For a wire `R/l = 1/2` and length of wire is `l = 5` cm . If potential differrence of 1V is applied across it current through wire will be : (R = Resistance)

Options

A. 40 A
B. 4A
C. 25 A
D. 2.5 A

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Understand the relationship between resistance, length, and the given ratio
We are given that \( \frac{R}{l} = \frac{1}{2} \), where \( R \) is the resistance and \( l \) is the length of the wire.

### Step 2: Calculate the resistance using the given length
We know the length of the wire \( l = 5 \) cm. First, we need to convert this length into meters:
\[ l = 5 \text{ cm} = 5 \times 10^{-2} \text{ m} \]

Now, using the relationship \( R = \frac{1}{2} l \):
\[ R = \frac{1}{2} \times (5 \times 10^{-2}) = \frac{5 \times 10^{-2}}{2} = 2.5 \times 10^{-2} \text{ Ohm} \]

### Step 3: Apply Ohm's Law to find the current
Ohm's Law states that \( V = IR \), where \( V \) is the potential difference, \( I \) is the current, and \( R \) is the resistance. We are given \( V = 1 \) V.

Rearranging the formula to find the current \( I \):
\[ I = \frac{V}{R} \]

Substituting the values we have:
\[ I = \frac{1 \text{ V}}{2.5 \times 10^{-2} \text{ Ohm}} = \frac{1}{0.025} = 40 \text{ A} \]

### Final Answer
The current through the wire is \( 40 \) A.

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