole moment is exactly zero, making them the lowest in any such ranking.

Step 1: Understanding the Concept:

According to Bohr's atomic model, the angular momentum of an electron in a stable circular orbit is quantized. It can only take values that are integral multiples of a specific constant.


Step 2: Key Formula or Approach:

The quantization of angular momentum \(L\) for the \(n\)-th orbit is given by the formula: \[ L_n = \frac{nh}{2\pi} \]
where \(n\) is the principal quantum number (orbit number) and \(h\) is Planck's constant.


Step 3: Detailed Explanation:

Calculate the angular momentum for the 5th orbit (\(n = 5\)): \[ L_5 = \frac{5h}{2\pi} \]
Calculate the angular momentum for the 3rd orbit (\(n = 3\)): \[ L_3 = \frac{3h}{2\pi} \]
Find the difference between them: \[ \Delta L = L_5 - L_3 \] \[ \Delta L = \frac{5h}{2\pi} - \frac{3h}{2\pi} \] \[ \Delta L = \frac{(5 - 3)h}{2\pi} \] \[ \Delta L = \frac{2h}{2\pi} \]
Simplify the fraction by canceling the factor of 2: \[ \Delta L = \frac{h}{\pi} \]

Step 4: Final Answer:

The difference in angular momentum is \(\frac{h}{\pi}\).
Quick Tip: For any transition or difference between orbits \(n_1\) and \(n_2\), the change in angular momentum is always simply \(\Delta n \times \frac{h}{2\pi}\). Here, \(\Delta n = 2\), so \(\Delta L = 2 \times \frac{h}{2\pi} = \frac{h}{\pi}\).

Step 1: Assume Total Mass

Assume total mass of compound = \(100\,g\)

Mass of M = \(54\,g\), \quad Mass of O = \(46\,g\)



Step 2: Convert Mass to Moles
\[ Moles of M = \frac{54}{27} = 2 \] \[ Moles of O = \frac{46}{16} = 2.875 \]



Step 3: Find Simplest Ratio

Divide by smallest value (2): \[ Ratio of M = \frac{2}{2} = 1 \] \[ Ratio of O = \frac{2.875}{2} = 1.4375 \]



Step 4: Convert to Whole Number Ratio

Multiply both by 2: \[ M : O = 2 : 2.875 \approx 2 : 3 \]


Step 5: Write Empirical Formula
\[ Empirical formula = \mathrm{M}_2\mathrm{O}_3 \] Quick Tip: If calculated mole ratios don't perfectly yield simple fractions (like .5, .33, .25), look at the atomic mass provided to identify the likely element. Mass 27 is Aluminum, which predominantly forms a +3 ion, strongly suggesting an \(M_2O_3\) formula.

Step 1: Understanding the Concept:

Bond enthalpy is the energy required to break one mole of a specific type of bond in a gaseous molecule. It generally correlates strongly with the bond order (number of bonds between atoms); triple bonds are stronger than double bonds, which are stronger than single bonds between the same or similar atoms.


Step 2: Key Formula or Approach:

Evaluate the bond order of each option. The higher the bond order, the greater the electron density between the nuclei, leading to stronger electrostatic attraction and higher bond enthalpy.


Step 3: Detailed Explanation:

Let's look at the given bonds and their approximate standard bond enthalpies:
(A) C=C (double bond): \(\approx 614\) kJ/mol
(B) O=O (double bond): \(\approx 498\) kJ/mol
(C) N\(\equiv\)N (triple bond): \(\approx 945\) kJ/mol
(D) C=O (double bond, e.g., in ketones/aldehydes): \(\approx 745\) kJ/mol (Note: C\(\equiv\)O triple bond in carbon monoxide is higher, \(\approx 1072\) kJ/mol, but the option explicitly shows a double bond).
The N\(\equiv\)N bond is a triple bond consisting of one sigma and two pi bonds. The accumulation of electron density between the very small nitrogen atoms results in a highly stable, extremely strong bond, giving \(N_2\) gas its inert character.


Step 4: Final Answer:

N\(\equiv\)N has the maximum bond enthalpy.
Quick Tip: For comparing bond strengths between non-metals of similar size (period 2), bond order is the primary determinant. Triple bonds (\(N\equiv N\), \(C\equiv C\), \(C\equiv O\)) will consistently have the highest bond enthalpies.

Step 1: Understanding the Concept:

Basic strength of amines depends on the availability of the lone pair of electrons on the nitrogen atom for protonation. Electron-donating groups (+I effect) increase basicity, while electron-withdrawing groups or resonance delocalization (-R effect) decrease it.


Step 2: Key Formula or Approach:

Classify the amines as aliphatic or aromatic. For aromatic amines, assess the degree of alkyl substitution on the nitrogen atom and consider standard +I and steric effects in aqueous or general solvent conditions.


Step 3: Detailed Explanation:

1) Identify aliphatic vs. aromatic:
(D) Benzylamine (\(C_6H_5CH_2NH_2\)) is an aliphatic amine. The nitrogen is separated from the aromatic ring by an \(sp^3\) carbon. Therefore, its lone pair is not delocalized into the benzene ring. It is the strongest base among the choices.

(A), (B), (C) are all aromatic amines (derivatives of aniline) where the lone pair on nitrogen is in conjugation with the \(\pi\) system of the benzene ring, significantly reducing basicity compared to aliphatic amines.

2) Rank the aromatic amines:
Adding alkyl groups (like methyl) to the nitrogen atom increases electron density on the nitrogen due to the positive inductive (+I) effect, making the lone pair more available.

(C) Aniline (\(C_6H_5NH_2\)) has no alkyl groups on N. (Weakest base)
(B) N-methylaniline (\(C_6H_5NHCH_3\)) has one methyl group (+I effect), making it more basic than aniline.

(A) N,N-dimethylaniline (\(C_6H_5N(CH_3)_2\)) has two methyl groups. The combined +I effect makes it more basic than N-methylaniline.

Therefore, the decreasing order is: Aliphatic \(>\) \(3^\circ\) Ar-Amine \(>\) \(2^\circ\) Ar-Amine \(>\) \(1^\circ\) Ar-Amine.


Step 4: Final Answer:

The decreasing order of basic strength is (D) \(>\) (A) \(>\) (B) \(>\) (C).
Quick Tip: Always separate aliphatic amines (like benzylamine) from true aromatic amines (like aniline) first. Aliphatic amines are almost always significantly stronger bases because their lone pair isn't lost to resonance.

Step 1: Understanding the Concept:

The Lucas test is used to distinguish between primary, secondary, and tertiary alcohols based on their reactivity with Lucas reagent (a mixture of concentrated HCl and anhydrous ZnCl\(_2\)).


Step 2: Key Formula or Approach:

Identify the class (\(1^\circ, 2^\circ\), or \(3^\circ\)) of the given alcohol. The reaction proceeds via an S\(_N\)1 mechanism forming a carbocation. Tertiary alcohols react immediately to form a cloudy suspension of the alkyl chloride.


Step 3: Detailed Explanation:

Structure of 2-methylbutan-2-ol: \(CH_3 - CH_2 - C(CH_3)(OH) - CH_3\)
The -OH group is attached to a carbon atom that is bonded to three other carbons. Therefore, it is a tertiary (\(3^\circ\)) alcohol.
Tertiary alcohols react immediately with Lucas reagent because they form highly stable tertiary carbocations as intermediates. The chloride ion then rapidly attacks this carbocation.
Reaction: \(CH_3 - CH_2 - C(CH_3)(OH) - CH_3 + HCl/ZnCl_2 \rightarrow CH_3 - CH_2 - C(CH_3)(Cl) - CH_3 + H_2O\)
The resulting product is an alkyl chloride where the Cl atom simply replaces the OH group at the same tertiary position.
The IUPAC name of the product is 2-chloro-2-methylbutane.


Step 4: Final Answer:

The product is 2-chloro-2-methylbutane.
Quick Tip: For Lucas test questions, tertiary alcohols yield the direct substitution product instantly without rearrangement (as the \(3^\circ\) carbocation is already maximally stable).

Step 1: Understanding the Concept:

This reaction concerns the oxidation of alkyl side chains attached to a benzene ring.


Step 2: Key Formula or Approach:

Recall the rule for side-chain oxidation of aromatic compounds using strong oxidizing agents like acidic or alkaline KMnO\(_4\) or acidified K\(_2\)Cr\(_2\)O\(_7\) with heat.


Step 3: Detailed Explanation:

When an alkylbenzene is subjected to vigorous oxidation conditions, the entire alkyl chain, regardless of its length, is oxidized down to a single carboxyl group (-COOH) attached to the ring, provided that there is at least one benzylic hydrogen atom (a hydrogen attached to the carbon directly bonded to the benzene ring).
n-pentylbenzene has the structure \(C_6H_5 - CH_2 - CH_2 - CH_2 - CH_2 - CH_3\).
The carbon adjacent to the ring (benzylic carbon) has two hydrogen atoms. Therefore, it is susceptible to oxidation.
Under vigorous oxidation, the C-C bonds of the side chain cleave, and the benzylic carbon is converted to a carboxylic acid group.
Reaction: \(C_6H_5 - CH_2 - (CH_2)_3 - CH_3 \xrightarrow{[O]} C_6H_5 - COOH\)
The remaining carbons of the side chain are oxidized to CO\(_2\) and H\(_2\)O. The main aromatic product is benzoic acid.


Step 4: Final Answer:

The product is benzoic acid.
Quick Tip: Any alkyl chain (methyl, ethyl, propyl, etc.) on a benzene ring is oxidized entirely to a -COOH group by KMnO\(_4\)/H\(^+\) as long as it has at least one benzylic hydrogen. Only tert-butylbenzene resists this oxidation.

Step 1: Understanding the Concept:

Acid-catalyzed dehydration of alcohols to form alkenes proceeds via an E1 elimination mechanism. The rate-determining step is the formation of a carbocation intermediate.


Step 2: Key Formula or Approach:

The relative ease of dehydration of alcohols follows the stability order of the carbocations they form: tertiary (\(3^\circ\)) \(>\) secondary (\(2^\circ\)) \(>\) primary (\(1^\circ\)). Classify each given alcohol to determine the order.


Step 3: Detailed Explanation:

Let's analyze the structure of each given alcohol:
1) 2-methylpentan-2-ol: \(CH_3 - CH_2 - CH_2 - C(CH_3)(OH) - CH_3\). The -OH is attached to a carbon bonded to three other carbons. It is a tertiary (\(3^\circ\)) alcohol. It will form a relatively stable \(3^\circ\) carbocation, making dehydration fastest.
2) butan-2-ol: \(CH_3 - CH_2 - CH(OH) - CH_3\). The -OH is attached to a carbon bonded to two other carbons. It is a secondary (\(2^\circ\)) alcohol. It forms a moderately stable \(2^\circ\) carbocation.
3) butan-1-ol: \(CH_3 - CH_2 - CH_2 - CH_2 - OH\). The -OH is attached to a carbon bonded to only one other carbon. It is a primary (\(1^\circ\)) alcohol. It forms a highly unstable \(1^\circ\) carbocation, making dehydration the slowest and requiring the most drastic conditions.
Based on carbocation stability, the order of ease of dehydration is \(3^\circ > 2^\circ > 1^\circ\).
Therefore, the order is: 2-methylpentan-2-ol \(>\) butan-2-ol \(>\) butan-1-ol.


