Step 1: Understanding the Concept:

The Doppler effect in sound describes the change in frequency perceived by an observer when there is relative motion between the source and the observer.

If both the source (the train's whistle) and the observer (the passenger inside the train) move with the same velocity, there is no relative motion between them.


Step 2: Key Formula or Approach:

The apparent frequency \( f' \) is given by:
\[ f' = f \left( \frac{v \pm v_o}{v \mp v_s} \right) \]

where:
\( v = \) velocity of sound in the medium
\( v_o = \) velocity of the observer
\( v_s = \) velocity of the source


Step 3: Detailed Explanation:

In this problem:

1. The passenger is inside the train, so they are moving with the train.

2. Therefore, the velocity of the observer \( v_o \) is equal to the velocity of the source \( v_s \) in both magnitude and direction (\( v_o = v_s = 10 ms^{-1} \)).

3. Plugging these into the formula (assuming the observer is ahead of the source in the direction of motion):
\[ f' = 400 \left( \frac{330 + 10}{330 + 10} \right) = 400 Hz \]

Since the relative velocity between the passenger and the whistle is zero, the passenger will hear the original frequency.


Step 4: Final Answer:

The frequency heard by the passenger is \( 400 Hz \).
Quick Tip: Doppler effect only occurs when there is \textbf{relative motion} between the source and the observer. If you are traveling inside the vehicle that produces the sound, you will always hear the true frequency regardless of the vehicle's speed.

Step 1: Understanding the Concept:

Since the temperature is constant throughout the lake, we can apply Boyle's Law (\( PV = constant \)) to the air bubble as it rises. As the bubble rises, the external hydrostatic pressure decreases, causing the volume to increase.


Step 2: Key Formula or Approach:

1. Boyle's Law: \( P_1 V_1 = P_2 V_2 \)

2. Total pressure at depth \( h \): \( P = P_{atm} + \rho gh \)


Step 3: Detailed Explanation:

At the bottom (State 1):

Depth \( h = 40 m \)
\( V_1 = 1 cm^3 \)
\( P_1 = P_{atm} + \rho gh = 10^5 + (1000 \times 10 \times 40) \)
\( P_1 = 1 \times 10^5 + 4 \times 10^5 = 5 \times 10^5 Pa \)


At the surface (State 2):
\( P_2 = P_{atm} = 1 \times 10^5 Pa \)


Applying \( P_1 V_1 = P_2 V_2 \):
\[ (5 \times 10^5 Pa) \times (1 cm^3) = (1 \times 10^5 Pa) \times V_2 \]
\[ V_2 = \frac{5 \times 10^5}{1 \times 10^5} \times 1 cm^3 = 5 cm^3 \]


Step 4: Final Answer:

The volume of the bubble at the surface is \( 5 cm^3 \).
Quick Tip: For every \( 10 m \) of water depth, the pressure increases by approximately \( 1 atm \) (\( 10^5 Pa \)). At \( 40 m \), the gauge pressure is \( 4 atm \), making the total pressure \( 1 (atm) + 4 = 5 atm \). Since pressure is \( 5 \) times higher at the bottom, volume must be \( 5 \) times larger at the surface.

Step 1: Understanding the Concept:

Statement I relates heat to temperature, while Statement II relates work to volume changes in a thermodynamic system. We evaluate these based on the First Law of Thermodynamics and the definition of work.


Step 2: Detailed Explanation:

Analysis of Statement I:

Heat added to a system (\( \Delta Q \)) does not always increase temperature. During a phase change (like boiling water or melting ice), heat is added but the temperature remains constant. Similarly, in an isothermal expansion, heat added is entirely converted to work, keeping \( \Delta T = 0 \). Thus, Statement I is false.


Analysis of Statement II:

The work done by a system is defined as \( W = \int P dV \). If the system does positive work (\( W > 0 \)), then \( dV \) must be positive (assuming pressure \( P \) is positive). This indicates an expansion, which means the volume must increase. Thus, Statement II is true.


Step 3: Final Answer:

Statement I is false and Statement II is true.
Quick Tip: Be careful with words like "must" in science. Isothermal processes and phase changes are standard exceptions to the idea that "heat equals temperature rise."

Step 1: Understanding the Concept:

Elastic potential energy is stored in a material when it is deformed. For a solid under tension or compression, this energy is quantified as energy density (energy per unit volume).


Step 2: Key Formula or Approach:

Energy density (\( u \)) stored in a wire is given by:
\[ u = \frac{1}{2} \times Stress \times Strain = \frac{1}{2} \times Y \times (Strain)^2 \]


Step 3: Detailed Explanation:

Given data:

Young's modulus \( Y = 7.0 \times 10^{10} N/m^2 \)

Strain \( \epsilon = 0.04% = \frac{0.04}{100} = 4 \times 10^{-4} \)


Substitute into the energy density formula:
\[ u = \frac{1}{2} \times (7.0 \times 10^{10}) \times (4 \times 10^{-4})^2 \]
\[ u = 0.5 \times 7.0 \times 10^{10} \times (16 \times 10^{-8}) \]
\[ u = 3.5 \times 16 \times 10^{(10 - 8)} \]
\[ u = 56 \times 10^2 = 5600 J/m^3 \]


Step 4: Final Answer:

The energy per unit volume stored is \( 5600 \) SI units.
Quick Tip: When dealing with percentages in strain, always convert to a decimal immediately (\( % / 100 \)) before squaring to avoid power-of-ten errors.

