Question

The value of ${\sum }_{i=1}^{10}$ r(i) is equal to

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

Answer 1

The correct option is C $5R[2-\frac{11k}{H}]$

Let $\theta$ be the semi-vertical angle of the cone
so that $\tan \theta =\frac{R}{H}$
Let the radius and heigth of the water cone at
time t be r and h respectively.
So $\tan \theta =\frac{r}{h}$
If $V$ is the volume of the water and $S$ is the surfave of the cone in contact
with at time t, then
$V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi r^2\cot \theta$ and $S=\pi r^2$
We are given $\frac{dv}{dt}\propto S$
$\Rightarrow \frac{dv}{dt}=-kS$ ($V$ is decreasing)
$\Rightarrow \frac{1}{3}\pi {3r}^2\cot \theta \frac{dr}{dt}=-k\pi r^2\Rightarrow \frac{dr}{dt}=-k\tan \theta$
Integrating, we get
$r=-\left(k\tan \theta \right)t+c$
When $t=0,r=k\therefore c=R$
Thus $r=-\left(k\tan \theta \right)t+R$
When cone is empty $r=0$
If $T$ is the time taken for the cone to be empty, then
$0+-\left(k\tan \theta \right)T+R\Rightarrow T=\frac{R}{k\tan \theta }=\frac{R}{k\frac{R}{H}}=\frac{H}{k}$
Hence the cone will be empty in time $\frac{H}{k}$
$r\left(1\right)=-k\tan \theta +R=-k\frac{R}{H}+R=R(1-\frac{k}{H})$
$\sum _{i=1}^{10}r\left(i\right)=10R-k\frac{R}{H}\sum _{i=1}^{10}i=10R-k\frac{R}{H}55$