# Engineering Physics Questions and Answers – Units and Measurements

This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Units and Measurements”.

1. Which of the following is an inferior planet to earth?
a) Mercury
b) Saturn
c) Pluto
d) Neptune

Explanation: The planets which are closer to the sun than earth are called inferior planets. Therefore Mercury and Venus are the inferior planets. The other planets are superior planets.

2. What is the method for determining the distance of an inferior planet?
a) Kepler’s third law of planetary motion
b) Triangulation method
c) Parallax method
d) Copernicus method

Explanation: Copernicus assumed circular orbits for the planets. The angle formed at the earth between the earth-planet direction and earth-sun direction is called the planet’s elongation.

3. What is the method for determining the distance of a superior planet?
a) Copernicus method
b) Kepler’s third law of planetary motion
c) Parallax method
d) Triangulation method

Explanation: The distance of a superior planet can be found using Kepler’s third law of planetary motion. This law states the square of the period of revolution of a planet around the sun is proportional to the cube of the semi-major axis of the orbit.

4. Which of the following is used to find the depth of the sea-bed?
a) Laser method
b) Sonar method
d) Reflection or echo method

Explanation: The word sonar stands for sound navigation and ranging. On a sonar ultrasonic waves are transmitted through the ocean. They are reflected by the submerged rocks and received by the receiver. By measuring the time delay of the receipt, the distance can be determined.

5. The shadow of a tower standing on a level plane is found to be 50m longer when the sun’s altitude is 30° that when it is 60°. Find the height if the tower.
a) 1.732 m
b) 43.3 m
c) 25 m
d) 25√3 m

Explanation: h = d/(cotθ2-cotθ1)
Here, d = 50m
θ1 = 60°
θ2 = 30°
h = 43.3 m.
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6. The moon is observed from two diametrically opposite points A and B on the earth. The angle θ subtended at the moon by the two directions of observation is 1°54’. Given the diameter of the earth to be 1.276×107 m, compute the distance of the moon from the earth.
a) 3.84×108 m
b) 1.276×107 m
c) 3.84m
d) 1.27m

Explanation: Here parallactic angle,
θ = 1°54’ = 114’ = (114×60)’’ = 114×60×4.85×10-6rad = 3.32×10-2 rad
Basic, b = AB = 1.276×107m
The distance of the moon from the earth, S=b/θ = 3.84×108m.

7. The angular diameter of the sun is 1920’’. If the distance of the sun from the earth is 1.5×1011 m, what is the linear diameter of the sun?
a) 4.85×10-6 m
b) 1.4×109m
c) 2.35×108 m
d) 1.4×108m

Explanation: Distance of the sun from the earth = 1.5×1011m
Angular diameter of the sun = 1920×4.85×10-6 rad
Linear diameter of the sun = 1.5×1011×1920×4.85×10-6 = 1.4×109 m.

8. In a submarine fitted with SONAR, the time interval between the generation of an ultrasonic wave and the receipt of its echo is 200s. What is the distance of the enemy submarine? The speed of the sound in water is 1.450km/s.
a) 2.811 km
b) 112.5 km
c) 145 km
d) 100 km

Explanation: The distance of the enemy submarine = (speed × time)/2 = 145 km.

9. A 35mm wide slide with a 24mm×36mm picture is projected on a screen placed 12cm from the slide. The image of the slide picture on the screen measures 1m×1.5m. What is the linear magnification of the projector-screen arrangement?
a) 150.6
b) 1736
c) 1524
d) 41.67

Explanation: Areal magnification = (1×1.5)/(24×10-3×36×10-3) = 1736
Linear magnification = √1736 = 41.67.

10. If the size of an atom (≅1A) were enlarged to the tip of a sharp pin (≅10-5), how large would the height of Mount Everest be?
a) 107m
b) 108m
c) 1010m
d) 109m

Explanation: Magnification = 105
Apparent height of Mount Everest = Actual height × magnification = 104 × 105 = 109m.

Sanfoundry Global Education & Learning Series – Engineering Physics.

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