The radius of a sphere is measured to be (7.50 ± 0.85) cm. Suppose the percentage error in its volume is x. The value of x, to the nearest x, is .

Question

The radius of a sphere is measured to be (7.50 ± 0.85) cm. Suppose the percentage error in its volume is x. The value of x, to the nearest x, is . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

∵ v =(4/3) π r 3 taking log then differentiate ( dV / V )=3 ( dr / r ) =(3 × 0.85/7.5) × 100 %=34 %

Q. The radius of a sphere is measured to be $(7.50 \pm 0.85) c m$ . Suppose the percentage error in its volume is $x$ . The value of $x$ , to the nearest $x$ , is _______.

$∵ v = \frac{4}{3} π r^{3}$
taking log then differentiate
$\frac{d V}{V} = 3 \frac{d r}{r}$
$= \frac{3 \times 0.85}{7.5} \times 100% = 34%$

Q. The radius of a sphere is measured to be (7.50±0.85)cm. Suppose the percentage error in its volume is x. The value of x, to the nearest x, is _______.

∵v=34​πr3 taking log then differentiate VdV​=3rdr​ =7.53×0.85​×100%=34%

Q. The radius of a sphere is measured to be $(7.50 \pm 0.85) c m$ . Suppose the percentage error in its volume is $x$ . The value of $x$ , to the nearest $x$ , is _______.

$∵ v = \frac{4}{3} π r^{3}$
taking log then differentiate
$\frac{d V}{V} = 3 \frac{d r}{r}$
$= \frac{3 \times 0.85}{7.5} \times 100% = 34%$