Chapter 7 · 10 hours

Motion and Force

SEE past exam questions

Past questions and answers

90 questions set from this chapter, 11 of them more than once; 5 are most repeated (set, or a close variant set, in 3 or more exams). Most repeated first.

Most repeated · 3 of 31 exams
Asked 3 times
2082 Asar · Grade increment · 2 marks
2082 Chaitra · Lumbini · 2 marks
2079 Chaitra · Lumbini · 3 marks

Where does an object have more weight between the place lying in lower altitude from the sea level like Janakpur and places lying higher altitude like Kathmandu?

Answer

An object has more weight at Janakpur (lower altitude).

Reason: $g=\frac{GM}{(R+h)^2}$ decreases as height $h$ above sea level increases. Janakpur is nearer to the centre of the Earth, so $g$ is greater there, and since $W=mg$ the weight is greater than at Kathmandu (higher altitude).

Answer checked by Claude Sonnet · Oct 21, 2026

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Most repeated · 3 of 31 exams
2081 Chaitra · Koshi · 2 marks

Check the correctness of the equation $s = ut + \frac{1}{2}at^2$ by using unit-wise analysis.

Similar questions: Check v²=u²+2as² by units · Check s=ut+½at by units (2082 Chaitra · Karnali) · Check v=u+at² by units (2082 Chaitra · Koshi)

Answer

Equation: $s=ut+\frac{1}{2}at^2$

L.H.S. $s$ : m
$ut$ : $\text{m s}^{-1}\times\text{s}=\text{m}$
$\frac12at^2$ : $\text{m s}^{-2}\times\text{s}^2=\text{m}$

Every term has the unit m, so the units on both sides are equal. The equation is correct.

Answer checked by Claude Sonnet · Oct 21, 2026

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Most repeated · 3 of 31 exams
2082 Chaitra · Karnali · 2 marks

Check the validity of equation $S = ut + \frac{1}{2}at$ , by unit wise analysis.

Similar questions: Check v=u+at² by units (2082 Chaitra · Koshi) · Check v²=u²+2as² by units · Check s=ut+½at² by units (2081 Chaitra · Koshi)

Answer

Equation: $S=ut+\frac12at$

L.H.S. $S$ : m
$ut$ : $\text{m s}^{-1}\times\text{s}=\text{m}$
$\frac12at$ : $\text{m s}^{-2}\times\text{s}=\text{m s}^{-1}$

The unit of $\frac12at$ ( $\text{m s}^{-1}$ ) differs from m. The equation is not valid (the correct one has $t^2$ ).

Answer checked by Claude Sonnet · Oct 21, 2026

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Most repeated · 3 of 31 exams
2082 Chaitra · Koshi · 2 marks

Check the validity of the equation $v = u + at^2$ by its unit wise analysis.

Similar questions: Check s=ut+½at by units (2082 Chaitra · Karnali) · Check v²=u²+2as² by units · Check s=ut+½at² by units (2081 Chaitra · Koshi)

Answer

Equation: $v=u+at^2$

L.H.S. $v$ : $\text{m s}^{-1}$
$u$ : $\text{m s}^{-1}$
$at^2$ : $\text{m s}^{-2}\times\text{s}^2=\text{m}$

The unit of $at^2$ (m) is not the same as that of $v$ ( $\text{m s}^{-1}$ ). The equation is not valid (the correct one is $v=u+at$ ).

Answer checked by Claude Sonnet · Oct 21, 2026

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Most repeated · 3 of 31 exams
SEE 2080 model question, Set 3, Science and Technology · 2 marks

Check the correctness of the equation $v^2 = u^2 + 2as^2$ by using unit wise analysis.

Similar questions: Check s=ut+½at² by units (2081 Chaitra · Koshi) · Check s=ut+½at by units (2082 Chaitra · Karnali) · Check v=u+at² by units (2082 Chaitra · Koshi)

Answer

Equation: $v^2=u^2+2as^2$

L.H.S. $v^2$ : $(\text{m s}^{-1})^2=\text{m}^2\text{s}^{-2}$
$u^2$ : $\text{m}^2\text{s}^{-2}$
$2as^2$ : $\text{m s}^{-2}\times\text{m}^2=\text{m}^3\text{s}^{-2}$

The unit of $2as^2$ ( $\text{m}^3\text{s}^{-2}$ ) is not the same as the units of the other terms ( $\text{m}^2\text{s}^{-2}$ ). The equation is not dimensionally correct (the correct one is $v^2=u^2+2as$ ).

