A $20 \,cm$ long tube is closed at one end. It is held vertically, and its open end is dipped in water until only half of it is outside the water surface. Consequently, water rises in it by height $h$ as shown in the figure. The value of $h$ is closest to .............. $\,m / s$ (assume that the temperature remains constant, $P _{\text {armosphere }}=10^5 \,N / m ^2$, density. of water $=10^3 \,kg / m ^3$, and acceleration due to gravity $g =10 \,m / s ^2$ )
KVPY 2021, Advanced
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(D)

Let final pressure inside tube be $P$

Apply $P_1 V_1=P_2 V_2$

$10^5 \times 20= P \times(20- h ) \ldots(1)$

Pressure Equation

$P = P _0+\rho g \frac{(10- h )}{100}$

$P =10^5+10^3 \times 10 \times \frac{(10- h )}{100}$

$P =10^5\left(1+\frac{10- h }{10^3}\right) \ldots(2)$

Substitute $(2)$ in $(1)$

$10^5 \times 20=10^5\left(1+\frac{10-h}{10^3}\right)(20- h )$

Solve for $h$

$h =0.2 \,cm$

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