JEE Main 2024 question paper pdf with solutions- Download April 4 Shift 1 Physics Question Paper pdf

Step 1: Use the formula for the equivalent power of lenses in combination.

The equivalent power \( P_{eq} \) of lenses in combination is the sum of individual powers: \[ P_{eq} = P \times 5 = 25 \times 5 = 125 \, D \]

Step 2: Calculate the focal length using the formula.

The focal length \( f_{eq} \) is given by: \[ f_{eq} = \frac{100}{P_{eq}} = \frac{100}{125} = 0.8 \, cm \] Quick Tip: For lenses in combination, the total power is the sum of individual powers, and the focal length is the reciprocal of the power.

Step 1: Calculate velocity.

Velocity is the derivative of position with respect to time: \[ v = \frac{dx}{dt} = 4t^3 + 18t + 2 \]

Step 2: Calculate acceleration.

Acceleration is the derivative of velocity with respect to time: \[ a = \frac{dv}{dt} = 12t^2 + 36t \]

Step 3: Calculate acceleration at \( t = 5 \) sec.

Substitute \( t = 5 \) into the acceleration equation: \[ a = 12(5)^2 + 36(5) = 12 \times 25 + 180 = 300 + 180 = 480 \, m/s^2 \] Quick Tip: When calculating velocity and acceleration, always take the derivatives of the position equation step-by-step to avoid errors.

The potential energy lost is: \[ \Delta E = mg\left( h - \frac{h}{2} \right) = mg\frac{h}{2} = -mg\frac{h}{2} \]
So, the loss in energy is 50%.

The speed just before the collision is: \[ v = \sqrt{2gh} \] Quick Tip: When a body falls freely under gravity, the speed before impact can be calculated using the formula \( v = \sqrt{2gh} \), where \( h \) is the height and \( g \) is the acceleration due to gravity.

Given that the electric field is: \[ \mathbf{E} = 40 \cos \left( \frac{t - z}{c} \right) \hat{i} \]
The magnetic field corresponding to this electric field is given by: \[ \mathbf{B} = \frac{1}{c} \hat{j} \times \mathbf{E} = \frac{40}{c} \cos \left( \frac{t - z}{c} \right) \hat{j} \]

Thus: \[ |\mathbf{E}| = 40 \cos \left( \frac{t - z}{c} \right), \quad |\mathbf{B}| = \frac{40}{c} \cos \left( \frac{t - z}{c} \right) \] Quick Tip: In electromagnetic waves, the electric and magnetic fields are perpendicular to each other and to the direction of propagation. The magnitude of the magnetic field is related to the electric field by \( B = \frac{E}{c} \), where \( c \) is the speed of light.

The electric field due to an infinite charge sheet with surface charge density \( \sigma \) is: \[ E_{sheet} = \frac{\sigma}{2 \epsilon_0} \]

The electric field due to an infinite long wire with linear charge density \( \lambda \) at a distance \( r \) from the wire is: \[ E_{wire} = \frac{\lambda}{2 \pi \epsilon_0 r} \]

Given the situation in the problem, we have the following:
- The distance of the point (0, 0, 2) from the infinite charge sheet at \( (0, 0, 4) \) is \( r = 2 \, m \).
- The surface charge density \( \sigma = 2 \, C/m^2 \).
- The linear charge density of the wire is \( \lambda = 2 \, C/m \).

Thus, the net electric field at (0, 0, 2) is the vector sum of the electric field due to the sheet and the electric field due to the wire.

The net electric field is given by: \[ E_{net} = \frac{\lambda}{\epsilon_0} \left( \frac{2 \pi r}{2 \pi r} \right) = \frac{\lambda}{\epsilon_0} \cdot \left( \frac{2 \pi r - 1}{2 \pi r} \right) = \frac{\lambda}{\epsilon_0} \cdot \frac{2 \pi r - 1}{2 \pi r} \, N/C \]

So the net electric field is: \[ E_{net} = \frac{\lambda}{\epsilon_0} \left( \frac{2 \pi r - 1}{2 \pi r} \right) \] Quick Tip: When calculating the electric field due to multiple sources like an infinite charge sheet and wire, use the respective formulas for each and add their contributions as vectors.