Answer checked by Claude Sonnet · Oct 21, 2026

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Asked 2 times
CDC model paper 2080, Science and Technology · 2 marks
2081 Chaitra · Lumbini · 2 marks
2082 Chaitra · Madhesh · 2 marks

Write any two effects of gravity.

Answer

Two effects of gravity:

It keeps objects and people on the surface of the Earth and gives them weight.
It makes freely released objects fall towards the Earth, and keeps the Moon and artificial satellites in orbit around the Earth (it also holds the planets in orbit around the Sun and causes tides).

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Chaitra · Karnali · 1 mark
2082 Chaitra · Sudurpashchim · 1 mark

What is the weight of body on the moon surface if its weight is 150 N on the earth surface?
(a) 20 N (b) 25 N (c) 30 N (d) 35 N

Answer

(b) 25 N

The Moon's gravity is about $\frac{1}{6}$ of the Earth's, so weight on the Moon $=\frac{150}{6}=25$ N.

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Chaitra · Bagmati · 1 mark
2080 Chaitra · Lumbini · 1 mark

If the Earth could be squeezed to the size of the Moon, what would be the effect on the value of acceleration due to gravity of the Earth?
(a) increases (b) decreases (c) becomes zero (d) remains the same

Answer

(a) increases

$g=\frac{GM}{R^2}$ . If the Earth is squeezed to the size of the Moon, its mass stays the same but the radius $R$ becomes much smaller, so $g$ increases.

Answer checked by Claude Sonnet · Oct 21, 2026

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Asked 2 times
2081 Chaitra · Bagmati · 1 mark
2080 Chaitra · Sudurpashchim · 1 mark

Which of the following group of fundamental units represent the derived unit of energy Joule (J)?
(a) $\text{kg ms}^{-2}$ (b) $\text{kg ms}^{-1}$ (c) $\text{kg m}^2\text{s}^{-2}$ (d) $\text{kg ms}^{-3}$

Answer

(c) $\text{kg m}^2\text{s}^{-2}$

Energy (work) = force $\times$ distance $=$ kg m s $^{-2}\times$ m $=$ kg m $^2$ s $^{-2}$ , so 1 J $=1$ kg m $^2$ s $^{-2}$ .

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Chaitra · Gandaki · 1 mark
2082 Asar · Grade increment · 1 mark

What is the relationship between weight of object (w) and acceleration due to gravity (g)?
(i) W = g (ii) W ∝ g (iii) W ∝ 1/g (iv) W = 2g

Answer

(ii) W ∝ g

Weight $W=mg$ . For a given mass, weight is directly proportional to the acceleration due to gravity.

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2081 Chaitra · Karnali · 4 marks
2082 Chaitra · Karnali · 2 marks

State Newton's Universal law of gravitation and prove that $F = \frac{Gm_1m_2}{d^2}$ with the help of appropriate diagram.

Answer

Newton's universal law of gravitation: Every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

Proof: Let two bodies of masses $m_1$ and $m_2$ be at a distance $d$ between their centres.

   (m1)<------------ d ------------>(m2)
    O---------------------------------O
   F <--                          --> F

The force is directly proportional to the product of the masses: $F\propto m_1m_2$ ... (i)
The force is inversely proportional to the square of the distance: $F\propto \frac{1}{d^2}$ ... (ii)

Combining (i) and (ii):

F\propto \frac{m_1m_2}{d^2}

F=G\frac{m_1m_2}{d^2}

where $G$ is the universal gravitational constant, $G=6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2}$ . Hence proved.

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Shrawan · Grade increment · 1 mark
2079 Chaitra · Lumbini · 2 marks

In which condition gravitational constant and gravitational force are equal?

Answer

The gravitational force equals the gravitational constant ( $F=G$ ) when two bodies of unit mass (1 kg each) are placed at unit distance (1 m) apart.

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2082 Chaitra · Bagmati · 2 marks
2079 Chaitra · Gandaki · 2 marks

Write any two differences between free fall and weightlessness.

Answer

Free fall	Weightlessness
The motion of a body falling only under gravity.	The state in which a body feels no weight (apparent weight is zero).
It is a motion with acceleration $g$ .	It is a condition/feeling that occurs during free fall (or far from any gravity).
It can be observed and measured, e.g. by the time taken to fall.	It is experienced, e.g. by astronauts in a satellite.

(Any two rows.)

