The length and breadth of a rectangle are (5.7± 0.1)cm and (3.4± 0.2)cm. Calculate area of the rectangle with error limits.

The correct answer is 19.4± 1.5 cm².

Let us solve the problem step-by-step to calculate the area of the rectangle along with its error limits.

Step 1: Identify the given values
Let the length of the rectangle be represented as $l\pm \Delta l$ and the breadth as $b\pm \Delta b$.
Here, we have:
Length, $l=5.7\text{cm}$
Absolute error in length, $\Delta l=0.1\text{cm}$
Breadth, $b=3.4\text{cm}$
Absolute error in breadth, $\Delta b=0.2\text{cm}$

Step 2: Calculate the value of the Area (A)
The area of a rectangle is given by the formula:
$A=l\times b$
Substituting the values:
$A=5.7\times 3.4=19.38{\text{cm}}^2$
Rounding off to two significant figures (consistent with the inputs):
$A\approx 19.4{\text{cm}}^2$

Step 3: Calculate the relative error and absolute error in Area
For a product of two quantities, the relative error in the result is the sum of the relative errors of the individual quantities:
$\frac{\Delta A}{A}=\frac{\Delta l}{l}+\frac{\Delta b}{b}$
Now, find the absolute error $\Delta A$:
$\Delta A=A\times \left(\frac{\Delta l}{l}+\frac{\Delta b}{b}\right)$
Substitute the values:
$\Delta A=19.38\times \left(\frac{0.1}{5.7}+\frac{0.2}{3.4}\right)$
$\Delta A=19.38\times \left(0.0175+0.0588\right)$
$\Delta A=19.38\times 0.0763\approx 1.48{\text{cm}}^2$
Rounding to one decimal place to match standard scientific precision conventions yields:
$\Delta A\approx 1.5{\text{cm}}^2$

Step 4: Write the final answer with error limits
Thus, the area of the rectangle with error limits is:
$A\pm \Delta A=\left(19.4\pm 1.5\right){\text{cm}}^2$
Note: The third option in the provided list contains a small typographical placement error representing the decimal point in the error limit ("19.4± 15 cm²" instead of "19.4± 1.5 cm²"). Based on the standard rules of error propagation, the correct mathematically derived limit is 1.5.