The length, breadth and height of a rectangular block of wood were measured to be : l=12.13±0.02cm; b=8.16±0.01cm; h = 3.46 ± 0.01 cm. Determine the percentage error in the volume of the block.

The correct option is 0.58%.

Step-by-Step Explanation:

We are given the dimensions of a rectangular block of wood with their respective absolute errors:
Length, $l=12.13\pm 0.02\text{cm}$ (so $l=12.13$ and $\Delta l=0.02$)
Breadth, $b=8.16\pm 0.01\text{cm}$ (so $b=8.16$ and $\Delta b=0.01$)
Height, $h=3.46\pm 0.01\text{cm}$ (so $h=3.46$ and $\Delta h=0.01$)

The volume $V$ of a rectangular block is given by the formula:
$V=l\cdot b\cdot h$

For a product of quantities, the relative error in the result is the sum of the relative errors of the individual quantities:
$\frac{\Delta V}{V}=\frac{\Delta l}{l}+\frac{\Delta b}{b}+\frac{\Delta h}{h}$

To find the percentage error in the volume, we multiply the relative error by 100:
$\text{Percentage Error in}V=\left(\frac{\Delta l}{l}+\frac{\Delta b}{b}+\frac{\Delta h}{h}\right)\times 100\%$

Now, let us calculate each fractional error term:
$\frac{\Delta l}{l}=\frac{0.02}{12.13}\approx 0.001649$

$\frac{\Delta b}{b}=\frac{0.01}{8.16}\approx 0.001225$

$\frac{\Delta h}{h}=\frac{0.01}{3.46}\approx 0.002890$

Summing these values gives:
$\frac{\Delta V}{V}\approx 0.001649+0.001225+0.002890=0.005764$

Expressing this as a percentage:
$\text{Percentage Error in}V\approx 0.005764\times 100\%=0.5764\%$

Rounding to two decimal places, we get:
$\text{Percentage Error}\approx 0.58\%$