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NEET PHYSICSMedium

Time intervals measured by a clock give the following readings: 1.25 s,1.24 s,1.27 s,1.21 s,1.28 s1.25 \text{ s}, 1.24 \text{ s}, 1.27 \text{ s}, 1.21 \text{ s}, 1.28 \text{ s}. What is the percentage relative error of the observations?

A

2%2\%

B

4%4\%

C

16%16\%

D

1.6%1.6\%

Step-by-Step Solution

Mean value of time tmean=1.25+1.24+1.27+1.21+1.285=6.255=1.25 st_{mean} = \frac{1.25 + 1.24 + 1.27 + 1.21 + 1.28}{5} = \frac{6.25}{5} = 1.25 \text{ s}. Absolute errors in the measurements are: Δt1=1.251.25=0.00 s\Delta t_1 = |1.25 - 1.25| = 0.00 \text{ s} Δt2=1.251.24=0.01 s\Delta t_2 = |1.25 - 1.24| = 0.01 \text{ s} Δt3=1.251.27=0.02 s\Delta t_3 = |1.25 - 1.27| = 0.02 \text{ s} Δt4=1.251.21=0.04 s\Delta t_4 = |1.25 - 1.21| = 0.04 \text{ s} Δt5=1.251.28=0.03 s\Delta t_5 = |1.25 - 1.28| = 0.03 \text{ s} Mean absolute error Δtmean=0.00+0.01+0.02+0.04+0.035=0.105=0.02 s\Delta t_{mean} = \frac{0.00 + 0.01 + 0.02 + 0.04 + 0.03}{5} = \frac{0.10}{5} = 0.02 \text{ s}. Percentage relative error =Δtmeantmean×100%=0.021.25×100%=1.6%= \frac{\Delta t_{mean}}{t_{mean}} \times 100\% = \frac{0.02}{1.25} \times 100\% = 1.6\%.

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