JEE Main 2024 question paper pdf with solutions- Download April 4 Shift 2 Physics Question Paper pdf

Step 1: Understanding the forces involved.

The block starts sliding when the component of gravitational force parallel to the incline exceeds the force of friction. The forces are:
- Gravitational force: \( mg \sin \theta \)
- Frictional force: \( \mu mg \cos \theta \)

Step 2: Equating the forces.

For the block to start sliding, the frictional force must be equal to the parallel component of gravitational force: \[ mg \sin \theta = \mu mg \cos \theta \]

Step 3: Simplifying the equation.

By canceling \( mg \) from both sides, we get: \[ \tan \theta = \mu \]

Step 4: Substituting \( \theta = 45^\circ \).

At \( \theta = 45^\circ \), we know that \( \tan 45^\circ = 1 \), so: \[ \mu = 1 \] Quick Tip: For a body to just start sliding on an incline, the friction coefficient is equal to the tangent of the angle of inclination.

Step 1: Understanding the setup.

The setup involves a rotating rod with magnetic field perpendicular to the plane of rotation. Points \( P \) and \( Q \) are at different locations on the rod, and we need to find the potential difference between them. Since the magnetic field is perpendicular to the plane of rotation, it causes a potential difference between these two points as the rod moves through the field.


Step 2: Using Faraday's Law.

In this case, the induced EMF is given by the equation: \[ EMF = B l \omega \]
where \( B \) is the magnetic field, \( l \) is the length of the rod, and \( \omega \) is the angular velocity of the rod. The potential difference between the points \( P \) and \( Q \) will be zero if they are at the same potential. Since \( P \) and \( Q \) are at the same potential, we conclude that: \[ V_P - V_Q = 0 \] Quick Tip: In a rotating conductor within a magnetic field, the potential difference between points on the conductor can be determined using Faraday’s Law of induction. If the points are at the same potential, the difference will be zero.

Step 1: Understanding the problem.

We are given a cube with charge \( Q \) placed at the centre of one of its faces. The total flux through the cube is calculated using Gauss’s Law.


Step 2: Applying Gauss’s Law.

By Gauss’s Law, the total electric flux through a closed surface is given by: \[ \Phi = \frac{q_{enc}}{\epsilon_0} \]
where \( q_{enc} \) is the charge enclosed by the surface and \( \epsilon_0 \) is the permittivity of free space.


Step 3: Understanding the cube configuration.

Since the charge \( Q \) is placed at the centre of one face of the cube, the charge enclosed in the cube will be half of \( Q \), because only half of the cube will contain the charge. Therefore, the charge enclosed is: \[ q_{enc} = \frac{Q}{2} \]

Step 4: Calculating the flux.

Thus, the total flux through the cube is: \[ \Phi = \frac{q_{enc}}{\epsilon_0} = \frac{Q}{2 \epsilon_0} \] Quick Tip: When using Gauss’s Law, the flux through a surface depends on the charge enclosed within that surface. If a charge is placed on a face of a cube, only part of the cube will enclose the charge.

Step 1: Understanding the formula for resistance.

The resistance \( R \) of a wire is given by the formula: \[ R = \rho \frac{L}{A} \]
where \( \rho \) is the resistivity, \( L \) is the length of the wire, and \( A \) is the cross-sectional area of the wire. The area \( A \) of a wire with radius \( r \) is given by: \[ A = \pi r^2 \]

Step 2: Using the given data.

Let the radius of wire A be \( r_A \) and that of wire B be \( r_B = 2r_A \). The resistance of A is given as \( R_A = 2 \, \Omega \). Using the formula for resistance, we can write the ratio of the resistances of wires B and A: \[ \frac{R_B}{R_A} = \frac{\rho \frac{L}{A_B}}{\rho \frac{L}{A_A}} = \frac{A_A}{A_B} = \frac{r_A^2}{r_B^2} = \frac{r_A^2}{(2r_A)^2} = \frac{1}{4} \]

Step 3: Calculating the resistance of B.

Since \( R_A = 2 \, \Omega \), we find: \[ R_B = R_A \times \frac{1}{4} = 2 \times 4 = 8 \, \Omega \] Quick Tip: Resistance of a wire is inversely proportional to the square of its radius. If the radius of a wire is doubled, the resistance increases by a factor of 4.

Step 1: Using energy conservation.

