Question

If a physical quality is P=a square b square/ c square and the percentage errors in the measurements of a b and c are 1% 2% 3% respectively ,find the maximum precentage error in the measurement of physical quanlity p.

Answer 1

Answer:

Given:

A physical Quantity P is denoted as :

$\boxed{ \red{P = \dfrac{ {a}^{2} {b}^{2} }{ {c}^{2} }}}$

Errors for a, b , c are in 1% , 2% , 3%.

To find:

Maximum percentage of error in calculation of P

Concept:

The maximum error could be calculated by considering each quantities' error and taking the exponential into product .

Calculation:

Max error in Calculation of P

$\dfrac{ \Delta P}{P} = \small{ (2 \times \dfrac{\Delta a}{a}) + (2 \times \dfrac{\Delta b}{b}) + (2 \times \dfrac{\Delta c}{c})}$

Putting the given data :

$\dfrac{ \Delta P}{P} = \:(2 \times 1) + (2 \times 2) + (2 \times 3)$

$\dfrac{ \Delta P}{P} = \:12\%$

So final answer :

$\: \: \: \: \: \: \: \: \: \: \: \boxed{ \large{ \sf{ \blue{max \: error = 12\%}}}}$

Simple Trick to solve this type of questions :

• Just multiply error of each quantity with the corresponding Exponential power.
• And always add the individual error for max total error.

Answer 2

Answer:

$\large\boxed{\sf{12\%}}$

Explanation:

Given a physical quantity such that

$p = \dfrac{ {a}^{2} {b}^{2} }{ {c}^{2} }$

Also, Percentage error in the measurements of a, b and c are 1% , 2% and 3% respectively.

Therefore, we have

$\dfrac{\triangle a}{a} = 1$

$\dfrac{\triangle b}{b} = 2$

$\dfrac{\triangle c}{c} = 3$

To find the Maximum Percentage error:

$= > \frac{\triangle p}{p} \times 100 = (2 \times \frac{\triangle a}{a} ) + (2 \times \frac{\triangle b}{b} ) + (2 \times \frac{\triangle c}{c} ) \\ \\ = > \frac{\triangle p}{p} \times 100 = (2 \times 1) + (2 \times 2) + (2 \times 3) \\ \\ = > \frac{\triangle p}{p} \times 100 = 2 + 4 + 6 \\ \\ = > \sf{(\frac{\triangle p}{p} \times 100)\% = 12\%}$

Hence, Maximum Percentage error is 12%