Question

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 10-6 Qx m, then find-
(i) Resistance of the conductor
(ii) Heat generated by the conductor in 5 seconds​

Answer 1

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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