Question

A student measures the time period of 100 oscillations of a simple pendulum four times ,
The data set is 0 s , 91 s , 92s and 95 s , if the minimum division in the measuring clock is 1s , then the reported mean time should be

a) => (92+,- 2)s

b)=> (82+,-5)s

c)=> (02 +,- 1.8)s

d) => (92 +,-3)s

Answer 1

Hello pari ,

we know that Arithmetic mean time of a oscillating simple pendulum is given by 

∑ Xi / N = > { 90s+92s+93s+95s }/4   => 92 s 

and we know that mean deviation of a simple pendulum is given by 

∑ | Xbar  -  Xi |/N    = >  {2+1+3+0} /4  = >  1.5 

Given minimum division in the measuring clock , ie . simple pendulum = > 1s

thus the reported mean time of a oscillating simple pendulum (92+,-2) s

so our required answer is A pari  

hopes this will help you !

@ engineer gopal khandelwal b-tech from IIT ROORKEY 

Answer 2

the answer is mentioned above