A copper wire has length L and cross-sectional area A. Its resistance is R. If we halved the length and halved the diameter of wire, then what will be the resistance of this wire?
Correct answer: C. 2R
- A. 4R
- B. 3R
- C. 2R
- D. R
Explanation
The correct answer is 2R. To understand why, we first consider that when the length of the wire is halved, the resistance is halved as well. Therefore, if the original resistance is R, halving the length gives R/2.Next, when the diameter of the wire is halved, the cross-sectional area decreases to a quarter of its original size (since area is proportional to the square of the diameter). This increase in resistance due to the diameter change causes the resistance to quadruple.Thus, the new resistance due to the change in diameter would be 4 times the reduced value from the halved length, which results in (4 * (R/2)) = 2R.The other options are incorrect because they do not account correctly for the effects of both changes on resistance. 4R and 3R incorrectly assume a different relationship between diameter and resistance, while R disregards the halving of the length.
Last updated
About Current Electricity
Steady current relates charge flow to potential difference, resistance and resistivity, including how temperature changes a conductor's resistance. Sources with internal resistance have terminal voltage below their emf when delivering current, and maximum power output occurs under a specific load condition rather than at zero resistance.
Practise Current Electricity
1,052 free Current Electricity MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Exams that ask Physics questions like this
Physics is on 13 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
Related questions
0.75 A current flows through an iron wire when a battery of 1.5 volts is connected across its ends. The length of the wire 5.0 m and its cross-sectional area is 2.5 x 10^-7 m2. What is the resistivity of iron?
1.6 mA current is flowing in conducting wire then the number of electrons flowing per second
1 kilo ohm = _ ohm
1 microvolt is
1/Req = 1/Rl + 1/R2 + 1/R3 + ..... +1/Rn is the combination in