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At the poles, a stretched wire of a given length vibrates in unison with a tuning fork. At the equator, for same setting to produce resonance with same fork, the
vibrating length of wire
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vibrating length of wire
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The correct answer is:
should be decreased.
(A)
Frequency of vibration $\quad \mathrm{n}=\frac{1}{2 \mathrm{e}} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$
If $\mathrm{T}=\mathrm{Mg}$, then $\mathrm{n}=\frac{1}{2 \mathrm{e}} \sqrt{\frac{\mathrm{Mg}}{\mathrm{m}}}$
If $\ell_{1}$ and $\mathrm{g}_{1}$ are length and acceleration due to gravity at the pole and $\ell_{2}$ and $\mathrm{g}_{2}$ at the equator then, if frequency is same, then
$n=\frac{1}{2 \ell_{1}} \sqrt{\frac{M g_{1}}{m}}=\frac{1}{2 \ell_{2}} \sqrt{\frac{M g_{2}}{m}}$
$\therefore \frac{\sqrt{g_{1}}}{\ell_{1}}=\frac{\sqrt{g_{2}}}{\ell_{2}}$
or $\frac{\ell_{2}}{\ell_{1}}=\sqrt{\frac{g_{2}}{g_{1}}}$
$\mathrm{g}_{2} < \mathrm{g}_{1} \quad \therefore \ell_{2} < \ell_{1}$
Frequency of vibration $\quad \mathrm{n}=\frac{1}{2 \mathrm{e}} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$
If $\mathrm{T}=\mathrm{Mg}$, then $\mathrm{n}=\frac{1}{2 \mathrm{e}} \sqrt{\frac{\mathrm{Mg}}{\mathrm{m}}}$
If $\ell_{1}$ and $\mathrm{g}_{1}$ are length and acceleration due to gravity at the pole and $\ell_{2}$ and $\mathrm{g}_{2}$ at the equator then, if frequency is same, then
$n=\frac{1}{2 \ell_{1}} \sqrt{\frac{M g_{1}}{m}}=\frac{1}{2 \ell_{2}} \sqrt{\frac{M g_{2}}{m}}$
$\therefore \frac{\sqrt{g_{1}}}{\ell_{1}}=\frac{\sqrt{g_{2}}}{\ell_{2}}$
or $\frac{\ell_{2}}{\ell_{1}}=\sqrt{\frac{g_{2}}{g_{1}}}$
$\mathrm{g}_{2} < \mathrm{g}_{1} \quad \therefore \ell_{2} < \ell_{1}$
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