Step 1: Understanding the Question:

The question asks to identify an element based on a given electronic configuration. However, the provided configuration `1s\(^2\)1p\(^4\)` is incorrect because the principal quantum number n=1 only allows for an angular momentum quantum number l=0 (the 's' orbital). There is no 'p' orbital (l=1) in the first energy level.


Step 2: Correcting the Configuration:

The solution provided in the source suggests the intended element is Sulfur. Let's verify this.

Sulfur (S) has an atomic number (Z) of 16. This means it has 16 protons and 16 electrons in a neutral atom.

Following the Aufbau principle, the correct electronic configuration for Sulfur is:
\[ 1s^2 2s^2 2p^6 3s^2 3p^4 \]

This configuration adds up to \(2+2+6+2+4 = 16\) electrons. The valence shell is the third shell (n=3), and it contains \(2+4=6\) valence electrons. The configuration ends in \(3p^4\).


Step 3: Final Answer:

Based on the likely typo in the question, the element with 16 electrons and the electron configuration \(1s^2 2s^2 2p^6 3s^2 3p^4\) is Sulfur.
Quick Tip: Always remember the rules for quantum numbers. For a principal quantum number 'n', the allowed values of 'l' are 0, 1, 2, ..., (n-1). For n=1, only l=0 (s orbital) is possible. For n=2, l=0 (s orbital) and l=1 (p orbital) are possible. This helps in quickly identifying invalid electronic configurations.

Step 1: Understanding the Question:

The question asks for the mathematical definition of enthalpy (H).


Step 2: Key Formula or Approach:

Enthalpy is a thermodynamic state function that represents the total heat content of a system. It is defined by the enthalpy equation.


Step 3: Detailed Explanation:

The expression for enthalpy (H) is given by:
\[ H = U + PV \]

Where:


H is the enthalpy of the system.

U is the internal energy of the system. Internal energy is the sum of all kinetic and potential energies of the particles within the system.

P is the pressure of the system.

V is the volume of the system.


The term PV represents the "pressure-volume work" or the energy required to establish the system's physical dimensions (its volume V) against the external pressure P. Thus, enthalpy accounts for both the internal energy and the work done to make room for the system. The change in enthalpy (\(\Delta H\)) at constant pressure is equal to the heat absorbed or released by the system (\(\Delta H = q_p\)).


Step 4: Final Answer:

The correct expression for enthalpy is H = U + PV.
Quick Tip: Remember that enthalpy (H) is particularly useful for processes occurring at constant pressure, as the change in enthalpy (\(\Delta H\)) directly equals the heat transferred. Internal energy (U) is more directly related to heat transfer at constant volume (\(\Delta U = q_v\)).

Step 1: Understanding the Question:

We are given the resistance (R) and voltage (V) in an electrical circuit and are asked to calculate the current (I).


Step 2: Key Formula or Approach:

This problem can be solved using Ohm's Law, which states the relationship between voltage, current, and resistance in a circuit.

The formula for Ohm's Law is:
\[ V = IR \]

To find the current, we can rearrange the formula to:
\[ I = \frac{V}{R} \]


Step 3: Detailed Explanation:

We are given the following values:


Voltage (V) = 3 V

Resistance (R) = 200 \(\Omega\)


Substitute these values into the rearranged Ohm's Law formula:
\[ I = \frac{3 V}{200 \Omega} \]
\[ I = 0.015 A \]

The question may also ask for the answer in milliamperes (mA). To convert from amperes to milliamperes, we multiply by 1000.
\[ I = 0.015 A \times 1000 \frac{mA}{A} = 15 mA \]


Step 4: Final Answer:

The current in the circuit is 0.015 A or 15 mA.
Quick Tip: Ohm's Law (V=IR) is a fundamental formula in electricity. Always ensure your units are in the standard SI form (Volts, Amperes, Ohms) before calculation to avoid errors.

Step 1: Understanding the Question:

The question asks about the relationship between the total energy of a particle undergoing Simple Harmonic Motion (SHM) and its physical properties like amplitude, frequency, etc.


