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Olympiad Test: Gravitation- 2 - Class 9 MCQ


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10 Questions MCQ Test - Olympiad Test: Gravitation- 2

Olympiad Test: Gravitation- 2 for Class 9 2024 is part of Class 9 preparation. The Olympiad Test: Gravitation- 2 questions and answers have been prepared according to the Class 9 exam syllabus.The Olympiad Test: Gravitation- 2 MCQs are made for Class 9 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Olympiad Test: Gravitation- 2 below.
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Olympiad Test: Gravitation- 2 - Question 1

A ball is thrown upward with the speed of 19.6 m/s, find the speed of ball after 3 seconds.

Detailed Solution for Olympiad Test: Gravitation- 2 - Question 1

- The initial speed (u) of the ball is 19.6 m/s, thrown upwards.
- The acceleration (a) due to gravity is -9.8 m/s² (negative because it acts downwards).
- After 3 seconds (t), use the formula: v = u + at.
- Substitute values: v = 19.6 + (-9.8) × 3.
- Calculate: v = 19.6 - 29.4 = -9.8 m/s.
- The negative sign indicates the ball is moving downward at 9.8 m/s after 3 seconds.

Olympiad Test: Gravitation- 2 - Question 2

A truck falls from a bridge into the river in 0.5 s. let g = 10 ms–2, what is the height of the bridge from the water surface?

Detailed Solution for Olympiad Test: Gravitation- 2 - Question 2

h = ut +(½)at2

Initial velocity , u = 0, t = 0.5s, a = 10 ms–2

h = 0 × 0.5 + (½)× 10 × (½) × (½)

= 1..25 m

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Olympiad Test: Gravitation- 2 - Question 3

Two objects of mass 200 kg and 800 kg separated by a distance of 50 m. Find the gravitational force between the two bodies.

Detailed Solution for Olympiad Test: Gravitation- 2 - Question 3

F =G× mM/r2

= 6.67 × 10–11 × {(200× 800)/(50×50)} 4.26×10-9 N

Olympiad Test: Gravitation- 2 - Question 4

When a ball is thrown vertically upwards, it goes through a distance of 19.6 m. Find the initial velocity of the ball.

Detailed Solution for Olympiad Test: Gravitation- 2 - Question 4

u = ?, v = 0, g = –9.8 m/s2 as ball goes up h 19.6 m

v2 = u2 + 2gh

02 = u2 + 2(–9.8 ) × 19.6

⇒ u2 = (19.6)2

⇒ u = 19.6 m/s

Olympiad Test: Gravitation- 2 - Question 5

Find the value of acceleration due to gravity on the surface of the moon, given mass of the moon = 7.4 ×1022 kg; radius of the moon = 1740 km; and G = 6.7 × 10–11 Nm2 /kg2

Detailed Solution for Olympiad Test: Gravitation- 2 - Question 5

g = GM/R2= 6.7× 10-11 ×7.4 × 1022 / (1.74×106 )2 ( R = 1740)

km = 1740 × 1000 m = 1.74 × 106 m

⇒ g = 1.63 m/s2

Olympiad Test: Gravitation- 2 - Question 6

If the distance between two bodies is doubled, the force of attraction F between them will be

Detailed Solution for Olympiad Test: Gravitation- 2 - Question 6

The force of attraction F is inversely proportional to the square of distance between two objects. Distance is doubled, force gets one–fourth.

Olympiad Test: Gravitation- 2 - Question 7

An object weighs 10 N in air, when immersed fully in a liquid it weighs only 8 N. The weight of the liquid displaced by the object will be

Detailed Solution for Olympiad Test: Gravitation- 2 - Question 7

The weight of liquid displaced = loss in weight = 10N – 8N = 2N

Olympiad Test: Gravitation- 2 - Question 8

The SI unit of relative density of any substance is

Detailed Solution for Olympiad Test: Gravitation- 2 - Question 8

Relative density is purely a ratio of two similar quantities (masses), so it has no units.

Olympiad Test: Gravitation- 2 - Question 9

What is the final velocity of a body moving against gravity when it attains the maximum height ?

Detailed Solution for Olympiad Test: Gravitation- 2 - Question 9

When a moving body attains the maximum height, final velocity becomes zero.

Olympiad Test: Gravitation- 2 - Question 10

A body whose weight is 120 kg on the earth. Find its weight on the surface of moon

Detailed Solution for Olympiad Test: Gravitation- 2 - Question 10

gmoon = (1/ 6) gearth

∴ Wmoon = (1/ 6) Wearth = (1/ 6) × 120 = 20

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