MCQ
If one end of a wire is fixed with a rigid support and the other end is stretched by a force of $10 \,N,$ then the increase in length is $0.5\, mm$. The ratio of the energy of the wire and the work done in displacing it through $1.5\, mm$ by the weight is
  • A
    $\frac{1}{3}$
  • B
    $\frac{1}{4}$
  • $\frac{1}{2}$
  • D
    $1$

Answer

Correct option: C.
$\frac{1}{2}$
c
(c) Work done in stretching a wire 

$W = \frac{1}{2}Fl = \frac{1}{2} \times 10 \times 0.5 \times {10^{ - 3}}$= $2.5 \times {10^{ - 3}}J$

Work done to displace it through $1.5 \,mm$

$W = F \times l = 5 \times {10^{ - 3}}J$

The ratio of above two work $= 1 : 2$

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