Step 2: Key Formula or Approach:

The total energy (E) of a simple harmonic oscillator is the sum of its kinetic energy (KE) and potential energy (PE). At any point in the oscillation, the total energy remains constant. The formula for the total energy is:
\[ E = KE + PE = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 \]

The total energy is most easily expressed at the extreme position, where displacement \(x = A\) (amplitude) and velocity \(v = 0\). At this point, the total energy is purely potential.
\[ E = \frac{1}{2}kA^2 \]

Where:


E is the total energy

k is the spring constant (or force constant)

A is the amplitude of the oscillation


We also know that the angular frequency \(\omega = \sqrt{k/m}\), so \(k = m\omega^2\). Substituting this into the energy equation gives another form:
\[ E = \frac{1}{2}m\omega^2A^2 \]


Step 3: Detailed Explanation:

From both forms of the total energy equation, \(E = \frac{1}{2}kA^2\) and \(E = \frac{1}{2}m\omega^2A^2\), we can see that the total energy (E) is directly proportional to the square of the amplitude (A\(^2\)), assuming k, m, and \(\omega\) are constant for a given system.
\[ E \propto A^2 \]

This means if you double the amplitude of the oscillation, the total energy becomes four times greater.


Step 4: Final Answer:

The total energy of simple harmonic oscillations is directly proportional to the square of the amplitude.
Quick Tip: Remember the key proportionalities for SHM total energy: \(E \propto A^2\) (square of amplitude) and \(E \propto f^2\) (square of frequency, since \(\omega = 2\pi f\)). These are common points of confusion.

Step 1: Understanding the Question:

The question asks to identify the catalyst used in the industrial production of sulfuric acid (H\(_2\)SO\(_4\)) via the Contact Process.


Step 2: Key Concept - The Contact Process:

The Contact Process is the modern industrial method for manufacturing sulfuric acid. It involves three main steps:


Burning sulfur or sulfide ores in air to produce sulfur dioxide (SO\(_2\)).

Converting sulfur dioxide (SO\(_2\)) to sulfur trioxide (SO\(_3\)) using a catalyst. This is the key step.

Absorbing the sulfur trioxide (SO\(_3\)) in concentrated sulfuric acid to produce oleum, which is then diluted with water to obtain sulfuric acid of the desired concentration.



Step 3: Detailed Explanation:

The critical step that requires a catalyst is the reversible oxidation of sulfur dioxide to sulfur trioxide:
\[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \quad (\Delta H = -197 kJ/mol) \]

This reaction is slow and needs a catalyst to proceed at an economically viable rate. The catalyst used for this specific reaction is Vanadium pentoxide (V\(_2\)O\(_5\)).

The reaction is typically carried out at a temperature of about 450 \(^\circ\)C and a pressure of 1-2 atm to achieve a high yield (around 99.5%).


Step 4: Final Answer:

The catalyst used in the Contact Process to prepare sulfuric acid is Vanadium pentoxide (V\(_2\)O\(_5\)).
Quick Tip: Memorize the key industrial processes and their catalysts. For example: Contact Process (H\(_2\)SO\(_4\)) uses V\(_2\)O\(_5\), Haber-Bosch Process (NH\(_3\)) uses Iron (Fe) with a promoter, and Ostwald Process (HNO\(_3\)) uses Platinum-Rhodium (Pt-Rh) gauze.

Step 1: Understanding the Question:

We need to find the ratio of the angular speed of a clock's second hand (\(\omega_s\)) to the angular speed of its minute hand (\(\omega_m\)).


Step 2: Key Formula or Approach:

Angular speed (\(\omega\)) is defined as the rate of change of angular displacement. For a uniformly rotating object, it can be calculated as the total angle traversed divided by the time taken. For one complete revolution, the angle is \(2\pi\) radians, and the time taken is the time period (T).
\[ \omega = \frac{\Delta\theta}{\Delta t} = \frac{2\pi}{T} \]


Step 3: Detailed Explanation:

Angular speed of the second hand (\(\omega_s\)):

The second hand completes one full rotation (\(2\pi\) radians) in 60 seconds.

