JEE Main 2024 question paper pdf with solutions- Download April 6 Shift 2 Physics Question Paper pdf

Step 1: Analyzing the forces acting on the car.

The forces acting on the car are the normal force (\(N\)), frictional force (\(F\)), and the weight of the car (\(mg\)). The frictional force provides additional centripetal force for the car to turn safely. The equation of motion can be written as: \[ N \sin \theta + \mu N \cos \theta = \frac{mv^2}{R} \]
And for vertical equilibrium: \[ N \cos \theta = \mu N \sin \theta = mg \]

Step 2: Deriving the equation for maximum speed.

From the above relations, we can derive the maximum speed equation as: \[ v_{max} = \sqrt{\frac{\tan \theta + \mu}{1 - \mu \tan \theta} Rg} \]

Step 3: Substituting the given values.

Given: \[ \mu = 0.2, \, \theta = 30^\circ, \, R = 300 \, m, \, g = 10 \, m/s^2 \]
Substitute the values into the equation: \[ v_{max} = \sqrt{\frac{0.2 + \frac{1}{\sqrt{3}}}{1 - 0.2 \times \frac{1}{\sqrt{3}}} \times 300 \times 10} \]

Step 4: Simplifying the expression.

First, calculate the trigonometric values: \[ \frac{1}{\sqrt{3}} \approx 0.577 \]
Now, simplify the expression: \[ v_{max} = \sqrt{\frac{0.2 + 0.577}{1 - 0.2 \times 0.577} \times 3000} \] \[ v_{max} = \sqrt{\frac{0.777}{0.884} \times 3000} \] \[ v_{max} = \sqrt{0.877 \times 3000} = \sqrt{2631} \approx 51.2 \, m/s \] Quick Tip: When solving for the safe speed of a car on a banked road, remember to account for both the angle of banking and the frictional force to ensure the car does not slip.

Step 1: Given data.

The block's weight: \( W = 200 \, N \)

The mass of the chain: \( m = 10 \, kg \)

The total weight at the topmost point is the sum of the weight of the block and the chain: \[ F = W_{block} + W_{chain} = 200 \, N + 100 \, N = 300 \, N \] Quick Tip: When calculating the tension in a chain, consider the weight of both the block and the chain.

Step 1: Relating kinetic energy and momentum.

Kinetic energy is given by: \[ K = \frac{p^2}{2m} \quad \Rightarrow \quad p = \sqrt{2mK} \]

Step 2: Finding the change in momentum.

Let the initial and final kinetic energies be \( K_i \) and \( K_f \), respectively. Then, the change in momentum \(\Delta p\) is: \[ \Delta p = \frac{p_f - p_i}{p_i} \times 100 \]
Substituting the expressions for momentum, we get: \[ \Delta p = \frac{\sqrt{2m K_f} - \sqrt{2m K_i}}{\sqrt{2m K_i}} \times 100 \] \[ \Delta p = \left( \frac{\sqrt{K_f}}{\sqrt{K_i}} - 1 \right) \times 100 \]

Step 3: Substituting the percentage increase in kinetic energy.

Given that the kinetic energy increases by 36%, we have: \[ \frac{K_f}{K_i} = 1.36 \]
Thus, \[ \Delta p = \left( \sqrt{1.36} - 1 \right) \times 100 = (1.166 - 1) \times 100 = 16.62% \] Quick Tip: When solving for momentum changes due to kinetic energy changes, remember the relationship \(K = \frac{p^2}{2m}\) and apply it to calculate the momentum change.

Step 1: Using the photoelectric equation.

The photoelectric equation is given by: \[ E_{photon} = \frac{hc}{\lambda} \]
Where: \( h = 6.626 \times 10^{-34} \, J s \), \( c = 3 \times 10^8 \, m/s \), \( \lambda = 300 \, nm = 300 \times 10^{-9} \, m \).

Step 2: Calculating the energy of the photon.

Substitute the values: \[ E_{photon} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}} = 12400 \, eV \]

Step 3: Finding the stopping potential.

