Laws of Motion

JEE Main level, single correct, medium

104 of 210

Question 104 of 210

As shown in the figure, forces of 105 N10^5 \mathrm{~N} each are applied in opposite directions, on the upper and lower faces of a cube of side 10 cm10 \mathrm{~cm}, shifting the upper face parallel to itself by 0.5 cm0.5 \mathrm{~cm}. If the side of another cube of the same material is, 20 cm20 \mathrm{~cm} then under similar conditions as above, the displacement will be:

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Answer: (B) 0.25 cm0.25 \mathrm{~cm}

For same material the ratio of stress to strain is same For first cube
 Stress 1= force 1 area 1=105(0.12)\text { Stress }_1=\frac{\text { force }_1}{\text { area }_1}=\frac{10^5}{\left(0.1^2\right)}
For second block,
 stress 2= force 2 area 2=105(0.22) strain 2= change in length 2 original length =x0.2\begin{aligned} &\text { stress }_2=\frac{\text { force }_2}{\text { area }_2}=\frac{10^5}{\left(0.2^2\right)} \\ &\text { strain }_2=\frac{\text { change in length }{ }_2}{\text { original length }}=\frac{x}{0.2} \end{aligned}
xx is the displacement for second block.
For same material,  stress 1 strain 1= stress 2 strain 2\frac{\text { stress }_1}{\text { strain }_1}=\frac{\text { stress }_2}{\text { strain }_2}
 or, 10.5(0.1)20.5×10−20.1=105(0.2)2x0.2\text { or, } \frac{\frac{10.5}{(0.1)^2}}{\frac{0.5 \times 10^{-2}}{0.1}}=\frac{\frac{10^5}{(0.2)^2}}{\frac{x}{0.2}}

 Solving we get, x=0.25 cm\text { Solving we get, } x=0.25 \mathrm{~cm}

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