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Question: The magnitude of force developed by raising the temperature from \[{{0}^{{}^\circ }}C\] to \[{{100}^...

The magnitude of force developed by raising the temperature from 0C{{0}^{{}^\circ }}C to 100C{{100}^{{}^\circ }}C of an iron bar 100cm long and 1cm21c{{m}^{2}} cross section when it is held so that it is not permitted to expand or bend is (α=12×106per C and Y=20×1011dyne/cm2)(\alpha =12\times {{10}^{-6}}per{{\text{ }}^{{}^\circ }}C\text{ and }Y=20\times {{10}^{11}}dyne/c{{m}^{2}})
A. 10N B. 102N C. 24×103N D. 104N \begin{aligned} & \text{A}\text{. 10N} \\\ & \text{B}\text{. 1}{{\text{0}}^{2}}N \\\ & \text{C}\text{. 24}\times \text{1}{{\text{0}}^{3}}N \\\ & \text{D}\text{. 1}{{\text{0}}^{4}}N \\\ \end{aligned}

Explanation

Solution

Hint: When a metal bar is heated it will expand. If we stopped it from further expansion then large forces will be set up within the bar. This force is equal to the force needed to compress the bar to its original length. This force of expansion is given by the formulaF=YAαΔTF=YA\alpha \Delta T, where Y is Young’s modulus of the bar, A is the area of cross section, α\alpha is the coefficient of linear expansion and ΔT\Delta Tis the change in temperature.
Formula used:
The Young’s modulus of a wire is given as
Y=FlAΔlY=\dfrac{Fl}{A\Delta l}
Where,
F is the force of expansion acting on a wire,
l is the length of wire
A represents the area of cross section of wire
Δl\Delta l is the change in length of wire

Complete step by step answer:
The Young’s modulus of a wire is given as
Y=FlAΔlY=\dfrac{Fl}{A\Delta l}
Where,
F is the force of expansion acting on a wire,
l is the length of wire
A represents the area of cross section of wire
Δl\Delta l is the change in length of wire
If α\alpha is the coefficient of linear expansion and ΔT\Delta Tis the change in temperature of wire then change in length can be written as
Δl=αlΔT\Delta l=\alpha l\Delta T
Therefore,
Y=FlAαlΔT Y=FAαΔT \begin{aligned} & Y=\dfrac{Fl}{A\alpha l\Delta T} \\\ & Y=\dfrac{F}{A\alpha \Delta T} \\\ \end{aligned}
Rearranging the terms,
F=YAαΔTF=YA\alpha \Delta T
Given that
α=12×106per C \alpha =12\times {{10}^{-6}}per{{\text{ }}^{{}^\circ }}C\text{ }
Y=20×1011dyne/cm2Y=20\times {{10}^{11}}dyne/c{{m}^{2}}
Area of cross section A= 1cm21c{{m}^{2}}
Increase in temperature = ΔT=100C0C=100C\Delta T={{100}^{{}^\circ }}C-{{0}^{{}^\circ }}C={{100}^{{}^\circ }}C
Substituting these values in the equation, we get,
F=20×1011×1×12×106×100 F=24×108dyne \begin{aligned} & F=20\times {{10}^{11}}\times 1\times 12\times {{10}^{-6}}\times 100 \\\ & F=24\times {{10}^{8}}dyne \\\ \end{aligned}
We have,
1dyne=105N1dyne={{10}^{-5}}N
Therefore,

& F=24\times {{10}^{8}}\times {{10}^{-5}}N \\\ & F=24\times {{10}^{3}}N \\\ \end{aligned}$$ Thus, the force developed by raising the temperature is $$24\times {{10}^{3}}N$$ Answer- C. $\text{24}\times \text{1}{{\text{0}}^{3}}N$ Note: From the given formula we can say that the force of expansion is directly proportional to the area of cross section and change in temperature. When the solids are heated, molecules within it start vibrating with greater amplitude and exerting more force. This force is responsible for the expansion of solids when they are heated. The degree of expansion is characterized by its coefficient of expansion factor. If a solid has more coefficient of expansion then it will expand more.