A completely filled hemispherical tank of radius R has an orifice of small area a at its bottom. Time required to completely empty the tank is (p √2 ⋅ π R5 / 2/q a √g). Find (p+q). (Assume that the top surface area of the liquid is always much greater than the orifice area)

Question

A completely filled hemispherical tank of radius R has an orifice of small area a at its bottom. Time required to completely empty the tank is (p √2 ⋅ π R5 / 2/q a √g). Find (p+q). (Assume that the top surface area of the liquid is always much greater than the orifice area) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/3a23034ee98252e2e3f842c5b416814f-.png" /> Q=a v=π x2(-(d h/d t)) a √2 g h d t=-π x2 d h x2=R2-(R-h)2

Q. A completely filled hemispherical tank of radius $R$ has an orifice of small area $a$ at its bottom. Time required to completely empty the tank is $\frac{p 2 \cdot π R ^{5/2}}{q a g}$ . Find $(p + q)$ .
(Assume that the top surface area of the liquid is always much greater than the orifice area)

$Q = a v = π x^{2} (- \frac{d h}{d t})$
$a 2 g h d t = - π x^{2} d h$
$x^{2} = R^{2} - (R - h)^{2}$

Q. A completely filled hemispherical tank of radius R has an orifice of small area a at its bottom. Time required to completely empty the tank is qag​p2​⋅πR5/2​. Find (p+q). (Assume that the top surface area of the liquid is always much greater than the orifice area)

Q=av=πx2(−dtdh​) a2gh​dt=−πx2dh x2=R2−(R−h)2

$Q = a v = π x^{2} (- \frac{d h}{d t})$
$a 2 g h d t = - π x^{2} d h$
$x^{2} = R^{2} - (R - h)^{2}$