A block of mass 500 kg is suspended by wire of length 70 cm. The area of cross-section of wire is 10 mm2. When the load is removed, the wire contracts by 0.5 cm. The Young’s modulus of the material of wire will be

Question

A block of mass 500 kg is suspended by wire of length 70 cm. The area of cross-section of wire is 10 mm2. When the load is removed, the wire contracts by 0.5 cm. The Young’s modulus of the material of wire will be (A) 10 × 1014 N / m 2 (B) 4 × 1014 N / m 2 (C) 8 × 1011 N / m 2 (D) 7 × 1010 N / m 2

Tardigrade Admin · Accepted Answer

Y=(F × L/A × Δ L) =(500 × 10 × 0.7/10 × 10-6 × 0.5 × 10-2) =(3500/5 × 10-8)=7 × 1010 N / m 2

Q. A block of mass $500 k g$ is suspended by wire of length $70 c m$ . The area of cross-section of wire is $10 m m^{2}$ . When the load is removed, the wire contracts by $0.5 c m$ . The Young’s modulus of the material of wire will be

$10 \times 1 0^{14} N / m^{2}$

$4 \times 1 0^{14} N / m^{2}$

$8 \times 1 0^{11} N / m^{2}$

$7 \times 1 0^{10} N / m^{2}$

$Y = \frac{F \times L}{A \times Δ L}$
$= \frac{500 \times 10 \times 0.7}{10 \times 1 0 ^{- 6} \times 0.5 \times 1 0 ^{- 2}}$
$= \frac{3500}{5 \times 1 0 ^{- 8}} = 7 \times 1 0^{10} N / m^{2}$

Q. A block of mass 500kg is suspended by wire of length 70cm. The area of cross-section of wire is 10mm2. When the load is removed, the wire contracts by 0.5cm. The Young’s modulus of the material of wire will be

10×1014N/m2

4×1014N/m2

8×1011N/m2

7×1010N/m2