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. $\sqrt{L_1L_2}$
• B. $\frac{L_1+L_2}{2}$
• C. 2 L 1 – L 2
• D. 3 L 1 – 2 L 2

Answer 1

Answer: (C)

2 L 1 – L 2

Answer 2

The correct option is(C): 2 L 1 – L 2

$y=\frac{FL}{A△L}$

$⇒△L=\frac{FL}{Ay}$

$⇒L_1=L+\frac{(1g)L}{Ay}....(i)$

and

$⇒L_2=L+\frac{(2g)L}{Ay}....(ii)$

⇒ L = 2 L 1 – L 2