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need solution for RD Sharma maths class 12 chapter Derivative As a Rate Measure exercise 12.2 question 19

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Answer: 0.64 m/min

Hint: \frac{dh}{dt}=\theta Here, Volume of the cone is V=\frac{1}{3} \pi r^{2} h

Given: The height of the cone is 10 meters and the radius of its base is 5 m and water is running into an inverted cone at a rate of \pi cubic meter per minute


Let r be the radius, h be height of the cone and v be the volume of cone at any time of t

Let O be the semi vertical angle of the cone CAB whose height CO is 10m and radius is OB = 5m

\begin{aligned} &\therefore \tan \theta=\frac{O B}{O C}=\frac{5}{10} \\\\ &\tan \theta=\frac{1}{2} \end{aligned}........(i)


Let v be the volume of water in the cone then,

\begin{aligned} &v=\frac{1}{3} \pi\left(O^{\prime} B^{\prime}\right)^{2}\left(C O^{\prime}\right) \\\\ &=\frac{1}{3} \pi(h \tan \theta)^{2}(h) \\\\ &=\frac{1}{3} \pi h^{3} \tan ^{2} \theta \end{aligned}

\begin{aligned} &\Rightarrow \frac{\pi h^{3}}{3}(\tan \theta)^{2} \\\\ &\Rightarrow \frac{\pi h^{3}}{3}\left(\frac{1}{2}\right)^{2}=\frac{\pi h^{3}}{3}\left(\frac{1}{4}\right) \quad\left[\tan \theta=\frac{1}{2}\right] \\\\ &\Rightarrow \frac{\pi h^{3}}{12} \end{aligned}


Differentiate w.r.t 't' then

\frac{d v}{d t}=\frac{\pi}{12} \frac{d}{d t}\left(h^{3}\right)=\frac{\pi}{12} 3 h^{2} \frac{d h}{d t}

\Rightarrow \pi=\frac{\pi}{12} 3 h^{2} \frac{d h}{d t}                    \left[\frac{d v}{d t}=\pi \frac{m^{3}}{\min }\right]

\begin{aligned} &\Rightarrow 1=\frac{h^{2}}{4} \frac{d h}{d t} \\\\ &\Rightarrow \frac{4}{h^{2}}=\frac{d h}{d t} \Rightarrow \frac{d h}{d t}=\frac{4}{h^{2}} \end{aligned}


Now, when the water stands 7.5 m below the base

\begin{aligned} &\text { i.e } 10-7.5=2.5 \mathrm{~m} \\\\ &\frac{d h}{d t}=\frac{4}{h^{2}}=\frac{4}{(2.5)^{2}}=0.64 m/min\end{aligned}


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