QJEEAdv16Ph7c7e: JEE Advanced 2016 Physics - Class 11 JEEAdv16Ph7c7e Solution: In an experiment to determine the acceleration due to gravity…

The time period is measured for five successive measurements as follows :

T(s)	0.52	0.56	0.57	0.54	0.59
$\Delta$ T	0.04	0	0.01	0.02	0.03

Therefore,

${T_{mean}} = {{0.52 + 0.56 + 0.57 + 0.54 + 0.59} \over 5}$

$= {{2.78} \over 5} = 0.556 \approx 0.56$

and we know $\Delta$ T(absolute error) = $\left| {{T_{mean}} - T} \right|$

Error in reading = $|{T_{mean}} - {T_1}| = 0.04$
$|{T_{mean}} - {T_2}| = 0.00$
$|{T_{mean}} - {T_3}| = 0.01$
$|{T_{mean}} - {T_4}| = 0.02$
$|{T_{mean}} - {T_5}| = 0.03$

$\Rightarrow \Delta {T_{mean}} = {{0.04 + 0 + 0.01 + 0.02 + 0.03} \over 5} = 0.02$

Thus, the error in the measurement of T is

${{\Delta T} \over T} \times 100 = {{0.02} \over {0.56}} \times 100 = 3.57\%$ .... (1)

Hence, option (B) is correct.

The error in the measurement of r is

${{\Delta r} \over r} \times 100 = {1 \over {10}} \times 100 = 10\%$ ..... (2)

Hence, option (A) is correct.

Now, it is given that

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

$\Rightarrow {T^2} = 4{\pi ^2}\left( {{{7(R - r)} \over {5g}}} \right)$

$\Rightarrow g = {{28{\pi ^2}} \over 5}\left( {{{R - r} \over {{T^2}}}} \right)$

Taking log and differentiating on both the sides of this equation, we get

$\ln g = \ln \left( {{{28{\pi ^2}} \over 5}} \right) + \ln (R - r) - 2\ln T$

$\Rightarrow {{\Delta g} \over g} = \left( {{{\Delta R + \Delta r} \over {R - r}}} \right) + {{2\Delta T} \over T}$

$= \left( {{{1 + 1} \over {50}}} \right) + 2 \times 0.0357$

$\Rightarrow {{\Delta g} \over g} \times 100 = \left( {{1 \over {25}} + 2 \times 0.0357} \right) \times 100$

$= (4 + 7.14)\% = 11.14\% = 11\%$

Therefore, the error in the determined value g is 11%.

Answer (A), (B), (D)