The length of wire, when M_1 is hung from it, | vedclass

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Question

(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

Vedclass · Accepted Answer

$\frac{M_1}{M_2}\left(l_1-l_2\right)+l_1$

Vedclass · Answer

(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

$I=\frac{M_1}{M_2}\left(l_1-l_2\right)+I_1$

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 ........

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