A rod of 2.0-m length and a square (2.0 mm X 2.0 mm) cross section is made of a material with a resistivity of 6.0 E–8 Ohm meter. If a potential difference of 0.50 V is placed across the ends of the rod, at what rate is heat generated in the rod? *

8.33*10^-16 Watt

Explanation:

Given that

Length of the rod, l = 2 m,

Area of the rod, A = 2 x 2 mm² = 4*10^-6 m²

resistivity of the rod, p = 6*10^-8 ohm metre,

Potential difference of the rod, V = 0.5 V

Let R be the resistance of the rod, then

R = p * l / A

R = (6*10^-8 * 2) / (4*10^-6)

R = 3*10^14 ohm

Heat generated per second = V² / R Heat = (0.5)² / (3*10^14)

Heat = 0.25 / 3*10^14

Heat = 8.33*10^-16 Watt

Therefore, the rate at which heat is generated is 8.33*10^-16 Watt

v 154.9= 0.9 length length s little over 154.9cm