Question

if error is in measurement of radius of sphere is 1% what will be the error in measurement of volume

Answer 1

Volume = 4πR³/3

ΔV/V × 100 = 3(∆R / R) × 100
= 3 × (1/100) × 100
= 3

∴ Percentage error in volume is 3%

Answer 2

The percentage error in volume is 3%

Given:

Measurement error = 1%

To find:

Volume error = ?

Solution:

For a sphere, the volume, $V=\left(\frac{4}{3}\right) \pi r^{3}$

Taking log, we get,

$\log V=3 \log r+\log \left(\frac{4}{3} \pi\right) \Rightarrow \log V=3 \log r+\log \left(\frac{4}{3} \pi\right)$

On differentiating both sides, we get,

$\frac{d V}{V}=3\left(\frac{d r}{r}\right)$

If dr = 0.01 = 1%

Then

$\frac{d V}{V}=3(0.01)$

${\text {Volume}=\frac{4}{3} \pi R^{3}}$

$\Rightarrow \frac{\Delta V}{V} \times 100=3\left(\frac{\Delta R}{R}\right) \times 100$

$=3 \times\left(\frac{1}{100}\right)$

$\therefore \frac{\Delta V}{V}=0.03=3 \%$