JEE Main 2023 Physics April 11 Shift 1 Question Paper with Solution PDF- Download Here

We are given that the radii of two planets 'A' and 'B' are \( R \) and \( 4R \), and their densities are \( \rho \) and \( \frac{\rho}{3} \), respectively. The formula for the acceleration due to gravity at the surface of a planet is:
\[ g = \frac{4\pi G R \rho}{3} \]

Since gravity is proportional to both the radius and density, the ratio of acceleration due to gravity at their surfaces can be written as:
\[ g_A : g_B = \frac{\frac{4\pi G R \rho}{3}}{\frac{4\pi G (4R) (\frac{\rho}{3})}{3}} = \frac{R \cdot \rho}{(4R) \cdot \frac{\rho}{3}} = \frac{1}{4} \cdot 3 = \frac{3}{4} \]

Thus, the correct ratio is \( g_A : g_B = 3:4 \). Quick Tip: The ratio of accelerations due to gravity can be determined by analyzing the dependence on both the radius and the density of the planets.

The frictional force is responsible for causing the coin to slip. This force is given by \( f_r = m \cdot \omega^2 r \), where \( \omega \) is the angular velocity and \( r \) is the distance from the center.

Given that the angular velocity is halved, we use the equation \( \omega^2 \cdot r^2 = constant \). When the angular velocity is halved, the distance \( r \) from the center at which the coin slips will be:
\[ r_2 = 4 \, cm \]

Thus, when the angular velocity is halved, the coin will slip at a distance of 4 cm from the center. Quick Tip: When the angular velocity is reduced, the distance at which the coin slips increases according to the square of the ratio of the angular velocities.

The root mean square speed \( V_{rms} \) is given by:
\[ V_{rms} = \sqrt{\frac{3RT}{m}} \]

Since the gases are at the same temperature and pressure, the root mean square speed depends on the molar mass \( m \). For neon (monoatomic), chlorine (diatomic), and uranium hexafluoride (polyatomic), the molar mass increases in the order:
\[ V_{rms} (mono) > V_{rms} (dia) > V_{rms} (poly) \]

Thus, the correct answer is \( V_{rms} (mono) > V_{rms} (dia) > V_{rms} (poly) \). Quick Tip: The root mean square speed decreases as the molar mass of the gas increases.

We are given that the half-life of A is the same as the average life of B. The average life \( \tau \) and half-life \( T \) are related to the decay constant \( \lambda \) by the equations:
\[ T = \frac{\ln 2}{\lambda} \quad and \quad \tau = \frac{1}{\lambda} \]

Since the half-life of A is the same as the average life of B, we have:
\[ \lambda_A = \lambda_B \ln 2 \]

Hence, the correct relation is \( \lambda_A = \lambda_B \ln 2 \). Quick Tip: When comparing the half-life and average life, use the relationship between the decay constant and the half-life to derive the necessary formula.

The given graph shows the variation of impedance with frequency.
- At low frequencies, the impedance of a capacitor \( X_C \) is low, and the impedance of an inductor \( X_L \) is high.
- At high frequencies, the impedance of an inductor increases while that of a capacitor decreases.

Thus, \( A \) represents the capacitive reactance \( X_C \), and \( B \) represents the inductive reactance \( X_L \). Quick Tip: Impedance of an inductive circuit increases with frequency, whereas the impedance of a capacitive circuit decreases with frequency.

Let \( R \) be the radius of the Earth, and \( h \) be the height of the antenna. The distance \( d \) between the two antennas is given by:
\[ d = \sqrt{2Rh} \]

We are given that \( d = 4 \, km \), and the radius of the Earth \( R = 6400 \, km \). Substituting these values into the equation:
\[ 4 = \sqrt{2 \times 6400 \times h} \]

Squaring both sides:
\[ 16 = 2 \times 6400 \times h \]
\[ h = \frac{16}{12800} = 1 \, m \]

Now, using the formula for the signal range:
\[ x = \frac{500}{4} = 125 \]

Thus, the value of \( x \) is 125. Quick Tip: For problems involving the height of an antenna and the line of sight, use the relation \( d = \sqrt{2Rh} \) to calculate the minimum height for the signal to be received.

