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QJEEM18Phe947: JEE Main 2018 Physics - Class 11 JEEM18Phe947 Solution: As shown in the figure, forces of 10 5 N each are applied in…

As shown in the figure, forces of 105 N each are applied in opposite directions, on the upper and lower faces of a cube of side 10 cm, shifting the upper face parallel to itself by 0.5 cm. If the side of another cube of the same material is 20 cm, then under similar conditions as above, the displacement will be :
JEE Main 2018 (Online) 15th April Evening Slot Physics - Properties of Matter Question 234 English

Multiple Choice Options

  • A 0.25 cm
  • B 0.37 cm
  • C 0.75 cm
  • D 1.00 cm

Step-by-Step Verified Solution

Given : Force applied F = 105 N; side of cube x = 10 m = 10 ×\times 10 −- 2 m; shift dx = 0.5 cm = 0.5 ×\times = 10 −- 2 m

JEE Main 2018 (Online) 15th April Evening Slot Physics - Properties of Matter Question 234 English Explanation

On cube of side x = 10 cm;

Stress = Fx2=105(10×10−2)2=107{F \over {{x^2}}} = {{{{10}^5}} \over {{{(10 \times {{10}^{ - 2}})}^2}}} = {10^7} N/m2

Strain = dxx=0.5×10−210×10−2=0.5{{dx} \over x} = {{0.5 \times {{10}^{ - 2}}} \over {10 \times {{10}^{ - 2}}}} = 0.5

On cube of side x' = 20 cm;

Stress' = Fx′2=105(20×10−2)2{F \over {x{'^2}}} = {{{{10}^5}} \over {{{(20 \times {{10}^{ - 2}})}^2}}}

Strain' = dx′x′=dx′20×10−2{{dx'} \over {x'}} = {{dx'} \over {20 \times {{10}^{ - 2}}}}

Now, it is given that the material is same for both cubes, therefore,

StressStrain=Stress′Strain′⇒1070.05=105(20×10−2)×20×10−2dx′{{Stress} \over {Strain}} = {{Stress'} \over {Strain'}} \Rightarrow {{{{10}^7}} \over {0.05}} = {{{{10}^5}} \over {(20 \times {{10}^{ - 2}})}} \times {{20 \times {{10}^{ - 2}}} \over {dx'}}

⇒107×1025=10520×10−21dx′\Rightarrow {{{{10}^7} \times {{10}^2}} \over 5} = {{{{10}^5}} \over {20 \times {{10}^{ - 2}}}}{1 \over {dx'}}

⇒dx′=10520×10−2×5107×102=5×10−220=0.25×10−2\Rightarrow dx' = {{{{10}^5}} \over {20 \times {{10}^{ - 2}}}} \times {5 \over {{{10}^7} \times {{10}^2}}} = {{5 \times {{10}^{ - 2}}} \over {20}} = 0.25 \times {10^{ - 2}} m

⇒dx′=0.25\Rightarrow dx' = 0.25 cm

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