Question 190: Properties of Matter

Question

As shown in the figure, forces of  10
5
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 :

Accepted Answer

Given : Force applied F = 10
5
N; side of cube x = 10 m = 10 $$	imes$$ 10
$$-$$2
m; shift dx = 0.5 cm = 0.5 $$	imes$$ = 10
$$-$$2
m
On cube of side x = 10 cm;
Stress = $${F \over {{x^2}}} = {{{{10}^5}} \over {{{(10 	imes {{10}^{ - 2}})}^2}}} = {10^7}$$ N/m
2
Strain = $${{dx} \over x} = {{0.5 	imes {{10}^{ - 2}}} \over {10 	imes {{10}^{ - 2}}}} = 0.5$$
On cube of side x' = 20 cm;
Stress' = $${F \over {x{'^2}}} = {{{{10}^5}} \over {{{(20 	imes {{10}^{ - 2}})}^2}}}$$
Strain' = $${{dx'} \over {x'}} = {{dx'} \over {20 	imes {{10}^{ - 2}}}}$$
Now, it is given that the material is same for both cubes, therefore,
$${{Stress} \over {Strain}} = {{Stress'} \over {Strain'}} \Rightarrow {{{{10}^7}} \over {0.05}} = {{{{10}^5}} \over {(20 	imes {{10}^{ - 2}})}} 	imes {{20 	imes {{10}^{ - 2}}} \over {dx'}}$$
$$ \Rightarrow {{{{10}^7} 	imes {{10}^2}} \over 5} = {{{{10}^5}} \over {20 	imes {{10}^{ - 2}}}}{1 \over {dx'}}$$
$$ \Rightarrow dx' = {{{{10}^5}} \over {20 	imes {{10}^{ - 2}}}} 	imes {5 \over {{{10}^7} 	imes {{10}^2}}} = {{5 	imes {{10}^{ - 2}}} \over {20}} = 0.25 	imes {10^{ - 2}}$$ m
$$ \Rightarrow dx' = 0.25$$ cm

Answer

0.25 cm

Answer

0.37 cm

Answer

0.75 cm

Answer

1.00 cm

Properties of Matter

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