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A cylindrical vessel of radius r is filled with a liquid to a height h such that the force exerted by the liquid on the sides of the vessel is equal to that exerted at the bottom. The ratio \mathrm{\frac{h}{r}} is

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

best_answer

Pressure at the bottom \mathrm{ =h \rho g}. Therefore, the force exerted at the bottom is \mathrm{ F_1=h \rho g \times \pi r^2.}
Now, since the mass of the liquid can be regarded as being concentrated at its centre of mass which is at a height \mathrm{ h / 2}, the pressure on the sides \mathrm{ =\frac{h}{2} \rho g}.

Therefore, force exerted on sides is

                     \mathrm{ \begin{aligned} F_2 & =\frac{h}{2} \rho g \times 2 \pi r h . \text { Given } F_1=F_2, \text { or } \\\\ h \rho g \times \pi r^2 & =\frac{h}{2} \rho g \times 2 \pi r h \end{aligned} }

which gives \mathrm{ r=h}. Hence the correct choice is (a).

Posted by

seema garhwal

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