JEE Main 2023 Physics April 6 Shift 2 Question Paper with Solution PDF- Download Here

Using \(v_{rms} = \sqrt{\frac{3RT}{m}}\)

At \(200 \, K\), the r.m.s. speed is \(v_0\): \[ v_0 = \sqrt{\frac{3R \times 200}{m}} \quad \cdots (1) \]
At \(800 \, K\), the r.m.s. speed is \(v'\): \[ v' = \sqrt{\frac{3R \times 800}{m}} \quad \cdots (2) \]
Dividing equation (2) by equation (1): \[ \frac{v'}{v_0} = \sqrt{\frac{800}{200}} = \sqrt{4} = 2 \]
Therefore, \(v' = 2v_0\). Quick Tip: When dealing with thermodynamics problems involving ideal gases, always start by using the ideal gas law and the relationship between temperature and the r.m.s. speed of gas molecules.

As we know the speed of light in a medium: \[ v = \frac{c}{\mu} \quad or \quad f \lambda = \frac{c}{\mu} \]
Therefore, \(\lambda \propto \frac{1}{\mu}\).

When light travels through two different mediums, their phase difference will change: \[ \Delta Q = \frac{2\pi}{\lambda} \]
Thus, statement R is the correct explanation of A. Quick Tip: For light wave problems, always consider the relationship between wavelength, refractive index, and phase difference when dealing with different media.

Given, modulation index \(m = 60% = 0.6\) and minimum amplitude \(A_c = 3 \, V\).

Using the formula for amplitude modulation: \[ A_m = mA_c \]
The maximum amplitude of the modulated wave is: \[ A_m + A_c = 0.6 \times 1 + 3 = 1.6 \, V \]

Now applying the amplitude formula: \[ A_m + A_c = \frac{1.6}{-0.4} = 12 \, V \] Quick Tip: When working with amplitude modulation problems, remember to use the formula for maximum and minimum amplitudes based on modulation index and carrier amplitude.

Using the formula for the speed of sound, \(v = \sqrt{\frac{\gamma RT}{m}}\), we can write for hydrogen and oxygen:
\[ \frac{v_{H_2}}{v_{O_2}} = \sqrt{\frac{m_{O_2}}{m_{H_2}}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4 : 1 \]

Thus, the ratio is \(4 : 1\). Quick Tip: Remember to apply the formula for speed of sound in gases, which depends on the molecular mass and temperature. The ratio of the molecular masses plays a crucial role in this type of problem.

In this case, since masses of both charges are not the same, we need to find center of mass (COM), about which dipole will oscillate and then we will find moment of inertia about this axis, to find torque and hence angular frequency \( \omega \).

Given, the mass of positive charge is \(m\), so the center of mass will be at a distance \(L\) from the negative charge: \[ \frac{L}{3} = \frac{2m}{3m} = \frac{2L}{3} \]
The moment of inertia about the axis is \(I = \frac{2mL^2}{3}\). Hence, angular frequency: \[ \omega = \sqrt{\frac{qE}{I}} = \frac{4qE}{3ml} \] Quick Tip: For dipole oscillations in electric fields, consider the center of mass for accurate calculation of moment of inertia, which affects the angular frequency.

As per Pascal’s law, when we apply pressure to an ideal liquid, it is equally distributed in the entire liquid and to the walls as well. The pressure change allows the toothpaste to be squeezed out.

Thus, both statements are true, and R explains A. Quick Tip: For problems related to fluids, always remember Pascal's law which applies to incompressible fluids and explains how pressure changes propagate within the fluid.

This question is based on conceptual clarity that we should connect ammeter in series and voltmeter in parallel to measure current and potential difference, respectively. Also, when using a galvanometer to create an ammeter, shunt resistance should be very small and should be in parallel.

When we create a voltmeter shunt should be large and in series with the galvanometer.

All these criteria are satisfied in option (2). Quick Tip: For circuit-based experiments, always ensure the correct placement of ammeters and voltmeters for accurate measurement of current and voltage. Remember that a shunt resistor is crucial for converting a galvanometer into an ammeter.

