Question

If ${{L = (20 \pm 0}}{\text{.01)m}}\,{\text{and}}\,{{B = (10 \pm 0}}{\text{.02)m}}$ then, $\dfrac{L}{B}$ is-
(A) $(2 \pm 0.03 )$ m
(B) $(2 \pm 0.015 )$ m
(C) $(2 \pm 0.01 )$ m
(D) $(2 \pm 0.005 )$ m

Answer 1

Hint
Firstly find the value of $\dfrac{L}{B}$ and then find out the maximum permissible error in $\dfrac{L}{B}$. Then represent the answer in the form of $A \pm \Delta A$.

Complete step by step answer
Given, Error in measurement of L = 0.01.
Error in measurement of B= 0.02.
Value of L = 20.
Value of B = 10.
We can easily find the value of $\dfrac{L}{B}$ which is $\dfrac{{20}}{{10}} = 2$.
Maximum Error in $\dfrac{L}{B}$ = Error in measurement of L + Error in measurement of B.
Maximum Error in $\dfrac{L}{B}$ = 0.01 + 0.02.
Maximum Error in = 0.03.
Hence the value of $\dfrac{L}{B}$ is $(2 \pm 0.03)$.

Note
Error is treated as a positive quantity. It means that errors are always added and never subtracted. This is done to obtain the maximum error. Let’s take an example of a product having errors in two components A and B both of which have an error of 5%. Now if we subtract the errors, the error becomes 0 which implies that there is no error in the product but we know in reality that there is an error. Hence we always add errors and never subtract them.