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If we add $3 \mathrm{~kg}$ load to the hanger of sonometer, the fundamental frequency
becomes two times its initial value. The initial load must be
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becomes two times its initial value. The initial load must be
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Verified Answer
The correct answer is:
$1 \mathrm{~kg}$
$\mathrm{n} \propto \sqrt{\mathrm{T}} \quad \frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\sqrt{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}}$
$\frac{1}{2}=\left(\frac{T}{T+3}\right)^{1 / 2} \Rightarrow \frac{1}{4}=\frac{T}{T+3}$
$1+\frac{3}{T}=4$
$\frac{3}{T}=3 \Rightarrow T=1 \mathrm{~kg}$
$\frac{1}{2}=\left(\frac{T}{T+3}\right)^{1 / 2} \Rightarrow \frac{1}{4}=\frac{T}{T+3}$
$1+\frac{3}{T}=4$
$\frac{3}{T}=3 \Rightarrow T=1 \mathrm{~kg}$
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