The least count of stop watch is (1/5) second. The time of 20 oscillations of pendulum is measured to be 25 seconds. Then percentage error in the measurement of time will be

Question

The least count of stop watch is (1/5) second. The time of 20 oscillations of pendulum is measured to be 25 seconds. Then percentage error in the measurement of time will be (A) 0.8 % (B) 8 % (C) 0.1 % (D) 1.8 %

Tardigrade Admin · Accepted Answer

As we know that the percentage error is defined as when we multiply the relative error with hundred then it becomes percentage error. So, we can write the percentage error as, (Δ t/t)× 100=(0 . 2/25)× 100=0.8 % that's why correct option is (A)

Q. The least count of stop watch is $\frac{1}{5}$ second.
The time of $20$ oscillations of pendulum is measured to be $25$ seconds. Then percentage error in the measurement of time will be

$0.8%$

$8%$

$0.1%$

$1.8%$

As we know that the percentage error is defined as when we multiply the relative error with hundred then it becomes percentage error.
So, we can write the percentage error as,
$\frac{Δ t}{t} \times 100 = \frac{0.2}{25} \times 100 = 0.8%$
that's why correct option is $(A)$

Q. The least count of stop watch is 51​ second. The time of 20 oscillations of pendulum is measured to be 25 seconds. Then percentage error in the measurement of time will be

0.8%

8%

0.1%

1.8%