JEE Main 2024 question paper pdf with solutions- Download April 5 Shift 1 Physics Question Paper pdf

Step 1: Understanding the problem.

The system has two masses, one hanging and one on a pulley. The forces acting on the system include gravity, tension, and the resulting acceleration. We need to calculate the tension \(T_1\) in the string using the equations of motion.


Step 2: Apply Newton's second law.

For the hanging mass: \[ T_1 - m_1 g = m_1 a \]
For the mass on the table: \[ T_2 = m_2 g \]
Solving these equations will give the tension \(T_1 = 144 N\).


Step 3: Conclusion.

Thus, the correct tension in the string is 144 N.
Quick Tip: In pulley systems, the tension in the string is the same throughout the string if the pulley is massless and frictionless.

Step 1: Formula for electrostatic force.

The electrostatic force is given by Coulomb’s law: \[ F_e = \frac{k_e \cdot e^2}{r^2} \]
where \(k_e = 9 \times 10^9 \, N m^2/C^2\), and \(e = 1.6 \times 10^{-19} \, C\).

Step 2: Formula for gravitational force.

The gravitational force is given by Newton's law of gravitation: \[ F_g = \frac{G \cdot m_e \cdot m_p}{r^2} \]
where \(G = 6.67 \times 10^{-11} \, N m^2/kg^2\), and the masses of the electron and proton are \(m_e = 9.1 \times 10^{-31} \, kg\) and \(m_p = 1.67 \times 10^{-27} \, kg\).

Step 3: Calculate the ratio.
\[ \frac{F_e}{F_g} = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{6.67 \times 10^{-11} \times 9.1 \times 10^{-31} \times 1.67 \times 10^{-27}} = 2.27 \times 10^{39} \]
Thus, the correct ratio is \(2.27 \times 10^{39}\).


Step 4: Conclusion.

The correct ratio is 2.27 \times 10^{39}.
Quick Tip: When calculating ratios, always check the units to ensure consistency in physical quantities.

Step 1: Formula for time period of pendulum.

The time period \(T\) of a simple pendulum is related to the acceleration due to gravity \(g\): \[ T \propto \frac{1}{\sqrt{g}} \]
At a distance \(R\) from the Earth's surface, the gravitational force is \(g_1\), and at a distance \(2R\), it is \(g_2 = \frac{g_1}{4}\).

Step 2: Apply the ratio.

Using the relation for time period: \[ \frac{T_2}{T_1} = \sqrt{\frac{g_1}{g_2}} = \sqrt{\frac{g_1}{\frac{g_1}{4}}} = 2 \]
Thus, the time period at distance \(2R\) is \(2 \times T_1 = 6\).


Step 3: Conclusion.

The time period of the pendulum at a distance \(2R\) is 6.
Quick Tip: The time period of a pendulum increases as the acceleration due to gravity decreases.

Step 1: Understanding the problem.

The problem describes a rotating rod with a magnetic field perpendicular to the plane of rotation. Points P and Q are on the rod, and we are asked to find the potential difference between them. The potential difference in a rotating system in a magnetic field can be calculated using the formula for induced emf in a rotating conductor.


Step 2: Apply the formula.

The potential difference is given by: \[ V_P = V_Q = \frac{e B_0 (L^2)}{4} \]
Since the field is perpendicular and the magnetic forces are symmetric, the potential difference between points P and Q is zero. Thus, \[ V_P = V_Q = 0 \]

Step 3: Conclusion.

The potential difference between points P and Q is 0.
Quick Tip: In rotating systems, the induced emf is zero when the magnetic field is symmetric and perpendicular to the plane of rotation.

Step 1: Understanding the cyclic process.

The work done in a cyclic process is given by the area enclosed by the curve on the P-V diagram. In this case, the cyclic process is a circle. The formula for the work done by an ideal gas in such a process is: \[ Work done = \pi r^2 \]
where \(r\) is the radius of the circle.


Step 2: Calculating the radius.

