$\mathrm{v}=\sqrt{\frac{\mathrm{T}}{\mu}}$ $...(I)$
where $\mu=\frac{\mathrm{m}}{1}=\frac{\text { mass of string }}{\text { length of string }}$
The tension $\mathrm{T}=\frac{\mathrm{m}}{\ell} \times \mathrm{x} \times \mathrm{g}$ $...(II)$
From $(a)$ and $(b)$
$\frac{d x}{d t}=\sqrt{g x}$
$\mathrm{x}^{-1 / 2} \mathrm{dx}=\sqrt{\mathrm{g}} \mathrm{dt} \quad \therefore \int_{0}^{\ell} \mathrm{x}^{-1 / 2} \mathrm{dx}-\sqrt{\mathrm{g}} \int_{0}^{\ell} \mathrm{dt}$
$2 \sqrt{l}=\sqrt{g} \times t \quad \therefore t=2 \sqrt{\frac{\ell}{g}}=2 \sqrt{\frac{20}{10}}=2 \sqrt{2}$
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| Column - $\mathrm{I}$ | Column - $\mathrm{II}$ |
| $(a)$Rain drops moves downwards with constant velocity. | $(i)$ Viscous liquids |
| $(b)$ Floating clouds at a height in air. | $(ii)$ Viscosity |
| $(iiii)$ Less density |
[Given: $\pi=22 / 7, g=10 ms ^{-2}$, density of water $=1 \times 10^3 kg m ^{-3}$, viscosity of water $=1 \times 10^{-3} Pa$-s.]
$(A)$ The work done in pushing the ball to the depth $d$ is $0.077 J$.
$(B)$ If we neglect the viscous force in water, then the speed $v=7 m / s$.
$(C)$ If we neglect the viscous force in water, then the height $H=1.4 m$.
$(D)$ The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is $500 / 9$.