JEE Main 2024 question paper pdf with solutions- Download April 6 Shift 1 Physics Question Paper pdf

Step 1: Escape velocity formula.

The escape velocity \( V_e \) is given by the formula: \[ V_e = \sqrt{\frac{2GM}{R}} \]

Step 2: Kinetic energy required.

The minimum kinetic energy required for the particle to escape is given by: \[ KE = \frac{1}{2} mv_e^2 \]
Substituting the value of \( V_e \), we get: \[ KE = \frac{1}{2} m \times \frac{2GM}{R} = \frac{GMm}{R} \]

Step 3: Conclusion.

Thus, the minimum kinetic energy required for the particle to escape the earth’s surface is \( \frac{GMm}{R} \).
Quick Tip: Escape velocity is the minimum velocity an object needs to escape from a celestial body’s gravitational influence, and it is derived from the gravitational potential energy and kinetic energy balance.

Step 1: Write the equation for acceleration.

The acceleration of the block is given by the formula: \[ a = \frac{(m_2 - m_1)}{(m_1 + m_2)} g \]

Step 2: Substitute the given values.

We are given that the acceleration of block \( m_1 \) is \( \frac{g}{\sqrt{2}} \). So, we substitute \( a = \frac{g}{\sqrt{2}} \) into the equation: \[ \frac{g}{\sqrt{2}} = \frac{(m_2 - m_1)}{(m_1 + m_2)} g \]

Step 3: Simplify the equation.

Canceling \( g \) from both sides: \[ \frac{1}{\sqrt{2}} = \frac{(m_2 - m_1)}{(m_1 + m_2)} \]

Step 4: Rearrange to solve for \( \frac{m_1}{m_2} \).

Cross-multiplying: \[ 1 = \sqrt{2} - \sqrt{2} \left( \frac{m_1}{m_2} \right) \]
Simplifying: \[ 1 + \frac{m_1}{m_2} = \sqrt{2} - \sqrt{2} \left( \frac{m_1}{m_2} \right) \]

Step 5: Solve for the ratio \( \frac{m_1}{m_2} \).

Solving this gives us: \[ \frac{m_1}{m_2} = \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \]

Step 6: Conclusion.

The ratio of \( m_1 \) to \( m_2 \) is: \[ m_1 : m_2 = 3 - 2\sqrt{2} \] Quick Tip: In pulley systems, the acceleration depends on the relative masses of the blocks. By applying Newton's laws, you can derive a relationship between the masses and the acceleration.

Step 1: Use the time period formula.

The time period \( T \) is related to the angular frequency \( \omega \) by the formula: \[ T = \frac{2\pi}{\omega} \]
Given that \( T = \pi \), we substitute this value into the equation: \[ \pi = \frac{2\pi}{\omega} \]

Step 2: Solve for \( \omega \).

Rearranging the equation: \[ \omega = 2 \, rad/s \]

Step 3: Find the maximum velocity.

The maximum velocity \( v_{max} \) in SHM is given by: \[ v_{max} = \omega A \]
Substituting the values of \( \omega = 2 \, rad/s \) and \( A = 0.6 \, m \), we get: \[ v_{max} = 2 \times 0.6 = 1.2 \, m/s \]

Step 4: Conclusion.

Thus, the maximum velocity is \( 1.2 \, m/s \).
Quick Tip: The maximum velocity in SHM is directly proportional to the amplitude and angular frequency.

Step 1: Use the formula for percentage change in kinetic energy.

The percentage change in kinetic energy is given by: \[ % \Delta K = \frac{K_f - K_i}{K_i} \times 100 \]

Step 2: Write the kinetic energy formula.

The kinetic energy is given by \( K = \frac{1}{2} mv^2 \). So, the initial and final kinetic energies are: \[ K_i = \frac{1}{2} m v_i^2 \quad and \quad K_f = \frac{1}{2} m v_f^2 \]
Thus, the percentage change becomes: \[ % \Delta K = \frac{\frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2}{\frac{1}{2} m v_i^2} \times 100 \]

Step 3: Simplify the expression.
\[ % \Delta K = \left( \frac{v_f^2}{v_i^2} - 1 \right) \times 100 \]

Step 4: Substitute the given values.

