Change in Resistance on Stretching the Wire Numerical Class-12 Nootan ISC Physics Ch-5 Electric-Resistance-and-Ohms-Law. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
Change in Resistance on Stretching the Wire Numerical Class-12 Nootan ISC Physics Ch-5 Electric-Resistance-and-Ohms-Law
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-5 | Electric Resistance and Ohm’s Law |
| Topics | Numericals on Change in Resistance on Stretching the Wire |
| Academic Session | 2025-2026 |
Numericals on Change in Resistance on Stretching the Wire
Class-12 Nootan ISC Physics Ch-5 Electric-Resistance-and-Ohms-Law.
Change in Resistance on Stretching the Wire:- When a wire is stretched, its resistance increases. This is because the increased length and decreased cross-sectional area both contribute to a higher resistance. The new resistance is proportional to the square of the stretching facto
Que-15. A thick wire of 10 Ω resistance is drawn into a thin wire so that its length becomes three times. Calculate its new resistance.
Ans-15 When a wire is stretched then R ∝ l^2
=> R1/R2 = l1^2/l2^2
=> 10/R2 = (l1/3l1)^2
=> 10/R2 = l1^2/9l1^2 = 1/9
=> R2 = 90 Ω
Que-16. A 1-kg piece of copper is drawn into a wire 1 mm thick and another piece into a wire 2 mm thick. Compare the resistances of these wires.
Ans-16 R = ρ(l/A)
R1/R2 = l1/l2 x A1/A2 {as ρ is constant}
but volume of wire is constant
i.e. l1A1 = l2A2
=> l1/l2 = A1/A2
=> R1/R2 = A2^2/A1^2 = (πr2^2)^2/(πr1^2)^2 = r2^4/r1^4
=> R1/R2 = (2/1)^4 = 16 : 1
— : End of Change in Resistance on Stretching the Wire Numerical Class-12 Nootan ISC Physics Ch-5 Electric-Resistance-and-Ohms-Law. :–
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