Question

refer to the attachment for the question​

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept\;:-}}}$

Here the Concept of Coefficient of Elasticity of the material has been used. We know that the coefficient of Elasticity is same for all solids made of same material. First we will find the Stress and Strain of First cube. Then assuming the change in dimension of second Cube, we will find Stress and Strain of second cube. After this, we will equate them of Coefficient of Elasticity and find our answer.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{Stress\;=\;\bf{\dfrac{Force}{Area}\;=\;N\:m^{-2}}}}$

$\\\;\boxed{\sf{Strain\;=\;\bf{\dfrac{Change\;in\;Dimension}{Original\;Dimension}}}}$

$\\\;\boxed{\sf{Coefficient\;of\;Elasticity\;=\;\bf{\dfrac{Stress}{Strain}\;=\;K\;\;\tt{(constant)}}}}$

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★ Solution :-

Given,

» Force applied on each cube = 10⁵ N

» Side of first cube = original dimension = side₁ = 10 cm = 0.1 m

» Side of second cube = original dimension = side₂ = 20 cm = 0.2 m

» Change in dimension of first cube = shift₁ = 0.5 cm = 0.5 × 10⁻² m

» Shape of side of cube = Square

• Let the change ib dimension of second cube that is shift₂ be 'x'. This x is also the displacement in the second cube.

Then,

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~ Stress and Strain of First Cube ::

→ Stress₁ -

$\\\;\;\sf{\odot\;\;Stress\;=\;\bf{\dfrac{Force}{Area}\;=\;N\:m^{-2}}}$

Since the side of cube is Square in shape. Then Area of Square = (Side)².

$\\\;\;\sf{:\rightarrow\;\;Stress_{1}\;=\;\bf{\dfrac{Force}{Area}}}$

$\\\;\;\sf{:\rightarrow\;\;Stress_{1}\;=\;\bf{\dfrac{10^{5}}{(0.1)^{2}}}}$

$\\\;\;\sf{:\rightarrow\;\;Stress_{1}\;=\;\bf{\dfrac{10^{5}}{0.01}}}$

$\\\;\;\underline{\sf{:\rightarrow\;\;Stress_{1}\;=\;\bf{10^{7}\;\;N\:m^{2}}}}$

→ Strain₁ -

$\\\;\;\sf{\odot\;\;Strain\;=\;\bf{\dfrac{Change\;in\;Dimension}{Original\;Dimension}}}$

$\\\;\;\sf{:\Longrightarrow\;\;Strain_{1}\;=\;\bf{\dfrac{0.5\;\times\;10^{-2}}{0.1}}}$

$\\\;\;\sf{:\Longrightarrow\;\;Strain_{1}\;=\;\bf{\dfrac{0.5\;\times\;10^{-1}}{1}}}$

$\\\;\;\underline{\sf{:\Longrightarrow\;\;Strain_{1}\;=\;\bf{0.05}}}$

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~ Stress and Strain of Second Cube ::

→ Stress₂ -

$\\\;\;\sf{\odot\;\;Stress\;=\;\bf{\dfrac{Force}{Area}\;=\;N\:m^{-2}}}$

$\\\;\;\sf{:\rightarrow\;\;Stress_{2}\;=\;\bf{\dfrac{Force}{Area}}}$

$\\\;\;\sf{:\rightarrow\;\;Stress_{2}\;=\;\bf{\dfrac{10^{5}}{(0.2)^{2}}}}$

$\\\;\;\sf{:\rightarrow\;\;Stress_{2}\;=\;\bf{\dfrac{10^{5}}{0.04}}}$

$\\\;\;\underline{\bf{:\rightarrow\;\;Stress_{2}\;=\;\bf{25\;\times\;10^{5}\;\;N\:m^{2}}}}$

→ Strain₂ -

$\\\;\;\sf{\odot\;\;Strain\;=\;\bf{\dfrac{Change\;in\;Dimension}{Original\;Dimension}}}$

$\\\;\;\underline{\bf{:\Longrightarrow\;\;Strain_{2}\;=\;\bf{\dfrac{x}{0.2}}}}$

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~ For displacement in Second Cube (x) ::

$\\\;\;\sf{\odot\;\;Coefficient\;of\;Elasticity\;=\;\bf{\dfrac{Stress}{Strain}\;=\;K\;\;\tt{(constant)}}}$

$\\\;\;\sf{:\mapsto\;\;Coefficient\;of\;Elasticity_{(First\;Cube)}\;=\;\bf{\dfrac{Stress_{1}}{Strain_{1}}\;=\;K\;\;\tt{(constant)}}}$

$\\\;\;\sf{:\mapsto\;\;Coefficient\;of\;Elasticity_{(Second\;Cube)}\;=\;\bf{\dfrac{Stress_{2}}{Strain_{2}}\;=\;K\;\;\tt{(constant)}}}$

Equating, these both we get,

$\\\;\;\sf{:\mapsto\;\;\dfrac{Stress_{1}}{Strain_{1}}\;=\;\bf{\dfrac{Stress_{2}}{Strain_{2}}}}$

$\\\;\;\sf{:\mapsto\;\;\dfrac{10^{7}}{0.05}\;=\;\bf{\dfrac{25\;\times\;10^{5}}{\bigg(\dfrac{x}{0.2}\bigg)}}}$

$\\\;\;\sf{:\mapsto\;\;\dfrac{10^{7}}{0.05}\;=\;\bf{\dfrac{25\;\times\;10^{5}\;\times\;0.2}{x}}}$

$\\\;\;\sf{:\mapsto\;\;x\;=\;\bf{\dfrac{25\;\times\;10^{5}\;\times\;0.2\;\times\;0.05}{10^{7}}}}$

$\\\;\;\sf{:\mapsto\;\;x\;=\;\bf{\dfrac{25000}{10000000}}}$

$\\\;\;\sf{:\mapsto\;\;x\;=\;\bf{0.0025\;\;m}}$

$\\\;\;\bf{:\mapsto\;\;x\;=\;\bf{0.25\;\;cm}}$

So, option c.) 0.25 cm is correct.

$\\\;\underline{\boxed{\tt{Displacement\;\;in\;\;Second\;\;Cube\;=\;\bf{0.25\;\;cm}}}}$

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★ More Formulas to know :-

$\\\;\tt{\leadsto\;\;Surface\;\;Tension\;=\;\dfrac{Force}{Length}}$

$\\\;\tt{\leadsto\;\;Surface\;\;Energy\;=\;\dfrac{Work}{Area}}$

$\\\;\tt{\leadsto\;\;Angle\;=\;\dfrac{Arc}{Radius}}$

$\\\;\tt{\leadsto\;\;Velocity\;\;Gradient\;=\;\dfrac{Velocity}{Distance}}$

$\\\;\tt{\leadsto\;\;Pressure\;\;Gradient\;=\;\dfrac{Pressure}{Distance}}$

Answer 2

Answer:

Here the Concept of Coefficient of Elasticity of the material has been used. We know that the coefficient of Elasticity is same for all solids made of same material. First we will find the Stress and Strain of First cube. Then assuming the change in dimension of second Cube, we will find Stress and Strain of second cube. After this, we will equate them of Coefficient of Elasticity and find our answer.