JEE Advanced 2019 · Paper 1 · Physics Q14

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PhysicsNumerical+3Question 14 of 54

A block of weight 100 N is suspended by copper and steel wires of same cross sectional area 0.5 cm² and, length $\sqrt3$ 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 $(\Delta l_C)$ and elongation in steel wire is $(\Delta l_S)$ , then the ratio $\dfrac{\Delta l_C}{\Delta l_S}$ is ______.

[Young's modulus for copper and steel are $1\times10^{11}\ \mathrm{N/m^2}$ and $2\times10^{11}\ \mathrm{N/m^2}$ , respectively.]

A block hangs from two wires attached to a horizontal ceiling: a steel wire of length 1 m making 60° with the ceiling on the left, and a copper wire of length √3 m making 30° with the ceiling on the right; the wires meet at a right angle at the block.

Your answer

Answer and worked solution

Official answer: 2

The wires make $60^\circ$ and $30^\circ$ with the ceiling, so they meet at a right angle at the block.

Tensions.

Horizontal: $T_S\cos60^\circ=T_C\cos30^\circ$ , so $T_S=\sqrt3\,T_C$ .
Vertical: $T_S\sin60^\circ+T_C\sin30^\circ=100$ , so $2T_C=100$ .

That gives $T_C=50$ N and $T_S=50\sqrt3$ N.

Elongations, $\Delta l=\dfrac{TL}{AY}$ (same $A$ ):
$\frac{\Delta l_C}{\Delta l_S}=\frac{T_CL_C/Y_C}{T_SL_S/Y_S}=\frac{50\sqrt3/10^{11}}{50\sqrt3/(2\times10^{11})}=2.$
Answer: 2

Topics:Hooke's law and elastic moduli Free body diagrams

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JEE Advanced 2019 · Paper 1, Physics question 14