A) (3.45 ± 0.2) ms-1
B) (3.45 ± 0.3) ms-1
C) (3.45 ± 0.4) ms-1
D) (3.45 ± 0.5) ms-1
Correct Answer: B
Solution :
Here, \[S=(13.8\pm 0.2)\]m and \[t=(4.0\pm 0.3)\] sec Expressing it in percentage error, we have, \[S=13.8\pm \frac{0.2}{13.8}\times 100%=13.8\pm 1.4%\] and \[t=4.0\pm \frac{0.3}{4}\times 100%=4\pm 7.5%\] \[\because V=\frac{s}{t}\]\[=\frac{13.8\pm 1.4}{4\pm 7.5}=(3.45\pm 0.3)\ m/s.\]
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