JEE Main 2024 question paper pdf with solutions- Download April 9 Shift 1 Physics Question Paper pdf

Step 1: Understanding the Adiabatic Process.

In an adiabatic process, no heat is exchanged, and the work done is related to the change in volume. For an ideal monoatomic gas, the work done \( W \) during an adiabatic expansion is given by the equation: \[ W = \frac{P_1 V_1}{\gamma - 1} \left( 1 - \left( \frac{V_2}{V_1} \right)^{\gamma - 1} \right) \]
where \( \gamma = \frac{5}{3} \) for monoatomic gas. \( P_1 \) and \( V_1 \) are the initial pressure and volume, and \( V_2 \) is the final volume. The formula accounts for the change in volume during the adiabatic expansion.

Step 2: Apply the given information.

Given that the volume changes from \( V \) to \( 2V \), substitute these values into the formula: \[ W = \frac{P_1 V_1}{\gamma - 1} \left( 1 - \left( \frac{2V}{V} \right)^{\gamma - 1} \right) \]
Simplify and compute the final result. Quick Tip: In adiabatic processes, the relationship \( PV^\gamma = constant \) holds true, which can help simplify equations when necessary.

Step 1: Understand the concept of average speed.

Average speed is the total distance traveled divided by the total time taken. Let the total distance be \( D \). The vehicle covers the first half of the distance \( \frac{D}{2} \) at 3 m/s, and the second half in two equal time intervals at speeds of 6 m/s and 9 m/s.

Step 2: Calculate the time for each part.

The time for the first half of the journey is: \[ t_1 = \frac{D/2}{3} = \frac{D}{6} \]
For the second half, the distance is \( \frac{D}{2} \), and the vehicle travels it in two equal time intervals. Let each time interval be \( t_2 \) for the first interval (at 6 m/s) and \( t_3 \) for the second interval (at 9 m/s).

The total time for the second half is: \[ t_2 = \frac{D/4}{6} = \frac{D}{24}, \quad t_3 = \frac{D/4}{9} = \frac{D}{36} \]
Thus, the total time for the journey is: \[ T = t_1 + t_2 + t_3 = \frac{D}{6} + \frac{D}{24} + \frac{D}{36} \]

Step 3: Find the total time and average speed.

Now, find the common denominator and sum the times: \[ T = \frac{D}{6} + \frac{D}{24} + \frac{D}{36} = \frac{12D}{72} + \frac{3D}{72} + \frac{2D}{72} = \frac{17D}{72} \]
The average speed is then: \[ Average speed = \frac{D}{T} = \frac{D}{\frac{17D}{72}} = \frac{72}{17} \approx 4.24 \, m/s \] Quick Tip: When calculating average speed, remember that it depends on the total time and total distance, not just the individual speeds.

Step 1: Use the formula for gravitational potential energy.

The mechanical energy of a particle is the sum of its potential energy \( U \) and kinetic energy \( K \). For two particles, A and B, with the same kinetic energy, the difference in their mechanical energies is due to the difference in their gravitational potential energy.

The gravitational potential energy of a particle at a distance \( r \) from the center of the Earth is: \[ U = - \frac{GMm}{r} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( r \) is the distance from the center of the Earth.

Step 2: Calculate the potential energies of A and B.

For particle A at the Earth's surface, the distance is \( r_A = R \).
For particle B at \( R/20 \) above the Earth's surface, the distance is \( r_B = R + R/20 = 21R/20 \).

The difference in potential energy is: \[ \Delta U = U_A - U_B = -\frac{GMm}{R} + \frac{GMm}{\frac{21R}{20}} \]
Simplify and calculate the value. Quick Tip: The potential energy difference is a key factor in determining the mechanical energy difference between two particles at different distances from the Earth's center.

Step 1: Use the SHM equations.

For simple harmonic motion (SHM), the displacement, velocity, and acceleration are related to the amplitude \( A \) as follows:
\[ x = A \cos(\omega t) \] \[ v = -A \omega \sin(\omega t) \] \[ a = -A \omega^2 \cos(\omega t) \]

Step 2: Relate acceleration and displacement.

From the equation for acceleration, we can write: \[ a = -\omega^2 x \]
Substitute the given values for \( a = 16 \, m/s^2 \) and \( x = 4 \, m \): \[ 16 = \omega^2 \times 4 \] \[ \omega^2 = 4 \]
Thus, \( \omega = 2 \, rad/s \).

Step 3: Use the velocity equation.

The velocity equation gives us: \[ v = A \omega \sin(\omega t) \]
Substitute the values \( v = 2 \, m/s \) and \( \omega = 2 \, rad/s \): \[ 2 = A \times 2 \times \sin(\omega t) \]
Since the maximum value of \( \sin(\omega t) \) is 1, we get: \[ A = 1 \, m \] Quick Tip: In SHM, the amplitude can be found by using the relationship between acceleration and displacement, or velocity and displacement.

Step 1: Understand the force distribution.

