Step 1: Understanding the Concept:

The motion of a satellite in a circular orbit around the Earth is governed by the gravitational pull of the Earth, which provides the necessary centripetal force.

The orbital parameters depend on the mass of the central body (Earth) and the radius of the orbit, but may or may not depend on the mass of the satellite itself.


Step 2: Key Formula or Approach:

The orbital speed \( v \) of a satellite at a distance \( r \) from the center of the Earth is given by:
\[ v = \sqrt{\frac{GM}{r}} \]

Where:
\( G \) is the Universal Gravitational Constant.
\( M \) is the mass of the Earth.
\( r \) is the radius of the orbit.


Step 3: Detailed Explanation:

1. Speed: From the formula \( v = \sqrt{\frac{GM}{r}} \), it is evident that the orbital speed depends only on the mass of the Earth and the orbital radius. Since both satellites A and B are in the same orbit (same \( r \)), their orbital speeds will be identical, regardless of their individual masses.

2. Kinetic Energy: \( KE = \frac{1}{2} m v^{2} \). Since the mass of A is twice that of B (\( m_{A} = 2m_{B} \)), satellite A will have twice the kinetic energy of satellite B.

3. Potential Energy: \( PE = -\frac{GMm}{r} \). Since \( PE \propto m \), the potential energy will differ for A and B.

4. Total Energy: \( E_{total} = -\frac{GMm}{2r} \). Since \( E_{total} \propto m \), the total mechanical energy will also be different for the two satellites.


Step 4: Final Answer:

Therefore, only the orbital speed remains the same for both satellites.
Quick Tip: Remember that orbital velocity and time period of a satellite are independent of the mass of the satellite. They only depend on the mass of the planet and the orbital radius.

Step 1: Understanding the Concept:

To match the items, we need to derive the dimensional formulas for each physical quantity based on their defining equations.


Step 2: Key Formula or Approach:

We use the fundamental units: Mass \([M]\), Length \([L]\), and Time \([T]\).


Step 3: Detailed Explanation:

A. Spring constant (\( k \)):

From Hooke's Law, \( F = kx \Rightarrow k = \frac{F}{x} \).

Dimensions of Force \( [F] = [MLT^{-2}] \).

Dimensions of displacement \( [x] = [L] \).

So, \( [k] = \frac{[MLT^{-2}]}{[L]} = [MT^{-2}] \). (Matches II)


B. Angular speed (\( \omega \)):
\( \omega = \frac{\theta}{t} \). Since angle \( \theta \) is dimensionless:
\( [\omega] = \frac{1}{[T]} = [T^{-1}] \). (Matches I)


C. Angular momentum (\( L \)):
\( L = mvr \).

Dimensions: \( [M] \times [LT^{-1}] \times [L] = [ML^{2}T^{-1}] \). (Matches IV)


D. Moment of Inertia (\( I \)):
\( I = mr^{2} \).

Dimensions: \( [M] \times [L]^{2} = [ML^{2}] \). (Matches III)


Step 4: Final Answer:

Matching the derived dimensions: A-II, B-I, C-IV, D-III.
Quick Tip: Angular quantities like frequency, angular velocity, and velocity gradient all share the same dimension \([T^{-1}]\). This is a common shortcut for dimensional analysis questions.

Step 1: Understanding the Concept:

Statement I relates to the Work-Energy Theorem. Statement II relates to the vector nature of velocity and acceleration.


Step 3: Detailed Explanation:

Analysis of Statement I:

According to the Work-Energy Theorem, the work done by the retarding force is equal to the change in kinetic energy:
\[ W = F \cdot d = \Delta KE \]

Since both vehicles have the same initial kinetic energy (\( KE \)) and are brought to rest (final \( KE = 0 \)), the change in kinetic energy is the same.

Given that the retarding force \( F \) is equal for both:
\[ d = \frac{KE}{F} \]

Since \( KE \) and \( F \) are the same for both the truck and the car, the stopping distance \( d \) must be equal.

Thus, Statement I is correct.


Analysis of Statement II:

Velocity is a vector quantity. It has both magnitude (speed) and direction.

When the car turns from East to North, even if the speed (magnitude) remains constant, the direction changes.

Acceleration is defined as the rate of change of velocity (\( \vec{a} = \frac{d\vec{v}}{dt} \)).

Since the velocity vector changes direction, the acceleration cannot be zero.

Thus, Statement II is incorrect.


Step 4: Final Answer:

Statement I is correct and Statement II is incorrect.
Quick Tip: In stopping distance problems, if Kinetic Energy is constant and Force is constant, distance is independent of mass. If momentum were constant instead, the lighter body would travel further.

Step 1: Understanding the Concept:

In Amplitude Modulation (AM), a high-frequency carrier wave has its amplitude varied in accordance with a low-frequency modulating (message) signal. The resulting AM wave consists of the carrier frequency and two sideband frequencies.


Step 2: Key Formula or Approach:

General equations:

Carrier signal: \( c(t) = A_{c} \sin(2\pi f_{c} t) \)

Modulating signal: \( m(t) = A_{m} \sin(2\pi f_{m} t) \)

Frequencies present in AM signal: \( f_{c} \), \( f_{c} - f_{m} \) (Lower Side Band), and \( f_{c} + f_{m} \) (Upper Side Band).


Step 3: Detailed Explanation:

1. Carrier Frequency (\( f_{c} \)):

From the expression \( 15 \sin(1000 \pi t) \):
\( 2\pi f_{c} = 1000\pi \Rightarrow f_{c} = \frac{1000}{2} = 500 Hz \). (Frequency A)


2. Modulating Frequency (\( f_{m} \)):

From the expression \( 10 \sin(4 \pi t) \):
\( 2\pi f_{m} = 4\pi \Rightarrow f_{m} = \frac{4}{2} = 2 Hz \).


