CBSE Class 10 Science Chapter 12: Electricity NCERT Solutions

The resistance of a wire is directly proportional to its length. Let the original resistance of the wire be R. When this wire is cut into five equal parts, the resistance of each part becomes R/5.

These five parts, each with resistance R/5, are then connected in parallel. The formula for the equivalent resistance (R') 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, each $r\_i\; =\; R/5$. So, the equation becomes:

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

Simplifying this, we get:

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

Therefore, the equivalent resistance R' is:

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

We need to find the ratio R/R'. Rearranging the equation for R', we get:

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

Thus, the ratio R/R' is 25. This corresponds to option (d).

Electrical power (P) in a circuit can be represented by several formulas derived from the basic definition $P\; =\; VI$ (where V is voltage and I is current) and Ohm's law ($V\; =\; IR$).

Let's examine each option:

1. $P\; =\; I^2R$: By substituting $V\; =\; IR$ into $P\; =\; VI$, we get $P\; =\; (IR)\; \backslash times\; I\; =\; I^2R$. This is a valid formula for power.
2. $P\; =\; VI$: This is the fundamental definition of electrical power. This is a valid formula for power.
3. $P\; =\; V^2/R$: By substituting $I\; =\; V/R$ (from Ohm's law) into $P\; =\; VI$, we get $P\; =\; V\; \backslash times\; (V/R)\; =\; V^2/R$. This is also a valid formula for power.
4. $P\; =\; IR^2$: This expression does not represent electrical power. If we try to derive it, it does not match the standard formulas. For instance, substituting $I\; =\; V/R$ into $P\; =\; VI$ gives $P\; =\; V(V/R)\; =\; V^2/R$, not $IR^2$. Similarly, substituting $V\; =\; IR$ into $P\; =\; VI$ gives $P\; =\; (IR)I\; =\; I^2R$, not $IR^2$.

Therefore, the term that does not represent electrical power in a circuit is $IR^2$.

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. The resistance (R) of the bulb's filament is a constant property (assuming it doesn't change significantly with temperature).

We can calculate the resistance of the bulb using the power formula $P\; =\; V^2/R$. Rearranging this formula to find R, we get:

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

Substituting the given values:

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

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

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

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

Calculating the value:

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

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