Substitute \( v_i = 100 \, m/s \) and \( v_f = 40 \, m/s \): \[ % \Delta K = \left( \frac{(40)^2}{(100)^2} - 1 \right) \times 100 = \left( \frac{1600}{10000} - 1 \right) \times 100 \]

Step 5: Final Calculation.
\[ % \Delta K = \left( \frac{4}{25} - 1 \right) \times 100 = \left( -\frac{21}{25} \right) \times 100 = -84% \]

Step 6: Conclusion.

Thus, the percentage change in kinetic energy is \( -84% \), indicating a decrease in kinetic energy.
Quick Tip: When calculating percentage change in kinetic energy, ensure to square the velocities and take the difference before calculating the percentage.

Step 1: Use Wheatstone Bridge condition.

Since the potential at B and D are the same, this is a Wheatstone bridge. In a Wheatstone bridge, the following condition holds: \[ R_{AB} \times R_{CD} = R_{BC} \times R_{AD} \]

Step 2: Substituting known values.

From the circuit, we have: \[ R_{AB} = 6 \, \Omega, \quad R_{BC} = \frac{1}{2} \, \Omega, \quad R_{CD} = 0.5 \, \Omega \]
Let \( R_{AD} = x + 3 \, \Omega \), as given in the problem. Substituting these values into the Wheatstone bridge condition: \[ 6 \times 0.5 = 0.5 \times (x + 3) \]

Step 3: Simplify the equation.
\[ 3 = 0.5 \times (x + 3) \]
Multiplying both sides by 2: \[ 6 = x + 3 \]

Step 4: Solve for \( x \).
\[ x = 3 \, \Omega \]

Step 5: Conclusion.

Thus, the value of \( x \) is \( 3 \, \Omega \).
Quick Tip: In a Wheatstone bridge, the condition for balanced bridge is \( \frac{R_1}{R_2} = \frac{R_3}{R_4} \).

Step 1: Understanding the problem.

The electric field due to an infinite sheet of charge with surface charge density \( \sigma \) is given by: \[ E = \frac{\sigma}{2 \epsilon_0} \]
For a non-conducting sheet, the electric field points away from the sheet if the charge density is positive, and towards the sheet if the charge density is negative.

Step 2: Electric field contributions.

The electric field due to the three sheets at point \( P \) can be determined by considering the contributions from each sheet individually. The field due to each sheet is constant and directed as follows:

- The sheet with charge density \( -\sigma \) produces an electric field directed towards the sheet.
- The sheet with charge density \( -2\sigma \) produces an electric field directed towards the sheet (but stronger).
- The sheet with charge density \( \sigma \) produces an electric field directed away from the sheet.

Step 3: Calculating the total electric field.

At point \( P \), the electric field due to each sheet adds up according to the principle of superposition:
\[ E_{net} = E_1 + E_2 + E_3 \]
Where: \[ E_1 = \frac{\sigma}{2 \epsilon_0}, \quad E_2 = \frac{2\sigma}{2 \epsilon_0}, \quad E_3 = \frac{\sigma}{2 \epsilon_0} \]
Adding these contributions: \[ E_{net} = \frac{\sigma}{2 \epsilon_0} + \frac{2\sigma}{2 \epsilon_0} + \frac{\sigma}{2 \epsilon_0} = \frac{4\sigma}{2 \epsilon_0} = \frac{2\sigma}{\epsilon_0} \]

Step 4: Conclusion.

Thus, the net electric field at point \( P \) is: \[ E_{net} = \frac{2\sigma}{\epsilon_0} \] Quick Tip: The electric field due to an infinite sheet of charge is constant in magnitude and direction, regardless of the distance from the sheet. The direction depends on the sign of the charge density.

Step 1: Use the formula for rms speed.

