Question

26. When a force is applied on a wire of uniform cross
Sectional area 3 x 10-6 m2 and length 4 m, the
increase in length is 1 mm. Energy stored in it will
be (Y=2x 1011 N/m?)
(1) 8250 J
(3) 0075J
(2) 0.177 J
(4) 0.150 J​

Answer 1

Answer:

option (3), 0.075 J is correct

Answer 2

(3)The energy stored in the wire will be 0.075 J.

Explanation:

Given the cross-sectional area, $A =3\times10^-^6m^2$.

The initial length of the wire, l = 4 m.

The increase in length,$\Delta l=1mm=1\times10^-^3m$.

The  young modulus, $Y=2\times10^1^1N/m$.

The energy stored in wire is given as

$U=\frac{1}{2} \frac{Y(\Delta l)^2A}{l}$.

Substitute the given values, we get

$U=\frac{1}{2}[ \frac{2\times10^1^1\times(10^-^3)^2\times3\times10^-^6}{4} ]$

U = 0.075 J.

Thus, the energy stored in the wire will be 0.075 J.

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