Method-1:

Using the exact law of cooling: \[ T - T_s = (T_0 - T_s) e^{-kt} \]
Case-I: \((40 - 10) = (60 - 10) e^{-7k}\) \[ 30 - 50e^{-7k} \quad \dots \, (1) \]
Case-II: \((T - 10) = (40 - 10) e^{-7k} \quad T - 10 = 30e^{-7k}\)
Dividing equation (2) by (1): \[ \frac{40 - T}{30} = \frac{20}{40 + T} \] \[ \Rightarrow T - 10 = 30 \times \frac{30}{50} = 18 \]
Thus, the temperature after 7 more minutes is \(T = 28°C\).


Method-2:

Using Newton's law of cooling: \[ \frac{T - T_s}{t} = k \left( \frac{T_1 + T_2}{2} - T_s \right) \]
Case-I: \[ \frac{60 - 40}{7} = k \left( \frac{40 + 10}{2} - 10 \right) \Rightarrow k = \frac{20}{7} \]
Case-II: \[ \frac{40 - 10}{7} = k \left( \frac{20 + T}{2} - 10 \right) \Rightarrow T = 28°C \] Quick Tip: Always remember that cooling problems can be solved using either the exact law of cooling or Newton's law of cooling. Choose the method that suits the given data.

By theory of electromagnetic waves: \[ U_E = \frac{1}{2} \epsilon_0 E^2 \quad and \quad U_B = \frac{1}{2} \mu_0 B^2 \] Quick Tip: For energy density in electromagnetic waves, remember that the energy is shared equally between the electric and magnetic fields.

Using Newton's formula: \[ F = \frac{GMm}{r^2} \]
At surface of the earth, \[ F = \frac{GMm}{R_e^2} \quad (eq. 1) \]
At \(r = R_e + \frac{R_e}{4} = \frac{5R_e}{4}\), \[ F' = \frac{GMm}{\left(\frac{5R_e}{4}\right)^2} = \frac{16GMm}{25R_e^2} \]
Thus, \[ F' = \frac{16}{25} F = \frac{16}{25} \times 100 = 64 \, N \] Quick Tip: Remember that gravitational force decreases with the square of the distance from the center of the Earth. When the distance increases, the force decreases.

Given alternating AC source \(E = 36 \sin (120 \pi t)\) V and capacitor \(C = 150 \mu F\), we can write: \[ Q = CV \quad and \quad i = \frac{dQ}{dt} = C E \omega \cos \omega t \]
Maximum value of current: \[ i_0 = C E_0 \omega = 150 \times 10^{-6} \times 36 \times 120 \pi = 2.03 A \] Quick Tip: In AC circuits with capacitors, the maximum current is found by multiplying the capacitance, maximum voltage, and angular frequency.

Based on the data provided: \[ U = 100 - 80 = 20 \, cm, \quad V = 180 - 100 = 80 \, cm \]
Using the formula: \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \quad or \quad f = \frac{uv}{u + v} \quad \Rightarrow f = \frac{20 \times 80}{20 + 80} = 16 \, cm \]
For error analysis: \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]
Differentiating: \[ \frac{Df}{f^2} = \frac{Du}{u^2} + \frac{Dv}{v^2} \]
Now: \[ \Delta u = 0.4 \, cm, \, \Delta v = 0.4 \, cm \]
Now, \[ \frac{\Delta f}{f} = \left[ \frac{16 \times 0.4}{(80)^2} + \frac{16 \times 0.4}{(20)^2} \right] \] \[ \Rightarrow \frac{\Delta f}{f} = 16 \times 0.4 \left( \frac{17}{400} \right) \quad \Rightarrow % \, Error = \frac{17 \times 0.4}{400} \times 1000 = 1.7 \] Quick Tip: When calculating percentage error, remember that any measurement errors in \( u \) and \( v \) can affect \( f \). Use differentials to find the total error in calculated quantities.

Let potential of the junction be \( x \) volts using junction law \( i_1 + i_2 + i_3 = 0 \). \[ x - 30 = \frac{x - 12}{10} \quad and \quad x - 2 = \frac{x - 12}{20} \]
Solving for \( x \): \[ x = 20 \, V, \quad current through \, 20 \, \Omega \, resistor = \frac{20 - 12}{20} = 0.4 \, A \] Quick Tip: For circuits with multiple components, apply Kirchhoff's current and voltage laws to simplify the analysis and solve for unknowns.