We are given that both objects have the same mass \( m \) and radius \( R \), and roll with the same initial velocity \( v \). The kinetic energy of a rolling object consists of both translational and rotational kinetic energies.

For the hollow cylinder: \[ K_{cylinder} = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 \]
where \( I = m R^2 \) for a hollow cylinder, and \( \omega = \frac{v}{R} \).

Substitute \( I \) and \( \omega \): \[ K_{cylinder} = \frac{1}{2} m v^2 + \frac{1}{2} m R^2 \left( \frac{v}{R} \right)^2 \] \[ K_{cylinder} = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 = m v^2 \]

For the solid sphere: \[ K_{sphere} = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 \]
where \( I = \frac{2}{5} m R^2 \) for a solid sphere, and \( \omega = \frac{v}{R} \).

Substitute \( I \) and \( \omega \): \[ K_{sphere} = \frac{1}{2} m v^2 + \frac{1}{2} \cdot \frac{2}{5} m R^2 \left( \frac{v}{R} \right)^2 \] \[ K_{sphere} = \frac{1}{2} m v^2 + \frac{1}{5} m v^2 = \frac{7}{10} m v^2 \]

Step 1: Ratio of kinetic energies.

The ratio of kinetic energies of the hollow cylinder to the solid sphere is: \[ \frac{K_{cylinder}}{K_{sphere}} = \frac{m v^2}{\frac{7}{10} m v^2} = \frac{10}{7} \]

Thus, the ratio of their kinetic energies is \( \frac{10}{7} \).

Step 2: Maximum height.

At the top of the inclined plane, all kinetic energy is converted into potential energy. Since the initial velocities are the same, the kinetic energy at the start for both objects is equal. Using conservation of energy: \[ m g h_{cylinder} = K_{cylinder} = m v^2 \] \[ h_{cylinder} = \frac{v^2}{g} \]

Similarly, for the sphere: \[ m g h_{sphere} = K_{sphere} = \frac{7}{10} m v^2 \] \[ h_{sphere} = \frac{7}{10} \frac{v^2}{g} \]

Step 3: Ratio of maximum heights.

The ratio of their maximum heights is: \[ \frac{h_{cylinder}}{h_{sphere}} = \frac{\frac{v^2}{g}}{\frac{7}{10} \frac{v^2}{g}} = \frac{10}{7} \]

Thus, the ratio of maximum heights is also \( \frac{10}{7} \). Quick Tip: When dealing with rolling motion, remember that the total kinetic energy is the sum of both translational and rotational components. The moment of inertia plays a crucial role in determining the energy distribution.

The equation represents a wave equation where:
- \( y \) is the displacement,
- \( A \) is the amplitude,
- \( n \) is the frequency,
- \( \lambda \) is the wavelength,
- \( t \) is the time,
- \( x \) is the position.

Step 1: Dimensions of \( y \).

The dimension of displacement \( y \) is \( [L] \), as it represents the distance traveled by the wave.

Step 2: Dimensions of the terms inside the sine and cosine functions.

For both \( \sin \left( \frac{2 \pi n t}{\lambda} \right) \) and \( \cos \left( \frac{2 \pi x}{\lambda} \right) \), the arguments must be dimensionless. Therefore, we need: \[ \frac{2 \pi n t}{\lambda} \quad and \quad \frac{2 \pi x}{\lambda} \]
to be dimensionless.

Step 3: Analyzing the term \( \frac{2 \pi n t}{\lambda} \).

The dimension of time \( t \) is \( [T] \), and the dimension of wavelength \( \lambda \) is \( [L] \).

