A wire of length L is hanging from a fixed support. The length changes to L1 and L2 when masses M1 and M2 are suspended respectively from its free end. Then L is equal to

Question

A wire of length L is hanging from a fixed support. The length changes to L1 and L2 when masses M1 and M2 are suspended respectively from its free end. Then L is equal to (A) (L1+L2/2) (B) √L1 L2 (C) (L1 M2+L2 M1/M1+M2) (D) (L1 M2-L2 M1/M2-M1)

Tardigrade Admin · Accepted Answer

Let L be the original length of the wire

When the mass

M_{1}

is suspended from the wire, change in length of wire is

Δ L_{1} = L_{1} - L

When the mass

M_{2}

is suspended from it, change in length of wire is

Δ L_{2} = L_{2} - L

From figure (b),

T_{1} = M_{1} g \dots (i)

From figure (c),

T_{2} = M_{2} g \dots (ii)

As Young's modulus,

Y = \frac{T _{1} L}{A Δ L _{1}} = \frac{T _{2} L}{A Δ L _{2}}

∴ \frac{T _{1}}{Δ L _{1}} = \frac{T _{2}}{Δ L _{2}} or \frac{T _{1}}{L _{1} - L} = \frac{T _{2}}{L _{2} - L}

or

\frac{M _{1} g}{L _{1} - L} = \frac{M _{2} g}{L _{2} - L}

(Using (i) and (ii))
or

M_{1} (L_{2} - L) = M_{2} (L_{1} - L)

or

M_{1} L_{2} - M_{1} L = M_{2} L_{1} - M_{2} L

or

L (M_{2} - M_{1}) = L_{1} M_{2} - L_{2} M_{1}

or

L = \frac{L _{1} M _{2} - L _{2} M _{1}}{M _{2} - M _{1}}

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

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

$L_{1} L_{2}$

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

$\frac{L _{1} M _{2} - L _{2} M _{1}}{M _{2} - M _{1}}$

Q. A wire of length L is hanging from a fixed support. The length changes to L1​ and L2​ when masses M1​ and M2​ are suspended respectively from its free end. Then L is equal to

2L1​+L2​​

L1​L2​​

M1​+M2​L1​M2​+L2​M1​​

M2​−M1​L1​M2​−L2​M1​​

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

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

$L_{1} L_{2}$

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

$\frac{L _{1} M _{2} - L _{2} M _{1}}{M _{2} - M _{1}}$