JEE Main 2021 · 18 March · morning · Single correct · +4 / −1
The time period of a simple pendulum is given by T = 2π √(l/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 :-
- (a) 2%
- (b) 3%
- (c) 5%
- (d) 4%
Answer and solution
Answer (b) 3%
Solution
Physics sir · 4 steps
Write g in terms of l and T, then add the fractional errors, with T counted twice because it is squared.
- Make g the subject. Squaring the formula gives g = (4π² l)/T². Here l has power 1 and T has power 2, and 4π² is an exact constant with no error.
(Δ g)/g = (Δ l)/l + 2 (Δ T)/T
- Length error. Δ l = 1 mm = 0.1 cm and l = 10 cm, so (Δ l)/l = 0.1/10 = 1%.
- Time error. The clock reads the total time t = 100 s to within 1 s. Since T = t/200, the fractional error in T equals that in t: (Δ T)/T = 1/100 = 1%.
- So. (Δ g)/g = 1% + 2 × 1% = 3%, so x = 3 and the answer is (b).
The idea to keep. For a product of powers, the percentage error is each percentage error times the absolute value of its power, all added. Timing many oscillations does not change the fractional error in T.
Answer from the NTA final answer key. Solution by PrepWiser.