## Class 9 Gravitational Force Problems with Solutions

Here are a few extra class 9 gravitational Force problems that will further help you in understanding the chapter. Practice these Gravitational Force Problems questions, most importantly try to solve on your own before looking at the solution given at the end of the questions.

### Gravitational Force Problems

Q 1. A block of wood is kept on a tabletop. The mass of the wooden block is 5 kg and its dimensions are 40 cm × 20 cm × 10 cm. Find the pressure exerted by the wooden block on the tabletop if it is made to lie on the tabletop with its sides of dimensions –

(a) 20 cm × 10 cm

(b) 40 cm × 20 cm

Thrust = F = m × g = 5 kg × 9.8 m/s² = 49 N

Pressure = thrust/Area

(a) Area = 20 cm × 10 cm = 0.02 m²

Pressure = 49/0.02 = 2450 Pa

(b) Area = 40 cm × 20 cm = 0.08 m²

Pressure = 49/0.08 = 612.5 Pa

Q 2. The mass of the earth is 6 × 1024 kg and that of the moon is 7.4 × 1022 kg. If the distance between the earth and the moon is 3.84×105 km, calculate the force exerted by the earth on the moon. (Take G = 6.7 × 10–11 N m²/kg²)

Use Formula F = GMm/r²

Q 3. Calculate the value of g, if universal gravitational constant (G) = 6.7 × 10–11 N m²/kg²; mass of the earth (M) = 6 × 1024 kg, and radius of the earth (R) = 6.4 × 106 m.

g = GM/r²

Q 4. The mass of an object is 10 kg. What is its weight on the earth?

Weight (W) = mass (m) × acceleration due to gravity (g)

W = m×g

Q 5. The volume of 50 g of a substance is 20 cm³. If the density of water is 1 g/cm³, will the substance float or sink?

Mass of packet (M) = 50 g

Volume of the packet (V) = 20 cm³

Density of packet (d) = m/v = 50 g/20 cm³ = 2.5 g/cm³

Since density of the body is greater than that of water, the body will sink.

Q 6. The volume of a 500 g sealed packet is 350 cm³. Will the packet float or sink in water if the density of water is 1 g/cm³? What will be the mass of the water displaced by this packet?

Mass of packet (M) = 500 g

Volume of the packet (V) = 350 cm³

Density of packet (d) = m/v = 500g/350 cm³ = 1.428g/cm³

Since the density of packet is more than the density of solution, packet will sink

Mass of water displaced = volume of packet × density of solution

= 350 × 1.2

= 420 g

Q 7. The weight of the man on earth is 150 N and on a certain planet is 25 N. Take g=10m/s² on earth

(a) Find the mass of the man on earth and planet

(b) Find the acceleration due to gravity on the planet

Read Also: Force and Laws of Motion Numerical Class 9

Weight on earth =150 N

So, mass of man on earth= 150/10 = 15 kg

Now mass does not vary and it will remain the same on earth and planet

Now weight on Planet = 25 N

Now as mass = 15 kg

Acceleration due to gravity on Planet = 25/15 = 1.66 m/s²

Read Also: CBSE NCERT Solutions & Revision Notes

Q 8. A solid weighs 50 gf in the air (where gf is gram force) and 44 gf when completely immersed in water. Calculate:

(a) the Upthrust

(b) the volume of the solid

(c) the relative density of the solid

Given the Density of water= 1000 kg/m³

(a) Upthrust = Weight in air – Weight in water = 50 – 44 = 6 gf = 0.0588 N

(b) Upthrust = V×ρw×g = V×1000×9.8 = 9800V

Now

9800V = 0.0588

or V = 0.000006 m³ = 6 cm³

Mass of Solid = 50 g

Volume = 6 cm³

Density=mass/volume = 50/6 = 8.33 gm/cm³

Now relative density of the solid=Density of Solid/Density of water = 8.33/1 = 8.33

Q 9. A body drops from the edge of the roof. It passes a 2 m high window in 0.1 s. How far is the roof above the window?

Let h be the height of the roof above the window

Applying Newton’s Equation above the window

S = ut + ½ gt²

h = ½ × 9.8 ×= 4.9t²

Now, considering motion from top to bottom of window

Time (t1) = t + 0.1, S = h+2

Put the values in

S = ut + ½ gt²

and solve for t = 1.9 seconds

h = 4.9t² = 4.9 × 1.9 × 1.9 = 19.41 m

Q 10. How far from the Earth must a body be along a line towards the sun so that the sun’s gravitational pull on it balances that of the earth? (Given: Distance b/w earth and sun’s center is 14.467×1010 km and Mass of the sun is 3.24 × 105 times the mass of the earth)

$\frac { Gm{ M }_{ e } }{ { x }^{ 2 } } =\frac { Gm{ M }_{ s } }{ { \left( r-x \right) }^{ 2 } }$
$\frac { { \left( r-x \right) }^{ 2 } }{ { x }^{ 2 } } =\frac { { M }_{ s } }{ { M }_{ e } } =3\cdot 24\times { 10 }^{ 5 }$
Solve the equation and get $x=2\cdot 62\times { 10 }^{ 10 }\quad m$