JEE Main 2023 Physics Question Paper Jan 29 Shift 2- Download PDF

For substance A:
\[ (N_0)_A = \frac{320}{16} = 20 \, moles \] \[ N_A = (N_0)_A \left(\frac{1}{2}\right)^2 = 20 \times \frac{1}{4} = 5 \, moles \]
For substance B:
\[ (N_0)_B = \frac{320}{32} = 10 \, moles \] \[ N_B = (N_0)_B \left(\frac{1}{2}\right)^4 = 10 \times \frac{1}{16} = 0.625 \, moles \]
Total moles = \(5 + 0.625 = 5.625\). The total number of atoms: \[ 5.625 \times 6.023 \times 10^{23} = 3.38 \times 10^{24} \] Quick Tip: For half-life problems, use \((1/2)^n\), where \(n\) is the number of half-lives elapsed.

The root mean square (rms) speed and average speed are related as: \[ \sqrt{\frac{3RT}{M}} = \frac{\alpha + 5}{\alpha} \sqrt{\frac{8RT}{\pi M}} \]
Squaring both sides and simplifying: \[ 3 = \frac{\alpha + 5}{\alpha} \cdot \frac{8}{\pi} \]
Substitute \(\pi = \frac{22}{7}\): \[ 3 = \frac{\alpha + 5}{\alpha} \cdot \frac{8 \times 7}{22} \] \[ 3 = \frac{\alpha + 5}{\alpha} \cdot \frac{28}{22} \] \[ \alpha = 28 \] Quick Tip: Understand the relationship between rms speed and average speed to solve such problems efficiently.

The de-Broglie wavelength is given by: \[ \lambda = \frac{h}{\sqrt{2mqV}} \]
For \(\alpha\)-particle and proton, the ratio: \[ \frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p}{m_\alpha}} \]
Mass of \(\alpha\)-particle = 4 \(\times\) mass of proton: \[ \frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{1}{4}} = \sqrt{8}, \, m = 8 \] Quick Tip: De-Broglie wavelength depends inversely on the square root of particle mass.

The given circuit consists of a combination of NOT, AND, and OR gates. The truth table is derived as follows:

1. The NOT gates invert the inputs \(A\) and \(B\).

2. These inverted values are then input into the AND gates, and the output of each AND gate is determined.

3. Finally, the outputs of the AND gates are fed into the OR gate to produce the final output \(X\).


Truth Table Analysis:


% Option
(A) When \(A = 0\) and \(B = 0\), the output \(X = 1\).
% Option
(B) When \(A = 0\) and \(B = 1\), the output \(X = 0\).
% Option
(C) When \(A = 1\) and \(B = 0\), the output \(X = 1\).
% Option
(D) When \(A = 1\) and \(B = 1\), the output \(X = 0\).


Hence, the correct truth table corresponds to Table 2. Quick Tip: To construct a truth table, process the inputs through each gate step-by-step while following the logic gate rules.

For the smooth inclined plane: \[ a_1 = g \sin\theta = \frac{g}{\sqrt{2}} \]
For the rough inclined plane: \[ a_2 = g \sin\theta - \mu g \cos\theta = \frac{g}{\sqrt{2}} - \mu \frac{g}{\sqrt{2}} \]
Using \(t_2 = n t_1\) and the equation of motion: \[ a_1 t_1^2 = a_2 t_2^2 \] \[ \frac{g}{\sqrt{2}} t_1^2 = \left(\frac{g}{\sqrt{2}} - \mu \frac{g}{\sqrt{2}}\right)(n t_1)^2 \]
Simplifying: \[ 1 = n^2 (1 - \mu) \quad \Rightarrow \quad \mu = 1 - \frac{1}{n^2} \] Quick Tip: Use the equations of motion with modified acceleration for rough and smooth inclined planes to calculate friction coefficients.

The velocity of the body after 20 s of applied force is: \[ S = v t \quad \Rightarrow \quad 50 = v \times 10 \quad \Rightarrow \quad v = 5 \, m/s \]
Using \(v = u + at\), where \(u = 0\): \[ v = a \times 20 \quad \Rightarrow \quad 5 = a \times 20 \quad \Rightarrow \quad a = \frac{1}{4} \, m/s^2 \]
The force is: \[ F = ma = 20 \times \frac{1}{4} = 5 \, N \] Quick Tip: For problems with changing forces, relate acceleration to the change in velocity over the time of applied force.