So, its time period, \(T_s = 60\) s.
\[ \omega_s = \frac{2\pi}{T_s} = \frac{2\pi}{60} rad/s \]


Angular speed of the minute hand (\(\omega_m\)):

The minute hand completes one full rotation (\(2\pi\) radians) in 60 minutes. We must convert this time to seconds for consistency.

So, its time period, \(T_m = 60 minutes \times 60 \frac{seconds}{minute} = 3600\) s.
\[ \omega_m = \frac{2\pi}{T_m} = \frac{2\pi}{3600} rad/s \]


Calculate the ratio:

We need to find the ratio \(\frac{\omega_s}{\omega_m}\).
\[ \frac{\omega_s}{\omega_m} = \frac{\frac{2\pi}{60}}{\frac{2\pi}{3600}} \]
\[ \frac{\omega_s}{\omega_m} = \frac{2\pi}{60} \times \frac{3600}{2\pi} \]

The \(2\pi\) terms cancel out.
\[ \frac{\omega_s}{\omega_m} = \frac{3600}{60} = \frac{60}{1} \]


Step 4: Final Answer:

The ratio of the angular speeds of the second hand to the minute hand is 60:1.
Quick Tip: When dealing with clock problems, it's crucial to convert all time periods to a common unit, usually seconds. The time periods for the second, minute, and hour hands are 60 s, 3600 s, and 43200 s (12 hours) respectively.

Step 1: Understanding the Question:

The question asks for the gravitational force on a particle at a certain position inside the Earth, given the force at another position. We assume the Earth is a sphere of uniform density.


Step 2: Key Formula or Approach:

The acceleration due to gravity (\(g'\)) at a distance \(r\) from the center of the Earth (where \(r < R\), the Earth's radius) is given by:
\[ g' = \frac{GMr}{R^3} \]

This shows that inside the Earth, the acceleration due to gravity is directly proportional to the distance from the center (\(g' \propto r\)).

The gravitational force (weight) on a particle of mass \(m\) is \(F = mg'\). Therefore, the force is also directly proportional to the distance from the center:
\[ F \propto r \]


Step 3: Detailed Explanation based on Interpretation:

The term 'depth d' is typically measured from the surface. However, this leads to a complex ratio. A common simplification in such problems is that 'd' is used to mean the distance from the center. Let's solve with this assumption.

Let the distance from the center be \(r\). So, \(F = kr\), where \(k\) is a constant of proportionality.

Case 1: At distance \(r_1 = d\), the force is \(F_1 = 300\) N.
\[ 300 = k \cdot d \quad \cdots(1) \]

Case 2: We need to find the force \(F_2\) at a distance \(r_2 = d/2\).
\[ F_2 = k \cdot (d/2) \quad \cdots(2) \]

Now, we can find the ratio of the forces:
\[ \frac{F_2}{F_1} = \frac{k(d/2)}{kd} = \frac{1}{2} \]
\[ F_2 = \frac{1}{2} F_1 = \frac{1}{2} \times 300 N = 150 N \]

This interpretation yields the answer 150 N, which is likely what was expected.

Alternative Interpretation (using 'd' as depth from surface):
\(r_1 = R - d\), so \(F_1 = k(R-d) = 300\).
\(r_2 = R - d/2\), so \(F_2 = k(R-d/2)\).

The ratio is \(\frac{F_2{300} = \frac{R-d/2}{R-d}\), which cannot be solved without knowing \(R\) and \(d\). This confirms the first interpretation is the intended one for a simple integer answer.


Step 4: Final Answer:

Assuming 'd' represents the distance from the center, the force at depth 'd/2' is 150 N.
Quick Tip: In gravitation problems, remember the two key variations of 'g': with altitude 'h' (approximated as \(g_h \approx g(1 - 2h/R)\)) and with depth 'd' (\(g_d = g(1 - d/R)\)). Inside the Earth, it's simpler to use the relation that gravity is proportional to the distance from the center, \(g' \propto r\).

