Question

A glass plate of length 3.5 cm and width 1.5 cm. what will be the maximum possible error in measuring the area of the plate?.

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Answer 1

Relative error in length

= ΔL/L

= 0.1/3.5

= 1/35

Relative error in width

= ΔW/W

= 0.1/1.5

= 1/15

Relative error in area

ΔΑ/A = ΔL/L + ΔW/W

= 1/35+1/15

= 2/21

Maximum possible Error in area

= ΔA/A × A

= 2/21 × (1.5 × 3.5)

= +0.5 or -0.5 cm^2

Answer 2

Answer:

The maximum possible error in measuring the area of the plate will be equal to ± 0.5cm².

Explanation:

The given glass plate is a rectangle.

We have given, the length of glass plate, L = 3.5cm

The width of the rectangular plate, W = 1.5cm

The relative error gets added, in the multiplication of quantities.

Area of rectangle = Length × width

Relative error in area = relative error in length + relative error in width

Firstly, relative error in length of glass plate:

$\frac{\triangle L}{L}=\frac{0.1}{3.5}=\frac{1}{35}$

Relative error in width of glass plate:

$\frac{\triangle W}{W}=\frac{0.1}{1.5}=\frac{1}{15}$

Relative error in area of glass plate:

$\frac{\triangle A}{A}=\frac{\triangle L}{L}+\frac{\triangle W}{W}$

$\frac{\triangle A}{A}=\frac{1}{35}+\frac{1}{15}$

$\frac{\triangle A}{A}=\frac{3+7}{105}$

$\frac{\triangle A}{A}=\frac{10}{105}$

$\frac{\triangle A}{A}=\frac{2}{21}$

Maximum possible error in  the area of the plate $=(\frac{\triangle A}{A})*A}$

$=\frac{2}{21}*(3.5*1.5)$

$=\frac{2}{21}*5.25$

$=\pm\; 0.5 cm^2$

Therefore, the maximum error in the area of glass plate is ± 0.5cm².