Question

The radius of sphere is (2.6+-0.1)cm . the percentage error in its volume?......want the process ......

Answer 1

here is the answer. If any problem occurs plz contact

Answer 2

Answer:

The percentage error in volume is 11.54 %.

Explanation:

Given that,

Radius of sphere r= (2.6±0.1) cm

The volume of sphere is defined as:

$V= \dfrac{4}{3}\pi\times r^3$

$V= \dfrac{4}{3}\times3.14\times(2.6)^3$

$V=73.6\ cm^3$

The percentage error in volume is

$\dfrac{\Delta V}{V}=3\times\dfrac{\Delta r}{r}$

$\Delta V=V\times 3\times\dfrac{\Delta r}{r}$

$\Delta V=73.6\times3\times\dfrac{0.1}{2.6}$

$\Delta V=8.49\ cm^3$

The percentage error in volume

$\dfrac{\Delta V}{V}\times100= \dfrac{8.49}{73.6}\times100$

$\dfrac{\Delta V}{V}$=11.54 %

Hence, The percentage error in volume is 11.54 %.