    Question

# In the experiment to determine the acceleration due to gravity g the formula used for the time period of a periodic motion is T=2π√7(R−r)5g. The values of R and r are measured to be (60±1) mm and (10±1) mm respectivley. 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?

A
The error in the measurement of r is 10%
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
The error in the measurement of T is 3.57%
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
The error in the measurement of T is 2%
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
The error in the determined value of g is 11%
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct options are A The error in the measurement of r is 10% B The error in the measurement of T is 3.57% D The error in the determined value of g is 11%S.NOTAbsolute error=|T−Tmean|10.520.0420.560.0030.570.0140.540.0250.590.03Tmean=2.785(ΔT)mean=0.02Tmean=0.56 % error in T=0.020.56×100=3.57% % error in r=drr×100=110×100=10% According to the question g∝T2R−r dgg=2dTT+dR+drR−r dgg=2(3.57%)+1+160−10×100% dgg=11%  Suggest Corrections  0    Join BYJU'S Learning Program
Select...  Related Videos   Rate of Change
MATHEMATICS
Watch in App  Explore more
Join BYJU'S Learning Program
Select...