3. Sideband Frequencies:

Lower Sideband (LSB) = \( f_{c} - f_{m} = 500 - 2 = 498 Hz \). (Frequency D)

Upper Sideband (USB) = \( f_{c} + f_{m} = 500 + 2 = 502 Hz \). (Frequency E)


Step 4: Final Answer:

The frequencies present are 500 Hz, 498 Hz, and 502 Hz. These correspond to A, D, and E.
Quick Tip: Bandwidth of an AM signal is always twice the modulating frequency (\( BW = 2f_{m} \)). Here, \( BW = 502 - 498 = 4 Hz \), which is indeed \( 2 \times 2 Hz \).

Step 1: Understanding the Concept:

These statements concern the resonance behavior and power characteristics of a series LCR circuit.


Step 3: Detailed Explanation:

Analysis of Statement I:

The impedance \( Z \) of a series LCR circuit is \( Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}} \).

Current \( I = \frac{V}{Z} \).

At low frequencies, \( X_{C} \) is very high, so \( Z \) is high and \( I \) is low.

As frequency increases, \( X_{L} - X_{C} \) decreases, reaching zero at resonance (\( f_{r} = \frac{1}{2\pi\sqrt{LC}} \)). At this point, \( Z = R \) (minimum), and current is maximum.

As frequency increases beyond resonance, \( X_{L} \) dominates and increases, causing \( Z \) to increase and \( I \) to decrease.

Thus, the current curve is bell-shaped (increases, peaks, then decreases). Statement I is true.


Analysis of Statement II:

Power factor is given by \( \cos \phi = \frac{R}{Z} \).

At resonance, \( X_{L} = X_{C} \), which means \( Z = R \).

Substituting this into the power factor formula: \( \cos \phi = \frac{R}{R} = 1 \).

Thus, Statement II is true.


Step 4: Final Answer:

Both statements are true.
Quick Tip: At resonance, a series LCR circuit behaves as a purely resistive circuit. The phase difference between voltage and current is zero (\( \phi = 0 \)), leading to a unity power factor.

Step 1: Understanding the Concept:

Escape velocity is the minimum speed needed for an object to break free from the gravitational attraction of a celestial body.


Step 2: Key Formula or Approach:

The formula for escape velocity \( v_{e} \) is:
\[ v_{e} = \sqrt{\frac{2GM}{R}} \]

Where \( M \) is the mass of the body and \( R \) is its radius.


Step 3: Detailed Explanation:

Let \( M_{e} \) and \( R_{e} \) be the mass and radius of the Earth.

The escape velocity of Earth is \( v_{e} = \sqrt{\frac{2GM_{e}}{R_{e}}} \).


For the planet:

Mass \( M_{p} = 16 M_{e} \)

Radius \( R_{p} = 4 R_{e} \)

Escape velocity of the planet \( v_{p} = \sqrt{\frac{2GM_{p}}{R_{p}}} \).


Taking the ratio:
\[ \frac{v_{p}}{v_{e}} = \frac{\sqrt{\frac{2GM_{p}}{R_{p}}}}{\sqrt{\frac{2GM_{e}}{R_{e}}}} = \sqrt{\frac{M_{p}}{M_{e}} \times \frac{R_{e}}{R_{p}}} \]

Substituting the given values:
\[ \frac{v_{p}}{v_{e}} = \sqrt{\frac{16 M_{e}}{M_{e}} \times \frac{R_{e}}{4 R_{e}}} \]
\[ \frac{v_{p}}{v_{e}} = \sqrt{\frac{16}{4}} = \sqrt{4} = 2 \]


Step 4: Final Answer:

The ratio is 2:1.
Quick Tip: If the density of the planet is the same as Earth, escape velocity is directly proportional to the radius (\( v_{e} \propto R \)). Here, density changes, so use the \( M \) and \( R \) formula directly.

Step 1: Understanding the Concept:

This problem uses Newton's Law of Cooling, which states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings.


Step 2: Key Formula or Approach:

The average form of Newton's Law of Cooling is:
\[ \frac{T_{1} - T_{2}}{t} = K \left( \frac{T_{1} + T_{2}}{2} - T_{0} \right) \]

Where \( T_{1} \) is initial temp, \( T_{2} \) is final temp, \( T_{0} \) is surrounding temp, \( t \) is time, and \( K \) is a constant.


Step 3: Detailed Explanation:

Case 1: Cooling from \( 80^{\circ}C \) to \( 60^{\circ}C \) in 5 min (\( 300 s \))
\[ \frac{80 - 60}{300} = K \left( \frac{80 + 60}{2} - 20 \right) \]
\[ \frac{20}{300} = K (70 - 20) \]
\[ \frac{1}{15} = 50K \Rightarrow K = \frac{1}{750} s^{-1} \]


Case 2: Cooling from \( 60^{\circ}C \) to \( 40^{\circ}C \) in time \( t \)
\[ \frac{60 - 40}{t} = K \left( \frac{60 + 40}{2} - 20 \right) \]
\[ \frac{20}{t} = \left( \frac{1}{750} \right) (50 - 20) \]
\[ \frac{20}{t} = \frac{1}{750} \times 30 \]
\[ \frac{20}{t} = \frac{1}{25} \]
\[ t = 20 \times 25 = 500 s \]


Step 4: Final Answer:

The time taken is 500 seconds.
Quick Tip: As the body's temperature approaches the surrounding temperature, the rate of cooling decreases. Therefore, cooling through the same temperature interval (\( 20^{\circ}C \) drop) at lower temperatures will always take more time.

Step 1: Understanding the Concept:

The maximum theoretical efficiency of any heat engine is defined by the Carnot efficiency, which depends only on the absolute temperatures of the hot and cold reservoirs. Real engines are always less efficient than Carnot engines.


