Question

From class 11 Physics​

Answer 1

Refer the attachment...

Answer 2

Answer:

(19 $\pm$ 1.5)$\rm cm^2$

Explanation:

Given,

Length of rectangle = $\rm (5.7 \pm 0.1) cm$

Therefore,

Error in length ($\Delta$l) = 0.1 cm

Breadth of rectangle = $\rm (3.4 \pm 0.2)cm$

Therefore,

Error in breadth ($\Delta$b) = 0.2 cm

We know that area of rectangle = Length x Breadth

Therefore,

Area of the rectangle = 5.7 x 3.4 = 19.38 $\rm cm^2$ = 19 $\rm cm^2$ (Rounding off to two significant figures)

We know that,

$\rm \dfrac{\Delta A}{A} = \pm\bigg(\dfrac{\Delta l}{l} + \dfrac{\Delta b}{b} \bigg)$

Sub the values of A, $\rm \Delta l, \Delta b$, l, b

$\rm \dfrac{\Delta A}{19} = \pm \bigg(\dfrac{0.1}{5.7} + \dfrac{0.2}{3.4} \bigg)$

$\rm \dfrac{\Delta A}{19} = \pm \bigg(\dfrac{3.4 \times 0.1 + 5.7 \times 0.2}{5.7 \times 3.4} \bigg )$

$\rm \dfrac{\Delta A}{19} = \pm \dfrac{1.48}{19.38}$

$\rm \Delta A = \pm 1.48$

Rounding off to two significant figures,

$\rm \Delta A = 1.5$

Therefore,

The area of the rectangle with error limits = (19 $\pm$ 1.5) $\rm cm^2$