Using Bohr’s postulate: angular momentum \( L = mvr = \frac{nh}{2\pi} \).
Therefore: \[ m r \omega^2 = \frac{nh}{2\pi} \] \[ \Rightarrow r \propto \sqrt{n} \] Quick Tip: Bohr's model applies quantization of angular momentum to describe the radii of orbits in atomic systems. Use this principle to relate radius to quantum number \( n \).

The diffusion current in a forward biased p-n junction is greater than the drift current, as the charge carriers move due to the gradient in the concentration. The direction of the diffusion current is from the n-side to the p-side, which is correctly explained by Reason R. Quick Tip: In p-n junctions, diffusion current dominates in forward bias conditions. Remember that drift current is due to an electric field, while diffusion current is due to concentration gradients.

Since planets revolve around the sun in an elliptical orbit, its linear speed is not constant, hence option (1) is incorrect. Other statements are correct as per theory.
Quick Tip: For planets in elliptical orbits, remember that their linear speed changes depending on their position along the orbit. The total energy of a planet remains constant in elliptical orbits.

Given, \(m = 5 \, kg\), \(R = 2 \, m\), time \(t\) for 1 rev \(= 3.14 \, sec\) or \(\pi \, sec\).
So, \(\theta\) for 1 rev \(= 2\pi \, rad\).
Therefore, \(\omega = \frac{\theta}{t} = \frac{2\pi}{\pi} = 2 \, rad/s\).
Centrifugal force \(F = mR\omega^2 = 5 \times 2 \times 2^2 = 40 \, N\).
Quick Tip: To calculate centrifugal force in rotational motion, use the formula \( F = mR\omega^2 \) where \(\omega\) is the angular velocity and \(R\) is the radius of the circular path.

Given, speed \(v = \pi \, m/s\), or \(R\omega = \pi\).
Therefore, \(\omega = \frac{\pi}{R} \, rad/s\).
Angular displacement \(\theta = 120^\circ = \frac{2\pi}{3}\).
Using \(\theta = \omega t\), we get: \[ t = \frac{\theta}{\omega} = \frac{2\pi/3}{\pi/R} = \frac{2R}{3}. \]
Linear displacement \(d = 2R \sin( \theta / 2 ) = 2R \sin(60^\circ) = 2R \times \frac{\sqrt{3}}{2} = R\sqrt{3}\).
Hence, average velocity \( = \frac{d}{t} = \frac{R \sqrt{3}}{2R/3} = 1.5 \sqrt{3} \, m/s\).
Quick Tip: To find the average velocity for circular motion, use the formula \( average velocity = \frac{d}{t} \) where \(d\) is the linear displacement and \(t\) is the time taken.

Using \( K.E_{max} = eV_s = hf - \phi_0 \), where \(\phi_0\) is work function, \(V_s\) is stopping potential, and \(f\) is frequency, we get: \[ V_s = \frac{h}{e} f - \frac{\phi_0}{e}. \]
Therefore, the slope \( m \) will be the same for all graphs and will be independent of \(\phi_0\).
Quick Tip: In photoelectric experiments, the slope of the stopping potential versus frequency plot is independent of the material’s work function.

Using the first equation of motion: \[ t = \frac{v - u}{a} \] \[ t = \frac{60 - 10}{2} = \frac{50}{2} = 25 \, sec \] Quick Tip: Remember, for uniform acceleration problems, use the first equation of motion: \( v = u + at \), where \( v \) is final velocity, \( u \) is initial velocity, \( a \) is acceleration, and \( t \) is time.

