JEE Main 2024 question paper pdf with solutions- Download April 5 Shift 2 Physics Question Paper pdf

Step 1: Understanding the circuit.

The given circuit involves two gates: an OR gate and an AND gate. To ensure that the output is 0, we need to check the combination of inputs \( A \) and \( B \) that will lead to this result.


Step 2: Simplifying the circuit expression.

The OR gate produces an output of \( A + B \), and the AND gate receives the output from the OR gate and \( B \) as inputs, so the expression becomes \( (A + B) \cdot B \).


Step 3: Determining the correct values.

To get the output as 0, we analyze the following possibilities:
- When \( A = 0 \) and \( B = 0 \), the expression simplifies to \( (0 + 0) \cdot 0 = 0 \), which satisfies the condition for the output being 0.


Step 4: Conclusion.

Thus, the correct values for \( A \) and \( B \) are 0, 0.
Quick Tip: In logical circuits, always simplify the Boolean expressions to find the combination of inputs that will yield the desired output.

Step 1: Power and velocity relation.

The power delivered is related to the velocity by the equation: \[ P = \frac{1}{2} mv^2 \]
Solving for \( v \), we get: \[ v = \sqrt{\frac{2Pt}{m}} \]

Step 2: Relating distance and velocity.

The distance travelled is the integral of velocity: \[ \frac{dx}{dt} = v = \sqrt{\frac{2Pt}{m}} \]

Step 3: Integration.

Integrating both sides from \( t = 0 \) to \( t \), and from \( x = 0 \) to \( x \), we get: \[ x = \int_0^t \sqrt{\frac{2Pt}{m}} \, dt \] \[ x = \frac{2}{3} \sqrt{\frac{2P}{m}} t^{3/2} \] Quick Tip: To find distance travelled under constant power, use the velocity equation and integrate over time.

Step 1: Maximum range formula.

The maximum range for projectile motion is given by: \[ R_{max} = \frac{u^2}{g} \]
where \( u \) is the initial velocity and \( g \) is the acceleration due to gravity.

Step 2: Effect of halving velocity.

When the velocity is halved, the new maximum range becomes: \[ R'_{max} = \frac{(u/2)^2}{g} = \frac{R_{max}}{4} \] \[ R'_{max} = \frac{64}{4} = 16 \, m \] Quick Tip: When the initial velocity is halved, the range of a projectile decreases by a factor of 4.

Step 1: Resistance of each part.

The total resistance of the wire is 20\(\Omega\), and it is divided into 10 equal parts, so the resistance of each part is: \[ R = \frac{20}{10} = 2\Omega \]

Step 2: Using the formula for parallel resistors.

The resistances are connected in parallel, and the formula for equivalent resistance \( R_{eq} \) of parallel resistors is: \[ \frac{1}{R_{eq}} = \frac{1}{R} + \frac{1}{R} + \cdots (10 times) \] \[ \frac{1}{R_{eq}} = \frac{10}{R} = \frac{10}{2} = 5 \]

Step 3: Calculate the equivalent resistance.

Thus, \[ R_{eq} = \frac{2}{10} = 0.2\Omega \] Quick Tip: When resistors are connected in parallel, the total resistance is always smaller than the smallest individual resistance.

Step 1: Electric force.

The force on the charge due to the electric field is given by: \[ \vec{F_1} = q\vec{E} \]

Step 2: Magnetic force.

The force on the charge due to the magnetic field is given by: \[ \vec{F_2} = q (\vec{v} \times \vec{B}) \] Quick Tip: The electric force on a charge is given by \( \vec{F} = q\vec{E} \), and the magnetic force is given by \( \vec{F} = q(\vec{v} \times \vec{B}) \), where \( \vec{v} \) is the velocity vector and \( \vec{B} \) is the magnetic field.

Step 1: Use the formula for least count.

The least count of the vernier scale is given as: \[ LC = 1 \, MSD - 1 \, VSD \]

Step 2: Calculate main scale division (MSD).

Given that 1 VSD = \( \frac{19}{20} \) MSD, we have: \[ 1 \, VSD = \frac{19}{20} \, MSD \]

Step 3: Find the least count.

Since 1 VSD = 0.1 mm, \[ 1 MSD = \frac{20}{19} \times 0.1 \, mm = 2 \, mm \] Quick Tip: The least count of a Vernier caliper can be calculated by subtracting 1 VSD from 1 MSD. The main scale division gives the value of one unit of measurement.

