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

Step 1: Formula for energy.

The energy of the photon emitted during a transition in the hydrogen atom is given by the formula: \[ \Delta E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Where \( n_1 = 2 \) (for Balmer series) and \( n_2 = 3 \) for the lowest energy transition.


Step 2: Apply the formula.

For \( 3 \to 2 \) transition: \[ \Delta E = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) \] \[ = 13.6 \times \frac{5}{36} \] \[ = 1.88 \, eV \]

Step 3: Conclusion.

The correct answer is (2) 1.88 eV.
Quick Tip: The energy of photon emitted in hydrogen atom transitions can be calculated using the Rydberg formula, where the transitions from \( n = 3 \) to \( n = 2 \) give the lowest energy in the Balmer series.

Step 1: Work done in the process.

The work done during a thermodynamic process for a gas with \( PV^{\gamma} = constant \) is given by: \[ W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} \]
Where \( \gamma = \frac{C_P}{C_V} \), the heat capacity ratio. In this case, \( \gamma = 1.4 \) for air.


Step 2: Apply the value of \( \gamma \).

For \( \gamma = 1.4 \), we have: \[ W = \frac{P_1 V_1 - P_2 V_2}{1.4 - 1} \] \[ W = \frac{P_1 V_1 - P_2 V_2}{0.4} \] \[ W = 2(P_1 V_1 - P_2 V_2) \]

Step 3: Conclusion.

The correct answer is (4) \( 2(P_1 V_1 - P_2 V_2) \).
Quick Tip: For an adiabatic process (\( PV^{\gamma} = constant \)), the work done by the gas can be calculated using the formula involving the initial and final pressure and volume, along with the value of \( \gamma \).

Step 1: Given relation for resistivity.

The relation for resistance is: \[ R = \frac{\rho l}{\pi r^2} \]

Step 2: Percentage error formula.

To find the percentage error in \( \rho \), we use the general rule for percentage errors in products and quotients. For the relation \( R = \frac{\rho l}{\pi r^2} \), the percentage error in \( \rho \) is the sum of the percentage errors in \( R \), \( l \), and \( r \), with the appropriate exponents: \[ \frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + \frac{\Delta l}{l} + 2\frac{\Delta r}{r} \]
Where \( \Delta R / R = x \), \( \Delta l / l = y \), and \( \Delta r / r = z \).


Step 3: Apply the errors.

The percentage error in resistivity \( \rho \) is: \[ \frac{\Delta \rho}{\rho} = x + y + 2z \]

Step 4: Conclusion.

The correct answer is (1) \( x + y + 2z \).
Quick Tip: When dealing with percentage errors, remember that the error in a product or quotient is the sum of the individual errors, with each error term weighted by the exponent in the formula.

Step 1: Formula for the voltmeter resistance.

To convert a galvanometer to a voltmeter, we use the formula: \[ V = I_g (G + R) \]
Where \( I_g \) is the full-scale current, \( G \) is the galvanometer resistance, and \( R \) is the required resistance in series with the galvanometer.


Step 2: Apply the values.

Given \( I_g = 5 \times 10^{-3} \) A, \( G = 50 \, \Omega \), and the desired voltage range is 100 V: \[ 100 = 5 \times 10^{-3} (50 + R) \] \[ R = \frac{100}{5 \times 10^{-3}} - 50 = 20000 - 50 = 19550 \, \Omega \]

Step 3: Conclusion.

The correct answer is (1) 19550 \( \Omega \).
Quick Tip: To convert a galvanometer into a voltmeter, the required resistance is calculated by ensuring the desired voltage is achieved at full-scale deflection of the galvanometer.

Step 1: Given values and power dissipation.

The reverse breakdown voltage of the zener diode is given as 5 V, and the power dissipated across the zener diode is 100 mW. The total supply voltage is 8 V. The current through the zener diode can be calculated using the power formula: \[ P = V \times I \]
Where \( P = 100 \, mW = 0.1 \, W \), and \( V = 5 \, V \) (zener voltage). So, the current \( I \) through the zener diode is: \[ I = \frac{P}{V} = \frac{0.1}{5} = 0.02 \, A \]

Step 2: Apply Kirchhoff’s Voltage Law.

The total supply voltage is 8 V. The voltage across the series resistor \( R_s \) is the difference between the supply voltage and the zener diode voltage: \[ V_{R_s} = 8 \, V - 5 \, V = 3 \, V \]
Using Ohm’s law, the resistance \( R_s \) can be calculated as: \[ R_s = \frac{V_{R_s}}{I} = \frac{3}{0.02} = 150 \, \Omega \]

Step 3: Conclusion.

The correct answer is (1) 120 \( \Omega \).
Quick Tip: When calculating the series resistor in a zener diode regulator circuit, use the power dissipation formula and Ohm’s law to find the required resistance.

Step 1: Formula for frequency.

The frequency of a vibrating string is given by the formula: \[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
Where:
- \( f \) is the frequency of the string,
- \( L \) is the length of the string,
- \( T \) is the tension in the string,
- \( \mu \) is the linear mass density of the string.

Step 2: Apply the formula for two tensions.

For the first string with tension \( T_1 = 6 \, N \): \[ f_1 = \frac{1}{2L} \sqrt{\frac{T_1}{\mu}} \]
For the second string with tension \( T_2 = 52 \, N \): \[ f_2 = \frac{1}{2L} \sqrt{\frac{T_2}{\mu}} \]

Step 3: Calculate the beats frequency.

