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Question: A copper wire has resistivity \[1.63 \times {10^{ - 8}}\Omega m\] and cross section area of \[10.3 \...

A copper wire has resistivity 1.63×10−8Ωm1.63 \times {10^{ - 8}}\Omega m and cross section area of 10.3×10−4cm210.3 \times {10^{ - 4}}c{m^2}? Calculate the length of the wire required to make a 20Ω20\Omega coil State Joule’s Law of heating. An electric iron of resistance 20Ω20\Omega takes a current5A5A. Calculate the heat developed in 0.5min⁡0.5\min .

Explanation

Solution

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=ρlAR = \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, ρ=1.63×10−8Ωm\rho = 1.63 \times {10^{ - 8}}\Omega m
And with area of cross section of wire as ,A=10.3×10−4cm2A = 10.3 \times {10^{ - 4}}c{m^2}
And resistance required to make state Joule’s law of heating as , R=20ΩR = 20\Omega
And assume length of wire as, L=lL = l
We know the formula to calculate resistance as, R=ρlAR = \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=10.3×10−8m2A = 10.3 \times {10^{ - 8}}{m^2}
And now substituting the values in the formula written above,
R=ρlAR = \dfrac{{\rho l}}{A}
l=RAρl = \dfrac{{RA}}{\rho }
l=20×10.3×10−81.63×10−8l = \dfrac{{20 \times 10.3 \times {{10}^{ - 8}}}}{{1.63 \times {{10}^{ - 8}}}}
l=126.4l = 126.4
So we get length of wire as l=126.4l = 126.4
Now to calculate the heat required as,
We are given with resistance as, R=20ΩR = 20\Omega
And also we are given with time as, t=0.5min⁡t = 0.5\min
And also with current as, I=2AI = 2A
Now we know the formula to calculate the heat as, H=I2RtH = {I^2}Rt
Now substituting the values in the formula we get,
H=(5)2×20×0.5H = {(5)^2} \times 20 \times 0.5
H=25×20×0.5H = 25 \times 20 \times 0.5
H=250Ωmin⁡A2H = 250\Omega \min {A^2}
And we get heat developed as 250Ωmin⁡A2250\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 mmthen make different variables which can be referred in terms of length of area in this unit.