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 string is increased to $1.5 x$ , the speed of the sound will be

1.22 V

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0.61 V

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1.50 V

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0.75 V

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Solution

The correct option is A $1.22 V$
$V = \frac{1}{2 l} \sqrt{\frac{T}{μ}} = \frac{1}{2 l} \sqrt{\frac{F A}{μ} (\frac{Δ l}{l})}$
$∴ V α \sqrt{\frac{Δ l}{l}}$
$∴ \frac{V_{1}}{V_{2}} = \frac{\sqrt{x}}{\sqrt{1.5 x}}$
$\Rightarrow v_{2} = 1.22 v$

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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 string is increased to 1.5x, the speed of the sound will be

The correct option is A 1.22VV=12l√Tμ=12l√FAμ(Δll)∴Vα√Δll∴V1V2=√x√1.5x⇒v2=1.22 v

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 string is increased to $1.5 x$ , the speed of the sound will be

The correct option is A $1.22 V$
$V = \frac{1}{2 l} \sqrt{\frac{T}{μ}} = \frac{1}{2 l} \sqrt{\frac{F A}{μ} (\frac{Δ l}{l})}$
$∴ V α \sqrt{\frac{Δ l}{l}}$
$∴ \frac{V_{1}}{V_{2}} = \frac{\sqrt{x}}{\sqrt{1.5 x}}$
$\Rightarrow v_{2} = 1.22 v$