For pendulum: \[ T_{max} = mg + \frac{mv^2}{L} \tag{1} \]
Given \( m = \frac{1}{4} \, kg, L = 1 \, m, g = 9.8 \, m/s^2 \) and amplitude \( A = \frac{1}{10} \, m \)

For SHM, \( K_{max} = \frac{1}{2} mv^2 = \frac{1}{2} m \omega^2 A^2 \)

Using \( \omega = \sqrt{\frac{g}{L}} \), we get: \[ mv^2 = m \left( \frac{g}{L} \right) A^2 = m g \frac{A^2}{L} \]
Substitute this into equation (1): \[ T_{max} = 2mg + \frac{mgA^2}{L^2} \]
Substituting the given values: \[ T_{max} = 2 \times \frac{1}{4} \times 9.8 + \frac{1}{4} \times 9.8 \times \frac{101}{100} \] \[ T_{max} = \frac{98.98}{40} \]
Therefore, \( x = 99 \) Quick Tip: In SHM, maximum tension is calculated by considering both gravitational force and the force due to the velocity of the bob at maximum displacement.

Using \( E = \frac{k e^2}{2r} \), we can find \( r \) as follows: \[ r = \frac{Re^2}{2E} \]
Given \( E = 12.8 \, eV = 12.8 \times 1.6 \times 10^{-19} \, J \), \[ r = \frac{2 \times 10^9 e^2}{9 \times 10^{-10}} = \frac{9 \times 10^{-10}}{(2 \times 12.8 / 1.6)} \times 10 \] \[ r = 16 \times 10^{-10} \, m \]

Therefore, \( x = 16 \) Quick Tip: The orbital radius of an electron can be calculated using the energy required to separate the proton and electron in a hydrogen atom, based on Coulomb's law.

Let \( n_1 \) maxima of 7000 \(\AA\) coincide with \( n_2 \) maxima of 5500 \(\AA\).
Therefore, \( n_1 \beta_1 = n_2 \beta_2 \), \[ \frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{5500}{7000} = \frac{11}{14} \]
Hence, the 11th maximum of 7000 \(\AA\) coincides with the 14th maximum of 5500 \(\AA\).

To find the least distance from this, \[ y = n_1 \beta_1 D / d \] \[ y = \frac{11 \times 7000 \times 10^{-10} \times 150 \times 10^{-2}}{2.5 \times 10^{-3}} = 462 \times 10^{-5} \, m \]

Thus, \( n = 462 \) Quick Tip: In interference problems, use the relationship between the wavelengths and the maxima to calculate the least distance where bright fringes coincide.

Given, \[ a = 1000 \, cm, \quad b = 1 \, cm \]
As the larger coil is taken as primary, \[ Mutual inductance M = \frac{\mu_0 N n b^2}{2a} \] \[ M = \frac{4\pi \times 10^{-7} \times 200 \times 10 \times \pi \times 1 \times 10^{-4}}{2 \times 1000 \times 10^{-2}} \] \[ M = 4 \times 10^{-8} \, H \]

Therefore, the value of mutual inductance is 4 \(\times 10^{-8}\) H.
Quick Tip: For mutual inductance, always remember the formula \(M = \frac{\mu_0 N n b^2}{2a}\) for concentric coils, where \(a\) is the radius of the larger coil and \(b\) is the radius of the smaller coil.

As per the arrangement given, the distance between the capacitor plates are \(a\) and \(b\) and \(a \neq b\). Using the diagram we can write \[ b = 5 - a - 1 = (4 - a) in mm \]
As we know the capacitance of a capacitor \(C = \dfrac{\varepsilon_0 A}{d}\) and in series arrangement \[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \] \[ \frac{1}{C_{eq}} = \frac{a A}{\varepsilon_0 A} + \frac{4 - a}{\varepsilon_0 A} = \frac{4 (in mm)}{\varepsilon_0 A} \]
or \[ C_{eq} = \frac{\varepsilon_0 A}{4(mm)} \]
Given \(A = 200 cm^2\), \[ C_{eq} = \varepsilon_0 \times 200 \times 10^{-4} = 4 \times 10^{-3} = 50 \times 10^{-1} \]
or \[ C_{eq} = 5 \varepsilon_0 farad \]
Therefore, \(C_{eq} = 5 \varepsilon_0\), so \(x = 5\).
Quick Tip: Royal genealogies and epics provide valuable evidence of the dominance of patriliny in early Indian society.

