Two identical wires of substances ' P ' and ' Q ' are subjected to equal stretching force along the length. If the elongation of ' Q ' is more than that of ' P ', then (a) both P and Q are equally elastic (b) P is more elastic than Q (c) P is plastic and Q is elastic (d) Q is more elastic than P

Answer: (b). Same force on identical wires: Δ L = FL/AY ∝ 1/Y. Δ LQ > Δ LP ⇒ YP > YQ. A larger Young's modulus means the material resists deformation more, i.e. it is more elastic. So the answer is (b) P is more elastic than Q.

Two identical wires of substances ' P ' and ' Q ' are subjected to… | MHT CET 2019 (2 May, Shift 1)

Q19MHT CET 2019 · 2 MAY · SHIFT 1Easy

Elastic Behaviour, Stress & Strain

Two identical wires of substances ' $P$ ' and ' $Q$ ' are subjected to equal stretching force along the length. If the elongation of ' $Q$ ' is more than that of ' $P$ ', then

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Answer: (b)

Same force on identical wires: $\Delta L = \frac{FL}{AY} \propto \frac{1}{Y}$ .
$\Delta L_Q > \Delta L_P \Rightarrow Y_P > Y_Q$ .
A larger Young's modulus means the material resists deformation more, i.e. it is more elastic.
So the answer is (b) $P$ is more elastic than $Q$ .

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