If an error of k% is made in measuring the radius of a sphere, then percentage error in its volume is

Question:

If an error of k% is made in measuring the radius of a sphere, then percentage error in its volume is

(a) $k \%$

(b) $3 k \%$

(c) $2 k \%$

 

(d) $k / 3 \%$

Solution:

(b) 3k%
Let x be the radius of the sphere and y be its volume.
Then,

$\frac{\Delta x}{x} \times 100=k$

Also, $y=\frac{4}{3} \pi x^{3}$

$\Rightarrow \frac{d y}{d x}=4 \pi x^{2}$

 

$\Rightarrow \frac{\Delta y}{y}=\frac{4 \pi x^{2}}{y} d x=\frac{4 \pi x^{2}}{\frac{4}{3} \pi x^{3}} \times \frac{k x}{100}$

$\Rightarrow \frac{\Delta y}{y} \times 100=3 k$

Hence, the error in the volume is $3 k \%$.

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