Question

Suppose that water is being emptied from a spherical tank of radius $10$ m. At a particular instance of time if the radius of the water in the tank is $6$ m and is deceasing at the rate of $2$ m/s then, at what rate is the depth of the water in the tank decreasing?

• A. $\frac{3}{2}$ m/s
• B. $-\frac{3}{2}$ m/s
• C. $\frac{3}{4}$ m/s
• D. $-\frac{3}{4}$ m/s

Answer 1

The correct option is A $\frac{3}{2}$ m/s

There are two cases. Level of water can be in the upper or lower half of sphere.

But the level cannot be in the upper half as the radius is decreasing.

Let the depth of water be $h\text{m}$
$OD=10,CD=h$
$OC=10-h$
In $\Delta OAC$
$(10-h{)}^2+r^2={10}^2$
$\Rightarrow h^2-20h+r^2=0$
when $r=6$
$h^2-20h+36=0\Rightarrow (h-18)(h-2)=0$
$\Rightarrow h=18$ or $h=2$
$h=18$ is not possible.
$h{}^{\prime }=\frac{dh}{dt},r{}^{\prime }=\frac{dr}{dt}$
$h^2-20h+r^2=0$
On dfferentiating the above equation, we have -
$2hh{}^{\prime }-20h{}^{\prime }+2rr{}^{\prime }=0$
$\Rightarrow 2h{}^{\prime }-10h{}^{\prime }+6(-2)=0$
$\Rightarrow -8h{}^{\prime }=12\Rightarrow h{}^{\prime }=-\frac{3}{2}\text{m/s}$
$\therefore$ Depth is decreasing at a rate of $\frac{3}{2}\text{m/s}$.