CBSE Class 10 Science Chapter 12: Electricity NCERT Solutions

Let the original resistance of the wire be R. The wire is cut into five equal parts. Therefore, the resistance of each part is $R/5$.

These five parts are then connected in parallel. The formula for the equivalent resistance ($R\text{'}$) of resistors connected in parallel is given by:

$$\backslash frac\{1\}\{R\text{'}\}\; =\; \backslash frac\{1\}\{R\_1\}\; +\; \backslash frac\{1\}\{R\_2\}\; +\; \backslash frac\{1\}\{R\_3\}\; +\; \backslash frac\{1\}\{R\_4\}\; +\; \backslash frac\{1\}\{R\_5\}$$

In this case, $R\_1\; =\; R\_2\; =\; R\_3\; =\; R\_4\; =\; R\_5\; =\; R/5$. Substituting these values:

$$\backslash frac\{1\}\{R\text{'}\}\; =\; \backslash frac\{1\}\{R/5\}\; +\; \backslash frac\{1\}\{R/5\}\; +\; \backslash frac\{1\}\{R/5\}\; +\; \backslash frac\{1\}\{R/5\}\; +\; \backslash frac\{1\}\{R/5\}$$

$$\backslash frac\{1\}\{R\text{'}\}\; =\; \backslash frac\{5\}\{R\}\; +\; \backslash frac\{5\}\{R\}\; +\; \backslash frac\{5\}\{R\}\; +\; \backslash frac\{5\}\{R\}\; +\; \backslash frac\{5\}\{R\}\; =\; \backslash frac\{25\}\{R\}$$

To find $R\text{'}$, we invert both sides:

$R\text{'}\; =\; \backslash frac\{R\}\{25\}$

The question asks for the ratio $R/R\text{'}$. We can rearrange the equation $R\text{'}\; =\; R/25$ to find this ratio:

$\backslash frac\{R\}\{R\text{'}\}\; =\; 25$

Therefore, the ratio $R/R\text{'}$ is 25. This corresponds to option (d).

Electrical power (P) in a circuit can be expressed in several ways, derived from the basic definition $P\; =\; VI$, where V is voltage and I is current.

Using Ohm's Law, $V\; =\; IR$, we can substitute V in the power formula:

$P\; =\; (IR)\; \backslash times\; I\; =\; I^2R$

This shows that $I^2R$ (option a) represents electrical power.

We can also express current I using Ohm's Law as $I\; =\; V/R$. Substituting this into the basic power formula:

$P\; =\; V\; \backslash times\; (V/R)\; =\; V^2/R$

This shows that $V^2/R$ (option d) also represents electrical power.

The term $VI$ (option c) is the fundamental definition of electrical power.

The term $IR^2$ (option b) is not a standard formula for electrical power in a circuit. Therefore, $IR^2$ does not represent electrical power.

The rating of the electric bulb is 220 V and 100 W. This means that when the bulb is operated at a voltage (V) of 220 V, it consumes a power (P) of 100 W.

We can use the formula for electrical power, $P\; =\; V^2/R$, to find the resistance (R) of the bulb. The resistance of the bulb is assumed to be constant.

Rearranging the formula to solve for R:

$R\; =\; \backslash frac\{V^2\}\{P\}$

Substituting the given values:

$R\; =\; \backslash frac\{(220\; \backslash text\{\; V\})^2\}\{100\; \backslash text\{\; W\}\}\; =\; \backslash frac\{48400\}\{100\}\; =\; 484\; \backslash Omega$

Now, the bulb is operated at a different voltage, $V\_\{new\}\; =\; 110$ V. The resistance of the bulb remains $R\; =\; 484\; \backslash Omega$.

We can calculate the new power consumed ($P\_\{new\}$) using the same power formula:

$$P\_\{new\}\; =\; \backslash frac\{V\_\{new\}^2\}\{R\}\; =\; \backslash frac\{(110\; \backslash text\{\; V\})^2\}\{484\; \backslash Omega\}$$

$$P\_\{new\}\; =\; \backslash frac\{12100\}\{484\}\; =\; 25\; \backslash text\{\; W\}$$

Therefore, when the bulb is operated on 110 V, the power consumed will be 25 W. This corresponds to option (d).