Step 1: Understanding the Question:

We need to determine the magnitude of the electric field at any point inside a sphere that has charge residing on its surface. This applies to both a hollow charged sphere and a solid conducting sphere.


Step 2: Key Formula or Approach:

The most fundamental way to solve this is by using Gauss's Law of electrostatics. Gauss's Law states that the net electric flux (\(\Phi_E\)) through any closed surface (called a Gaussian surface) is equal to the net charge enclosed (\(q_{enc}\)) by the surface divided by the permittivity of free space (\(\epsilon_0\)).
\[ \Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\epsilon_0} \]


Step 3: Detailed Explanation:


Consider a point P inside the charged sphere at a distance \(r\) from the center, where \(r < R\) (R is the radius of the sphere).

To find the electric field at this point, we construct a spherical Gaussian surface of radius \(r\) that is concentric with the charged sphere and passes through point P.

The question states that all the charge resides on the surface of the sphere. Therefore, the net charge enclosed by our Gaussian surface (which is inside the sphere) is zero.

\[ q_{enc} = 0 \]

According to Gauss's Law:

\[ \oint \vec{E} \cdot d\vec{A} = \frac{0}{\epsilon_0} = 0 \]

For a spherical surface, due to symmetry, the electric field \(\vec{E}\) must be constant in magnitude and directed radially. Therefore, the integral simplifies to \(E \cdot (4\pi r^2)\).

\[ E \cdot (4\pi r^2) = 0 \]

Since the area of the Gaussian surface (\(4\pi r^2\)) is not zero, the magnitude of the electric field (E) must be zero.

\[ E = 0 \]


This holds true for any point inside the sphere. For a conducting sphere, any net charge given to it will automatically reside on its outer surface, and the electric field inside the conductor in electrostatic equilibrium is always zero.


Step 4: Final Answer:

The electric field inside a sphere with charge only on its surface is zero.
Quick Tip: Remember this fundamental result from Gauss's Law: The electric field inside any charged spherical shell or a solid conducting sphere is always zero. The electric potential, however, is constant and non-zero inside.

Step 1: Understanding the Question:

The question asks for the condition under which a series LCR (Inductor-Capacitor-Resistor) circuit experiences resonance.


Step 2: Key Concepts and Formulas:

In a series LCR circuit connected to an AC source, the total opposition to current flow is called impedance (Z), given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]

Where:


R = Resistance

\(X_L = \omega L\) = Inductive Reactance (proportional to frequency)

\(X_C = \frac{1}{\omega C}\) = Capacitive Reactance (inversely proportional to frequency)

\(\omega\) = Angular frequency of the AC source



Step 3: Detailed Explanation:

Resonance is a special condition in an LCR circuit where the current in the circuit reaches its maximum possible value. According to Ohm's law for AC circuits, \(I = V/Z\). The current (I) will be maximum when the impedance (Z) is minimum.

Looking at the impedance formula, \(Z = \sqrt{R^2 + (X_L - X_C)^2}\), the impedance Z will be at its minimum value when the term \((X_L - X_C)^2\) is zero. This occurs when:
\[ X_L - X_C = 0 \]

or
\[ X_L = X_C \]

This is the fundamental condition for resonance. At this point, the inductive and capacitive reactances cancel each other out, and the impedance of the circuit becomes equal to the resistance (\(Z_{min} = R\)). The circuit behaves like a purely resistive circuit.

We can also express this condition in terms of frequency. Substituting the formulas for \(X_L\) and \(X_C\):
\[ \omega_0 L = \frac{1}{\omega_0 C} \]

Solving for the resonant angular frequency, \(\omega_0\):
\[ \omega_0^2 = \frac{1}{LC} \implies \omega_0 = \frac{1}{\sqrt{LC}} \]


Step 4: Final Answer:

The correct condition for an LCR circuit to be at resonance is that the inductive reactance equals the capacitive reactance (\(X_L = X_C\)).
Quick Tip: At resonance, remember these key features: \(X_L = X_C\), impedance is minimum (\(Z=R\)), current is maximum (\(I_{max}=V/R\)), and the voltage and current are in phase (\(\phi = 0\)).

