Physics Test

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Physics Test - 17

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Question 1
1 / -0

Which of the following quantities is unitless as well as dimensionless?

A
Angle

B
Solid angle

C
Mechanical equivalent of heat

D
Refractive index

Solution

In optics, the refractive index or index of refraction of a substance or medium is a measure of the speed of light in that medium. It is expressed as the ratio of the speed of light in vacuum relative to that in the considered medium. This can be written mathematically as:
n = speed of light in a vacuum/speed of light in a medium
Since refractive index is the ratio of same quantities, so it has no unit and is a dimensionless quantity.
Question 2
1 / -0

Which of the following magnitudes is expressed most accurately?

A
1.3 x 10^-1 m

B
1.30 x 10^-2 m

C
1.3000 x 10^-4 m

D
1.3000 x 10^-6m

Solution

The result is expressed in maximum number of significant figures in the 4th option, so this option is the answer.
Question 3
1 / -0

Match the following:

A
A - 4, B - 3, C - 1, D - 2

B
A - 1, B - 4, C - 2, D - 3

C
A - 3, B - 1, C - 4, D - 2

D
A - 2, B - 4, C - 1, D - 3

Solution

Unit of magnetic field intensity is Am^-1.

Unit of magnetic flux is Wb (weber)

Unit of magnetic potential is Wbm^-1.

Unit of magnetic induction is Wnm^-2.
Question 4
1 / -0

Time period 'T' of a simple pendulum may depend on 'm', the mass of the bob, 'l', the length of the string, and 'g', the acceleration due to gravity. This means that 'T' ∝ m^a l^bg^c. What are the respective values of 'a', 'b' and 'c'?

A
1/2, 0 and 1/2

B
1/2, 0 and - 1/2

C
0, 1/2 and - 1/2

D
0, 1/2 and 1/2

Solution
Question 5
1 / -0

A
[M⁰L⁰T⁰]

B
[ML⁰T⁰]

C
[M⁰LT⁰]

D
[M⁰L⁰T]

Solution
Question 6
1 / -0

A particle has an initial velocity of 9 m/s due east and a constant acceleration of 2 m/s2 due west. The distance covered by the particle in the fifth second of its motion is

A
0 m

B
0.5 m

C
2 m

D
None of these

Solution

The displacement is indeed 0 but not the distance. the distance covered is 0.5 m.
Actually, the car is changing the direction of motion at 4.5 seconds. As such displacement between 4 and 5 seconds is 0 but still the car covers some distance.
The distance is the mod of displacement made by the car between 4 and 4.5 seconds + mod of the displacement between 4.5 and 5 seconds.
Displacement b/w 4 to 4.5 seconds = (9)(0.5) - (4.5² - 4²) = 4.5 - 4.25 = 0.25
Displacement b/w 4.5 to 4 seconds = (9)(0.5) - (5² - 4.5²) = 4.5 - 4.75 = -0.25
Displacement = 0, Distance = 0.5 m
Question 7
1 / -0

A bird flies for 4 s with a velocity of |t - 2| m/s in a straight line, where t = time in seconds. It covers a distance of

A
2 m

B
4 m

C
6 m

D
8 m

Solution
Question 8
1 / -0

A balloon starts rising from the ground with an acceleration of 1.25 m/s2. After 8 s, a stone is released from the balloon. The stone will, after release from the balloon,

A
cover a distance of 40 m

B
have a displacement of 50 m

C
reach the ground in 4 s

D
begin to move down

Solution
Question 9
1 / -0

A car accelerates from rest at a constant rate ∝ for some time. Afterwards, it decelerates at a constant rate β to come to rest. If the total time elapsed is t , the maximum velocity acquired by the car is given by

A

B

C

D

Solution
Question 10
1 / -0

From the top of a tower, a stone is thrown up and reaches the ground in time t₁ = 9 s. A second stone is thrown down with the same speed and reaches the ground in time t₂ = 4 s. A third stone is released from rest and reaches the ground in time t₃, which is equal to

A
6.5 s

B
6 s

C
65 s

D
64 s

Solution
Question 11
1 / -0

A body of mass 1 kg is projected with velocity 40 m/s at an angle of projection 30° with the horizontal. The magnitude of change in momentum from t = 0 to t = 2 sec is

A
20 Ns

B
30 Ns

C
10 Ns

D
10√3 Ns

Solution
Question 12
1 / -0

A body is projected at time t = 0 from a certain point on a planet’s surface with a certain velocity at a certain angle with the planet’s surface (assumed horizontal). The horizontal and vertical displacements x and y (in metres), respectively vary with time t (in seconds) as x = 10 √3 t and y = 10 t – t².

What is the value of acceleration due to the gravity on the planet’s surface?

A
1.0 ms^–2

B
2 ms^–2

C
4 ms^–2

D
9.8 ms^–2

Solution
Question 13
1 / -0

A body is projected at time t = 0 from a certain point on a planet’s surface with a certain velocity at a certain angle with the planet’s surface (assumed horizontal). The horizontal and vertical displacements x and y (in metres), respectively vary with time t (in seconds) as x = 10 √3 t and y = 10 t – t².

What is the magnitude and direction of the velocity with which the body is projected?

A
20 ms^–1 at an angle of 30° with the horizontal

B
20 ms^–1 at an angle of 60° with the horizontal

C
10 ms^–1at an angle of 30° with the horizontal

D
10 ms^–1 at an angle of 60° with the horizontal

Solution
Question 14
1 / -0

A stone weighing 0.5 Kg tied to a rope of length 0.5 m revolves along a circular path in a vertical plane. The tension of the rope at the bottom point of the circle is 45 N. To what height will the stone rise if the rope breaks the moment the velocity is directed upwards? (g = 10m/s²)

A
2 m

B
4 m

C
6 m

D
8 m

Solution
Question 15
1 / -0

A body of mass 0.4 kg is whirled in a vertical circle making 2 rev/s. If the radius of the circle is 1.2 m, then the tension in the string when the body is at the top of the circle is

A
71.78 N

B
89.86 N

C
109.86 N

D
115.86 N

Solution