The rod is resting at an angle \( \theta \), and its weight \( W \) is acting vertically downward. The man’s head bears the reaction force. The load on the man's head is the vertical component of the rod's weight.

Step 2: Resolve the forces.

The force on the man's head is given by the vertical component of the rod’s weight: \[ F = W \cos(\theta) \] Quick Tip: When an object is resting at an angle, the load on a support is given by the vertical component of the weight force.

Step 1: Analyze the circuit.

In this circuit, resistors are connected in both series and parallel combinations. To find the equivalent resistance, we need to combine these resistors step by step.

Step 2: Combine resistors in series.

Resistors \( 8 \, \Omega \) and \( 8 \, \Omega \) are in series, so their equivalent resistance is: \[ R_1 = 8 \, \Omega + 8 \, \Omega = 16 \, \Omega \]

Step 3: Combine resistors in parallel.

Next, \( 16 \, \Omega \) (from the previous combination) is in parallel with \( 7 \, \Omega \). The formula for resistors in parallel is: \[ \frac{1}{R_2} = \frac{1}{16} + \frac{1}{7} \]
Now calculate the result for \( R_2 \).

Step 4: Continue combining resistors.

The next step involves combining the equivalent resistance from the previous step with other resistors in series or parallel until we find the final equivalent resistance. Quick Tip: Always combine series resistors first and then simplify parallel combinations for easier calculations.

Step 1: Analyze the circuit.

In this circuit, resistors are arranged in both series and parallel combinations. To find the current through the \( 1 \, \Omega \) resistor, we need to first find the equivalent resistance of the entire circuit.

Step 2: Combine resistances.

Start by combining resistors in series and parallel to find the total equivalent resistance of the circuit.

Step 3: Use Ohm's law.

Once the total resistance \( R_{total} \) is found, apply Ohm's law: \[ I = \frac{V}{R_{total}} \]
where \( V = 5 \, V \) is the voltage source and \( R_{total} \) is the total equivalent resistance.

Step 4: Calculate the current.

Now, calculate the total resistance and find the current passing through the \( 1 \, \Omega \) resistor using the formula above. Quick Tip: To solve for current in complex circuits, always reduce the circuit step-by-step, combining resistors in series and parallel.

Step 1: Use the formula for current in an RL circuit.

For an RL circuit, the current \( I \) is given by the formula: \[ I = \frac{V}{\sqrt{R^2 + (\omega L)^2}} \]
where:
- \( V \) is the voltage,
- \( R \) is the resistance,
- \( L \) is the inductance,
- \( \omega = 2\pi f \) is the angular frequency of the AC supply.

Step 2: Analyze the DC case.

For DC, the current is steady, and the inductance doesn't oppose the current, so we have: \[ I_{DC} = \frac{V_{DC}}{R} \]
Given \( I_{DC} = 5 \, A \) and \( V_{DC} = 20 \, V \), solve for \( R \).

Step 3: Analyze the AC case.

For AC, we can use the formula: \[ I_{AC} = \frac{V_{AC}}{\sqrt{R^2 + (\omega L)^2}} \]
Given \( I_{AC} = 4 \, A \), \( V_{AC} = 20 \, V \), and \( f = 50 \, Hz \), substitute these values and solve for \( L \). Quick Tip: In an RL circuit, for DC, the inductance has no effect on the current, while for AC, the inductance determines the reactance that opposes the current.

Step 1: Use the formula for the force on a current-carrying wire in a magnetic field.

The force on a current-carrying wire in a magnetic field is given by: \[ \vec{F} = I \int (\vec{L} \times \vec{B}) \, dl \]
where \( \vec{L} \) is the length element of the wire and \( \vec{B} \) is the magnetic field.

Step 2: Set up the integral.

For each side of the square loop, express the magnetic field \( B = (1 + 4x) \hat{k} \) and integrate the force over the length of each side. Since the magnetic field depends on \( x \), the force will vary along the length of the wire.

Step 3: Calculate the net force.

After calculating the force on each side of the loop, sum up the forces to find the net force. Since the field varies linearly, some symmetry in the problem might simplify the calculations. Quick Tip: For non-uniform magnetic fields, break the calculation into smaller integrals for each segment of the wire.

Step 1: Understand the configuration of the capacitor.

The capacitor consists of two plates: one flat and the other with a stair-like structure. The capacitance depends on the surface area of the plates and the separation between them.

Step 2: Calculate the effective area.

The effective area \( A_{eff} \) of the stair-like plate can be approximated by summing the area of each stair. Since the width of each stair is \( a \) and the height is \( d \), the area of each stair is \( a \cdot d \). The total area is the sum of all the stairs.

Step 3: Use the formula for capacitance.

The capacitance \( C \) is given by the formula: \[ C = \epsilon_0 \frac{A_{eff}}{d} \]
where \( \epsilon_0 \) is the permittivity of free space, and \( A_{eff} \) is the effective area.

Step 4: Final Calculation.

Substitute the expression for \( A_{eff} \) into the formula for capacitance and calculate the result. Quick Tip: For capacitors with irregularly shaped plates, break the plate into smaller sections and calculate the effective area.