Step 1: Understanding the Question:

We need to find the packing efficiency for silver (Ag). Packing efficiency is the percentage of total space within a crystal lattice that is occupied by atoms. First, we need to know the crystal structure of silver. Silver crystallizes in a face-centered cubic (FCC) lattice, which is also known as cubic close-packed (ccp).


Step 2: Key Formula or Approach:

The formula for packing efficiency (PE) is:
\[ PE = \frac{Volume of atoms in one unit cell}{Total volume of the unit cell} \times 100% \]
\[ PE = \frac{Z \times \frac{4}{3}\pi r^3}{a^3} \times 100% \]

Where:


Z = number of atoms per unit cell

r = radius of the atom

a = edge length of the unit cell



Step 3: Detailed Explanation for FCC Lattice:


Find Z for FCC: In an FCC unit cell, there are atoms at the 8 corners and at the center of the 6 faces.

Contribution from corners: \(8 \times \frac{1}{8} = 1\) atom.

Contribution from faces: \(6 \times \frac{1}{2} = 3\) atoms.

Total atoms, \(Z = 1 + 3 = 4\).

Find the relationship between 'a' and 'r': In an FCC lattice, the atoms touch along the face diagonal. The length of the face diagonal is \(\sqrt{2}a\). This length is equal to four times the atomic radius (\(4r\)).

\[ \sqrt{2}a = 4r \implies a = \frac{4r}{\sqrt{2}} = 2\sqrt{2}r \]

Calculate Packing Efficiency:

Volume of atoms in unit cell = \(Z \times \frac{4}{3}\pi r^3 = 4 \times \frac{4}{3}\pi r^3 = \frac{16}{3}\pi r^3\).

Total volume of unit cell = \(a^3 = (2\sqrt{2}r)^3 = 16\sqrt{2}r^3\).

Now, substitute these into the PE formula:

\[ PE = \frac{\frac{16}{3}\pi r^3}{16\sqrt{2}r^3} \times 100% \]

The \(16\) and \(r^3\) terms cancel out.

\[ PE = \frac{\pi}{3\sqrt{2}} \times 100% \]

Using \(\pi \approx 3.14159\) and \(\sqrt{2} \approx 1.414\):

\[ PE \approx \frac{3.14159}{3 \times 1.414} \times 100% \approx 0.7404 \times 100% \approx 74% \]


This value is a constant for all elements that crystallize in an FCC/CCP structure.


Step 4: Final Answer:

The packing efficiency of silver metal is 74%.
Quick Tip: It is highly recommended to memorize the packing efficiencies for the three major cubic lattices: Simple Cubic (SC) = 52.4%, Body-Centered Cubic (BCC) = 68%, and Face-Centered Cubic (FCC/CCP) = 74%. These are very frequently asked.

Step 1: Understanding the Question:

We need to find the derivative of the given inverse trigonometric function with respect to x. The expression inside the `arccot` can be simplified first to make differentiation easier.


Step 2: Key Formula or Approach:

We will use trigonometric identities and properties of inverse trigonometric functions.


Trigonometric identity: \(\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\)

Inverse trigonometric property: \(\cot^{-1}(y) = \frac{\pi}{2} - \tan^{-1}(y)\)

Derivative property: \(\frac{d}{dx}(\tan^{-1}(\tan(f(x)))) = f'(x)\)



Step 3: Detailed Explanation:

Let \(y = \cot^{-1}\left(\frac{3+4\tan x}{4-3\tan x}\right)\).