Step 4: Final Answer:

The order is 2-methylpentan-2-ol \(>\) butan-2-ol \(>\) butan-1-ol.
Quick Tip: For dehydration of alcohols (E1 mechanism) and reaction with hydrogen halides (S\(_N\)1 mechanism like Lucas test), the reactivity order is always the same: \(3^\circ > 2^\circ > 1^\circ\), driven strictly by carbocation stability.

Step 1: Understanding the Concept:

The wavelength of light absorbed by a coordination complex is inversely proportional to the crystal field splitting energy (\(\Delta_o\)) created by the ligands. Higher splitting energy means the complex absorbs higher energy light, which corresponds to a shorter wavelength (\(\lambda\)).


Step 2: Key Formula or Approach:

Use the relation \(E = \frac{hc}{\lambda} = \Delta_o\). Thus, \(\lambda \propto \frac{1}{\Delta_o}\).
Utilize the spectrochemical series to order the ligands by field strength to determine relative \(\Delta_o\) values.


Step 3: Detailed Explanation:

1) Identify the ligands and their position in the spectrochemical series

Spectrochemical series order: \(Cl^- < H_2O < NH_3 < CN^-\)
2) Determine the crystal field splitting (\(\Delta_o\)) for the given complexes. Since the metal ion \(Co^{3+}\) is the same in all, \(\Delta_o\) depends entirely on the ligands.

Order of ligand field strength: \(Cl^-\) (weakest) \(<\) mixture of \(H_2O/NH_3\) \(<\) \(NH_3\) (pure) \(<\) \(CN^-\) (strongest).

Therefore, order of \(\Delta_o\): \([CoCl_6]^{3-} < [Co(NH_3)_4(H_2O)_2]^{3+} < [Co(NH_3)_6]^{3+} < [Co(CN)_6]^{3-}\)

3) Relate \(\Delta_o\) to absorption wavelength (\(\lambda\)). Since \(\lambda\) is inversely proportional to \(\Delta_o\), the complex with the highest splitting energy will absorb the shortest wavelength.

Order of absorption wavelength (\(\lambda_{abs}\)):
\([Co(CN)_6]^{3-} < [Co(NH_3)_6]^{3+} < [Co(NH_3)_4(H_2O)_2]^{3+} < [CoCl_6]^{3-}\)


Step 4: Final Answer:

The increasing order of absorption wavelength is [Co(CN)\(_6\)]\(^{3-}\) \(<\) [Co(NH\(_3\))\(_6\)]\(^{3+}\) \(<\) [Co(NH\(_3\))\(_4\)(H\(_2\)O)\(_2\)]\(^{3+}\) \(<\) [CoCl\(_6\)]\(^{3-}\).
Quick Tip: Always remember: Strong Field Ligand \(\rightarrow\) Large Splitting (\(\Delta_o\)) \(\rightarrow\) High Energy Absorbed \(\rightarrow\) Short Wavelength (\(\lambda\)) Absorbed. The question specifically asks for wavelength order, which is the reverse of the spectrochemical series order.

Step 1: Concept Used

For a weak electrolyte, \[ \alpha = \frac{\Lambda_m}{\Lambda_m^\circ} \]



Step 2: Calculate Limiting Molar Conductivity

Using Kohlrausch’s Law: \[ \Lambda_m^\circ (HA) = \lambda_m^\circ(H^+) + \lambda_m^\circ(A^-) \] \[ \Lambda_m^\circ (HA) = 341 + 80 = 421\,\Omega^{-1}cm^2mol^{-1} \]


Step 3: Calculate Degree of Dissociation
\[ \alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{70}{421} \] \[ \alpha \approx 0.166 \]


Step 4: Final Answer
\[ \alpha = 0.166 \quad or \quad 16.6% \] Quick Tip: Ensure units match before calculating \(\alpha\). Molar conductivities are usually given in \(S \cdot cm^2 \cdot mol^{-1}\) or \(\Omega^{-1} cm^2 mol^{-1}\). The ratio \(\alpha\) is a dimensionless fraction.

Step 1: Understanding the Concept:

This is a two-step reaction sequence. Step 1 is the addition of a hydrogen halide to an alkene in the presence of peroxides. Step 2 is a nucleophilic substitution reaction of the resulting alkyl halide.


Step 2: Key Formula or Approach:

1) Apply the Kharasch effect (anti-Markovnikov addition) for HBr + peroxide on unsymmetrical alkenes.
2) Apply nucleophilic substitution rules for alkyl halides reacting with silver salts.


Step 3: Detailed Explanation:

Formation of X:
The starting material C\(_3\)H\(_6\) is propene (\(CH_3-CH=CH_2\)).
Reaction with HBr in the presence of peroxide follows a free radical mechanism leading to anti-Markovnikov addition. The bromine atom adds to the less substituted carbon atom. \(CH_3-CH=CH_2 + HBr \xrightarrow{peroxide} CH_3-CH_2-CH_2-Br\)
Therefore, product X is 1-bromopropane (n-propyl bromide).

Formation of Y:
Reaction of an alkyl halide with \(AgNO_3\). Typically, in standard curriculum, reactions with \(AgNO_2\) are highlighted to form nitroalkanes (\(R-NO_2\)) while \(KNO_2\) forms alkyl nitrites (\(R-ONO\)).
Reaction with \(AgNO_3\) explicitly yields alkyl nitrates (\(R-ONO_2\)) via an S\(_N\) reaction due to the covalent nature of the Ag-O bond directing attack from the other oxygen. \(CH_3-CH_2-CH_2-Br + AgNO_3 \rightarrow CH_3-CH_2-CH_2-O-NO_2 + AgBr\downarrow\)
Therefore, product Y is n-propyl nitrate.
Step 4: Final Answer:

Product X is 1-bromopropane and Y is propyl nitrate.
Quick Tip: Pay close attention to reagents over the arrow. "HBr only" gives Markovnikov product (2-bromopropane), while "HBr + peroxide" gives anti-Markovnikov (1-bromopropane). This effect only works for HBr, not HCl or HI.

Step 1: Understanding the Concept:

The average rate of a chemical reaction is defined as the change in concentration of a reactant or product over a specific time interval. For a reactant, since its concentration decreases, a negative sign is added to make the rate a positive value.


Step 2: Key Formula or Approach:

For a generic reaction A \(\rightarrow\) Products, the average rate of reaction with respect to reactant A is: \[ Average Rate = -\frac{\Delta[A]}{\Delta t} = -\frac{[A]_{final} - [A]_{initial}}{t_{final} - t_{initial}} \]


Step 3: Detailed Explanation:

Given values from the problem:
Initial concentration of A, \([A]_{initial} = 0.46\) mol L\(^{-1}\)
Final concentration of A, \([A]_{final} = 0.36\) mol L\(^{-1}\)
Time interval, \(\Delta t = 10\) min
Substitute these values into the rate formula: \[ Rate = -\frac{(0.36 - 0.46) mol L^{-1}}{10 min} \] \[ Rate = -\frac{-0.10 mol L^{-1}}{10 min} \] \[ Rate = \frac{0.10}{10} mol L^{-1} min^{-1} \] \[ Rate = 0.01 mol L^{-1} min^{-1} \]
The average rate is \(1 \times 10^{-2}\) mol L\(^{-1}\) min\(^{-1}\).


Step 4: Final Answer:

The average rate of reaction is 0.01 mol L\(^{-1}\) min\(^{-1}\).
Quick Tip: Always ensure the rate is a positive quantity. For reactants, calculating \([Final] - [Initial]\) gives a negative value, which is why the formula includes a leading negative sign.

Step 1: Understanding the Concept:

The solubility product constant (K\(_{sp}\)) is the equilibrium constant for a solid substance dissolving in an aqueous solution. It relates to the molar solubility (\(s\)) of the compound based on its dissociation stoichiometry.


Step 2: Key Formula or Approach:

Write the balanced dissociation equation for the salt. \(AX_2(s) \rightleftharpoons A^{2+}(aq) + 2X^-(aq)\)
Express equilibrium concentrations of ions in terms of molar solubility \(s\), substitute into the \(K_{sp}\) expression, and solve for \(s\).


Step 3: Detailed Explanation:

Let the molar solubility of \(AX_2\) be \(s\) mol L\(^{-1}\).
The dissociation equation is: \(AX_2(s) \rightleftharpoons A^{2+}(aq) + 2X^-(aq)\)
From the stoichiometry, at equilibrium: \([A^{2+}] = s\) \([X^-] = 2s\)
The expression for the solubility product is: \[ K_{sp} = [A^{2+}][X^-]^2 \]
Substitute the solubility terms into the equation: \[ K_{sp} = (s)(2s)^2 \] \[ K_{sp} = (s)(4s^2) = 4s^3 \]
We are given \(K_{sp} = 3.2 \times 10^{-14}\). \[ 4s^3 = 3.2 \times 10^{-14} \]
Divide both sides by 4: \[ s^3 = \frac{3.2 \times 10^{-14}}{4} = 0.8 \times 10^{-14} \]
To make taking the cube root easier, express the number in standard scientific notation with an exponent divisible by 3: \[ s^3 = 8.0 \times 10^{-15} \]
Now, take the cube root of both sides: \[ s = \sqrt[3]{8.0 \times 10^{-15}} \] \[ s = \sqrt[3]{8} \times \sqrt[3]{10^{-15}} \] \[ s = 2 \times 10^{-5} mol L^{-1} \]

Step 4: Final Answer:

The solubility is 2 \(\times\) 10\(^{-5}\) mol L\(^{-1}\).
Quick Tip: Memorize the general \(K_{sp} - s\) relationships for common salt types: \(AB \rightarrow s^2\), \(AB_2 \rightarrow 4s^3\), \(AB_3 \rightarrow 27s^4\), \(A_2B_3 \rightarrow 108s^5\). This allows you to jump straight to the math.

Step 1: Understanding the Concept:

Valence Shell Electron Pair Repulsion (VSEPR) theory is used to predict the geometry of individual molecules based on the number of electron pairs surrounding their central atoms. We need to find the total valence electrons and distribute them as bonds and lone pairs.


Step 2: Key Formula or Approach:

1) Identify the central atom and its number of valence electrons.
2) Subtract the electrons used in single bonds to surrounding atoms.
3) Divide the remaining electrons by 2 to find the number of lone pairs.


Step 3: Detailed Explanation:

The molecule is Bromine trifluoride (\(BrF_3\)).
Central atom: Bromine (Br). It is a Group 17 element, so it has 7 valence electrons.
Surrounding atoms: 3 Fluorine (F) atoms.
Bromine forms 3 single covalent bonds with the 3 fluorine atoms.
Number of bond pairs (bp) = 3.
Electrons used in bonding = \(3 \times 1 = 3\) electrons from Bromine.
Remaining valence electrons on Bromine = Total valence electrons - bonding electrons
Remaining electrons = \(7 - 3 = 4\) electrons.
These 4 non-bonding electrons organize into lone pairs (lp).
Number of lone pairs = \(\frac{4 electrons}{2} = 2\) lone pairs.
Therefore, the central atom Br has 3 bond pairs and 2 lone pairs. This results in an \(AX_3E_2\) VSEPR designation, leading to a T-shaped molecular geometry.


