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

JEE Main · Physics · Mechanical Properties of Solids

A compressive force is applied to a uniform rod of rectangular...

A compressive force is applied to a uniform rod of rectangular cross-section so that its length decreases by 1%1 \%. If the Poisson's ratio for the material of the rod be 0.2,0.2, which of the following statements is correct? "The volume approximately ...........

Options

  1. Adecreases by 1%1 \%
  2. Bdecreases by 0.8%0.8 \%
  3. Cdecreases by 0.6%0.6 \%
  4. Dincreases by 0.2%0.2 \%

Correct answer

C. decreases by 0.6%0.6 \%

Step-by-step solution

Given, Decrement in the length =1%=1 \% Poisson's ratio for material of the rod, σ=0.2\sigma=0.2 As we know that, Volume, v=πr2l,[v=\pi r^{2} l,[ where ,l, l is the length of the rod ]ΔVV=2Δrr+Δll...(]\frac{\Delta V}{V}=\frac{2 \Delta r}{r}+\frac{\Delta l}{l}...(i) since, Poisson's ratio, σ=−ΔD/DΔL/L=−Δr/rΔL/L\sigma=\frac{-\Delta D / D}{\Delta L / L}=\frac{-\Delta r / r}{\Delta L / L}So, Eq. (i) can be written as, ΔVV=Δll(1−2σ)ΔVV×100=(Δll×100)×(1−2σ)=−1×[1−2×(0.2)]=−1×[1−0.4]=−0.6%\begin{aligned}\frac{\Delta V}{V}=\frac{\Delta l}{l}(1-2 \sigma)\\ \frac{\Delta V}{V} \times100 &= \left(\frac{\Delta l}{l} \times100\right)\times(1-2 \sigma)\\ &=-1 \times[1-2 \times(0.2)] \\ &=-1 \times[1-0.4] \\ &=-0.6 \% \end{aligned}Here, negative sign shows the decrement in the volume.$
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