Since the particle is released from rest, we can use the principle of conservation of mechanical energy. The total mechanical energy at the initial position is equal to the total mechanical energy at the final position, neglecting air resistance. The equation for energy conservation is: \[ Initial Potential Energy + Initial Kinetic Energy = Final Potential Energy + Final Kinetic Energy \]
At the initial position, the kinetic energy is zero because the particle is at rest. So, the equation becomes: \[ mg(R + R \cos 45^\circ) = \frac{1}{2}mv^2 \]
where \( m = 2 \, kg \), \( g = 10 \, m/s^2 \), \( R = 14 \, m \), and \( v \) is the velocity at point B.


Step 2: Solving for velocity.

Simplifying the equation: \[ 2g \left( 1 + \frac{1}{\sqrt{2}} \right) = v^2 \]
Substitute the values: \[ 2 \times 10 \times 14 \times \left( 1 + \frac{1}{\sqrt{2}} \right) = v^2 \] \[ = 280 + \frac{280}{\sqrt{2}} \] \[ = 280 + 200 = 480 \]
Thus, the velocity at point B is: \[ v = \sqrt{480} = 4\sqrt{30} \, m/s \] Quick Tip: In problems involving circular motion and energy conservation, the sum of potential and kinetic energies is constant, and you can use this principle to solve for unknown quantities such as velocity.

Step 1: Using the power formula.

The power consumed by an electrical instrument is related to the voltage and resistance by the formula: \[ P = \frac{V^2}{R} \quad \Rightarrow \quad P \propto V^2 \]

Step 2: Applying the formula for two different voltages.

Let the power at 200 volts be \( P_1 = 500 \, W \) and the power at 100 volts be \( P_2 \). From the equation, we have: \[ \frac{P_2}{P_1} = \left( \frac{V_2}{V_1} \right)^2 \]
Substitute \( V_1 = 200 \) and \( V_2 = 100 \): \[ \frac{P_2}{500} = \left( \frac{100}{200} \right)^2 = \frac{1}{4} \]

Step 3: Calculating the power consumed at 100 volts.

Thus, we find: \[ P_2 = \frac{500}{4} = 125 \, W \] Quick Tip: Power consumed by an electrical instrument is proportional to the square of the voltage applied.

Step 1: Formula for magnetic field due to a current-carrying wire.

The magnetic field at a distance \( r \) from an infinitely long straight wire carrying current \( I \) is given by: \[ B = \frac{\mu_0 I}{2 \pi r} \]
where \( \mu_0 \) is the permeability of free space, and \( r \) is the distance from the wire.


Step 2: Applying the formula for points A and B.

Let the distances from the wires to points A, B, and C be \( r_A \), \( r_B \), and \( r_C \), respectively. The magnetic fields at points A and B are given by: \[ B_A = \frac{\mu_0 I}{2 \pi r_A}, \quad B_B = \frac{\mu_0 I}{2 \pi r_B} \]
The total magnetic field at point A and B is the sum of the contributions from the two wires. At point A, the field is due to the current in both wires, and similarly for point B.

Step 3: Finding the ratio of magnetic fields.

The total magnetic field at point A is: \[ B_A = \frac{\mu_0 I}{2 \pi r_A} + \frac{\mu_0 I}{2 \pi r_C} \]
And at point B: \[ B_B = \frac{\mu_0 I}{2 \pi r_B} + \frac{\mu_0 I}{2 \pi r_C} \]

Thus, the ratio \( \frac{B_A}{B_C} \) is: \[ \frac{B_A}{B_B} = \frac{5 \times 3}{7 \times 3} = \frac{5}{7} \] Quick Tip: The magnetic field at a point due to multiple current-carrying wires is the vector sum of the individual fields from each wire. The distance from each wire affects the magnitude of the field.

Step 1: Understanding the situation.

The particle travels along a circular path from point P to point S. The radius of the circle is 2 meters. We are asked to find the displacement of the particle.


Step 2: Calculating the displacement.

The displacement is the shortest distance between the initial and final positions of the particle. Since the particle moves along a circle, the displacement is the straight line between points P and S, which is the chord of the circle.


Step 3: Applying the formula.

The displacement \( \left| \overrightarrow{PS} \right| \) can be calculated using the Pythagorean theorem. The distance from P to S is the diagonal of a square formed by the radius of the circle. The formula for the displacement is: \[ |PS| = \sqrt{r^2 + r^2} = \sqrt{2r \times 2r} = \sqrt{2 \times 2 \times 2} = 2.82 \, m \] Quick Tip: When the particle moves along a circular path, the displacement is always the shortest distance between the starting and ending points, which is the chord of the circle.

