There is a hole of area $\frac{1}{25}{\mathrm{cm}}^2$ in the bottom of a cylindrical vessel containing fluid up to height h, The liquid flows out in time t. If the liquid were filled in the vessel up to height 4h,then it would flow out in time

Let A and a be the cross-sectional areas of the vessel and hole respectively. Let h be the height of water in the vessel at time. Let $(-\frac{\mathrm{dh}}{\mathrm{dt}})$
fall of level.
Then, $\mathrm{A}(-\frac{\mathrm{dh}}{\mathrm{dt}}) = \mathrm{\alpha v} = \mathrm{a}\sqrt{2\mathrm{gh}}$

or $-\frac{\mathrm{dh}}{\sqrt{\mathrm{h}} } = \frac{\mathrm{\alpha }\sqrt{2\mathrm{g}}}{\mathrm{A}}\mathrm{dt}$

$-{\int }_{\mathrm{A}}^0\frac{1}{\sqrt{\mathrm{h}}}\mathrm{dh} = \frac{\mathrm{a}\sqrt{2\mathrm{g}}}{\mathrm{A}}{\int }_0^{\mathrm{g}}\mathrm{dt}$

$-(-2\sqrt{\mathrm{h}}) = \frac{\mathrm{a}\sqrt{2\mathrm{g}}}{\mathrm{A}}\mathrm{t}$

or t = $\frac{\mathrm{A}}{\mathrm{\alpha }}\frac{1}{\sqrt{2}\mathrm{g}}\times 2\sqrt{\mathrm{h}} \mathrm{or} \mathrm{t} = \frac{\mathrm{A}}{\mathrm{\alpha }}\sqrt{\frac{2\mathrm{h}}{\mathrm{g}}}$

Now, t $\mathrm{\alpha } \sqrt{\mathrm{h}}$
When & is quadrupled, r is doubled