The radius of a sphere is (5.3 ± 0.1) cm. The percentage error in its volume is

The correct option is (3x0.1/5.3)x100.

Step-by-Step Explanation:

1. Volume of a Sphere:
The volume $V$ of a sphere of radius $r$ is given by the formula:
$V=\frac{4}{3}\pi r^3$
In this formula, $\frac{4}{3}$ and $\pi$ are constants and do not contribute to the measurement error.

2. Error Propagation:
When a quantity is raised to a power, its relative (fractional) error is multiplied by the power. Since $V\propto r^3$, the fractional error in volume $\frac{\Delta V}{V}$ is related to the fractional error in radius $\frac{\Delta r}{r}$ by:
$\frac{\Delta V}{V}=3\times \frac{\Delta r}{r}$

3. Percentage Error:
The percentage error in volume is obtained by multiplying the fractional error by 100:
$\text{Percentage Error in Volume}=\frac{\Delta V}{V}\times 100=\left(\frac{3\times \Delta r}{r}\right)\times 100$

4. Substituting the Given Values:
The measured radius of the sphere is given as:
$r\pm \Delta r=(5.3\pm 0.1)\text{cm}$
This gives:
- Radius, $r=5.3\text{cm}$
- Absolute error in radius, $\Delta r=0.1\text{cm}$

Substituting these values into the percentage error formula:
$\text{Percentage Error in Volume}=\left(\frac{3\times 0.1}{5.3}\right)\times 100$