To solve the problem step by step, we will follow these steps:
### Step 1: Calculate the total resistance of the bridge wire
The resistance of the bridge wire is given as \(0.1 \, \Omega/\text{cm}\) and the length of the wire is \(100 \, \text{cm}\).
\[
\text{Total resistance of the wire} = \text{Resistance per cm} \times \text{Length} = 0.1 \, \Omega/\text{cm} \times 100 \, \text{cm} = 10 \, \Omega
\]
### Step 2: Understand the ratio in which the wire is divided
The null point divides the wire in the ratio \(2:3\). This means that the left segment of the wire (where the unknown resistance \(X\) is connected) is \(2\) parts, and the right segment (where \(6 \, \Omega\) is connected) is \(3\) parts.
Let the total length of the wire be \(L = 100 \, \text{cm}\).
- Length of left segment = \( \frac{2}{2+3} \times 100 \, \text{cm} = \frac{2}{5} \times 100 \, \text{cm} = 40 \, \text{cm} \)
- Length of right segment = \( \frac{3}{2+3} \times 100 \, \text{cm} = \frac{3}{5} \times 100 \, \text{cm} = 60 \, \text{cm} \)
### Step 3: Calculate the resistance of each segment of the wire
Using the resistance per cm, we can calculate the resistance of each segment:
- Resistance of left segment (40 cm):
\[
R_1 = 0.1 \, \Omega/\text{cm} \times 40 \, \text{cm} = 4 \, \Omega
\]
- Resistance of right segment (60 cm):
\[
R_2 = 0.1 \, \Omega/\text{cm} \times 60 \, \text{cm} = 6 \, \Omega
\]
### Step 4: Apply the Wheatstone bridge condition
In a balanced meter bridge, the ratio of the resistances is given by:
\[
\frac{X}{6} = \frac{R_1}{R_2} = \frac{4}{6}
\]
From this, we can find the unknown resistance \(X\):
\[
X = 6 \times \frac{4}{6} = 4 \, \Omega
\]
### Step 5: Calculate the equivalent resistance of the circuit
Now we have:
- Resistance \(X = 4 \, \Omega\)
- Resistance of the wire segments: \(R_1 = 4 \, \Omega\) and \(R_2 = 6 \, \Omega\)
The total resistance in the circuit is the sum of the resistances in series:
\[
R_{\text{total}} = R_1 + R_2 = 4 \, \Omega + 6 \, \Omega = 10 \, \Omega
\]
### Step 6: Calculate the current drawn from the battery
Using Ohm's law, the current \(I\) drawn from the battery can be calculated as:
\[
I = \frac{V}{R_{\text{total}}} = \frac{5 \, \text{V}}{10 \, \Omega} = 0.5 \, \text{A}
\]
### Final Answer:
The current drawn from the battery is \(0.5 \, \text{A}\).
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