Step 1: Understanding the Concept:

For a satellite of mass \( m \) revolving around Earth (mass \( M \)) at radius \( r \), we analyze the relationships between Potential Energy (\( U \)), Kinetic Energy (\( K \)), and Total Energy (\( E \)).


Step 2: Key Formula or Approach:

The energy relations are:

1. Potential Energy \( U = -\frac{GMm}{r} \)

2. Kinetic Energy \( K = \frac{GMm}{2r} \)

3. Total Energy \( E = K + U = -\frac{GMm}{2r} \)


From these, we derive:
\( K = -E = |E| \)
\( U = 2E \)


Step 3: Detailed Explanation:

Analysis of Statement I:

Potential energy \( U \) is \( 2E \). The statement says it is \( \frac{E}{2} \). Therefore, Statement I is incorrect.


Analysis of Statement II:

Kinetic energy \( K \) is equal to the magnitude of the total energy (\( K = |E| \)). The statement says it is half the magnitude (\( \frac{|E|}{2} \)). Therefore, Statement II is incorrect.


Step 4: Final Answer:

Both Statement I and Statement II are incorrect.
Quick Tip: Remember the \( 1:2:-1 \) ratio for satellite energies: \( K : U : E = 1 : -2 : -1 \). This means \( U \) is always twice \( E \) and \( K \) is always equal in magnitude to \( E \).

Step 1: Understanding the Concept:

According to Newton's Second Law, the force acting on a particle is the product of its mass and acceleration (\( \vec{F} = m\vec{a} \)). Acceleration is the time derivative of velocity (\( \vec{a} = \frac{d\vec{v}}{dt} \)).


Step 2: Key Formula or Approach:

1. \( \vec{a} = \frac{d\vec{v}}{dt} \)

2. \( \vec{F} = m\vec{a} \)


Step 3: Detailed Explanation:

Given:
\( \vec{v} = 2t\hat{i} + 3t^2\hat{j} \)

Mass \( m = 500 g = 0.5 kg \)


Step 1: Find acceleration vector \( \vec{a} \).
\[ \vec{a} = \frac{d}{dt}(2t\hat{i} + 3t^2\hat{j}) = 2\hat{i} + 6t\hat{j} \]


Step 2: Find acceleration at \( t = 1 s \).
\[ \vec{a}_{t=1} = 2\hat{i} + 6(1)\hat{j} = 2\hat{i} + 6\hat{j} ms^{-2} \]


Step 3: Find force vector \( \vec{F} \).
\[ \vec{F} = m\vec{a} = 0.5 \times (2\hat{i} + 6\hat{j}) = 1\hat{i} + 3\hat{j} N \]


Comparing with the given force \( (\hat{i} + x\hat{j}) \), we find \( x = 3 \).


Step 4: Final Answer:

The value of \( x \) is \( 3 \).
Quick Tip: Always convert mass to SI units (kg) before calculating force in Newtons. Failing to do so is a common source of calculation error in physics exams.

Step 1: Understanding the Concept:

The resultant of two vectors \( \vec{P} \) and \( \vec{Q} \) is found using the parallelogram law. When the vectors are perpendicular (\( \theta = 90^\circ \)), the magnitude of the resultant simplifies to the Pythagorean theorem.


Step 2: Key Formula or Approach:

For perpendicular vectors:
\[ R = \sqrt{P^2 + Q^2} \]


Step 3: Detailed Explanation:

Given:

Magnitude of first force \( P = A \)

Magnitude of second force \( Q = \frac{A}{2} \)


Substitute into the resultant formula:
\[ R = \sqrt{A^2 + \left(\frac{A}{2}\right)^2} \]
\[ R = \sqrt{A^2 + \frac{A^2}{4}} \]
\[ R = \sqrt{\frac{4A^2 + A^2}{4}} = \sqrt{\frac{5A^2}{4}} \]
\[ R = \frac{\sqrt{5}A}{2} \]


Step 4: Final Answer:

The magnitude of the resultant is \( \frac{\sqrt{5}A}{2} \).
Quick Tip: For any perpendicular vectors with a ratio \( 1 : 1/2 \), the resultant is always \( \sqrt{1^2 + (1/2)^2} = \sqrt{1.25} \approx 1.118 \) times the larger vector.

Step 1: Understanding the Concept:

The acceleration due to gravity (\( g \)) varies with depth inside the Earth. As you go deeper, the mass of the Earth effectively contributing to the gravitational pull decreases linearly.


