A wire of length L is hanging from a fixed support. The length changes to L 1 and L 2 when masses 1 kg and 2 kg are suspended respectively from its free end. Then the value of L is equal to:

Question

A wire of length L is hanging from a fixed support. The length changes to L 1 and L 2 when masses 1 kg and 2 kg are suspended respectively from its free end. Then the value of L is equal to: (A) √ L 1 L 2 (B) ( L 1+ L 2/2) (C) 2 L 1- L 2 (D) 3 L1-2 L2

Tardigrade Admin · Accepted Answer

By Hooke's Law so F α Δ L ( F 1/ F 2)=(Δ L 1/Δ L 2) (10/20)=(( L 1- L )/( L 2- L )) L =2 L 1- L 2

Q. A wire of length $L$ is hanging from a fixed support. The length changes to $L_{1}$ and $L_{2}$ when masses $1 k g$ and $2 k g$ are suspended respectively from its free end. Then the value of $L$ is equal to:

$L_{1} L_{2}$

$\frac{L _{1} + L _{2}}{2}$

$2 L_{1} - L_{2}$

$3 L_{1} - 2 L_{2}$

By Hooke's Law
so $F α Δ L$
$\frac{F _{1}}{F _{2}} = \frac{Δ L _{1}}{Δ L _{2}}$
$\frac{10}{20} = \frac{( L _{1} - L )}{( L _{2} - L )}$
$L = 2 L_{1} - L_{2}$

Q. A wire of length L is hanging from a fixed support. The length changes to L1​ and L2​ when masses 1kg and 2kg are suspended respectively from its free end. Then the value of L is equal to:

L1​L2​​

2L1​+L2​​

2L1​−L2​

3L1​−2L2​