A uniform wire of resistance `(50 Omega)`is cut into 5 equal parts. These parts are now connected in parallel.the Equivalent resistance of the combination is

Options

A. (a)(2 Omega)
B. (b)(10 Omega)
C. (c )(250 Omega)
D. (d) (6250 Omega)

Answer

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

### Step 1: Understand the initial resistance
We start with a uniform wire that has a total resistance of \( R_0 = 50 \, \Omega \).

### Step 2: Cut the wire into equal parts
The wire is cut into 5 equal parts. Since resistance is proportional to length, cutting the wire into 5 equal parts means each part will have a resistance that is one-fifth of the original resistance.

### Step 3: Calculate the resistance of each part
The resistance of each part can be calculated as:
\[ R_{\text{individual}} = \frac{R_0}{5} = \frac{50 \, \Omega}{5} = 10 \, \Omega \]

### Step 4: Connect the parts in parallel
When resistors are connected in parallel, the equivalent resistance \( R_{\text{eq}} \) can be calculated using the formula for \( n \) equal resistances:
\[ R_{\text{eq}} = \frac{R_{\text{individual}}}{n} \]
where \( n \) is the number of resistors in parallel.

### Step 5: Substitute the values
Here, \( R_{\text{individual}} = 10 \, \Omega \) and \( n = 5 \):
\[ R_{\text{eq}} = \frac{10 \, \Omega}{5} = 2 \, \Omega \]

### Final Answer
The equivalent resistance of the combination is \( R_{\text{eq}} = 2 \, \Omega \).

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