A spherical ball of radius $1 \times 10^{-4} \mathrm{~m}$ and of density 104 $\mathrm{kg} \mathrm{m}^{-3}$…

A spherical ball of radius $1 \times 10^{-4} \mathrm{~m}$ and of density 104 $\mathrm{kg} \mathrm{m}^{-3}$ falls freely under gravity through a distance ' $h$ ' before entering a tank of water. After' entering water if the velocity of the ball does not change, then ' $h$ ' is
  1. 20.4 cm
  2. 20.4 mm
  3. 20.4 m
  4. 10.2 m

Solution

$\mathrm{r}=1 \times 10^{-4} \mathrm{~m}, \sigma=10^4 \mathrm{~kg} \mathrm{~m}^{-3}$
Terminal speed, $v=\frac{2}{9} \frac{\mathrm{gr}^2}{\eta}(\sigma-\delta)$ $\begin{aligned} & =\frac{2}{9} \times \frac{9.8 \times\left(10^{-4}\right)^2}{9.8 \times 10^{-6}}\left(10^4-10^3\right) \\ & =20 \mathrm{~m} / \mathrm{s} \end{aligned}$
By conservation of mechanical energy, $\begin{aligned} & \frac{1}{2} \mathrm{mv}^2=\mathrm{mgh} \\ & \Rightarrow \mathrm{~h}=\frac{\mathrm{v}^2}{2 \mathrm{~g}}=\frac{(20)^2}{2 \times 9.8}=20.4 \mathrm{~m} \end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 2)

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