Step 2: Key Formula or Approach:

Carnot Efficiency \( \eta = 1 - \frac{T_{L}}{T_{H}} \)

Where temperatures must be in Kelvin (\( K \)).


Step 3: Detailed Explanation:

1. Identify Temperatures:

Boiling point of water \( T_{H} = 100^{\circ}C = 100 + 273 = 373 K \).

Freezing point of water \( T_{L} = 0^{\circ}C = 0 + 273 = 273 K \).


2. Calculate Carnot Efficiency:
\[ \eta_{Carnot} = 1 - \frac{273}{373} = \frac{373 - 273}{373} = \frac{100}{373} \]
\[ \eta_{Carnot} \approx 0.268 or 26.8% \]


3. Evaluate Statements:

A. Efficiency more than 27%: False (Maximum possible is 26.8%).

B. Efficiency less than Carnot: True (For any real engine).

C. Efficiency equal to 27%: False.

D. Efficiency less than 27%: True (Since \( \eta \leq 26.8% \), it is definitely less than 27%).


Step 4: Final Answer:

Statements B and D are correct.
Quick Tip: No engine can have an efficiency of 100% according to the second law of thermodynamics. The Carnot engine represents the ideal, reversible case which provides the upper limit.

Step 1: Understanding the Concept:

Root Mean Square (RMS) speed of gas molecules depends on the absolute temperature and the molar mass of the gas.


Step 2: Key Formula or Approach:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]

At constant temperature \( T \), \( v_{rms} \propto \frac{1}{\sqrt{M}} \).


Step 3: Detailed Explanation:

Let \( v_{Cl} \) and \( M_{Cl} \) be the speed and mass of Chlorine.

Let \( v_{Ar} \) and \( M_{Ar} \) be the speed and mass of Argon.

Given: \( v_{Cl} = 490 m/s \), \( M_{Cl} = 70.9 u \), \( M_{Ar} = 39.9 u \).

\[ \frac{v_{Ar}}{v_{Cl}} = \sqrt{\frac{M_{Cl}}{M_{Ar}}} \]
\[ v_{Ar} = v_{Cl} \times \sqrt{\frac{70.9}{39.9}} \]
\[ v_{Ar} = 490 \times \sqrt{1.7769} \]
\[ v_{Ar} \approx 490 \times 1.333 \]
\[ v_{Ar} \approx 653.17 m/s \]

The closest option provided is 651.7 m/s.


Step 4: Final Answer:

The r.m.s speed of argon molecules is approximately 651.7 m/s.
Quick Tip: Lighter gas molecules move faster than heavier ones at the same temperature. Since Argon (39.9) is lighter than Chlorine (70.9), its speed must be significantly higher than 490 m/s. This helps eliminate option A immediately.

Step 1: Understanding the Concept:

In SHM, the total energy is conserved and alternates between potential energy (maximum at extremes) and kinetic energy (maximum at mean position).


Step 2: Key Formula or Approach:

Potential Energy (\( PE \)) at displacement \( x \): \( PE = \frac{1}{2} k x^{2} \)

Kinetic Energy (\( KE \)) at displacement \( x \): \( KE = \frac{1}{2} k (A^{2} - x^{2}) \)

Where \( A \) is the amplitude and \( k \) is the force constant.


Step 3: Detailed Explanation:

Given displacement \( x = \frac{A}{2} \).


1. Potential Energy:
\[ PE = \frac{1}{2} k \left( \frac{A}{2} \right)^{2} = \frac{1}{2} k \frac{A^{2}}{4} = \frac{1}{8} k A^{2} \]


2. Kinetic Energy:
\[ KE = \frac{1}{2} k \left( A^{2} - \left( \frac{A}{2} \right)^{2} \right) = \frac{1}{2} k \left( A^{2} - \frac{A^{2}}{4} \right) \]
\[ KE = \frac{1}{2} k \left( \frac{3A^{2}}{4} \right) = \frac{3}{8} k A^{2} \]


3. Ratio \( PE : KE \):
\[ \frac{PE}{KE} = \frac{\frac{1}{8} k A^{2}}{\frac{3}{8} k A^{2}} = \frac{1}{3} \]


Step 4: Final Answer:

The ratio is 1:3.
Quick Tip: At \( x = \frac{A}{\sqrt{2}} \), \( PE = KE \), and the ratio is 1:1. At \( x = \frac{A}{2} \), the energy is \( \frac{1}{4} \) potential and \( \frac{3}{4} \) kinetic.

Step 1: Understanding the Concept:

According to Gauss's Law in electrostatics, the net electric flux \( \phi \) through any closed surface is proportional to the net charge \( q_{enclosed} \) enclosed by that surface.

The relation is given by:
\[ \phi = \frac{q_{enclosed}}{\epsilon_{0}} \]


Step 3: Detailed Explanation:

An electric dipole consists of two charges: \( +q \) and \( -q \), separated by a small distance.

The net charge of an electric dipole is:
\[ q_{net} = (+q) + (-q) = 0 \]

Since the electric dipole is enclosed by the surface, the total charge enclosed \( q_{enclosed} \) is zero.

Applying Gauss's Law:
\[ \phi = \frac{0}{\epsilon_{0}} = 0 \]

Therefore, the net flux coming out of the surface is zero.

Assertion A is true.

Reason R states that a dipole consists of two equal and opposite charges, which is the exact reason why the net charge is zero and consequently the flux is zero.

Thus, Reason R is true and correctly explains Assertion A.


Step 4: Final Answer:

Both Assertion and Reason are correct, and the Reason is the correct explanation for the Assertion.
Quick Tip: For any closed surface, if the total enclosed charge is zero, the net flux is always zero, regardless of the complexity or magnitude of the dipole moment inside.

Step 1: Understanding the Concept:

When a wire is melted and reshaped, its volume remains constant.

