# Class 11 Physics MCQ – Speed of a Travelling Wave

This set of Class 11 Physics Chapter 15 Multiple Choice Questions & Answers (MCQs) focuses on “Speed of a Travelling Wave”.

1. What is the expression for wave speed in the given equation: y = Asin(kx-wt + π)?
a) dy/dt
b) w/k
c) 2πw/k
d) 2πk/w

Explanation: The distance moved by a particular value of displacement per unit time is called wave speed. Basically, we have to study the motion of a fixed phase along the direction of wave propagation. So, kx – wt = constant. On differentiating we get: kdx – wdt = 0
dx/dt = w/k, which is the wave speed.

2. A string has a mass of 100gm & is under a tension of 10N. What should be its length if the speed of transverse waves is to be 10m/s on the string.
a) 1.8m
b) 10m
c) 1m
d) 2m

Explanation: v = √(T/μ),
μ = M/L.
V2 = TL/M
Or L = v2M/T
= 100 * 0.1/10
= 1m.

3. Why is the speed of longitudinal waves sound in solids greater than that in liquids?
a) Solids have lesser bulk modulus
b) Gases have lesser density
c) Solids have greater bulk modulus and density than gases but increase in bulk modulus dominantes over increase in density
d) Gases don’t have lesser particles in a given volume and therefore transfer sound energy slowly

Explanation: Speed is expressed as √(B/ρ).
For solids B and ρ is greater than that of gases but an increase in B compensates over the increase in ρ, and thus speed is greater in solids.

4. The formula v = √(P/ρ) can be used for any gas considering isothermal process. True or False?
a) True
b) False

Explanation: Speed of longitudinal waves is v = √(B/ρ), where B is bulk modulus.
For an isothermal process PV=const.
On differentiating, we get: PdV + VdP = 0.
∴ P = -dP/(dV/V) = B.
∴ v = √(B/ρ) = √(P/ρ).
We have used the ideal gas equation, so this is only valid for an ideal gas.

5. The speed of sound in air is 330m/s. What is the mass of 1 mole of air? Assume it is an ideal gas.
a) 28gm
b) 38gm
c) 20.56gm
d) 29gm

Explanation: 1 mole of air occupies 22.4L at STP.
Thus, density ρ = m/22.4*10-3.
Using equation v = √(P/ρ), we get:
v2 = (P/m)*22.4*10-3.
Pressure at STP = 105Pa.
∴ m = (P/v2)*22.4*10-3
= 105/3302* 22.4*10-3
= 20.56gm.
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6. What is the ratio of speed of sound in air when measured after laplace correction to that measured by Newton’s formula?
a) 1.18
b) 0.84
c) 1.4
d) 0.71

Explanation: By Newton’s formula v = √(P/ρ) & after laplace’s correction v = √(P/ρ), where is ratio of specific heats of air.
The ratio of speeds = √γ.
For air γ = 7/5
& the ratio of speeds = √(7/5) = 1.18.

7. Laplace correction makes use of which of the following processes?
a) Isothermal
c) Isochoric
d) Isobaric

Explanation: Laplace said that during propagation of sound energy there isn’t enough time for heat to flow and maintain temperature, so he considered the process to be adiabatic.

Sanfoundry Global Education & Learning Series – Physics – Class 11.

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