Step 2: Key Formula or Approach:

Acceleration due to gravity at depth \( d \):
\[ g_d = g \left(1 - \frac{d}{R}\right) \]

Weight at depth \( d \): \( W_d = m g_d = W \left(1 - \frac{d}{R}\right) \)


Step 3: Detailed Explanation:

Given:

Weight on surface \( W = 400 N \)

Depth \( d = \frac{R}{2} \) (where \( R \) is the radius of Earth)


Calculate weight at depth \( d \):
\[ W_d = 400 \left(1 - \frac{R/2}{R}\right) \]
\[ W_d = 400 \left(1 - \frac{1}{2}\right) \]
\[ W_d = 400 \times \frac{1}{2} = 200 N \]


Step 4: Final Answer:

The weight of the body at the given depth is \( 200 N \).
Quick Tip: At the center of the Earth (\( d = R \)), the weight is zero. Since the variation with depth is strictly linear, going halfway to the center reduces the weight by exactly half.

Step 1: Understanding the Concept:

The range of a projectile is the horizontal distance it covers during its flight. It depends on both the initial speed and the angle of projection.


Step 2: Key Formula or Approach:

Horizontal Range \( R = \frac{u^2 \sin(2\theta)}{g} \)


Step 3: Detailed Explanation:

For Projectile A:
\( u_A = 40 m/s \), \( \theta_A = 30^\circ \)
\( R_A = \frac{40^2 \sin(2 \times 30^\circ)}{g} = \frac{1600 \sin(60^\circ)}{g} = \frac{1600 \times (\sqrt{3}/2)}{g} \)


For Projectile B:
\( u_B = 60 m/s \), \( \theta_B = 60^\circ \)
\( R_B = \frac{60^2 \sin(2 \times 60^\circ)}{g} = \frac{3600 \sin(120^\circ)}{g} = \frac{3600 \times (\sqrt{3}/2)}{g} \)

(Note: \( \sin 120^\circ = \sin(180^\circ - 60^\circ) = \sin 60^\circ \))


Calculating the ratio:
\[ \frac{R_A}{R_B} = \frac{1600 \times (\sqrt{3}/2) / g}{3600 \times (\sqrt{3}/2) / g} \]
\[ \frac{R_A}{R_B} = \frac{1600}{3600} = \frac{16}{36} = \frac{4}{9} \]


Step 4: Final Answer:

The ratio of their ranges is \( 4 : 9 \).
Quick Tip: Complementary angles (\( \theta \) and \( 90^\circ - \theta \)) result in the same value for \( \sin(2\theta) \). Therefore, if projection speeds were equal, the ranges would be equal. Here, the ratio simplifies solely to the square of the ratio of their velocities.

Step 1: Understanding the Concept:

Density (\( \rho \)) is mass divided by volume. For a cylinder, volume is \( \pi r^2 l \). Errors in measurements propagate through the formula according to error combination rules for products and quotients.


Step 2: Key Formula or Approach:

Density \( \rho = \frac{m}{\pi r^2 l} \)

Maximum relative error:
\[ \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l} \]


Step 3: Detailed Explanation:

Given:
\( \frac{\Delta m}{m} = \frac{0.01}{0.4} = \frac{1}{40} = 2.5% \)
\( \frac{\Delta l}{l} = \frac{0.04}{8} = \frac{4}{800} = 0.5% \)
\( \frac{\Delta r}{r} = \frac{0.03}{6} = \frac{3}{600} = 0.5% \)


Substituting into the relative error formula:
\[ % error in \rho = (2.5%) + 2(0.5%) + (0.5%) \]
\[ % error in \rho = 2.5% + 1.0% + 0.5% = 4% \]


Step 4: Final Answer:

The maximum error in density is \( 4% \).
Quick Tip: Remember that the power in the formula becomes a coefficient in the error term. Since radius \( r \) is squared (\( r^2 \)), its relative error is doubled.

Step 1: Understanding the Concept:

The coverage range of a transmitting antenna is limited by the curvature of the Earth. The range is the distance to the horizon from the top of the antenna. The area covered is the area of the circle with this range as the radius.


Step 2: Key Formula or Approach:

1. Range \( d = \sqrt{2Rh} \)

2. Area \( A = \pi d^2 = \pi (2Rh) \)


Step 3: Detailed Explanation:

Given:

Height of antenna \( h = 98 m = 0.098 km \)

Radius of Earth \( R = 6400 km \)


Calculate the covered area:
\[ A = \pi \times 2 \times 6400 \times 0.098 \]
\[ A = 3.1416 \times 12800 \times 0.098 \]
\[ A = 3.1416 \times 1254.4 \]
\[ A \approx 3940.8 km^2 \]


Rounding to the nearest option gives \( 3942 km^2 \).


Step 4: Final Answer:

The surface area covered is approximately \( 3942 km^2 \).
Quick Tip: Avoid calculating the range \( d \) separately if you only need the area. Using \( A = 2\pi Rh \) directly saves time and reduces rounding errors.

Step 1: Understanding the Concept:

The given circuit consists of three NAND gates.

The first two NAND gates have their inputs shorted, which makes them behave as NOT gates.

The outputs of these NOT gates are then fed into a third NAND gate.


Step 2: Key Formula or Approach:

Let the inputs be \( A \) and \( B \).

Output of first gate \( = \overline{A} \).

Output of second gate \( = \overline{B} \).

Final output \( Y = \overline{\overline{A} \cdot \overline{B}} \).

