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(D) Initial state: The air column length is $L_1 = 8 \ cm$ and the pressure is atmospheric,$P_1 = 76 \ cm$ of $Hg$.Final state: The tube is raised by

Vedclass · Accepted Answer

$16$

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(D) Initial state: The air column length is $L_1 = 8 \ cm$ and the pressure is atmospheric,$P_1 = 76 \ cm$ of $Hg$.Final state: The tube is raised by $46 \ cm$. Let the new length of the air column be $L_2$. Let the height of the mercury column inside the tube above the external mercury level be $x$. The total length of the tube above the external mercury level is $8 + 46 = 54 \ cm$. Thus,$L_2 = 54 - x$.The pressure of the trapped air in the final state is $P_2 = P_0 - x = 76 - x$.Using Boyle's Law $(P_1 V_1 = P_2 V_2)$ and assuming constant cross-sectional area $A$:$76 	imes A 	imes 8 = (76 - x) 	imes A 	imes (54 - x)$$608 = 4104 - 76x - 54x + x^2$$x^2 - 130x + 3496 = 0$Solving the quadratic equation using the formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{130 \pm \sqrt{16900 - 13984}}{2} = \frac{130 \pm \sqrt{2916}}{2} = \frac{130 \pm 54}{2}$$x_1 = 92$ (impossible as $x Therefore,the length of the air column is $L_2 = 54 - 38 = 16 \ cm$.

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