The energy of the photon is related to the stopping potential by the equation: \[ E_{photon} = eV_s + \phi \]
Where: \( e = 1.6 \times 10^{-19} \, C \), \( V_s = ? \), \( \phi = 2.4 \, eV \).
\[ V_s = \frac{E_{photon} - \phi}{e} = \frac{12400 \, eV - 2.4 \, eV}{1.6 \times 10^{-19} \, C} \approx 1.73 \, V \] Quick Tip: When calculating the stopping potential, subtract the work function from the photon energy and divide by the charge of the electron.

Step 1: Analyzing the motion of each particle.


For the first particle, which is thrown vertically upwards: \[ -H = u t_1 - \frac{1}{2} g t_1^2 \quad ... (i) \]
For the second particle, which is thrown vertically downwards: \[ -H = u t_2 - \frac{1}{2} g t_2^2 \quad ... (ii) \]
For the third particle, which is released from rest: \[ -H = 0 - \frac{1}{2} g t_3^2 \quad ... (iii) \]

Step 2: Multiply the equations.

Multiply equation (i) by \(t_1\), and equation (ii) by \(t_2\), then add the equations: \[ -H t_1 - H t_2 = -\frac{1}{2} g t_1^2 - \frac{1}{2} g t_2^2 \]
\[ H = \frac{1}{2} g t_1 t_2 \]

Step 3: Relating \(t_3\).

Now, substitute the value for \(H\) from equation (iii): \[ \frac{1}{2} g t_1 t_2 = \frac{1}{2} g t_3^2 \]

Simplifying: \[ t_3^2 = t_1 t_2 \]

Thus, \[ t_3 = \sqrt{t_1 t_2} \] Quick Tip: In problems involving motion under gravity, carefully apply the kinematic equations for each particle and solve step-by-step.

Step 1: Specific Heat Formula.

Specific heat is defined as the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius or Kelvin. The formula is: \[ dQ = mS dT \] \[ [S] = \frac{[dQ]}{m \, [dT]} = \frac{M L^2 T^{-2}}{M K} = ML^2 T^{-2} K^{-1} \]

Step 2: Latent Heat Formula.

Latent heat is the amount of heat required to change the state of a unit mass of a substance at constant temperature. The formula is: \[ dQ = mL \] \[ [L] = \frac{[dQ]}{m} = \frac{M L^2 T^{-2}}{M} = L^2 T^{-2} \] Quick Tip: To find the dimensional formula of physical quantities, use the fundamental quantities of mass, length, time, and temperature, and apply the appropriate formula.

Step 1: Weight at the surface.

At the surface of the Earth, the weight of the object is given as: \[ W = mg \quad where \quad g = g_s \quad is the acceleration due to gravity at the surface. \]
Given: \[ W = 300 \, N \quad \Rightarrow \quad 300 = mg \quad ... (i) \]

Step 2: Gravity at depth.

At a depth \(d = R/4\) from the surface of the Earth, the acceleration due to gravity changes. The formula for gravity at a depth is: \[ g' = g_s \left[ 1 - \frac{d}{R} \right] \]
Substitute \(d = \frac{R}{4}\): \[ g' = g_s \left[ 1 - \frac{R/4}{R} \right] = g_s \left[ 1 - \frac{1}{4} \right] = g_s \left[ \frac{3}{4} \right] \]
Thus, the new gravity is: \[ g' = \frac{3}{4} g_s \]

Step 3: Weight at depth.

Now, the weight of the object at the depth is: \[ W' = mg' = m \left(\frac{3}{4} g_s\right) \]
Using equation (i), where \( mg = 300 \, N \), we get: \[ W' = 300 \times \frac{3}{4} = 225 \, N \] Quick Tip: At a depth inside the Earth, the acceleration due to gravity decreases, and the weight of an object is reduced proportionally to the depth.

Step 1: Using the formula for the internal energy of an ideal gas.

The internal energy \(U\) for an ideal gas is given by the equation: \[ U = \frac{3}{2} nRT \]
Where: \(n = 10\) moles, \(R = 8.31 \, J/mol K\), \(T\) is the temperature.

