Question

A5 metre long wire is fixed to the ceiling
weight of 10 kg is hung at the lower end an
I metre above the floor. The wire was clong
by 1 mm. The energy stored in the wire du
stretching is
(1) Zero
(2) 0.05 joule
(3) 100 joule​

Answer 1

Answer:

0.05 joules

Explanation:

0.05 joules

Answer 2

(2)The energy stored in a stretched wire is 0.05 J.

Explanation:

The energy stored in a stretched wire is given as

$W=\frac{1}{2} F\times l$

Here, F is force applied on the wire and l is elongation in the wire.

Given weight hung at lower end of the wire which is force on the wire F = 10 kg and elongation l = 1 mm = 0.001 m

Substitute the given values, we get

$W=\frac{1}{2}\times10kg\times0.001 m$

W = 0.05 J.

Thus, the energy stored in a stretched wire is 0.05 J.

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Topic : energy stored

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