Oil is leaking at the rate of 16 mL/s from a vertically kept cylindrical drum containing oil. If the radius of the drum is 7 cm and its height is 60 cm, find the rate at which the level of the oil is changing when the oil level is 18 cm

Options

A. `(-16)/(49 pi)cms^(-1)`
B. `(-16)/(48pi)cms^(-1)`
C. `16/(49)picms^(-1)`
D. `(-16)/(47pi)cms^(-1)`

Answer

A

Explanation

Let `h` be the height of oil level at any instant `t` and `V` be the volume of oil in cylindrical drum.
Given `h=60cm, r=7cm`
and `(dV)/(dt)=-16cm^(3)//s`
`:'V=pir^(2)himplies(dV)/(dt)=pir^(2)(dh)/(dt)=2pir(dr)/(dt)`
(Since `r` is constant all the time i.e. `(dr)/(dt)=0`)

`implies-16=pi(7)^(2)(dh)/(dt)implies(dh)/(dt)=-16/(49pi)`
At `h=18, ((dh)/(dt))_(h=18)=-16/(49 pi)`
So height of oil is decreasing at the rate fo `(16/49pi)cms^(-1)`