Q33.
3
A conductor of length 50 cm and diameter 0.07cm is connected to a source of 220
V. If resistivity of the conductor is 4.4 X 10- Qx m, then find-
(i) Resistance of the conductor
(ii) Heat generated by the conductor in 5 seconds
Answers
Given : A conductor of length 50 cm and diameter 0.07cm is connected to a source of 220 V
resistivity of the conductor is 4.4 X 10- ⁶
To Find : (i) Resistance of the conductor
(ii) Heat generated by the conductor in 5 seconds
Solution:
R = Resistance
ρ = Resistivity = 4.4 x 10- ⁶ Ω-m
L = Length = 50 cm = 0.5 m
A = Area of Cross-section
A = πr²
r = radius = diameter /2
diameter = 0.07 cm = 7 x 10⁻² cm = 7 x 10⁻⁴ m
r = = 7/2 x 10⁻⁴ m = 3.5 x 10⁻⁴ m
A = π (3.5 x 10⁻⁴)²
=> A = 3.848 x 10⁻⁷ m²
R = 4.4 x 10- ⁶ x 0.5 / 3.848 x 10⁻⁷
=> R = 5.717 Ω
Resistance of the conductor = 5.717 Ω
Heat generated by the conductor in 5 seconds
H = I²Rt = (V/R)²Rt = V²t/R
= 220² x 5/ 5.717
= 42,329.9 Joule
= 42.33 kJoule
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