A thermometer bulb has a volume of 10^{-6} ,m | vedclass

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(B) The change in volume of the mercury is given by $\Delta V = V_0 \gamma \Delta T$. Given: $V_0 = 10^{-6} \,m^3$, $\gamma = 18 	imes 10^{-5} /{ }^{

Vedclass · Accepted Answer

$9$ cm

Vedclass · Answer

(B) The change in volume of the mercury is given by $\Delta V = V_0 \gamma \Delta T$. Given: $V_0 = 10^{-6} \,m^3$, $\gamma = 18 	imes 10^{-5} /{ }^{\circ} C$, and $\Delta T = 100^{\circ} C - 0^{\circ} C = 100^{\circ} C$. Substituting the values: $\Delta V = (10^{-6} \,m^3) 	imes (18 	imes 10^{-5} /{ }^{\circ} C) 	imes (100^{\circ} C) = 18 	imes 10^{-9} \,m^3$. The cross-sectional area of the stem $A = 0.002 \,cm^2 = 0.002 	imes 10^{-4} \,m^2 = 2 	imes 10^{-7} \,m^2$. The change in volume is also equal to the area times the length of the mercury column: $\Delta V = A 	imes L$. Therefore, $L = \frac{\Delta V}{A} = \frac{18 	imes 10^{-9} \,m^3}{2 	imes 10^{-7} \,m^2} = 9 	imes 10^{-2} \,m = 9 \,cm$.

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