Question

A solid cube is being measured. Percentage error in the measurement of side 3% and Mass 4%, find the percentage error in density
of cube.

Answer 1

$\huge{\bold{\underline{\underline{Solution:-}}}}$

$\large{\bold{\underline{To find}}}$

Percentage error in density = ?.

$\large{\bold{\underline{\underline{Step - by - Step - Explanation}}}}$

We know that ${Density = \frac{Mass}{Volume}}$

${D = \frac{M}{V}}$

${ \therefore D = \frac{M}{{L}^{3}}}$

L = Length of side of the cube.

In terms of percentage error.

${ \frac{ \triangle D}{D} \times 100 = ( \frac{ \triangle M}{M} \times 100) + 3( \frac{ \triangle L}{L} \times 100)}$

given,

percentage error in M = 4%

percentage error in L = 3%

${ \frac{ \triangle D}{D} \times 100 = 4percent + 3 \times 3percent}$

${ \frac{ \triangle D}{D} \times 100 = 4percent + 9percent}$

${ \frac{ \triangle D}{D} \times 100 = 13 percent}$

$\huge{\boxed{\boxed{ \frac{ \triangle D}{D} \times 100 = 13 percent}}}}$

So,the percentage error is 13%.