Using Bernoulli’s principle: \[ P_2 A - P_1 A = 5.4 \times 10^5 \times g \quad \Rightarrow \quad P_2 - P_1 = \frac{5.4 \times 10^5 \times 10}{500} = 10.8 \times 10^3 \, N/m^2 \]
The pressure difference is related to the velocity difference: \[ P_2 - P_1 = \frac{1}{2} \rho (V_1^2 - V_2^2) = \rho (V_1 - V_2)(V_1 + V_2) \]
Substituting: \[ 10.8 \times 10^3 = 1.2 \times (V_1 - V_2) \times 600 \] \[ V_1 - V_2 = 30 \, m/s \quad \Rightarrow \quad \frac{V_1 - V_2}{V_1} \times 100 = \frac{30}{300} \times 100 = 10% \] Quick Tip: Apply Bernoulli's principle and continuity equations for airfoil problems to relate pressure and velocity differences.

- (B) Work done by gravity is negative because the gravitational force opposes the upward motion.

- (E) Work done by air resistance is always negative because it opposes motion.

- Other statements are incorrect because:

- (A) Work done by a man is positive as the applied force is in the direction of displacement.

- (C) Work done by friction is negative, not positive.

- (D) Work done is zero only if there is no displacement, not due to roughness. Quick Tip: For work-related problems, analyze the direction of force relative to displacement.

The object is moving in a circular path with constant speed. At \(x = +2 \, m\), the velocity is \(-4\hat{j} \, m/s\), indicating a downward motion.

At \(x = -2 \, m\), the velocity will be upward, i.e., \(+4\hat{j} \, m/s\).

The acceleration is centripetal and always directed toward the center of the circle. At \(x = -2 \, m\), the acceleration will point along the positive x-axis: \[ a = 8\hat{i} \, m/s^2 \] Quick Tip: In uniform circular motion, velocity is tangential, and acceleration is centripetal, always pointing toward the center of the circle.

Work done by the electric field is given by: \[ W = q E \cdot d \]
where:

\(q = 2 \times 10^{-2} \, C\),
\(E = 30 \, N/C\),
\(d\) is the displacement along the direction of the electric field.


The displacement along the x-axis is: \[ \Delta x = 1 \, m. \]

Substitute the values: \[ W = (2 \times 10^{-2}) \cdot 30 \cdot 1 = 0.6 \, J = 600 \, mJ. \]

Since the charge is moved opposite to the direction of the field, the work done is negative: \[ W = -600 \, mJ. \] Quick Tip: The work done by an electric field is negative if the charge moves opposite to the direction of the field.

The modulation index (\(\mu\)) is given by: \[ \mu = \frac{A_{max} - A_{min}}{A_{max} + A_{min}}. \]

Substitute the values: \[ \mu = \frac{14 - 6}{14 + 6} = \frac{8}{20}. \]

Simplify: \[ \mu = 0.4. \] Quick Tip: The modulation index determines the extent of modulation in an AM wave. It is typically less than 1 for under-modulated waves.

The magnetic field at the center of a circular coil is given by: \[ B = \frac{\mu_0 I N}{2R}, \]
where \(N\) is the number of turns and \(R\) is the radius of the coil.

Step 1: Magnetic Field with 4 Turns

Initially, \(N = 4\) and the magnetic field is: \[ B = 32 \, T. \]

Step 2: Magnetic Field with 1 Turn

When the coil is rewound into a single turn (\(N' = 1\)), the radius increases proportionally (\(R' = 4R\)). The new magnetic field becomes: \[ B' = \frac{\mu_0 I (1)}{2 (4R)} = \frac{B}{4}. \]

Step 3: Calculate New Magnetic Field

Substitute \(B = 32 \, T\): \[ B' = \frac{32}{4} = 8 \, T. \] Quick Tip: For the same wire length, the radius of the coil is inversely proportional to the number of turns, which directly affects the magnetic field.