Using De Morgan's Law: \( \overline{\overline{A} \cdot \overline{B}} = \overline{\overline{A}} + \overline{\overline{B}} = A + B \).

Thus, the circuit acts as an OR gate.


Step 3: Detailed Explanation:

Now, let's analyze the input waveforms to determine the output \( Y = A OR B \):

1. For interval \( 0 \le t < 1 \): \( A = 0, B = 0 \). Output \( Y = 0 OR 0 = 0 \).

2. For interval \( 1 \le t < 2 \): \( A = 0, B = 1 \). Output \( Y = 0 OR 1 = 1 \).

3. For interval \( 2 \le t < 3 \): \( A = 1, B = 0 \). Output \( Y = 1 OR 0 = 1 \).

4. For interval \( 3 \le t < 4 \): \( A = 1, B = 1 \). Output \( Y = 1 OR 1 = 1 \).

The resulting waveform is 0 for \( t < 1 \) and 1 for \( t > 1 \).


Step 4: Final Answer:

The output waveform matches Option (A).
Quick Tip: A NAND gate with tied inputs is a NOT gate.
NAND followed by NOT inputs is equivalent to an OR gate (Bubbled NAND = OR).

Step 1: Understanding the Concept:

The Liquid Drop Model provides the Semi-Empirical Mass Formula (SEMF) for nuclear binding energy.

The total binding energy is given by \( BE = E_v - E_s - E_c - E_a \pm E_p \).


Step 2: Key Formula or Approach:

1. Volume Energy: \( E_v = a_v A \). (Directly proportional to nucleon count).

2. Surface Energy: \( E_s = a_s A^{2/3} \). (Reduces binding energy; proportional to surface area \( \propto R^2 \propto A^{2/3} \)).

3. Coulomb Energy: \( E_c = a_c \frac{Z(Z-1)}{A^{1/3}} \). (Repulsive force between protons reduces binding energy).


Step 3: Detailed Explanation:

Analysis of statements:

A. Incorrect. The surface energy contribution to binding energy is \( -a_s A^{2/3 \). Surface energy per nucleon would be \( \propto A^{-1/3 \).

B. Correct. This is the standard Coulomb term representing the work done against electrostatic repulsion.

C. Correct. Volume energy is the dominant term proportional to volume (\( \propto A \)).

D. Correct. Surface nucleons have fewer neighbors, reducing total binding energy. This effect is proportional to the surface area of the "liquid drop".

E. Incorrect. The "12 nucleons" interaction assumption is for bulk properties in crystal lattices, not a general requirement for SEMF surface energy estimation.


Step 4: Final Answer:

Statements B, C, and D are correct.
Quick Tip: Remember the dependencies on A:
Volume \( \propto A \), Surface \( \propto A^{2/3} \), Coulomb \( \propto A^{-1/3} \).

Step 1: Understanding the Concept:

The de-Broglie wavelength \( \lambda \) of a particle is related to its linear momentum \( p \) by the relation \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant.


Step 2: Detailed Explanation:

Given that the de-Broglie wavelengths are equal:
\[ \lambda_p = \lambda_e \]

Substituting the formula:
\[ \frac{h}{p_p} = \frac{h}{p_e} \]

This directly implies:
\[ p_p = p_e \]

Therefore, the ratio of their momentum \( \frac{p_p}{p_e} \) must be \( 1 : 1 \).

The mass ratio provided in the question is irrelevant for the momentum ratio when the wavelengths are explicitly stated as the same.


Step 3: Final Answer:

The ratio of their momentum is 1 : 1.
Quick Tip: If wavelength is constant, momentum is constant regardless of the mass of the particle. If kinetic energy was constant, then mass would matter.

Step 1: Understanding the Concept:

Reflecting telescopes (like the Cassegrain or Newtonian) use a large primary parabolic mirror to gather light.

Because the focal point of the primary mirror is inside the path of incoming light, a secondary mirror is needed to redirect the light.


Step 2: Detailed Explanation:

1. Light enters the tube and reflects off the primary mirror at the back.

2. Without a secondary mirror, the observer would have to sit at the focal point inside the tube, blocking incoming light.

3. A secondary mirror (flat in Newtonian, convex in Cassegrain) reflects the light to a location where it can be conveniently viewed by an eyepiece located outside the primary light path (on the side or through a hole in the primary).


Step 3: Final Answer:

The secondary mirror is primarily used to move the eyepiece outside the telescopic tube for easy viewing.
Quick Tip: Reflecting telescopes naturally have zero chromatic aberration because mirrors don't refract light. Parabolic mirrors also eliminate spherical aberration.

Step 1: Understanding the Concept:

From Maxwell's equations for electromagnetic waves in vacuum, the speed of light \( c \) is fundamentally related to the permittivity \( \epsilon_0 \) and permeability \( \mu_0 \) of free space.


Step 2: Key Formula or Approach:

The speed of light is given by:
\[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \]


Step 3: Detailed Explanation:

Squaring both sides of the speed equation:
\[ c^2 = \frac{1}{\mu_0 \epsilon_0} \]

We know the dimension of speed \( [c] = [L T^{-1}] \).