The root mean square (rms) speed \( v_{rms} \) of molecules is given by: \[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the gas constant, \( T \) is the temperature, and \( M \) is the molar mass of the gas.

Step 2: Relationship between rms speeds of He and O\(_2\).

Since both gases are at the same temperature, the ratio of their rms speeds is: \[ \frac{v_{rms, He}}{v_{rms, O_2}} = \sqrt{\frac{M_{O_2}}{M_{He}}} \]

Step 3: Substituting the molar masses.

The molar mass of \( O_2 \) is \( 32 \, g/mol \), and the molar mass of He is \( 4 \, g/mol \). So: \[ \frac{v_{rms, He}}{v_{rms, O_2}} = \sqrt{\frac{32}{4}} = \sqrt{8} = 2\sqrt{2} \]

Step 4: Conclusion.

Thus, the ratio of the rms speeds is \( \sqrt{2} \).
Quick Tip: For gases at the same temperature, the rms speed is inversely proportional to the square root of the molar mass.

Step 1: Torque.
The dimensional formula for torque is \( [M L^2 T^{-2}] \), which corresponds to option (a).


Step 2: Magnetic field.
The dimensional formula for magnetic field is \( [M A^{-1} T^{-2}] \), which corresponds to option (b).


Step 3: Magnetic moment.
The dimensional formula for magnetic moment is \( [M L^2 T^0 A] \), which corresponds to option (c).


Step 4: Permeability.
The dimensional formula for permeability is \( [M L^3 T^{-2} A^{-2}] \), which corresponds to option (d).


Step 5: Conclusion.
Thus, the correct matching is: \[ (i) \to (a), \, (ii) \to (b), \, (iii) \to (c), \, (iv) \to (d) \] Quick Tip: To match quantities with their dimensional formulas, remember that torque, magnetic field, and magnetic moment all have specific units that relate to mass, length, time, and current.

The circuit contains the following logic gates:
- Inverter (NOT gate) on input \( A \),
- AND gate for \( A \) and \( B \),
- OR gate for the output of the NOT gate and the AND gate.

Step 1: Truth Table Analysis.

For \( A = 0 \) and \( B = 0 \),
- \( A = 0 \),
- \( \overline{A} = 1 \),
- \( A \cdot B = 0 \),
- Final output = \( 1 \).

For \( A = 0 \) and \( B = 1 \),
- \( A = 0 \),
- \( \overline{A} = 1 \),
- \( A \cdot B = 0 \),
- Final output = \( 1 \).

For \( A = 1 \) and \( B = 0 \),
- \( A = 1 \),
- \( \overline{A} = 0 \),
- \( A \cdot B = 0 \),
- Final output = \( 0 \).

For \( A = 1 \) and \( B = 1 \),
- \( A = 1 \),
- \( \overline{A} = 0 \),
- \( A \cdot B = 1 \),
- Final output = \( 1 \).

Step 2: Truth Table.
Quick Tip: In logic circuits, the output of the NOT gate is the opposite of the input, the AND gate outputs 1 only when both inputs are 1, and the OR gate outputs 1 when at least one input is 1.

Step 1: Given relation.

The time period is given by: \[ T = 2\pi \sqrt{\frac{m}{K}} \]

Step 2: Differentiation with respect to time.

Differentiating both sides with respect to time: \[ \frac{dT}{T} = \frac{1}{2} \left( \frac{dm}{m} - \frac{dK}{K} \right) \]

Step 3: Substitute given values.

We are given that \( m \) decreases by 1% and \( T \) increases by 2%. So: \[ \frac{dm}{m} = -1% \quad and \quad \frac{dT}{T} = +2% \]

Substitute these into the equation: \[ 2% = \frac{1}{2} \left( -1% - \frac{dK}{K} \right) \]

Step 4: Solve for \( \frac{dK}{K} \).
\[ 2% = -\frac{1}{2} \left( 1% + \frac{dK}{K} \right) \]
Simplifying: \[ 2% = -\frac{1}{2} \times 1% - \frac{1}{2} \times \frac{dK}{K} \] \[ 2% + \frac{1}{2} \times 1% = -\frac{1}{2} \times \frac{dK}{K} \] \[ 2% + 0.5% = -\frac{1}{2} \times \frac{dK}{K} \] \[ 2.5% = -\frac{1}{2} \times \frac{dK}{K} \] \[ \frac{dK}{K} = -5% \]

Step 5: Conclusion.

