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%.
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 ?

Three students $S_1$, $S_2$ and $S_3$ perform an experiment for determining the acceleration due to gravity (g) using a simple pendulum. They use different lengths of pendulum and record time for different number of oscillations. The observations are as shown in the table.

(least count of length =0.1 cm
least count for time =0.1s )
If $E_1$, $E_2$ and $E_3$ are the percentage errors in 'g' for students 1, 2 and 3 respectively, then the minimum percentage error is obtained by student no. __________.
The diameter of a wire measured by a screw gauge of least count $0.001\text{ cm}$ is $0.08\text{ cm}$. The length measured by a scale of least count $0.1\text{ cm}$ is $150\text{ cm}$. When a weight of $100\text{ N}$ is applied to the wire, the extension in length is $0.5\text{ cm}$, measured by a micrometer of least count $0.001\text{ cm}$. The error in the measured Young's modulus is $\alpha \times 10^9\text{ N/m}^2$. The value of $\alpha$ is ________.
(Ignore the contribution of the load to Young's modulus error calculation)
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 ?
