Question

If error in measurement of radius is 2%. Find error in measurement of volume of sphere

Answer 1

ANSWER:

Error in measurement of volume of sphere = 6%

GIVEN:

• Error in measurement of radius is 2%.

TO FIND:

• Error in measurement of volume of sphere.

EXPLANATION:

$\boxed{ \large{ \bold{Volume \: of \: the \: sphere = \frac{4}{3} \pi {r}^{3} }}}$

$\dfrac{\Delta r}{r} \times 100 = 2 \%$

NEGLECT 4/3 π AS IT IS A CONSTANT AND DOESN'T CHANGE WITH CHANGE IN VOLUME OR RADIUS.

$\dfrac{\Delta V}{V} = 3 \times \dfrac{\Delta r}{r}$

$\dfrac{\Delta V}{V} \times 100 = 3 \times \dfrac{\Delta r}{r} \times 100$

$\dfrac{\Delta V}{V} \times 100 = 3 \times 2$

$\dfrac{\Delta V}{V} \times 100 = 6\%$

HENCE THE ERROR IN THE VOLUME OF THE SPHERE WILL BE 6%.

Answer 2

★Question:–

If error in measurement of radius is 2%. Find error in measurement of volume of sphere.

★Answer:–

Given:

measurements of radius is 2%

Solution:

Percentage error in radius is given as 2% i.e.∆r / r×100=2 %

Volume of sphere VV4/3πr³

Percentage error in volume

ΔV/V ×100=3× Δr/r×100=3×2=6 %