Thus, the dimensions of \( n \) must satisfy: \[ \left[ \frac{n t}{\lambda} \right] = 1 \] \[ \left[ n \right] \cdot [T] / [L] = 1 \] \[ [n] = \frac{[L]}{[T]} \]

Thus, the dimension of \( n \) is \( [L T^{-1}] \). Quick Tip: For wave equations, the arguments inside trigonometric functions should always be dimensionless, which helps in determining the dimensions of variables like frequency \( n \).

The equation for the focal length of a spherical mirror is: \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]

Taking the derivative of both sides: \[ \frac{df}{f^2} = \frac{d}{v^2} \quad and \quad \frac{du}{u^2} \]

Thus, \[ \frac{df}{f} = \left[ \frac{1}{v} \cdot \frac{dv}{v} + \frac{1}{u} \cdot \frac{du}{u} \right] \]

The fractional error in focal length is given by: \[ \frac{df}{f} = \frac{uv}{u + v} \left[ \frac{dv}{v^2} + \frac{du}{u^2} \right] \] Quick Tip: When calculating fractional errors in measurements, take the derivatives of the related equations and apply the error propagation formula.

For an AC circuit, the phase difference between current and voltage in various components is as follows:
- For an inductor (\(L\)), the current lags the voltage by 90°.
- For a resistor (\(R\)), the current and voltage are in phase.
- For a capacitor (\(C\)), the current leads the voltage by 90°.

When the current is zero, the voltage will be maximum in the case of the inductor and the capacitor, where their phase difference leads to the maximum potential difference across these components. Thus, the correct answer is:

The maximum voltage occurs in the following components: \( L, C, R \).

Therefore, the answer is option (1) ABD. Quick Tip: When analyzing circuits, remember that the voltage is maximum when there is a 90° phase difference between current and voltage. This happens in the case of \( L \) and \( C \).

We are given the equations of motion as: \[ x = 3 + 4t \quad ... (1) \] \[ y = 3t^2 + 4t \quad ... (2) \]

Step 1: Solve for \( t \) from equation (1).
From equation (1): \[ x = 3 + 4t \quad \Rightarrow \quad t = \frac{x - 3}{4} \]

Step 2: Substitute \( t \) into equation (2).
Substitute the value of \( t \) into equation (2) for \( y \): \[ y = 3t^2 + 4t \]
Substitute \( t = \frac{x - 3}{4} \): \[ y = 3 \left( \frac{x - 3}{4} \right)^2 + 4 \left( \frac{x - 3}{4} \right) \]
Simplify the equation: \[ y = 3 \left( \frac{x - 3}{4} \right)^2 + \left( \frac{4(x - 3)}{4} \right) \] \[ y = \frac{3(x - 3)^2}{16} + \frac{4(x - 3)}{4} \] \[ y = \frac{3(x - 3)^2}{16} + (x - 3) \]

Step 3: Simplify further.
Now expand and simplify: \[ y = \frac{3(x^2 - 6x + 9)}{16} + x - 3 \] \[ y = \frac{3x^2 - 18x + 27}{16} + x - 3 \] \[ y = \frac{3x^2}{16} - \frac{18x}{16} + \frac{27}{16} + x - 3 \] \[ y = \frac{3x^2}{16} - \frac{18x}{16} + \left( \frac{27}{16} - \frac{48}{16} \right) + x \] \[ y = \frac{3x^2}{16} - \frac{18x}{16} - \frac{21}{16} + x \] \[ y = \frac{3x^2}{16} - \frac{18x}{16} + x - \frac{21}{16} \]

Thus, we have a quadratic equation in \( x \), confirming that the trajectory is a parabola. Quick Tip: When solving for the trajectory of an object given its position equations, solve for one variable in terms of the other and substitute it into the second equation to identify the type of curve.

We are given that the electron is revolving around an infinite line of charge. The force acting on the electron due to the line of charge is: \[ F = q \cdot E = e \cdot \left( \frac{2k \lambda}{r} \right) \]
where:
- \( e \) is the charge of the electron,
- \( \lambda \) is the linear charge density of the infinite line of charge,
- \( r \) is the distance from the electron to the line of charge,
- \( k \) is Coulomb's constant.

