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

  • Option 1)

    92 ± 2 s

  • Option 2)

    92 ± 5.0 s

  • Option 3)

    92 ± 1.8 s

  • Option 4)

    92 ± 3 s

     

 

Answers (1)

As we learnt in

Time period of oscillation of simple pendulum -

T=2\pi \sqrt{\frac{l}{g}}

- wherein

l = length of pendulum 

g = acceleration due to gravity.

 

Sum of all observation = 90 + 91 + 95 + 92 = 368

Average =\frac{368}{4}=92

Sum of modulus observations = 2 + 1 + 3 = 6

\therefore   Average error =\frac{6}{4}=1.5,  rounded of 2.

\therefore  Final answer =92\pm 2

Correct option is 1.

 

 


Option 1)

92 ± 2 s

This is the correct option.

Option 2)

92 ± 5.0 s

This is an incorrect option.

Option 3)

92 ± 1.8 s

This is an incorrect option.

Option 4)

92 ± 3 s

 

This is an incorrect option.

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