On increasing the length by 0.5 mm in a steel wire of length 4 m and area of cross section 5 mm 2, the force required is (Y=11.90 × 1012 N / m 2)

Question

On increasing the length by 0.5 mm in a steel wire of length 4 m and area of cross section 5 mm 2, the force required is (Y=11.90 × 1012 N / m 2) (A) 4 × 103 N (B) 2 × 103 N (C) 7.4 × 103 N (D) 103 N

Tardigrade Admin · Accepted Answer

Y=(F × L/A × Δ L) ⇒ F=(Y A Δ L/L) ∴ F=(11.9 × 1012 × 5 × 10-6 × 5 × 10-4/4) =(297.5 × 102/4) =74.375 × 102 N F =7.4 × 10+3 N

Q. On increasing the length by $0.5 mm$ in a steel wire of length $4 m$ and area of cross section $5 m m^{2},$ the force required is $(Y = 11.90 \times 1 0^{12} N / m^{2})$

$4 \times 1 0^{3} N$

$2 \times 1 0^{3} N$

$7.4 \times 1 0^{3} N$

$1 0^{3} N$

$Y = \frac{F \times L}{A \times Δ L}$
$\Rightarrow F = \frac{Y A Δ L}{L}$
$∴ F = \frac{11.9 \times 1 0 ^{12} \times 5 \times 1 0 ^{- 6} \times 5 \times 1 0 ^{- 4}}{4}$
$= \frac{297.5 \times 1 0 ^{2}}{4}$
$= 74.375 \times 1 0^{2} N$
$F = 7.4 \times 1 0^{+ 3} N$

Q. On increasing the length by 0.5mm in a steel wire of length 4m and area of cross section 5mm2, the force required is (Y=11.90×1012N/m2)

4×103N

2×103N

7.4×103N

103N