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Q.

A sonometer wire has a length l and tension T. If on reducing the tension to half of its original value and changing the length, the second harmonic becomes equal to the fundamental frequency of the first case, then the new length of the wire is:

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a

${"mathml":"<math class="image_resized" style="width:40px;height:22px"><span style="font-family:$

b

${"mathml":"<math class="image_resized" style="width:31px;height:22px"><span style="font-family:$

c

${"mathml":"<math class="image_resized" style="width:20px;height:11px"><span style="font-family:$

d

${"mathml":"<math class="image_resized" style="width:12px;height:11px"><span style="font-family:$

answer is B.(Unlock A.I Detailed Solution for FREE)

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Detailed Solution

Concept- We will change the parameters in the sonometer formula to resolve this issue.A sonometer operating formula is required. i.e.

f n = n 2 l T m

, where,

f n

is the frequency of n t h mode, n is the mode number, l is the wire's length, t is its tension, and m is the linear mass density of the wire, or its mass per unit length.
If the lengthens and the tension are cut in half

l 1

then from the above condition,

1 2 l T m = 2 2 l 1 T 2 m

n was not included in the formula because it is a constant number.

⇒ l 1 = 12

.
Hence, the correct option is 2.

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