From the P-V curve, the radius of the circle can be calculated as the difference between the pressures along the horizontal axis (in \( kPa \)) and the vertical axis (in \( cc \)): \[ r = \frac{(P_{max} - P_{min}) \times (V_{max} - V_{min})}{4} \]
Using the given values of the pressure and volume, we find the radius and then compute the work done as follows: \[ Work done = \pi \times \left( 240 \right)^2 \, \frac{kPa \times cc}{4} = \pi \times (24) \times (24) \times (100) \times kPa \times (0.001) \, J \]

Step 3: Final calculation.
\[ Work done = \frac{144 \pi}{10} \, J \]

Step 4: Conclusion.

Thus, the work done by the gas in the cyclic process is \( \frac{144 \pi}{10} \, J \).
Quick Tip: In cyclic processes, the work done by the gas is equal to the area enclosed by the P-V curve. For circular processes, this is \( \pi r^2 \).

Step 1: Understanding the logic gate system.

In the given system, the logic gates are connected in a series to control the bulb. The bulb will glow when the potential difference between the two points is 1. The circuit's behavior depends on the truth values for the inputs \( A \), \( B \), and \( C \).


Step 2: Expression for potential difference.

The potential difference is given by \( Potential difference = X - Y \), where \( X \) and \( Y \) are the outputs of the logic gates. The potential difference for the given circuit is: \[ Potential difference = \left( A + (B \cdot C) \right) \]
where \( A \), \( B \), and \( C \) are the input values, and \( B \cdot C \) represents the AND operation.

Step 3: Analyzing the circuit.

By evaluating the output for each combination of \( A \), \( B \), and \( C \) from the truth table, we determine that the bulb will glow when the potential difference is 1, which happens in case (D).


Step 4: Conclusion.

The bulb will glow in case (D), where the output of the logic gates results in a potential difference of 1.
Quick Tip: In logic gate circuits, the output is determined by the combination of input values and the logical operations (AND, OR, NOT). Evaluate each combination to find the correct output.

Step 1: Magnetic flux.

The magnetic flux \( \phi \) through the inner loop due to the current in the outer loop is given by: \[ \phi = B \cdot \pi a^2 \]
where \( B \) is the magnetic field produced by the outer loop at the location of the inner loop.


Step 2: Expression for mutual inductance.

The magnetic field at the position of the inner loop due to the current in the outer loop is: \[ B = \frac{\mu_0 I}{2b} \]
Thus, the mutual inductance \( M \) is given by: \[ M = \frac{\mu_0}{2b} \pi a^2 \]

Step 3: Conclusion.

The mutual inductance of the given system is \( \frac{\mu_0}{2b} \pi a^2 \).
Quick Tip: For concentric loops, the mutual inductance is directly proportional to the area of the inner loop and inversely proportional to the distance between the loops.

Step 1: Use the formula for distance covered in the \(n\)-th second.

The distance covered by the particle in the \(n\)-th second is given by: \[ S_n = u + \frac{a(2n - 1)}{2} \]
where \(u = 0\) (initial velocity), and \(a\) is the constant acceleration. Thus: \[ S_n = \frac{a(2n - 1)}{2} \]

Step 2: Distance covered in \((n-1)\)-th second.

The distance covered in the \((n-1)\)-th second is given by: \[ S_{n-1} = \frac{a(2(n-1) - 1)}{2} = \frac{a(2n - 3)}{2} \]

Step 3: Ratio of distances.

The ratio of distances covered in the \(n\)-th second to the \((n-1)\)-th second is: \[ \frac{S_n}{S_{n-1}} = \frac{2n - 1}{2n - 3} \]

Step 4: Conclusion.

Thus, the ratio of distances is \( \frac{2n-1}{2n-3} \).
Quick Tip: For a particle starting from rest with constant acceleration, the distance covered in the \(n\)-th second is directly related to the acceleration and the time interval \(n\).

Step 1: Dimension of energy density (\( \mu \)).

Energy density \( \mu \) has the dimension of energy per unit volume. The dimension of energy is \( ML^2T^{-2} \), and the dimension of volume is \( L^3 \). Hence, the dimension of \( \mu \) is: \[ [\mu] = \frac{ML^2T^{-2}}{L^3} = ML^{-1}T^{-2} \]

Step 2: Dimension of gravitational constant (\( G \)).

The gravitational constant \( G \) has the dimension: \[ [G] = \frac{M^{-1}L^3}{T^2} \]

Step 3: Dimension of \( \sqrt{\mu G} \).

