Question

The length of a rectangular plate is measured by a meter scale and is found to be 10.0 cm. Its width is measured by vernier calipers as 1.00 cm. The least count of the meter scale and vernier calipers is 0.1 cm and 0.01 cm respectively. Minimum possible error in area measurement is:
(A) $\pm 0.02c{m^2}$
(B) $\pm 0.01c{m^2}$
(C) $\pm 0.03c{m^2}$
(D) Zero

Answer 1

Hint Least count of an instrument is defined as the minimum quantity it can measure without any error. For eg, if the length of the plate is 10.03cm, the scale won't be able to measure it properly and it will show the reading as 10 cm only. here the error would be0.03cm. Similarly, we need to find the minimum error that can occur in the measurement when using these equipment.
Use the formula $\Delta A = \pm \left( {\dfrac{{\Delta L}}{L} + \dfrac{{\Delta B}}{B}} \right) \times A$ to get the maximum error in area measurement.

Complete step by step solution
Let the length of the rectangle be L and the breadth be B. The formula for area (A) of a rectangle is $A = L \times B$. Then,
$\Rightarrow A \pm \Delta A = \left( {L \pm \Delta L} \right)\left( {B \pm \Delta B} \right) \\\ \Rightarrow A \pm \Delta A = LB \pm L\Delta B \pm B\Delta L \pm \Delta L\Delta B \\\$
Dividing left hand side and right hand side by A,

$$\Rightarrow \dfrac{{A \pm \Delta A}}{A} = \dfrac{{LB \pm L\Delta B \pm B\Delta L \pm \Delta L\Delta B}}{A} \\\ \Rightarrow 1 \pm \dfrac{{\Delta A}}{A} = \dfrac{{LB \pm L\Delta B \pm B\Delta L \pm \Delta L\Delta B}}{A} \\\$$

Now we will substitute A=LB on the right hand side,

$$\Rightarrow 1 \pm \dfrac{{\Delta A}}{A} = \dfrac{{LB \pm L\Delta B \pm B\Delta L \pm \Delta L\Delta B}}{{LB}} \\\ \Rightarrow 1 \pm \dfrac{{\Delta A}}{A} = 1 \pm \dfrac{{\Delta B}}{B} \pm \dfrac{{\Delta L}}{L} \pm \dfrac{{\Delta L}}{L}\dfrac{{\Delta B}}{B} \\\ \Rightarrow \dfrac{{\Delta A}}{A} = \pm \left( {\dfrac{{\Delta B}}{B} + \dfrac{{\Delta L}}{L} + \dfrac{{\Delta L}}{L}\dfrac{{\Delta B}}{B}} \right) \\\$$

Since, $\Delta L$and$\Delta B$ are small quantities, their product$\Delta L\Delta B$will be a very small quantity and can be neglected from the above equation.
$\Rightarrow \dfrac{{\Delta A}}{A} = \pm \left( {\dfrac{{\Delta B}}{B} + \dfrac{{\Delta L}}{L}} \right)$
Hence, the maximum relative error $\Rightarrow \dfrac{{\Delta A}}{A} = \pm \left( {\dfrac{{\Delta L}}{L} + \dfrac{{\Delta B}}{B}} \right)$
And, maximum possible error $\Rightarrow \Delta A = \pm \left( {\dfrac{{\Delta L}}{L} + \dfrac{{\Delta B}}{B}} \right)A$ -----(1)
Now, we are given that the length of the rectangular plate is 10.0 cm.
$\Rightarrow L = 10.0cm$
We are also given that this length is measured by a meter scale. The least count of the meter scale is given to be 0.1 cm. The maximum possible uncertainty in the measurement of length will be equal to the least count of the meter scale.
$\Rightarrow \Delta L = 0.1cm$
Similarly, we are given that the breadth of the rectangular plate is measured to be 1.00 cm.
$\Rightarrow B = 1.00cm$
It is measured by vernier calipers whose least count is 0.01 cm. Hence, the maximum possible uncertainty in the measurement of breadth will be 0.01 cm.
$\Rightarrow \Delta B = 0.01cm$
Putting these values in the formula for maximum possible error given by equation (1),
$\Rightarrow \Delta A = \pm \left( {\dfrac{{0.1}}{{10.0}} + \dfrac{{0.01}}{{1.00}}} \right)\left( {10.0 \times 1.00} \right) \\\ \Rightarrow \Delta A = \pm \left( {0.01 + 0.01} \right)\left( {10.0 \times 1.00} \right) \\\ \Rightarrow \Delta A = \pm \left( {0.02} \right)\left( {10.0} \right) \\\ \Rightarrow \Delta A = \pm \left( {0.2} \right)c{m^2} \\\$
Therefore, the maximum possible error in measurement of area will become $\pm \left( {0.2} \right)c{m^2}$. But we are asked for the minimum possible error in the measurement. $\left| {\Delta A} \right|$ will be minimum when it is zero.

Hence, option (D) is correct.

Note: This solution is a general solution to explain how the errors can be determined. However, for a quick alternative, one can easily see that the minimum possible error in any measurement or experiment will be zero always. Zero errors would signify that there were no instrumental, technical or personal errors while the measurement was done.