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A uniform wire of length l and radius r ...

A uniform wire of length l and radius r has a resistance of `100 Omega`. It is recast into wire of radius `(R )/(2)`

A

`1600 Omega`

B

`400 Omega`

C

`200 Omega`

D

`100 Omega`

Text Solution

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The correct Answer is:
To solve the problem step by step, we will follow these steps: ### Step 1: Understand the initial conditions We have a uniform wire of length \( L \) and radius \( R \) with a resistance of \( 100 \, \Omega \). ### Step 2: Calculate the volume of the original wire The volume \( V \) of the original wire can be calculated using the formula for the volume of a cylinder: \[ V = \text{Cross-sectional Area} \times \text{Length} = \pi R^2 L \] ### Step 3: Determine the new dimensions after recasting The wire is recast into a new wire with a radius of \( \frac{R}{2} \). Let the new length of the wire be \( L' \). The volume of the new wire is: \[ V' = \pi \left(\frac{R}{2}\right)^2 L' = \pi \frac{R^2}{4} L' \] ### Step 4: Set the volumes equal Since the wire is recast, the volume remains constant: \[ \pi R^2 L = \pi \frac{R^2}{4} L' \] Cancelling \( \pi R^2 \) from both sides gives: \[ L = \frac{L'}{4} \] Thus, we can express \( L' \) in terms of \( L \): \[ L' = 4L \] ### Step 5: Calculate the new resistance The resistance \( R \) of a wire is given by the formula: \[ R = \frac{\rho L}{A} \] where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area. For the original wire: \[ R = \frac{\rho L}{\pi R^2} = 100 \, \Omega \] For the new wire: \[ R' = \frac{\rho L'}{A'} \] where \( A' = \pi \left(\frac{R}{2}\right)^2 = \pi \frac{R^2}{4} \). Substituting \( L' = 4L \) into the resistance formula for the new wire: \[ R' = \frac{\rho (4L)}{\pi \frac{R^2}{4}} = \frac{16 \rho L}{\pi R^2} \] ### Step 6: Relate new resistance to original resistance We know from the original resistance: \[ R = \frac{\rho L}{\pi R^2} = 100 \, \Omega \] Thus, \[ R' = 16 \times R = 16 \times 100 = 1600 \, \Omega \] ### Final Answer The new resistance \( R' \) of the wire after recasting is: \[ \boxed{1600 \, \Omega} \]

To solve the problem step by step, we will follow these steps: ### Step 1: Understand the initial conditions We have a uniform wire of length \( L \) and radius \( R \) with a resistance of \( 100 \, \Omega \). ### Step 2: Calculate the volume of the original wire The volume \( V \) of the original wire can be calculated using the formula for the volume of a cylinder: \[ ...
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A uniform wire of length 1 and radius r has resistance 100 Omega . it is recast into a thin wire of (i) length 2l (ii) radius r//2 . Calculate the resistance of new wire in each case.

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Knowledge Check

  • A uniform wire of length / and radius r has a resistance of 100 Omega . It is recast into a wire of radius (r )/(2) . The resistance of new wire will be:

    A
    `1600 Omega`
    B
    `400 Omega`
    C
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    D
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    A
    `1.25m`
    B
    `2.5m`
    C
    `10m`
    D
    `20m`
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