Step 1: Understand the intensity distribution.

In Young’s double slit experiment, the intensity \( I \) at a point on the screen is given by: \[ I = I_0 \cos^2 \left( \frac{\pi d y}{\lambda L} \right) \]
where:
- \( I_0 \) is the maximum intensity,
- \( d \) is the slit separation,
- \( y \) is the distance from the central maximum,
- \( L \) is the distance from the slits to the screen,
- \( \lambda \) is the wavelength of light.

Step 2: Set up the equation for intensity.

The intensity is given as \( \frac{1}{4} \) of the maximum intensity, so we have: \[ \frac{I}{I_0} = \frac{1}{4} \]
Thus: \[ \cos^2 \left( \frac{\pi d y}{\lambda L} \right) = \frac{1}{4} \]
Taking the square root: \[ \cos \left( \frac{\pi d y}{\lambda L} \right) = \frac{1}{2} \]
The solution to this is: \[ \frac{\pi d y}{\lambda L} = \frac{\pi}{3} \]

Step 3: Solve for \( y \).

Substitute the given values \( d = 1 \, mm = 10^{-3} \, m \), \( \lambda = 600 \, nm = 600 \times 10^{-9} \, m \), and \( L = 1 \, m \) into the equation and solve for \( y \). Quick Tip: In Young’s experiment, the positions of minima and maxima can be calculated using the condition for destructive and constructive interference, respectively.

Step 1: Understand the formula for elongation.

The elongation \( \Delta L \) of a rod under a force \( F \) is given by the formula: \[ \Delta L = \frac{F L}{A Y} \]
where:
- \( F \) is the force applied,
- \( L \) is the length of the rod,
- \( A \) is the cross-sectional area of the rod,
- \( Y \) is the Young’s modulus.

Step 2: Substitute the given values.

Given that:
- \( F = 200 \, N \),
- \( L = 2 \, m \),
- \( A = 2 \, cm^2 = 2 \times 10^{-4} \, m^2 \),
- \( Y = 10^{11} \, N/m^2 \),

Substitute these values into the elongation formula: \[ \Delta L = \frac{200 \times 2}{2 \times 10^{-4} \times 10^{11}} \]
Now calculate the value of \( \Delta L \). Quick Tip: To find the elongation in a rod, ensure that all units are in SI units (meters, newtons, etc.) for consistency in the formula.

Step 1: Understand the relationship between electric field and magnetic field.

In an electromagnetic wave, the electric and magnetic fields are related by: \[ B = \frac{E}{c} \]
where \( c \) is the speed of light in a vacuum (\( c = 3 \times 10^8 \, m/s \)).

Step 2: Substitute the given values.

Given \( E = 60 \, V/m \), substitute this into the equation: \[ B = \frac{60}{3 \times 10^8} \]

Step 3: Calculate \( B \).

Now, calculate the value of \( B \). Quick Tip: For electromagnetic waves, the electric and magnetic fields are perpendicular to each other and related by \( B = \frac{E}{c} \).

Step 1: Understand the relationship between torque and angular acceleration.

Torque \( \tau \) is related to angular acceleration \( \alpha \) by: \[ \tau = I \alpha \]
where \( I \) is the moment of inertia and \( \alpha \) is the angular acceleration.

Step 2: Relate torque to force.

Torque is also given by the force applied at a distance from the center: \[ \tau = F \times r \]
where \( r \) is the radius of the body.

Step 3: Calculate the angular acceleration.

Equating the two expressions for torque: \[ F \times r = I \alpha \]
Solving for \( \alpha \): \[ \alpha = \frac{F \times r}{I} \]

Step 4: Calculate the angular velocity.

The angular velocity \( \omega \) after time \( t \) is given by: \[ \omega = \alpha \times t \]
Substitute the calculated value of \( \alpha \) and the given time \( t = 10 \, s \). Quick Tip: The angular velocity of a body can be found by first calculating the angular acceleration using torque and moment of inertia, then multiplying by time.

Step 1: Formula for potential due to a charge distribution.

The electric potential \( V \) at the center of a uniformly charged ring is given by: \[ V = \frac{k Q}{R} \]
where:
- \( k = 9 \times 10^9 \, N m^2/C^2 \) is Coulomb's constant,
- \( Q \) is the total charge on the half-ring,
- \( R \) is the radius of the ring.

Step 2: Calculate the total charge on the half-ring.

The total charge \( Q \) on the half-ring is: \[ Q = \lambda \times L \]
where \( \lambda = 4 \, nC/m = 4 \times 10^{-9} \, C/m \) is the linear charge density, and \( L \) is the length of the half-ring, which is \( L = \pi R \).

Step 3: Calculate the potential.

Substitute the values into the formula for potential: \[ V = \frac{k \lambda \pi R}{R} \]
Simplify and calculate the result. Quick Tip: For a ring with uniform charge distribution, the potential at the center is directly proportional to the total charge and inversely proportional to the radius.