Elongation due to a suspended weight

JEE Advanced 2019 Paper 1, Question 14

A block of weight 100 N is suspended by copper and steel wires of same cross sectional area 0.5 cm^2 and, length \sqrt{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^{\circ} and 60^{\circ}, respectively.
If elongation in copper wire is \left(\Delta l_{C}\right) and elongation in steel wire is \left(\Delta l_{S}\right), then the ratio \frac{\Delta l_{C}}{\Delta 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.]

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We will label the tension in the steel and copper wires by F_S and F_C respectively (see figure below). For equilibrium, the horizontal component of the forces must balance each other out. That is,

(1)   \begin{equation*}   F_S \cos 60^\circ = F_C \cos 30^\circ \implies F_S = \sqrt{3} F_C . \end{equation*}

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The Young’s modulus of a material is the ratio of stress (F/A) to strain (\Delta l/l). That means

(2)   \begin{equation*}   Y = \frac{F/A}{\Delta l / l} \implies \Delta l = \frac{F l}{A Y}. \end{equation*}

where A is the cross-sectional area of the wire, which is the same for both wires in our problem. Thus,

(3)   \begin{equation*}   \frac{\Delta l_C}{\Delta l_S}     = \frac{F_C l_C}{A Y_C} \times \frac{A Y_S}{F_S l_S}     = \frac{F_C \times \sqrt{3} \, {\rm m}}{1 \times 10^{11} \, {\rm N/m}^2} \times       \frac{2 \times 10^{11} \, {\rm N/m}^2}{(\sqrt{3} F_C) \times  1 \, {\rm m}}     = 2 . \end{equation*}

Notice that we did not have to determine the exact value of the forces at all, only the ratio between the forces.

General tip: In order to save time, find the symbolic expression for the quantity you are asked to calculate, and then figure out what parameters you actually need. The problem might give you more information than you actually require.

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