The density ($\rho$) of a uniform cylinder is given by the formula:
$ \rho = \frac{\text{Mass}}{\text{Volume}} = \frac{m}{V} $
The volume ($V$) of a cylinder is calculated using its length ($l$) and radius ($r$):
$ V = \pi r^2 l $
Since the diameter ($d$) is given, the radius is $r = \frac{d}{2}$. Substituting this into the volume formula:
$ V = \pi \left(\frac{d}{2}\right)^2 l = \frac{\pi d^2 l}{4} $
Now, substitute the volume formula back into the density formula:
$ \rho = \frac{m}{\frac{\pi d^2 l}{4}} = \frac{4m}{\pi d^2 l} $
To find the fractional error in density ($\frac{\Delta \rho}{\rho}$), we use the error propagation rules for multiplication and division. Constants like $4$ and $\pi$ do not contribute to the error. The formula becomes:
$ \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2 \frac{\Delta d}{d} + \frac{\Delta l}{l} $
We are given the following measurements:
$ \frac{\Delta m}{m} = \frac{0.02}{97.42} \approx 0.000205 $
$ \frac{\Delta d}{d} = \frac{0.02}{20.20} \approx 0.000990 $
$ \frac{\Delta l}{l} = \frac{0.05}{8.35} \approx 0.005988 $
Now, substitute these values into the total fractional error formula:
$ \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2 \frac{\Delta d}{d} + \frac{\Delta l}{l} $
$ \frac{\Delta \rho}{\rho} \approx 0.000205 + 2 \times (0.000990) + 0.005988 $
$ \frac{\Delta \rho}{\rho} \approx 0.000205 + 0.001980 + 0.005988 $
$ \frac{\Delta \rho}{\rho} \approx 0.008173 $
To find the percentage fractional error, multiply by 100%:
$ \text{Percentage Error} = \frac{\Delta \rho}{\rho} \times 100\% \approx 0.008173 \times 100\% \approx 0.8173\% $
Rounding to two decimal places gives 0.82%.
| 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. $\sqrt{\frac{Gh}{c^{3}}}$ |