Step 1: Formula for gravity at height \( h \).

The gravitational force at height \( h \) from the Earth’s surface is given by: \[ g = g_s \left( \frac{1}{1 + \frac{h}{R}} \right)^2 \]
where \( g_s \) is the gravitational acceleration at the surface of the Earth, and \( R \) is the radius of the Earth. The weight of the man at height \( h = 2R \) is given by: \[ g = g_s \left( \frac{1}{1 + \frac{2R}{R}} \right)^2 = g_s \left( \frac{1}{3} \right)^2 = \frac{g_s}{9} \]

Step 2: Calculating the weight of the man.

Since the mass of the man is 90 kg, the weight at the height of \( 2R \) will be: \[ W = \frac{W_s}{9} \]
where \( W_s = 90 \times 9.8 = 882 \, N \), so: \[ W = \frac{882}{9} = 98 \, N \]

Step 3: Conclusion.

Thus, the weight of the man at height \( 2R \) is \( 10 \, kg-wt \). Quick Tip: At a height \( h \) above the Earth's surface, the weight of an object decreases as \( g \) decreases with the square of the distance from the Earth's center.

Step 1: Understanding the Center of Mass Formula.

The center of mass \( d_{com} \) of a system of particles is given by the equation: \[ d_{com} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]
where:
- \( m_1 = 3 \, kg \) (mass of first particle)
- \( m_2 = 2 \, kg \) (mass of second particle)
- \( x_1 = 2 \, cm \) (displacement of the 3 kg mass towards the 2 kg mass)
- \( x_2 \) is the displacement of the 2 kg mass (to be calculated) towards the 3 kg mass such that the center of mass remains at the same point.

Step 2: Apply the condition that the center of mass remains stationary.

We want the center of mass to remain at the same point, which means: \[ d_{com} = 0 \]
Substitute the values in the center of mass formula: \[ 0 = \frac{3(2) + 2(-x)}{3 + 2} \]
Here, \( x \) is the displacement of the 2 kg mass towards the 3 kg mass. We take the displacement as negative because it is towards the 3 kg mass.


Step 3: Solve for \( x \).

Simplify the equation: \[ 0 = \frac{6 - 2x}{5} \]
Multiply both sides by 5 to eliminate the denominator: \[ 0 = 6 - 2x \]
Now solve for \( x \): \[ 2x = 6 \quad \Rightarrow \quad x = 3 \, cm \] Quick Tip: The center of mass is calculated by taking the weighted average of the positions of the masses. When the system is in equilibrium, the displacements must balance out such that the center of mass remains unchanged.

Step 1: Applying the equation for adiabatic process.

The equation for an adiabatic process is given by: \[ P V^{\gamma} = constant \]
Given that \( \gamma = \frac{3}{2} \), initial volume \( V_1 = 20 \, L \), final volume \( V_2 = 60 \, L \), and initial pressure \( P_1 = 5 \, atm \), we can use the equation to find the final pressure \( P_2 \).

Step 2: Calculate final pressure.

Using the equation: \[ P_1 V_1^{\gamma} = P_2 V_2^{\gamma} \] \[ \Rightarrow 5(60)^{3/2} = P_2 (20)^{3/2} \] \[ \Rightarrow P_2 = \frac{5 (60)^{3/2}}{(20)^{3/2}} = 5 \times 3^{3/2} \] \[ \Rightarrow P_2 = 15 \sqrt{3} \, atm \]

Step 3: Calculating the work done.

The work done in an adiabatic process is given by the formula: \[ W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} \]
Substitute the values: \[ W = \frac{5 \times 60 - 15 \sqrt{3} \times 20}{\frac{3}{2} - 1} \] \[ = \frac{300 - 15 \sqrt{3} \times 20}{\frac{1}{2}} = 20 \times 5 \left[ 3 - \sqrt{3} \right] \] \[ = 200 \times \left( 3 - \sqrt{3} \right) = 600 \left( 1 - \sqrt{3} \right) \, atm Litre \] Quick Tip: In an adiabatic process, the work done is calculated using the relationship between pressure and volume, and the value of \( \gamma \).

Step 1: Understanding the phase relationships.

