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The radius and height of a cylinder are measured at 5cm and 10cm there is an error of 0.02 cm in both the measurements then the error in its volume is
$\left( {\text{A}} \right){\text{ }}2.5\pi c.c$
$\left( {\text{B}} \right){\text{ }}25\pi c.c$
$\left( {\text{C}} \right){\text{ }}0.25\pi c.c.$
$\left( {\text{D}} \right){\text{ }}0.025\pi c.c.$

Last updated date: 13th Jun 2024
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Hint: In this question, we have to find two things.
First, we have to find the volume of the cylinder by using the formula.
Also, we have to find the error in volume; by the way of finding the error is by partially differentiating the required volume.
Finally, we get the required answer.

Formula used:
Volume of the cylinder, $V = \pi {r^2}h...\left( 1 \right)$
Where $r$ is the radius and $h$ is the height.

Complete step by step answer:
It is given that radius is $5cm$ and height is $10cm$
Now, we have to find the volume of the given cylinder.
Here we use the formula for the volume of the cylinder,
So we can write it as, $V = \pi {r^2}h$,
Substitute the given data in the formula we get,
$ \Rightarrow V = \pi \times {(5)^2} \times 10$
On squaring the terms and doing multiplication we get
$ \Rightarrow V = 250\pi c{m^3}$
Now we have to find an error in volume,
So we have to do partial differentiation on the formula for the volume of the cylinder,
So, we can write it as,
The formula for volume of the cylinder, $V = \pi {r^2}h$
We will do partial differentiation on both sides,
$ \Rightarrow \dfrac{{\Delta v}}{v} = \left( {2\dfrac{{\Delta r}}{r} + \dfrac{{\Delta h}}{h}} \right)....\left( 2 \right)$
It is given that errors in both measurements are radius and height is $0.02cm$
So we can write it as $\Delta r = \Delta h = 0.02$
It is given that radius is $5cm$ and height is $10cm$
Putting these values in the above equation$\left( 2 \right)$, we get
$ \Rightarrow \dfrac{{\Delta v}}{v} = \left( {2\dfrac{{0.02}}{5} + \dfrac{{0.02}}{{10}}} \right)$
On multiplying the term we get,
$ \Rightarrow \dfrac{{\Delta v}}{v} = \left( {\dfrac{{0.04}}{5} + \dfrac{{0.02}}{{10}}} \right)$
Let us divide the bracket term and we get,
\[ \Rightarrow \dfrac{{\Delta v}}{v} = \left( {0.008 + 0.002} \right)\]
On adding the two terms we get
$\Rightarrow \dfrac{{\Delta v}}{v} = 0.01$
The volume of the cylinder we calculated in the above steps is $250c{m^3} = 250c.c$, where $c.c$ are cubic centimeters.
At the place of $v$ we will substitute this value, we get
$\Rightarrow \dfrac{{\Delta v}}{{250\pi c.c}} = 0.01$
By cross multiplication,
$\Rightarrow \Delta v = 2.5\pi c.c$

Hence, the error in volume, $\Delta v = 2.5\pi c.c$. Therefore, the correct option is the option (A).

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