The resistance of a wire depends on its resistivity \( \rho \), length \( L \), and cross-sectional area \( A \).


Step 2: Key Formula or Approach:

Resistance \( R = \rho \frac{L}{A} \).

Since Volume \( V = A \cdot L \) is constant, we can write \( A = \frac{V}{L} \).

Substituting this into the resistance formula:
\[ R = \rho \frac{L}{V/L} = \frac{\rho L^{2}}{V} \]

This implies \( R \propto L^{2} \) for a wire of constant volume.


Step 3: Detailed Explanation:

Let the initial resistance be \( R_{1} = 160 \Omega \) and initial length be \( L_{1} \).

The new length is \( L_{2} = \frac{1}{4} L_{1} \).

Using the proportionality \( R \propto L^{2} \):
\[ \frac{R_{2}}{R_{1}} = \left( \frac{L_{2}}{L_{1}} \right)^{2} \]
\[ \frac{R_{2}}{160} = \left( \frac{L_{1}/4}{L_{1}} \right)^{2} \]
\[ \frac{R_{2}}{160} = \left( \frac{1}{4} \right)^{2} = \frac{1}{16} \]
\[ R_{2} = \frac{160}{16} = 10 \Omega \]


Step 4: Final Answer:

The new resistance of the wire is \(10 \Omega\).
Quick Tip: If a wire is stretched or compressed to \( n \) times its length, the new resistance becomes \( n^{2} R \). Here, \( n = 1/4 \), so the resistance becomes \( (1/4)^{2} = 1/16 \) of the original.

Step 1: Understanding the Concept:

Diamagnetism is a fundamental property of all matter, arising from the orbital motion of electrons.

Magnetic susceptibility \( \chi \) measures how a material responds to an external magnetic field.


Step 3: Detailed Explanation:

Analysis of Statement I:

Diamagnetic materials have negative susceptibility.

Unlike paramagnetic and ferromagnetic materials, which follow Curie's Law (\( \chi \propto 1/T \)), diamagnetism is largely independent of temperature (except for certain specialized cases like superconductors).

Therefore, Statement I is incorrect.


Analysis of Statement II:

When a diamagnetic material is placed in an external magnetic field, the orbital motion of electrons is adjusted such that it produces an induced magnetic moment in a direction opposite to the applied field (Lenz's Law on an atomic scale).

This causes the material to be weakly repelled by the magnetic field.

Therefore, Statement II is true.


Step 4: Final Answer:

Statement I is false, but Statement II is true.
Quick Tip: Remember: "Dia" is independent of T, "Para" and "Ferro" depend on T. This is a very frequent theoretical concept in entrance exams.

Step 1: Understanding the Concept:

For an object thrown vertically upwards, the velocity at any time \( t \) can be determined using the first equation of motion under gravity.


Step 2: Key Formula or Approach:

Velocity \( v = u - gt \)

Where:
\( u = 150 m/s \) (initial velocity)
\( g = 10 m/s^{2} \) (acceleration due to gravity)


Step 3: Detailed Explanation:

Calculate velocity at \( t = 3 s \) (\( v_{1} \)):
\[ v_{1} = 150 - (10 \times 3) = 150 - 30 = 120 m/s \]


Calculate velocity at \( t = 5 s \) (\( v_{2} \)):
\[ v_{2} = 150 - (10 \times 5) = 150 - 50 = 100 m/s \]


Calculate the ratio:
\[ \frac{v_{1}}{v_{2}} = \frac{120}{100} = \frac{6}{5} \]


According to the question, the ratio is given as \( \frac{x+1}{x} \).

So, we set up the equation:
\[ \frac{x+1}{x} = \frac{6}{5} \]

Cross-multiplying:
\[ 5(x+1) = 6x \]
\[ 5x + 5 = 6x \]
\[ x = 5 \]


Step 4: Final Answer:

The value of \( x \) is 5.
Quick Tip: When solving kinematic ratios, simplify the calculated values immediately to their lowest fraction form to make comparison with the given variable expression easier.

Step 1: Understanding the Concept:

When an object is at rest under the action of multiple forces, the vector sum of all forces (net force) must be zero.
\[ \vec{F}_{net} = \vec{F}_{1} + \vec{F}_{2} + \vec{F}_{3} = 0 \]


Step 3: Detailed Explanation:

1. Initial Equilibrium State:

The particle is at rest, meaning the net force is zero.
\[ \vec{F}_{1} + (\vec{F}_{2} + \vec{F}_{3}) = 0 \Rightarrow \vec{F}_{2} + \vec{F}_{3} = -\vec{F}_{1} \]

This implies that the resultant of \( \vec{F}_{2} \) and \( \vec{F}_{3} \) must be equal in magnitude and opposite in direction to \( \vec{F}_{1} \).

Check magnitude of resultant of \( F_{2} \) and \( F_{3} \) (since they are perpendicular):
\[ R_{23} = \sqrt{F_{2}^{2} + F_{3}^{2}} = \sqrt{8^{2} + 6^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10 N \]

This matches the magnitude of \( F_{1} = 10 N \).


2. Removing \( F_{1} \):

If force \( F_{1} \) is removed, the only forces remaining are \( F_{2} \) and \( F_{3} \).

The new net force \( \vec{F}_{new} = \vec{F}_{2} + \vec{F}_{3} \).

The magnitude of this net force is \( 10 N \).


3. Calculation of Acceleration:

According to Newton's Second Law (\( F = ma \)):
\[ a = \frac{F_{net}}{m} = \frac{10}{5} = 2 ms^{-2} \]


Step 4: Final Answer:

The acceleration of the particle is \(2 ms^{-2}\).
Quick Tip: If a body is in equilibrium and one force is removed, the net force on the body becomes equal in magnitude to the removed force.

