The voltage across a lamp is (6.0± 0.1) volt and the current passing through it is (4.0± 0.2) ampere. Find the power consumed.

The correct option is (24.0± 1.6) watt.

Step 1: Calculate the nominal power consumed.
The power $P$ consumed by an electrical component is given by the formula:
$P=V\times I$
where:
- Voltage, $V=6.0\text{V}$
- Current, $I=4.0\text{A}$

Substituting the nominal values into the equation:
$P=6.0\times 4.0=24.0\text{watt}$

Step 2: Calculate the error in power.
For a product of two quantities, the relative error in the result is the sum of the relative errors of the individual quantities:
$\frac{\Delta P}{P}=\frac{\Delta V}{V}+\frac{\Delta I}{I}$
where:
- Absolute error in voltage, $\Delta V=0.1\text{V}$
- Absolute error in current, $\Delta I=0.2\text{A}$

Substitute the values to find the fractional error in power:
$\frac{\Delta P}{P}=\frac{0.1}{6.0}+\frac{0.2}{4.0}$

Find a common denominator to sum the fractions:
$\frac{\Delta P}{P}=\frac{1}{60}+\frac{2}{40}=\frac{1}{60}+\frac{1}{20}$

$\frac{\Delta P}{P}=\frac{1+3}{60}=\frac{4}{60}=\frac{1}{15}$

Now, solve for the absolute error in power $\Delta P$:
$\Delta P=P\times \frac{1}{15}$

$\Delta P=24.0\times \frac{1}{15}=1.6\text{watt}$

Step 3: Express the final result with error limits.
Combining the nominal power and its absolute error:
$\mathrm{Power}=(P\pm \Delta P)=(24.0\pm 1.6)\text{watt}$