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Class 11 Physics Mcqs

Q:

In case of electromagnetic waves, doppler shifts are independent of whether source moves or observer provided relative speed is the same. True or False?

A) True B) False
 
Answer & Explanation Answer: A) True

Explanation: Whether the source moved or the observer would only matter if the waves required a material medium to propagate because then the relative motion between medium and observer would matter.

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Q:

Two longitudinal waves: s1 = 3cos(11t) & s2 = 3cos(10t) interfere. What is the amplitude of the resulting wave?

A) 6cos(0.5t) B) 6sin(0.5t)
C) 6cos(10.5t) D) 6sin(10.5t)
 
Answer & Explanation Answer: A) 6cos(0.5t)

Explanation: s = s1+s2 = 2(3)*cos(1/2)t*cos(21/2)t
= 6cos(0.5t)cos(10.5t).
The term 6cos(0.5t) corresponds to the amplitude.

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Q:

Let’s say we can hear a beat frequency of 2Hz from two tuning forks. The first tuning fork has a frequency of 100Hz. If wax is applied on the 2nd tuning fork, the beat frequency becomes 4Hz, what is the frequency of the 2nd tuning fork?

A) 100Hz B) 98Hz
C) 106Hz D) 104Hz
 
Answer & Explanation Answer: B) 98Hz

Explanation: As the initial beat frequency is 2Hz, the 2nd tuning fork can have a frequency of 98Hz or 102Hz. When wax is applied to the second tuning fork its frequency should decrease. On doing so the beat frequency has also increased, so the initial value of the frequency of the 2nd tuning fork must be lower than that of the first tuning fork.

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Q:

Two interfering waves have a frequency of 2Hz & 6Hz. What is the beat frequency?

A) 8Hz B) 4Hz
C) 2Hz D) 0
 
Answer & Explanation Answer: B) 4Hz

Explanation: Beat frequency is equal to the difference between frequencies of interfering waves. In this case, beat frequency
= 6-2 = 4Hz.

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Q:

Consider the phenomenon of beats & select the correct statement.

A) It occurs when frequencies are close or equal B) It correspond to the change in intensity we hear due to interference of 2 waves having close frequencies
C) It corresponds to a constant intensity sound heard during wave interference D) Beat frequency is variable
 
Answer & Explanation Answer: B) It correspond to the change in intensity we hear due to interference of 2 waves having close frequencies

Explanation: When two waves with close, but not equal, frequencies interfere, beats are produced. They have a frequency equal to the average frequency of the two interfering waves and we recognize them as the variable intensity sound we hear.

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Q:

Standing waves can be produced when two identical waves, having a phase difference of ?, are travelling in the same direction. True or False?

A) True B) False
 
Answer & Explanation Answer: B) False

Explanation: The equation of standing waves is given by; y = Asin(kx)cos(wt).
This is formed by adding two waves of the form asin(kx-wt) & bsin(kx+wt)
which shows that they both must be travelling in opposite directions.

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Q:

A pipe open at both ends has a fundamental frequency of 2Hz. If the pipe is closed at one end and half filled with water, what will be the fundamental frequency?

A) 0.5Hz B) 2Hz
C) 4Hz D) 1Hz
 
Answer & Explanation Answer: B) 2Hz

Explanation: The fundamental frequency in the first case is given by: f1 = v/2L.
In the second case length becomes half & one end is closed,
so fundamental frequency f2 = v/(4L/2)
= v/2L = f1.
So the value of f2 = 2Hz.

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Q:

Consider standing waves in an air column with one end closed. What is a pressure node?

A) Pressure variation is maximum B) Displacement variation is minimum
C) Same as displacement node D) Least pressure change
 
Answer & Explanation Answer: D) Least pressure change

Explanation: The point where there is an antinode corresponds to maximum amplitude of displacement. This also implies that here, pressure changes are minimum and it is therefore called a pressure node.

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