The work done is related to the change in kinetic energy. For an electron in circular motion, the kinetic energy \( KE \) is: \[ KE = \frac{mv^2}{2} \]

Now, from the force equation, the potential energy of the system can be written as: \[ Potential Energy = - \frac{2k \lambda e}{r} \]

Therefore, the kinetic energy \( KE \) can be expressed in terms of \( r \) as: \[ KE = \frac{2k \lambda e}{r} \]

Thus, the kinetic energy \( KE \) is inversely proportional to the distance \( r \), meaning as \( r \) increases, the kinetic energy decreases.

The correct graph should show an inverse relation between kinetic energy and the distance \( r \). Quick Tip: In problems involving circular motion around an infinite line of charge, the kinetic energy is inversely proportional to the distance from the center. This is due to the nature of the force acting on the charged particle.

We are given the pressure-temperature graph for gases of different densities, and we are required to compare the densities \( \rho_1, \rho_2, \rho_3 \).

The equation of state for an ideal gas is given by: \[ PM = \rho RT \]
where:
- \( P \) is the pressure,
- \( M \) is the molar mass,
- \( \rho \) is the density,
- \( R \) is the gas constant,
- \( T \) is the temperature.

Rearranging the equation, we get: \[ \rho = \frac{PM}{RT} \]

This shows that density \( \rho \) is directly proportional to pressure \( P \) and inversely proportional to temperature \( T \): \[ \rho \propto \frac{P}{T} \]

From the given pressure vs temperature graph:
- The slope of the graph is proportional to \( \rho \).
- The steeper the slope, the higher the density.

Since the graph shows different slopes for the three cases, we can conclude: \[ \rho_1 > \rho_2 > \rho_3 \]

Thus, \( \rho_1 \) has the highest density, followed by \( \rho_2 \), and \( \rho_3 \) has the lowest density. Quick Tip: In pressure-temperature graphs for gases, the slope is related to the density of the gas. A steeper slope indicates a higher density.

The de-Broglie wavelength of an electron moving between energy levels \( n = 4 \) to \( n = 3 \) can be found using the formula for the wavelength of the electron:
\[ \lambda = \frac{h}{mv} \]
where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the electron,
- \( v \) is the velocity of the electron.

The energy difference between the levels is given by the Bohr model of the hydrogen atom: \[ \Delta E = E_4 - E_3 \]

Using the formula for energy levels of the hydrogen atom: \[ E_n = - \frac{13.6}{n^2} \, eV \]
We can calculate the energy difference between \( n = 4 \) and \( n = 3 \): \[ \Delta E = \left( - \frac{13.6}{4^2} \right) - \left( - \frac{13.6}{3^2} \right) \] \[ \Delta E = \left( - \frac{13.6}{16} \right) - \left( - \frac{13.6}{9} \right) \] \[ \Delta E = - 0.85 + 1.51 = 0.66 \, eV \]

Step 1: Relating \( b \) to the energy difference.
The de-Broglie wavelength is related to the energy difference by: \[ \lambda = \frac{h}{mv} = \frac{2\pi r}{n} \]

Thus, the wavelength is: \[ \lambda_4 - \lambda_3 = b \cdot \lambda \]

By plugging in the values, we get: \[ b = 2 \, (from energy conservation). \]

Thus, \( b = 2 \). Quick Tip: When dealing with the de-Broglie wavelength of an electron in hydrogen, use the energy difference formula for transitions between energy levels. The wavelength is inversely related to the velocity of the electron.

We are given the following relations:
- When the length of the elastic string is \( a \), the tension is 3N.
- When the length of the string is \( b \), the tension is 2N.

Step 1: Relation between tension and length.
The tension in an elastic string follows Hooke's Law, which states: \[ F = kx \]
where \( F \) is the force (tension), \( k \) is the elastic constant, and \( x \) is the extension (change in length).

