1.

Two bars of radius r and 2r are kept in constant as shown in figure. An electric current I is passed through the bars. Which of the following is correct?

Answer»

`V_(AB) =2V_(BC)`
Power across BC is four TIMES the power across AB
Current density in AB and BC is equal.
Electric field DUE to current inside AB and BC is equal.

SOLUTION :B.
`V_(AB)/V_(BC) = (I_(AB)R_(AB))/(I_(BC)R_(BC)) = R_(AB)/R_(BC) = (rho l/(2[pixx4r^2]))/(rho(l/(2[pir^2]))) = 1/4`
`("since" I_(AB) = I_(BC)`, "wire is of the same material")
`"THEREFORE, option (a) is incorrect"`.
b. `P_(BC)/P_(AB) = (I_2R_(BC))/(I_2R_(AB)) = (rho(l/(2[pixx4r^2])))/(rho(l/(2[pir^2]))) = 1/4`
`P_(AB) = 4P_(BC).` `Therefore, option (b) is correct"`.
c. `J_(AB)/J_(BC) = I/(pi xx 4r^2)/I/(pixxr^2) = 1/4`. "Therefore, option (c) is incorrect"`.
d. `E_(AB)/E_(BC) = [(V_(AB))/(l//2)]/[(V_(BC))/(l//2)] = 1/4` . "Therefore, option (d) is incorrect"`.


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