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2080 Chaitra · Koshi · 4 marks
2079 Chaitra · Sudurpashchim · 3 marks

A man found 90 N weight of an object of 9 kg mass at a place. From the same place the object was dropped down for 2 s to reach on the ground assuming air resistance is zero;
(i) From what height did the object drop?
(ii) If a stone of 500 g is dropped down from the same heights, how long does it take to reach the ground? Explain with reason.

Answer

Given: $W=90$ N, $m=9$ kg, $t=2$ s, $u=0$ .

g=\frac Wm=\frac{90}{9}=10\ \text{m/s}^2

(i) Height:

h=ut+\frac12gt^2=0+\frac12\times10\times2^2=20\ \text{m}

(ii) The 500 g stone will also take 2 s. In free fall the time depends only on height and $g$ ( $t=\sqrt{2h/g}$ ), not on mass, because all bodies fall with the same acceleration (no air resistance).

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Chaitra · Sudurpashchim · 2 marks
2079 Chaitra · Bagmati · 2 marks

Write any two differences between acceleration due to gravity (g) and gravitational constant (G).

Answer

Acceleration due to gravity ( $g$ )	Gravitational constant ( $G$ )
Its value is $9.8\ \text{m/s}^2$ on the Earth's surface and varies with place, height and planet.	Its value is $6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2}$ everywhere, i.e. it is a universal constant.
It is a vector quantity; unit $\text{m/s}^2$ .	It is a scalar quantity; unit $\text{N m}^2\text{kg}^{-2}$ .
$g=\frac{GM}{R^2}$	Defined by $G=\frac{Fd^2}{m_1m_2}$

(Any two rows.)

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2081 Chaitra · Sudurpashchim · 2 marks

The value of acceleration due to gravity is more in the polar region than the equatorial region, why?

Answer

The value of $g$ is more at the poles than at the equator because:

The Earth is not a perfect sphere; it is flattened at the poles, so the polar radius is smaller than the equatorial radius. Since $g=\frac{GM}{R^2}$ , a smaller $R$ gives a larger $g$ at the poles.
The Earth's rotation is greatest at the equator, which reduces the effective $g$ there, while at the poles it has no such effect.

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2080 Chaitra · Gandaki · 4 marks

The weight of an object is more at polar region than at equator. Give any two reasons. If the value of 'g' on the surface of earth is $9.8\ \text{m/s}^2$ , at what height from the surface of earth the value of g becomes $2.5\ \text{m/s}^2$ . [ $M=6\times10^{24}$ kg, $R=6400$ km]

Answer

Two reasons why weight is more at the poles than at the equator:

The polar radius of the Earth is smaller than the equatorial radius (the Earth is flattened at the poles), so $g=\frac{GM}{R^2}$ is greater at the poles.
Because of the Earth's rotation, the effect on weight is maximum at the equator and zero at the poles. Hence the weight is greater at the poles ( $W=mg$ ).

Numerical:

Given: $g'=2.5\ \text{m/s}^2$ , $M=6\times10^{24}$ kg, $R=6400$ km $=6.4\times10^{6}$ m, $G=6.67\times10^{-11}$ . Find $h$ .

g'=\frac{GM}{(R+h)^2}

(R+h)^2=\frac{6.67\times10^{-11}\times6\times10^{24}}{2.5}=1.6\times10^{14}

R+h=1.265\times10^{7}\ \text{m}

h=1.265\times10^{7}-6.4\times10^{6}\approx6.25\times10^{6}\ \text{m}\approx6250\ \text{km}

The value of $g$ becomes $2.5\ \text{m/s}^2$ at a height of about 6250 km from the Earth's surface.

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2080 Chaitra · Bagmati · 1 mark

What is the weight of a body having 50 kg mass on the moon? (i) 8.35 N (ii) 83.5 N (iii) 835 N (iv) 8355 N

Answer

(ii) 83.5 N

Taking $g$ on the Moon as $1.67\ \text{m/s}^2$ : $W=mg=50\times1.67=83.5$ N. (Mass stays 50 kg; only weight changes.)

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2081 Chaitra · Karnali · 1 mark

What could be the weight of a body on the Earth if its weight on the Moon is 5 N?
(a) $\frac{5}{6}$ N (b) 5 N (c) 30 N (d) 50 N

Answer

(c) 30 N

Weight on the Moon is $\frac{1}{6}$ of that on the Earth, so weight on the Earth $=5\times6=30$ N.

