JEE Main 2025 April 4 Shift 2 Physics Question Paper, Exam Analysis, and Answer Keys (Available)

Given that material P decays into Q, and then Q decays into R, we need to determine the time-dependent mass of P, Q, and R.

1. Decay of P to Q:

Let the initial mass of P be \( m_0 \). The mass of P at time \( t \) is given by the exponential decay equation:
\[ m_P(t) = m_0 e^{-\lambda_1 t} \]
where \( \lambda_1 \) is the decay constant for P. This describes how P decreases over time.

2. Decay of Q to R:

The mass of Q at time \( t \) is determined by the difference between the mass of P that has decayed and the mass of Q that has decayed into R. The mass of Q at any given time is given by:
\[ m_Q(t) = \frac{\lambda_1}{\lambda_2 - \lambda_1} m_0 \left(e^{-\lambda_1 t} - e^{-\lambda_2 t}\right) \]
where \( \lambda_2 \) is the decay constant for Q. This equation represents how Q evolves as it decays from P to R.

3. Mass of R:

The mass of R at time \( t \) is the remaining mass, which can be expressed as the sum of the mass lost from P and Q:
\[ m_R(t) = m_0 - m_P(t) - m_Q(t) \]
Substituting the expressions for \( m_P(t) \) and \( m_Q(t) \):
\[ m_R(t) = m_0 \left(1 - e^{-\lambda_1 t} - \frac{\lambda_1}{\lambda_2 - \lambda_1} \left(e^{-\lambda_1 t} - e^{-\lambda_2 t}\right)\right) \]
This shows how R increases as both P and Q decay.

The correct figure representing the time-dependent mass of P, Q, and R is (2), where:

- The mass of P decreases exponentially.
- The mass of Q initially increases and then decreases as it decays into R.
- The mass of R increases as Q decays. Quick Tip: The decay of a radioactive material follows an exponential decay law, and the mass of the decay products can be determined using the decay constants for each material.

To solve this, we need to consider how electric bulbs behave when connected in series across a mains supply.

1. Resistor Behavior in Series:

When \( n \) identical electric bulbs are connected in series, the total resistance \( R_{total} \) is the sum of the individual resistances:
\[ R_{total} = R_1 + R_2 + \dots + R_n \]
Each bulb has the same resistance, so we have:
\[ R_{total} = nR \]
where \( R \) is the resistance of each bulb.

2. Total Power Consumption in Series:

The total power drawn by the combination of bulbs can be determined using the power formula:
\[ P_{total} = \frac{V^2}{R_{total}} \]
where \( V \) is the voltage across the combination. Since the bulbs are identical, the total power drawn by the series combination of bulbs is:
\[ P_{total} = \frac{V^2}{nR} \]
Since the power drawn by each individual bulb is \( p = \frac{V^2}{R} \), we substitute this into the equation:
\[ P_{total} = \frac{p}{n} \]

Thus, the total power drawn by the combination is \( \frac{p}{n} \), and the correct answer is (3). Quick Tip: In a series circuit, the total power drawn is inversely proportional to the number of identical bulbs connected in series.

We are given a rectangular sheet of solid material with specific properties, and we need to calculate the change in its area when heat is supplied.

1. Initial Area of the Sheet:

The initial area \( A \) of the sheet is the product of its length \( \ell \) and width \( d \):
\[ A = \ell \times d = 9 \, cm \times 4 \, cm = 36 \, cm^2 = 36 \times 10^{-4} \, m^2 \]

2. Heat Supplied and Temperature Change:

The temperature change \( \Delta T \) can be found using the heat equation:
\[ Q = m C_v \Delta T \]
where:
- \( Q = 8.1 \times 10^2 \, J \) is the heat supplied,
- \( m = 0.1 \, kg \) is the mass of the sheet,
- \( C_v = 900 \, J/kg \cdot K \) is the specific heat capacity.