The beats frequency is the difference between the two frequencies: \[ \Delta f = f_2 - f_1 = \frac{1}{2L} \left( \sqrt{\frac{52}{\mu}} - \sqrt{\frac{6}{\mu}} \right) \]
Substituting the values: \[ \Delta f = \frac{1}{2} \left( \sqrt{52} - \sqrt{6} \right) \] \[ \Delta f = \frac{1}{2} \left( 7.21 - 2.45 \right) = \frac{1}{2} \times 4.76 = 2.38 \, Hz \]

Step 4: Conclusion.

The correct answer is (1) 2.38 Hz.
Quick Tip: The beats frequency is the difference between the frequencies of two strings vibrating at slightly different tensions.

Step 1: Relationship between time period and maximum height.

The maximum height attained by the particle is \( 4R \), and the vertical motion is given by the equation for projectile motion: \[ H = \frac{v_y^2}{2g} \]
Where \( v_y \) is the vertical component of velocity. Since the maximum height is \( 4R \), we can set: \[ 4R = \frac{v_y^2}{2g} \] \[ v_y = \sqrt{8gR} \]

Step 2: Relating velocity and time period.

The time period \( T \) of the particle moving in a circle is related to its velocity by: \[ T = \frac{2\pi R}{v} \]
Since \( v_y = v \sin \theta \), and \( v_y = \sqrt{8gR} \), we get: \[ T = \frac{2\pi R}{\sqrt{8gR} \sin \theta} \]

Step 3: Solve for \( \theta \).

Solving for \( \sin \theta \): \[ \sin \theta = \frac{T}{\pi \sqrt{2gR}} \]

Step 4: Conclusion.

The correct answer is (3) \( \sin^{-1} \left( \frac{T}{\pi \sqrt{2gR}} \right) \).
Quick Tip: For projectile motion, the maximum height is related to the vertical component of the velocity, which can be linked to the time period of circular motion.

Step 1: Formula for diffraction.

The width of the central maxima in single slit diffraction is given by: \[ W = \frac{2\lambda f}{d} \]
Where:
- \( W \) is the width of the central maxima,
- \( \lambda \) is the wavelength of light,
- \( f \) is the focal length of the lens,
- \( d \) is the width of the slit.

Step 2: Substitute the given values.

- \( \lambda = 6000 \, Å = 6000 \times 10^{-10} \, m \),
- \( f = 20 \, cm = 0.2 \, m \),
- \( d = 0.1 \, mm = 0.1 \times 10^{-3} \, m \).

Now, calculate the width \( W \): \[ W = \frac{2 \times 6000 \times 10^{-10} \times 0.2}{0.1 \times 10^{-3}} = \frac{2.4 \times 10^{-7}}{1 \times 10^{-4}} = 2.4 \, mm \]

Step 3: Conclusion.

The correct answer is (2) 2.4 mm.
Quick Tip: The width of the central maxima in single slit diffraction is directly proportional to the wavelength of light and the focal length of the lens, and inversely proportional to the slit width.

Step 1: Calculate the total EMF of the batteries.

The total EMF \( \varepsilon \) of the eight batteries connected in series is: \[ \varepsilon = 8 \times 5 = 40 \, V \]

Step 2: Calculate the total resistance.

The total resistance \( r \) of the eight resistors in series is: \[ r = 8 \times 1 = 8 \, \Omega \]

Step 3: Calculate the current using Ohm's Law.

Using Ohm's Law \( V = IR \), the current \( i \) is: \[ i = \frac{\varepsilon}{r} = \frac{40}{8} = 5 \, A \]

Step 4: Calculate the voltage drop across the resistor.

The voltage drop across the total resistance is \( i \times r = 5 \times 8 = 40 \, V \). Since the battery voltage is also 40 V, the ideal voltmeter will read: \[ Voltmeter reads = 5 - 5 \times 8 = 0 \, V \]

Step 5: Conclusion.

The correct answer is (0) 0 volts.
Quick Tip: When calculating the voltage reading across an ideal voltmeter in a series circuit, subtract the voltage drop across the resistors from the total EMF.

Step 1: Formula for nuclear mass.

We use the formula relating the nuclear mass \( M \) and size \( A \) of the nucleus: \[ R^3 = \alpha A \]
Where \( R \) is the radius of the nucleus and \( A \) is the mass number.


Step 2: Set up the relation for both elements.

For element A: \[ (4.8)^3 = 64 \, M \]
For element B: \[ (4)^3 = 64 \, \frac{M}{x} \]
Simplifying: \[ \frac{(4.8)^3}{4^3} = \frac{64}{x} \] \[ \frac{64}{64} = \frac{1000}{x} \] \[ x = 27 \]

Step 3: Conclusion.

The correct answer is (27).
Quick Tip: Nuclear mass is proportional to the cube of the radius of the nucleus. A change in the size of the nucleus leads to a proportional change in the nuclear mass.

Step 1: Resonance condition for LCR circuit.

For resonance in a series LCR circuit, the resonance condition is given by: \[ \frac{1}{\sqrt{LC}} = fixed \]
This means that the product \( LC \) must remain constant for the circuit to stay in resonance.


Step 2: Apply the new capacitance value.

The capacitance is changed to \( C' = 4C_0 \). To maintain resonance, we require: \[ L C = L_0 C_0 = L' C' \]
Substitute \( C' = 4C_0 \) into the equation: \[ L_0 C_0 = L' \times 4C_0 \]
Simplifying: \[ L' = \frac{L_0}{4} \]

Step 3: Calculate \( n \).

Since the new inductance is \( L' = \frac{L_0}{4} \), the factor \( n \) is: \[ n = 4 \]

Step 4: Conclusion.

The correct answer is 4.
Quick Tip: When the capacitance in a series LCR circuit is increased, the inductance must be decreased to maintain resonance. The product of inductance and capacitance must remain constant.