Question

13.
Amass of 0.5 kg is suspended from wire, then length
of wire increase by 3 mm then find out work done :
(1) 4.5 x10-3 J
(2) 7.3 x10-3 J
(3) 9.3 x 10-2 J
(4) 2.5 x 10-2 J​

Answer 1

$\underline{ \boxed{ \bold{ \huge{ \purple{Answer}}}}} \\ \\ \dagger \rm \: \red{Given} \\ \\ \leadsto \rm \: mass \: of \: body \: m = 0.5 \: kg \\ \\ \leadsto \rm \: change \: in \: length \: (dx) = 3 \: mm \\ \\ \dagger \rm \: \red{To \: Find} \\ \\ \leadsto \rm \: work \: done \: for \: streching \: a \: wire \\ \\ \dagger \rm \: \red{Formula} \\ \\ \leadsto \rm \: W = \frac{F \: (dx)}{2} = \frac{mg \: (dx)}{2} \\ \\ \dagger \rm \: \red{Calculation} \\ \\ \leadsto \rm \: W = \frac{0.5 \times 9.8 \times 3 \times {10}^{ - 3} }{2} = 7.35 \times {10}^{ - 3} \: J \\ \\ \ast \ast \: \boxed{ \rm{ \orange{W = 7.3 \times {10}^{ - 3} \: J}}} \: \ast \ast$