The energy density in an electromagnetic wave is given by the formula:
\[ u = \frac{1}{2} \epsilon_0 E_0^2 \]

Where \( E_0 \) is the peak electric field. Substituting the given values:
\[ u = \frac{1}{2} \times 8.85 \times 10^{-12} \times (20)^2 = 8.85 \times 10^{-13} \, J/m^3 \]

The total energy contained in a volume \( V = 5 \times 10^4 \, m^3 \) is:
\[ Energy = u \times V = 8.85 \times 10^{-13} \times 5 \times 10^4 = 8.85 \times 10^{-13} \, J \]

Thus, the energy is \( 8.85 \times 10^{-13} \, J \). Quick Tip: For electromagnetic waves, use the formula for energy density \( u = \frac{1}{2} \epsilon_0 E_0^2 \) and multiply by volume to find the total energy.

The power dissipated in a resistor is given by the formula:
\[ P = \frac{V^2}{R} \]

For two identical resistors, the heat produced in parallel and series configurations is as follows:

For parallel connection, the effective resistance is \( R_{eff} = \frac{R}{2} \), and the power is:
\[ P_{parallel} = \frac{V^2}{R/2} = 2 \frac{V^2}{R} \]

For series connection, the effective resistance is \( R_{eff} = 2R \), and the power is:
\[ P_{series} = \frac{V^2}{2R} \]

Now, the ratio of heat produced in parallel to series is:
\[ \frac{P_{parallel}}{P_{series}} = \frac{2 \frac{V^2}{R}}{\frac{V^2}{2R}} = 4 \]

Thus, the ratio of heat produced is \( 4 : 1 \). Quick Tip: The heat produced in a resistor depends on the square of the voltage and inversely on the resistance. Parallel and series connections affect the total resistance and hence the power.

The energy stored in a capacitor is given by the formula:
\[ E = \frac{1}{2} C V^2 \]

For a single capacitor with capacitance \( C = 2 \, F \), the energy is:
\[ E_1 = \frac{1}{2} \times 2 \times V^2 = V^2 \]

Now, the two capacitors are connected in parallel, and the total capacitance becomes:
\[ C_{total} = C + C = 2C = 4 \, F \]

The energy in the parallel combination is:
\[ E_5 = \frac{1}{2} \times 4 \times V^2 = 2V^2 \]

Now, the ratio of the energies is:
\[ \frac{E_5}{E_1} = \frac{2V^2}{V^2} = 2 \]

Thus, the ratio of the energies is \( 2 : 1 \). Quick Tip: When capacitors are connected in parallel, the total capacitance is the sum of individual capacitances. The energy stored in the combination increases due to the increased capacitance.

The force exerted by the machine gun is related to the rate of change of momentum of the bullets:
\[ F = \frac{d(mv)}{dt} \]

Since \( m = 10 \, g = 0.01 \, kg \) and \( v = 250 \, m/s \), the momentum of each bullet is:
\[ mv = 0.01 \times 250 = 2.5 \, kg m/s \]

Now, the rate of change of momentum gives the force:
\[ F = \frac{d(mv)}{dt} = \frac{10 \times 250}{1000} \, N \]

Thus, the number of bullets fired per second \( n \) is:
\[ n = \frac{125}{\frac{10 \times 250}{1000}} = 50 \]

Thus, the number of bullets fired per second is 50. Quick Tip: The force required to keep the machine gun in position is equal to the rate of change of momentum. Use this principle to find the number of bullets fired per second.

Displacement is the area under the graph and distance is the total area covered, including all positive and negative regions of the graph.

Displacement = Area of graph with sign \[ = \left(\frac{1}{2} \times 10 \times 5 \right) + \left(10 \times 5 \right) = 25 + 50 = 75 \, m \]

Distance = Area of graph with positive value \[ = \left(\frac{1}{2} \times 5 \times 20 \right) + \left(5 \times 20 \right) = 25 + 50 = 250 \, m \]

Now, the ratio of distance to displacement is:
\[ \frac{250}{75} = \frac{2}{5} \]

Thus, the ratio is \( \frac{2}{5} \). Quick Tip: When finding distance, always consider the absolute value of the displacement, i.e., sum the areas without sign.

