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(D) The thermal force $F$ produced in a wire when it is prevented from expanding due to a temperature change $\Delta 	heta$ is given by the formula:$

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$3/2$

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(D) The thermal force $F$ produced in a wire when it is prevented from expanding due to a temperature change $\Delta 	heta$ is given by the formula:$F = YA\alpha \Delta 	heta$where $Y$ is Young's modulus,$A$ is the area of cross-section,$\alpha$ is the coefficient of linear expansion,and $\Delta 	heta$ is the change in temperature.Given that $Y$,$A$,and $\Delta 	heta$ are the same for both wires $A$ and $B$,the force $F$ is directly proportional to the coefficient of linear expansion $\alpha$:$F \propto \alpha$Therefore,the ratio of the forces is:$\frac{F_A}{F_B} = \frac{\alpha_A}{\alpha_B}$Given that $\alpha_A = \frac{3}{2} \alpha_B$,we have:$\frac{F_A}{F_B} = \frac{3}{2}$Thus,the ratio of the forces produced in the two wires is $3/2$.

$S$.No.	$l (m)$	$\Delta T (^{\circ}C)$	$\Delta l (m)$
$(1)$	$2$	$10$	$4 \times 10^{-4}$
$(2)$	$1$	$10$	$4 \times 10^{-4}$
$(3)$	$2$	$20$	$2 \times 10^{-4}$
$(4)$	$3$	$10$	$6 \times 10^{-4}$

Two wires $A$ and $B$ of the same length,same area of cross-section,and having the same Young's modulus are heated to the same temperature range. If the coefficient of linear expansion of wire $A$ is $3/2$ times that of wire $B$,what is the ratio of the thermal forces produced in the two wires?

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