Step 1: Understanding the Concept:

When an object is viewed through a denser medium, it appears closer than it actually is. This is known as apparent depth.


Step 2: Key Formula or Approach:

The relationship between real depth (\( d_{real} \)), apparent depth (\( d_{app} \)), and refractive index (\( \mu \)) is:
\[ \mu = \frac{d_{real}}{d_{app}} \Rightarrow d_{real} = \mu \times d_{app} \]


Step 3: Detailed Explanation:

Let the ice cube have a side length \( L = 24 cm \).

Let the actual distance of the bubble from one side be \( x \).

Then, the actual distance of the bubble from the opposite side is \( L - x = 24 - x \).


For the first side:

Apparent depth \( d_{app1} = 12 cm \).

Real depth \( x = \mu \times 12 \).


For the opposite side:

Apparent depth \( d_{app2} = 4 cm \).

Real depth \( 24 - x = \mu \times 4 \).


Adding the two real depths to eliminate \( x \):
\[ x + (24 - x) = (\mu \times 12) + (\mu \times 4) \]
\[ 24 = 16\mu \]
\[ \mu = \frac{24}{16} = \frac{3}{2} \]


Step 4: Final Answer:

The refractive index of the ice cube is \( \frac{3}{2} \) or \( 1.5 \).
Quick Tip: For a slab of thickness \( T \) with an object inside, the refractive index is always the sum of real depths divided by the sum of apparent depths: \( \mu = \frac{Total Thickness}{\sum Apparent Depths} \).

Step 1: Understanding the Concept:

The de-Broglie wavelength \( \lambda \) of a particle depends on its momentum. When a charged particle is accelerated through a potential difference \( V \), it gains kinetic energy \( qV \).


Step 2: Key Formula or Approach:
\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mKE}} = \frac{h}{\sqrt{2mqV}} \]

Where \( h \) is Planck's constant, \( m \) is mass, \( q \) is charge, and \( V \) is accelerating potential.


Step 3: Detailed Explanation:

Let subscript \( p \) denote the proton and \( \alpha \) denote the \( \alpha \)-particle.

1. Proton parameters:

Mass \( m_{p} \), Charge \( q_{p} = e \), Potential \( V_{p} = 2 V \).

2. \( \alpha \)-particle parameters:

Mass \( m_{\alpha} = 4m_{p} \), Charge \( q_{\alpha} = 2e \), Potential \( V_{\alpha} = 4 V \).


Ratio of wavelengths:
\[ \frac{\lambda_{p}}{\lambda_{\alpha}} = \frac{\frac{h}{\sqrt{2m_{p}q_{p}V_{p}}}}{\frac{h}{\sqrt{2m_{\alpha}q_{\alpha}V_{\alpha}}}} = \sqrt{\frac{m_{\alpha}q_{\alpha}V_{\alpha}}{m_{p}q_{p}V_{p}}} \]

Substituting the values:
\[ \frac{\lambda_{p}}{\lambda_{\alpha}} = \sqrt{\frac{(4m_{p}) \times (2e) \times 4}{m_{p} \times e \times 2}} \]
\[ \frac{\lambda_{p}}{\lambda_{\alpha}} = \sqrt{\frac{32}{2}} = \sqrt{16} = \frac{4}{1} \]


Step 4: Final Answer:

The ratio of their de-Broglie wavelengths is 4:1.
Quick Tip: Always remember the relative mass and charge values: \( m_{\alpha} = 4m_{p} \) and \( q_{\alpha} = 2q_{p} \). Misremembering these is the most common cause of error in atomic physics ratios.

Step 1: Understanding the Concept:

When gaseous hydrogen is bombarded with electrons, the atoms absorb energy and jump to higher energy levels (excited states). Spectral lines are produced when they transition back to lower states.


Step 3: Detailed Explanation:

1. Find the highest energy level (\( n \)) reached:

The ground state energy of Hydrogen is \( E_{1} = -13.6 eV \).

Maximum energy of the atom after collision = \( -13.6 + 12.5 = -1.1 eV \).

Energy levels of Hydrogen:
\( E_{2} = \frac{-13.6}{2^{2}} = -3.4 eV \)
\( E_{3} = \frac{-13.6}{3^{2}} = -1.51 eV \)
\( E_{4} = \frac{-13.6}{4^{2}} = -0.85 eV \)

Since \( -1.1 eV \) is between \( E_{3} \) and \( E_{4} \), the electrons can excite atoms to the \( n=3 \) level but not \( n=4 \).


2. Calculate the number of spectral lines:

For atoms excited to level \( n \), the number of possible emission lines is given by:
\[ N = \frac{n(n-1)}{2} \]

Substituting \( n = 3 \):
\[ N = \frac{3(3-1)}{2} = \frac{3 \times 2}{2} = 3 \]

The possible transitions are \( 3 \rightarrow 1 \), \( 3 \rightarrow 2 \), and \( 2 \rightarrow 1 \).


Step 4: Final Answer:

The number of spectral lines emitted is 3.
Quick Tip: Memorize the energy levels of Hydrogen (\(-13.6\), \(-3.4\), \(-1.51\), \(-0.85 eV\)) to quickly determine excitation levels without re-calculating during the exam.

Step 1: Understanding the Concept:

In a common emitter configuration, the current gain (denoted as \( \beta \)) is the ratio of the change in collector current to the change in base current.


