A 20 ,cm long tube is closed at one end. It i | vedclass

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(D) Let the initial pressure inside the tube be $P_0 = 10^5 \,N/m^2$ and the initial volume be $V_1 = A 	imes 20 \,cm$,where $A$ is the cross-section

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$0.2$

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(D) Let the initial pressure inside the tube be $P_0 = 10^5 \,N/m^2$ and the initial volume be $V_1 = A 	imes 20 \,cm$,where $A$ is the cross-sectional area of the tube.When the tube is dipped such that $10 \,cm$ is outside the water,the air column inside the tube is compressed. Let the new length of the air column be $L = (20 - h) \,cm$. The new volume is $V_2 = A 	imes (20 - h) \,cm$.Since the temperature is constant,we apply Boyle's Law: $P_1 V_1 = P_2 V_2$.$10^5 	imes 20 = P 	imes (20 - h) \quad \dots(1)$The pressure $P$ inside the tube at the water level is given by the atmospheric pressure plus the pressure due to the water column outside the tube relative to the air-water interface inside the tube. The depth of the interface below the outer water surface is $(10 - h) \,cm = \frac{10 - h}{100} \,m$.$P = P_0 + ho g \Delta y = 10^5 + 10^3 	imes 10 	imes \frac{10 - h}{100} = 10^5 + 100(10 - h) = 10^5 + 1000 - 100h \quad \dots(2)$Substituting $(2)$ into $(1)$:$20 	imes 10^5 = (10^5 + 1000 - 100h)(20 - h)$$20 	imes 10^5 = 20 	imes 10^5 - 10^5 h + 20000 - 1000h - 2000h + 100h^2$$0 = 100h^2 - 103000h + 20000$Dividing by $100$:$h^2 - 1030h + 200 = 0$Since $h$ is very small,$h^2 \approx 0$,so $1030h \approx 200 \implies h \approx \frac{200}{1030} \approx 0.194 \,cm$.This is closest to $0.2 \,cm$.

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