Class 9 Science Chapter 9 Question Answers - Gravitation

Q1: Weight of a girl is 294 N. Find her mass.
Ans: 
W = mg  
294 = m × 9.8  m
Q2: Calculate the force of gravitation due to a child of mass 25 kg on his mother of mass 75 kg if the distance between their centres is 1 m from each other. Given G = (20/3) × 10^–11 Nm² kg^–2
Ans: Here m₁ = 25 kg; m₂ = 75 kg; d = 1 m;
Using,

F = 12,500 × 10^–11
or F = 1.25 × 10^–7 N

Q3: A sealed tin of Coca-Cola of 400 g has a volume of 300 cm³. Calculate the density of the tin.
Ans: Here,
mass of tin, M = 400 g
Volume of tin, V = 300 cm³
Density of tin,


Q4: A sealed can of mass 600 g has a volume of 500 cm3. Will this can sink in water? Density of water is 1 g cm^–3.
Ans: Here,
mass of can, M = 600 g
Volume of can, V = 500 cm³
Density of can,
Since, density of the can is greater than the density of water, so the can will sink in water.

Q5: The gravitational force between two objects is 49 N. How much distance between these objects be decreased so that the force between them becomes double?
Ans: Let ‘r’ be the distance between the object of mass m₁ and m₂

Now, the distance is reduced to ‘x’ so that the force become twice, then

Dividing eq. (i) by (ii)

So, the distance must decrease by  times the original distance.

Q6: A force of 200 N is applied perpendicular to its surface having area 4 square metres. Calculate the pressure.
Ans:
Thrust = 200 N
Area = 4 m2  
Pressure = ?  
Pressure = Thrust / Area =
= 50 Nm^–2 = 50 Pa

Q7: The density of water is 1000 kg m³. If relative density of iron is 7.874, then calculate the density of iron.
Ans:  
Density of water = 1000 kg/m³
Relative density (R.D.) of iron = 7.874
Using, R.D. of iron we get  
Density of iron = R.D. of iron   × density of water  
= 7.874 × 1000 kg/m³
= 7874 kg/m³.

Q8: What is the force of gravitation between two point masses of 1 kg and 2 kg kept 1 m apart?
Ans:
m = 1 kg, m₂ = 2 kg, r = 1 m

= 13.34 x 10^-11N
This is an extremely small force.

Q9: A particle is thrown up vertically with a velocity of 50 m/s. (a) What will be its velocity at the highest point of its journey? (b) How high would the particle rise? (c) What time would it take to reach the highest point?
Ans: 
At the highest point the velocity will be zero. Considering activity A to B
Using
 = u + at
0 = 50 – 9.8 × t
t = 5.1 sec
Also v² – u² = 2as s = 127.5 m

Q10: A ball thrown up is caught by the thrower after 4 s. With what velocity was it thrown up? How high did it go? Where was it after 3 s ? (g = 9.8 ms^–2) 
Ans:

In case of upward motion of the ball from A to B

• Initial velocity, u = ?
• Final velocity, v = 0 (at maximum height)
• Time taken by the ball to reach the highest point = 2s (time of ascent = time of descent)
• Acceleration due to gravity g = -9.8 m/s² (upward motion)

Finding the initial velocity of the ball

Using the first equation of motion, v = u + gt:

v = u - gt

0 = u - 9.8 × 2

u = 19.6 m/s

The initial velocity of the ball is 19.6 m/s.

Finding the maximum height (h) attained by the ball

Using the second equation of motion, h = ut + 1/2 gt²:

h = 19.6 × 2 - 1/2 × 9.8 × (2)²

h = 39.2 - 19.6

h = 19.6 m

Let the ball be at C after t = 3 sec

Consider motion from A to C

• u = 19.6 m/s
• t = 3 s
• g = -9.8 m/s²
• s = h'

s = ut + 1/2 gt²

h' = 19.6 × 3 - 1/2 × 9.8 × (3)²

h' = 58.8 - 44.1 = 14.7 m

Distance from top

x = h - h'

x = 19.6 - 14.7 = 4.9 m

Hence, the ball goes to the maximum height of 19.6 m, the velocity at which it was thrown is 19.6 m/s, and the distance below its highest point after 3 sec is 4.9 m.

