Question

The radius of a right circular cylinder increases at a constant rate. Its altitude is a linear function of the radius and increases three times as fast as radius. When the radius is $1 cm$ the altitude is $6 cm$. When the radius is $6 cm$, the volume is increasing at the rate of $1 Cu cm / sec$. When the radius is $36 cm$, the volume is increasing at a rate of $n cu. cm / sec$. The value of ' $n$ ' is equal to:

• A. 12
• B. 22
• C. 30
• D. 33

Answer 1

Answer: (D)

$ \frac{ dr }{ dt }= c \text { and } h = ar + b ; \text { also } \frac{ dh }{ dt }=3 \frac{ dr }{ dt } \text { (given) }$
$\therefore a \frac{ dr }{ dt }=3 \frac{ dr }{ dt } \Rightarrow a =3 $
$\text { hence } h =3 r + b$
$\text { when } r =1 ; h =6 \Rightarrow 6=3+ b \Rightarrow b =3 $
$\therefore h =3( r +1) $
$V =\pi r ^2 h =3 \pi r ^2( r +1)=3 \pi\left( r ^3+ r ^2\right) $
$\frac{ dV }{ dt }=3 \pi\left(3 r ^2+2 r \right) \frac{ dr }{ dt } $

$\text { where } r =6 ; \frac{ dV }{ dt }=1 \,cc / \sec$
$\therefore 1=3 \pi(108+12) \frac{ dr }{ dt } \Rightarrow 360 \pi \frac{ dr }{ dt }=1$
$\text { again when } r =36, \frac{ dV }{ dt }= n $
$n =3 \pi\left((3.36)^2+2.36\right) \frac{ dr }{ dt }$
$n =3 \pi \cdot 36(110) \cdot \frac{1}{360 \pi} $
$n =33 \Rightarrow \text { (D) } $