Step 2: Key Formula or Approach:
\[ \beta = \frac{\Delta I_{C}}{\Delta I_{B}} \]


Step 3: Detailed Explanation:

1. Calculate change in collector current (\( \Delta I_{C} \)):
\[ I_{C1} = 5 mA \]
\[ I_{C2} = 16 mA \]
\[ \Delta I_{C} = 16 - 5 = 11 mA = 11 \times 10^{-3} A \]


2. Calculate change in base current (\( \Delta I_{B} \)):
\[ I_{B1} = 100 \muA \]
\[ I_{B2} = 200 \muA \]
\[ \Delta I_{B} = 200 - 100 = 100 \muA = 100 \times 10^{-6} A = 0.1 mA \]


3. Calculate Current Gain (\( \beta \)):
\[ \beta = \frac{11 mA}{0.1 mA} = 110 \]


Step 4: Final Answer:

The current gain of the transistor is 110.
Quick Tip: Ensure units are consistent before dividing. It is often easiest to convert everything to mA or \(\mu\)A. Here, \(11 mA = 11000 \muA\), so \(11000 / 100 = 110\).

Step 1: Understanding the Concept:

Electromagnetic (EM) waves are categorized in the EM spectrum based on their wavelength (\( \lambda \)) and frequency (\( f \)). Energy \( E = hf = \frac{hc}{\lambda} \).


Step 3: Detailed Explanation:

Analysis of Assertion A:

Optical communication typically uses visible light or infrared waves.

Microwaves are used in Radar.

Order of the EM spectrum (increasing wavelength):

Gamma rays \( < \) X-rays \( < \) UV \( < \) Visible \( < \) Infrared (Optical) \( < \) Microwaves \( < \) Radio waves.

Since Microwaves come after Infrared/Visible in the spectrum, Microwaves have longer wavelengths.

Thus, Assertion A is false.


Analysis of Reason R:

Energy \( E \propto \frac{1}{\lambda} \).

As shown above, Infrared has a shorter wavelength than Microwaves.

Therefore, Infrared waves have higher frequency and higher energy than microwaves.

Thus, Reason R is true.


Step 4: Final Answer:

Assertion A is false, but Reason R is true.
Quick Tip: Use the mnemonic "R-M-I-V-U-X-G" (Radio, Micro, Infra, Visible, Ultra, X-ray, Gamma) to remember the spectrum in order of increasing frequency and decreasing wavelength.

Step 1: Understanding the Concept:

A compass needle oscillates in the horizontal plane due to the horizontal component of the Earth's magnetic field (\(B_H\)).

The time period of oscillation \(T\) is given by \(T = 2\pi\sqrt{\frac{I}{MB_H}}\), where \(I\) is the moment of inertia and \(M\) is the magnetic moment.

The frequency of oscillation \(f\) is therefore \(f = \frac{1}{T} = \frac{1}{2\pi}\sqrt{\frac{MB_H}{I}}\).

Thus, \(f^2 \propto B_H\).


Step 2: Key Formula or Approach:

The horizontal component of the magnetic field is \(B_H = B \cos \delta\), where \(B\) is the total magnetic field and \(\delta\) is the angle of dip.

From the proportionality, we have:
\[ \left(\frac{f_1}{f_2}\right)^2 = \frac{B_1 \cos \delta_1}{B_2 \cos \delta_2} \]


Step 3: Detailed Explanation:

Given:
\(f_1 = 20 oscillations/min\), \(\delta_1 = 30^{\circ}\)
\(f_2 = 30 oscillations/min\), \(\delta_2 = 60^{\circ}\)

Ratio of total fields \(\frac{B_1}{B_2} = \frac{4}{\sqrt{x}}\)


Substituting the values:
\[ \left(\frac{20}{30}\right)^2 = \frac{B_1 \cos 30^{\circ}}{B_2 \cos 60^{\circ}} \]
\[ \frac{4}{9} = \frac{B_1}{B_2} \cdot \frac{\sqrt{3}/2}{1/2} \]
\[ \frac{4}{9} = \frac{B_1}{B_2} \cdot \sqrt{3} \]
\[ \frac{B_1}{B_2} = \frac{4}{9\sqrt{3}} \]


Comparing this with the given ratio \(\frac{4}{\sqrt{x}}\):
\[ \sqrt{x} = 9\sqrt{3} \]

Squaring both sides:
\[ x = 81 \times 3 = 243 \]


Step 4: Final Answer:

The value of \(x\) is 243.
Quick Tip: Always remember that a standard compass needle responds only to the horizontal component of the Earth's magnetic field. If a dip needle were used, it would respond to the total magnetic field.

Step 1: Understanding the Concept:

To maintain a constant speed, the net force on the bus must be zero.

This means the driving force provided by the engine must exactly balance the retarding force of friction.


Step 2: Key Formula or Approach:

Frictional force \(f = \mu N = \mu mg\).

Work done \(W = F_{engine} \times s \cos \theta\), where \(s\) is distance and \(\theta = 0^{\circ}\) because force and displacement are in the same direction.


Step 3: Detailed Explanation:

Given:

Mass \(m = 500 kg\)

Distance \(s = 4 km = 4000 m\)

Coefficient of friction \(\mu = 0.04\)

Acceleration due to gravity \(g = 10 m/s^2\) (Standard assumption)


Calculating the force of friction:
\[ f = 0.04 \times 500 \times 10 = 200 N \]

Since speed is constant, \(F_{engine} = f = 200 N\).


Calculating work done:
\[ W = 200 N \times 4000 m = 800,000 J \]

Converting to KJ:
\[ W = \frac{800,000}{1000} = 800 KJ \]


Step 4: Final Answer:

The work done by the engine is 800 KJ.
Quick Tip: In problems where speed is "maintained", the change in kinetic energy is zero. Thus, the work done by the engine is entirely used to overcome the work done by friction (\(W_{ext} + W_{friction} = 0\)).

Step 1: Understanding the Concept:

A rolling object possesses both translational kinetic energy (\(K_t\)) and rotational kinetic energy (\(K_r\)).

For pure rolling, \(v = R\omega\).


