Length and breadth of a rectangular sheet are 16.2 cm and 10.1 CM respectively the area of the sheet in appropriate significant figures and error is
Answers
Answer:
Explanation:
Given Length and breadth of a rectangular sheet are 16.2 cm and 10.1 CM respectively the area of the sheet in appropriate significant figures and error is
Let Δa = error in measurement of a
Δb = error in measurement of b
Δx = error in finding x
Now maximum fractional error in x is
So Δx /x = (Δa/a + Δb/b)
Now given length = (16. cm^2 ± 0.1) cm
Breadth = b = (10.1 ± 0.1) cm
Area = l x b
= 16.2 x 10.1
= 163.62 sq cm
Now rounding off the area we get area = 164 sq cm
If ΔA is error in area, then relative error is given as ΔA/A = Δl/l ± Δb/b
= 0.1/16.2 ± 0.1/10.1
= 1.01 ± 1.62 / 163.62
= 2.63 / 163.62
It implies ΔA = A x 2.63 / 163.62 sq cm
= 163.62 x 2.63 / 163.62
= 2.63 sq cm
By rounding off to one significant figure we get
ΔA = 3 sq cm
Now Area = A ± ΔA
= (164 ± 3) sq cm
Given:
Length of a rectangular sheet = 16.2 cm
Breadth of a rectangular sheet = 10.1 cm
Assume,
Δa = error in measurement of a
Δb = error in measurement of b
Δx = error in finding x
So, the maximum fractional error in x is
So Δx /x = (Δa/a + Δb/b)
Now given length = (16. cm^2 ± 0.1) cm
Breadth = b = (10.1 ± 0.1) cm
Area = l x b
= 16.2 x 10.1
= 163.62 sq cm
Now rounding off the area we get area = 164 sq cm
If ΔA is error in area, then relative error is given as ΔA/A = Δl/l ± Δb/b
= 0.1/16.2 ± 0.1/10.1
= 1.01 ± 1.62 / 163.62
= 2.63 / 163.62
It implies ΔA = A x 2.63 / 163.62 sq cm
= 163.62 x 2.63 / 163.62
= 2.63 sq cm
By rounding off to one significant figure we get
ΔA = 3 sq cm
Now Area = A ± ΔA
= (164 ± 3) sq cm