Question

Power developed in a uniform wire when
connected to a certain cell of negligible
internal resistance is P.If the wire is melted
and recast in a wire of length double that of
the original and the new wire is connected to
the same cell, then the power developed in
the wire would be:​

Answer 1

Answer:

same

Explanation:

because internal resistance is negligible

Answer 2

Answer:

If the new wire is connected to the same cell, then the power developed in the wire is reduced by one-fourth of the original.

Explanation:

Given, power in a cell of negligible internal resistance is P

     For a cell of emf, V and a resistor of resistance R, the power is

                                 $P=\frac{V^{2} }{R}$  

For a wire of resistivity, $\rho$, length $l$ and cross-sectional area, the resistance is given by

                                    $R=\frac{\rho l}{A}$

Given, the wire is melted and recast with double the length that of the original.

As the new wire is made of same material, the resistivity do not change.

And also the volume remains same.

   Therefore, from $V=l\times A$

      if the length of new wire is doubled, then area is halved.

Hence, the resistance of new wire is given by

                                    $R_{n} =\frac{\rho \times 2l}{\frac{A}{2} }=\frac{4\rho l}{A }=4R$

If the new wire is connected to the same cell, then the power developed in the wire is

                      $P_{n} =\frac{V^{2} }{R_{n} }$

                           $=\frac{V^{2} }{4R}=\frac{P }{4}$

The power developed in the wire is reduced by one-fourth of the original.