Step 2: Key Formula or Approach:

Translational KE: \(K_t = \frac{1}{2}mv^2\)

Rotational KE: \(K_r = \frac{1}{2}I\omega^2\)

Total KE: \(K_{total} = K_t + K_r = \frac{1}{2}mv^2 \left(1 + \frac{k^2}{R^2}\right)\)


Step 3: Detailed Explanation:

For a spherical shell (hollow sphere), the moment of inertia is \(I = \frac{2}{3}mR^2\).

Thus, the radius of gyration squared is \(k^2 = \frac{2}{3}R^2\), giving \(\frac{k^2}{R^2} = \frac{2}{3}\).


Rotational Kinetic Energy:
\[ K_r = \frac{1}{2} \left(\frac{2}{3}mR^2\right) \left(\frac{v}{R}\right)^2 = \frac{1}{3}mv^2 \]


Total Kinetic Energy:
\[ K_{total} = \frac{1}{2}mv^2 + \frac{1}{3}mv^2 = \frac{5}{6}mv^2 \]


Ratio of Rotational KE to Total KE:
\[ Ratio = \frac{K_r}{K_{total}} = \frac{\frac{1}{3}mv^2}{\frac{5}{6}mv^2} = \frac{1}{3} \times \frac{6}{5} = \frac{2}{5} \]


Given ratio is \(\frac{x}{5}\), so \(x = 2\).


Step 4: Final Answer:

The value of \(x\) is 2.
Quick Tip: The fraction of total energy that is rotational is given by \(\frac{k^2/R^2}{1 + k^2/R^2}\). For a solid sphere (\(2/5\)), it is \(2/7\). For a hollow sphere (\(2/3\)), it is \(2/5\).

Step 1: Understanding the Concept:

For a non-viscous, incompressible fluid, we use the Principle of Continuity and Bernoulli's Equation.


Step 2: Key Formula or Approach:

Continuity: \(A_1 v_1 = A_2 v_2 = Q\)

Bernoulli's: \(P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2\)


Step 3: Detailed Explanation:

Given:
\(\rho = 1.25 \times 10^3 kg/m^3 = 1250 kg/m^3\)
\(A_1 = 10 cm^2\), \(A_2 = 5 cm^2 \Rightarrow A_1 = 2 A_2\)

From continuity: \(v_2 = \frac{A_1}{A_2} v_1 = 2 v_1\)

Pressure drop \(\Delta P = P_1 - P_2 = 3 N/m^2\)


Applying Bernoulli's equation:
\[ P_1 - P_2 = \frac{1}{2}\rho (v_2^2 - v_1^2) \]
\[ 3 = \frac{1}{2}(1250) ((2v_1)^2 - v_1^2) \]
\[ 3 = 625 (3v_1^2) \]
\[ 1 = 625 v_1^2 \Rightarrow v_1 = \sqrt{\frac{1}{625}} = \frac{1}{25} = 0.04 m/s \]


Rate of flow \(Q = A_1 v_1\):
\(A_1 = 10 cm^2 = 10 \times 10^{-4} m^2 = 10^{-3} m^2\)
\[ Q = 10^{-3} \times 0.04 = 4 \times 10^{-5} m^3/s \]

Comparing with \(x \times 10^{-5}\), we get \(x = 4\).


Step 4: Final Answer:

The value of \(x\) is 4.
Quick Tip: Flow rate can be calculated using the Venturi meter formula directly: \(Q = A_1 A_2 \sqrt{\frac{2 \Delta P}{\rho (A_1^2 - A_2^2)}}\). This saves time in manipulating variables.

Step 1: Understanding the Concept:

The frequency ratio 1:3:5 indicates that only odd harmonics are present. This characterizes a closed organ pipe (one end closed, one end open).


Step 2: Key Formula or Approach:

For a closed organ pipe, the frequency of the \(n\)-th harmonic is \(f_n = n \frac{v}{4L}\), where \(n\) is an odd integer (\(n = 1, 3, 5, \dots\)).


Step 3: Detailed Explanation:

Given:

Ratio 1:3:5 \(\implies\) Closed organ pipe.

Fifth harmonic (\(n=5\)) frequency \(f_5 = 405 Hz\).

Speed of sound \(v = 324 m/s\).


Applying the formula:
\[ f_5 = 5 \times \frac{v}{4L} \]
\[ 405 = 5 \times \frac{324}{4L} \]

Dividing both sides by 5:
\[ 81 = \frac{324}{4L} \]

Rearranging to solve for \(L\):
\[ 4L = \frac{324}{81} \]
\[ 4L = 4 \]
\[ L = 1 m \]


Step 4: Final Answer:

The length of the organ pipe is 1 m.
Quick Tip: In a closed pipe, the term "5th harmonic" and "3rd resonance" refer to the same frequency (\(n=5\)). In an open pipe, the "5th harmonic" would correspond to \(n=5\) with even harmonics present as well.

Step 1: Understanding the Concept:

When drops combine, the total volume and total charge are conserved. Potential depends on charge and radius (\(V = \frac{kQ}{R}\)).


Step 2: Key Formula or Approach:

For \(n\) drops combining:

New radius \(R = n^{1/3} r\)

New charge \(Q = nq\)

New potential \(V_{big} = n^{2/3} V_{small}\)


Step 3: Detailed Explanation:

Given:

Number of drops \(n = 64\)

Initial potential \(V_{small} = 10 mV\)


Using the potential scaling formula:
\[ V_{big} = (64)^{2/3} \times V_{small} \]
\[ V_{big} = (64^{1/3})^2 \times 10 \]
\[ V_{big} = (4)^2 \times 10 \]
\[ V_{big} = 16 \times 10 = 160 mV \]


Step 4: Final Answer:

The potential of the bigger drop is 160 mV.
Quick Tip: Quick scaling factors for \(n\) drops: Radius \(\propto n^{1/3}\), Charge \(\propto n^1\), Potential \(\propto n^{2/3}\), Capacitance \(\propto n^{1/3}\), Energy \(\propto n^{5/3}\).

