You measure two quantities as A=1.0 m ± 0.2 m , B=2.0 m ± 0.2 m . We should report correct value for A B as

Correct answer: D. 1.4 m ± 0.2 m Given, A=1.0 m ± 0.2 m, B=2.0 m ± 0.2 m Let Y= A B = (1.0)(20) =1.414 m Rounding off to two significant digits aligned Y & =1.4 m (Δ Y)/(Y) & =(1)/(2)[(Δ A)/(A)+(Δ B)/(B)] & =(1)/(2)[(0.2)/(1.0)+(0.2)/(2.0)] & =(0.6)/(2 × 20) Δ Y & =(0.6 Y)/(2 × 20)=(0.6 × 1.4)/(2 × 20)=0.212 aligned Rounding off to one significant digit, Δ Y=0.2 m

BITSAT 2024

Exam: BITSAT
Year: 2024
Subject: Physics
Chapter: Mathematics in Physics

BITSAT 2024 · Physics · Mathematics in Physics

You measure two quantities as A=1.0 m ± 0.2 m , B=2.0 m ± 0.2 m ....

You measure two quantities as

A=1.0 \mathrm{ m} \pm 0.2 \mathrm{ m}

B=2.0 \mathrm{ m} \pm 0.2 \mathrm{ m}

. We should report correct value for

\sqrt{A B}

Options

A $1.4 \mathrm{ m} \pm 0.4 \mathrm{ m}$
B $1.41 \mathrm{ m} \pm 0.15 \mathrm{ m}$
C $1.4 \mathrm{ m} \pm 0.3 \mathrm{ m}$
D $1.4 \mathrm{ m} \pm 0.2 \mathrm{ m}$

Correct answer

D. $1.4 \mathrm{ m} \pm 0.2 \mathrm{ m}$

Step-by-step solution

Given,

A=1.0 \mathrm{ m} \pm 0.2 \mathrm{ m}, B=2.0 \mathrm{ m} \pm 0.2 \mathrm{ m}

Let

Y=\sqrt{A B}=\sqrt{(1.0)(20)}=1.414 \mathrm{ m}

Rounding off to two significant digits

\begin{aligned} Y & =1.4 \mathrm{ m} \\ \frac{\Delta Y}{Y} & =\frac{1}{2}\left[\frac{\Delta A}{A}+\frac{\Delta B}{B}\right] \\ & =\frac{1}{2}\left[\frac{0.2}{1.0}+\frac{0.2}{2.0}\right] \\ & =\frac{0.6}{2 \times 20} \\ \Rightarrow \quad \Delta Y & =\frac{0.6 Y}{2 \times 20}=\frac{0.6 \times 1.4}{2 \times 20}=0.212 \end{aligned}

Rounding off to one significant digit,

\Delta Y=0.2 \mathrm{ m}

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