Question

In a physical quantity p=a.b. c²/d³ is determinant by measuring a, b, c, d separately with the percentage error 2%, 3%, 2%, and1% respectively minimum amount of error is contribute by measurement of

Answer 1

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ Given :- }}}}$

• In a physical quantity , p = a.b.c²/d³ is determinant by measuring a, b, c, d .
• The percentage error 2%, 3%, 2%, and1% respectively .

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ To Find :- }}}}$

• The minimum amount of error is contributed by which quantity .

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ Solution :- }}}}$

The given equation to find the physical quantity is ,

$\sf\longrightarrow p = \dfrac{ a.b.c^2}{d^3}$

Therefore , the maximum possible relative error in p will be ,

$\sf\longrightarrow \dfrac{\Delta p }{p}= \bigg( \dfrac{\Delta a }{a}\bigg)^1 + \bigg( \dfrac{\Delta b }{b}\bigg)^1+\bigg( \dfrac{\Delta c }{c}\bigg)^2+\bigg( \dfrac{\Delta d}{d}\bigg)^{-3}$

And , the percentage error will be ,

$\sf\longrightarrow \dfrac{\Delta p }{p}\times 100= \bigg\{\bigg( \dfrac{\Delta a }{a}\bigg)^1 + \bigg( \dfrac{\Delta b }{b}\bigg)^1+\bigg( \dfrac{\Delta c }{c}\bigg)^2+\bigg( \dfrac{\Delta d}{d}\bigg)^{-3} \bigg\}\times 100 \\\\\\\sf\longrightarrow \dfrac{\Delta p }{p}\times 100=\bigg( \dfrac{\Delta a}{a} +\dfrac{\Delta b}{b}+\dfrac{ 2\Delta c}{c}+\dfrac{3\Delta d}{d}\bigg)\\\\\\\sf\longrightarrow \dfrac{\Delta p }{p}\times 100=2\% + 3\% + 2*2\% + 3*1\%\\\\\\\sf\longrightarrow \dfrac{\Delta p }{p}\times 100= 2\% + 3\% + 4\% + 3\%$

Hence we can see that the minimum amount of percentage error is contributed by a .

$\sf\longrightarrow \underset{\blue{\sf Required\ Answer }}{\underbrace{\boxed{\pink{\frak{ Minimum \ Error \ contributed = a }}}}}$