Physics Test 254

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Physics Test 254

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Question 1
4 / -1

In a circuit for finding the resistance of a galvanometer by half deflection method, a 6 V battery and a high resistance of 11 kΩ are used. The figure of merit of the galvanometer is 60 μA/division. In the absence of shunt resistance, the galvanometer produces a deflection of θ = 9 divisions, when current flows in the circuit. The value of the shunt resistance that can cause the deflection of θ/2 is closest to:

A
550 Ω

B
220 Ω

C
55 Ω

D
110 Ω

Solution

For deflection of θ/2, current also reduces to I/2 with shunt resistance S.

Hence,
Question 2
4 / -1

A current of 1 A is flowing on the sides of an equilateral triangle of side 4.5 × 10^-2 m. The magnetic field at the centroid of the triangle will be:

A
8 × 10^-5 Wb/m²

B
4 × 10^-5 Wb/m²

C
Zero

D
2 × 10^-5 Wb/m²

Solution
Question 3
4 / -1

A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?

A
1.639

B
2.639

C
3.639

D
4.639

Solution

The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus, the process is adiabatic.

As we know that, the value of the ratio of specific heat for standard gas = 1.40.
Question 4
4 / -1

The figure shows a 2.0 V potentiometer used for the determination of the internal resistance of a 1.5 V cell. The balance point of the cell in the open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm of the potentiometer wire. Determine the internal resistance of the cell.

A
1.63Ω

B
1.66Ω

C
1.68Ω

D
1.72Ω

Solution

The internal resistance of the cell = r

Balance point of the cell in open circuit, I₁ = 76.3cm

An external resistance (R.) is connected to the circuit with R = 9.5Ω

New balance point of the circuit, I₂ = 64.8cm

Current flowing through the circuit = 1

Using the relation connecting resistance and emf is,
Question 5
4 / -1

What speed should a galaxy move with respect to us so that the sodium line at 589.0 nm is observed at 589.6 nm?

A
1.06 x 10⁵ m/s

B
2.06 x 10⁵m/s

C
3.06 x 10⁵ m/s

D
4.06 x 10⁵ m/s

Solution

Given: Wavelength of sodium line (λ₁) = 589nm Wavelength at which sodium line is observed (λ₂) = 589.6nm

Change in wavelengths is given by Δλ = λ₂ - λ₁

Substituting the values in above equation we get, Δλ 589.6 - 589 = 0.6nm
Question 6
4 / -1

Light from a point source in the air falls on a spherical glass surface (μ=1.5 and radius of curvature 20cm). The distance of the light source from the glass surface is 100cm. The position where the image is formed is:

A
50 cm

B
100 cm

C
125 cm

D
25 cm

Solution

As per the given criteria, Refractive index of air, μ₁ = 1, Refractive index of glass, μ₂ = 1.5, Radius of curvature, R = 20 cm, Object distance, u = -100 cm

We know that,
Question 7
4 / -1

A metal block of area 0.10 m² is connected to a 0.01 kg mass via a string that passes over a massless and frictionless 0.01 kg pulley as shown in the figure. A liquid with a film thickness of 0.3 mm is placed between the block and the table. When released the block moves to the right with a constant speed of 0.085 ms^-1. The coefficient of viscosity of the liquid is: (Take g = 10 ms^-2)

A
0.46 × 10^-3 Pa s

B
1.46 × 10^-3 Pa s

C
2.46 × 10^-3 Pa s

D
3.46 × 10^-3 Pa s

Solution

Here, m = 0.01 kg, l = 0.3 mm = 0.3 × 10^-3 m, g = 10 ms^-2 V= 0.085 ms^-1, A = 0.1 m²

The metal block moves to the right because of the tension in the string. The tension T is equal in magnitude to the weight of the suspended mass m.

Thus, the shear force F is: F = T = mg = 0.010 kg x 9.8 ms^-2= 9.8 x 10^-2 N
Question 8
4 / -1

Assume that the light of wavelength is 6000Å coming from a star. What is the time resolution of a telescope whose objective has a diameter of 100 inch?

A
1.83 × 10⁷rad

B
7.32 × 10⁷rad

C
6.00 × 10⁷rad

D
2.9 × 10^-7rad

Solution

Resolving power (R.P.) of the astronomical telescope is its ability to form separate images of two neighbouring astronomical objects like stars etc.
Question 9
4 / -1

The magnetic field in a plane electromagnetic wave is given by By = (2 x 10^-7) T Sin (0.5 x 10³x + 1.5 x 10¹¹t). This electromagnetic wave is:

A
Visible light

B
Infrared

C
Microwave

D
Radiowave

Solution

The magnetic field in a plane of the electromagnetic wave is given by, B = 2 x 10^-7 sin (0.5 × 10³x + 1.5 × 10¹¹t)

Comparing this equation with the standard wave equation: B_y = B₀ sin[Kx + ωt]

From the above equation, K = 0.5 x 10³ = propagation constant

The wavelength range of microwaves is 10^-3 m to 0.3 m. The wavelength of this wave lies between 10^-3m to 0.3 m, so the equation represents microwaves.
Question 10
4 / -1

A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz.

A
7 A

B
14 A

C
28 A

D
42 A

Solution