This set of Heat Transfer Multiple Choice Questions & Answers (MCQs) focuses on “Intensity Of Radiations”.

1. The solid angle is defined by a region by the rays of a sphere, and is measured as

a) A_{n}/r ^{2}

b) A_{n}/r

c) A_{n}/r ^{3}

d) A_{n}/r ^{4}

View Answer

Explanation: Solid angle is represented by α. Where, A

_{n}is projection of incident surface normal to line of propagation.

2. The plane angle is defined by a region by the rays of a circle, and is measured as

a) 3 L/ r

b) 2 L/ r

c) L/ r

d) 4 L / r

View Answer

Explanation: It is the ratio of arc of length on the circle to the radius of the circle. Where L is the arc of length and r is the radius of the circle.

3. When the incident surface is a sphere, the projection of surface normal to the line of propagation is the silhouette disk of the sphere which is a circle of the diameter of

a) Parabola

b) Sphere

c) Triangle

d) Hyperbola

View Answer

Explanation: It must be a circle of a diameter of a sphere.

4. If I _{n }denotes the normal intensity and I _{α} represents the intensity at angle α, then

a) I _{α }= 2 I _{n} cos α

b) I _{α }= 3 I _{n} cos α

c) I _{α }= 4 I _{n} cos α

d) I _{α }= I _{n} cos α

View Answer

Explanation: The intensity of radiation in a direction from the normal is proportional to cosine of the angle.

5. The intensity of normal radiation I _{n }is how much times the emissive power?

a) 1/π

b) 2/ π

c) 3/ π

d) 4/ π

View Answer

Explanation: I

_{n }= σ T

^{ 4}/ π and E = σ T

^{ 4}.

6. A small surface emits diffusively, and measurements indicate that the total intensity associated with emission in the normal direction I _{n }= 6500 W/square m sr. The emitted radiation is intercepted by three surfaces. Mark calculations for intensity associated with emission

a) 3500 W/m^{2} sr

b) 4500 W/m^{2} sr

c) 5500 W/m^{2} sr

d) 6500 W/m^{2} sr

View Answer

Explanation: For a diffusion emitter, the intensity of the emitted radiation is independent of direction.

7. Consider a deep-space probe constructed as 1 m diameter polished aluminum sphere. Estimate the equilibrium temperature that the probe reaches if the solar energy received is 300 W/m^{2}. For solar radiation, absorptivity of aluminum is 0.3 and the average emissivity appropriate for aluminum at low temperature is 0.04

a) 415.67 K

b) 315.67 K

c) 215.67 K

d) 115.67 K

View Answer

Explanation: Q

_{in }= α q A

_{P}= 70.7 W. Q

_{out }= E σ

_{b }A T

^{4}.

8. The total emissive power of the emitter with area d A and temperature T is given by

a) E = 2 σ T^{ 4} d A

b) E = 3 σ T^{ 4} d A

c) E = σ T^{ 4} d A

d) E = ½ σ T^{ 4} d A

View Answer

Explanation: E = I

_{n }π d A.

9. A black body of 0.2 m^{2} area has an effective temperature of 800 K. Calculate the intensity of normal radiations

a) 1234.65 W/m^{2} sr

b) 7396.28 W/m^{2} sr

c) 3476.74 W/m^{2} sr

d) 8739.43 W/m^{2} sr

View Answer

Explanation: I

_{n }= α T

^{4}/π = 7396.28 W/m

^{2}sr.

10. The energy radiated out decreases with increases in α and becomes zero at an angle of

a) 45

b) 30

c) 0

d) 90

View Answer

Explanation: I

_{α }= I

_{n}cos α. So at 90 degree it becomes zero.

**Sanfoundry Global Education & Learning Series – Heat Transfer.**

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