The sensitivity of the potentiometer is defined as the smallest potential difference that can be measured. This sensitivity is:

Directly proportional to the length of the potentiometer wire (\(A\)): Longer wires provide higher sensitivity.
Inversely proportional to the potential gradient (\(C\)): Lower gradients increase sensitivity. Quick Tip: To increase the sensitivity of a potentiometer, increase the wire length and decrease the potential gradient.

The resolving power (\(P\)) of a microscope is given by: \[ P = \frac{2 \mu \sin \theta}{1.22 \lambda}, \]
where:

\(\mu\) is the refractive index of the medium,
\(\lambda\) is the wavelength of light.


To improve resolving power:

Increasing \(\mu\) (refractive index) improves \(P\),
Decreasing \(\lambda\) also increases \(P\). Quick Tip: To enhance resolving power, use a higher refractive index medium and shorter wavelength light.

- Statement I: True, as electromagnetic waves consist of oscillating electric and magnetic fields that do not experience deflection in external electric or magnetic fields.

- Statement II: False, because the relation is \(E_0 = c B_0\), not \(E_0 = \sqrt{\mu_0 / \epsilon_0} B_0\). Quick Tip: Electromagnetic waves are self-propagating and do not require external forces for their motion.

Step 1: Calculate the heat required to raise the temperature of ice from \(-12^\circ C\) to \(0^\circ C\):
\[ Q_1 = m S \Delta T \]
Substitute the values: \[ Q_1 = 0.600 \, kg \times 2222.3 \, J/kg/^\circ C \times 12^\circ C \] \[ Q_1 = 16000.56 \, J \]

Step 2: Find the remaining heat available after raising the ice to \(0^\circ C\):
\[ \Delta Q_1 = 184000 - 16000.56 = 167999.44 \, J \]

Step 3: Check if the remaining heat is sufficient to melt all the ice:

Heat required to completely melt \(0.600 \, kg\) of ice: \[ Q_2 = m L = 0.600 \, kg \times 336000 \, J/kg \] \[ Q_2 = 201600 \, J \]
Since \(167999.44 \, J < 201600 \, J\), only part of the ice melts.

Step 4: Calculate the mass of ice melted:
\[ Mass of melted ice = \frac{\Delta Q_1}{L} = \frac{167999.44}{336000} \approx 0.4999 \, kg \]

Step 5: Determine the ratio of ice to water:
\[ Mass of remaining ice = 0.600 - 0.4999 = 0.1001 \, kg \] \[ Ice:Water ratio = 0.1001 : 0.4999 \approx 1:5 \]

Final system: Mixture of ice and water at \(0^\circ C\) with a ratio of 1:5. Quick Tip: To solve heat exchange problems, calculate the heat required for each phase change and ensure total energy conservation.

Step 1: Impedance in Circuit (a):

Circuit (a) consists only of a resistor. The total impedance is: \[ Z_a = R \]
This results in the maximum possible rms current for a given voltage: \[ I_a = \frac{V}{R} \]

Step 2: Impedance in Circuit (b):

Circuit (b) consists of a resistor and reactive components (inductor and capacitor). The total impedance is: \[ Z_b = \sqrt{R^2 + (X_L - X_C)^2} \]
Since \(|X_L - X_C| \geq 0\), the impedance in circuit (b) is always greater than or equal to \(R\): \[ Z_b \geq Z_a \]

Step 3: Compare the rms currents:

The rms current in circuit (b) is: \[ I_b = \frac{V}{Z_b} \]
Since \(Z_b \geq Z_a\), it follows that: \[ I_b \leq I_a \]
Hence, the rms current in circuit (b) can never exceed that in circuit (a). Quick Tip: In AC circuits, the impedance determines the current. Adding reactance increases impedance, reducing current.

Step 1: Relationship between time period and radius:

The orbital time period (\(T\)) is related to the orbital radius (\(R\)) as: \[ T^2 \propto R^3 \]

Step 2: Given change in radius:

The new radius is: \[ R_2 = \frac{R_1}{4} \]

Step 3: Relate the time periods:
\[ \left(\frac{T_2}{T_1}\right)^2 = \left(\frac{R_2}{R_1}\right)^3 \] \[ \left(\frac{T_2}{24}\right)^2 = \left(\frac{1}{4}\right)^3 = \frac{1}{64} \] \[ T_2 = 24 \times \frac{1}{8} = 3 \, hours \] Quick Tip: For satellite motion, remember that \(T^2 \propto R^3\) allows you to easily relate time periods and orbital radii.

