JEE MAIN - Physics (2021 - 24th February Evening Shift - No. 10)
The period of oscillation of a simple pendulum is $$T = 2\pi \sqrt {{L \over g}} $$. Measured value of 'L' is 1.0 m from meter scale having a minimum division of 1 mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01 s resolution. The percentage error in the determination of 'g' will be :
1.30%
1.33%
1.13%
1.03%
Explanation
Given, $$T = 2\pi \sqrt {{L \over g}} $$ .... (i)
where, time period, T = 1.95 s
Length of string, l = 1 m
Acceleration due to gravity = g
Error in time period, $$\Delta$$T = 0.01 s = 10$$-$$2 s
Error in length, $$\Delta$$L = 1 mm = 1 $$\times$$ 10$$-$$3 m
Squaring Eq. (i) on both sides, we get
$${T^2} = 4{\pi ^2}{L \over g}$$
$$ \Rightarrow g = 4{\pi ^2}{L \over {{T^2}}}$$
$$ \Rightarrow {{\Delta g} \over g} = {{\Delta L} \over L} + {{2\Delta T} \over T} = {{{{10}^{ - 3}}} \over 1} + {{2 \times {{10}^{ - 2}}} \over {1.95}}$$
$$ = {10^{ - 3}} + 1.025 \times {10^{ - 2}}$$
$$ = {10^{ - 3}} + 10.25 \times {10^{ - 3}}$$
$$ = 11.25 \times {10^{ - 3}}$$
$$\because$$ $$\Delta g/g \times 100 = 11.25 \times {10^{ - 3}} \times {10^2}$$
$$ = 1.125\% \simeq 1.13\% $$
where, time period, T = 1.95 s
Length of string, l = 1 m
Acceleration due to gravity = g
Error in time period, $$\Delta$$T = 0.01 s = 10$$-$$2 s
Error in length, $$\Delta$$L = 1 mm = 1 $$\times$$ 10$$-$$3 m
Squaring Eq. (i) on both sides, we get
$${T^2} = 4{\pi ^2}{L \over g}$$
$$ \Rightarrow g = 4{\pi ^2}{L \over {{T^2}}}$$
$$ \Rightarrow {{\Delta g} \over g} = {{\Delta L} \over L} + {{2\Delta T} \over T} = {{{{10}^{ - 3}}} \over 1} + {{2 \times {{10}^{ - 2}}} \over {1.95}}$$
$$ = {10^{ - 3}} + 1.025 \times {10^{ - 2}}$$
$$ = {10^{ - 3}} + 10.25 \times {10^{ - 3}}$$
$$ = 11.25 \times {10^{ - 3}}$$
$$\because$$ $$\Delta g/g \times 100 = 11.25 \times {10^{ - 3}} \times {10^2}$$
$$ = 1.125\% \simeq 1.13\% $$
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