Now, we calculate the dimension of \( \sqrt{\mu G} \): \[ [\sqrt{\mu G}] = \sqrt{[\mu] \cdot [G]} = \sqrt{\left( ML^{-1}T^{-2} \right) \cdot \left( \frac{M^{-1}L^3}{T^2} \right)} \]
Simplifying: \[ [\sqrt{\mu G}] = \sqrt{L^2T^{-4}} = LT^{-2} \]

Step 4: Conclusion.

Thus, the dimension of \( \sqrt{\mu G} \) is \( LT^{-2} \).
Quick Tip: When dealing with dimensions, always break down the units of each term and simplify to find the required dimension.

Step 1: Use the formula for resistance.

The resistance of a wire is given by: \[ R = \rho \frac{l}{A} \]
where \( R \) is the resistance, \( \rho \) is the resistivity, \( l \) is the length, and \( A \) is the cross-sectional area of the wire. The cross-sectional area \( A \) of a wire is given by: \[ A = \pi r^2 \]
Thus, the resistance \( R \) is proportional to the inverse of the square of the radius: \[ R \propto \frac{1}{r^2} \]

Step 2: Calculate the new resistance.

If the radius is halved, then the new resistance \( R_2 \) is related to the original resistance \( R_1 \) by: \[ \frac{R_1}{R_2} = \left( \frac{r_2}{r_1} \right)^2 = \left( \frac{2}{4} \right)^2 = \frac{1}{4} \]
Thus, \[ R_2 = 4 \times R_1 = 4 \times 2 \, \Omega = 8 \, \Omega \]

Step 3: Conclusion.

The resistance of the new wire is \( 8 \, \Omega \).
Quick Tip: When the radius of a wire is halved, the resistance quadruples because the area of the wire is proportional to \( r^2 \).

Step 1: Analyze the circuit.

In the given circuit, we need to find the total resistance across the cell. The resistors are connected in series and parallel, so we will use the formulas for combined resistance for series and parallel connections.


Step 2: Apply the formula for equivalent resistance.

For resistors in series, the equivalent resistance \( R_{eq} \) is given by: \[ R_{eq} = R_1 + R_2 \]
For resistors in parallel, the equivalent resistance \( R_{eq} \) is given by: \[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]

After applying these formulas step by step, the equivalent resistance comes out to be 1 ohm.


Step 3: Conclusion.

The equivalent resistance is \( 1 \, \Omega \), and the current across the circuit is 1 A as per Ohm's law \( I = \frac{V}{R} \).
Quick Tip: In circuits with resistors in both series and parallel, always simplify the circuit step by step using the appropriate formulas.

Step 1: Understanding the forces.

The block is moving downward with the platform, so the effective acceleration due to gravity is modified by the downward acceleration of the platform. The normal force \( N \) is the force exerted by the platform on the block.


Step 2: Force equation.

The force equation is given by: \[ F_{net} = m \cdot a \]
where \( m \) is the mass of the block and \( a \) is the acceleration relative to the platform. The weight of the block is \( W = m \cdot g \), where \( g = 10 \, m/s^2 \) is the acceleration due to gravity. The net force is the difference between the weight and the normal force because the platform is moving downward.

Step 3: Calculation.

For the block, the equation becomes: \[ 30g - N = 30 \times 2 \quad (where \( g = 10 \, \text{m/s^2 \) and \( a = 2 \, m/s^2 \))} \]
Substituting the values: \[ 300 - N = 60 \]
Solving for \( N \): \[ N = 300 - 60 = 240 \, N \]

Step 4: Conclusion.

The normal force exerted on the block by the platform is \( 240 N \).
Quick Tip: When a system is in motion with acceleration, the normal force is adjusted by the acceleration, particularly if the platform is accelerating.

Step 1: Understanding the wavefront.

When a point source is placed at the first principal focus of a convex lens, the light rays emerging from the lens are parallel and form a planar wavefront. This is because the convex lens converges the light rays in such a way that the wavefronts are parallel after emerging from the lens.


Step 2: Conclusion.

Thus, the wavefront emerging from the convex lens is a planar wavefront.
Quick Tip: For a point source at the focus of a convex lens, the light emerging from the lens forms a parallel beam, which corresponds to a planar wavefront.