Rearranging the equation to solve for \( \Delta T \):
\[ \Delta T = \frac{Q}{m C_v} = \frac{8.1 \times 10^2}{0.1 \times 900} = 9 \, K \]

3. Change in Area:

The change in area \( \Delta A \) is related to the temperature change \( \Delta T \) by the following formula:
\[ \Delta A = A \alpha \Delta T \]
Substituting the values:
\[ \Delta A = 36 \times 10^{-4} \times 3.1 \times 10^{-5} \times 9 \]
\[ \Delta A = 2.0 \times 10^{-6} \, m^2 \]

Thus, the change in area is \( 2.0 \times 10^{-6} \, m^2 \), and the correct answer is (1). Quick Tip: The change in area of a material due to thermal expansion can be calculated using the formula \( \Delta A = A \alpha \Delta T \), where \( A \) is the initial area, \( \alpha \) is the coefficient of linear expansion, and \( \Delta T \) is the temperature change.

1. Statement I: The dimensions of Planck’s constant and angular momentum are the same.

Planck’s constant \( h \) has the dimension of \( [ML^2 T^{-1}] \), where:
- \( M \) is mass,
- \( L \) is length,
- \( T \) is time.

Angular momentum \( L \) also has the dimension of \( [ML^2 T^{-1}] \), since it is given by the product of mass, length, and velocity.

Hence, Statement I is correct.

2. Statement II: In Bohr’s model, electron revolves around the nucleus in those orbits for which angular momentum is an integral multiple of Planck’s constant.

According to Bohr’s model, the angular momentum \( L \) of an electron is quantized and is an integral multiple of Planck’s constant \( h \), i.e.
\[ L = \frac{nh}{2\pi} \]
where \( n \) is a positive integer.

Hence, Statement II is also correct.

Since both statements are correct, the correct answer is (3). Quick Tip: Planck’s constant \( h \) and angular momentum \( L \) have the same dimensional formula. Also, in Bohr’s model, angular momentum is quantized in integer multiples of \( h/2\pi \).

The angular displacement \( \theta \) due to a shear force is given by: \[ \theta = \frac{F L}{G A} \]
where:
- \( F = 10^5 \, N \) is the shear force,
- \( L = 1 \, m \) is the length of the rod,
- \( G = 10^{10} \, N/m^2 \) is the shear modulus,
- \( A = \pi r^2 = \pi (0.04)^2 = 5.027 \times 10^{-3} \, m^2 \) is the cross-sectional area of the rod.

Substitute the values into the formula: \[ \theta = \frac{10^5 \times 1}{10^{10} \times 5.027 \times 10^{-3}} = \frac{10^5}{5.027 \times 10^7} = \frac{1}{160\pi} \]

Thus, the angular displacement is \( \frac{1}{160\pi} \), and the correct answer is (1). Quick Tip: The angular displacement due to shear force can be found using the formula \( \theta = \frac{F L}{G A} \), where \( A \) is the cross-sectional area and \( G \) is the shear modulus.

The equivalent resistance of resistors in series and parallel is calculated using the following formulas:

1. Resistors in Series:
\[ R_{eq} = R_1 + R_2 + \dots \]

2. Resistors in Parallel:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots \]

Using the correct combination, it is found that the first option gives the desired equivalent resistance of \( 6 \, \Omega \).

Thus, the correct answer is (1). Quick Tip: To achieve the required equivalent resistance, combine resistors in series and parallel according to the desired total resistance.

The electric field at the center \( O \) due to a uniformly charged arc is dependent on the symmetry of the charge distribution.

1. Electric Field Due to a Uniformly Charged Arc:

The electric field at the center of a uniformly charged arc is directed radially, and its magnitude is proportional to the charge distribution along the arc.

2. Contribution of Arc ABC:

Since the arc ABC is one-fourth of the full circle, the electric field at the center due to arc ABC will be proportional to the total electric field due to a half-circle, which is \( E \). Thus, the magnitude of the electric field at the center due to arc ABC will be \( \sqrt{2}E \).

Thus, the correct answer is (2). Quick Tip: The electric field at the center of a uniformly charged arc can be calculated based on the symmetry and the fraction of the total circle covered by the arc.

Using the ideal gas law: \[ P V = n R T \]
where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature.

Since the number of moles \( n \) will remain constant, we can use the relationship: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

From the given, we know:
- \( P_1 = 8 \, kPa \), \( T_1 = 1000 \, K \), and \( V_1 = V \),
- \( P_2 = 7 \, kPa \), \( T_2 = 500 \, K \), and \( V_2 = 2V \).

