If the absolute errors in two physical quantities $A$ and $B$ are $a$ and $b$ respectively, then the absolute error in the value of $A B$ is:

[Disclaimer: The options in the original question have been changed to match with the actual solution / answer.]

b B - a A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a A + b B

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a B + b A

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

a B - b A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $a B + b A$
Given : $Δ A = a$ $Δ B = b$
The quantity given $X = A B$
Let absolute error in $X$ be $Δ X$
$∴$ $X + Δ X = (A + Δ A) (B + Δ B)$
OR $X + Δ X = (A + a) (b + B) = A B + a B + b A + a b$
As $a$ and $b$ are very small, thus we can neglect the term $a b$ .
$⟹$ $X + Δ X = A B + a B + b A$
$⟹$ $Δ X = a B + b A$ $(∵ X = A B)$

Suggest Corrections

Similar questions

9000 = 5^{a} \times a^{b} \times b^{a}

a^{a} + b^{b}

\frac{s i n (x - α)}{s i n (x - β)} = \frac{a}{b}, \frac{c o s (x - α)}{c o s (x - β)} = \frac{A}{B}

\neq

c o s (α - β) = \frac{a A + b B}{a B + b A} .

∣ ∣ ∣ ∣ \begin{matrix} a & b - c & c + b a + c & b & c - a a - b & b + a & c \end{matrix} ∣ ∣ ∣ ∣

Join BYJU'S Learning Program