JEE MAIN - Physics (2021 - 18th March Morning Shift - No. 25)
The time period of a simple pendulum is given by $$T = 2\pi \sqrt {{l \over g}} $$. The measured value of the length of pendulum is 10 cm known to a 1mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1s resolution. The percentage accuracy in the determination of 'g' using this pendulum is 'x'. The value of 'x' to be nearest integer is :-
2%
3%
5%
4%
Explanation
$$T = 2\pi \sqrt {{l \over g}} $$
$${T^2} = 2\pi \left( {{l \over g}} \right)$$
$$g = 2\pi {l \over {{T^2}}}$$
$${{\Delta g} \over g} = {{\Delta l} \over l} + {{2\Delta T} \over T}$$
$${{\Delta g} \over g} = {{1 \times {{10}^{ - 3}}} \over {1 \times {{10}^{ - 2}}}} + {{2 \times 1} \over {100}}$$
$${{\Delta g} \over g} = 0.02 + 0.01 = 0.03$$
$$100 \times {{\Delta g} \over g} = 0.03 \times 100 = 3\% $$
$${{\Delta g} \over g} \times 100 = 3\% $$
$${T^2} = 2\pi \left( {{l \over g}} \right)$$
$$g = 2\pi {l \over {{T^2}}}$$
$${{\Delta g} \over g} = {{\Delta l} \over l} + {{2\Delta T} \over T}$$
$${{\Delta g} \over g} = {{1 \times {{10}^{ - 3}}} \over {1 \times {{10}^{ - 2}}}} + {{2 \times 1} \over {100}}$$
$${{\Delta g} \over g} = 0.02 + 0.01 = 0.03$$
$$100 \times {{\Delta g} \over g} = 0.03 \times 100 = 3\% $$
$${{\Delta g} \over g} \times 100 = 3\% $$
Comments (0)