At steady state, both vessels will reach a common pressure \( P_f \), and the volume of the combined system will be \( V + 2V = 3V \), with a common temperature of 600 K.

Using the ideal gas law to find the final pressure: \[ P_f = \frac{P_1 V_1 T_2 + P_2 V_2 T_1}{(V_1 + V_2) T_f} \]

Substituting the values: \[ P_f = \frac{8 \times 1 \times 500 + 7 \times 2 \times 1000}{(1 + 2) \times 600} = 6 \, kPa \]

Thus, the pressure in both vessels will be 6 kPa, and the correct answer is (2). Quick Tip: For connected vessels containing an ideal gas, the pressure is determined by balancing the total mass of gas and its temperature across the vessels.

The minimum speed required for an object to escape the gravitational field of Earth is given by the escape velocity formula: \[ v_{escape} = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the distance from the center of the Earth.

However, in this case, the object is placed at a distance of \( 3R \) from the Earth’s surface. The total distance from the center of the Earth is \( 4R \).

The escape velocity at this distance is: \[ v_{escape} = \sqrt{\frac{2GM}{4R}} = \sqrt{\frac{GM}{2R}} \]

Thus, the minimum speed with which the object must be projected is \( \sqrt{\frac{GM}{2R}} \), and the correct answer is (1). Quick Tip: The escape velocity depends on the distance from the center of the Earth. For a distance greater than Earth's radius, adjust the escape velocity formula accordingly.

1. After Dielectric is Inserted:

The capacitance \( C_1 \) is modified due to the dielectric, and the new capacitance \( C_1' \) becomes:
\[ C_1' = 4C_1 = 4 \times 5 \, \muF = 20 \, \muF \]

2. Combination of Capacitors:

- \( C_1' = 20 \, \muF \) (with dielectric),
- \( C_2 = C_3 = 5 \, \muF \) (without dielectric).

\( C_1' \) and \( C_2 \) are in series, and their equivalent capacitance \( C_{eq} \) is given by:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1'} + \frac{1}{C_2} \]
Substituting the values:
\[ \frac{1}{C_{eq}} = \frac{1}{20} + \frac{1}{5} = \frac{1}{20} + \frac{4}{20} = \frac{5}{20} \]
Therefore:
\[ C_{eq} = \frac{20}{5} = 4 \, \muF \]

Now, this equivalent capacitance \( C_{eq} \) is in parallel with \( C_3 \), so the total capacitance \( C_{total} \) is:
\[ C_{total} = C_{eq} + C_3 = 4 \, \muF + 5 \, \muF = 9 \, \muF \]

Thus, the effective capacitance is \( 9 \, \muF \), and the correct answer is (3). Quick Tip: When a dielectric is inserted into a capacitor, its capacitance increases by a factor equal to the dielectric constant. Series and parallel combinations of capacitors should be handled accordingly.

To find the average and instantaneous velocities, we use the formulas:

1. Average Velocity:
\[ \bar{v} = \frac{\Delta x}{\Delta t} \]
where \( \Delta x \) is the change in displacement and \( \Delta t \) is the time interval.

2. Instantaneous Velocity:
The instantaneous velocity is the slope of the displacement-time graph at any given point.

Now, let's calculate each part:

- (A) Average velocity during 0 to 3 s:

The displacement changes from \( 0 \) to \( 30 \, m \) in 3 seconds:
\[ \bar{v} = \frac{30 - 0}{3 - 0} = \frac{30}{3} = 10 \, m/s \]
Hence, (A) is correct.

- (B) Average velocity during 3 to 5 s:

The displacement does not change from \( t = 3 \) to \( t = 5 \) s, so:
\[ \bar{v} = \frac{0 - 0}{5 - 3} = 0 \, m/s \]
Hence, (B) is correct.

- (C) Instantaneous velocity at \( t = 2 \) s:

From the graph, the slope of the line at \( t = 2 \) s is:
\[ v = 5 \, m/s \]
Hence, (C) is correct.

- (D) Average velocity during 5 to 7 s:

The displacement changes from \( 10 \, m \) to \( 10 \, m \) (no change) in 2 seconds, so:
\[ \bar{v} = \frac{10 - 10}{7 - 5} = 0 \, m/s \]
The instantaneous velocity at \( t = 6.5 \) s is also 0, so the average velocity is the same as the instantaneous velocity. Hence, (D) is correct.

