The length and width of a rectangular room are measured to be 3.95 ± 0.05 ~m and 3.05 ± 0.05 ~m , respectively, the area of the floor is

Correct answer: C. 12.05 ± 0.34 ~m^ 2 A= Bd A= dB+B d d A A = dB B + Bd B d A A = dB B + d =(0.05)/(3.05)+(0.05)/(3.95) =0.016+0.012 =0.028 × 12.05dA=0.3312.05 ± 0.34 Ans. (C)

JEE Main

Exam: JEE Main
Subject: Physics
Chapter: Mathematics in Physics

JEE Main · Physics · Mathematics in Physics

The length and width of a rectangular room are measured to be 3.95...

The length and width of a rectangular room are measured to be

3.95 \pm 0.05 \mathrm{~m}

and

3.05 \pm 0.05 \mathrm{~m}

, respectively, the area of the floor is

Options

A $12.05 \pm 0.01 \mathrm{~m}^{2}$
B $12.05 \pm 0.005 \mathrm{~m}^{2}$
C $12.05 \pm 0.34 \mathrm{~m}^{2}$
D $12.05 \pm 0.40 \mathrm{~m}^{2}$

Correct answer

C. $12.05 \pm 0.34 \mathrm{~m}^{2}$

Step-by-step solution

\mathrm{A}=\ell \mathrm{B}\mathrm{d} \mathrm{A}=\ell \mathrm{dB}+\mathrm{B} \mathrm{d} \ell\frac{\mathrm{d} \mathrm{A}}{\mathrm{A}}=\frac{\ell \mathrm{dB}}{\ell \mathrm{B}}+\frac{\mathrm{Bd} \ell}{\ell \mathrm{B}}\frac{\mathrm{d} \mathrm{A}}{\mathrm{A}}=\frac{\mathrm{dB}}{\mathrm{B}}+\frac{\mathrm{d} \ell}{\ell}\quad=\frac{0.05}{3.05}+\frac{0.05}{3.95}\quad=0.016+0.012\quad=0.028 \times 12.05\mathrm{dA}=0.3312.05 \pm 0.34

Ans. (C)

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