Explanation
The terminal velocity $$V$$ of the body of radius $$r$$, density $$\rho$$ falling through a medium of density $$\rho_o$$ is given by $$V=\displaystyle\frac{2r^2(\rho-\rho_o)g}{9 \eta}$$ where $$\eta$$ is the coefficient of viscosity of medium. clearly $$V \propto r^2 $$$$\Rightarrow V \propto (volume)^{2/3}$$$$\Rightarrow V \propto (m)^{2/3}$$$$\Rightarrow \dfrac{V_2}{V_1} =\left (\frac{m_2}{m_1} \right)^{2/3} $$$$\Rightarrow \dfrac{V_2}{V} =\left (\frac{8m}{m} \right)^{2/3} $$$$\Rightarrow V_2=4V$$
The tube is wider earlier and it becomes narrow as it progresses. Pressure in the thin tube will be more compared to that of the thick region of the tube. So, there would be enough pressure due to the thin region, so that liquid will move in that vertical part. But as the liquid is moving forward, its velocity will decrease. So, figure D is incorrect. And height in the second vertical part will be less compared to the height in the first vertical tube.
$$Answer:$$
Hence, option A is the correct answer.
Please disable the adBlock and continue. Thank you.