To increase the length by 0.5 mm in a steel wire of length 2 m and area of cross-section 2 mm 2, the force required is [ Y text steel =2.2 × 1011 N / m 2]

Question

To increase the length by 0.5 mm in a steel wire of length 2 m and area of cross-section 2 mm 2, the force required is [ Y text steel =2.2 × 1011 N / m 2] (A) 1.1 × 105 N (B) 1.1 × 104 N (C) 1.1 × 103 N (D) 1.1 × 102 N

Tardigrade Admin · Accepted Answer

F=(Y A e/l)=(2.2 × 1011 × 2 × 10-6 × 5 × 10-4/2) =11 × 101=1.1 × 102 N

Q. To increase the length by $0.5 mm$ in a steel wire of length $2 m$ and area of cross-section $2 m m^{2}$ , the force required is $[Y_{steel} = 2.2 \times 1 0^{11} N / m^{2}]$

$1.1 \times 1 0^{5} N$

$1.1 \times 1 0^{4} N$

$1.1 \times 1 0^{3} N$

$1.1 \times 1 0^{2} N$

$F = \frac{Y A e}{l} = \frac{2.2 \times 1 0 ^{11} \times 2 \times 1 0 ^{- 6} \times 5 \times 1 0 ^{- 4}}{2}$
$= 11 \times 1 0^{1} = 1.1 \times 1 0^{2} N$

Q. To increase the length by 0.5mm in a steel wire of length 2m and area of cross-section 2mm2, the force required is [Ysteel ​=2.2×1011N/m2]

1.1×105N

1.1×104N

1.1×103N

1.1×102N