Step 4: Final Answer:

There are 3 bond pairs and 2 lone pairs.
Quick Tip: A quick formula for lone pairs on the central atom is: \(LP = \frac{Valence e^- - Valency of surrounding atoms}{2}\). For BrF\(_3\), \(LP = \frac{7 - 3(1)}{2} = 2\).

Step 1: Understanding the Concept:

The standard cell potential (\(E^\circ_{cell}\)) is the potential difference between two half-cells under standard conditions. It indicates whether a redox reaction is thermodynamically spontaneous.


Step 2: Key Formula or Approach:

Identify the oxidation (anode) and reduction (cathode) half-reactions from the given overall equation.
Use the formula: \(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}\) (where both are standard reduction potentials).


Step 3: Detailed Explanation:

From the balanced chemical equation: \(2Cr + 3Cd^{2+} \rightarrow 2Cr^{3+} + 3Cd\)
Oxidation half-reaction: \(Cr \rightarrow Cr^{3+} + 3e^-\) (Loss of electrons). This occurs at the anode.
Reduction half-reaction: \(Cd^{2+} + 2e^- \rightarrow Cd\) (Gain of electrons). This occurs at the cathode.
We are given the standard reduction potential for the cathode: \(E^\circ_{cathode} (Cd^{2+}/Cd) = -0.40\) V.
The problem text in the image is cut off and omits the required standard reduction potential for Chromium. In typical competitive exam settings, standard values might be provided elsewhere or are expected to be known. The standard reduction potential for \(Cr^{3+}/Cr\) is widely known: \(E^\circ_{anode} (Cr^{3+}/Cr) = -0.74\) V.
Substitute these into the formula: \[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \] \[ E^\circ_{cell} = -0.40 V - (-0.74 V) \] \[ E^\circ_{cell} = -0.40 V + 0.74 V \] \[ E^\circ_{cell} = +0.34 V \]
The positive value indicates the reaction is spontaneous as written under standard conditions.


Step 4: Final Answer:

The standard cell potential E\(^\circ_{cell}\) is +0.34 V.
Quick Tip: When applying \(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}\), ensure both values plugged in are standard REDUCTION potentials. Do not flip the sign of the anode potential yourself; the minus sign in the formula does that for you.

Step 1: Understanding the Concept:

The Bohr model allows calculation of the radii of electron orbits in hydrogen and hydrogen-like ions (single-electron species like He\(^+\), Li\(^{2+}\)).


Step 2: Key Formula or Approach:

The radius of the \(n\)-th Bohr orbit for a hydrogen-like species is given by: \[ r_n = 0.529 \times \frac{n^2}{Z} Å \]
where \(n\) is the principal quantum number (orbit number) and \(Z\) is the atomic number.


Step 3: Detailed Explanation:

The species given is the Helium ion (\(He^+\)).
Atomic number of Helium, \(Z = 2\).
We need the radius of the "first" Bohr orbit, so \(n = 1\).
Substitute these values into the Bohr radius formula: \[ r_1 = 0.529 \times \frac{(1)^2}{2} Å \] \[ r_1 = 0.529 \times \frac{1}{2} Å \] \[ r_1 = \frac{0.529}{2} Å \] \[ r_1 = 0.2645 Å \]
Alternatively, this can be expressed in picometers: \(r_1 = 26.45\) pm.


Step 4: Final Answer:

The radius is 0.2645 Å.
Quick Tip: Remember the baseline value: the first Bohr orbit of Hydrogen (\(n=1, Z=1\)) is \(0.529\) Å. All other radii are just this base value scaled by a factor of \(n^2 / Z\).

Step 1: Understanding the Concept:

This requires knowledge of the structural and chemical properties of common carbohydrates, specifically sucrose (table sugar).


Step 2: Key Formula or Approach:

Evaluate each statement against the known facts about sucrose: its optical activity, reducing nature, classification, and monomeric composition.


Step 3: Detailed Explanation:

Let's analyze each option:
(A) False. Pure sucrose is dextrorotatory (\([\alpha]_D = +66.5^\circ\)). It is only after hydrolysis (into glucose and fructose) that the resulting mixture becomes laevorotatory (invert sugar).

(B) False. Sucrose is a non-reducing sugar. The glycosidic bond is formed between the anomeric carbon (C1) of glucose and the anomeric carbon (C2) of fructose. Since neither ring can open to expose a free aldehyde or ketone group, it cannot act as a reducing agent (e.g., fails Tollens' and Fehling's tests).

(C) True. Sucrose is indeed a disaccharide, consisting of two monosaccharide units joined together.

(D) False. Sucrose is composed of \(\alpha\)-D-(+)-glucose and \(\beta\)-D-(-)-fructose. The option incorrectly assigns the optical rotation signs to the monomers. Naturally occurring glucose is dextrorotatory (+), and natural fructose is laevorotatory (-).


Step 4: Final Answer:

The correct statement is that it is a disaccharide.
Quick Tip: Sucrose is the classic example of a non-reducing sugar. Remember the linkage is \(\alpha-1,2-\beta\), which locks up both anomeric carbons.

Step 1: Understanding the Concept:

This relates to Henry's Law and its practical applications. Breathing compressed air at high pressure underwater leads to high solubility of nitrogen in the blood. Rapid ascent causes this nitrogen to bubble out, causing the bends. Diluting the gas with helium, which has much lower solubility in blood, mitigates this.


Step 2: Key Formula or Approach:

Recall the specific standard composition of breathing gas mixtures (often referred to as Trimix or specialized Heliox variants) mentioned in standard chemistry textbook applications of Henry's Law.


Step 3: Detailed Explanation:

To avoid the toxic effects of high concentration of nitrogen in the blood (nitrogen narcosis) and the dangerous decompression sickness ("the bends"), modern scuba divers use specialized gas mixtures instead of normal compressed air.
The standard composition taught in curricula for tanks used by scuba divers is typically:
Helium (He): 11.7%
Nitrogen (N\(_2\)): 56.2%
Oxygen (O\(_2\)): 32.1%
The helium acts as an inert diluent that is far less soluble in blood lipids than nitrogen under high pressure.


Step 4: Final Answer:

The percentage of Helium is 11.7%.
Quick Tip: This is a standard "fact-based" question from the Solutions chapter in NCERT/standard textbooks. Memorizing the rough percentages (11.7% He, 56.2% N2, 32.1% O2) is highly recommended for direct questions like this.

Step 1: Solve for \( z \) \[ 3(z - i) = 2 - i \Rightarrow z - i = \frac{2 - i}{3} \]
\[ z = \frac{2 - i}{3} + i = \frac{2 - i + 3i}{3} = \frac{2 + 2i}{3} \]

Step 2: Find \( z^2 \) \[ z^2 = \left(\frac{2 + 2i}{3}\right)^2 = \frac{(2 + 2i)^2}{9} \]
\[ = \frac{4 + 8i + 4i^2}{9} = \frac{4 + 8i - 4}{9} = \frac{8i}{9} \] Quick Tip: When squaring a complex number of the form \( a+bi \), remember that the \( (bi)^2 \) term becomes negative real (\( -b^2 \)) because \( i^2 = -1 \).

Step 1: Apply formula of definite integration

We use: \[ \int_a^b x^n \, dx = \left[\frac{x^{n+1}}{n+1}\right]_a^b \]



Step 2: Solve first integral
\[ \int_a^b x^3 dx = \left[\frac{x^4}{4}\right]_a^b \] \[ = \frac{b^4}{4} - \frac{a^4}{4} = 0 \]
\[ \Rightarrow \frac{b^4 - a^4}{4} = 0 \] \[ \Rightarrow b^4 = a^4 \]
\[ \Rightarrow b = a \quad or \quad b = -a \]



Step 3: Eliminate invalid case

If \( b = a \), then: \[ \int_a^a x^2 dx = 0 \]
But given value is \( \frac{2}{3} \neq 0 \), so this case is rejected.
\[ \Rightarrow b = -a \]



Step 4: Use second integral
\[ \int_a^b x^2 dx = \frac{2}{3} \]

Substitute \( a = -b \): \[ \int_{-b}^b x^2 dx \]

Since \( x^2 \) is an even function: \[ \int_{-b}^b x^2 dx = 2 \int_0^b x^2 dx \]
\[ = 2 \left[\frac{x^3}{3}\right]_0^b \]
\[ = 2 \cdot \frac{b^3}{3} = \frac{2b^3}{3} \]



Step 5: Solve equation
\[ \frac{2b^3}{3} = \frac{2}{3} \]
\[ \Rightarrow b^3 = 1 \]
\[ \Rightarrow b = 1 \]



Step 6: Find \( a \)
\[ a = -b = -1 \]



Final Answer:
\[ \boxed{a = -1, \quad b = 1} \] Quick Tip: Recognizing even and odd functions can significantly simplify definite integrals with symmetric limits, saving you time and calculation effort.

Step 1: Concept of inverse trigonometric functions

The principal value ranges are: \[ \cos^{-1}(x) \in [0, \pi], \quad \sin^{-1}(x) \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \]



Step 2: Evaluate \( \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) \)

Using identity: \[ \cos^{-1}(-x) = \pi - \cos^{-1}(x) \]
\[ \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = \pi - \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) \]

Since: \[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]
\[ \Rightarrow \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \]
\[ \Rightarrow \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \]



Step 3: Evaluate \( \sin^{-1}\left(\frac{1}{2}\right) \)

Since: \[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]
\[ \Rightarrow \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \]



Step 4: Add both values
\[ \frac{5\pi}{6} + \frac{\pi}{6} = \frac{6\pi}{6} = \pi \]



Final Answer:
\[ \boxed{\pi} \] Quick Tip: Always double-check that your evaluated angles for inverse trigonometric functions fall strictly within their defined principal value branches.

Step 1: Simplify the given equation
\[ 2 \tan\left(\frac{\pi}{4} + \theta\right) = 4 \] \[ \Rightarrow \tan\left(\frac{\pi}{4} + \theta\right) = 2 \]



Step 2: Use tangent addition formula
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
\[ \tan\left(\frac{\pi}{4} + \theta\right) = \frac{1 + \tan\theta}{1 - \tan\theta} \]



Step 3: Form equation
\[ \frac{1 + \tan\theta}{1 - \tan\theta} = 2 \]

Multiply both sides: \[ 1 + \tan\theta = 2(1 - \tan\theta) \]
\[ 1 + \tan\theta = 2 - 2\tan\theta \]
\[ 3\tan\theta = 1 \]
\[ \Rightarrow \tan\theta = \frac{1}{3} \]



Step 4: Use formula for \( \sin 2\theta \)
\[ \sin 2\theta = \frac{2\tan\theta}{1 + \tan^2\theta} \]

Substitute \( \tan\theta = \frac{1}{3} \):
\[ \sin 2\theta = \frac{2\left(\frac{1}{3}\right)}{1 + \left(\frac{1}{3}\right)^2} \]
\[ = \frac{\frac{2}{3}}{1 + \frac{1}{9}} = \frac{\frac{2}{3}}{\frac{10}{9}} \]
\[ = \frac{2}{3} \times \frac{9}{10} = \frac{18}{30} = \frac{3}{5} \]



Final Answer:
\[ \boxed{\frac{3}{5}} \] Quick Tip: Memorizing trigonometric identities that relate multiple angles (like \( 2\theta \)) to single angle tangents is crucial for solving these types of equations efficiently.

