The volume ($V$) of a sphere is given by the formula: $V = \frac{4}{3}\pi r^3$, where $r$ is the radius.
Since the radius ($r$) is half the diameter ($d$), $r = \frac{d}{2}$. Substituting this into the volume formula gives the volume in terms of diameter:
$V = \frac{4}{3}\pi \left(\frac{d}{2}\right)^3 = \frac{4}{3}\pi \frac{d^3}{8} = \frac{\pi}{6}d^3$.
This shows that the volume ($V$) is proportional to the cube of the diameter ($d^3$), meaning $V \propto d^3$.
When a quantity depends on a power of a variable, the percentage error propagates accordingly. If $V \propto d^n$, the relationship between the percentage errors is:
$ \frac{\Delta V}{V} \times 100\% = n \left( \frac{\Delta d}{d} \times 100\% \right) $
In this problem, the volume depends on the cube of the diameter ($V \propto d^3$), so the exponent $n=3$. We are given the percentage error in the diameter measurement:
$ \frac{\Delta d}{d} \times 100\% = 2\% $
Now, we apply the error propagation rule:
$ \text{Percentage Error in Volume} = 3 \times (\text{Percentage Error in Diameter}) $
$ \frac{\Delta V}{V} \times 100\% = 3 \times (2\%) $
$ \frac{\Delta V}{V} \times 100\% = 6\% $
Therefore, the percentage error in the calculated volume of the sphere is $6\%$.
| List - I | List - II |
| A. Meter (L) | I. $\sqrt{\frac{hc}{G}}$ |
| B. Second (S) | II. $\sqrt{\frac{Gh}{c^{5}}}$ |
| C. Kilogram (M) | III. $\sqrt{\frac{K^{2}L^{2}c^{3}}{Gh}}$ |
| D. Kelvin (K) | IV. $\sqrt{\frac{Gh}{c^{3}}}$ |
A man in a car at location Q on a straight highway is moving with speed v. He decides to reach a point P in qa field at a distance d from the highway (point m) as shown in the figure. Speed of the car in the field is half to that on the highway. What should be the distance RM, so that the time taken to reach P is minimum ?

| List - I | List - II |
| A. Meter (L) | I. $\sqrt{\frac{hc}{G}}$ |
| B. Second (S) | II. $\sqrt{\frac{Gh}{c^{5}}}$ |
| C. Kilogram (M) | III. $\sqrt{\frac{K^{2}L^{2}c^{3}}{Gh}}$ |
| D. Kelvin (K) | IV. $\sqrt{\frac{Gh}{c^{3}}}$ |