Question

The radius of a sphere is measured to be $7.50 \pm 0.85\mathrm{cm}$. Suppose the percentage error in its volume is $x$. The value of $x$, to the nearest $x$, is

Answer 1

Hint: On the surface of a sphere, you can discover a point that is equally distant from any other point. The diameter of the sphere is the length of the straightest line that circles its centre. It is double the sphere's radius in length.

Complete answer:
A sphere is a three-dimensional circular object with perfect symmetry. The radius of the square is the line that extends from the centre to the edge. On the surface of a sphere, you can discover a point that is equally distant from any other point. The diameter of the sphere is the length of the straightest line that circles its centre. It is twice as long as the sphere's radius.

Given that,
Radius of the sphere $=(7.50\pm 0.85)$ cm
$\therefore \quad r=7.50$ and $d r=0.85$

Every point on the volume's surface will be equally distant from the volume's centre if the surface area and diameter are doubled. The following mathematical formula is used to determine a sphere's volume:
the volume of a sphere$V=\dfrac{4 \pi r^{3}}{3}$, where r denotes the sphere's radius

Now that we are aware of a sphere's volume,
$V=\dfrac{4 \pi r^{3}}{3}$
Taking log on both sides,
$\ln V=\ln \left(\dfrac{4 \pi}{3}\right)+3 \ln r$

Differentiating both sides,
$\dfrac{d V}{V}=\dfrac{3 d r}{r}$

Percentage error in volume,
$\dfrac{d V}{V} \times 100=\dfrac{3 d r}{r} \times 100=3 \times \dfrac{0.85}{7.50} \times 100=34 \%$
$=\dfrac{3dr}{r}\times 100$

Substitute the values
$=3\times \dfrac{0.85}{7.50}\times 100$
$=34%$

So the answer is $34%$.

Note: According to Cavalieri's Principle, the volumes of the cylinder/cone and hemisphere are equal because they both have the same height. The volume of the cylinder is $\pi R^{3}$, the volume of the cone is, and the volume of the hemisphere is $\dfrac{2}{3} \pi R^{3}$. As a result, the sphere's radius $R$ has volume$\dfrac{4}{3} \pi R^{3}$.