The relative error in the determination of the surface area of a sphere is $\alpha$. Then the relative error in the determination of its volume is

The mass of an object is measured as $225 \pm 0.05 \, g$. Find the percentage error in this measurement. (in $\%$)

The relative error in resistivity of a material where resistance $= 1.05 \pm 0.01 \, \Omega$,diameter $= 0.60 \pm 0.01 \, mm$,and length $= 75.3 \pm 0.1 \, cm$ is:

The International Avogadro Coordination project created the world's most perfect sphere using Silicon in its crystalline form. The diameter of the sphere is $9.4 \,cm$ with an uncertainty of $0.2 \,nm$. The atoms in the crystals are packed in cubes of side $a$. The side is measured with a relative error of $2 \times 10^{-9}$,and each cube has $8$ atoms in it. Then,the relative error in the mass of the sphere is closest to (assume molar mass of Silicon and Avogadro's number to be known precisely)

The resistance $R = \frac{V}{I}$ where $V = (100 \pm 5) \text{ V}$ and $I = (10 \pm 0.2) \text{ A}$. The percentage error in $R$ is: (in $\%$)

The error in the measurement of resistance, when $(10 \pm 0.5) \text{ A}$ current passing through it produces a potential difference of $(100 \pm 6) \text{ V}$ across it, is (in $\%$)