A block of weight 100 N is suspended by copper and steel wires of same cross sectional area and, length 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

Question

A block of weight 100 N is suspended by copper and steel wires of same cross sectional area $0.5 {cm}^{2}$ and, length 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 $(Δ l_{c})$ and elongation in steel wire is $(Δ l_{s})$ , then the ratio $\frac{Δ l_{c}}{Δ l_{s}}$ is _______.

[Young's modulus for copper and steel are $1 \times 10^{11} {N/m}^{2}$ and $2 \times 10^{11}$ ${N/m}^{2}$ , respectively.] [2019]

Smart Achievers · Accepted Answer

(2)

$Given : l_{c} = \sqrt{3} m, l_{s} = 1 m, Y_{c} = 1 \times 10^{11} {N/m}^{2}$ $and Y_{s} = 2 \times 10^{11} {N/m}^{2}$

$At equilibrium, T_{s} \cos 60 ° = T_{c} \cos 30 °$

$\Rightarrow \frac{T_{s}}{2} = \frac{T_{c} \sqrt{3}}{2}$

$\Rightarrow T_{s} = \sqrt{3} T_{c} \Rightarrow \frac{T_{c}}{T_{s}} = \frac{1}{\sqrt{3}}$

$∴$ $\frac{l_{c}}{l_{s}} = \frac{\sqrt{3}}{1}$

$and \frac{Y_{c}}{Y_{s}} = \frac{1 \times 10^{11}}{2 \times 10^{11}} = \frac{1}{2}$

$From, Y = \frac{F l}{A Δ l} \Rightarrow Δ l = \frac{F l}{A Y}$

$Here, A_{s} = A_{c}$

$∴$ $\frac{Δ l_{c}}{Δ l_{s}} = (\frac{T_{c}}{T_{s}}) \times (\frac{l_{c}}{l_{s}}) \times (\frac{Y_{s}}{Y_{c}}) = (\frac{1}{\sqrt{3}}) \times (\frac{\sqrt{3}}{1}) \times (\frac{2}{1}) = 2$

Solution

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Solution

Ans. (2) Given : lc=3 m, ls=1 m, Yc=1×1011 N/m2 and Ys=2×1011 N/m2 At equilibrium, Tscos60°=Tccos30° ⇒Ts2=Tc32 ⇒Ts=3 Tc⇒TcTs=13 ∴ lcls=31 and YcYs=1×10112×1011=12 From, Y=FlAΔl⇒Δl=FlAY Here, As=Ac ∴ ΔlcΔls=(TcTs)×(lcls)×(YsYc)=(13)×(31)×(21)=2

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