Question

A cylindrical tank of a radius \[10\]m is being filled with wheat at the rate of \[314\] cubic meter per hour. Then the depth of the wheat is increasing at the rate of
(A \[1\;{m^3}/h\]
(B) \[0.1\;{m^3}/h\]
(C)\[1.1\;{m^3}/h\]
(D) None of these

Answer 1

Hint: To handle such types of questions we will use the underneath steps with the goal that we can make our answer as easy as could be expected so that it will be helpful to save our critical time.
Write down the volume of cylinder to differentiate with respect to time here radius is given So assume it as constant and substitute the value \[\dfrac{{dv}}{{dt}} = 314\] and obtain the value of rate of change in depth.

Complete step-by-step answer:
As we know that the volume of the cylindrical tank
\[V = \pi {r^2}h\]
r = radius of tank
h = depth of tank
Now differentiate the V with respect to t
\[\dfrac{{dv}}{{dt}} = \pi {r^2}\dfrac{{dh}}{{dt}}\]
Now substitute
\[
\Rightarrow \dfrac{{dv}}{{dt}} = 314,\,\,r = 10 \\
\Rightarrow 314 = \pi {(10)^2} \times \dfrac{{dh}}{{dt}} \\
\Rightarrow \dfrac{{dh}}{{dt}} = \dfrac{{314}}{{3.14 \times 100}} \\
\Rightarrow \dfrac{{dh}}{{dt}} = \dfrac{{100}}{{100}} = 1 \;
 \]
Hence, we can say that depts is increasing at \[1\;{m^3}/h\]
So, the correct answer is “Option A”.

Note: In this kind of problems we have to deal with the numerous things and some of them are referenced here which will be truly useful to comprehend the concept:
We have to utilize right formula so as to not turn out to be excessively complex solution:
We need to focus on the fact that radius is constant here.
Don't use \[\pi = \dfrac{{22}}{7}\] or \[\pi = 3.14\] will be cancelled out by\[\dfrac{{dv}}{{dt}} = 314\].