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A block of weight $100 N$ is suspended by copper and steel wires of same cross sectional area $0.5 cm ^2$ and, length $\sqrt{3} m$ and $1 m$, respectively. Their other ends are fixed on a ceiling as shown in figure. The angles subtended by copper and steel wires with ceiling are $30^{\circ}$ and $60^{\circ}$, respectively. If elongation in copper wire is $\left(\Delta \ell_{ C }\right)$ and elongation in steel wire is $\left(\Delta \ell_{ s }\right)$, then the ratio $\frac{\Delta \ell_{ C }}{\Delta \ell_{ S }}$ is. . . . . .
[Young's modulus for copper and steel are $1 \times 10^{11} N / m ^2$ and $2 \times 10^{11} N / m ^2$ respectively]
$1$
$0$
$2$
$3$
Solution
Let $T_S=$ tension in steel wire $T _{ C }=$ Tension in copper wire in $x$ direction
$T _{ C } \cos 30^{\circ}= T _{ S } \cos 60^{\circ}$
$T _{ C } \times \frac{\sqrt{3}}{2}= T _{ S } \times \frac{1}{2}$
$\sqrt{3} T _{ C }= T _{ S } \ldots . \text { (i) }$
in $y$ direction
$T _{ C } \sin 30^{\circ}+ T _{ S } \sin 60^{\circ}=100$
$\frac{ T _{ C }}{2}+\frac{ T _{ S } \sqrt{3}}{2}=100 \ldots . \text { (ii) }$
Solving equation $(i)$ & $(ii)$
$T _{ C }=50 N$
$T _{ S }=50 \sqrt{3} N$
We know
$\Delta L =\frac{ FL }{ AY }$
$=\frac{\Delta L _{ C }}{\Delta L _{ S }}=\frac{ T _{ C } L _{ C }}{ A _{ C } Y _{ C }} \times \frac{ A _{ S } Y _{ S }}{ T _{ S } L _{ S }}$
On solving above equation
$\frac{\Delta L _{ C }}{\Delta L _{ S }}=2$
Ans. $2.00$
