11Ans. : B
g = 4Î 2L/T2
Λg/g = ΛL/L + 2 ΛT/T
E1 = [ 0.1/64 + 2*0.1/128 ]*100 = 5/16 %
E2 = [ 0.1/64 + 2*0.1/64 ]*100 = 15/32 %
E3 = [ 0.1/20 + 2*0.1/36 ]*100 = 19/18 %
Hence ans is B
10 Answers
11One doubt bhaiyya....
If I have two readings of which one has more precision & the other has more accuracy , which has more error ????
1111thnx bhaiyya.
One more thing...
can we give the ans to this ques. directly by saying that no. of oscillations in I is maximum , so error will be minimum ????
11one more thing
so number of trials never comes into picture unless there is same value of percentage error for two different cases?????
11But bhaiyya.....isnt it true that more no. of observations minimises error.
62it does rkrish...
but that cannot alone be the reason.. suppose the errror in the length was too large
i willl give a very hypothetical case
the length of the pendulum was as small as its error... then what would you have done?