## IIT JEE Main 2016 Physics Question No 3

A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90s, 91s, 95 s and 92 s. If The minimum division in the measuring clock is 1s, then the reported mean time should be :

(1) 92 +/- 1.8 s

(2) 92 +/- 3 s

(3) 92 +/- 2 s

(4) 92 +/- 5.0 s

Solution

Mean value of observations i.e. time is obtained by,

T_{mean} = (90 + 91 + 95 + 92)/4

=> T_{mean} = 92s

Also absolute errors are as follows:

E_{1} = |92 – 90| = 2

E_{2} = |92 – 91| = 1

E_{3} = |92 – 95| = 3

E_{4} = |92 – 92| = 0

Therefore, mean absolute error = (2 + 1 + 3 + 0)/4

= 1.5 s, but since the least count of the clock is 1 second, so it can’t count 1.5 s. Hence, it will count 2 seconds, as an error.

(92 +/- 2) seconds is the correct option.

Correct answer to given problem No 3 of IIT JEE Main Physics paper of 2016 is option (3).