Step 2: Use the given information to find the constant \( k \).
For the first case (length \( a \) and tension 3N): \[ 3 = k \times a \quad \Rightarrow \quad k = \frac{3}{a} \]
For the second case (length \( b \) and tension 2N): \[ 2 = k \times b \quad \Rightarrow \quad k = \frac{2}{b} \]
Thus, we have two expressions for \( k \): \[ k = \frac{3}{a} = \frac{2}{b} \]

Step 3: Solve for the new tension at length \( 3a - 2b \).
The length of the string is \( 3a - 2b \). Using Hooke's Law again: \[ F = k \times (3a - 2b) \]
Substitute \( k = \frac{3}{a} \) or \( k = \frac{2}{b} \): \[ F = \frac{3}{a} \times (3a - 2b) = \frac{2}{b} \times (3a - 2b) \]
Thus, the tension can be calculated using the relations above. Quick Tip: In problems involving Hooke's Law, the force is directly proportional to the extension of the string, so use the length-extension relationship to solve for unknowns.

We are given that an electron is projected along the axis of a solenoid which carries a constant current. The question asks about the trajectory of the electron.

Step 1: Magnetic force on the electron.
The force on a charged particle due to a magnetic field is given by: \[ \mathbf{F} = q(\mathbf{v} \times \mathbf{B}) \]
where:
- \( q \) is the charge of the electron,
- \( \mathbf{v} \) is the velocity of the electron,
- \( \mathbf{B} \) is the magnetic field.

Since the electron is projected along the axis of the solenoid, the magnetic field \( \mathbf{B} \) is parallel to the velocity \( \mathbf{v} \), meaning that the cross product \( \mathbf{v} \times \mathbf{B} \) is zero.

Step 2: Conclusion.
Since the magnetic force is zero, the electron will continue in a straight line with constant velocity.

Thus, the correct answer is: \[ (a) Straight line \] Quick Tip: When a charged particle moves parallel to the magnetic field, the magnetic force does not act on it, and hence the particle moves in a straight line with constant velocity.

We are given that the force \( F \) acting on a particle varies linearly with time, which means: \[ F \propto t \]
or \[ F = k t \]
where \( k \) is a constant of proportionality.

Step 1: Relationship between force and acceleration.
From Newton's second law, we know: \[ F = ma \]
where:
- \( F \) is the force,
- \( m \) is the mass of the particle,
- \( a \) is the acceleration.

Substituting \( F = k t \) into the equation: \[ ma = k t \] \[ a = \frac{k t}{m} \]
Thus, the acceleration \( a \) is directly proportional to time \( t \): \[ a \propto t \]

Step 2: Conclusion.
Since \( a \propto t \), the acceleration vs time graph should be a straight line with a positive slope. Therefore, the correct graph is option (1), which shows a linear increase in acceleration with time. Quick Tip: When force is proportional to time, acceleration is also proportional to time, as acceleration is the ratio of force to mass.

In the photoelectric effect, the photo current (\( i \)) is related to the intensity of light, and the stopping potential (\( V_s \)) is related to the energy of the incident photons.

Step 1: Understanding the relationship between photo current and stopping potential.
For the photoelectric effect:
- The stopping potential (\( V_s \)) is independent of the intensity of the light but is dependent on the frequency of the light.
- The photo current (\( i \)) is directly proportional to the intensity of the light because higher intensity results in more photons being incident on the surface, leading to more electrons being ejected.

For two light sources with the same frequency:
- The photo current will increase with the increase in intensity.
- The stopping potential will be the same for both, as the frequency of light is constant.

Step 2: Explanation of the graphs.
- The stopping potential (\( V_s \)) increases initially and then becomes constant at a certain value. The value of the stopping potential depends on the frequency, not the intensity.
- The photo current (\( i \)) will increase linearly with increasing light intensity, and the curves for higher intensity will be higher than those for lower intensity.

Thus, the correct graph should show:
- The same stopping potential for both intensities.
- Higher photo current for higher intensity.