The given equation of the circle is: \[ (x - At)^2 + \left(y - \frac{t}{B}\right)^2 = a^2. \]

Step 1: Dimensional Analysis of the First Term:

The term \((x - At)\) must have the same dimensions as \(x\), which is \([L]\) (length): \[ [At] = [L]. \]
Since \(t\) has dimensions of \([T^{-1}]\), the dimensions of \(A\) are: \[ [A] = \frac{[L]}{[T^{-1}]} = [L] \cdot [T] = [LT]. \]

Step 2: Dimensional Analysis of the Second Term:

The term \((y - \frac{t}{B})\) must have the same dimensions as \(y\), which is \([L]\): \[ \frac{t}{B} = [L]. \]
Substituting the dimensions of \(t\) as \([T^{-1}]\), the dimensions of \(B\) are: \[ [B] = \frac{[T^{-1}]}{[L]} = [L^{-1}] \cdot [T^{-1}] = [L^{-1} T^{-1}]. \]

Step 3: Finalize the Dimensions:

From the above analysis: \[ A = [LT], \quad B = [L^{-1} T^{-1}]. \]

Thus, the correct dimensions are \(A = [LT]\) and \(B = [L^{-1} T^{-1}]\). Quick Tip: Always ensure that the dimensions of each term in the equation match the physical quantities they represent. For terms involving multiplication or division, combine dimensions accordingly.

Step 1: Write the formula for work done:

The work done (\(W\)) in pulling the loop is related to the induced emf and the resistance of the loop: \[ W = F \cdot l = \frac{B^2 v l^2}{R} \]
where:

\(B = 40 \, T\) (magnetic field strength),
\(v = 0.05 \, m/s\) (velocity of pulling the loop),
\(l = 0.05 \, m\) (length of one side of the loop, calculated as \(\sqrt{Area} = \sqrt{25 \, cm^2}\)),
\(R = 10 \, \Omega\) (resistance of the loop).


Step 2: Substitute the values into the formula:
\[ W = \frac{40^2 \cdot 0.05 \cdot 0.05^2}{10} \] \[ W = \frac{1600 \cdot 0.05 \cdot 0.0025}{10} \] \[ W = \frac{1600 \cdot 0.000125}{10} = \frac{0.2}{10} = 0.001 \, J \]

Step 3: Convert to millijoules:
\[ W = 1.0 \times 10^{-3} \, J \] Quick Tip: For induced emf and work problems, always ensure the velocity of motion and loop dimensions are consistent with the magnetic field direction.

Using the meter bridge principle: \[ \frac{R_1 + R_2}{10} = \frac{60}{40} \implies R_1 + R_2 = 15 \, \Omega. \]
For the parallel combination: \[ \frac{R_1 R_2}{R_1 + R_2} = 3 \implies R_1 R_2 = 30 \, \Omega^2. \] Quick Tip: In meter bridge problems, use the ratio of the lengths to find equivalent resistance in the left and right gaps.

The angular momentum about the starting point is: \[ L = m \cdot v_x \cdot h, \]
where:

\(m = 0.1 \, kg\),
\(v_x = v \cos 45^\circ = 20 \cdot \frac{\sqrt{2}}{2} = 10\sqrt{2} \, ms^{-1}\),
\(h = vertical displacement at t = 2 \, s\).

The vertical displacement is: \[ h = v_y t - \frac{1}{2} g t^2 = 20 \cdot \frac{\sqrt{2}}{2} \cdot 2 - \frac{1}{2} \cdot 10 \cdot 2^2 = 20\sqrt{2} - 20 = 20(\sqrt{2} - 1) \, m. \]

Substitute values into \(L\): \[ L = (0.1) \cdot (10\sqrt{2}) \cdot [20(\sqrt{2} - 1)] = 20\sqrt{2} \cdot (\sqrt{2} - 1) = \sqrt{800} \, kgm^2/s. \] Quick Tip: Angular momentum is calculated about the point of projection using \(L = m \cdot v_x \cdot h\), where \(h\) is the height at the given time.