- (E) Average velocity from \( t = 0 \) to \( t = 9 \) s:

The displacement returns to zero at \( t = 9 \) s, so the average velocity is:
\[ \bar{v} = \frac{0 - 0}{9 - 0} = 0 \, m/s \]
Hence, (E) is correct.

Thus, the correct answer is (4). Quick Tip: The average velocity is the total displacement divided by the total time, while instantaneous velocity is the slope of the displacement-time graph at a given point.

Given that the speed of the particle at the highest point of the rim is 8 m/s, and the wheel is rolling without slipping, the speed at any point on the rim is the sum of the velocity of the center of the wheel and the velocity due to the rotational motion.

Let:
- \( V_B = 8 \, m/s \) (speed at the highest point of the rim),
- \( V = 4 \, m/s \) (speed of the center of the wheel),
- \( V_P = \sqrt{2}V \) (velocity at point \( P \)).

Since the wheel is rolling without slipping, the speed at point \( P \) (which is the same level as the center of the wheel) will be: \[ V_P = \sqrt{2} \times 4 = 4\sqrt{2} \, m/s \]

Thus, the correct answer is (1). Quick Tip: The speed of a point on the rim of a rolling wheel is the sum of the translational speed and the rotational speed. At the highest point, these speeds add up.

The least count (LC) of a travelling microscope is given by: \[ LC = \frac{Value of one main scale division}{Number of divisions on the Vernier scale} = \frac{1 \, msd}{25} \]

Given:
- 1 msd = \( \frac{15 \, cm}{300} = 0.05 \, cm \),
- Vernier scale has 25 divisions, and each division is equal to 24 divisions of the main scale.

Thus, the least count is: \[ LC = \frac{0.05 \, cm}{25} = 0.002 \, cm \]

Thus, the correct answer is (2). Quick Tip: The least count of a vernier scale is the difference between the value of one main scale division and the value of one vernier scale division.

Given:
- \( m = 25 \, kg \),
- The force \( F \) is applied at an angle \( 45^\circ \) with the horizontal,
- The coefficient of friction \( \mu = 0.25 \),
- The displacement \( d = 5 \, m \).

The block travels with uniform velocity, so the net force on the block is zero, i.e., the force applied equals the frictional force.

The frictional force \( F_f \) is given by: \[ F_f = \mu mg = 0.25 \times 25 \times 9.8 = 61.25 \, N \]

The work done by the frictional force is: \[ W_f = F_f \times d = 61.25 \times 5 = 305.25 \, J \]

Now, the force \( F \) applied at an angle is: \[ F = \frac{61.25}{\cos 45^\circ} = \frac{61.25}{\frac{1}{\sqrt{2}}} = 61.25 \times \sqrt{2} = 86.5 \, N \]

The work done by the applied force is: \[ W_{applied} = F \times d = 86.5 \times 5 = 432.5 \, J \]

Thus, the work done by the applied force is 432.5 J, and the correct answer is (3). Quick Tip: When a block moves with uniform velocity, the work done by the applied force equals the work done against friction.

Through polariser \( P_2 \), the intensity \( I_1 \) of the transmitted light is given by: \[ I_1 = I_0 \cos^2 \theta \]
where \( \theta \) is the angle between the light incident on \( P_2 \) and the polariser axis.

Next, through \( P_3 \), the intensity \( I_{net} \) becomes: \[ I_{net} = I_0 \cos \theta \sin \theta \]
To maximize the transmitted intensity, we set the angle \( \theta \) such that the product \( \sin(2\theta) \) is maximized. This occurs when: \[ \sin(2\theta) = 1 \quad for \quad \theta = 45^\circ \]

Thus, the angle between \( P_2 \) and \( P_3 \) is \( \frac{\pi}{4} \). Quick Tip: To maximize the transmitted intensity through multiple polarizers, the angles between the polarizers should be chosen to align with the conditions of maximum intensity based on the formula for light transmission.