First, simplify the expression inside the parentheses. Divide the numerator and the denominator by 4:
\[ \frac{3+4\tan x}{4-3\tan x} = \frac{\frac{3}{4}+\tan x}{1-\frac{3}{4}\tan x} \]

This expression is in the form of the \(\tan(A+B)\) identity. Let \(\tan A = \frac{3}{4}\). Then the expression becomes:
\[ \frac{\tan A + \tan x}{1 - \tan A \tan x} = \tan(A+x) \]

So, our function simplifies to:
\[ y = \cot^{-1}(\tan(A+x)) \]

Now, we can use the property \(\cot^{-1}(z) = \tan^{-1}(1/z)\). Another way is to use \(\tan(\theta) = \cot(\pi/2 - \theta)\).
\[ y = \cot^{-1}\left(\cot\left(\frac{\pi}{2} - (A+x)\right)\right) \]
\[ y = \frac{\pi}{2} - A - x \]

Now, we can differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{2} - A - x \right) \]

Since \(\frac{\pi}{2}\) is a constant, its derivative is 0. Also, \(A = \tan^{-1}(3/4)\) is a constant, so its derivative is also 0.
\[ \frac{dy}{dx} = 0 - 0 - 1 = -1 \]


Step 4: Final Answer:

The differentiation of the given expression is -1.
Quick Tip: Before applying complex differentiation rules like the chain rule or quotient rule to inverse trigonometric functions, always check if the inner expression can be simplified using trigonometric identities. This can often reduce the problem to differentiating a simple linear function.

Step 1: Understanding the Question:

The question asks how the concentration of charge carriers (both majority and minority) in a p-n junction diode changes when the temperature is raised.


Step 2: Key Concepts - Charge Carriers in a p-n Junction:


p-type semiconductor: Majority carriers are holes, minority carriers are electrons.

n-type semiconductor: Majority carriers are electrons, minority carriers are holes.

Majority carriers primarily come from dopant atoms and their concentration is relatively stable with temperature.

Minority carriers are generated by thermal energy breaking covalent bonds, creating electron-hole pairs. This process is highly sensitive to temperature.



Step 3: Detailed Explanation:

When the temperature of a semiconductor (and thus the p-n junction) is increased, the atoms in the crystal lattice vibrate more vigorously. This increased thermal energy becomes sufficient to break more covalent bonds throughout the material.

Each broken bond releases a free electron and creates a hole, generating an electron-hole pair. This process is called thermal generation.


Effect on Minority Carriers: The concentration of minority carriers is very low at room temperature. The thermally generated electrons in the p-side and holes in the n-side add to the existing small number of minority carriers. Because the initial number is so small, this addition causes a very large \textit{percentage increase in their concentration. The concentration of minority carriers approximately doubles for every 10\(^\circ\)C rise in temperature.

Effect on Majority Carriers: The concentration of majority carriers is determined by the high level of doping and is already very large. The number of thermally generated carriers is usually insignificant compared to the number of carriers provided by the dopant atoms. Therefore, the concentration of majority carriers increases only by a very small, often negligible, amount.


This significant increase in minority carriers leads to an increase in the reverse saturation current of the diode.


Step 4: Final Answer:

When the temperature of a p-n junction diode is increased, the concentration of minority charge carriers increases significantly, while the concentration of majority charge carriers shows a negligible increase.
Quick Tip: A key takeaway for semiconductor devices is that temperature primarily affects the minority carriers. This is why leakage currents (reverse saturation current) in diodes and transistors are highly dependent on temperature.

Step 1: Understanding the Question:

The question asks for a formula that relates the height (h) of a conical pendulum to its time period (T). A conical pendulum consists of a mass (bob) attached to a string, moving in a horizontal circle at a constant speed. The height 'h' is the vertical distance from the point of suspension to the plane of the circular motion.


Step 2: Key Formula or Approach:

We can derive the relationship by analyzing the forces acting on the pendulum bob.

Let:


\(m\) = mass of the bob

\(L\) = length of the string

\(\theta\) = angle the string makes with the vertical

\(T_{str}\) = tension in the string

\(r\) = radius of the horizontal circle

\(h\) = height of the pendulum


The forces acting on the bob are tension (\(T_{str}\)) and gravity (\(mg\)). We resolve the tension into vertical and horizontal components.


