Question

If error in measuring the edge of a cube is $k$% then the percentage error in estimating its volume is

• A. $k$
• B. $3k$
• C. $\frac{k}{3}$
• D. none of these

Answer 1

Answer: (B)

$3k$

Let the actual length of the cube be a.

Therefore the measured length of the cube will be

$=a(1\pm 0.0k)$

$=a(1\pm \frac{k}{100})$

Considering positive error,

$a^{\prime }=a(1+\frac{k}{100})$

$V^{\prime }=a^3(1+\frac{k}{100}{)}^3$

$=a^3(1+3(\frac{k}{100})+3(\frac{k}{100}{)}^2+\left(\frac{k}{100}{)}^3\right)$

Since $\frac{k}{100}<<1$, hence we neglect the higher order terms.

Thus

$V^{\prime }=a^3(1+3(\frac{k}{100}\left)\right)$

Actual volume V

$V=a^3$

Therefore

$V^{\prime }-V=a^3(1+\frac{3k}{100})-a^3$

$=a^3\left(\frac{3k}{100}\right)$

$\frac{V^{\prime }-V}{V}=\frac{a^3\frac{3k}{100}}{a^3}$

$=\frac{3k}{100}$

$=\frac{3k}{100}$

$\frac{V^{\prime }-V}{V}\times 100=3k$

Therefore percentage error in volume is $3k$.