Engineering Questions with Answers - Multiple Choice Questions

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# Transverse Waves – 1

A source of sound frequency 600Hz is placed inside water. The speed of sound in water is 1500 m/s and in air it is 300 m/s. The frequency of sound is recorded by an observer, who is standing in the air is?

a) 200 Hz

b) 3000 Hz

c) 120 Hz

d) 600 Hz

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Answer: d

Explanation: The frequency of sound does not change during its refraction from water into the air.

The ratio of the speed of sound in nitrogen gas to that in the helium gas at 300 K is?

a) √(2/7)

b) √(1/7)

c) √3/5

d) √6/5

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Answer: c

Explanation: v(N2)/vHe =√(γ(N2)/γHe ×MHe/M(N2))=√((7/5)/(5/3)×4/28)=√3/5.

Two monatomic ideal gases 1 and 2 of molecular masses M1 and M2 respectively are enclosed in separate containers kept at the same temperature. The ratio of the speed pf spund in gas 1 to that in gas 2 is given by?

a) √(M1/M2)

b) √(M2/M1)

c) M1/M2

d) M2/M1

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Answer: b

Explanation: v=√(3RT/M)

v1/v2 = √(M2/M1).

The extension in a string, obeying Hooke’s law, is x. The speed of sound in the stretched string is v. If the extension in the spring is increased to 1.5x, the speed of sound will be ___________

a) 1.22v

b) 0.16v

c) 1.50v

d) 0.75v

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Answer: a

Explanation: According to Hooke’s law,

Tension (T) is proportional to Extension (x)

Also, speed of sound,

v∝√T

v‘∝√1.5x

v‘/v=√1.5=1.22

v‘=1.22v.

A travelling wave in a stretched string is described by the equation y = Asin(kx-ωt). The maximum particle velocity is ___________

a) Aω

b) ω/k

c) dω/dk

d) x/t

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Answer: a

Explanation: y=Asin(kx-ωt)

u=dy/dt=-ωAcos(kx-ωt)

umax=ωA.

A wave represented by the equation y=acos(kx-ωt) is superposed with another wave to form a stationary wave such that point x=0 is a node. The equation for the other wave is?

a) acos(kx-ωt)

b) -acos(kx-ωt)

c) -acos(kx+ωt)

d) -asin(kx-ωt)

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Answer: c

Explanation: For the formation of a stationary wave, two identical waves travelling in opposite directions must separate each other. At x=0, resultant y should be zero for getting a node. Hence -acos(kx+ωt) is the correct answer.

Two vibrating springs of the same material but lengths L and 2L have radii 2r and r respectively. They are stretched under the same tension. Both the strings vibrate in their fundamental modes, one of the lengths L with frequency f1 and the other with frequency f2. The ration is given by?

a) 2

b) 4

c) 8

d) 1

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Answer: d

Explanation: f=1/2l×√(T/m)=1/2l×√(T/(πR2 ρ))=1/2lR×√(T/πρ)

As T and ρ are the same for both strings,

f∝1/lR

For the first string,

f1∝1/L2r

For the second string

f2∝1/2Lr

Hence

f1/f2 =1.

In the experiment to determine the speed of sound using a resonance column ___________

a) Prongs of the tuning fork are kept in a vertical plane

b) Prongs of the tuning fork are kept in a horizontal plane

c) In one of the two resonances observed, the length of the resonating air column is close to the wavelength of sound in air

d) In one of the two resonances observed, the length of the resonating air column is close to half of the wavelength of sound in air

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Answer: a

Explanation: The prongs of the vibrating tuning fork are kept in a vertical plane just above the opening of the resonance tube.

In a resonance tube with tuning fork of frequency 512Hz, first resonance occurs at water level equal to 30.3cm and second resonance occurs at 63.7cm. The maximum possible error in the speed of sound is?

a) 51.2 cm/s

b) 102.4m/s

c) 204.8cm/s

d) 1536cm/s

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Answer: a

Explanation: The maximum possible error in the speed of sound is 51.2 cm/s.

In the experiment for the determination of the speed of sound in air using the resonance column method, the length of the air column that resonates in the fundamental mode, with a tuning fork is 0.1m. When this length is changed to 0.35m, the same tuning fork resonates with the first overtime. Calculate the end correction.

a) 0.012m

b) 0.025m

c) 0.05m

d) 0.024m

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Answer: b

Explanation: End correction=(l2-3l1)/2

=(0.35-3×0.1)/2=0.025m.