The refractive index is: \[ \mu = \frac{Real thickness}{Apparent thickness} = \frac{5.25}{5.00}. \]

For uncertainty in \(\mu\): \[ \frac{\Delta \mu}{\mu} = \frac{\Delta h}{h} + \frac{\Delta h'}{h'}, \]
where \(\Delta h = 0.01 \, mm\) and \(\Delta h' = 0.01 \, mm\): \[ \Delta \mu = \mu \left(\frac{\Delta h}{h} + \frac{\Delta h'}{h'}\right). \]

Substitute values: \[ \Delta \mu = \frac{5.25}{5.00} \cdot \left(\frac{0.01}{5.25} + \frac{0.01}{5.00}\right) = 41 \cdot 10^{-3}. \] Quick Tip: Uncertainty in measurements is propagated through addition of relative uncertainties when dividing two quantities.

The electric field is related to potential as: \[ E = -\frac{dV}{dr} = -\frac{d}{dr}(2ar^2 + b) = -4ar. \]

The charge density is: \[ \rho = \varepsilon_0 \cdot \nabla \cdot \mathbf{E} = \varepsilon_0 \cdot \frac{1}{r^2} \frac{d}{dr}(r^2 E). \]

Substitute \(E = -4ar\): \[ \rho = \varepsilon_0 \cdot \frac{1}{r^2} \frac{d}{dr}(-4ar^3) = \varepsilon_0 \cdot (-12a). \]

Thus: \[ \lambda = 12. \] Quick Tip: For spherically symmetric charge distributions, use Gauss's law and the relation \(E = -dV/dr\).

Step 1: Relating Tangential and Centripetal Acceleration

Given: \[ v_0 = 54 \, km/hr = 15 \, m/s, \quad R = 600 \, m. \]
The tangential acceleration is \(a_t = \frac{dv}{dt}\), and the centripetal acceleration is \(a_c = \frac{v^2}{R}\). The problem states that: \[ a_t = a_c \implies \frac{dv}{dt} = \frac{v^2}{R}. \]

Step 2: Differential Equation for Velocity

Rearrange the equation: \[ \frac{dv}{v^2} = \frac{dt}{R}. \]
Integrate both sides: \[ \int_{v_0}^{v} \frac{dv}{v^2} = \int_{0}^{t} \frac{dt}{R}. \]
The integration gives: \[ -\frac{1}{v} \Big|_{v_0}^v = \frac{t}{R}. \]
Simplify: \[ \frac{1}{v} - \frac{1}{v_0} = \frac{t}{R}. \]

Step 3: Solve for Velocity as a Function of Time

Rearrange to find \(v(t)\): \[ \frac{1}{v} = \frac{1}{v_0} + \frac{t}{R}. \]
Thus: \[ v(t) = \frac{1}{\frac{1}{v_0} + \frac{t}{R}}. \]

Step 4: Angular Displacement for Quarter Revolution

The angular displacement \(\theta\) for circular motion is related to velocity by: \[ d\theta = \frac{v}{R} dt. \]
Substitute \(v(t)\): \[ d\theta = \frac{1}{R \left( \frac{1}{v_0} + \frac{t}{R} \right)} dt. \]

Integrate from \(\theta = 0\) to \(\pi/2\) for the quarter revolution: \[ \int_{0}^{\pi/2} d\theta = \int_{0}^{t} \frac{1}{R \left( \frac{1}{v_0} + \frac{t}{R} \right)} dt. \]

Solve the integral: \[ \theta = -\ln \left( 1 - \frac{t R}{v_0} \right). \]

For \(\theta = \pi/2\), substitute and solve: \[ t = 40 (1 - e^{-\pi/2}) \, s. \]

Thus, the time for the first quarter revolution is \(t = 40 \, s\). Quick Tip: For problems involving circular motion with equal tangential and centripetal accelerations, use the relationship \(\frac{dv}{v^2} = \frac{dt}{R}\) and integrate to find velocity or time.

