Question

A copper wire has resistivity \[1.63 \times {10^{ - 8}}\Omega m\] and cross section area of \[10.3 \times {10^{ - 4}}c{m^2}\]? Calculate the length of the wire required to make a \[20\Omega \] coil State Joule’s Law of heating. An electric iron of resistance \[20\Omega \] takes a current\[5A\]. Calculate the heat developed in \[0.5\min \].

Answer 1

Hint In the given question we are provided with values of resistance, resistivity and area and we know have to find length so we know the formula \[R = \dfrac{{\rho l}}{A}\], after substituting the values we can easily calculate the length of wire.

Complete step by step answer As given in the question we are given with resistivity of copper wire as, \[\rho = 1.63 \times {10^{ - 8}}\Omega m\]
And with area of cross section of wire as ,\[A = 10.3 \times {10^{ - 4}}c{m^2}\]
And resistance required to make state Joule’s law of heating as , \[R = 20\Omega \]
And assume length of wire as, \[L = l\]
We know the formula to calculate resistance as, \[R = \dfrac{{\rho l}}{A}\]
So substituting the values in it and make sure that you are substituting the values in same units so that we get right answer
So changing the units of area we get, \[A = 10.3 \times {10^{ - 8}}{m^2}\]
And now substituting the values in the formula written above,
\[R = \dfrac{{\rho l}}{A}\]
\[l = \dfrac{{RA}}{\rho }\]
\[l = \dfrac{{20 \times 10.3 \times {{10}^{ - 8}}}}{{1.63 \times {{10}^{ - 8}}}}\]
\[l = 126.4\]
So we get length of wire as \[l = 126.4\]
Now to calculate the heat required as,
We are given with resistance as, \[R = 20\Omega \]
And also we are given with time as, \[t = 0.5\min \]
And also with current as, \[I = 2A\]
Now we know the formula to calculate the heat as, \[H = {I^2}Rt\]
Now substituting the values in the formula we get,
\[H = {(5)^2} \times 20 \times 0.5\]
\[H = 25 \times 20 \times 0.5\]
\[H = 250\Omega \min {A^2}\]
And we get heat developed as \[250\Omega \min {A^2}\]

Note In the given question we given values of different variables in different units and in the formula to calculate the answer we have to put the variables in same units that can be altered example if we are having something in \[m\]then make different variables which can be referred in terms of length of area in this unit.