Therefore, the dimension of \( c^2 \) is:
\[ [c^2] = [L T^{-1}]^2 = [L^2 T^{-2}] \]

In fraction notation, this is \( L^2/T^2 \).


Step 4: Final Answer:

The dimensions of \( \frac{1}{\mu_0 \epsilon_0} \) are \( L^2/T^2 \).
Quick Tip: The term \( \frac{1}{\sqrt{\mu_0 \epsilon_0}} \) has dimensions of velocity. Any power of this term will have the corresponding power of velocity dimensions.

Step 1: Understanding the Concept:

This question refers to the damping mechanism in a galvanometer. When a metallic core (like aluminum) is used, eddy currents are induced in it as the coil moves.


Step 2: Detailed Explanation:

1. When current flows, the coil rotates, and if it has a metallic frame or core, the flux through it changes.

2. According to Lenz's law, induced eddy currents produce a torque that opposes the motion.

3. This is called electromagnetic damping.

4. The result is "dead-beat" behavior, where the pointer stops at its final position without oscillating for a long time.


Step 3: Final Answer:

The function is to bring the coil to rest quickly using electromagnetic damping.
Quick Tip: Damping turns kinetic energy into heat via eddy currents. Soft iron cores make the field radial, but "non-magnetic metallic" specifically points to damping.

Step 1: Understanding the Concept:

The motion of a charge \( q \) in a magnetic field \( \vec{B} \) is determined by the Lorentz force \( \vec{F} = q(\vec{v} \times \vec{B}) \).


Step 2: Detailed Explanation:

Let the velocity have two components: \( v_{\parallel} \) (parallel to B) and \( v_{\perp} \) (perpendicular to B).

1. Force due to \( v_{\parallel} \): \( F_{\parallel} = q v_{\parallel} B \sin(0^\circ) = 0 \). The particle continues to move linearly in the direction of B with constant speed \( v_{\parallel} \).

2. Force due to \( v_{\perp} \): \( F_{\perp} = q v_{\perp} B \sin(90^\circ) = q v_{\perp} B \). This force is perpendicular to both \( v_{\perp} \) and B, causing the particle to move in a circle in the plane perpendicular to B.

3. Combined Motion: The combination of a linear translation along B and a circular motion perpendicular to B results in a helix.

The "axis" of the helix is the direction about which the particle rotates, which is the direction of the magnetic field B.


Step 3: Final Answer:

The path is a helical path with the axis along magnetic field B.
Quick Tip: If \( \theta = 0^\circ or 180^\circ \), path is a straight line.
If \( \theta = 90^\circ \), path is a circle.
For any other \( \theta \), path is a helix.

Step 1: Understanding the Concept:

In a steady-state DC circuit, capacitors act as open circuits (infinite resistance). No current flows through branches containing capacitors.


Step 2: Key Formula or Approach:

First, calculate the current through the resistive part of the circuit (the galvanometer branch).

Let nodes be A (positive terminal), B (junction before G), C (junction after G), and D (negative terminal).


Step 3: Detailed Explanation:

1. Resistors in series: \( R_{total} = R_{AB} + R_G + R_{CD} = 6 \Omega + 2 \Omega + 8 \Omega = 16 \Omega \).

2. Current in circuit: \( I = \frac{V}{R} = \frac{4 V}{16 \Omega} = 0.25 A \).

3. Potentials at nodes (assuming \( V_D = 0 \)):

- \( V_A = 4 V \).

- \( V_B = V_A - I \cdot R_{AB} = 4 - (0.25 \cdot 6) = 2.5 V \).

- \( V_C = V_B - I \cdot R_G = 2.5 - (0.25 \cdot 2) = 2.0 V \).

4. Potential difference across capacitors:

- \( C_1 \) is between A and C: \( \Delta V_1 = V_A - V_C = 4 - 2.0 = 2.0 V \).

- \( C_2 \) is between B and D: \( \Delta V_2 = V_B - V_D = 2.5 - 0 = 2.5 V \).

5. Ratio: \( \frac{\Delta V_1}{\Delta V_2} = \frac{2.0}{2.5} = \frac{20}{25} = \frac{4}{5} \).


Step 4: Final Answer:

The ratio of potential difference across \( C_1 \) and \( C_2 \) is \( 4/5 \).
Quick Tip: Treat capacitors as broken wires when solving for potentials in DC steady state. Only find the voltage at the nodes they are connected to.

Step 1: Understanding the Concept:

We use Gauss's Law to find the electric field inside and outside a non-conducting solid sphere with uniform volume charge density \( \rho \).


Step 2: Key Formula or Approach:

1. Inside (\( r \le R \)): \( E = \frac{\rho r}{3\epsilon_0} \). Thus, \( E \propto r \) (Linear).

2. Outside (\( r > R \)): \( E = \frac{kQ}{r^2} \). Thus, \( E \propto \frac{1}{r^2} \) (Inverse Square).


Step 3: Detailed Explanation:

- At the center (\( r = 0 \)), the electric field is zero.

- As we move from the center to the surface, the field increases linearly with distance.

- At the surface (\( r = R \)), the field reaches its maximum value \( E_{max} = \frac{kQ}{R^2} \).

