Question

A cylindrical tank of radius 10m is being filled with wheat at the rate of 314 cubic metre per hour. Then the depth of the wheat is increasing at the rate of
A. \[1\dfrac{{{m^3}}}{h}\]
B. \[0.1\dfrac{{{m^3}}}{h}\]
C. \[1.1\dfrac{{{m^3}}}{h}\]
D. \[0.5\dfrac{{{m^3}}}{h}\]
E. None of these

Answer 1

Hint: The derivative is expressed as the rate of change of some quantity. Here in this question the increase rate of change of depth or sides with respect to the time of a cylindrical tank by applying the application of derivative we can find the increase rate of depth with respect to the time using volume of cylindrical tank \[V = \pi {r^2}h\]

Complete step by step solution:
The rate of change of a quantity is one of the applications of derivatives. By applying the differentiation properties and formulas we determine the rate of change of the quantity.
The radius of a cylindrical tank is 10 m. Given the increased rate of change of volume with respect to time i.e., \[\dfrac{{dV}}{{dt}} = 314cubic.m\] per hour. Now we have to find the increased rate of depth? that is \[\dfrac{{dh}}{{dt}}\]
We know the
volume of cylindrical tank i.e., \[V = \pi {r^2}h\] --- (1)
Here V represents the volume, r represents the radius and the h represents the height
Differentiate the equation (1) with respect to the time we have
\[ \Rightarrow \dfrac{{dV}}{{dt}} = \pi {r^2}\dfrac{{dh}}{{dt}}\]
Substituting the value to the above equation we have
\[ \Rightarrow 314 = \pi {(10)^2}\dfrac{{dh}}{{dt}}\]
As we know the value of \[\pi = 3.14\] and writing the equation for \[\dfrac{{dh}}{{dt}}\]
\[ \Rightarrow \dfrac{{dh}}{{dt}} = \dfrac{{314}}{{3.14{{(10)}^2}}}\]
On simplifying we have
\[ \Rightarrow \dfrac{{dh}}{{dt}} = 1\dfrac{{{m^3}}}{h}\]
Therefore the depth of the wheat is increasing at the rate of \[1\dfrac{{{m^3}}}{h}\]

So, the correct answer is “Option A”.

Note: The derivative is known as rate. In application of derivatives we have different topics like rate of change, increasing rate, decreasing rate, tangents, normal etc., Hence by applying the differentiation to the equation then we can obtain the required solution for the question.