Thus, \( K \) decreases by 5%.
Quick Tip: When calculating percentage changes in variables like \( m \), \( K \), or \( T \), use the method of differentiation to find the relationship between changes in variables.

Step 1: Understand the relationship between the radii.

The surface energy of a droplet is proportional to the square of its radius. If a big drop is made from 1000 small drops, the ratio of surface energy is related to the ratio of the radii. We can write: \[ \frac{1000 \times \frac{4}{3} \pi r^2}{\frac{4}{3} \pi R^2} = \frac{10}{x} \]

Step 2: Simplify the equation.
\[ \frac{r^3}{R^3} = \frac{10}{x} \]
Given that the big drop radius \( R = 10r \), we can substitute this into the equation: \[ \left( \frac{r}{R} \right)^3 = \left( \frac{1}{10} \right)^3 \]
Thus: \[ R = 10r \]

Step 3: Surface energy comparison.

Now, we calculate the ratio of the surface energies: \[ \frac{surface energy final}{surface energy initial} = \frac{T(4 \pi R^2)}{1000 \times T(4 \pi r^2)} = \frac{10}{x} \]
Substituting the value for \( R \), we get: \[ \frac{(10r)^2}{1000 r^2} = \frac{10}{x} \]
Simplifying further: \[ \frac{100}{1000} = \frac{10}{x} \]
Thus, \[ x = 100 \]

Step 4: Conclusion.

Therefore, \( x = 100 \).
Quick Tip: When combining small droplets into a bigger droplet, the radius of the larger droplet increases, but the total surface area does not increase proportionally due to the surface energy relationship.

Step 1: Use the photoelectric equation.

The energy of the incident light is related to the work function \( \varphi \) and the maximum kinetic energy of the emitted electrons by the equation: \[ E = K.E_{max} + \varphi \]
where \( E \) is the incident energy and \( K.E_{max} \) is the maximum kinetic energy of the electrons.

Step 2: Use the stopping potential.

The maximum kinetic energy \( K.E_{max} \) is related to the stopping potential \( V_s \) by: \[ K.E_{max} = eV_s \]
Substitute \( V_s = 0.5 \, V \) into the equation: \[ K.E_{max} = (0.5) \, eV = 0.5 \, eV \]

Step 3: Solve for the work function.

Substitute into the photoelectric equation: \[ 2.48 \, eV = 0.5 \, eV + \varphi \]
Solving for \( \varphi \): \[ \varphi = 2.48 \, eV - 0.5 \, eV = 1.98 \, eV \]

Step 4: Conclusion.

Thus, the work function is \( \varphi = 1.98 \, eV \).
Quick Tip: The work function is the minimum energy required to release an electron from the surface of a metal. It can be calculated using the energy of incident light and the stopping potential.

Step 1: Statement 1 - Inductor at resonance frequency.

In an LRC circuit, at the resonance frequency, the inductive reactance (\(X_L\)) and capacitive reactance (\(X_C\)) are equal and cancel each other out. Hence, the current is maximum at this frequency because the total impedance is at its minimum. Therefore, Statement 1 is correct.


Step 2: Statement 2 - Current in purely resistive circuit.

In a purely resistive circuit, the current is determined solely by the resistance, and it cannot be less than the current in a series LRC circuit where the total impedance is greater due to the inductive and capacitive reactance. Therefore, Statement 2 is also correct.


Step 3: Conclusion.

Thus, both statements are correct, and the correct answer is option (3).
Quick Tip: In an LRC circuit, resonance occurs when the inductive reactance equals the capacitive reactance, resulting in maximum current at resonance frequency.