Discussion
A metallic wire of certain length ‘L’ and diameter ‘d’ is stretched in such a way that its length is increased to ‘3L’ and its diameter is reduced to ‘d/2’. What will be the change in resistance of the wire, if the resistivity remains same?
Answer: D
The resistance increases 12 times. The resistance of a wire is R = ρL/A, and its area depends on the square of the diameter. When the length becomes 3L and the diameter becomes d/2, the area becomes one fourth.
\[ R' = \frac{\rho (3L)}{A/4} = 12\,\frac{\rho L}{A} = 12R \]
A longer wire increases the resistance (by 3 times) and a thinner wire increases it again (by 4 times), so the effect is multiplied, 3 x 4 = 12, and it cannot decrease. So the fourth option is correct.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development