Question

A \(1\) kW bulb radiates light uniformly in all directions. The intensity at a point on the surface of the surrounding sphere of area \(200\,\text{m}^2\) is (in \(\text{W m}^{-2}\))

• A. \(2\)
• B. \(4\)
• C. \(5\)
• D. \(6\)
• E. \(8\)

Hint:

For isotropic sources (uniform in all directions), always use \(I=\frac{P}{4\pi r^2}\) or directly \(P/A\) if area is given.

Answer 1

The Correct Option is C

Step 1: Understand the concept of intensity.
Intensity is defined as the power per unit area: \[ I=\frac{P}{A} \] where \(P\) is total power radiated and \(A\) is the area over which it is distributed.

Step 2: Identify the given quantities.
Power of the bulb: \[ P=1\,\text{kW}=1000\,\text{W} \] Area of the sphere: \[ A=200\,\text{m}^2 \]

Step 3: Note uniform radiation.
Since the bulb radiates uniformly in all directions, the power is evenly distributed over the surface of the sphere.

Step 4: Apply the intensity formula.
\[ I=\frac{1000}{200} \]

Step 5: Simplify the expression.
\[ I=5\,\text{W m}^{-2} \]

Step 6: Interpret physically.
This means each square meter of the sphere receives \(5\) watts of power.

Step 7: Final answer.
\[ \boxed{5} \] which matches option \((3)\).