In AC circuits, the phase relationship between the current and voltage determines the behavior of the component. The four types of components are:
- Capacitive: Current leads voltage by 90 degrees.
- Inductive: Current lags voltage by 90 degrees.
- Resistive: Current and voltage are in phase (i.e., no phase difference).


Step 2: Matching the options.

- (A) The first graph shows current leading voltage, which represents a capacitive circuit.
- (B) The second graph shows current lagging voltage, representing an inductive circuit.
- (C) The third graph shows current and voltage in phase, which corresponds to a resistive circuit.
- (D) The fourth graph shows current lagging voltage again, which is inductive.


Step 3: Final answer.

The correct matching is: \[ A \to (i) \, Capacitive, \quad B \to (ii) \, Inductive, \quad C \to (iii) \, Resistive, \quad D \to (ii) \, Inductive \] Quick Tip: In AC circuits: - Capacitive circuits have current leading voltage. - Inductive circuits have current lagging voltage. - Resistive circuits have current and voltage in phase.

Step 1: Starting with the formula for the time period of a satellite.

The time period \( T \) of a satellite in orbit is given by the formula: \[ T^2 \propto \frac{R^3}{GM} \]
where:
- \( R \) is the radius of the orbit (distance from the center of the Earth to the satellite),
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth.


Step 2: Dimensional analysis.

From the dimensional formula of the time period \( [T] \), we can use the relation: \[ [T] = [L] \left[ m \cdot L^{-1} \cdot T^{-2} \cdot [M] \right]^1 \]
where the exponents of the dimensional formula give us the values to calculate the relation between \( T \), \( R \), \( G \), and \( M \).


Step 3: Conclusion.

Using dimensional analysis, we find that: \[ T^2 \propto \frac{R^3}{GM} \] Quick Tip: The time period of a satellite is related to the radius of its orbit and the mass of the central body (Earth) by the equation \( T^2 \propto \frac{R^3}{GM} \).

Step 1: Using the formula for angular momentum.

The angular momentum \( L \) of an electron in orbit in Bohr’s model is given by: \[ L = mvr = \frac{nh}{2\pi} \]
where \( n \) is the principal quantum number (in this case, \( n = 4 \)), \( h \) is Planck’s constant, and \( m \) is the mass of the electron.


Step 2: Substituting the values.

For \( n = 4 \), we have: \[ L = \frac{4h}{2\pi} = \frac{2h}{\pi} \] Quick Tip: In Bohr's model, the angular momentum of an electron is quantized and is given by \( L = \frac{nh}{2\pi} \), where \( n \) is the principal quantum number.

Step 1: Understanding the assertion.

The assertion states that the number of photons increases with an increase in the frequency of light. However, this is not true. The number of photons depends on the intensity of light, not just its frequency. Higher frequency light does not necessarily increase the number of photons; it increases the energy per photon.


Step 2: Understanding the reason.

The reason given is true. According to the photoelectric equation, the maximum kinetic energy of the emitted electrons is given by: \[ K_{max} = h\nu - \phi \]
where \( h\nu \) is the energy of the incident photon and \( \phi \) is the work function of the material. As the frequency of light increases, the kinetic energy of the emitted electrons increases, provided the frequency is above the threshold frequency.


Step 3: Conclusion.

Thus, the assertion is false, but the reason is true. Quick Tip: The number of photons is related to the intensity of light, not its frequency. The energy of each photon increases with frequency, which in turn increases the kinetic energy of emitted electrons in the photoelectric effect.

Step 1: Using the formula for work done.

The work done to bring a magnetic moment \( \mu \) in a magnetic field \( B \) from one position to another is given by the formula: \[ W_{ext} = - \Delta U \]
where \( U = -\mu \cdot B \) is the potential energy.


Step 2: Calculating the change in potential energy.

The work done is: \[ W_{ext} = - \left( (\mu B)_{final} - (\mu B)_{initial} \right) \]
When the magnetic moment is in the most stable position, the angle \( \theta = 0^\circ \), and when it is in the least stable position, the angle \( \theta = 180^\circ \). Therefore: \[ W_{ext} = MB \left[ \cos(180^\circ) - \cos(0^\circ) \right] \] \[ W_{ext} = 40 \times 10^{-18} \times (-1 - 1) \] \[ W_{ext} = 8 \times 10^{-17} \, J \] Quick Tip: The work done to move a magnetic moment in a magnetic field depends on the change in potential energy, which is related to the angle between the magnetic moment and the magnetic field.