Given: \[ B = \frac{\pi}{2} \times 10^{-3} \, T, \quad K.E. = 2.0 \, eV, \quad m = 1.67 \times 10^{-27} \, kg, \quad Charge on proton = 1.6 \times 10^{-19} \, C \]
From the kinetic energy formula \( K.E. = \frac{1}{2} m v^2 \), we can solve for the velocity \(v\) as follows:



\[ v = \sqrt{\frac{2 K.E.}{m}} \]
The pitch of the helical path is given by: \[ Pitch = v \cos 60^\circ \times time period of one rotation \] \[ v = v \cos 60^\circ = \frac{2\pi r}{e B} \] \[ = \sqrt{\frac{2 \times 2 \times 10^{-19}}{1.6 \times 10^{-27}}} \times \cos 60^\circ = \frac{2 \pi \times 1.6 \times 10^{-19}}{1.6 \times 10^{-27} \times \frac{\pi}{2} \times 10^{-3}} \] \[ = 2 \times 10^4 \times \frac{1}{2} \times 4 \times 10^{-5} = 4 \times 10^1 cm \]

Thus, the pitch of the helical path is 40 cm.
Quick Tip: To calculate the pitch of a proton's helical path, use the relationship between velocity, magnetic field, and the angle between them.

Let \(u\) and \(v\) be the speeds, just before and after the body strikes the ground.

Given \( \frac{u}{v} = 4 \).

We know that the loss in kinetic energy is: \[ \Delta K.E. = \frac{1}{2} m u^2 - \frac{1}{2} m v^2 \]
Thus, \[ \Delta K.E. = \frac{1}{2} m \left( u^2 - v^2 \right) \]
Substitute the values: \[ \Delta K.E. = \frac{1}{2} m \left( \left( 4v \right)^2 - v^2 \right) \] \[ = \frac{1}{2} m \left( 16v^2 - v^2 \right) = \frac{1}{2} m \times 15v^2 \]
Thus, the percentage loss is: \[ \frac{15}{16} \times 100 = 93% \]
Therefore, the percentage loss in kinetic energy is 93%.
Quick Tip: This problem involves kinetic energy calculations and how the velocity changes with the body's rebound after striking the ground.

Given radius of gyration is same for ring and solid sphere: \[ K_R = K_{ss} \]
For ring and sphere, the radius of gyration is: \[ R_R = \sqrt{\frac{2}{5}} R_{ss} \]
Thus, \[ \frac{R_R}{R_{ss}} = \sqrt{\frac{2}{5}} \]
Therefore, \(x = 5\).
Quick Tip: The radius of gyration for different objects depends on their mass distribution. For a ring and a solid sphere, the ratio depends on how their mass is spread relative to their axis of rotation.

Method-I:
The equivalent resistance for the given circuit is:



\[ R_{eq} = 2 + \frac{5R}{5 + R} \]
Using Ohm's law: \[ i = \frac{3}{R_{eq}} = \frac{3}{5 + R} \]
The current in the circuit is: \[ i = 3 \times \frac{5R}{5 + R} \]
From the voltage readings, we calculate \(R = 20 \, \Omega\).


Method-II:
Given potential across 5 \(\Omega\) and voltmeter is 2V, to find the resistance \(R\) of the voltmeter: \[ i = \frac{2}{5} = \frac{1}{2} \]
Using junction law: \[ i = i_1 + i_2 \]
Hence, \(R = 20 \, \Omega\).
Quick Tip: In voltage divider circuits, the resistance of the voltmeter plays a crucial role in the reading and must be taken into account when calculating the equivalent resistance.

Using stress = \(\frac{force}{area} = \frac{mg}{A}\)
\[ \Rightarrow m = \frac{S \times A}{g} \] \[ m = \frac{7 \times 10^5 \times \pi \times (7 \times 10^{-3})^2}{9.8} \] \[ m = \frac{7 \times 10^5 \times \frac{22}{7} \times (7 \times 10^{-3})^2}{9.8} \] \[ m = 11 \, kg \]
(Note: \(14 \, mm\) is diameter)
Quick Tip: The formula for stress is \(Stress = \frac{Force}{Area}\), where area is the cross-sectional area of the wire, and force is due to the weight of the block.