Olympiad Test: Gravitation- 2 - Class 9 MCQ

- 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.

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

F =G× mM/r²

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

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

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

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

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

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

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

g_moon = (1/ 6) g_earth

∴ W_moon = (1/ 6) W_earth = (1/ 6) × 120 = 20