Multiple choice

A hemispherical tank of radius $R$ is completely filled with water. Now an orifice of small area '$a$' is made at the bottom of tank. The time required to empty the tank is

  1. $\displaystyle \frac{14\pi R^{\tfrac 52}}{15\sqrt{2}a\sqrt{g}}$
  2. $\displaystyle \frac{14\pi R^{\tfrac 52}}{\sqrt{2}a\sqrt{g}}$
  3. $\displaystyle \frac{14\pi R^{\tfrac 52}}{5\sqrt{2}a\sqrt{g}}$
  4. $\displaystyle \frac{4\pi R^{\tfrac 52}}{15\sqrt{2}a\sqrt{g}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The time to empty a tank is given by t = (A_tank / a) * sqrt(2/g) * integral(sqrt(y) dy) from 0 to R. For a hemisphere, A_tank = pi * (R^2 - y^2). The integration leads to the standard formula t = (14 * pi * R^(5/2)) / (15 * a * sqrt(2g)).

AI explanation

To find the time to empty the hemispherical tank, we integrate Torricelli's law. The volume of the water is V = (pi)(2Rh^2 - h^3)/3, so dV/dt = (pi)(4Rh - 3h^2)(dh/dt). The rate of outflow is a * sqrt(2gh), leading to dh/dt = -a * sqrt(2gh) / (pi(4Rh - 3h^2)). Integrating this from h = R to h = 0 gives the time T as (14 * pi * R^(5/2)) / (15 * sqrt(2) * a * sqrt(g)).

Keep practicing — related questions