For maximum current, the circuit must be in resonance, i.e., the inductive reactance equals the capacitive reactance: \[ \frac{1}{2 \pi f C} = 2 \pi f L. \]

Rearrange to find \(C\): \[ C = \frac{1}{4 \pi^2 f^2 L}. \]

Substitute values: \[ f = 7 \, kHz = 7 \times 10^3 \, Hz, \quad L = 2 \times 10^{-6} \, H. \]
\[ C = \frac{1}{4 \pi^2 (7 \times 10^3)^2 (2 \times 10^{-6})}. \]

Simplify: \[ C = \frac{1}{4 \times (3.14)^2 \times 49 \times 10^6 \times 2 \times 10^{-6}} = \frac{1}{3872}. \]

Thus, \(x = 3872\). Quick Tip: For resonance in an LCR circuit, equate the inductive reactance and capacitive reactance to find the resonant capacitance.

Using the equation for viscous force: \[ F = \eta A \frac{\Delta v}{\Delta h}. \]

Substitute: \[ 0.1 = (5 \times 10^{-3}) (0.2) \frac{v}{0.25 \times 10^{-3}}. \]

Solve for \(v\): \[ v = \frac{0.1 \cdot 0.25 \times 10^{-3}}{5 \times 10^{-3} \cdot 0.2}. \]

Simplify: \[ v = 25 \times 10^{-3} \, m/s. \]

Thus, the speed is \(25 \times 10^{-3} \, m/s\). Quick Tip: For objects moving in viscous fluids, use the formula \(F = \eta A (\Delta v / \Delta h)\) to relate force, viscosity, and velocity.

The force equation for SHM is: \[ F = -kx \quad and \quad a = -\omega^2 x. \]

Equate: \[ \omega^2 = \frac{k}{m}. \]

Given \(F = -25x\), so \(k = 25 \, N/m\), and \(m = 0.25 \, kg\): \[ \omega^2 = \frac{25}{0.25} = 100 \implies \omega = 10 \, rad/s. \]

The maximum speed is: \[ v_{max} = A \omega. \]

Substitute \(v_{max} = 4 \, m/s\) and \(\omega = 10 \, rad/s\): \[ A = \frac{v_{max}}{\omega} = \frac{4}{10} = 0.4 \, m = 40 \, cm. \] Quick Tip: The amplitude in SHM can be calculated using the maximum speed formula \(v_{max} = A \omega\).

The condition for reflected and refracted rays to be perpendicular is known as Brewster's law. The angle of incidence satisfies: \[ \tan \theta = \sqrt{\frac{\mu_2}{\mu_1}}, \]
where \(\mu_1 = \sqrt{\varepsilon_1}\) and \(\mu_2 = \sqrt{\varepsilon_2}\). Given \(\varepsilon_1 = 2.8\) and \(\varepsilon_2 = 6.8\): \[ \tan \theta = \sqrt{\frac{6.8}{2.8}}. \]

Simplify: \[ \tan \theta = \sqrt{2.43} \approx 1.56. \]

Take the arctangent: \[ \theta = \tan^{-1}(1.56) \approx 7^\circ. \] Quick Tip: At the Brewster angle, reflected and refracted rays are perpendicular, and \(\tan \theta = \sqrt{\mu_2 / \mu_1}\) applies for dielectric media.

Let \(\varepsilon\) be the emf of the cell and \(r\) be its internal resistance. Using the potentiometer principle, we write: \[ \frac{\varepsilon}{r + 5} = 200k \quad and \quad \frac{\varepsilon}{r + 15} = 300k, \]
where \(k\) is the potential gradient.

Dividing these two equations: \[ \frac{\varepsilon / (r + 5)}{\varepsilon / (r + 15)} = \frac{200}{300}. \]

Simplify: \[ \frac{r + 15}{r + 5} = \frac{300}{200} = \frac{3}{2}. \]

Cross-multiply: \[ 2(r + 15) = 3(r + 5). \]

Expand and simplify: \[ 2r + 30 = 3r + 15 \implies r = 15 \, \Omega - 30 = 5 \, \Omega. \] Quick Tip: In potentiometer problems, use the ratio of null lengths to relate the emf and resistances in the circuit.