Question

The radius of a sphere is measured to be $(7.50\pm 0.85)cm$.

Suppose the percentage error in its volume is $x$.

The value of $x$, to the nearest integer $x$, is

Answer 1

Calculate the percentage error of volume and its nearest integer

We know that,

Let $V$ be the volume and $r$ be the radius.

So the volume of a sphere is $V=\frac{4}{3}\mathrm{\pi }{r}^3$

Percentage error is defined as the difference between estimated value and the actual value in comprised to the actual value

So, its percentage error in volume is

$\begin{array}{rcl}& \Rightarrow & \frac{∆V}{V}\times 100=3\frac{∆\mathrm{r}}{\mathrm{r}}\times 100\\ & =& 3\left(\frac{0.85}{7.50}\right)\times 100\\ & =& \left(\frac{2.55}{7.50}\right)\times 100\\ & =& 0.34\times 100\\ & =& 34\end{array}$

Hence, the nearest integer is $34$