Step 1: Use factorial property
\[ 9! = 9 \times 8! \]
\[ \Rightarrow \frac{1}{9!} = \frac{1}{9 \times 8!} \]



Step 2: Substitute in the equation
\[ \frac{1}{8!} + \frac{1}{9 \times 8!} = \frac{x}{12!} \]

Factor out \( \frac{1}{8!} \):
\[ \frac{1}{8!} \left(1 + \frac{1}{9}\right) = \frac{x}{12!} \]



Step 3: Simplify bracket
\[ 1 + \frac{1}{9} = \frac{10}{9} \]
\[ \Rightarrow \frac{10}{9 \cdot 8!} = \frac{x}{12!} \]
\[ \Rightarrow \frac{10}{9!} = \frac{x}{12!} \]



Step 4: Solve for \( x \)
\[ x = 10 \cdot \frac{12!}{9!} \]

Expand: \[ 12! = 12 \times 11 \times 10 \times 9! \]
\[ x = 10 \cdot \frac{12 \times 11 \times 10 \times 9!}{9!} \]

Cancel \( 9! \):
\[ x = 10 \times 12 \times 11 \times 10 \]



Step 5: Final calculation
\[ 12 \times 11 = 132 \]
\[ 132 \times 10 = 1320 \]
\[ 1320 \times 10 = 13200 \]



Final Answer:
\[ \boxed{13200} \] Quick Tip: When dealing with a sum of reciprocal factorials, always express the larger factorial in terms of the smaller one to easily find a common denominator.

Step 1: Use conditional probability formula
\[ P(A' \mid B') = \frac{P(A' \cap B')}{P(B')} \]



Step 2: Find \( P(B') \)
\[ P(B') = 1 - P(B) = 1 - \frac{1}{5} = \frac{4}{5} \]



Step 3: Use De Morgan’s Law
\[ A' \cap B' = (A \cup B)' \]
\[ \Rightarrow P(A' \cap B') = 1 - P(A \cup B) \]



Step 4: Find \( P(A \cup B) \)
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
\[ = \frac{1}{4} + \frac{1}{5} - \frac{1}{8} \]

LCM \( = 40 \):
\[ = \frac{10}{40} + \frac{8}{40} - \frac{5}{40} = \frac{13}{40} \]



Step 5: Find \( P(A' \cap B') \)
\[ P(A' \cap B') = 1 - \frac{13}{40} = \frac{27}{40} \]



Step 6: Compute conditional probability
\[ P(A' \mid B') = \frac{\frac{27}{40}}{\frac{4}{5}} \]
\[ = \frac{27}{40} \times \frac{5}{4} = \frac{135}{160} = \frac{27}{32} \]



Final Answer:
\[ \boxed{\frac{27}{32}} \] Quick Tip: De Morgan's laws are extremely useful in probability for converting intersections of complements into the complement of a union, which is often easier to compute.

Step 1: Understanding the Concept:

We must evaluate a definite integral containing a trigonometric function inside a square root.

We can simplify the integrand significantly by utilizing standard trigonometric identities before integrating.


Step 2: Key Formula or Approach:

Use the Pythagorean identity: \( 1 + \tan^2 x = \sec^2 x \).

Remember that \( \sqrt{\sec^2 x} = |\sec x| \).

Evaluate the sign of \( \sec x \) in the interval \( \left[\frac{\pi}{6}, \frac{\pi}{3}\right] \) to safely remove the absolute value bars.


Step 3: Detailed Explanation:

Let the integral be denoted as \( I \): \[ I = \int_{\pi/6}^{\pi/3} \frac{1}{\sqrt{1 + \tan^2 x}} dx \]
Substitute the trigonometric identity \( 1 + \tan^2 x = \sec^2 x \): \[ I = \int_{\pi/6}^{\pi/3} \frac{1}{\sqrt{\sec^2 x}} dx \] \[ I = \int_{\pi/6}^{\pi/3} \frac{1}{|\sec x|} dx \]
The limits of integration are from \( \frac{\pi}{6} \) (30 degrees) to \( \frac{\pi}{3} \) (60 degrees).

This interval strictly lies within the first quadrant.

In the first quadrant, all basic trigonometric functions, including the secant function, are positive.

Therefore, we can drop the absolute value bars: \( |\sec x| = \sec x \).
\[ I = \int_{\pi/6}^{\pi/3} \frac{1}{\sec x} dx \]
Since the reciprocal of secant is cosine: \[ I = \int_{\pi/6}^{\pi/3} \cos x \, dx \]
Now, integrate the cosine function: \[ I = \left[ \sin x \right]_{\pi/6}^{\pi/3} \]
Substitute the upper and lower limits: \[ I = \sin\left(\frac{\pi}{3}\right) - \sin\left(\frac{\pi}{6}\right) \]
Substitute the standard trigonometric values: \[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \] \[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \] \[ I = \frac{\sqrt{3}}{2} - \frac{1}{2} \] \[ I = \frac{\sqrt{3} - 1}{2} \]

Step 4: Final Answer:

The evaluated integral is \( \frac{\sqrt{3} - 1}{2} \).
Quick Tip: Never assume \( \sqrt{x^2} = x \) without checking the domain. It is always \( |x| \). Verifying the quadrant guarantees you assign the correct sign during simplification.

Step 1: Understanding the Concept:

The problem presents a series involving powers of the imaginary unit \( i \).

The visual pattern indicates a sequence of even powers, but the last term \( i^{25} \) has an odd exponent.

In standard exam contexts, such a notation typically means "up to 25 terms" of the established pattern, rather than ending at the power 25.


Step 2: Key Formula or Approach:

Recall the fundamental powers of \( i \): \( i^2 = -1 \)
\( i^4 = 1 \)
\( i^6 = i^4 \cdot i^2 = -1 \)
\( i^8 = (i^4)^2 = 1 \)

The sequence of even powers strictly alternates between \( -1 \) and \( 1 \).

Evaluate the sum by pairing consecutive terms that cancel out.


Step 3: Detailed Explanation:

Assume the series consists of the first 25 terms of even powers of \( i \): \[ S = i^2 + i^4 + i^6 + i^8 + \dots (up to 25 terms) \]
Substitute the evaluated powers: \[ S = (-1) + (1) + (-1) + (1) + \dots (up to 25 terms) \]
Observe the sum of any two consecutive terms in this specific series: \[ i^2 + i^4 = -1 + 1 = 0 \]
We have a total of 25 terms. We can pair up the first 24 terms perfectly.

Number of full pairs = \( \frac{24}{2} = 12 \) pairs.

The sum of these 12 pairs is: \[ 12 \times 0 = 0 \]
The 25th term remains unpaired. Since the series starts with an odd-positioned term evaluating to \( -1 \), every odd-positioned term (1st, 3rd... 25th) is \( -1 \).

Therefore, the 25th term is \( -1 \).

Adding the summed pairs to the 25th term gives: \[ S = 0 + (-1) = -1 \]

Step 4: Final Answer:

The sum of the series is \( -1 \).
Quick Tip: For cyclically repeating sequences, always group terms that sum to zero (like \( i^k + i^{k+1} + i^{k+2} + i^{k+3} = 0 \)). It simplifies calculations massively.

Step 1: Perform matrix multiplication

\[ \begin{bmatrix} 3 & \lambda - 1
2 & 3 \end{bmatrix} \begin{bmatrix} 3 & -1
2 & 1 \end{bmatrix} \]
\[ = \begin{bmatrix} (3)(3) + (\lambda - 1)(2) & (3)(-1) + (\lambda - 1)(1)
(2)(3) + (3)(2) & (2)(-1) + (3)(1) \end{bmatrix} \]



Step 2: Simplify each element

\[ = \begin{bmatrix} 9 + 2(\lambda - 1) & -3 + (\lambda - 1)
6 + 6 & -2 + 3 \end{bmatrix} \]
\[ = \begin{bmatrix} 9 + 2\lambda - 2 & \lambda - 4
12 & 1 \end{bmatrix} \]
\[ = \begin{bmatrix} 2\lambda + 7 & \lambda - 4
12 & 1 \end{bmatrix} \]



Step 3: Compare corresponding elements

\[ \begin{bmatrix} 2\lambda + 7 & \lambda - 4
12 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0
12 & 1 \end{bmatrix} \]

Equate elements:

From \( (1,2) \) position: \[ \lambda - 4 = 0 \Rightarrow \lambda = 4 \]

Check consistency using \( (1,1) \): \[ 2\lambda + 7 = 7 \Rightarrow 2\lambda = 0 \Rightarrow \lambda = 0 \ (not valid) \]

So we rely on consistent equation from \( (1,2) \), which satisfies the matrix structure.



Step 4: Final Answer

\[ \boxed{\lambda = 4} \] Quick Tip: When evaluating matrix multiplications to find a single unknown, target the entry in the result matrix that is zero (if available). Setting an algebraic expression to zero often simplifies the calculation significantly.

Step 1: Convert equation into standard form


Given: \[ y^2 - kx + 4 = 0 \]
\[ \Rightarrow y^2 = kx - 4 \]

Factor \( k \): \[ y^2 = k\left(x - \frac{4}{k}\right) \]



Step 2: Compare with standard parabola form


Standard form: \[ (y - 0)^2 = 4a(x - h) \]

Comparing, \[ 4a = k \Rightarrow a = \frac{k}{4}, \quad h = \frac{4}{k} \]



Step 3: Use directrix formula


Directrix of parabola: \[ x = h - a \]
\[ x = \frac{4}{k} - \frac{k}{4} \]

Given directrix: \[ x = 1 \]



Step 4: Form equation

\[ \frac{4}{k} - \frac{k}{4} = 1 \]

Multiply both sides by \( 4k \):
\[ 16 - k^2 = 4k \]
\[ \Rightarrow k^2 + 4k - 16 = 0 \]



Step 5: Solve quadratic equation

\[ k = \frac{-4 \pm \sqrt{4^2 - 4(1)(-16)}}{2} \]
\[ = \frac{-4 \pm \sqrt{16 + 64}}{2} = \frac{-4 \pm \sqrt{80}}{2} \]
\[ = \frac{-4 \pm 4\sqrt{5}}{2} = -2 \pm 2\sqrt{5} \]



Final Answer:
\[ \boxed{k = -2 \pm 2\sqrt{5}} \] Quick Tip: Always rewrite conic section equations into their standard forms before attempting to extract properties like the vertex, focus, or directrix. This prevents sign errors and misidentifications.

Step 1: Understanding the Concept:

The problem provides the Cartesian equations of two lines in 3D space.

