Question

The volume $V$ and depth $x$ of water in a vessel are connected by the relation $V=5x-\frac{x^2}{6}$ and the volume of water is increasing at the rate of $5$ $\raisebox{1ex}{$c{m}^3$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$, when $x=2cm$.

The rate at which the depth of water is increasing is

• A. $\frac{5}{18}$ $\raisebox{1ex}{$cm$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$
• B. $\frac{1}{4}$$\raisebox{1ex}{$cm$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$
• C. $\frac{5}{16}$$\raisebox{1ex}{$cm$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option:

Given that,

The volume $V$ and depth $x$ of water in a vessel are connected by the relation $V=5x-\frac{x^2}{6}$

The volume of water is increasing at the rate of $5$ $\raisebox{1ex}{$c{m}^3$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$, when $x=2cm$

The volume of water in the vessel is given as $V=5x-\frac{x^2}{6}$

Differentiating it with respect to time $\left(t\right)$ we get

$\frac{dV}{dt}=5\frac{dx}{dt}-2\times \frac{x}{6}\times \frac{dx}{dt}$ $\left[\because \frac{d}{dx}\left(x^2\right)=2x\right]$

$\Rightarrow$$\frac{dV}{dt}=\frac{dx}{dt}\left(5-\frac{x}{3}\right)$

$\Rightarrow$$\frac{dx}{dt}=\frac{{\displaystyle \frac{dV}{dt}}}{\left(5-{\displaystyle \frac{x}{3}}\right)}$

where, $\frac{dx}{dt}$ is the rate of change of depth and $\frac{dV}{dt}$ is the rate of change of volume

At $x=2cm$ it is given that $\frac{dV}{dt}=$$5$ $\raisebox{1ex}{$c{m}^3$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$

Substituting these values we get

$\Rightarrow$$\frac{dx}{dt}=\frac{{\displaystyle 5}}{\left(5-{\displaystyle \frac{2}{3}}\right)}$

$\Rightarrow$$\frac{dx}{dt}=\frac{{\displaystyle 5}}{{\displaystyle \frac{13}{3}}}$

$\Rightarrow$$\frac{dx}{dt}=\frac{{\displaystyle 15}}{{\displaystyle 13}}$ $\raisebox{1ex}{$cm$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$

Thus the depth of the water in the vessel is increasing at $\frac{15}{13}$ $\raisebox{1ex}{$cm$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$ when the depth is $2cm$.

Hence, option $\left(D\right)$ i.e. None of these is the correct option