Step 1: Relationship between collision frequency and temperature.

The collision frequency \( f \) is directly proportional to the square root of temperature \( T \): \[ f \propto \sqrt{T} \]

Step 2: Given values.

The collision frequency at \( 27^\circ C \) (300K) is given as 2, and we are asked to find the collision frequency at \( 127^\circ C \) (400K). The ratio of the collision frequencies is: \[ \frac{f'}{f} = \frac{\sqrt{400}}{\sqrt{300}} = \frac{2}{\sqrt{3}} \]
Thus, the collision frequency at \( 127^\circ C \) is: \[ f' = 2 \times \frac{2}{\sqrt{3}} = \frac{8}{\sqrt{3}} \]

Step 3: Conclusion.

Thus, the collision frequency at \( 127^\circ C \) is \( \frac{8}{\sqrt{3}} \).
Quick Tip: The collision frequency of gas particles is proportional to the square root of the temperature in Kelvin.

Step 1: Escape velocity.

The escape velocity \( v_e \) from a body is given by: \[ v_e = \sqrt{\frac{2GM}{r}} \]
Thus, the corresponding energy is \( \frac{GMm}{2a} \), so the correct match is \( (1) \rightarrow (c) \).


Step 2: Kinetic energy.

The kinetic energy \( K.E. \) is given by: \[ K.E. = \frac{GMm}{2a} \]
Thus, the correct match is \( (2) \rightarrow (a) \).


Step 3: Gravitational potential energy.

The gravitational potential energy \( U \) is given by: \[ U = -\frac{GMm}{r} \]
Thus, the correct match is \( (3) \rightarrow (b) \).


Step 4: Total energy.

The total energy \( E \) is the sum of kinetic and potential energies: \[ E = K.E. + U = -\frac{GMm}{2r} \]
Thus, the correct match is \( (4) \rightarrow (d) \).


Step 5: Conclusion.

The correct matching is: \[ (1) \rightarrow (c), \, (2) \rightarrow (a), \, (3) \rightarrow (b), \, (4) \rightarrow (d) \] Quick Tip: In gravitational systems, the escape velocity is derived from the gravitational potential energy and kinetic energy. The total energy is the sum of the two.

Step 1: Work done when the block is moved up the wedge.

In the first case, when the block is moved up the wedge, the work done by the gravitational force is: \[ W_1 = mgh \]
where \( m \) is the mass of the block, \( g \) is the acceleration due to gravity, and \( h \) is the vertical height the block is moved.


Step 2: Work done when the block is moved vertically upward.

In the second case, when the block is moved vertically upward, the work done by the gravitational force is: \[ W_2 = mgh \]
Again, the work is the same because the height moved is the same, and only the path of movement is different.


Step 3: Ratio of work done.

Since the work done in both cases is the same, we can write the ratio as: \[ \frac{W_1}{W_2} = \frac{mgh}{mgh} = 1 \]

Step 4: Conclusion.

The ratio of work done by gravitational force in both the cases is \( 1 \).
Quick Tip: The work done by gravitational force depends only on the vertical displacement, so the path does not affect the work done as long as the height is the same.

Step 1: Use the formula for displacement current.

The displacement current \( I_d \) is given by: \[ I_d = C \frac{dv}{dt} \]
where \( C \) is the capacitance, and \( \frac{dv}{dt} \) is the rate of change of voltage with respect to time.


Step 2: Given values.

The given values are: \[ C = 12 \, \muF, \quad v = 40 \, V, \quad f = 40 \, KHz = 40 \times 10^3 \, Hz \]
The rate of change of voltage is: \[ \frac{dv}{dt} = 2 \pi f v \]
Substituting the given values: \[ \frac{dv}{dt} = 2 \pi \times 40 \times 10^3 \times 40 = 3200 \pi \]

Step 3: Calculate the displacement current.

Now, the displacement current is: \[ I_d = 12 \times 10^{-6} \times 3200 \pi \approx 0.0381 \, A \]

Step 4: Conclusion.

Thus, the displacement current is \( 0.0381 \, A \).
Quick Tip: The displacement current is related to the change in voltage and capacitance. It can be calculated by the formula \( I_d = C \frac{dv}{dt} \).