Question

A physical quantity is determined $P=\frac{\sqrt{ab{c}^2}}{d^3{e}^{1/3}}$ measuring a, b, c, d and e separately with the percentage error of 2%, 3%, 2%, 1% and 6% respectively. Minimum amount of error is contributed by the measurement of

• A. b
• B. a
• C. d
• D. e

Answer 1

The correct option is B a

$P=\left[\sqrt{\frac{ab{c}^2}{d^3{e}^{\frac{1}{3}}}}\right]$
% error of following quantity
$a=\frac{1}{2}\times 2=1\%,b=\frac{1}{2}\times 3=1.5\%$
$c=\frac{2\times 1\times 2}{2}\%=2\%,d=3\times 1\%=3\%e=\frac{1}{3}\times 6=2\%$
‘a’ measurement is measured with minimum amount of error.