From the given information, the temperature scale \( X \) has the following relationships:

The boiling point of water is 65° X and the freezing point is 15° X, so the scale has a linear relation between these two points. Let's use the formula for converting between temperature scales:
\[ F = X \times \frac{9}{5} + 32 \]

Now, to convert 95° X to Fahrenheit:
\[ F = 95 \times \frac{9}{5} + 32 = 171 + 32 = 148° F \]

Thus, the equivalent temperature is 148° F. Quick Tip: To convert from a custom temperature scale to Fahrenheit, use the linear conversion formula \( F = \left(\frac{9}{5} \times X\right) + 32 \).

We know the formula for the magnetic field inside a toroid is:
\[ B = \frac{\mu_0 N I}{2 \pi r} \]

Where \( \mu_0 \) is the permeability of free space and \( N \) is the number of turns. When a material with susceptibility \( \chi \) is added, the magnetic field increases by the factor:
\[ B_{new} = B_0 (1 + \chi) \]

Given that \( \chi = 2 \times 10^3 \), the percentage increase in the magnetic field is:
\[ Percentage increase = \left( \frac{B_{new} - B_0}{B_0} \right) \times 100 = 2% \]

Thus, the correct answer is 2%. Quick Tip: The susceptibility of a material increases the magnetic field inside a toroid by the factor \( (1 + \chi) \), where \( \chi \) is the susceptibility.

The change in internal energy at constant pressure during a phase change can be calculated using the formula:
\[ \Delta U = mL \]

Where:
- \( m \) is the mass of the substance (1 kg),
- \( L \) is the latent heat of vaporization (\( 2257 \, kJ/kg \)).

Thus, the change in internal energy is:
\[ \Delta U = 1 \times 2257 = 2257 \, kJ \]

Now, the volume change from liquid to steam is:
\[ \Delta V = V_2 - V_1 = (1.671 \times 10^{-3}) - (1.00 \times 10^{-3}) = 0.671 \times 10^{-3}\, m^3 \]

Using the given atmospheric pressure of \( 1 \times 10^5 \, Pa \), we can calculate the work done:
\[ W = P \Delta V = (1.0 \times 10^5) \times (0.671 \times 10^{-3}) = 67.1 \, J \]

Thus, the total change in internal energy is:
\[ \Delta U + W = 2257 + 67.1 = 2426.1 \, kJ \]

So, the change in internal energy is approximately \( 2426 \, kJ \). Quick Tip: When converting liquid to gas at constant pressure, the change in internal energy is mainly due to the latent heat of vaporization.

The formula for the power of a lens is given by:
\[ P = \frac{1}{f} \]

Where \( f \) is the focal length of the lens. The focal length is related to the radius of curvature (\( R \)) and refractive index \( n \) by the lens-maker's formula:
\[ \frac{1}{f} = (n - 1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

For a convex lens, \( R_1 = R \) and \( R_2 = -R \), so the focal length in air is:
\[ \frac{1}{f_{air}} = (1.8 - 1) \left( \frac{1}{R} + \frac{1}{R} \right) = 0.8 \times \frac{2}{R} \]

Thus:
\[ f_{air} = \frac{R}{1.6} \]

Now, when the lens is immersed in a liquid of refractive index 1.5, the power changes. The new focal length is given by:
\[ \frac{1}{f_{liquid}} = (1.8 - 1.5) \left( \frac{1}{R} + \frac{1}{R} \right) = 0.3 \times \frac{2}{R} \]

Thus:
\[ f_{liquid} = \frac{R}{0.6} \]

The ratio of the power in air to the power in liquid is:
\[ Ratio = \frac{P_{air}}{P_{liquid}} = \frac{1/f_{air}}{1/f_{liquid}} = \frac{R/1.6}{R/0.6} = 2 \]

Thus, the ratio of the power in air to the power in liquid is 2. Quick Tip: When a lens is immersed in a medium with a different refractive index, the power of the lens changes based on the ratio of the refractive indices.