Step 1: Understanding the Concept:

Assuming constant voltage \(V\), current \(I\) is inversely proportional to resistance \(R\). Resistance varies linearly with temperature: \(R_t = R_0(1 + \alpha t)\).


Step 3: Detailed Explanation:

Let the source voltage be \(V\).

At \(0^{\circ}C\): \(R_0 = \frac{V}{2}\)

At \(100^{\circ}C\): \(R_{100} = \frac{V}{1.2} = R_0(1 + 100\alpha)\)


Substituting \(R_0\):
\[ \frac{V}{1.2} = \frac{V}{2} (1 + 100\alpha) \]
\[ \frac{2}{1.2} = 1 + 100\alpha \implies \frac{5}{3} = 1 + 100\alpha \]
\[ 100\alpha = \frac{2}{3} \implies 50\alpha = \frac{1}{3} \]


At \(50^{\circ}C\):
\[ R_{50} = R_0(1 + 50\alpha) \]
\[ R_{50} = \frac{V}{2} \left(1 + \frac{1}{3}\right) = \frac{V}{2} \cdot \frac{4}{3} = \frac{2V}{3} \]


Current at \(50^{\circ}C\):
\[ I_{50} = \frac{V}{R_{50}} = \frac{V}{2V/3} = 1.5 A \]

Converting to mA:
\[ 1.5 A = 1500 mA = 15 \times 10^2 mA \]


Step 4: Final Answer:

The value is 15.
Quick Tip: Note that current does not vary linearly with temperature, but resistance does. Always convert to resistance first when dealing with temperature coefficients.

Step 1: Understanding the Concept:

According to Faraday's Law, induced emf is the rate of change of magnetic flux (\(\epsilon = \left|\frac{d\Phi}{dt}\right|\)). Flux \(\Phi = BA \cos \theta\).


Step 2: Key Formula or Approach:

Here, \(\theta = 0^{\circ}\) (plane \(\perp\) field means normal is \(\parallel\) field).
\(\Phi = B \pi r^2\)
\(\epsilon = \frac{d}{dt}(B \pi r^2) = B \pi (2r \frac{dr}{dt})\)


Step 3: Detailed Explanation:

Given:
\(B = 0.4 T\)

Rate of expansion \(\frac{dr}{dt} = 1 mm/s = 10^{-3} m/s\)

Instantaneous radius \(r = 2 cm = 2 \times 10^{-2} m\)


Calculating induced emf:
\[ \epsilon = 0.4 \times \pi \times 2 \times (2 \times 10^{-2}) \times 10^{-3} \]
\[ \epsilon = 1.6 \pi \times 10^{-5} V \]
\[ \epsilon = 16 \pi \times 10^{-6} V = 16 \pi \muV \]

Using \(\pi \approx 3.14\):
\[ \epsilon \approx 16 \times 3.14 = 50.24 \muV \]


Step 4: Final Answer:

The magnitude is approximately 50 \(\muV\) (or 16\(\pi\)).
Quick Tip: If the field is constant, the emf is "motional" in nature. For an expanding loop, it is equivalent to a rod of length \(2\pi r\) moving with speed \(v\) in a magnetic field (\(\epsilon = Blv\)).

Step 1: Understanding the Concept:

For a system of lenses, the image formed by the first lens acts as the object for the second lens.


Step 3: Detailed Explanation:

1. First Lens:

For a distant object, \(u_1 = \infty\).

Since it is a convex lens, the image is formed at the focus.
\(v_1 = f_1 = 20 cm\) (to the right of the first lens).


2. Second Lens:

The separation is \(d = 60 cm\).

The position of the first image relative to the second lens is \(u_2 = v_1 - d = 20 - 60 = -40 cm\).

Using lens formula \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\):
\[ \frac{1}{v_2} - \frac{1}{-40} = \frac{1}{20} \]
\[ \frac{1}{v_2} = \frac{1}{20} - \frac{1}{40} = \frac{1}{40} \]
\[ v_2 = 40 cm \]

This image is 40 cm to the right of the second lens.


3. Distance from first lens:

Total distance = Separation + \(v_2\)
\[ Distance = 60 + 40 = 100 cm \]


Step 4: Final Answer:

The image is at 100 cm from the first lens.
Quick Tip: Always draw a quick schematic. If the image from the first lens falls between the two lenses, the object for the second lens is real (\(u\) is negative). If it falls beyond, the object is virtual (\(u\) is positive).

Step 1: Understanding the Concept:

The energy released (\(Q\)-value) in a nuclear reaction is equal to the mass defect (\(\Delta m\)) multiplied by the energy equivalent of one atomic mass unit.


Step 2: Key Formula or Approach:
\[ Q = [M_{parent} - (M_{daughter} + M_{\alpha})] \times 931.5 MeV \]


Step 3: Detailed Explanation:

Mass of reactants:
\(M_U = 238.05060 u\)


Mass of products:
\(M_{Th} + M_{He} = 234.04360 + 4.00260 = 238.04620 u\)


Mass defect \(\Delta m\):
\[ \Delta m = 238.05060 - 238.04620 = 0.00440 u \]


Calculating Energy released (\(Q\)):
\[ Q = 0.00440 \times 931.5 MeV \]
\[ Q = 4.0986 MeV \]


Rounding to one decimal place as is common in such numeric entries:
\[ Q \approx 4.1 MeV \]


Step 4: Final Answer:

The energy released is 4.1 MeV.
Quick Tip: In \(\alpha\)-decay, the \(Q\)-value is shared as kinetic energy between the \(\alpha\)-particle and the daughter nucleus. Because the \(\alpha\)-particle is much lighter, it carries away most (\(\approx 98%\)) of this energy.