Physics And Measurement Question 192

Question: While measuring the acceleration due to gravity by a simple pendulum, a student makes a positive error of 1% in the length of the pendulum and a negative error of 3% in the value of time period. His percentage error in the measurement of g by the relation g=4π2(l/T2) will be

Options:

A) 2%

B) 4%

C) 7%

D) 10%

Show Answer

Answer:

Correct Answer: C

Solution:

[c] Given that Δll× 100=+1%and ΔTT× 100=-3%

Percentage error in the measurement of g is

[4π2lT2]=100×Δll2×ΔTT×100 =1%-2[-3%]=+7%



NCERT Chapter Video Solution

Dual Pane