1. Given relation

   $T = 2\pi \sqrt{\frac{7(R-r)}{5g}}$

   Rearranging for $g$:

   $g = \frac{28\pi^2 (R-r)}{5T^2}$

   So, for percentage error:

   $\frac{\Delta g}{g} = \frac{\Delta (R-r)}{(R-r)} + 2\frac{\Delta T}{T}$

2. Error in measurement of $r$

   Given:

   $r = (10 \pm 1)\text{ mm}$

   Hence percentage error in $r$ is

   $\frac{1}{10}\times 100 = 10\%$

   Therefore, Option A is correct.

3. Mean value of time period $T$

   The five readings are:

   $0.52,\ 0.56,\ 0.57,\ 0.54,\ 0.59\text{ s}$

   Mean:

   $\bar T = \frac{0.52+0.56+0.57+0.54+0.59}{5}$

   $\bar T = \frac{2.78}{5} = 0.556\text{ s}$

4. Error in $T$

   Mean absolute error:

   $\Delta T_{\text{mean}} = \frac{|0.52-0.556|+|0.56-0.556|+|0.57-0.556|+|0.54-0.556|+|0.59-0.556|}{5}$

   $= \frac{0.036+0.004+0.014+0.016+0.034}{5}$

   $= \frac{0.104}{5} = 0.0208\text{ s}$

   Least count error of watch = $0.01\,$s.

   We take the total error in $T$ as the larger/measured observational error, i.e. about $0.02\,$s.

   Thus percentage error in $T$:

   $\frac{0.0208}{0.556}\times 100 \approx 3.74\%$

   This is very close to $3.57\%$, which is obtained if we round mean time as $0.56\,$s and error as $0.02\,$s:

   $\frac{0.02}{0.56}\times 100 = 3.57\%$

   Therefore, Option B is correct and Option C is incorrect.

5. Error in $(R-r)$

   Given:

   $R = (60\pm1)\text{ mm}, \qquad r=(10\pm1)\text{ mm}$

   So,

   $R-r = 60-10 = 50\text{ mm}$

   Maximum absolute error in $(R-r)$:

   $\Delta(R-r)=\Delta R + \Delta r = 1+1=2\text{ mm}$

   Hence percentage error:

   $\frac{2}{50}\times100 = 4\%$

6. Error in $g$

   $\frac{\Delta g}{g} = \frac{\Delta(R-r)}{R-r} + 2\frac{\Delta T}{T}$

   $= 4\% + 2(3.57\%)$

   $= 4\% + 7.14\% = 11.14\%$

   Approximately,

   $\boxed{11\%}$

   Therefore, Option D is correct.

Final evaluation of options

• A: True
• B: True
• C: False
• D: True

So the correct options are:

$\boxed{A,\ B,\ D}$

7. Comparison with stored answer

   Stored correct answer: B, D, A

   Our derived answer: A, B, D

   These are the same set of options.