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

Question

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

Tardigrade Admin · Accepted Answer

Least count =Δ T=(1/5) s =0.2 s T=25 s Percentage error =(Δ T/T) × 100 % =(0.2/25) × 100 %=0.8 %

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

$0.1%$

$0.8%$

$1.8%$

$8%$

Least count $= Δ T = \frac{1}{5} s = 0.2 s$
$T = 25 s$
Percentage error $= \frac{Δ T}{T} \times 100%$
$= \frac{0.2}{25} \times 100% = 0.8%$

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

0.1%

0.8%

1.8%

8%