Step 1: Calculate inductive reactance.

The inductive reactance \( X_L \) is given by: \[ X_L = 2\pi f L = 2\pi \times 50 \times 10 \times 10^{-3} = \pi \, \Omega \]

Step 2: Calculate impedance of the circuit.

The total impedance \( Z \) of the series RL circuit is: \[ Z = \sqrt{R^2 + X_L^2} = \sqrt{14^2 + \pi^2} = \sqrt{196 + 9.87} = \sqrt{205.87} \approx 14.36 \, \Omega \]

Step 3: Calculate RMS current.

The RMS value of the current \( I_{rms} \) is given by: \[ I_{rms} = \frac{V_{rms}}{Z} = \frac{220}{\sqrt{14^2 + \pi^2}} = \frac{220}{14.36} \approx 15.33 \, A \] Quick Tip: In an RL circuit, the RMS value of current can be found using the formula \( I_{rms} = \frac{V_{rms}}{Z} \), where \( Z = \sqrt{R^2 + X_L^2} \) is the impedance.

Step 1: Understanding the equation for stopping potential.

The stopping potential is given by: \[ K.E_{max} = hv - \phi \] \[ \Rightarrow e v_s = hv - \phi \] \[ \Rightarrow v_s = \frac{h}{e} \left( \frac{v}{c} \right) - \frac{\phi}{e} \]
Thus, stopping potential depends on frequency of light and the nature of the material, but it does not depend on intensity.
Quick Tip: The stopping potential in the photoelectric effect is independent of light intensity, but it depends on the frequency of the incident light and the nature of the material.

The time period \( T \) of a satellite orbiting around a planet is given by: \[ T = 2 \pi \sqrt{\frac{r^3}{G M}} \]

For a geostationary satellite, we know the time period is 6 hrs, so: \[ T = 6 \, hrs = 6 \times 60 \times 60 \, seconds \]

Step 1: Solve for \( r \).

Rearranging the formula, we get: \[ r = \left( \frac{G m_e T^2}{4 \pi^2} \right)^{1/3} \] Quick Tip: The orbital radius of a satellite can be calculated using the formula derived from the universal law of gravitation and centripetal force.

Using the formula for force and area of a piston: \[ \frac{F_2}{A_2} = \frac{F_1}{A_1} \]

Step 1: Calculate areas.

The area of a piston is given by: \[ A = \pi r^2 \]

Thus, \[ A_1 = \pi \left( \frac{d_1}{2} \right)^2 = \pi \times 0.7^2 \] \[ A_2 = \pi \left( \frac{d_2}{2} \right)^2 = \pi \times 7^2 \]

Step 2: Solve for \( F_2 \).

Now, using the formula: \[ F_2 = F_1 \times \frac{A_2}{A_1} = 10 \times \frac{\pi \times 7^2}{\pi \times 0.7^2} = 1000 \, N \] Quick Tip: When dealing with hydraulic systems, the force is related to the areas of the pistons. The force on the large piston can be found using the ratio of areas.

Step 1: Electric field at point A.

The electric field at point A due to the electric dipole is given by: \[ E_A = \frac{2kp}{r^3} \]
where \( k \) is Coulomb's constant, \( p \) is the dipole moment, and \( r \) is the distance from the dipole to point A.

Step 2: Electric field at point B.

The electric field at point B due to the dipole is given by: \[ E_B = \frac{k p}{(2r)^3} = \frac{k p}{8r^3} \]

Step 3: Find the ratio of electric fields.

The ratio of the electric field at point A to the electric field at point B is: \[ \frac{E_A}{E_B} = \frac{\frac{2kp}{r^3}}{\frac{kp}{8r^3}} = \frac{2}{\frac{1}{8}} = 16 \] Quick Tip: The electric field due to a dipole at any point depends on the distance from the dipole and the dipole moment. The field at point A is stronger than the field at point B due to the difference in distances.

Step 1: Understanding the wavelength ranges.

- Infrared has wavelengths greater than \( 10^{-3} \, nm \), which corresponds to (iv).
- \(\gamma\)-rays have the shortest wavelengths, less than \( 10^{-3} \, nm \), corresponding to (i).
- X-rays have wavelengths from \( 1 \, nm \) to \( 300 \, nm \), corresponding to (iii).
- UV rays have wavelengths from \( 300 \, nm \) to \( 600 \, nm \), corresponding to (ii).

