A load of 4.0 kg is suspended from a ceiling through a steel wire of length 2.0 m and radius 2.0 mm. It is found that the length of the wire increases by 0.031 mm as equilibrium is achieved. Taking, g=3.1 π ms -2, the Young's modulus of steel is

Question

A load of 4.0 kg is suspended from a ceiling through a steel wire of length 2.0 m and radius 2.0 mm. It is found that the length of the wire increases by 0.031 mm as equilibrium is achieved. Taking, g=3.1 π ms -2, the Young's modulus of steel is (A) 2.0 × 108 Nm -2 (B) 2.0 × 109 Nm -2 (C) 2.0 × 1011 Nm -2 (D) 2.0 × 1013 Nm -2

Tardigrade Admin · Accepted Answer

As, Y=(M g L/π r2 × l)=(4 ×(3.1 π) ×(2.0)/ .π ×(2 × 10-3)2 × 0.031 × 10-3)) =2 × 1011 Nm -2

Q. A load of $4.0 k g$ is suspended from a ceiling through a steel wire of length $2.0 m$ and radius $2.0 mm$ . It is found that the length of the wire increases by $0.031 mm$ as equilibrium is achieved. Taking, $g = 3.1 πm s^{- 2}$ , the Young's modulus of steel is

$2.0 \times 1 0^{8} N m^{- 2}$

$2.0 \times 1 0^{9} N m^{- 2}$

$2.0 \times 1 0^{11} N m^{- 2}$

$2.0 \times 1 0^{13} N m^{- 2}$

As, $Y = \frac{M gL}{π r ^{2} \times l} = \frac{4 \times ( 3.1 π ) \times ( 2.0 )}{π \times ( 2 \times 1 0 ^{- 3} ) ^{2} \times 0.031 \times 1 0 ^{- 3} )}$
$= 2 \times 1 0^{11} N m^{- 2}$

Q. A load of 4.0kg is suspended from a ceiling through a steel wire of length 2.0m and radius 2.0mm. It is found that the length of the wire increases by 0.031mm as equilibrium is achieved. Taking, g=3.1πms−2, the Young's modulus of steel is

2.0×108Nm−2

2.0×109Nm−2

2.0×1011Nm−2

2.0×1013Nm−2