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Question: A public park, in the form of a square, has an area of\(\left( {100 \pm 0 \cdot 2} \right){m^2}\). T...

A public park, in the form of a square, has an area of(100±0⋅2)m2\left( {100 \pm 0 \cdot 2} \right){m^2}. The side of park is:
A) (10±0⋅01)m\left( {10 \pm 0 \cdot 01} \right)m
B) (10±0⋅1)m\left( {10 \pm 0 \cdot 1} \right)m
C) (10±0⋅02)m\left( {10 \pm 0 \cdot 02} \right)m
D) (10±0⋅2)m\left( {10 \pm 0 \cdot 2} \right)m

Explanation

Solution

Error is defined as the deviation of the actual value from the standard value. The given area of the public park is 100m2100{m^2} with an error of 0⋅2m20 \cdot 2{m^2} in the area of the public park this means that the length of the park will also have some error.

Formula used:
The formula of the error in the area is given by,
ΔAA=2Δll\dfrac{{\Delta A}}{A} = 2\dfrac{{\Delta l}}{l}
Where ΔA\Delta A is the error in area AA is the original area Δl\Delta l is the error in the area and ll is the length

Complete step by step answer:
It is given in the problem that the public park is in the form of a square that has an area of (100±0⋅2)m2\left( {100 \pm 0 \cdot 2} \right){m^2} and we need to find the value of the side of the public park.
The area of the square park without the error would be an area of the square equal to,
⇒A=l2\Rightarrow A = {l^2}
⇒100=l2\Rightarrow 100 = {l^2}
⇒l=100\Rightarrow l = \sqrt {100}
⇒l=10m\Rightarrow l = 10m
So the length of the park in an ideal situation would be l=10ml = 10m.
Since the formula of the error in the area is given by,
ΔAA=2Δll\dfrac{{\Delta A}}{A} = 2\dfrac{{\Delta l}}{l}
Where ΔA\Delta A is the error in area A is the original area Δl\Delta l is the error in the area and ll is the length
As the error in the area is equal to ΔA=0⋅2m2\Delta A = 0 \cdot 2{m^2} the area is equal to A=100m2A = 100{m^2} and the value of length is given by l=10ml = 10m.
Therefore, replace the value of the error in the area, the value of the given area of the park, and the length of the side of the park if no error is made in the formula given above.
⇒ΔAA=2Δll\Rightarrow \dfrac{{\Delta A}}{A} = 2\dfrac{{\Delta l}}{l}
On substituting the corresponding values, we get
⇒0⋅2100=2(Δl10)\Rightarrow \dfrac{{0 \cdot 2}}{{100}} = 2\left( {\dfrac{{\Delta l}}{{10}}} \right)
⇒0⋅210=2⋅(Δl)\Rightarrow \dfrac{{0 \cdot 2}}{{10}} = 2 \cdot \left( {\Delta l} \right)
On simplifications,
⇒0⋅110=(Δl)\Rightarrow \dfrac{{0 \cdot 1}}{{10}} = \left( {\Delta l} \right)
⇒Δl=0⋅01\Rightarrow \Delta l = 0 \cdot 01.

Therefore the value of the length of the public park is equal to l=(100±0⋅01)ml = \left( {100 \pm 0 \cdot 01} \right)m. The correct answer for this problem is option A.

Note:
We cannot measure everything with great accuracy. We will always have some sort of error in the measurement and if the error occurs in the measurement of the length then that error will be reflected in the area or the volume of the particular solid.