Question

(Mg) ms
4. A wire of length l is stretched by a mass m by an equal length. Then:
1
1) The loss of energy as heat is 5mgl
2) The gain in elastic P.E. is 1/2mgl

3) Loss of energy is mgl
4) Neither (A) nor (B) are correct​

Answer 1

Given info : A wire of length l is stretched by a mass m by an equal length.

To check : the correct option (s) :

1) The loss of energy as heat is 5mgl

2) The gain in elastic P.E. is 1/2mgl

3) Loss of energy is mgl

4) Neither (A) nor (B) are correct

solution : mass of an object by which wire is stretched = m

length of wire = l

elongation of wire = l

so potential energy stored in the wire = 1/2 × stress × strain × volume.

= 1/2 × (weight of mass)/area × ∆l/l × Al

= 1/2 × (mg)/A × l/l × Al

= 1/2 mgl

Therefore the gain in elastic potential energy is 1/2 mgl. i.e., correct option is (2).

Answer 2

Answer:

2)The gain in elastic P.E. is 1/2mgl

Explanation:

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