# A wire of length L is hanging from a fixed support. The length changes to L1 and L2 when masses …

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

(a) $\displaystyle \frac{L_1 + L_2}{2}$

(b) $\displaystyle \sqrt{L_1 L_2}$

(c) $\displaystyle \frac{L_1 M_2 + L_2 M_1}{M_1 + M_2}$

(d) $\displaystyle \frac{L_1 M_2 – L_2 M_1}{M_2 – M_1}$

Ans: (d)

Sol: Let Y be the Young’s modulus of elasticity

When mass M1 is suspended ,

$\displaystyle Y = \frac{M_1 g L}{A(L_1 – L)}$ …(i)

When mass M2 is suspended ,

$\displaystyle Y = \frac{M_2 g L}{A(L_2 – L)}$ …(ii)

From equation (i) & (ii)

$\displaystyle \frac{M_1 g L}{A(L_1 – L)} = \frac{M_2 g L}{A(L_2 – L)}$

$\displaystyle M_1 (L_2 – L) = M_2 (L_1 – L)$

$\displaystyle M_1 L_2 – M_1 L = M_2 L_1 – M_2 L$

$\displaystyle M_2 L_1 – M_1 L_2 = (M_2 – M_1) L$

$\displaystyle L = \frac{M_2 L_1 – L_2 M_1}{M_2 -M_1}$