The area of cross-section of a steel wire (Y=2.0 × 1011 N / m 2) is 0.1 cm 2. The force required to double its length will be

Question

The area of cross-section of a steel wire (Y=2.0 × 1011 N / m 2) is 0.1 cm 2. The force required to double its length will be (A)  2× 1012N  (B)  2× 1011N  (C)  2× 1010N (D)  2× 106N

Tardigrade Admin · Accepted Answer

Young's modulus of elasticity Y=(F L/A l) To double the length l=L ∴ Y =(F/A) ⇒ F=Y A =2 × 1011 × 0.1 × 10-4 =2 × 106 N

Q. The area of cross-section of a steel wire $(Y = 2.0 \times 1 0^{11} N / m^{2})$ is $0.1 c m^{2}$ . The force required to double its length will be

$2 \times 1 0^{12} N$

$2 \times 1 0^{11} N$

$2 \times 1 0^{10} N$

$2 \times 1 0^{6} N$

Young's modulus of elasticity
$Y = \frac{F L}{A l}$
To double the length $l = L$
$∴ Y = \frac{F}{A} \Rightarrow F = Y A$
$= 2 \times 1 0^{11} \times 0.1 \times 1 0^{- 4}$
$= 2 \times 1 0^{6} N$

Q. The area of cross-section of a steel wire (Y=2.0×1011N/m2) is 0.1cm2. The force required to double its length will be

2×1012N

2×1011N

2×1010N

2×106N