Estimate the radiation pressure due to a bulb that emits 25 W of EM radiation at a distance of 9.5 cm from the center of the bulb. Assume that light is completely absorbed. Express your answer to two significant figures and include the appropriate units.

Given Information:  

Power of bulb = w = 25 W atts

distance = d = 9.5 cm = 0.095 m

Required Information:  

Radiation Pressure = ?

Radiation Pressure =7.34x10⁻⁷ N/m²

Explanation:

We know that radiation pressure is given by

P = I/c

Where I is the intensity of radiation and is given by

I = w/4πd²

Where w is the power of the bulb in watts and d is the distance from the center of the bulb.

So the radiation pressure becomes

P = w/c4πd²

Where c = 3x10⁸ m/s is the speed of light

P = 25/(3x10⁸*4*π*0.095²)

P = 7.34x10⁻⁷ N/m²

Therefore, the radiation pressure due to a 25 W bulb at a distance of 9.5 cm from the center of the bulb is 7.34x10⁻⁷ N/m²

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .

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