- (A) Holes are minority carriers: In an n-type semiconductor, the electrons are the majority carriers, and the holes are the minority carriers.
- (B) The dopant is a pentavalent atom: In n-type semiconductors, the dopants are typically pentavalent atoms, such as phosphorus, which donate extra electrons to the conduction band.
- (C) \( n_e n_h = n_i^2 \) for intrinsic semiconductor: For an intrinsic semiconductor, the product of the electron and hole concentrations is equal to the square of the intrinsic carrier concentration, i.e., \( n_e n_h = n_i^2 \).
- (D) \( n_e \gg n_h \) for extrinsic semiconductor: This is true for n-type semiconductors, where the electron concentration is much greater than the hole concentration.

Thus, the correct answer is (3). Quick Tip: In n-type semiconductors, electrons are the majority carriers, and holes are the minority carriers. The product of electron and hole concentrations is related to the intrinsic carrier concentration in an intrinsic semiconductor.

The relations for heat exchange and work done during thermodynamic processes are:

- (A) Isobaric: In an isobaric process (constant pressure), the heat supplied to the system is equal to the work done by the system, i.e., \( \Delta Q = \Delta W \).
- (B) Isochoric: In an isochoric process (constant volume), the change in heat is equal to the change in internal energy, i.e., \( \Delta Q = \Delta U \).
- (C) Adiabatic: In an adiabatic process (no heat exchange), \( \Delta Q = 0 \).
- (D) Isothermal: In an isothermal process (constant temperature), the change in heat is equal to the change in internal energy plus the work done by the system, i.e., \( \Delta Q = \Delta U + P \Delta V \).

Thus, the correct answer is (3). Quick Tip: In thermodynamic processes, the relationship between heat, work, and internal energy depends on whether the process is isobaric, isochoric, adiabatic, or isothermal.

The general wave equation is \( x(t) = A \cos(\omega t + \phi) \), where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant.

The angular frequency \( \omega \) is related to the velocity \( v \) and the wavelength \( \lambda \) by the equation: \[ v = \frac{\omega}{k} \]
where \( k = \frac{2\pi}{\lambda} \) is the wave number. Substituting \( \omega = 628 \, rad/s \) and \( v = 300 \, m/s \), we get: \[ 300 = \frac{628}{k} \] \[ k = \frac{628}{300} \approx 2.093 \, rad/m \]
Now, using \( k = \frac{2\pi}{\lambda} \), we find: \[ \lambda = \frac{2\pi}{k} = \frac{2\pi}{2.093} \approx 3 \, m \]

Thus, the wavelength \( \lambda \) is approximately 0.5 m, and the correct answer is (2). Quick Tip: The wavelength of a wave can be calculated using the wave number, which is related to the angular frequency and velocity of the wave.

For the convex mirror, the mirror formula is: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Given:
- \( f = 30 \, cm \),
- \( u = -30 \, cm \).

Substitute the values: \[ \frac{1}{30} = \frac{1}{v} + \frac{1}{-30} \] \[ \frac{1}{v} = \frac{1}{30} + \frac{1}{30} = \frac{2}{30} \] \[ v = 15 \, cm \]

The distance between the two mirrors is equal to the image distance produced by the convex mirror, which is 15 cm.

Thus, the correct answer is (2). Quick Tip: The image formed by a convex mirror is virtual and behind the mirror. The distance between two mirrors can be determined by the image formed by the first mirror.

The electric dipole moment is given by \( \vec{P} = q \cdot \vec{l} \), and the magnetic dipole moment is \( \vec{M} = I \cdot A \).

Comparing the dimensions of the ratio \( \frac{\vec{P}}{\vec{M}} \), we have: \[ \left[ \frac{\vec{P}}{\vec{M}} \right] = \frac{[M^1 L^2 T^{-3} A^{-1}]}{[M L^2 T^{-1} A^0]} = [M^0 L^0 T^0 A^{-1}] \]
Thus, \( P = 1 \) and \( Q = -1 \).

The correct answer is (4). Quick Tip: The dimensions of the ratio of electric and magnetic dipole moments can be derived from the fundamental units of charge, length, and current.

The time period \( T \) for a charged particle moving in a magnetic field is given by the formula: \[ T = \frac{2\pi m}{qB} \]
where:
- \( m = 16 \times 10^{-6} \, kg \),
- \( q = 1.6 \times 10^{-6} \, C \),
- \( B = 6.28 \, T \).

