A) \[92\pm 2s\]
B) \[92\pm 5.0s\]
C) \[92\pm 1.8s\]
D) \[92\pm 3s\]
Correct Answer: A
Solution :
[a] Measured time period of 100 oscillations are 90 sec, 91 sec, 95 sec and 92 sec. Mean value of time= \[{{t}_{m}}=\frac{90+91+95+92}{4}=92\sec \] Absolute error in measurement \[\left| \Delta {{t}_{1}} \right|=\left| {{t}_{m}}-{{t}_{1}} \right|=2\sec \] \[\left| \Delta {{t}_{2}} \right|=\left| {{t}_{m}}-{{t}_{2}} \right|=1\sec \] \[\left| \Delta {{t}_{3}} \right|=\left| {{t}_{m}}-{{t}_{3}} \right|=3\sec \] \[\left| \Delta {{t}_{4}} \right|=\left| {{t}_{m}}-{{t}_{4}} \right|=0\sec \] Mean absolute error \[\Delta {{t}_{mean}}=\frac{2+1+3+0}{4}=1.5\sec \] But the least count of the measuring clock is 1 sec, so it cannot measure up to 0.5 second, so we have to round it off. So mean error will be 2 second. Hence men time\[(92\pm 2sec)\].
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