Explanation
23.023 = 5 significant figure
0.0003 = 1 significant figure
2.1 × 10-3 = 2 significant figure
We will do addition of 3.8 ×10-6 and 42 ×10-6 and we get 45.8 ×10-6 or
4.58 ×10-5.Least Number of significant figure = 2.Therefore, Result is 4.6 ×10-5.
0.04580 has 4 significant figures
m = 3.513 kg = 4 significant figure
v = 5.00 ms-1 = 3 significant figure
Therefore, Momentum is given by P = mv
= 17.565
P must have 3 significant figure as m has 4 significant figure and v has 3 significant figure
Given Least Count L.C = 1/(2×50) = 1/100 Zero error, ZE = 0.03 Main Scale Reading = 3 Circular Scale Reading = 35
Diameter = MSR + CSR × LC – zero error
= 3 + 35 × 1/2 × 50 - (- 0.03) = 3.38 mm.
The second is duration of 9192631770 period of the radiation corresponding to the transition between the two hyperfine levels of the ground state of cesium-133 atom. 1 ns of 10-9 s of Cs-clock of 9192631770 oscillations.
Please disable the adBlock and continue. Thank you.