Step 2: Substituting the given values.

Substitute \(n = 10\) into the formula: \[ U = \frac{3}{2} \times 10 \times RT = 15RT \] Quick Tip: For an ideal gas, the internal energy is directly proportional to the temperature and the number of moles. For a monoatomic ideal gas, the factor \(\frac{3}{2}\) comes from its degrees of freedom.

Step 1: Using the formula for the intensity of an EM wave.

The intensity \(I\) of an electromagnetic wave is given by: \[ I = \frac{1}{2} \epsilon_0 E_0^2 c \]
Where: \(\epsilon_0 = 8.85 \times 10^{-12} \, C^2/Nm^2\) (permittivity of free space), \(E_0 = 600 \, V/m\) (maximum electric field), \(c = 3 \times 10^8 \, m/s\) (speed of light).

Step 2: Substituting the given values.

Substitute \(E_0 = 600\), \(\epsilon_0 = 8.85 \times 10^{-12}\), and \(c = 3 \times 10^8\) into the formula: \[ I = \frac{1}{2} \times 8.85 \times 10^{-12} \times (600)^2 \times 3 \times 10^8 \] \[ I = \frac{1}{2} \times 8.85 \times 10^{-12} \times 360000 \times 10^8 \] \[ I = \frac{1}{2} \times 8.85 \times 6 \times 10^6 = 477.9 \, watt/m^2 \] Quick Tip: The intensity of an electromagnetic wave is proportional to the square of the maximum electric field and the speed of light. Use the formula \(I = \frac{1}{2} \epsilon_0 E_0^2 c\) for quick calculations.

Step 1: Initial setup.

The force between two identical conducting shells with the same charge is given by Coulomb’s law: \[ F = \frac{kQ^2}{r^2} = 16 \, N \]
Where \(Q\) is the charge on each shell, \(r\) is the distance between the centers, and \(k\) is Coulomb's constant.

Step 2: Charge distribution after touching the first shell.

When the uncharged identical conducting shell is touched to the first shell, the charge gets distributed equally between both the shells. Therefore, the charge on each shell becomes: \[ Q_1 = Q_2 = \frac{Q}{2} \]

Step 3: Touching the second shell.

Now, the uncharged shell is touched to the second shell, and the charge gets redistributed again. The new charge on the shells is: \[ Q_1' = \frac{Q}{2} + \frac{Q}{4} = \frac{3Q}{4}, \quad Q_2' = \frac{3Q}{4} \]

Step 4: Final force calculation.

Now, the new force between the shells is given by Coulomb’s law: \[ F' = \frac{k \left(\frac{3Q}{4}\right)^2}{r^2} = \frac{9}{16} \times \frac{kQ^2}{r^2} \]
Substitute the given value of the initial force: \[ F' = \frac{9}{16} \times 16 = 6 \, N \] Quick Tip: When two conductors touch, charge gets redistributed equally. After touching, use Coulomb's law to find the new force by adjusting the charges on each conductor.

Step 1: Analyzing List-1:

- (A) The magnetic field decreases with distance from the wire and behaves like a curve, which matches with (IV) from List-2.
- (B) The magnetic field drops off rapidly with increasing distance from the wire and has a characteristic shape, matching (II) from List-2.
- (C) The magnetic field inside a solenoid increases with distance from the centre, which matches (III) from List-2.
- (D) The magnetic susceptibility increases linearly with intensity of magnetisation, corresponding to (I) from List-2. Quick Tip: Magnetic fields around wires decrease with distance, and for solenoids, the field increases linearly with distance from the centre. Magnetic susceptibility vs. intensity of magnetisation is a linear relation.

Step 1: Equivalent resistance of the ammeter.

The equivalent resistance \(R_{eq}\) of the ammeter is given by: \[ R_{eq of A} = \frac{2400}{250} = \frac{48}{5} = 9.6 \, \Omega \]

Step 2: Equivalent resistance of the circuit.

