Question

The value of $\sum _{i=1}^{20}r\left(i\right)$ is equal to

• A. $5R[4-\frac{42k}{H}]$
• B. $4R[2-\frac{11k}{H}]$
• C. $10R[2-\frac{k}{H}]$
• D. $5R[2-\frac{11k}{H}]$

Answer 1

The correct option is A $5R[4-\frac{42k}{H}]$

Let$\theta$ be semi-vertical angle of the cone so that $\tan \theta =\frac{R}{H}$
Let the radius and height of water cone at time $t$ be $r$ and $h$ respectively. So,$\tan \theta =\frac{r}{h}$

If $V$ is the volume of water and $S$ is surface of the cone in contact with air at time $t$, then
$V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi r^3\cot \theta$ and $S=\pi r^2$

We are given that \dfrac{dV}{dt}\propto S\)

$\frac{dV}{dt}=kS$(V is decreasin)

$\Rightarrow \frac{1}{3}\pi \left(3r^2\right)\cot \theta \frac{dr}{dt}=-k\pi r^2$

$\frac{dr}{dt}=-k\tan \theta$

$\therefore r=-\left(k\tan \theta \right)t+C$

$\therefore t=0,r=R$

$r=(-k\tan \theta )t+R$

When cone is empty, $r=0$

$0=(-k\tan \theta )T+R$

$T=\frac{R}{k\tan \theta }=\frac{R}{k\left(R/H\right)}=\frac{H}{k}$

$r\left(2\right)=-2k\tan \theta +R=-k\frac{2R}{H}+R$

$=R(1-\frac{2k}{H})$

$\sum _{i=1}^{20}r\left(i\right)=20R\frac{Rk}{H}\sum _{i=1}^{20}i$

$=20R\frac{Rk}{H}\left(210\right)=5R(4-\frac{42k}{H})$