Resistance and Resistivity 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.
Resistance and Resistivity 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 Resistance and Resistivity |
| Academic Session | 2025-2026 |
Numericals on Resistance and Resistivity
Que-6. Find out the resistivity of a conductor in which a current density of 2.5 A-m^-2 is found when an electric field of 15 V-m^-1 is applied on it.
Ans-6 j = σ E => j = E/ρ
=> ρ = E/j => 15/2.5 = 6 Ω-m
Que-7. A very small current is sent through a wire of length 15 m and cross-sectional area 6.0 x 10^-7 m^-2 m Its resistance is measured to be 5.0 Ω. Find the specific resistance of the material of the wire at the temperature of the experiment. What will be the specific conductance?
Ans-7 R = ρ(l/A)
=> ρ = RA/l = 5 x 6 x 10^-7 / 15 = 2 x 10^-7 Ω-m
Specific Conductance
σ = 1/ρ = 1 / 2 x 10^-7 = 5 x 10^6 (Ω-m)^-1
Que-8. What length of a wire of diameter 0.46 mm and specific resistance 50 x 10^-6 ohm-cm will be required to prepare a coil of 10 ohm?
Ans-8 R = ρ(l/A)
l = RA/ρ = 10 x π x (0.23 x 10^-3)^2 / 50 x 10^-8
=> (10 x 3.14 x 0.23 x 0.23 x 10^-6 / 50 x 10^-8) x 10^2
=> 3.32 m = 332 cm
Que-9. The specific resistance of nichrome is 100 μΩ-cm. What will be the resistance of a nichrome wire of 5 m length and 0.01 cm^2 area of cross-section?
Ans-9 R = ρ(l/A) = 100 x 10^-8 x 5 / 0.01 x 10^-4 {100 μΩ-cm = 100 x 10^-8 Ω-m}
=> 10000 x 10^-4 x 5 {0.01 cm^2 = .01 x 10^-4 m^2}
=> 5 Ω
Que-10. A piece of copper of mass 50 g is drawn into a wire of length 50 m. Find the resistance of the wire. The density and the specific resistance of copper are 8.9 x 10^3 kg m^-3 and 1.7 x 10^-8 Ω-m respectively.
Ans-10 R = ρ(l/A) = ρl^2/Al = ρl^2/V {A x l = volume}
R = ρl^2/(M/d) = ρdl^2/M {d = M/V => V = M/d}
=> 1.7 x 10^-8 x 8.9 x 10^3 x (50)^2 / 50 x 10^-3
=> 7.6 Ω
Que-11. Two copper wires A and B of equal masses are taken. The length of A is double the length of B. If the resistance of the wire A be 160 Ω, then calculate the resistance of the wire B.
Ans-11 for equal mass
R1/R2 = l1^2/l2^2
=> RA/RB = (lA/lB)^2
=> 160 Ω / RB = (2lB/lB)^2 {lA = 2lB}
=> RB = 160/4 = 40 Ω
Que-12. A 200 V bulb takes a current of 0.2 A. The length of its filament is 22 cm and the specific resistance of the material of the filament is 7 × 10^5 Ω-cm. Find the radius of the filament (π = 22/7).
Ans-12 R = ρ(l/A) = ρ(l/πr^2)
=> r^2 = ρl/πR => r = √ρl/πR
=> r = √ρl x I / π V {V=IR}
=> r = √22 x 10^-2 x 7 x 10^-7 x 0.2 x 7 / 22 x 200
=> r = √7 x 7 x 10^-10 / 100
=> r = 7 x 10^-6 m = 7 x 10^-4 cm
Que-13. A rectangular block of iron has dimensions 1.2 cm x 1.2 cm x 15 cm. Compute the resistance of the block (a) between the two square ends, (b) between the two opposite rectangular faces. The specific resistance of iron at room temperature is 9.68 × 10^-8 Ω m.
Ans-13 Resistance between two square ends
(i) R1 = ρ(l/A)
=> R1 = 9.68 x 10^-8 x 15 x 10^-2 / (1.2 x 10^-2)^2
=> 100 x 10^-6 Ω = 100 μC
(ii) R2 = 9.68 x 1.2 x 10^-2 x 10^-8 / 15 x 10^-2 x 1.2 x 10^-2
=> 0.65 μΩ
Que-14. Two wires of different materials; of lengths 50 cm and 75 cm and respective diameters 0.2 mm and 0.1 mm, have resistances of 0.8 Ω and 1.5 Ω respectively. What is the ratio of the specific resistances of the materials?
Ans-14 R = ρ(l/A) = ρ(l/πr^2)
=> ρ = πR.r^2/l
=> ρ1/ρ2 = R1/R2 x l2/l1 x r1^2/r2^2
=> ρ1/ρ2 = R1/R2 x l2/l1 x (D1/D2)^2
=> ρ1/ρ2 = 0.8/1.5 x 75/50 x (0.2/0.1)^2
=> ρ1/ρ2 = 16/5 = 16 : 5
— : End of Resistance and Resistivity Numerical Class-12 Nootan ISC Physics Ch-5 Electric-Resistance-and-Ohms-Law. :–
Return to : – Nootan Solutions for ISC Class-12 Physics
Thanks
Please share with your friends