- For points outside the sphere, it behaves as a point charge located at the center, and the field decreases following the inverse square law.

The graph must start at the origin, go up as a straight line to \( R \), and then curve down towards zero.


Step 4: Final Answer:

This behavior is correctly depicted in the graph of Option (D).
Quick Tip: For a hollow conducting sphere, \( E_{inside} = 0 \). For a solid insulating sphere, \( E_{inside} \) is linear. This is a common point of confusion.

Step 1: Understanding the Concept:

An organ pipe open at both ends supports standing waves where antinodes are formed at the open ends.

The fundamental frequency corresponds to \(n=1\), and the second harmonic corresponds to \(n=2\).


Step 2: Key Formula or Approach:

For an open organ pipe, the frequency of the \(n^{th}\) harmonic is given by:
\[ f_n = \frac{nv}{2L} \]

where:
\(n =\) harmonic number
\(v =\) speed of sound
\(L =\) length of the pipe


Step 3: Detailed Explanation:

Given:
\(L = 40 cm = 0.4 m\)
\(v = 360 ms^{-1}\)

For the second harmonic, \(n = 2\):
\[ f_2 = \frac{2 \times 360}{2 \times 0.4} \]
\[ f_2 = \frac{360}{0.4} \]
\[ f_2 = 900 Hz \]


Step 4: Final Answer:

The frequency of the second harmonic is 900 Hz.
Quick Tip: For an open-open pipe, the second harmonic is simply the speed of sound divided by the length of the pipe (\(v/L\)). This saves calculation time during the exam.

Step 1: Understanding the Concept:

When an air bubble rises steadily, the net force acting on it is zero. The upward buoyant force is balanced by the downward viscous drag (Stokes' drag) and weight. Since the density of air is neglected, the upward force is buoyancy and the downward force is the viscous force.


Step 2: Key Formula or Approach:

The terminal velocity \(v_t\) of a sphere of radius \(r\) in a liquid of viscosity \(\eta\) is:
\[ v_t = \frac{2r^2 (\rho - \sigma)g}{9\eta} \]

Since density of air (\(\sigma\)) is neglected:
\[ \eta = \frac{2r^2 \rho g}{9v_t} \]


Step 3: Detailed Explanation:

Given:

Diameter \(d = 6 mm \Rightarrow\) Radius \(r = 3 mm = 3 \times 10^{-3} m\)

Density of solution \(\rho = 1750 kg/m^3\)

Terminal velocity \(v_t = 0.35 cm/s = 0.35 \times 10^{-2} ms^{-1}\)
\(g = 10 ms^{-2}\)

Substituting the values:
\[ \eta = \frac{2 \times (3 \times 10^{-3})^2 \times 1750 \times 10}{9 \times (0.35 \times 10^{-2})} \]
\[ \eta = \frac{2 \times 9 \times 10^{-6} \times 17500}{9 \times 3.5 \times 10^{-3}} \]
\[ \eta = \frac{2 \times 10^{-6} \times 17500}{3.5 \times 10^{-3}} \]
\[ \eta = \frac{35000 \times 10^{-6}}{3.5 \times 10^{-3}} \]
\[ \eta = \frac{3.5 \times 10^4 \times 10^{-6}}{3.5 \times 10^{-3}} \]
\[ \eta = 10^{4 - 6 + 3} = 10^1 = 10 Pas \]


Step 4: Final Answer:

The coefficient of viscosity is 10 Pas.
Quick Tip: Always check the units carefully. Converting \(0.35 cm/s\) to \(m/s\) and \(3 mm\) to \(m\) correctly is crucial for numerical questions.

Step 1: Understanding the Concept:

The moment of inertia (\(I\)) of a body about an axis is the sum of the products of the mass of each particle and the square of its distance from the axis.


Step 2: Key Formula or Approach:
\[ I = \int r^2 dm \]


Step 3: Detailed Explanation:

For a semicircular ring, every small element of mass \(dm\) is located at exactly the same distance \(R\) from the center of the ring.

If the axis passes through the center and is perpendicular to the plane of the ring, the distance \(r\) for every element is constant and equal to \(R\).
\[ I = \int R^2 dm \]

Since \(R\) is constant:
\[ I = R^2 \int dm \]

The integral of \(dm\) over the entire semicircular ring is its total mass \(M\).
\[ I = MR^2 \]

Comparing this with the given expression \(\frac{1}{x}MR^2\):
\(\frac{1}{x} = 1 \Rightarrow x = 1\).


Step 4: Final Answer:

The value of \(x\) is 1.
Quick Tip: For any thin circular arc (full, half, or even a quadrant) of mass \(M\) and radius \(R\), the moment of inertia about its center of curvature (perpendicular to its plane) is always \(MR^2\) because all mass is at distance \(R\).

Step 1: Understanding the Concept:

Kinetic energy and linear momentum of a body are related through its mass. When momentum changes, kinetic energy changes proportionally to the square of the momentum change.


Step 2: Key Formula or Approach:

The relationship between Kinetic Energy (\(K\)) and Momentum (\(p\)) is:
\[ K = \frac{p^2}{2m} \]


Step 3: Detailed Explanation:

Let the initial momentum be \(p_1\) and initial kinetic energy be \(K_1 = \frac{p_1^2}{2m}\).