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2082 Chaitra · Bagmati · 1 mark

What will be the weight of stone on the surface of the Earth if it weighs 78.4 N on the surface of the Moon? (i) 380 N (ii) 490 N (iii) 590 N (iv) 690 N

Answer

(ii) 490 N

Weight on the Earth $\approx6\times$ weight on the Moon $=6\times78.4\approx470$ N, which is closest to option (ii) 490 N.

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2082 Chaitra · Koshi · 1 mark

If the weight of an object on the surface of the Moon is 66 N, what might be the weight of the object on Earth? (Acceleration due to gravity of Moon is 1.62 m/s $^2$ )
(a) 396 N (b) 66 N (c) 11 N (d) 6.6 N

Answer

(a) 396 N

Mass $m=\frac{W_m}{g_m}=\frac{66}{1.62}\approx40.7$ kg. Weight on the Earth $=mg=40.7\times9.8\approx399$ N, which is about $6\times66=396$ N.

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2081 Chaitra · Koshi · 1 mark

If the weight of a person on the surface of the Earth is 294 N, what might be the weight of the person on the moon? (a) 6 N (b) 9.8 N (c) 49 N (d) 60 N

Answer

(c) 49 N

Weight on the Moon $=\frac{1}{6}\times$ weight on the Earth $=\frac{294}{6}=49$ N.

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2082 Chaitra · Gandaki · 1 mark

What will be the weight of a man on the Moon, if his weight on the Earth is 750 N? (Acceleration due to gravity of the Moon = 1.63 m/s $^2$ )
(a) 124.7 N (b) 125 N (c) 126.8 N (d) 127.8 N

Answer

(a) 124.7 N

Mass $m=\frac{750}{9.8}=76.5$ kg. Weight on the Moon $=mg_m=76.5\times1.63\approx124.7$ N.

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2082 Chaitra · Lumbini · 1 mark

What will be the weight of an object on the planet having acceleration due to gravity 2.5 m/s $^2$ if the object weighs 700 N on the Earth?
(a) 240 N (b) 1750 N (c) 175 N (d) 700 N

Answer

(c) 175 N

Mass $m=\frac{700}{9.8}\approx71.4$ kg (mass is the same on every planet). Weight on the planet $=71.4\times2.5\approx178.6$ N, which is closest to option (c) 175 N.

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2081 Chaitra · Lumbini · 2 marks

What mass can be lifted by a person on the Moon if he can lift 100 kg mass on the Earth? Calculate.

Answer

Given: mass lifted on the Earth $=100$ kg, $g_e=9.8\ \text{m/s}^2$ , $g_m=\frac{g_e}{6}$ .

The maximum force (weight) the person can lift is the same:

mg_m=100\times g_e

m=\frac{100\times g_e}{g_e/6}=600\ \text{kg}

The person can lift 600 kg on the Moon.

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2082 Chaitra · Sudurpashchim · 2 marks

Test the validity of power (P) = $\frac{W}{t}$ by unit wise analysis.

Answer

Equation: $P=\frac{W}{t}$

L.H.S. power $P$ : watt $=\text{kg m}^2\text{s}^{-3}$
R.H.S. $\frac{W}{t}$ : $\frac{\text{J}}{\text{s}}=\frac{\text{kg m}^2\text{s}^{-2}}{\text{s}}=\text{kg m}^2\text{s}^{-3}$

Both sides have the same units, so the equation is valid.

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2080 Chaitra · Karnali · 4 marks

Prove that 'Watt' is derived unit showing the proper relation among the fundamental units involved in it. Check unit wise homogeneity in
$s = ut^2 + \tfrac{1}{2}at^2$

Answer

Watt is a derived unit:

P=\frac{W}{t}=\frac{F\times d}{t}=\frac{\text{kg m s}^{-2}\times\text{m}}{\text{s}}=\text{kg m}^2\text{s}^{-3}

So 1 watt $=1\ \text{kg m}^2\text{s}^{-3}$ . It is formed by combining the fundamental units kg, m and s, hence it is a derived unit.

Homogeneity of $s=ut^2+\frac12at^2$ :

L.H.S. $s$ : m
$ut^2$ : $\text{m s}^{-1}\times\text{s}^2=\text{m s}$
$\frac12at^2$ : $\text{m s}^{-2}\times\text{s}^2=\text{m}$

The terms do not have the same unit (m s differs from m), so the equation is not homogeneous.