The induced emf in the plate is given by Faraday's Law of Induction:
\[ emf = - \frac{d\Phi}{dt} \]

Where \( \Phi \) is the magnetic flux given by:
\[ \Phi = B \times A \]

Here, \( A = 4 \, m^2 \), and \( \frac{dB}{dt} \) is the slope of the \( B \)-time graph from \( t = 2 \, s \) to \( t = 4 \, s \).

From the graph, the change in \( B \) is \( B_2 - B_1 = 8 - 4 = 4 \, T \) and the change in time is \( \Delta t = 4 - 2 = 2 \, s \).

Now, the induced emf is:
\[ emf = \frac{dB}{dt} \times A = \frac{4 \, T}{2 \, s} \times 4 \, m^2 = 8 \, V \]

Thus, the magnitude of the induced emf is \( 8 \, V \). Quick Tip: The induced emf in a conductor can be found using Faraday's law, where the emf is proportional to the rate of change of the magnetic field through the area.

The force in a wire is given by Hooke's Law:
\[ F = kx \]

Where:
- \( F \) is the applied force,
- \( k \) is the spring constant, and
- \( x \) is the elongation in the wire.

From the problem statement, we are given that:
\[ F = k \Delta x \]

Now, for tensions \( F_1 = 100 \, N \) and \( F_2 = 120 \, N \), the corresponding lengths of the wire will be:

For \( F_1 \):
\[ 100 = k (l_1 - l_0) \quad (1) \]

For \( F_2 \):
\[ 120 = k (l_2 - l_0) \quad (2) \]

Now, let the natural length of the wire be \( l_0 \). From the given information, \( l_1 = 11l_0 \).

From equation (1), we can write:
\[ 100 = k(11l_0 - l_0) \Rightarrow k(10l_0) = 100 \]

Thus,
\[ k = \frac{10}{l_0} \]

Substituting into equation (2):
\[ 120 = \frac{10}{l_0} (l_2 - l_0) \]
\[ l_2 - l_0 = \frac{120 l_0}{10} = 12 l_0 \]

Thus,
\[ l_2 = 13 l_0 \]

The ratio of \( l_2 \) to \( l_1 \) is:
\[ \frac{l_2}{l_1} = \frac{13 l_0}{11 l_0} = \frac{13}{11} \]

Therefore, \( x = 2 \). Quick Tip: To determine the natural length of a wire under tension, use Hooke’s Law and consider the elongation for different applied forces.

Given \( h = 4.25 \times 10^{-15} \, eV \cdot s \), the frequency of the incident light is related to the energy absorbed by the hydrogen atoms using the equation:
\[ E = h \times f \]

Where:
- \( E \) is the energy absorbed,
- \( h \) is Planck's constant,
- \( f \) is the frequency of the incident light.

The total emission lines are 6, so the electron must have absorbed energy and jumped from \( n = 1 \) to \( n = 4 \). The energy difference between these levels can be expressed as:
\[ \Delta E = h \times f \]

Using the formula for the frequency, we calculate:
\[ \Delta E = 13.6 \left[ \frac{1}{4^2} - \frac{1}{2^2} \right] \, eV \]

Now solving for the frequency:
\[ f = \frac{12.75 \times 10^{-15}}{4.25 \times 10^{-15}} = 3 \times 10^{15} \, Hz \]

Thus, \( x = 12 \). Quick Tip: The frequency of the incident light can be determined using the energy differences between electron energy levels in the hydrogen atom.

The work done by a force is given by the integral:
\[ w = \int_{x_1}^{x_2} F(x) \, dx \]

Where \( F(x) = 2 + 3x \). Substituting into the integral:
\[ w = \int_0^4 (2 + 3x) \, dx \]

Solving this:
\[ w = \int_0^4 2 \, dx + \int_0^4 3x \, dx = [2x]_0^4 + \left[ \frac{3x^2}{2} \right]_0^4 \]
\[ w = (2 \times 4) + \left( \frac{3 \times 4^2}{2} \right) = 8 + \left( \frac{3 \times 16}{2} \right) = 8 + 24 = 32 \, J \]

Thus, the work done is 32 J. Quick Tip: To calculate the work done by a variable force, integrate the force with respect to displacement over the given limits.