The momentum is increased by 50%:
\(p_2 = p_1 + 0.5p_1 = 1.5p_1\).

The new kinetic energy \(K_2\) is:
\[ K_2 = \frac{p_2^2}{2m} = \frac{(1.5p_1)^2}{2m} = 2.25 \frac{p_1^2}{2m} \]
\[ K_2 = 2.25 K_1 \]

The fractional increase in kinetic energy is:
\[ \frac{K_2 - K_1}{K_1} = \frac{2.25 K_1 - K_1}{K_1} = 1.25 \]

Percentage increase \(= 1.25 \times 100 = 125%\).


Step 4: Final Answer:

The percentage increase in kinetic energy is 125%.
Quick Tip: If momentum increases by \(x%\), use the multiplier \(k = (1 + x/100)\). The percentage change in \(K\) is \((k^2 - 1) \times 100\). For 50%, \(1.5^2 = 2.25\), and \(2.25 - 1 = 1.25\) or 125%.

Step 1: Understanding the Concept:

The energy released in a nuclear reaction (gain in binding energy) is the difference between the total binding energy of the products and the total binding energy of the reactants.


Step 2: Key Formula or Approach:

Total Binding Energy \(= (Binding Energy per nucleon) \times (Mass number)\)


Step 3: Detailed Explanation:

Initial State (Reactant):

Mass number \(A = 242\)

Binding energy per nucleon \(= 7.6 MeV\)

Total initial B.E. \(= 242 \times 7.6 = 1839.2 MeV\)


Final State (Products):

Two fragments, each with \(A = 121\).

Binding energy per nucleon for each fragment \(= 8.1 MeV\)

Total final B.E. \(= 2 \times (121 \times 8.1) = 242 \times 8.1 = 1960.2 MeV\)


Gain in B.E. \(= Total final B.E. - Total initial B.E.\)

Gain in B.E. \(= 1960.2 - 1839.2 = 121 MeV\)

Alternatively:

Gain \(= A \times (BE_{final} - BE_{initial}) = 242 \times (8.1 - 7.6) = 242 \times 0.5 = 121 MeV\)


Step 4: Final Answer:

The total gain in binding energy is 121 MeV.
Quick Tip: Since the total mass number is conserved, calculating the gain as (Mass Number \(\times\) Change in BE per nucleon) is much faster and less prone to calculation errors.

Step 1: Understanding the Concept:

When an object is placed between parallel mirrors, multiple images are formed due to successive reflections. Each image acts as an object for the other mirror.


Step 2: Detailed Explanation:

Distance between mirrors \(d = 10 cm\).

Object distance from mirror A (\(x\)) = 2 cm.

Object distance from mirror B (\(y\)) = \(10 - 2 = 8 cm\).


Let's trace the images formed behind mirror A:

1. First image in mirror A (\(I_{A1}\)): It is the direct reflection of object O in A.

Distance behind A = \(x = 2 cm\).

2. To find the second image behind A (\(I_{A2}\)), we look for the image formed by A reflecting an image already formed in B.

The first image in mirror B (\(I_{B1}\)) is at distance \(y = 8 cm\) behind B.

This \(I_{B1}\) is at a distance of (\(d + y = 10 + 8 = 18 cm\)) from mirror A.

Reflection of \(I_{B1}\) in mirror A gives the second image \(I_{A2}\) at a distance of 18 cm behind A.

3. Let's check other sequences:

The first image in A (\(I_{A1}\) at 2 cm behind A) is reflected in B to form \(I_{B2}\).
\(I_{A1}\) is at distance (\(d + x = 10 + 2 = 12 cm\)) from mirror B. So \(I_{B2}\) is at 12 cm behind B.

This \(I_{B2}\) will be reflected in A to form the third image \(I_{A3}\) at a distance of (\(d + 12 = 10 + 12 = 22 cm\)) behind A.


Summary of distances behind mirror A: 2 cm, 18 cm, 22 cm, ...

The second nearest image behind mirror A is at 18 cm.


Step 3: Final Answer:

The distance of the second nearest image behind mirror A is 18 cm.
Quick Tip: For parallel mirrors, images behind mirror A are at: \(2x, 2d-2x, 2d+2x, 4d-2x...\) wait, the simple sequence is: \(a, 2d-a, 2d+a, 4d-a, ...\) where \(a\) is distance from that mirror. Here \(a=2\), so: \(2, 20-2=18, 20+2=22...\)

Step 1: Understanding the Concept:

In an LC oscillation, energy oscillates between the electric field of the capacitor and the magnetic field of the inductor. Energy is conserved.


