Question

A Vernier Caliper with a least count of 0.01cm was used to measure diameter of a cylinder as 4cm and a scale (0-15cm) with the least count of 1mm was used to measure a length of 5cm. The % error in the measurement of volume of the cylinder is
$
A)3.0 \\
B)4.0 \\
C)5.0 \\
D)2.5 \\
$

Answer 1

Hint- In this question we will proceed by partial differentiating the formula of volume of cylinder by applying the triple product rule to solve the equation i.e. $\pi {r^2}l$ then we will use the formula for finding the error in the measurement of volume of the cylinder that is $\dfrac{{\Delta V}}{V} \times 100$ . This will help to approach the solution.

Formula used-
$\pi {r^2}l$
$\dfrac{{\Delta V}}{V} \times 100$

Complete step-by-step solution -
Now, we have given that
Least count to measure the diameter of cylinder=0.01cm
Diameter of cylinder=4cm
Length of cylinder=5cm
Least count to measure the length of cylinder=1mm
And we have to find the % error in the measurement of the volume of the cylinder.
Now we know that
Volume of cylinder$ = \pi {r^2}l$
Also we can replace r with $\dfrac{d}{2}$ , As radius is half of diameter.
Therefore,
Volume of cylinder$ = \pi {\left( {\dfrac{d}{2}} \right)^2}l$
Or $V = \dfrac{\pi }{4}{d^2}l$ ----------(i)
Now partially differentiate (i) by applying triple product rule, we get
Now, we know that in partial differentiation the differentiation of constant is 0 .
Thus,
$\dfrac{{\Delta V}}{V} = 0\left( {{d^2}} \right)l + 2 \times \dfrac{{\Delta d}}{d} \times \dfrac{{\partial l}}{{\partial d}} + \dfrac{{\Delta l}}{l} \times \dfrac{{\partial d}}{{\partial l}}$
$\dfrac{{\Delta V}}{V} = 2 \times \dfrac{{\Delta d}}{d} + \dfrac{{\Delta l}}{l}$ (ii) (As in partial differentiation the differentiation of other variable is 1 )
Now, it is given that
$
  \Delta d = 0.01cm \\
  d = 4cm \\
  \Delta l = 1mm = 0.1cm \\
  l = 5cm \\
 $
Putting these values in (ii) we get,
$\dfrac{{\Delta V}}{V} = 2 \times \dfrac{{0.01}}{4} + \dfrac{{0.1}}{5}$
 $ \Rightarrow \dfrac{{\Delta V}}{V} = \dfrac{{0.01}}{2} + \dfrac{{0.1}}{5}$
$\Rightarrow \dfrac{{\Delta V}}{V} = \dfrac{1}{{200}} + \dfrac{1}{{50}}$
$\Rightarrow \dfrac{{\Delta V}}{V} = \dfrac{{1 + 4}}{{200}}$
$\Rightarrow \dfrac{{\Delta V}}{V} = \dfrac{5}{{200}}$
$\Rightarrow \dfrac{{\Delta V}}{V} = \dfrac{1}{{40}}$
Now, we have to find % error of volume thus me multiply both sides by 100.
$ \Rightarrow $ % error$ = \dfrac{{\Delta V}}{V} \times 100 = \dfrac{1}{{40}} \times 100$
% error $ = \dfrac{{10}}{4}$
= 2.5
Thus the correct option is (D).

Note- In order to solve this question one should know what a Vernier caliper is. A Vernier caliper is an instrument which is used to measure the internal dimensions, outside dimensions and depth. It consists of a main scale with two jaws one each on the upper and lower portions.