The radius of a sphere is $7 \text{ cm}$. If an error of $0.08 \text{ cm}^2$ is made in measuring its surface area, then the approximate error (in $\text{cm}^3$) found in its volume is:

The approximate value of $\log _{10} 1002$ is (Given $\log _{10} e = 0.4343$)

If an error of $0.02 \text{ cm}^2$ is found in the surface area of a sphere when its radius is measured as $10 \text{ cm}$, then the approximate error that occurs in the volume of the sphere, in cubic centimetres, is

The approximate value of $\cot ^{-1}(1.001)$ is

The approximate value of $\sin(60^{\circ} 0^{\prime} 10^{\prime \prime})$ is (given that $\sqrt{3}=1.732, 1^{\circ}=0.0175^{c}$):

If the radius of a sphere is measured as $7 \text{ m}$ with an error of $0.02 \text{ m}$, then the approximate error in calculating its volume is (in $\pi \text{ m}^3$)