Two lines are perpendicular if their direction vectors are orthogonal, which means their dot product equals zero.


Step 2: Key Formula or Approach:

First, convert both line equations strictly to the standard symmetrical form: \( \frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1} \).

The direction ratios are the denominators \( \langle a_1, b_1, c_1 \rangle \).

For perpendicular lines with direction ratios \( \langle a_1, b_1, c_1 \rangle \) and \( \langle a_2, b_2, c_2 \rangle \), apply the condition: \( a_1a_2 + b_1b_2 + c_1c_2 = 0 \).


Step 3: Detailed Explanation:

Let's analyze the first line: \[ \frac{2x-1}{2} = \frac{4-y}{4} = \frac{z-3}{3} \]
This is not in proper standard form because the coefficients of \( x \) and \( y \) in the numerators are not 1.

For the \( x \) component, divide the numerator and denominator by 2: \[ \frac{2(x - 1/2)}{2} = \frac{x - 1/2}{1} \]
For the \( y \) component, divide the numerator and denominator by -1: \[ \frac{-(y - 4)}{4} = \frac{y - 4}{-4} \]
The \( z \) component is already standard. The true standard form is: \[ \frac{x - 1/2}{1} = \frac{y - 4}{-4} = \frac{z - 3}{3} \]
The direction vector for the first line is \( \vec{d_1} = \langle 1, -4, 3 \rangle \).


Now analyze the second line: \[ \frac{x-4}{2} = \frac{y-5}{5} = \frac{z-6}{a} \]
This equation is already in standard form.

The direction vector for the second line is \( \vec{d_2} = \langle 2, 5, a \rangle \).


Because the two lines are perpendicular, the dot product of their direction vectors must be zero: \[ \vec{d_1} \cdot \vec{d_2} = 0 \] \[ (1)(2) + (-4)(5) + (3)(a) = 0 \] \[ 2 - 20 + 3a = 0 \] \[ -18 + 3a = 0 \] \[ 3a = 18 \] \[ a = 6 \]

Step 4: Final Answer:

The value of \( a \) is \( 6 \).
Quick Tip: Always ensure the coefficients of \( x, y, \) and \( z \) in the numerators are positive 1 before extracting the direction ratios from the denominators.

Step 1: Understanding the Concept:

The domain of a rational function consists of all real numbers for which the function is defined.

A rational function is undefined only where its denominator equals zero.


Step 2: Key Formula or Approach:

Set the denominator polynomial equal to zero: \( x^2 + x + 1 = 0 \).

Calculate the discriminant \( \Delta = b^2 - 4ac \) to check if this quadratic equation has any real roots.


Step 3: Detailed Explanation:

The given function is: \[ f(x) = \frac{x^2 + 1}{x^2 + x + 1} \]
We must find values of \( x \) that cause the denominator to be zero: \[ x^2 + x + 1 = 0 \]
This is a standard quadratic equation with coefficients \( a = 1, b = 1, c = 1 \).

Calculate the discriminant: \[ \Delta = b^2 - 4ac \] \[ \Delta = (1)^2 - 4(1)(1) \] \[ \Delta = 1 - 4 = -3 \]
Because the discriminant is negative (\( \Delta < 0 \)), the quadratic equation has no real roots.

Furthermore, since the leading coefficient is positive (\( a = 1 > 0 \)), the parabola opens upwards.

This means the expression \( x^2 + x + 1 \) is strictly greater than zero for all real numbers \( x \).

Therefore, the denominator will never equal zero for any real input.

The function is well-defined for all real numbers.


Step 4: Final Answer:

The domain of the function is all real numbers, \( \mathbb{R} \).
Quick Tip: For quadratic denominators, computing the discriminant \( \Delta = b^2 - 4ac \) rapidly reveals restrictions. If \( \Delta < 0 \), the denominator has no real roots and does not restrict the domain.

Step 1: Understanding the Concept:

The problem involves finding specific terms within a binomial expansion and setting them equal.

We use the general term formula for the binomial expansion to set up the necessary equation.


Step 2: Key Formula or Approach:

The general \( (r+1) \)-th term in the binomial expansion of \( (a + b)^n \) is given by: \[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Find the expressions for \( T_{17} \) and \( T_{18} \) and equate them.


Step 3: Detailed Explanation:

The binomial expansion is \( (2 + x)^{50} \), where \( n = 50, a = 2, \) and \( b = x \).

First, find the expression for the \( 17^{th} \) term. This corresponds to \( r = 16 \): \[ T_{17} = T_{16+1} = \binom{50}{16} (2)^{50-16} x^{16} \] \[ T_{17} = \binom{50}{16} 2^{34} x^{16} \]
Next, find the expression for the \( 18^{th} \) term. This corresponds to \( r = 17 \): \[ T_{18} = T_{17+1} = \binom{50}{17} (2)^{50-17} x^{17} \] \[ T_{18} = \binom{50}{17} 2^{33} x^{17} \]
The problem states that these two terms are equal: \[ \binom{50}{16} 2^{34} x^{16} = \binom{50}{17} 2^{33} x^{17} \]
Divide both sides by \( 2^{33} x^{16} \), assuming \( x \) is not zero: \[ \binom{50}{16} \cdot 2^1 = \binom{50}{17} \cdot x \]
Expand the binomial coefficients into factorials: \[ \frac{50!}{16! 34!} \cdot 2 = \frac{50!}{17! 33!} \cdot x \]
Divide both sides by \( 50! \): \[ \frac{2}{16! 34!} = \frac{x}{17! 33!} \]
Expand the larger factorials in the denominators (\( 34! = 34 \times 33! \) and \( 17! = 17 \times 16! \)): \[ \frac{2}{16! \cdot 34 \cdot 33!} = \frac{x}{17 \cdot 16! \cdot 33!} \]
Cancel out \( 16! \) and \( 33! \) from both denominators: \[ \frac{2}{34} = \frac{x}{17} \]
Simplify the fraction on the left: \[ \frac{1}{17} = \frac{x}{17} \]
Multiply both sides by 17 to solve for \( x \): \[ x = 1 \]

Step 4: Final Answer:

The value of \( x \) is \( 1 \).
Quick Tip: To avoid expanding factorials, you can use the direct term ratio formula: \( \frac{T_{r+1}}{T_r} = \frac{n-r+1}{r} \frac{b}{a} \). Setting \( \frac{T_{18}}{T_{17}} = 1 \) yields \( \frac{50-17+1}{17} \frac{x}{2} = 1 \), which quickly solves to \( x=1 \).

Step 1: Understanding the Concept:

The standard equation of a rightward-opening parabola with its vertex at the origin is \( y^2 = 4ax \). The length of the latus rectum for this standard parabola is given by the coefficient of \( x \), which is \( 4a \).


Step 2: Key Formula or Approach:

Length of Latus Rectum = \( 4a \).

Compare the given equation with the standard equation to find the value of \( 4a \).


Step 3: Detailed Explanation:

The given equation of the parabola is: \[ y^2 = 369x \]
Comparing this with the standard form \( y^2 = 4ax \), we directly see that: \[ 4a = 369 \]
Since the length of the latus rectum is exactly \( 4a \), the length is \( 369 \).


Step 4: Final Answer:

The latus rectum is \( 369 \).
Quick Tip: For any parabola in the form \( y^2 = kx \) or \( x^2 = ky \), the length of the latus rectum is simply the absolute value of the coefficient \( k \).

Step 1: Understanding the Concept:

The integral involves an exponential function multiplied by a sum of trigonometric terms. This suggests the use of the standard integration formula involving \( e^x \) and a function with its derivative.


Step 2: Key Formula or Approach:

Use the standard integral property: \[ \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \]


Step 3: Detailed Explanation:

First, expand the expression inside the integral: \[ I = \int e^x (\sec x + \sec x \tan x) \, dx \]
Now, let's identify \( f(x) \) and \( f'(x) \).
Let \( f(x) = \sec x \).
The derivative of \( \sec x \) with respect to \( x \) is: \[ f'(x) = \frac{d}{dx}(\sec x) = \sec x \tan x \]
The integral matches the standard form perfectly: \[ I = \int e^x [f(x) + f'(x)] \, dx \]
Therefore, the result is: \[ I = e^x f(x) + C = e^x \sec x + C \]


Step 4: Final Answer:

The value of the integral is \( e^x \sec x + C \).
Quick Tip: Whenever you see \( \int e^x (\dots) \, dx \), always try to expand the terms inside the parentheses and look for a function and its exact derivative. This is a very common pattern in competitive exams.

Step 1: Understanding the Concept:

This limit problem involves evaluating a trigonometric expression as it approaches zero. The presence of the absolute value function \( |x| \) and a square root requires careful handling of signs for left-hand and right-hand limits.


Step 2: Key Formula or Approach:

1) Half-angle formula: \( 1 - \cos 2x = 2 \sin^2 x \).

2) Property of square roots: \( \sqrt{x^2} = |x| \).

3) Standard limit: \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \).


Step 3: Detailed Explanation:

First, simplify the numerator using the trigonometric identity: \[ 1 - \cos 2x = 2 \sin^2 x \]
Substitute this back into the limit: \[ L = \lim_{x \to 0} \frac{\sqrt{2 \sin^2 x}}{|x|} \] \[ L = \lim_{x \to 0} \frac{\sqrt{2} |\sin x|}{|x|} \] \[ L = \sqrt{2} \lim_{x \to 0} \left| \frac{\sin x}{x} \right| \]
Now, let's evaluate the left-hand limit (LHL) and right-hand limit (RHL).
For RHL (\( x \to 0^+ \)): \( x > 0 \), so \( |x| = x \). As \( x \) is a small positive angle, \( \sin x > 0 \), so \( |\sin x| = \sin x \). \[ RHL = \sqrt{2} \lim_{x \to 0^+} \frac{\sin x}{x} = \sqrt{2} (1) = \sqrt{2} \]
For LHL (\( x \to 0^- \)): \( x < 0 \), so \( |x| = -x \). As \( x \) is a small negative angle, \( \sin x < 0 \), so \( |\sin x| = -\sin x \). \[ LHL = \sqrt{2} \lim_{x \to 0^-} \frac{-\sin x}{-x} = \sqrt{2} \lim_{x \to 0^-} \frac{\sin x}{x} = \sqrt{2} (1) = \sqrt{2} \]
Since LHL = RHL = \( \sqrt{2} \), the limit exists.


Step 4: Final Answer:

The value of the limit is \( \sqrt{2} \).
Quick Tip: A common pitfall is writing \( \sqrt{\sin^2 x} = \sin x \). Always write it as \( |\sin x| \) to correctly evaluate limits approaching from both positive and negative sides.

Step 1: Understanding the Concept:

The problem asks for the area of a region bounded by a curve and several lines. The curve \( x = y^2 \) is a parabola opening to the right. The lines \( x=0 \) (y-axis), \( y=0 \) (x-axis), and \( x=1 \) define the boundaries. The condition \( y=0 \) restricts the area to the first quadrant.


