GATE AG 2015 • Q30 NAT • 1 Mark

Section 3: Farm Power

Farm Power

Subtopic: I.C. engine systems – fuel, cooling, lubrication, ignition, electrical, intake and exhaust

Question Statement:

The conversion efficiency of a solar cell is 12%. For a maximum power output of 9×10⁻³ W at an incident solar radiation of 250 W m⁻², the required surface area of the solar cell in mm² will be _____.

Official Answer Key:

300

Step-by-Step Solution & Explanation

Efficiency η = P_out/(Radiation × Area) ⇒ Area = P_out/(η×Radiation). Area = 9×10⁻³/(0.12×250) = 9×10⁻³/30 = 3×10⁻⁴ m². Converting to mm² (1 m² = 10⁶ mm²): Area = 3×10⁻⁴ × 10⁶ = 300 mm². Matches the key exactly.

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