Thus, the matching is: \[ \gamma \rightarrow (i), \, X-rays \rightarrow (iii), \, UV-rays \rightarrow (ii), \, Infrared \rightarrow (iv) \] Quick Tip: The wavelength ranges for various types of electromagnetic radiation vary and can be identified by matching them to the correct categories.

Step 1: Air drag on the tanker.

As the tanker moves, air drag causes charges to be induced on the surface of the tanker. This happens due to the friction between the tanker and the surrounding air.

Step 2: Electrostatic charge accumulation.

If the tanker is carrying inflammable liquid, the buildup of static charge can be hazardous, as it could lead to sparks that may ignite the flammable liquid.

Step 3: Role of the metal chain.

To neutralize the accumulated charge, a metal chain is suspended at the rear of the tanker. The chain touches the ground, allowing the extra electrons to flow from the tanker to the earth, thereby neutralizing the charge. Quick Tip: Inflammable liquids in tankers can accumulate static charge, which can be neutralized by grounding the tanker using a metal chain.

Step 1: Understanding the setup.

We are given the following:
- Galvanometer resistance: \( R_g = 100 \, \Omega \)
- Series resistance: \( R = 400 \, \Omega \)
- The galvanometer is designed to measure a maximum voltage of 10V, and now it is being converted into an ammeter to measure a maximum current of 10A.

The goal is to find the value of the shunt resistance \( R_s \) that will allow the device to measure the current up to 10A.

Step 2: Formula for voltage across the galvanometer.

The voltage across the series combination of the galvanometer and the shunt resistance is given by the formula: \[ V = I_{max} \times (R_s + R_g) \]
Since the galvanometer is designed to measure a maximum voltage of 10V, the total voltage across the ammeter is also 10V. This voltage is dropped across both the galvanometer and the shunt resistance.
\[ V = 10 \, V \]

Step 3: Current through the galvanometer.

The current \( I_g \) through the galvanometer is related to the maximum current \( I_{max} \) by the following relation: \[ I_g = \frac{1}{50} \, A \]
This is because the galvanometer can measure a maximum of 1A, and the total current in the circuit is 10A. Thus, the current through the galvanometer will be \( 1/50 \) of the total current.

Step 4: Voltage across the galvanometer.

Now, using Ohm’s law, the voltage across the galvanometer is: \[ V_g = I_g \times R_g = \frac{1}{50} \times 100 = 2 \, V \]

Step 5: Voltage drop across the shunt resistance.

The remaining voltage, which is \( 10 - 2 = 8 \, V \), is dropped across the shunt resistance \( R_s \).

Step 6: Apply the formula for the shunt resistance.

Using Ohm’s law for the shunt resistance, the voltage drop across \( R_s \) is related to the current through it: \[ V_s = I_s \times R_s \]
where \( I_s = 10 - I_g = 10 - \frac{1}{50} \). Thus: \[ R_s = \frac{8}{I_s} = \frac{8}{10 - \frac{1}{50}} = 0.2 \, \Omega \] Quick Tip: To convert a galvanometer into an ammeter, use the shunt resistance in parallel with the galvanometer. The value of the shunt resistance is calculated by using the voltage drop across the shunt and the maximum current.

Step 1: Understanding the concept of mean free path.

The mean free path \( \lambda \) is the average distance a gas molecule travels between collisions with other molecules. It depends on the number density \( n \) (the number of molecules per unit volume) and the diameter \( d \) of the molecules.

Step 2: Formula for mean free path.

The formula for the mean free path is derived from the kinetic theory of gases. It is given by: \[ \lambda = \frac{1}{\sqrt{2} \, n d^2} \]
where:
- \( n \) is the number density of the gas molecules,
- \( d \) is the diameter of the gas molecules,
- \( \sqrt{2} \) is a constant factor arising from the relative motion of the molecules.

Step 3: Explanation of the formula.

The mean free path is inversely proportional to the number density and the square of the molecular diameter. This means that as the number of molecules increases or the size of the molecules increases, the mean free path decreases, indicating more frequent collisions between molecules. Quick Tip: The mean free path is an important quantity in understanding the behavior of gases. It is inversely proportional to the molecular density and the square of the molecular diameter.