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KEAM 2014 Math Paper

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Question : 35 of 120

Marks: +1, -0

The radius of a cylinder is increasing at the rate of 5 cm/min so that its volume is constant.When its radius is 5 cm and height is 3 cm the rate of decreasing of its height is

6 cm/min
3 cm/min
4 cm/min
5 cm/min
2 cm/min

Solution:

Given,

\frac{dr}{dt}=5\ \text{cm}/\text{min}

\because

Volume of cylinder

V=\pi r^{2} h

....(i)

On differentiating both sides w.r.t.

t

, we get

\frac{dV}{dt}=2\pi r \frac{dr}{dt} \cdot h

.....(ii)

Again, differentiating Eq. (i) w.r.t.

h,

we get

\frac{dV}{dt}=\pi r^{2} \frac{dh}{dt}

....(iii)

From Eqs. (ii) and (iii),

2\pi r \frac{dr}{dt} \cdot h=\pi r^{2} \frac{dh}{dt}

\Rightarrow \ 2 \frac{dr}{dt} \cdot h = r \frac{dh}{dt}

\Rightarrow \ 2 \cdot 5 \cdot 3 = 5 \frac{dh}{dt}

\therefore \ \frac{dh}{dt}=6 \text{ cm}/\text{min}

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