JEE Advance - Physics (2016 - Paper 2 Offline - No. 6)
In an experiment to determine the acceleration due to gravity g, the formula used for the time period of a
periodic motion is $$T = 2\pi \sqrt {{{7\left( {R - r} \right)} \over {5g}}} $$. The values of R and r are measured to be (60 $$ \pm $$ 1) mm and (10 $$ \pm $$ 1)
mm, respectively. In five successive measurements, the time period is found to be 0.52 s, 0.56 s, 0.57 s,
0.54 s and 0.59 s. The least count of the watch used for the measurement of time period is 0.01 s. Which of
the following statement(s) is(are) true?
The error in the measurement of r is 10 %
The error in the measurement of T is 3.57 %
The error in the measurement of T is 2 %
The error in the determined value of g is 11 %
Penjelasan
The time period is measured for five successive measurements as follows :
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)
| 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)