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Physics Questions & Solutions

Question 27. The work functions of Caesium (Cs), Potassium (K) and Sodium

(Na) are 2.14 eV, 2.30 eV and 2.75 eV respectively. If incident electromagnetic

radiation has an incident energy of 2.20 eV, which of these photosensitive

surfaces may emit photoelectrons?

(1) Cs only (2) Both Na and K (3) K only (4) Na only

Answer. (1) Cs only

Solution. The photoelectric effect occurs when the energy of the incident

photons is equal to or greater than the work function of the material. If the

incident energy is less than the work function, no photoelectrons will be emitted.

In this case, the incident energy is 2.20 eV. Comparing this to the work functions:

- The work function of caesium (Cs) is 2.14 eV. Since the incident energy is

slightly greater than the work function, Cs can emit photoelectrons. - The work

function of potassium (K) is 2.30 eV. The incident energy is slightly less than the

work function, so K may not emit photoelectrons. - The work function of sodium

(Na) is 2.75 eV. The incident energy is significantly less than the work function,

so Na may not emit photoelectrons. Therefore, the photosensitive surface that

may emit photoelectrons in this scenario is Cs only. So the answer is (1) Cs only.

Question 34. The net magnetic flux through any closed surface is

(1) Zero (2) Positive (3) Infinity (4) Negative

Answer. (1) Zero

Solution. The correct answer is (2) Zero. According to Gauss's law for

magnetism, the net magnetic flux through any closed surface is always zero. This

means that the total magnetic field passing through a closed surface is balanced

by an equal amount of magnetic field leaving the surface. In other words, the

magnetic field lines are always closed loops and do not have a net divergence.

This result is in contrast to Gauss's law for electric fields, where the net electric

flux through a closed surface can be nonzero if there are electric charges inside

the surface. Therefore, the correct choice is (2) Zero.

Question 8. A 12 V, 60 W lamp is connected to the secondary of a step-down

transformer, whose primary is connected to ac mains of 220 V. Assuming the

transformer to be ideal, what is the current in the primary winding?

(1) 0.27 A (2) 2.7 A (3) 3.7 A (4) 0.37 A

Answer. (1) 0.27 A

Solution. To find the current in the primary winding of the transformer, we can

use the power equation: Power (P) = Voltage (V) × Current (I) In this case, the

power of the lamp is given as 60 W, and the voltage across the lamp is 12 V. We

can calculate the current in the lamp using the power equation: 60 W = 12 V ×

I_lamp Solving for I_lamp: I_lamp = 60 W / 12 V I_lamp = 5 A Since we are

assuming the transformer to be ideal, the power in the primary winding is equal

to the power in the secondary winding. Therefore, the current in the primary

winding (I_primary) can be calculated using the same power equation: P_primary

= V_primary × I_primary The primary voltage is given as 220 V, and we need to

solve for I_primary: 60 W = 220 V × I_primary Solving for I_primary: I_primary =

60 W / 220 V I_primary ≈ 0.273 A Therefore, the current in the primary winding of

the transformer is approximately 0.273 A. The correct answer is (1) 0.27 A.

Question 9. A full wave rectifier circuit consists of two p-n junction diodes, a

centre-tapped transformer, capacitor and a load resistance. Which of these

components remove the ac ripple from the rectified output?

(1) A centre-tapped transformer (2) p-n junction diodes (3) Capacitor (4) Load

resistance

Answer. (3) Capacitor

Solution. The component that removes the AC ripple from the rectified output in

a full wave rectifier circuit is the capacitor. In a full wave rectifier circuit, the p-n

junction diodes are responsible for converting the AC input signal into a pulsating

DC output. However, the output still contains some ripple, which is the fluctuation

or variation in the DC voltage due to the AC component. To reduce this ripple and

obtain a smoother DC output, a capacitor is connected in parallel to the load

resistance. The capacitor acts as a filter and charges up during the peaks of the

rectified waveform and discharges during the troughs. This smoothing action

helps to even out the variations in the rectified output, reducing the ripple. The

load resistance is connected in series with the capacitor and the output, and it is

responsible for providing the desired load for the circuit. However, it does not

specifically remove the AC ripple. The centre-tapped transformer is used to

provide the necessary AC voltage input and to perform the rectification process

using the diodes. While it is an essential component of the full wave rectifier

circuit, it does not directly remove the AC ripple. Therefore, the correct choice is

(3) Capacitor.

Question 22. In a plane electromagnetic wave travelling in free space, the

electric field component oscillates sinusoidally at a frequency of 2.0 × 1010 Hz

and amplitude 48 V m–1 . Then the amplitude of oscillating magnetic field is

(Speed of light in free space = 3 × 108 m s–1 )

(1) 1.6 × 10–9 T (2) 1.6 × 10–8 T (3) 1.6 × 10–7 T (4) 1.6 × 10–6 T

Answer. (3) 1.6 × 10–7 T

Solution. In a plane electromagnetic wave propagating in free space, the

relationship between the electric field amplitude (E) and the magnetic field

amplitude (B) is given by: E = c * B where c is the speed of light in free space. In

this case, the electric field amplitude is given as 48 V/m. We can use this

information to find the magnetic field amplitude (B). B = E / c Substituting the

given values: B = (48 V/m) / (3 × 10^8 m/s) B = 1.6 × 10^(-7) T Therefore, the

amplitude of the oscillating magnetic field is 1.6 × 10^(-7) T. The correct answer

is (3) 1.6 × 10^(-7) T.

Question 21. A metal wire has mass (0.4 ± 0.002) g, radius (0.3 ± 0.001) mm

and length (5 ± 0.02) cm. The maximum possible percentage error in the

measurement of density will nearly be

(1) 1.2% (2) 1.3% (3) 1.6% (4) 1.4%

Answer. (3) 1.6%

Solution. To determine the maximum possible percentage error in the

measurement of density, we need to find the fractional error in the density

calculation. The fractional error is given by dividing the maximum error in the

result by the measured value.

Let's calculate the maximum error in the density calculation:

Mass: (0.4 ± 0.002) g

Radius: (0.3 ± 0.001) mm = (0.03 ± 0.0001) cm

Length: (5 ± 0.02) cm

Density (ρ) = Mass (m) / (π * Radius² * Length)

Maximum error in density = (Maximum error in mass + Maximum error in radius +

Maximum error in length) / (π * Radius² * Length)

Maximum error in mass = 0.002 g

Maximum error in radius = 0.001 cm

Maximum error in length = 0.02 cm

Plugging in the values:

Maximum error in density = (0.002 g + 0.001 cm + 0.02 cm) / (π * (0.03 cm)² * 5

cm)

= 0.023 cm / (π * 0.0009 cm² * 5 cm)

≈ 0.0159 cm / (π * 0.0009 cm² * 5 cm)

≈ 0.0159 / (π * 0.0009 * 5)

≈ 0.0159 / (0.0028274333882)

≈ 0.0056311615