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(C) Initially,the air column length $L_i = 0.05 \, m = 5 \, cm$. The pressure of the air $P_i = P_{atm} = 76 \, cm$ of $Hg$.When the tube is raised by

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$0.1$

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(C) Initially,the air column length $L_i = 0.05 \, m = 5 \, cm$. The pressure of the air $P_i = P_{atm} = 76 \, cm$ of $Hg$.When the tube is raised by $0.43 \, m = 43 \, cm$,let the new length of the air column be $x \, cm$. The mercury level inside the tube rises by $(48 - x) \, cm$ relative to the initial position,where $48 \, cm$ is the total displacement of the tube relative to the mercury surface.The final pressure of the air inside the tube is $P_f = P_{atm} - (48 - x) = 76 - 48 + x = (28 + x) \, cm$ of $Hg$.Since the temperature remains constant,we apply Boyle's Law: $P_i V_i = P_f V_f$.Assuming the cross-sectional area $A$ is constant:$76 	imes 5 	imes A = (28 + x) 	imes x 	imes A$$380 = 28x + x^2$$x^2 + 28x - 380 = 0$Solving the quadratic equation:$x = \frac{-28 \pm \sqrt{28^2 - 4(1)(-380)}}{2} = \frac{-28 \pm \sqrt{784 + 1520}}{2} = \frac{-28 \pm \sqrt{2304}}{2} = \frac{-28 \pm 48}{2}$Taking the positive root: $x = \frac{20}{2} = 10 \, cm = 0.1 \, m$.

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