The point charge is in equilibrium at rest. Hence, the forces on the point charge must balance out. The force due to the gravitational pull is counteracted by the electrostatic force.

The forces acting on the point charge are:
- The electrostatic force, \( F_e \), due to the other charge.
- The gravitational force, \( mg \), acting downward.

Since the system is in equilibrium, the electrostatic force is balanced by the component of gravitational force along the plane:
\[ F_e = mg \sin \theta \]

Now, we know the formula for the electrostatic force between two point charges:
\[ F_e = \frac{k q_0^2}{r^2} \]

Where:
- \( k = 9 \times 10^9 \, N \cdot m^2 \cdot C^{-2} \),
- \( q_0 = 2 \times 10^{-6} \, C \) (the charge),
- \( r \) is the distance between the charges, which is related to \( h \) by the geometry of the inclined plane, where \( r = \frac{h}{\sin \theta} \).

Substituting the values into the equation:
\[ \frac{k q_0^2}{r^2} = mg \sin 30^\circ \]

Substituting the known values:
\[ \frac{9 \times 10^9 \times (2 \times 10^{-6})^2}{\left( \frac{h}{\sin 30^\circ} \right)^2} = 20 \times 10^{-3} \times 10 \times \sin 30^\circ \]

Solving this equation:
\[ h^2 = 9 \times 10^{10} \Rightarrow h = 300 \times 10^{-3} \, m \]

Thus, the value of \( x \) is 300. Quick Tip: In problems involving point charges on an inclined plane, use equilibrium conditions and equate the electrostatic force to the component of the gravitational force along the plane.

The moment of inertia of a solid sphere about its diameter is given by:
\[ I = \frac{2}{5} m r^2 \]

The moment of inertia of the sphere about its tangent (using parallel axis theorem) is:
\[ I_t = \frac{2}{5} m r^2 + m r^2 = \frac{7}{5} m r^2 \]

The angular momentum about the diameter is:
\[ L_{diameter} = I \cdot \omega = \frac{2}{5} m r^2 \cdot \omega \]

Now, the angular momentum about the tangent is:
\[ L_{tangent} = I_t \cdot \omega = \frac{7}{5} m r^2 \cdot \omega \]

The ratio of the angular momentum about the tangent to the diameter is:
\[ \frac{L_{tangent}}{L_{diameter}} = \frac{\frac{7}{5} m r^2 \cdot \omega}{\frac{2}{5} m r^2 \cdot \omega} = \frac{7}{2} \]

This is given as \( x \times 10^2 \). Thus:
\[ x \times 10^2 = \frac{7}{2} \]

Solving for \( x \):
\[ x = \frac{7}{2} \times 10^2 = 35 \]

Thus, the value of \( x \) is 35. Quick Tip: Use the parallel axis theorem to calculate the moment of inertia about a tangent and then apply the ratio to find the desired value.

The equation of the wave is given as:
\[ Y = 10^2 \sin 2 \pi \left( (60t - 0.5x + \frac{\pi}{4}) \right) \]

The general form of the wave equation is:
\[ Y = A \sin \left( 2 \pi \left( \frac{x}{\lambda} - \frac{t}{T} \right) \right) \]

Here, \( \lambda \) is the wavelength, \( T \) is the period, and the wave number \( k = \frac{2\pi}{\lambda} \) and the angular frequency \( \omega = \frac{2\pi}{T} \).

From the given equation:
\[ \omega = 60 \quad and \quad k = 0.5 \]

The speed of the wave is given by:
\[ v = \frac{\omega}{k} \]

Substituting the values:
\[ v = \frac{60}{0.5} = 120 \, m/s \]

Now, converting the speed into km/h:
\[ v = 120 \times \frac{18}{5} = 1152 \, km/h \]

Thus, the speed of the wave is \( 1152 \, km/h \). Quick Tip: To find the speed of a wave, use the relation \( v = \frac{\omega}{k} \), and convert the units from m/s to km/h by multiplying by \( \frac{18}{5} \).