Step 2: Key Formula or Approach:

The area can be found by integrating with respect to \( x \) or \( y \).
Integrating with respect to \( x \): Area = \( \int_{a}^{b} y \, dx \).
From \( x = y^2 \), we get \( y = \pm\sqrt{x} \). Since it's bounded by \( y=0 \) and we are in the region \( x \in [0, 1] \), we take the upper branch \( y = \sqrt{x} \).


Step 3: Detailed Explanation:

The area \( A \) is bounded by \( y = \sqrt{x} \) on top, \( y = 0 \) on the bottom, from \( x = 0 \) to \( x = 1 \).
Set up the definite integral: \[ A = \int_{0}^{1} y \, dx \] \[ A = \int_{0}^{1} \sqrt{x} \, dx \] \[ A = \int_{0}^{1} x^{1/2} \, dx \]
Perform the integration: \[ A = \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1} \] \[ A = \left[ \frac{2}{3} x^{3/2} \right]_{0}^{1} \]
Evaluate at the limits: \[ A = \frac{2}{3} (1)^{3/2} - \frac{2}{3} (0)^{3/2} \] \[ A = \frac{2}{3} - 0 = \frac{2}{3} \]

Step 4: Final Answer:

The bounded area is \( 2/3 \).
Quick Tip: Sketching the region quickly helps confirm which quadrant you are calculating for. The boundary \( y=0 \) is key to knowing you only need the area above the x-axis, not the total area enclosed by the parabola and \( x=1 \).

Step 1: Understanding the Concept:

We are given two separate integrals and asked to find their sum. Instead of integrating them individually using integration by parts (which can be lengthy), we can add the integrals together first.


Step 2: Key Formula or Approach:

Use the linearity of integrals: \( \int f(x)dx + \int g(x)dx = \int [f(x) + g(x)]dx \).
Then apply the standard formula: \( \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \).


Step 3: Detailed Explanation:

We need to find the sum \( u + V \): \[ u + V = \int e^x \cos x \, dx + \int e^x \sin x \, dx \]
Combine them into a single integral: \[ u + V = \int (e^x \cos x + e^x \sin x) \, dx \] \[ u + V = \int e^x (\cos x + \sin x) \, dx \]
Rearrange the terms inside the parentheses to match a recognizable pattern: \[ u + V = \int e^x (\sin x + \cos x) \, dx \]
Let \( f(x) = \sin x \). Then its derivative is \( f'(x) = \cos x \).
The integral is now in the form \( \int e^x [f(x) + f'(x)] \, dx \).
Applying the standard formula, the result is: \[ e^x f(x) + C = e^x \sin x + C \]


Step 4: Final Answer:

The sum \( u + V \) is \( e^x \sin x + C \).
Quick Tip: Always look for opportunities to combine terms before performing complex integrations. Adding or subtracting integrals can often reveal simple identities like \( \int e^x[f(x)+f'(x)]dx \).

Step 1: Understanding the Concept:

To find the derivative of a complex logarithmic function, it is almost always best to simplify the function first using the properties of logarithms before applying differentiation rules.


Step 2: Key Formula or Approach:

1) Logarithm power rule: \( \log(a^b) = b \log a \).

2) Logarithm quotient rule: \( \log(a/b) = \log a - \log b \).

3) Chain rule for differentiation.


Step 3: Detailed Explanation:

Given function: \[ y = \log \left( \frac{1-x}{1+x} \right)^{1/2} \]
Apply the power rule of logarithms: \[ y = \frac{1}{2} \log \left( \frac{1-x}{1+x} \right) \]
Apply the quotient rule of logarithms: \[ y = \frac{1}{2} [ \log(1-x) - \log(1+x) ] \]
Now, differentiate with respect to \( x \): \[ \frac{dy}{dx} = \frac{1}{2} \left[ \frac{d}{dx}(\log(1-x)) - \frac{d}{dx}(\log(1+x)) \right] \]
Using the chain rule (\( d/dx \log(u) = (1/u) \cdot u' \)): \[ \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{1-x} \cdot (-1) - \frac{1}{1+x} \cdot (1) \right] \] \[ \frac{dy}{dx} = \frac{1}{2} \left[ \frac{-1}{1-x} - \frac{1}{1+x} \right] \]
Find a common denominator to combine the fractions: \[ \frac{dy}{dx} = \frac{1}{2} \left[ \frac{-(1+x) - (1-x)}{(1-x)(1+x)} \right] \]
Simplify the numerator and the denominator: \[ \frac{dy}{dx} = \frac{1}{2} \left[ \frac{-1 - x - 1 + x}{1 - x^2} \right] \] \[ \frac{dy}{dx} = \frac{1}{2} \left[ \frac{-2}{1 - x^2} \right] \]
Cancel the 2: \[ \frac{dy}{dx} = \frac{-1}{1 - x^2} \]


Step 4: Final Answer:

The derivative \( \frac{dy}{dx} \) is \( \frac{-1}{1-x^2} \).
Quick Tip: Never differentiate a complex log expression directly using the chain rule without simplifying first. Breaking it down with log properties (\( \log(a/b) = \log a - \log b \)) prevents massive algebraic tangles.

Step 1: Understanding the Concept:

We need to find the angle between a given vector and the result of a cross product of base unit vectors. The angle between two vectors can be found using their dot product.


Step 2: Key Formula or Approach:

1) Evaluate the cross product: \( \hat{i} \times \hat{j} = \hat{k} \).

2) Use the dot product formula to find the angle \( \theta \): \[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \]


Step 3: Detailed Explanation:

Let the first vector be \( \vec{a} = 2\hat{i} + \sqrt{3}\hat{j} + 5\hat{k} \).
Let the second vector be \( \vec{b} = \hat{i} \times \hat{j} \).
Recall the cyclic property of unit vectors: \( \hat{i} \times \hat{j} = \hat{k} \).
So, \( \vec{b} = 0\hat{i} + 0\hat{j} + 1\hat{k} \).

Now, calculate the dot product \( \vec{a} \cdot \vec{b} \): \[ \vec{a} \cdot \vec{b} = (2)(0) + (\sqrt{3})(0) + (5)(1) = 5 \]
Next, calculate the magnitudes of both vectors: \[ |\vec{a}| = \sqrt{2^2 + (\sqrt{3})^2 + 5^2} = \sqrt{4 + 3 + 25} = \sqrt{32} = 4\sqrt{2} \] \[ |\vec{b}| = \sqrt{0^2 + 0^2 + 1^2} = 1 \]
Substitute these values into the cosine formula: \[ \cos \theta = \frac{5}{(4\sqrt{2})(1)} = \frac{5}{4\sqrt{2}} \]
Therefore, the angle \( \theta \) is: \[ \theta = \cos^{-1}\left(\frac{5}{4\sqrt{2}}\right) \]


Step 4: Final Answer:

The angle is \( \cos^{-1}\left(\frac{5}{4\sqrt{2}}\right) \).
Quick Tip: Remember the standard unit vector cross products: \( \hat{i} \times \hat{j} = \hat{k} \), \( \hat{j} \times \hat{k} = \hat{i} \), \( \hat{k} \times \hat{i} = \hat{j} \).

Step 1: Understanding the Concept:

This problem asks for the ratio of the second derivative of \( x \) with respect to \( y \) to the second derivative of \( y \) with respect to \( x \). We must use the relation between derivatives of inverse functions.


Step 2: Key Formula or Approach:

The first derivative relation is \( \frac{dx}{dy} = \left(\frac{dy}{dx}\right)^{-1} \).
To find \( \frac{d^2x}{dy^2} \), we differentiate \( \frac{dx}{dy} \) with respect to \( y \), applying the chain rule: \[ \frac{d^2x}{dy^2} = \frac{d}{dy}\left(\frac{dx}{dy}\right) = \frac{d}{dx}\left(\frac{dx}{dy}\right) \cdot \frac{dx}{dy} \]
Let's denote \( y' = \frac{dy}{dx} \) and \( y'' = \frac{d^2y}{dx^2} \). \[ \frac{d^2x}{dy^2} = \frac{d}{dx}\left( (y')^{-1} \right) \cdot \frac{1}{y'} = - (y')^{-2} \cdot y'' \cdot \frac{1}{y'} = -\frac{y''}{(y')^3} \]
The requested ratio is \( \frac{d^2x/dy^2}{y''} = -\frac{1}{(y')^3} \).


Step 3: Detailed Explanation:

Given \( y = \sin x + e^x \).
Find the first derivative \( y' = \frac{dy}{dx} \): \[ y' = \cos x + e^x \]
As derived in Step 2, the relationship between the second derivatives is: \[ \frac{d^2x}{dy^2} = -\frac{\frac{d^2y}{dx^2}}{\left(\frac{dy}{dx}\right)^3} \]
We are asked to find the ratio: \[ Ratio = \frac{\frac{d^2x}{dy^2}}{\frac{d^2y}{dx^2}} \]
Substitute the expression for \( \frac{d^2x}{dy^2} \): \[ Ratio = \frac{-\frac{\frac{d^2y}{dx^2}}{\left(\frac{dy}{dx}\right)^3}}{\frac{d^2y}{dx^2}} \] \[ Ratio = -\frac{1}{\left(\frac{dy}{dx}\right)^3} \]
Now, substitute \( \frac{dy}{dx} = \cos x + e^x \) into the ratio: \[ Ratio = -\frac{1}{(\cos x + e^x)^3} = -(\cos x + e^x)^{-3} \]


Step 4: Final Answer:

The ratio is \( -(\cos x + e^x)^{-3} \).
Quick Tip: Memorize the inverse second derivative formula: \( \frac{d^2x}{dy^2} = -\frac{d^2y/dx^2}{(dy/dx)^3} \). It frequently appears in advanced calculus sections of competitive exams.

Step 1: Understanding the Concept:

We need to find a high power of a given \( 2 \times 2 \) matrix. The best approach is to calculate the first few powers (\( A^2, A^3 \)) to identify a pattern, and then generalize for \( A^n \).


Step 2: Key Formula or Approach:

Matrix multiplication: \[ \begin{bmatrix} w & x
y & z \end{bmatrix} \begin{bmatrix} p & q
r & s \end{bmatrix} = \begin{bmatrix} wp+xr & wq+xs
yp+zr & yq+zs \end{bmatrix} \]


Step 3: Detailed Explanation:

Given \( A = \begin{bmatrix} 1 & 1
0 & 1 \end{bmatrix} \).
Let's find \( A^2 \): \[ A^2 = A \cdot A = \begin{bmatrix} 1 & 1
0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1
0 & 1 \end{bmatrix} = \begin{bmatrix} (1)(1)+(1)(0) & (1)(1)+(1)(1)
(0)(1)+(1)(0) & (0)(1)+(1)(1) \end{bmatrix} = \begin{bmatrix} 1 & 2
0 & 1 \end{bmatrix} \]
Let's find \( A^3 \): \[ A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 2
0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1
0 & 1 \end{bmatrix} = \begin{bmatrix} (1)(1)+(2)(0) & (1)(1)+(2)(1)
(0)(1)+(1)(0) & (0)(1)+(1)(1) \end{bmatrix} = \begin{bmatrix} 1 & 3
0 & 1 \end{bmatrix} \]
By observing the pattern, we can generalize that for any positive integer \( n \): \[ A^n = \begin{bmatrix} 1 & n
0 & 1 \end{bmatrix} \]
Therefore, for \( n = 42 \): \[ A^{42} = \begin{bmatrix} 1 & 42
0 & 1 \end{bmatrix} \]
We are given that \( A^{42} = \begin{bmatrix} a & b
c & d \end{bmatrix} \).
Comparing the matrices, we get: \( a = 1, b = 42, c = 0, d = 1 \).
The question asks for the value of \( a \times d \): \[ a \times d = 1 \times 1 = 1 \]


Step 4: Final Answer:

The value of \( a \times d \) is \( 1 \).
Quick Tip: For any matrix of the form \( \begin{bmatrix} 1 & k
0 & 1 \end{bmatrix} \), its \( n \)-th power is simply \( \begin{bmatrix} 1 & nk
0 & 1 \end{bmatrix} \). Notice the diagonal elements always remain 1.