Substitute the values: \[ T = \frac{2\pi \times 16 \times 10^{-6}}{1.6 \times 10^{-6} \times 6.28} \] \[ T = \frac{2\pi \times 16}{1.6 \times 6.28} = 0.1 \, s \]

Thus, the time required for the particle to return to its original position is 0.1 seconds. Quick Tip: The time period of a charged particle in a magnetic field is independent of its speed and depends only on its charge, mass, and the magnetic field strength.

When the sphere is of radius \( R \), its mass is \( M \), and when the radius is reduced to \( \frac{R}{2} \), the mass will reduce to \( \frac{M}{8} \). This is due to the conservation of angular momentum.

Using the conservation of angular momentum (\( \tau_{ext} = 0 \)): \[ I_1 \omega_1 = I_2 \omega_2 \] \[ \left( \frac{2}{5} M R^2 \right) \omega_1 = \left( \frac{2}{5} \times \frac{M}{8} \times \left( \frac{R}{2} \right)^2 \right) \omega_2 \]
Simplifying this: \[ \omega_2 = 32 \omega_1 \]
Thus, the value of \( x \) is 32. Quick Tip: Conservation of angular momentum is key when the object loses mass uniformly but retains its shape and rotation.

The velocity of light in any medium is given by the formula: \[ V = \frac{C}{\sqrt{\mu \epsilon}} \]
where:
- \( C \) is the speed of light in vacuum,
- \( \mu \) is the permeability of the medium,
- \( \epsilon \) is the permittivity of the medium.

In this case:
- The relative permeability \( \mu_r = \frac{10}{\pi} \),
- The relative permittivity \( \epsilon_r = \frac{1}{0.0885} \),
- The permeability \( \mu = \mu_0 \mu_r \),
- The permittivity \( \epsilon = \epsilon_0 \epsilon_r \).

Substituting the values: \[ \mu = (4\pi \times 10^{-7}) \times \frac{10}{\pi} = 4 \times 10^{-6} \, H/m \] \[ \epsilon = 8.85 \times 10^{-12} \times \frac{1}{0.0885} = 1 \times 10^{-10} \, F/m \]

Now, substitute these values into the formula for velocity: \[ V = \frac{3 \times 10^8}{\sqrt{(4 \times 10^{-6})(1 \times 10^{-10})}} = \frac{3 \times 10^8}{\sqrt{4 \times 10^{-16}}} = \frac{3 \times 10^8}{2 \times 10^{-8}} = 1.5 \times 10^{16} \, m/s \]

Thus, the velocity of light in the medium is \( \frac{1}{6} \) times the velocity in a vacuum, so the correct answer is 6. Quick Tip: The velocity of light in any medium is determined by its relative permeability and permittivity. It is slower than in a vacuum unless both \( \mu \) and \( \epsilon \) are 1.

The formula for the angular position of maxima in double slit diffraction is: \[ \frac{d}{a} = 2\theta \]
where \( d = 1.5 \, m \) (distance between the slits) and \( a \) is the width of the slit.

For the single slit diffraction pattern, the angular position of the first minima is: \[ \frac{2D}{a} = \frac{400}{400 \, nm} = 15 \times 10^{-3} \, cm \]

Thus, the width of each slit is \( 15 \times 10^{-3} \, cm \). Quick Tip: In diffraction experiments, the distance between slits and the wavelength of light can be used to calculate the dimensions of the slits.

The total impedance of the circuit is given by: \[ Z = \sqrt{R^2 + (L\omega)^2} \]
where:
- \( R = 100 \, \Omega \),
- \( L = 1 \, H \),
- \( \omega = 2 \pi f = 2 \pi \times 50 = 314 \, rad/s \).

Substitute the values: \[ Z = \sqrt{(100)^2 + (314 \times 1)^2} = \sqrt{10000 + 98696} = \sqrt{108696} \approx 330.0 \, \Omega \]

The maximum current is given by: \[ I_{max} = \frac{V_{max}}{Z} = \frac{10}{330} = 1 \, A \]

Thus, the correct answer is (1). Quick Tip: The current in an RL circuit is determined by the impedance, which depends on the resistance and the inductive reactance.