A 5 -metre long wire is fixed to the ceiling. A weight of 10 kg is hung at the lower end and is 1 metre above the floor. The wire was elongated by 1 mm. The energy stored in the wire due to stretching is

Question

A 5 -metre long wire is fixed to the ceiling. A weight of 10 kg is hung at the lower end and is 1 metre above the floor. The wire was elongated by 1 mm. The energy stored in the wire due to stretching is (A) Zero (B) 0.05 J (C) 100 J (D) 500 J

Tardigrade Admin · Accepted Answer

W =(1/2) × F × l =(1/2) m g l=(1/2) × 10 × 10 × 1 × 10-1 =0.05 J

Q. A $5$ -metre long wire is fixed to the ceiling. A weight of $10 k g$ is hung at the lower end and is $1$ metre above the floor. The wire was elongated by $1 mm$ . The energy stored in the wire due to stretching is

Zero

0.05 J

100 J

500 J

$W = \frac{1}{2} \times F \times l$
$= \frac{1}{2} m g l = \frac{1}{2} \times 10 \times 10 \times 1 \times 1 0^{- 1}$
$= 0.05 J$

Q. A 5 -metre long wire is fixed to the ceiling. A weight of 10kg is hung at the lower end and is 1 metre above the floor. The wire was elongated by 1mm. The energy stored in the wire due to stretching is