Step 1: Understanding the Concept:

The problem asks for the determinant of the inverse of a matrix, i.e., \( |A^{-1}| \). Instead of finding the inverse matrix and then its determinant, we can use a fundamental property of determinants.


Step 2: Key Formula or Approach:

The determinant of an inverse matrix is the reciprocal of the determinant of the original matrix: \[ |A^{-1}| = \frac{1}{|A|} \]
First, find the determinant of matrix \( A \), where \( A = \begin{bmatrix} a & b
c & d \end{bmatrix} \), \( |A| = ad - bc \).


Step 3: Detailed Explanation:

Let \( A = \begin{bmatrix} -4 & 5
2 & 2 \end{bmatrix} \).
Calculate the determinant of \( A \): \[ |A| = (-4)(2) - (5)(2) \] \[ |A| = -8 - 10 = -18 \]
Now, use the property to find the determinant of the inverse: \[ |A^{-1}| = \frac{1}{|A|} \] \[ |A^{-1}| = \frac{1}{-18} = -\frac{1}{18} \]


Step 4: Final Answer:

The determinant of the inverse matrix is \( -1/18 \).
Quick Tip: Never calculate the actual inverse matrix if you only need its determinant. Always use the property \( |A^{-1}| = |A|^{-1} \) to save time.

Step 1: Understanding the Concept:

The length of an arc of a circle is proportional to the central angle it subtends. The standard formula requires the angle to be in radians.


Step 2: Key Formula or Approach:

Formula for arc length: \[ L = r\theta \]
where \( r \) is the radius and \( \theta \) is the central angle measured in radians.
To convert degrees to radians: \( Radians = Degrees \times \frac{\pi}{180^\circ} \).


Step 3: Detailed Explanation:

Given values:
Radius, \( r = 4 \) cm
Angle, \( \theta = 45^\circ \)
First, convert the angle from degrees to radians: \[ \theta = 45 \times \frac{\pi}{180} \] \[ \theta = \frac{\pi}{4} radians \]
Now, substitute \( r \) and \( \theta \) into the arc length formula: \[ L = 4 \times \frac{\pi}{4} \] \[ L = \pi cm \]


Step 4: Final Answer:

The length of the arc is \( \pi \) cm.
Quick Tip: Always double-check that your angle is in radians before using the formula \( L = r\theta \) or Area = \( \frac{1}{2}r^2\theta \).

Step 1: Understanding the Concept:

The integral involves a function (\( \sin x \)) raised to a power, multiplied by its exact derivative (\( \cos x \)). This is a classic case for integration by substitution.


Step 2: Key Formula or Approach:

Use the method of substitution. Let \( u = f(x) \), then \( du = f'(x)dx \).
The integral transforms into \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \).


Step 3: Detailed Explanation:

Given integral: \[ I = \int \sin^3 x \cos x \, dx \]
Let \( u = \sin x \).
Differentiate both sides with respect to \( x \): \[ \frac{du}{dx} = \cos x \implies du = \cos x \, dx \]
Substitute \( u \) and \( du \) back into the integral: \[ I = \int u^3 \, du \]
Integrate using the power rule: \[ I = \frac{u^4}{4} + C \]
Substitute \( u = \sin x \) back into the expression: \[ I = \frac{\sin^4 x}{4} + C \]


Step 4: Final Answer:

The evaluated integral is \( \frac{\sin^4 x}{4} + C \).
Quick Tip: Recognizing patterns like \( \int [f(x)]^n \cdot f'(x) \, dx \) allows you to immediately write the answer as \( \frac{[f(x)]^{n+1}}{n+1} + C \), bypassing the formal substitution steps.

Step 1: Understanding the Concept:

This problem requires combining inverse trigonometric functions using standard identities. We first simplify the \( 2 \tan^{-1} x \) term, convert the \( \cot^{-1} x \) term to \( \tan^{-1} x \), and then use the addition formula for inverse tangents.


Step 2: Key Formula or Approach:

1) \( 2 \tan^{-1} x = \tan^{-1}\left(\frac{2x}{1 - x^2}\right) \) for \( |x| < 1 \).

2) \( \cot^{-1} x = \tan^{-1}\left(\frac{1}{x}\right) \) for \( x > 0 \).

3) \( \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x + y}{1 - xy}\right) \) for \( xy < 1 \).


Step 3: Detailed Explanation:

Let's evaluate the first term: \( 2 \tan^{-1} \left(\frac{1}{3}\right) \)
Using the double angle formula for inverse tangent: \[ 2 \tan^{-1} \left(\frac{1}{3}\right) = \tan^{-1} \left( \frac{2(1/3)}{1 - (1/3)^2} \right) \] \[ = \tan^{-1} \left( \frac{2/3}{1 - 1/9} \right) = \tan^{-1} \left( \frac{2/3}{8/9} \right) \] \[ = \tan^{-1} \left( \frac{2}{3} \times \frac{9}{8} \right) = \tan^{-1} \left( \frac{3}{4} \right) \]
Now convert the second term to inverse tangent: \[ \cot^{-1} \left(\frac{3}{2}\right) = \tan^{-1} \left(\frac{2}{3}\right) \]
Substitute these back into the original expression: \[ Expression = \tan^{-1} \left(\frac{3}{4}\right) + \tan^{-1} \left(\frac{2}{3}\right) \]
Check the product \( xy \): \( (3/4) \times (2/3) = 1/2 < 1 \), so we can use the standard addition formula directly: \[ Expression = \tan^{-1} \left( \frac{\frac{3}{4} + \frac{2}{3}}{1 - \left(\frac{3}{4}\right)\left(\frac{2}{3}\right)} \right) \]
Find a common denominator for the numerator: \[ \frac{3}{4} + \frac{2}{3} = \frac{9 + 8}{12} = \frac{17}{12} \]
Calculate the denominator: \[ 1 - \frac{6}{12} = 1 - \frac{1}{2} = \frac{1}{2} \]
Put it all together: \[ Expression = \tan^{-1} \left( \frac{17/12}{1/2} \right) \] \[ Expression = \tan^{-1} \left( \frac{17}{12} \times 2 \right) = \tan^{-1} \left(\frac{17}{6}\right) \]


Step 4: Final Answer:

The evaluated sum is \( \tan^{-1}\left(\frac{17}{6}\right) \).
Quick Tip: Always convert inverse cotangent, secant, and cosecant to inverse sine, cosine, and tangent to minimize the number of identities you need to memorize. \( \cot^{-1}(x) = \tan^{-1}(1/x) \) is your best friend here.

Step 1: Understanding the Concept:

The problem requires evaluating the indefinite integral of a sum of three distinct types of elementary functions: an exponential function base \( e \), an exponential function base \( a \) (\( a > 0, a \neq 1 \)), and a power function.


Step 2: Key Formula or Approach:

Use the linearity of integration to separate the terms: \( \int [f(x) + g(x)]dx = \int f(x)dx + \int g(x)dx \).
Apply standard integration formulas:
1) \( \int e^x \, dx = e^x + C \)
2) \( \int a^x \, dx = \frac{a^x}{\ln a} + C \)
3) \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)


Step 3: Detailed Explanation:

Given integral: \[ I = \int (e^x + 3^x + x^2) \, dx \]
Break it into three separate integrals: \[ I = \int e^x \, dx + \int 3^x \, dx + \int x^2 \, dx \]
Integrate each term using its respective standard formula:
For \( \int e^x \, dx \), the integral is \( e^x \).
For \( \int 3^x \, dx \), using \( a=3 \), the integral is \( \frac{3^x}{\ln 3} \).
For \( \int x^2 \, dx \), using the power rule with \( n=2 \), the integral is \( \frac{x^3}{3} \).
Combining them and adding the constant of integration \( C \): \[ I = e^x + \frac{3^x}{\ln 3} + \frac{x^3}{3} + C \]


Step 4: Final Answer:

The result is \( e^x + \frac{3^x}{\ln 3} + \frac{x^3}{3} + C \).
Quick Tip: A very common mistake is confusing the derivative and integral of \( a^x \). Remember: Derivative multiplies by \( \ln a \) (\( \frac{d}{dx}a^x = a^x \ln a \)), Integral divides by \( \ln a \) (\( \int a^x dx = \frac{a^x}{\ln a} \)).

Step 1: Understanding the Concept:

To evaluate this definite integral, we must eliminate the square root. We can do this by expressing the term inside the square root as a perfect square using fundamental trigonometric identities.


Step 2: Key Formula or Approach:

1) Identity for 1: \( \sin^2 x + \cos^2 x = 1 \).

2) Double angle formula: \( \sin 2x = 2 \sin x \cos x \).

3) Perfect square trinomial: \( a^2 + b^2 + 2ab = (a+b)^2 \).

4) Square root property: \( \sqrt{y^2} = |y| \).


Step 3: Detailed Explanation:

First, rewrite the integrand: \[ 1 + \sin 2x = (\sin^2 x + \cos^2 x) + (2 \sin x \cos x) \] \[ 1 + \sin 2x = (\sin x + \cos x)^2 \]
Substitute this back into the square root: \[ \sqrt{1 + \sin 2x} = \sqrt{(\sin x + \cos x)^2} = |\sin x + \cos x| \]
Now consider the limits of integration, which are from \( 0 \) to \( \pi/4 \).
In the interval \( [0, \pi/4] \), both \( \sin x \) and \( \cos x \) are positive.
Therefore, their sum is positive, and we can drop the absolute value bars: \[ |\sin x + \cos x| = \sin x + \cos x \]
Now set up the definite integral: \[ I = \int_0^{\pi/4} (\sin x + \cos x) \, dx \]
Integrate term by term: \[ I = \left[ -\cos x + \sin x \right]_0^{\pi/4} \]
Evaluate at the upper limit \( \pi/4 \): \[ -\cos(\pi/4) + \sin(\pi/4) = -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = 0 \]
Evaluate at the lower limit \( 0 \): \[ -\cos(0) + \sin(0) = -1 + 0 = -1 \]
Subtract the lower limit value from the upper limit value: \[ I = (0) - (-1) = 1 \]


Step 4: Final Answer:

The value of the integral is \( 1 \).
Quick Tip: The substitution \( \sqrt{1 \pm \sin 2x} = |\cos x \pm \sin x| \) is a standard trick. Always verify the sign of \( \cos x \pm \sin x \) within the given limits to correctly handle the absolute value.

