Physics Test 194

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Physics Test 194

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Question 1
4 / -1

The circular motion of a particle with constant speed is

A
Simple harmonic but not periodic

B
Periodic and simple harmonic

C
Neither periodic nor simple harmonic

D
Periodic but not simple harmonic

Solution

In a circular motion particle repeats after equal intervals of time. So particle motion on a circular path is periodic but not simple harmonic as it does not execute to and fro motion about a fixed point.
Question 2
4 / -1

A body is vibrating in simple harmonic motion. If its acceleration is 12 cm s⁻² at a displacement 3 cm, then time period is

A
6.28 s

B
3.14 s

C
1.57 s

D
2.57 s

Solution
Question 3
4 / -1

A simple pendulum is released from A shown.

If m and l represent the mass of the bob and Length of the pendulum, the gain kinetic energy at B is

A
mgl/2

B
mgl/√2

C
√3/2mgl

D
2/√3mgl

Solution

Loss of potential energy in coming from A to B

= mgh

= mglcos30° = √3 / 2 mgl

Kinetic energy gained = loss of potential energy

= √3 / 2 mgl
Question 4
4 / -1

For Simple Harmonic Oscillator, the potential energy is equal to kinetic energy

A
Once during each cycle

B
twice during each cycle

C
when x = 0

D
when x = a

Solution

potential energy U and kinetic energy K in SHM are given by

hence When a particle executes SHM, there will be two position in each cycle where the PE is equal to KE of the body in SHM
Question 5
4 / -1

A particle is vibrating in simple harmonic motion with an amplitude of 4 cm. At what displacement from the equilibrium position is its energy half potential and half kinetic?

A
10 cm

B
√2 cm

C
2 cm

D
2√2 cm

Solution
Question 6
4 / -1

If the initial tension on a stretched string is doubled, then the ratio of the initial and final speeds of a transverse wave along the string is

A
1 : 1

B
√2 : 1

C
1 : √2

D
1 : 2

Solution
Question 7
4 / -1

In a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency 6Hz. When tension in B is slightly decreased, the beat frequency increases to 7Hz. If the frequency of A is 530Hz, the original frequency of B will be

A
524Hz

B
536Hz

C
537Hz

D
523Hz

Solution
Question 8
4 / -1

A tuning fork is used to produce resonance in a glass tube. The length of the air column in this tube can be adjusted by a variable piston. At room temperature of 27^∘C two successive resonances are produced at 20cm and 73cm of column length. If the frequency of the tuning fork is 320Hz, the velocity of sound in air at 27^∘C is

A
330ms⁻¹

B
339ms⁻¹

C
350ms⁻¹

D
300ms⁻¹

Solution

The velocity of sound in air at 27°C is v = 2(v)[L₂− L₁]; where v = frequency of tuning fork and L₁, L₂ are the successive column length.

∴ v = 2 × 320[73 − 20] × 10⁻²

= 339.2ms⁻¹ ≈ 339ms⁻¹.
Question 9
4 / -1

The two nearest harmonics of a tube closed at one end and open at other end are 220 Hz and 260 Hz. What is the fundamental frequency of the system?

A
20 Hz

B
30 Hz

C
40 Hz

D
10 Hz

Solution

Nearest harmonics of an organ pipe closed at one end is differ by twice of its fundamental frequency.

∴ 260 − 220 = 2υ, υ = 20 Hz
Question 10
4 / -1

The second overtone of an open organ pipe has the same frequency as the first overtone of a closed pipe L metre long. The length of the open pipe will be

A
L

B
2L

C
1/L

D
4L

Solution

Second overtone of an open organ pipe

=Third harmonic = 3 × υ′₀ = 3 × v/2L′

First overtone of a closed organ pipe

= Third harmonic = 3 × υ₀= 3 × v/4L

According to question,

3υ′₀ = 3υ₀⇒3 × v/2L′ = 3 × v/4L

⇒ L′ = 2L