STATEMENT-1 `:` In a chain of 50 bulbs connected in series, one bulb is taken out remaining 49 bulbs are again connected in series across the same supply then light gets decreased in the room.
and
STATEMENT-2 `:` More resistance in the circuit means lesser current drawn source.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1
B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1
C. Statement-1 is True, Statement-2 is True
D. Statement-1 is False , Statement-2 is True

Answer

To solve the problem, we will analyze both statements step by step.

### Step 1: Analyze Statement 1
In the first statement, we have a chain of 50 bulbs connected in series. When one bulb is removed, we are left with 49 bulbs connected in series across the same voltage supply.

- **Initial Resistance (R_initial)**:
- Each bulb has a resistance of \( R \).
- Therefore, the total resistance with 50 bulbs is:
\[ R_{\text{initial}} = 50R \]

- **Final Resistance (R_final)**:
- After removing one bulb, the total resistance with 49 bulbs is:
\[ R_{\text{final}} = 49R \]

### Step 2: Power Calculation
Now, we will calculate the power consumed in both scenarios using the formula for power:
\[ P = \frac{V^2}{R} \]
where \( V \) is the constant voltage supplied.

- **Initial Power (P_initial)**:
\[ P_{\text{initial}} = \frac{V^2}{R_{\text{initial}}} = \frac{V^2}{50R} \]

- **Final Power (P_final)**:
\[ P_{\text{final}} = \frac{V^2}{R_{\text{final}}} = \frac{V^2}{49R} \]

### Step 3: Compare Initial and Final Power
Now we compare \( P_{\text{initial}} \) and \( P_{\text{final}} \):
- Since \( 49R < 50R \), it follows that:
\[ P_{\text{final}} > P_{\text{initial}} \]
This means that the power (and thus the light) actually increases when one bulb is removed, contrary to what Statement 1 claims.

### Step 4: Analyze Statement 2
The second statement claims that more resistance in the circuit means lesser current drawn from the source.

Using Ohm's Law:
\[ V = I \times R \implies I = \frac{V}{R} \]
- If resistance \( R \) increases, the current \( I \) decreases, confirming that Statement 2 is true.

### Conclusion
- **Statement 1** is false because removing one bulb increases the light (power).
- **Statement 2** is true because increasing resistance leads to a decrease in current.

Thus, the final conclusion is:
- Statement 1 is false.
- Statement 2 is true.

### Final Answer
- The correct option is that Statement 1 is false and Statement 2 is true.

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