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$2l$ લંબાઇનો તાર બે દિવાલ વચ્ચે જડિત છે.તેના મઘ્યબિંદુ પર $W$ વજન લગાવવાથી તે $x $ જેટલું નીચે ખસે છે. $(X<< l )$ તો $m$ $=$___
$\frac{1}{2}\frac{{YA{x^2}}}{{g{L^2}}}$
$\frac{1}{2}\frac{{YA{L^2}}}{{g{x^2}}}$
$\frac{{YA{x^3}}}{{g{L^3}}}$
$\frac{{YA{L^3}}}{{g{x^2}}}$
Solution
$2T\sin \theta = mg$ $ \Rightarrow T = \frac{{mg}}{{2\sin \theta }} = \frac{{mgL}}{{2x}}$ [ $\theta $ $\sin \theta = \frac{x}{L}$]
$l = AC – AB$$ = \sqrt {{L^2} + {x^2}} – L$$ = {({L^2} + {x^2})^{1/2}} – L$ $ = L\left[ {{{\left( {1 + \frac{{{x^2}}}{{{L^2}}}} \right)}^{1/2}} – 1} \right]$
$ = L\left[ {1 + \frac{1}{2}\frac{{{x^2}}}{{{L^2}}} – 1} \right]$$ = \frac{{{x^2}}}{{2L}}$
$Y = \frac{T}{A}\frac{L}{l}$
$\therefore$ $T = \frac{{YAl}}{L}$
$\frac{{mgL}}{{2x}} = \frac{{YA}}{L}.\frac{{{x^2}}}{{2L}}$
$\therefore m = \frac{{YA{x^3}}}{{g{L^3}}}$