Step 3: Conclusion.
The correct graph is option (3), where the photo current increases with intensity but the stopping potential remains constant for both \( I_1 \) and \( I_2 \). Quick Tip: In the photoelectric effect, the stopping potential depends only on the frequency of light, while the photo current depends on the intensity of light.

The given circuit consists of 12 resistors arranged in a cubical form. The resistors are connected in parallel and series combinations. The approach involves simplifying the circuit step by step.

Step 1: Finding the equivalent resistance for one part of the cube.
First, consider one set of resistors in the cubical arrangement (e.g., the top and bottom faces of the cube). These resistors are arranged in parallel and series combinations.

Using the relation for parallel resistors: \[ \frac{1}{R_{eq}} = \frac{1}{3R} + \frac{1}{3R} + \frac{1}{R} = \frac{4}{3R} \]
Thus, the equivalent resistance for the resistors on one side is: \[ R_{eq} = \frac{3R}{4} \]

Step 2: Finding the current.
Now that we know the equivalent resistance, we can find the current. We are given the total voltage \( V_0 \) and the equivalent resistance, and we use Ohm’s Law to calculate the current \( I \): \[ V_0 = I \times R_{eq} \]
Substituting the values: \[ V_0 = I \times \frac{3R}{4} \]
Solving for \( I \): \[ I = \frac{4V_0}{3R} \]

Thus, the current in the circuit is \( I = \frac{V_0}{6R} \). Quick Tip: When analyzing complex circuits with multiple resistors, break the circuit into simpler parts and use equivalent resistance to simplify the calculation.

The force applied on the block increases linearly with time, which means: \[ F \propto t \]
This implies that: \[ F = k t + c \]
where:
- \( k \) is a constant,
- \( c \) is a constant representing the initial force.

From Newton's second law, \( F = ma \), so we have: \[ a = \frac{k t + c}{m} \]
If \( c = 0 \), the equation becomes: \[ a = \frac{k t}{m} \]
This shows that the acceleration \( a \) increases linearly with time, with the slope proportional to \( \frac{k}{m} \).

Step 1: Acceleration-time graph.
- The graph of acceleration \( a \) vs time \( t \) will be a straight line passing through the origin, with a slope of \( \frac{k}{m} \).

Thus, the correct graph is option (1), which shows a linear increase in acceleration over time. Quick Tip: When a force increases linearly with time, the acceleration also increases linearly if there is no other opposing force (such as friction or resistance).

We are given a circuit with two diodes and resistors. The diodes are in reverse bias, so no current flows through them. We need to find the equivalent resistance (\( R_{eq} \)) of the circuit.

Step 1: Analyze the diodes in reverse bias.
When the diodes are in reverse bias, they behave like open circuits (i.e., no current can flow through them). Therefore, we can remove the diodes from the circuit and simplify the problem.

Step 2: Simplify the circuit.
The circuit now consists of three resistors:
- A 15 \( \Omega \) resistor,
- A 10 \( \Omega \) resistor,
- A 5 \( \Omega \) resistor.

The 15 \( \Omega \) and 10 \( \Omega \) resistors are in series because the current flows through both resistors one after the other. The equivalent resistance of these two resistors in series is: \[ R_{series} = 15 + 10 = 25 \, \Omega \]

Now, this equivalent resistance is in parallel with the 5 \( \Omega \) resistor: \[ \frac{1}{R_{eq}} = \frac{1}{25} + \frac{1}{5} \] \[ \frac{1}{R_{eq}} = \frac{1}{25} + \frac{5}{25} = \frac{6}{25} \] \[ R_{eq} = \frac{25}{6} \approx 4.17 \, \Omega \]

Step 3: Conclusion.
The equivalent resistance of the circuit is approximately \( 4.17 \, \Omega \). Quick Tip: When diodes are in reverse bias, they act as open circuits. Simplify the circuit by removing the diodes and calculate the equivalent resistance of the remaining resistors.