Physics Test 195

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Physics Test 195

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Weekly Quiz Competition

Question 1
4 / -1

A siren emitting a sound of frequency 800 Hz moves away from an observer towards a cliff at a speed of 15ms⁻¹. Then, the frequency of sound that the observer hears in the echo reflected from the cliff is

(Take velocity of sound in air=330ms⁻¹)

A
838 Hz

B
885 Hz

C
765 Hz

D
800 Hz

Solution

Here, frequency of sound emitted by siren,

υ₀ = 800Hz

Speed of source, v_s = 15ms⁻¹

Speed of sound in air,v = 330ms⁻¹

Apparent frequency of sound at the cliff = frequency heard by observer = υ
Question 2
4 / -1

A source of sound S emitting waves of frequency 100 Hz and an observer O are located at some distance from each other. The source is moving with a speed of 19.4 m s'1 at an angle of 60° with the source observer line as shown in the figure. The observer is at rest.

The apparent frequency observed by the observer(velocity of sound in air 330ms⁻¹), is

A
106 Hz

B
97 Hz

C
100 Hz

D
103 Hz

Solution
Question 3
4 / -1

The number of possible natural oscillations of air column in a pipe closed at one end of length 85 cm whose frequencies lie below 1250 Hz are

(Velocity of sound = 340ms⁻¹)

A
4

B
5

C
7

D
6

Solution

The natural frequencies of the closed organ pipe will be

υ_n = (2n − 1)υ = υ, 3υ, 5υ, 7υ, 9υ, 11υ, 13υ,.............

=100Hz, 300Hz, 500Hz, 700Hz, 900Hz, 1100Hz, 1300Hz,.... and so on.

Thus, the natural frequencies lies below the 1250 Hz is 6.
Question 4
4 / -1

If we study the vibration of a pipe open at both ends, then the following statement is not true.

A
All harmonics of the fundamental frequency will be generated

B
Pressure change will be maximum at both ends.

C
Open end will be antinode.

D
Odd harmonics of the fundamental frequency will be generated.

Solution

Pressure change will be minimum at both ends.
Question 5
4 / -1

Two sources of sound placed close to each other, are emitting progressive waves given by y₁ = 4sin600πt and y₂ = 5sin608πt An observer located near these two sources of sound will hear

A
4 beats per second with intensity ratio 25 : 16 between waxing and waning.

B
8 beats per second with intensity ratio 25 : 16 between waxing and waning.

C
8 beats per second with intensity ratio 81 :1 between waxing and waning

D
4 beats per second with intensity ratio 81 :1 between waxing and waning.

Solution
Question 6
4 / -1

A geostationary satellite is orbiting the earth at a height of 5R above that surface of the earth, R being the radius of the earth. The time period of another satellite in hours at a height of 2R from the surface of the earth is

A
5

B
10

C
6√2

D
6 / √2

Solution
Question 7
4 / -1

A particle of mass m is thrown upwards from the surface of the earth, with a velocity u. The mass and the radius of the earth are, respectively, M and R.G is gravitational constant and g is acceleration due to gravity on the surface of the earth. The minimum value of u so that the particle does not return back to earth, is

A

B

C

D

Solution
Question 8
4 / -1

A simple pendulum bob has a mass "m" and length "L". The bob is drawn aside such that the string is horizontal and then it is released. The velocity of the bob while it crosses the equilibrium position is

A
√gL

B
√2gL

C
√5gL

D
√3gL

Solution

mgL = 1/2mv²
Question 9
4 / -1

A simple pendulum is made by attaching a 1 kg bob to a 5 m long copper wire. Its period is T. Now, if 1 kg bob is replaced by 10 kg bob, the period of oscillation

A
Remain T

B
Becomes greater than T

C
Becomes less than T

D
Any of above depends on locality

Solution

For simple pendulum T = 2π / lg when weight suspended on elastic wire is increased, its length increases, so time period increases.
Question 10
4 / -1

A body executes simple harmonic motion. The potential energy (PE), the kinetic energy (KE) and total energy (TE) are measured as function of displacement x. Which of the following statement is true?

A
KE is maximum when x = 0

B
TE is zero when x = 0

C
KE is maximum when x is maximum

D
PE is maximum when x = 0

Solution

In S.H.M., at mean position i.e. at x = 0. Kinetic energy will be maximum and the body will be at the lowest potential energy. So, PE will be minimum. Total energy is always constant.