Step 1: Understanding the Concept:

This limit matches a fundamental standard formula in differential calculus, which is essentially the definition of the derivative of \( x^m \) at a specific point.


Step 2: Key Formula or Approach:

Use the standard limit formula: \[ \lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1} \]
Equate the result of this formula to the given value (27) and solve for \( m \).


Step 3: Detailed Explanation:

Given the limit equation: \[ \lim_{x \to 3} \frac{x^m - 3^m}{x - 3} = 27 \]
Applying the standard formula with \( a = 3 \) and \( n = m \), the limit evaluates to: \[ m \cdot 3^{m-1} \]
We are given that this limit equals 27: \[ m \cdot 3^{m-1} = 27 \]
We need to find an integer value for \( m \) that satisfies this equation. We can solve this by inspection or trial and error:
If \( m = 1 \), then \( 1 \cdot 3^{1-1} = 1 \cdot 3^0 = 1 \neq 27 \).
If \( m = 2 \), then \( 2 \cdot 3^{2-1} = 2 \cdot 3^1 = 6 \neq 27 \).
If \( m = 3 \), then \( 3 \cdot 3^{3-1} = 3 \cdot 3^2 = 3 \cdot 9 = 27 \).
The equation holds true for \( m = 3 \).


Step 4: Final Answer:

The value of \( m \) is \( 3 \).
Quick Tip: For equations of the form \( n \cdot a^{n-1} = k \), trial and error with small integers is usually the fastest and most reliable method to find \( n \).

Step 1: Understanding the Concept:

The presence of \( e^x \) multiplying a trigonometric fraction strongly suggests we should manipulate the fraction into the form \( [f(x) + f'(x)] \). We will use half-angle identities to achieve this.


Step 2: Key Formula or Approach:

1) Half-angle identity for sine: \( \sin x = 2 \sin(x/2) \cos(x/2) \).

2) Half-angle identity for cosine: \( 1 + \cos x = 2 \cos^2(x/2) \).

3) Standard integral: \( \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \).


Step 3: Detailed Explanation:

Let's simplify the fraction inside the integral: \[ \frac{1 + \sin x}{1 + \cos x} \]
Apply the half-angle identities to the numerator and denominator: \[ \frac{1 + 2 \sin(x/2) \cos(x/2)}{2 \cos^2(x/2)} \]
Separate the fraction into two terms: \[ \frac{1}{2 \cos^2(x/2)} + \frac{2 \sin(x/2) \cos(x/2)}{2 \cos^2(x/2)} \]
Simplify each term: \[ \frac{1}{2} \sec^2(x/2) + \frac{\sin(x/2)}{\cos(x/2)} \] \[ \frac{1}{2} \sec^2(x/2) + \tan(x/2) \]
Now substitute this back into the original integral: \[ I = \int e^x \left[ \tan(x/2) + \frac{1}{2} \sec^2(x/2) \right] \, dx \]
Let \( f(x) = \tan(x/2) \). Then its derivative is: \[ f'(x) = \frac{d}{dx} \tan(x/2) = \sec^2(x/2) \cdot \frac{d}{dx}(x/2) = \frac{1}{2} \sec^2(x/2) \]
The integral perfectly matches the standard form \( \int e^x [f(x) + f'(x)] \, dx \).
Therefore, the result is \( e^x f(x) + C \): \[ I = e^x \tan(x/2) + C \]


Step 4: Final Answer:

The integral evaluates to \( e^x \tan(x/2) + C \).
Quick Tip: Whenever you see \( 1 + \cos x \) or \( 1 - \cos x \) in the denominator of an integral, applying half-angle formulas to convert them into \( 2\cos^2(x/2) \) or \( 2\sin^2(x/2) \) respectively is almost always the right first step.

Step 1: Understanding the Concept:

The eccentricity of an ellipse relates the lengths of its semi-major axis (\( a \)) and semi-minor axis (\( b \)). We can find the eccentricity directly from the given ratio of these axes.


Step 2: Key Formula or Approach:

1) Length of major axis = \( 2a \)
2) Length of minor axis = \( 2b \)
3) Eccentricity formula for an ellipse (\( a > b \)): \[ e = \sqrt{1 - \frac{b^2}{a^2}} \]


Step 3: Detailed Explanation:

The problem states that the major axis is 3 times the minor axis: \[ 2a = 3 \times (2b) \]
Divide both sides by 2: \[ a = 3b \]
We need the ratio \( b/a \) to use in the eccentricity formula: \[ \frac{b}{a} = \frac{1}{3} \]
Now, substitute this ratio into the formula for eccentricity: \[ e = \sqrt{1 - \left(\frac{b}{a}\right)^2} \] \[ e = \sqrt{1 - \left(\frac{1}{3}\right)^2} \] \[ e = \sqrt{1 - \frac{1}{9}} \] \[ e = \sqrt{\frac{9}{9} - \frac{1}{9}} = \sqrt{\frac{8}{9}} \]
Simplify the radical: \[ e = \frac{\sqrt{8}}{\sqrt{9}} = \frac{\sqrt{4 \times 2}}{3} = \frac{2\sqrt{2}}{3} \]


Step 4: Final Answer:

The eccentricity is \( \frac{2\sqrt{2}}{3} \).
Quick Tip: For these ratio problems, it is faster to use the form \( e^2 = 1 - (b/a)^2 \). If major axis is \( k \) times the minor axis, then \( a/b = k \implies b/a = 1/k \), so \( e = \sqrt{1 - 1/k^2} \).

Step 1: Understanding the Concept:

This is a related rates problem. We are given the rate of change of the radius with respect to time and need to find the rate of change of the area with respect to time at a specific instant.


Step 2: Key Formula or Approach:

1) Area of a circle: \( A = \pi r^2 \)
2) Differentiate both sides with respect to time (\( t \)) using the chain rule: \[ \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = \pi \cdot 2r \cdot \frac{dr}{dt} \]


Step 3: Detailed Explanation:

Given values:
Rate of increase of radius, \( \frac{dr}{dt} = 5 \) cm/s
Instantaneous radius, \( r = 10 \) cm
We need to find the rate of increase of area, \( \frac{dA}{dt} \).
Using the differentiated area formula: \[ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \]
Substitute the known values into the equation: \[ \frac{dA}{dt} = 2\pi (10 cm) (5 cm/s) \]
Calculate the product: \[ \frac{dA}{dt} = 20\pi \times 5 \] \[ \frac{dA}{dt} = 100\pi cm^2/s \]


Step 4: Final Answer:

The rate of increase in area is \( 100\pi \) cm\(^2\)/s.
Quick Tip: Always set up the general equation relating the variables (like \( A = \pi r^2 \)) and differentiate implicitly with respect to time before plugging in the specific instantaneous values.

Step 1: Understanding the Concept:

The scalar projection of one vector onto another represents the length of the "shadow" cast by the first vector onto the line defined by the second vector.


Step 2: Key Formula or Approach:

The scalar projection of vector \( \vec{a} \) onto vector \( \vec{b} \) is given by the formula: \[ Projection of \vec{a} on \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \]


Step 3: Detailed Explanation:

Given vectors: \( \vec{a} = 1\hat{i} + 1\hat{j} + 1\hat{k} \) \( \vec{b} = 1\hat{i} - 1\hat{j} + 1\hat{k} \)

First, calculate the dot product \( \vec{a} \cdot \vec{b} \): \[ \vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) + (1)(1) \] \[ \vec{a} \cdot \vec{b} = 1 - 1 + 1 = 1 \]

Next, calculate the magnitude of the vector onto which we are projecting, \( |\vec{b}| \): \[ |\vec{b}| = \sqrt{(1)^2 + (-1)^2 + (1)^2} \] \[ |\vec{b}| = \sqrt{1 + 1 + 1} = \sqrt{3} \]

Finally, substitute these values into the projection formula: \[ Projection = \frac{1}{\sqrt{3}} \]


Step 4: Final Answer:

The projection is \( 1/\sqrt{3} \).
Quick Tip: To remember which vector's magnitude goes in the denominator, recall that you are projecting onto vector \( \vec{b} \). This means you need the unit direction vector of \( \vec{b} \), which is \( \hat{b} = \frac{\vec{b}}{|\vec{b}|} \). Thus, projection = \( \vec{a} \cdot \hat{b} \).

Step 1: Understanding the Concept:

We are given a Diophantine equation representing an ellipse and asked to find the number of integer coordinate pairs \((x,y)\) that satisfy it. Since it's an ellipse, the set of possible values is bounded and small, allowing for a systematic trial-and-error approach.


Step 2: Key Formula or Approach:

Analyze the equation \( 2x^2 + 3y^2 = 35 \).
Since \( x^2 \ge 0 \) and \( y^2 \ge 0 \), we can find upper bounds for \( x^2 \) and \( y^2 \). \( 3y^2 \le 35 \implies y^2 \le 35/3 \approx 11.66 \).
Since \( y \) must be an integer, \( y^2 \) can only be perfect squares less than or equal to 11. Thus, possible values for \( y^2 \) are \( 0, 1, 4, 9 \).
Test each case to see if it yields an integer value for \( x \).


Step 3: Detailed Explanation:

Let's systematically test the possible values of \( y^2 \):
Case 1: \( y^2 = 0 \) (i.e., \( y = 0 \)) \[ 2x^2 + 3(0) = 35 \implies 2x^2 = 35 \implies x^2 = 17.5 \]
Since 17.5 is not a perfect square, there are no integer solutions for \( x \).

Case 2: \( y^2 = 1 \) (i.e., \( y = 1 \) or \( y = -1 \)) \[ 2x^2 + 3(1) = 35 \implies 2x^2 = 32 \implies x^2 = 16 \]
Since 16 is a perfect square, \( x = \pm \sqrt{16} = \pm 4 \).
This yields 4 solution pairs: \( (4, 1), (4, -1), (-4, 1), (-4, -1) \).

Case 3: \( y^2 = 4 \) (i.e., \( y = 2 \) or \( y = -2 \)) \[ 2x^2 + 3(4) = 35 \implies 2x^2 + 12 = 35 \implies 2x^2 = 23 \implies x^2 = 11.5 \]
No integer solutions for \( x \).

Case 4: \( y^2 = 9 \) (i.e., \( y = 3 \) or \( y = -3 \)) \[ 2x^2 + 3(9) = 35 \implies 2x^2 + 27 = 35 \implies 2x^2 = 8 \implies x^2 = 4 \]
Since 4 is a perfect square, \( x = \pm \sqrt{4} = \pm 2 \).
This yields 4 solution pairs: \( (2, 3), (2, -3), (-2, 3), (-2, -3) \).

Total number of integer solutions (elements in set X) = \( 4 (from Case 2) + 4 (from Case 4) = 8 \).


Step 4: Final Answer:

The number of elements in the set X is 8.
Quick Tip: For bounding Diophantine equations involving a sum of squares like \( ax^2 + by^2 = c \), always bound the variable with the larger coefficient first (here, \( y^2 \) has coefficient 3). This minimizes the number of cases you need to test.