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JEE Main
Exam
JEE Main
Subject
Physics
Chapter
Mechanical Properties of Solids

JEE Main · Physics · Mechanical Properties of Solids

A 20 ~g copper block is suspended by a vertical spring causing 1...

A 20 g20 \mathrm{~g} copper block is suspended by a vertical spring causing 1 cm1 \mathrm{~cm} elongation over the natural length of spring. If a beaker of water is placed below the block so that the copper block is completely immersed in the liquid, the elongation of the spring is (Density of copper 9000 kg m−39000 \mathrm{~kg} \mathrm{~m}^{-3}, Density of water 1000 kg\mathrmm−3, g=10 ms−2\mathrm{kg} \mathrmm^{-3}, \mathrm{~g}=10 \mathrm{~ms}^{-2} )

Options

  1. A0.25 cm0.25 \mathrm{~cm}
  2. B0.15 cm0.15 \mathrm{~cm}
  3. C0.78 cm0.78 \mathrm{~cm}
  4. D0.89 cm0.89 \mathrm{~cm}

Correct answer

D. 0.89 cm0.89 \mathrm{~cm}

Step-by-step solution

Mass of Block, m=20 g=20×10−3 kg\mathrm{m}=20 \mathrm{~g}=20 \times 10^{-3} \mathrm{~kg} Elongation, x=1 cm=0.01 mx=1 \mathrm{~cm}=0.01 \mathrm{~m} Spring elongation after immersed in water, \mathrmx′=\mathrmx^{\prime}= ? Density, ρ=9000 kg/m3\rho=9000 \mathrm{~kg} / \mathrm{m}^3 Density of water, ρω=1000 kg/m3\rho_\omega=1000 \mathrm{~kg} / \mathrm{m}^3 At equilibrium, kx=mgk=mgx=0.002×100.01=\mathrm{kx}=\mathrm{mg}\mathrm{k}=\frac{\mathrm{mg}}{\mathrm{x}}=\frac{0.002 \times 10}{0.01}=2  N\mathrm{~N}/ m\mathrm{m}when mass is immersed in watermg=kx′+(ρωV)gmg=kx′+ρω[mρc]g20×10−3×10=2×\mathrmx′+10009000×20×10−3×100.2=2\mathrmx′+19×0.20.110=\mathrmx′+190\mathrmx′=110−190=0.89\mathrmx′=0.89\begin{aligned}& \mathrm{mg}=\mathrm{kx}^{\prime}+\left(\rho_\omega \mathrm{V}\right) \mathrm{g}\\ & \mathrm{mg}=\mathrm{kx}^{\prime}+\rho_\omega\left[\frac{\mathrm{m}}{\rho_{\mathrm{c}}}\right] \mathrm{g}\\ & 20 \times10^{-3} \times10=2 \times \mathrmx^{\prime}+\frac{1000}{9000} \times20 \times10^{-3} \times10 \\ & 0.2=2 \mathrmx^{\prime}+\frac{1}{9} \times0.2 \\ & \frac{0.1}{10}=\mathrmx^{\prime}+\frac{1}{90}\\ & \mathrmx^{\prime}=\frac{1}{10}-\frac{1}{90}=0.89 \\ & \mathrmx^{\prime}=0.89 \end{aligned}
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