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8.Mechanical Properties of Solids
hard
The length of wire, when $M_1$ is hung from it, is $I_1$ and is $I_2$ with both $M_1$ and $M_2$ hanging. The natural length of wire is ........
A
$\frac{M_1}{M_2}\left(l_1-l_2\right)+l_1$
B
$\frac{M_2 l_1-M_1 l_2}{M_1+M_2}$
C
$\frac{l_1+l_2}{2}$
D
$\sqrt{l_1 l_2}$
Solution
(a)
Let the natural length of wire be $=1$
When only $M_1$ hanging
Using $\Delta l=\frac{F L}{A Y}$
$\left(l_1-l \right)=\frac{M_1 g \cdot l}{A Y \ldots(1)}$
When both $M_1, M_2$ hanging
$\left(l_2-l\right)=\frac{\left(M_1+M_2\right) g \cdot l}{A Y} \ldots(2)$
Dividing $(1)$ by $(2)$
$\frac{l_1-l}{l_2-l}=\frac{M_1}{M_1+M_2}$
Solving this we get
$I=\frac{M_1}{M_2}\left(l_1-l_2\right)+I_1$
Standard 11
Physics