Step 2: Key Formula or Approach:

Maximum magnetic energy = Maximum electrical energy:
\[ \frac{1}{2} L I_{max}^2 = \frac{1}{2} \frac{Q_{max}^2}{C} \]

Which gives:
\[ I_{max} = \frac{Q_{max}}{\sqrt{LC}} = \omega Q_{max} \]


Step 3: Detailed Explanation:

Given:
\(L = 75 mH = 75 \times 10^{-3} H\)
\(C = 1.2 \(\mu\)F = 1.2 \times 10^{-6 \text{ F\)
\(Q_{max = 2.7 \(\mu\)C = 2.7 \times 10^{-6 \text{ C\)

Calculate \(\sqrt{LC\):
\[ LC = 75 \times 10^{-3} \times 1.2 \times 10^{-6} = 90 \times 10^{-9} = 9 \times 10^{-8} \]
\[ \sqrt{LC} = \sqrt{9 \times 10^{-8}} = 3 \times 10^{-4} s \]

Now calculate maximum current:
\[ I_{max} = \frac{2.7 \times 10^{-6}}{3 \times 10^{-4}} \]
\[ I_{max} = 0.9 \times 10^{-2} A \]
\[ I_{max} = 9 \times 10^{-3} A = 9 mA \]


Step 4: Final Answer:

The maximum current in the circuit is 9 mA.
Quick Tip: Using \(\omega = 1/\sqrt{LC}\) makes the calculation structured. Ensure you convert prefixes like 'milli' (\(10^{-3}\)) and 'micro' (\(10^{-6}\)) correctly before solving.

Step 1: Understanding the Concept:

Magnetic intensity (\(H\)) for a solenoid is defined by the number of turns per unit length and the current flowing through it. It is independent of the permeability of the core.


Step 2: Key Formula or Approach:

For a long solenoid:
\[ H = nI \]

where:
\(n =\) number of turns per unit length
\(I =\) current


Step 3: Detailed Explanation:

Given:
\(H = 1.6 \times 10^3 Am^{-1}\)

Number of turns \(n' = 8 per cm\)

Convert \(n'\) to SI units (\(per meter\)):
\(n = 8 turns/cm \times 100 cm/m = 800 turns/m\)

Using the formula \(H = nI\):
\[ 1.6 \times 10^3 = 800 \times I \]
\[ I = \frac{1600}{800} \]
\[ I = 2 A \]


Step 4: Final Answer:

The current flowing through the solenoid is 2 A.
Quick Tip: Distinguish between magnetic field (\(B = \mu_0 nI\)) and magnetic intensity (\(H = nI\)). Magnetic intensity depends only on geometry (\(n\)) and current (\(I\)).

Step 1: Understanding the Concept:

Drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field. It is related to the current and the density of charge carriers.


Step 2: Key Formula or Approach:

The relationship between current (\(I\)) and drift velocity (\(v_d\)) is:
\[ I = nAev_d \]

where:
\(n =\) number density of electrons
\(A =\) area of cross-section
\(e =\) charge of an electron


Step 3: Detailed Explanation:

Given:
\(I = 2 A\)
\(A = 25.0 mm^2 = 25 \times 10^{-6} m^2\)
\(n = 2.0 \times 10^{28} m^{-3}\)
\(e = 1.6 \times 10^{-19} C\)

Substituting the values to find \(v_d\):
\[ v_d = \frac{I}{nAe} \]
\[ v_d = \frac{2}{2 \times 10^{28} \times 25 \times 10^{-6} \times 1.6 \times 10^{-19}} \]
\[ v_d = \frac{1}{25 \times 1.6 \times 10^{28 - 6 - 19}} \]
\[ v_d = \frac{1}{40 \times 10^3} \]
\[ v_d = \frac{1}{4} \times 10^{-4} = 0.25 \times 10^{-4} ms^{-1} \]
\[ v_d = 25 \times 10^{-6} ms^{-1} \]


Step 4: Final Answer:

The drift velocity is 25 \(\times 10^{-6} ms^{-1}\).
Quick Tip: Be careful with the conversion of \(mm^2\) to \(m^2\). \(1 mm^2 = (10^{-3} m)^2 = 10^{-6} m^2\).

Step 1: Understanding the Concept:

Work is done on a dipole when it is rotated in a uniform electric field because the field exerts a torque on it. The work done is equal to the change in potential energy of the dipole.


Step 2: Key Formula or Approach:

Work done \(W = pE(\cos\theta_1 - \cos\theta_2)\)

where:
\(p =\) dipole moment
\(E =\) electric field
\(\theta_1 =\) initial angle
\(\theta_2 =\) final angle


Step 3: Detailed Explanation:

Given:
\(p = 6.0 \times 10^{-6} Cm\)
\(E = 1.5 \times 10^3 NC^{-1}\)

Initially, the dipole is along the field: \(\theta_1 = 0^\circ\).

Finally, it is rotated by \(180^\circ\): \(\theta_2 = 180^\circ\).
\[ W = pE(\cos 0^\circ - \cos 180^\circ) \]
\[ W = pE(1 - (-1)) = 2pE \]
\[ W = 2 \times 6.0 \times 10^{-6} \times 1.5 \times 10^3 \]
\[ W = 18.0 \times 10^{-3} J \]
\[ W = 18 mJ \]


Step 4: Final Answer:

The work done is 18 mJ.
Quick Tip: Rotating a dipole from stable equilibrium (\(\theta=0^\circ\)) to unstable equilibrium (\(\theta=180^\circ\)) always requires \(2pE\) amount of work.