Question

The error in the measurement of the
sides of a rectangle is 1%. The error in
the measurement of its area is
(A) 1% (B) 1/2%
(C) 2% (D) None of the above.​

Answer 1

Let the length and the breadth of the rectangle be l and b respectively.

Here the length and the breadth of the rectangle has same percentage error, i.e., 1%. So,

$\dfrac{\Delta l}{l}\times 100\ =\ \dfrac{\Delta b}{b}\times 100\ =\ 1\%$

Let the area be A.

$A=lb$

We have to find the percentage error in A.

We know that in a quantity X,

$X=a_1^{b_1}\cdot a_2^{b_2}\cdot\[\]\dots\dots\cdot a_n^{b_n}\ \implies\ \dfrac{\Delta X}{X}=b_1\dfrac{\Delta a_1}{a_1}+b_2\dfrac{\Delta a_2}{a_2}+ \dots\dots +b_n\dfrac{\Delta a_n}{a_n}\\ \\ \\ \text{OR}\begin{center}\displaystyle X=\prod_{k=1}^{n}a_k^{b_k}\ \implies\ \dfrac{\Delta X}{X}=\sum_{k=1}^{n}b_k\dfrac{\Delta a_k}{a_k}\end{center}$

where  $a_1,\ a_2,\ \dots\dots,\ a_n$  are some quantities on which X depends and  $b_1,\ b_2,\ \dots\dots,\ b_n$  are some exponents of each quantity.

So, in our case,

$A=lb\ \implies\ \dfrac{\Delta A}{A}=\dfrac{\Delta l}{l}+\dfrac{\Delta b}{b}$

This is just relative error. Percentage error will be,

$\dfrac{\Delta A}{A}\times 100=\left(\dfrac{\Delta l}{l}+\dfrac{\Delta b}{b}\right)100\\ \\ \\ \implies\ \dfrac{\Delta A}{A}\times 100=\dfrac{\Delta l}{l}\times 100+\dfrac{\Delta b}{b}\times 100\\ \\ \\ \implies\ \dfrac{\Delta A}{A}\times 100=1\%+1\%\\ \\ \\ \implies\ \dfrac{\Delta A}{A}\times 100=\mathbf{2\%}$

Hence the answer is (C) 2%.