Vertical component: \(T_{str} \cos\theta\). This balances the weight: \(T_{str} \cos\theta = mg \quad \cdots(1)\)

Horizontal component: \(T_{str} \sin\theta\). This provides the necessary centripetal force (\(mv^2/r\) or \(m\omega^2r\)): \(T_{str} \sin\theta = m\omega^2r \quad \cdots(2)\)


From the geometry of the pendulum, we have \(h = L\cos\theta\) and \(r = L\sin\theta\).


Step 3: Detailed Explanation:

Divide equation (2) by equation (1):
\[ \frac{T_{str} \sin\theta}{T_{str} \cos\theta} = \frac{m\omega^2r}{mg} \]
\[ \tan\theta = \frac{\omega^2r}{g} \]

From trigonometry, \(\tan\theta = \frac{r}{h}\). Substitute this into the equation:
\[ \frac{r}{h} = \frac{\omega^2r}{g} \]

Cancel \(r\) from both sides:
\[ \frac{1}{h} = \frac{\omega^2}{g} \implies \omega^2 = \frac{g}{h} \implies \omega = \sqrt{\frac{g}{h}} \]

Now, relate the angular frequency (\(\omega\)) to the time period (T):
\[ \omega = \frac{2\pi}{T} \]

Substitute this into our derived equation:
\[ \frac{2\pi}{T} = \sqrt{\frac{g}{h}} \]

To solve for the height (h), we square both sides:
\[ \frac{4\pi^2}{T^2} = \frac{g}{h} \]

Rearranging the terms to make 'h' the subject:
\[ h = \frac{gT^2}{4\pi^2} \]


Step 4: Final Answer:

The height of a conical pendulum is related to its time period by the formula \(h = \frac{gT^2}{4\pi^2}\).
Quick Tip: Notice that the time period of a conical pendulum, \(T = 2\pi\sqrt{h/g}\), depends only on its vertical height 'h', not its length L or the angle \(\theta\) directly. This is similar to a simple pendulum's period, \(T = 2\pi\sqrt{L/g}\).

Step 1: Understanding the Question:

The question asks for the specific rotation of glucose. Glucose is a chiral molecule and is optically active, meaning it rotates the plane of polarized light. Specific rotation, \([\alpha]_D\), is a standardized measure of this rotation.


Step 2: Key Concepts - Anomers and Mutarotation:

In nature, glucose exists predominantly as D-glucose. When crystalline D-glucose is dissolved in water, it exists in two cyclic hemiacetal forms, called anomers:


\(\alpha\)-D-glucose

\(\beta\)-D-glucose


These two anomers have different initial specific rotations. However, in an aqueous solution, they slowly interconvert via the open-chain form until a stable equilibrium is reached. This change in optical rotation over time is called mutarotation.


Step 3: Detailed Explanation:

The specific rotation values are:


Pure \(\alpha\)-D-glucose has an initial specific rotation of \([\alpha]_D = +112.2^{\circ}\).

Pure \(\beta\)-D-glucose has an initial specific rotation of \([\alpha]_D = +18.7^{\circ}\).


When either anomer is dissolved in water, mutarotation occurs, and the specific rotation of the solution changes until it reaches a constant equilibrium value.

The equilibrium mixture consists of approximately 36% \(\alpha\)-D-glucose and 64% \(\beta\)-D-glucose (with a very small amount of the open-chain form). The specific rotation of this equilibrium mixture is:
\[ [\alpha]_D = +52.7^{\circ} \]

When a question asks for "the" specific rotation of glucose without specifying the anomer, it is standard convention to refer to this final, stable equilibrium value. The positive sign indicates that glucose is dextrorotatory, which is why it is also known as dextrose.


Step 4: Final Answer:

The specific rotation of an equilibrium solution of D-glucose is +52.7\(^{\circ}\).
Quick Tip: Remember that the general term "glucose" usually refers to D-glucose. The positive sign (+) in its specific rotation means it's dextrorotatory. This is a common fact-based question in the biomolecules chapter.