A family uses 8 kW of power.
(a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW?
(b) Compare this area to that of the roof of a typical house.

Question

A body of mass \(4\) kg is placed at a point \(P\) having coordinates \( (3,4) \) m. Under the action of force \( \mathbf{F} = (2\hat{i} + 3\hat{j}) \) N, it moves to a new point \(Q\) having coordinates \( (6,10) \) m in \(4\) sec. The average power and instantaneous power at the end of \(4\) sec are in the ratio:

Answer 1

(a) The area needed is 200 square meters.

To calculate, the useful power per square meter is 200 W/m² × 20% = 40 W/m². Thus, area = 8000 W / 40 W/m² = 200 m².

[1] Part (b): Roof Comparison

A typical house roof covers 100-200 m², roughly matching the required solar area. For instance, a 1200 sq ft (111 m²) house has about that footprint, while steeper roofs exceed it slightly. The solar panels could feasibly cover most of a standard home's roof.

Solution and Explanation

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