When a force is applied on a wire of uniform crosssectional area 3 × 10-6 m 2 and length 4 m, the increase in length is 1 mm. Energy stored in it will be (Y=2 × 1011 N / m 2)

Question

When a force is applied on a wire of uniform crosssectional area 3 × 10-6 m 2 and length 4 m, the increase in length is 1 mm. Energy stored in it will be (Y=2 × 1011 N / m 2) (A) 6250 J (B) 0.177 J (C) 0.075 J (D) 0.150 J

Tardigrade Admin · Accepted Answer

U =(1/2) × (Y A l2/L) =(1/2) × (2 × 1011 × 3 × 10-6 ×(1 × 10-3)2/4) =0.075 J

Q. When a force is applied on a wire of uniform crosssectional area $3 \times 1 0^{- 6} m^{2}$ and length $4 m$ , the increase in length is $1 mm$ . Energy stored in it will be $(Y = 2 \times 1 0^{11} N / m^{2})$

6250 J

0.177 J

0.075 J

0.150 J

$U = \frac{1}{2} \times \frac{Y A l ^{2}}{L}$
$= \frac{1}{2} \times \frac{2 \times 1 0 ^{11} \times 3 \times 1 0 ^{- 6} \times ( 1 \times 1 0 ^{- 3} ) ^{2}}{4}$
$= 0.075 J$

Q. When a force is applied on a wire of uniform crosssectional area 3×10−6m2 and length 4m, the increase in length is 1mm. Energy stored in it will be (Y=2×1011N/m2)