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Thermal Expansion

The length of...

Question

The length of wire, when $M_{1}$ is hung from it is $l_{1}$ and is $l_{2}$ with both $M_{1}$ and $M_{2}$ hanging. The natural length of wire is

\frac{M_{1}}{M_{2}} (l_{1} - l_{2}) + l_{1}

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\frac{M_{2} l_{1} - M_{1} l_{2}}{M_{1} + M_{2}}

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\frac{l_{1} + L_{2}}{2}

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\sqrt{l_{1} L_{2}}

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Solution

The correct option is A $\frac{M_{1}}{M_{2}} (l_{1} - l_{2}) + l_{1}$
Let the natural length of wire be $l$ .

When only $M_{1}$ is hanging, the external force acting on the wire is

$F = M_{1} g$

Now using Hooke's law,

$Δ L = \frac{F L}{A Y}$

$\Rightarrow (l_{1} - l) = \frac{(M_{1} g) l}{A Y} . . . . . . . . . (1)$

Here, $Y =$ Young's modulus
$A =$ cross-sectional area of wire

When both $M_{1}$ and $M_{2}$ are hanging, external force will be

$F_{e x t} = (M_{1} + M_{2}) g$

Again using formula of elastic constant

$Δ l = \frac{F_{e x t} L}{A Y}$

$\Rightarrow (l_{2} - l) = \frac{(M_{1} + M_{2}) g l}{A Y} . . . . . (2)$

Here, $l_{1}$ and $l_{2}$ are length of wire when $M_{1}$ and $(M_{1} + M_{2})$ are hanging.

From Eqs. $(1) & (2)$ ,

$\frac{l_{2} - l}{l_{1} - l} = \frac{M_{1} + M_{2}}{M_{1}}$

$\Rightarrow M_{1} l_{2} - M_{1} l = l_{1} (M_{1} + M_{2}) - l (M_{1} + M_{2})$

$\Rightarrow l = \frac{M_{1}}{M_{2}} (l_{1} - l_{2}) + l_{1}$

So, option $(a)$ is correct.

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Thermal Expansion

Standard XII Physics

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The length of wire, when M1 is hung from it is l1 and is l2 with both M1 and M2 hanging. The natural length of wire is

The length of wire, when $M_{1}$ is hung from it is $l_{1}$ and is $l_{2}$ with both $M_{1}$ and $M_{2}$ hanging. The natural length of wire is