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A thin rod of length $\mathrm{l}$ is suspended from a point from on its length which is at a distance $\mathrm{x}$ from its centre, for what value of $\mathrm{x}$ time period of oscillation will be minimum -

Option: 1
$\mathrm{\frac{l}{\sqrt{2}}}$

Option: 2
$\mathrm{\frac{l}{\sqrt{10}}}$

Option: 3
$\mathrm{\frac{l}{\sqrt{5}}}$

Option: 4
$\mathrm{\frac{l}{\sqrt{12}}}$

Answers (1)

best_answer

Time period is given by-

$\mathrm{T=2 \pi \sqrt{\frac{I}{m g d}}}$ ......(1)

if $\mathrm{d=x}$

then, $\mathrm{\quad I=I_c+m x^2}$

$\mathrm{ I=\frac{m l^2}{12}+m x^2 }$

$\mathrm{ \therefore }$ For find value of $\mathrm{ T }$ from equation (1)

$\mathrm{ T=2 \pi \sqrt{\frac{\frac{m l^2}{12}+m x^2}{m g x}} }$

for minimum time period. $\mathrm{ \frac{d T}{d x}=0 }$

$\mathrm{ \frac{d}{d x}\left[2 \pi \sqrt{\frac{m l^2+12 m x^2}{12 m g x}}\right]=0 }$

$\mathrm{ \frac{d}{d x}\left[2 \pi \sqrt{\left(\frac{m l^2}{12 m g x}+\frac{12 m x^2}{12 m g x}\right.}\right)=0 }$

After solving, $\mathrm{ \sqrt{12} x-l =0 }$

$\mathrm{ l =\sqrt{12} x }$

$\mathrm{ x =l / \sqrt{12} }$

Hence option 4 is correct.

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