Question

A Physical quantity P is related to four observables a, b, c and d as follows:
$P = \frac {a^{3} a^{2}} {( \sqrt{c.d} )}$
The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2% respectively. What is the percentage error in the quantity p? If the value of p calculated using above relation turns out to be 3.763, to what value should you round off the result.

Answer 1

Hii dear,

# Answer -
a) Percentage error in P is 13%.
b) P should be rounded off to 3.8

# Given-
P = 3.763 units
Δa/a=1%
Δb/b=3%
Δc/c=4%
Δd/d=2%

# Solution-
a) From given formula
ΔP/P = 3Δa/a + 2Δb/b + 0.5Δc/c + 0.5Δd/d
ΔP/P = 3×1+2×3+0.5×4+0.5×2
ΔP/P = 13%
Hence, Percentage error in P is 13%.

b) Rounding off value of P to first decimal place it would be P=3.8

Hope this helped you...