Dividing $(ii)$ by $(i)$ we get ${\left( {\frac{7}{6}} \right)^2} = \frac{{m + m'}}{m} = \frac{{49}}{{36}}$
$\frac{{m + m'}}{m} - 1 = \frac{{49}}{{36}} - 1 \Rightarrow \frac{{m'}}{m} = \frac{{13}}{{36}}$ $⇒$ $m' = \frac{{13m}}{{36}}$
Also $\frac{k}{m} = \frac{{4{\pi ^2}}}{{{{(0.6)}^2}}}$
Desired extension $ = \frac{{m'g}}{k}$$ = \frac{{13}}{{36}} \times \frac{{mg}}{k}$
$ = \frac{{13}}{{36}} \times 10 \times \frac{{0.36}}{{4{\pi ^2}}} \approx 3.5\;cm$
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| List-$I$ | List-$II$ |
| $(A)$ A force thatrestores anelastic body of unit area to its original state | $(I)$ Bulkmodulus |
| $(B)$ Two equal andopposite forcesparallel toopposite faces | $(II)$Young'smodulus |
| $(C)$Forcesperpendiculareverywhere tothe surface perunit areasameeverywhere | $(III)$ Stress |
| $(D)$Two equal andopposite forceperpendicular toopposite faces | $(IV)$ Shearmodulus |
Choose the correct answer from the options given below:
$y(x,t)\, = \,0.6\,\sin \,\left( {\frac{{2\pi }}{3}x} \right)\,\cos \,(120\,\pi t)$
where $x$ and $y$ are in $metre$ and $t$ in $second$ . The length of the string is $1.5\,m$ and its mass is $3.0\times 10^{-2}\,kg$ the tension in the string will be .... $N$