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A steel rod $25\;{\text{cm}}$ long has a cross sectional area of $0.8\;{\text{c}}{{\text{m}}^{\text{2}}}$ . Force required to stretch this rod by the same amount as the by the expansion produced by heating it through ${10^\circ }$ is
(Coefficient of linear expansion of steel is ${10^{ - 5}}{/^\circ }{\text{C}}$ and Young’s modulus of steel is $2 \times {10^{10}}\;{\text{N/}}{{\text{m}}^{\text{2}}}$ ).
A) ${\text{160}}\;{\text{N}}$
B) ${\text{360}}\;{\text{N}}$
C) ${\text{106}}\;{\text{N}}$
D) ${\text{260}}\;{\text{N}}$

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Last updated date: 27th Feb 2024
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IVSAT 2024
Answer
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Hint:- The ratio of the change in length to the original length can be termed as the strain. The coefficient of linear expansion is proportional to strain and inversely proportional to the change in temperature. From the expression for the Young’s modulus of the steel rod, the force can be found.

Complete Step by step answer:
Given the length of the steel rod is $l = 25\;{\text{cm = 25}} \times {\text{1}}{{\text{0}}^{ - 2}}\;{\text{m}}$, cross sectional area is$A = 0.8\;{\text{c}}{{\text{m}}^{\text{2}}} = 0.8\; \times {10^{ - 4}}\;{{\text{m}}^{\text{2}}}$, coefficient of linear expansion is $\alpha = {10^{ - 5}}{/^\circ }{\text{C}}$, Young’s modulus of steel is $Y = 2 \times {10^{10}}\;{\text{N/}}{{\text{m}}^{\text{2}}}$ and change in temperature is $\Delta t = {10^\circ }{\text{C}}$.
The expression for coefficient of linear expansion is given as,
$
  \alpha = \dfrac{{\Delta l}}{{l \times \Delta t}} \\
  \dfrac{{\Delta l}}{l} = \alpha \times \Delta t \\
$
Where, $\alpha$ is the coefficient of linear expansion, $\Delta t$ is the change in temperature and $\dfrac{{\Delta l}}{l}$ is the ratio of change in length to the actual length. This can be called the strain.
Substituting the values in the above expression,
$
  \dfrac{{\Delta l}}{l} = {10^{ - 5}}{/^\circ }{\text{C}} \times {10^\circ }{\text{C}} \\
  {\text{ = 1}}{{\text{0}}^{ - 4}} \\
$
The stain is obtained as ${\text{1}}{{\text{0}}^{ - 4}}$.
The expression for the young’s modulus is given as,
$Y = \dfrac{{F \times l}}{{A \times \Delta l}}$
Where, $F$ is the force required to stretch the steel rod of cross sectional area $A$ and length $l$ by $\Delta l$ .
From the above expression,
$
  F = \dfrac{{Y \times A \times \Delta l}}{l} \\
   = Y \times A \times \dfrac{{\Delta l}}{l} \\
$
Substituting the values in the above expression,
$
  F = 2 \times {10^{10}}\;{\text{N/}}{{\text{m}}^{\text{2}}} \times 0.8\; \times {10^{ - 4}}\;{{\text{m}}^{\text{2}}} \times {10^{ - 4}} \\
   = 160\;{\text{N}} \\
$
Therefore the force needed to stretch the steel rod is $160\;{\text{N}}$.

The answer is option A.

Note: We have to note that when the strain is a larger value, the force applied to stretch will be a larger value. And the measure of elasticity is the young’s modulus of the material.