The total resistance of the circuit is: \[ R_{eq of circuit} = 140.4 \, \Omega + \frac{48}{5} = 140.4 + 9.6 = 150 \, \Omega \]

Step 3: Current through the circuit.

The current \(i_b\) is given by: \[ i_b = \frac{24}{150} = 0.16 \, A \]

Thus, the reading of the ammeter is: \[ A = \frac{24}{150} \] Quick Tip: For an ammeter with a shunt resistance, the equivalent resistance is calculated by adding the shunt resistance in parallel with the resistance of the galvanometer coil.

Step 1: Using the resonance condition.

At resonance, the inductive reactance \(X_L\) equals the capacitive reactance \(X_C\). Thus, \[ X_L = X_C \]

Step 2: Formula for the reactance.

The inductive reactance is given by: \[ X_L = \omega L \]
The capacitive reactance is given by: \[ X_C = \frac{1}{\omega C} \]

Step 3: Solving for the angular frequency.

Equating \(X_L\) and \(X_C\), we get: \[ \omega L = \frac{1}{\omega C} \]
Solving for \(\omega\): \[ \omega^2 = \frac{1}{LC} \] \[ \omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{100 \times 2.5 \times 10^{-9}}} \] \[ \omega = \frac{1}{\sqrt{250 \times 10^{-9}}} = \frac{1}{5 \times 10^{-5}} = 2000 \, rad/sec \] Quick Tip: In an LRC circuit, the frequency at resonance is determined by the relationship \(\omega = \frac{1}{\sqrt{LC}}\).

Step 1: Using the formula for wavelength in the Bohr model.

The wavelength of a line in the hydrogen spectrum is given by: \[ \frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Where: \( R_H = 1.09 \times 10^7 \, m^{-1} \), \( n_1 = 3 \) (for the Paschen series), \( n_2 = 4 \) (maximum wavelength corresponds to the transition from \(n = 4\) to \(n = 3\)).

Step 2: Substituting the known values.

Substitute the values of \(R_H\), \(n_1\), and \(n_2\) into the equation: \[ \frac{1}{\lambda} = [1.09 \times 10^7] \times \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \] \[ \frac{1}{\lambda} = [1.09 \times 10^7] \times \left( \frac{1}{9} - \frac{1}{16} \right) \] \[ \frac{1}{\lambda} = [1.09 \times 10^7] \times \left( \frac{7}{144} \right) \] \[ \frac{1}{\lambda} = 0.053 \times 10^7 \]

Step 3: Final calculation of the wavelength.

Thus, \[ \lambda = \frac{1}{0.053 \times 10^7} = 18.867 \times 10^{-7} \, m \] Quick Tip: The maximum wavelength in the Paschen series corresponds to the transition from \(n_2 = 4\) to \(n_1 = 3\). The formula for the wavelength in the Bohr model is useful for determining wavelengths in hydrogen's spectral lines.

Step 1: Using the formula for maximum and minimum intensity in interference.

The maximum intensity \( I_{max} \) is given by: \[ I_{max} = ( \sqrt{I_1} + \sqrt{I_2} )^2 \]
Substitute \( I_1 = 4I \) and \( I_2 = I \): \[ I_{max} = \left( \sqrt{4I} + \sqrt{I} \right)^2 = (2\sqrt{I} + \sqrt{I})^2 = 9I \]

Step 2: Minimum intensity.

The minimum intensity \( I_{min} \) is given by: \[ I_{min} = ( \sqrt{I_1} - \sqrt{I_2} )^2 \]
Substitute \( I_1 = 4I \) and \( I_2 = I \): \[ I_{min} = \left( \sqrt{4I} - \sqrt{I} \right)^2 = (2\sqrt{I} - \sqrt{I})^2 = I \]

Step 3: Ratio of maximum to minimum intensity.

The ratio of maximum intensity to minimum intensity is: \[ \frac{I_{max}}{I_{min}} = \frac{9I}{I} = 9 \]

Thus, the ratio of maximum to minimum intensity is \( 8I \). Quick Tip: For interference patterns, the maximum intensity occurs when the two waves are in phase, and the minimum intensity occurs when they are out of phase.