A block of weight 100 N is suspended by copper and steel wires of same cross sectional area 0.5 cm2 and, length √3 m and 1 m, respectively. Their other ends are fixed on a ceiling as shown in figure. The angles subtended by copper and steel wires with ceiling are 30° and 60°, respectively. If elongation in copper wire is (Δ lc) and elongation in steel wire is (Δ ls), then the ratio (Δ lc/Δ lc) is [Youngâs modulus for copper and steel are 1×1011 N/m2 and 2×1011 N/m2, respectively.]

Question

A block of weight 100 N is suspended by copper and steel wires of same cross sectional area 0.5 cm2 and, length √3 m and 1 m, respectively. Their other ends are fixed on a ceiling as shown in figure. The angles subtended by copper and steel wires with ceiling are 30° and 60°, respectively. If elongation in copper wire is (Δ lc) and elongation in steel wire is (Δ ls), then the ratio (Δ lc/Δ lc) is [Youngâs modulus for copper and steel are 1×1011 N/m2 and 2×1011 N/m2, respectively.] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(T1/2)=(T2√3/2) T1=√3T2 (Δ l2/Δ l1)=(T2l2/A2Y2)×(A1Y1/T1Y1) (Δ l2/Δ l1)=(√3T2/T1)×(2/1)=2

2T1​​=2T2​3​​ T1​=3​T2​ Δl1​Δl2​​=A2​Y2​T2​l2​​×T1​Y1​A1​Y1​​ Δl1​Δl2​​=T1​3​T2​​×12​=2

$\frac{T _{1}}{2} = \frac{T _{2} 3}{2}$
$T_{1} = 3 T_{2}$
$\frac{Δ l _{2}}{Δ l _{1}} = \frac{T _{2} l _{2}}{A _{2} Y _{2}} \times \frac{A _{1} Y _{1}}{T _{1} Y _{1}}$
$\frac{Δ l _{2}}{Δ l _{1}} = \frac{3 T _{2}}{T _{1}} \times \frac{2}{1} = 2$