Question

If radius of the sphere is (5.3 $\pm$ 0. 1) cm. Then percentage error in its volume will be

• A. $3 + 6.01 \times \frac{ 100 }{ 5.3 }$
• B. $\frac{ 1}{ 3} \times 0.01 \times \frac{100 }{ 5.3 }$
• C. $\bigg( \frac{ 3 \times 0.1 }{ 5.3} \bigg) \times 100$
• D. $\frac{ 0.1 }{ 5.3} \times 100$

Answer 1

Answer: (C)

$\bigg( \frac{ 3 \times 0.1 }{ 5.3} \bigg) \times 100$

Answer 2

Volume of sphere (V) = $\frac{ 4 }{ 3} \pi r^3$ % error m volume = $\frac{ 3 \times \triangle r }{ r } \times 100$ = $\bigg( \frac{ 3 \times 0.1 }{ 5.3 } \bigg) \times 100$