Question

If,R=a^2√b/c^4d and percentage error in measurement of a,b,c and d are 0.1%,0.2% ,0.3% and 0.4% respectively ,then the maximum percentage error in R is (a) 1.9%
(b) 3.1%
(c) 2.9%
(d)4.0%

Answer 1

Answer:

• The Percentage error is 1.9 %

Given:

1. Error % in a = 0.1 %
2. Error % in b = 0.2 %
3. Error % in c = 0.3 %
4. Error % in d = 0.4 %

Explanation:

$\rule{300}{1.5}$

Given equation is,

$\longrightarrow\qquad\qquad\boxed{\boxed{\sf R=\dfrac{a^{2}\sqrt{b}}{c^{4}\;d}}}$

Therefore, For finding the percentage error,

$\bigstar\;\underline{\boxed{\sf \dfrac{\Delta R}{R}=2\Bigg\lgroup\dfrac{\Delta a}{a}\Bigg\rgroup+\dfrac{1}{2}\Bigg\lgroup\dfrac{\Delta b}{b}\Bigg\rgroup+4\Bigg\lgroup\dfrac{\Delta c}{c}\Bigg\rgroup+\Bigg\lgroup\dfrac{\Delta d}{d}\Bigg\rgroup}}$

Now, For percentage we need to multiply it with 100.

$\longrightarrow\sf \dfrac{\Delta R}{R}\times100=2\Bigg\lgroup\dfrac{\Delta a\times100}{a}\Bigg\rgroup+\dfrac{1}{2}\Bigg\lgroup\dfrac{\Delta b\times100}{b}\Bigg\rgroup+4\Bigg\lgroup\dfrac{\Delta c\times100}{c}\Bigg\rgroup+\Bigg\lgroup\dfrac{\Delta d\times100}{d}\Bigg\rgroup$

Substituting the values,

$\displaystyle\longrightarrow\sf \dfrac{\Delta R}{R}\times 100 = \bigg[2\times 0.1\bigg] + \left[\dfrac{1}{2} \times0.2\right] +\bigg[4\times 0.3\bigg] +\bigg[1\times 0.4\bigg]\\\\\\\longrightarrow\sf \dfrac{\Delta R}{R}\times 100 = 0.2+0.1+1.2+0.4\\\\\\\longrightarrow\sf \dfrac{\Delta R}{R}\times 100 = 1.2+0.7\\\\\\\longrightarrow\sf \dfrac{\Delta R}{R}\times 100=1.9\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf \dfrac{\Delta R}{R}\times 100=1.9\%}}}}$

∴ The Percentage error is 1.9 %.

Hence, 1st Option is correct !

$\rule{300}{1.5}$

Answer 2

$\huge\underline\mathbb\purple{ANSWER:-}$

•The Percentage error is 1.9 %