Q11: How much force should be applied on an area of 1 cm² to get a pressure of 15 Pa?
Ans: 
Here,
Area, A = 1 cm² = 10–4 m²
Pressure (P) = 15 Pa = 15 N/m²
As F = P × A  
= (15 N/m²) × (10–4 m²)  
= 1.5 × 10^–3 N

Q12: A force of 20N acts upon a body whose weight is 9.8N. What is the mass of the body and how much is its acceleration? Given, Force = 20 N, Weight W = 9.8 N. We know, W = mg; 9.8 = m × 9.8 m = 1 kg 
Ans: 
We know
F = ma
20 = 1 × a  
a = 20 m/s²

Q13: An object is thrown vertically upwards and reaches a height of 78.4 m. Calculate the velocity at which the object was thrown? (g = 9.8 m/s²)
Ans:
Given,
h = 78.4 m  
v = 0  
g = –9.8 m/s²
Now, v² = u² – 2gh

u = 39.2 m/s²

Q14: Two bodies A and B having masses 2 kg and 4 kg respectively are separated by 2 m. Where should a body of mass 1 kg be placed so that the gravitational force on this body due to bodies A and B is zero?
Ans:  
Mass of body a is M_a = 2 kg
Mass of body b is M_b = 4 kg
Mass of body c is M_c = 1 kg
Separation between a and b = 2 m
Let the body C be placed at a distance d from body A Gravitational force between A and C

Gravitational force between B and C is

For body C the gravitational force is 0.

Hence, F_AC = F_BC

 d = 0.83

Q15: Let us find force of attraction between two blocks lying 1 m apart. Let the mass of each block is 40 kg.
Ans:
F = ?
m₁ = 40 kg
m₂ = 40 kg
d = 1 m
G = 6.67 × 10–11 Nm²kg^–2

= 1.0672 x 10^-7N

Q1: A ball thrown up is caught by the thrower after 4 s. With what velocity was it thrown up? How high did it go? Where was it after 3 s ? (g = 9.8 ms^–2) 
Ans: Since the time taken to go up is the same as the time taken to come down, the time to go up is half of the total time. So, the time to go up is $\frac{4}{2}=2$ seconds. Let the initial upward velocity be $\mathrm{u. }\mathrm{<br\; style="box-sizing:\; border-box;\; word-break:\; break-word;\; color:\; black;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; letter-spacing:\; inherit\; !important;\; user-select:\; initial\; !important;\; font-size:\; 20px;">}\mathrm{<br\; style="box-sizing:\; border-box;\; word-break:\; break-word;\; color:\; black;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; letter-spacing:\; inherit\; !important;\; user-select:\; initial\; !important;\; font-size:\; 20px;">}\mathrm{<br\; style="box-sizing:\; border-box;\; word-break:\; break-word;\; color:\; black;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; letter-spacing:\; inherit\; !important;\; user-select:\; initial\; !important;\; font-size:\; 20px;">}\mathrm{<br\; style="box-sizing:\; border-box;\; word-break:\; break-word;\; color:\; black;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; letter-spacing:\; inherit\; !important;\; user-select:\; initial\; !important;\; font-size:\; 20px;">}\mathrm{<br\; style="box-sizing:\; border-box;\; word-break:\; break-word;\; color:\; black;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; letter-spacing:\; inherit\; !important;\; user-select:\; initial\; !important;\; font-size:\; 20px;">}\mathrm{<br\; style="box-sizing:\; border-box;\; word-break:\; break-word;\; color:\; black;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; letter-spacing:\; inherit\; !important;\; user-select:\; initial\; !important;\; font-size:\; 20px;">}$

Given: u = ?; a = – 9.8 ms^–2; t = 2 s; v = 0 (at the top); s = h
Using v = u + at

we get = u – 9.8 × 2

or u = 19.6 ms^–1

Again v² – u² = 2as  

0 – (19.6)² = 2 (–9.8) h

h = 19.6 m

After 2 s, it starts coming downwards (starting with u = 0). Considering downward motion,  

u = 0;

a = 9.8 ms^–2;

t = 3 – 2= 1s;  

s = ?

= 4.9 m from top.