A load of 1 kg weight is a attached to one end of a steel wire of area of cross-section 3 mm2 and Young's modulus 1011 N/m2. The other end is suspended vertically from a hook on a wall, then the load is pulled horizontally and released. When the load passes through its lowest position the fractional change in length is (g = 10 m/s2)

Question

A load of 1 kg weight is a attached to one end of a steel wire of area of cross-section 3 mm2 and Young's modulus 1011 N/m2. The other end is suspended vertically from a hook on a wall, then the load is pulled horizontally and released. When the load passes through its lowest position the fractional change in length is (g = 10 m/s2) (A) 0.3 × 10-4 (B) 0.3 × 10-3 (C) 0.3 × 103 (D) 0.3 × 104

Tardigrade Admin · Accepted Answer

Y=(m g l/A Δ l) ⇒ (Δ l/l)=(m g/A Y) ∴ (Δ l/l)=(1 × 10/3 × 10-6 × 1011) =0.3 × 10-4

$0.3 \times 1 0^{- 4}$

$0.3 \times 1 0^{- 3}$

$0.3 \times 1 0^{3}$

$0.3 \times 1 0^{4}$

$Y = \frac{m g l}{A Δ l}$
$\Rightarrow \frac{Δ l}{l} = \frac{m g}{A Y}$
$∴ \frac{Δ l}{l} = \frac{1 \times 10}{3 \times 1 0 ^{- 6} \times 1 0 ^{11}}$
$= 0.3 \times 1 0^{- 4}$

0.3×10−4

0.3×10−3

0.3×103

0.3×104

Y=AΔlmgl​ ⇒lΔl​=AYmg​ ∴lΔl​=3×10−6×10111×10​ =0.3×10−4

$0.3 \times 1 0^{- 4}$

$0.3 \times 1 0^{- 3}$

$0.3 \times 1 0^{3}$

$0.3 \times 1 0^{4}$

$Y = \frac{m g l}{A Δ l}$
$\Rightarrow \frac{Δ l}{l} = \frac{m g}{A Y}$
$∴ \frac{Δ l}{l} = \frac{1 \times 10}{3 \times 1 0 ^{- 6} \times 1 0 ^{11}}$
$= 0.3 \times 1 0^{- 4}$