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2080 Chaitra · Bagmati · 1 mark

Which fundamental units of following represents the proper relation for Newton? (i) $kg\,m^{-2}s^{-2}$ (ii) $kg\,m^{2}s^{-1}$ (iii) $kg\,m^{2}s^{2}$ (iv) $kg\,m\,s^{-2}$

Answer

(iv) $kg\,m\,s^{-2}$

Force $=$ mass $\times$ acceleration $=\text{kg}\times\text{m s}^{-2}$ , so 1 N $=1\ \text{kg m s}^{-2}$ .

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2081 Chaitra · Lumbini · 1 mark

Write the fundamental units included in the unit of force.

Answer

The fundamental units in the unit of force (newton) are kilogram (kg), metre (m) and second (s): $1\ \text{N}=1\ \text{kg m s}^{-2}$ .

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2081 Shrawan · Grade increment · 1 mark

Why is watt called derived unit?

Answer

Watt is called a derived unit because it is obtained by combining fundamental units: $1\ \text{W}=1\ \text{kg m}^2\text{s}^{-3}$ .

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2081 Shrawan · Grade increment · 2 marks

Prove that unit of the work is (J) = $kg\,m^2/s^2$ .

Answer

Work $=$ force $\times$ displacement

W=F\times s

Force $=$ mass $\times$ acceleration $=\text{kg}\times\text{m s}^{-2}=\text{kg m s}^{-2}$ (newton).

W=\text{kg m s}^{-2}\times\text{m}=\text{kg m}^2\text{s}^{-2}

Hence the unit of work, joule (J) $=\text{kg m}^2/\text{s}^2$ .

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2081 Chaitra · Madhesh · 2 marks

Prove that, fundamental units of power is $kgm^2/s^3$ .

Answer

Power $=\frac{\text{work}}{\text{time}}$

Work $=\text{kg m}^2\text{s}^{-2}$ (joule), so

P=\frac{\text{kg m}^2\text{s}^{-2}}{\text{s}}=\text{kg m}^2\text{s}^{-3}

Hence the fundamental units of power (watt) are $\text{kg m}^2/\text{s}^3$ .

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2080 Chaitra · Koshi · 2 marks

Prove that unit of gravitational constant is derived unit.

Answer

From $F=\frac{Gm_1m_2}{d^2}$ ,

G=\frac{Fd^2}{m_1m_2}

Unit of $G=\frac{\text{N}\times\text{m}^2}{\text{kg}\times\text{kg}}=\frac{\text{kg m s}^{-2}\times\text{m}^2}{\text{kg}^2}=\text{m}^3\text{kg}^{-1}\text{s}^{-2}$

(also written $\text{N m}^2\text{kg}^{-2}$ ). Since it is made by combining the fundamental units kg, m and s, the unit of $G$ is a derived unit.

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2080 Grade increment · 1 mark

Keeping the mass of two bodies constant, the distance between them is doubled. What will be the gravitational force exerted between them?
(a) $F=G\frac{m_1 m_2}{d^2}$ N (b) $F=G\frac{m_1 m_2}{2d^2}$ N (c) $F=2G\frac{m_1 m_2}{d^2}$ N (d) $F=G\frac{m_1 m_2}{4d^2}$ N

Answer

(d) $F=G\frac{m_1m_2}{4d^2}$ N

New distance is $2d$ , so $F'=G\frac{m_1m_2}{(2d)^2}=\frac{1}{4}G\frac{m_1m_2}{d^2}$ . The force becomes one-fourth.

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2082 Chaitra · Sudurpashchim · 2 marks

What happens to the gravitational force when masses of two objects kept constant and the distance between them is made half? Explain with suitable equation.

Answer

The gravitational force becomes four times.

Originally $F=\frac{Gm_1m_2}{d^2}$ . When the distance is halved, $d'=\frac d2$ :

F'=\frac{Gm_1m_2}{(d/2)^2}=\frac{4Gm_1m_2}{d^2}=4F

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2080 Chaitra · Koshi · 1 mark

What change occurs in the gravitational force between two bodies when their masses are kept constant and the distance between their centres is reduced by 3 times?

Answer

Since $F\propto\frac{1}{d^2}$ , when the distance is reduced to $\frac13$ , the gravitational force becomes $3^2=$ 9 times the original.

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2080 Chaitra · Gandaki · 1 mark

Two bodies of unit mass are kept at unit distance. What is the gravitational force produced between the bodies? (a) $6.66\times10^{-9}$ N (b) $6.67\times10^{-10}$ N (c) $6.67\times10^{-11}$ N (d) $6.67\times10^{-12}$ N

Answer

(c) $6.67\times10^{-11}$ N

$F=\frac{Gm_1m_2}{d^2}$ with $m_1=m_2=1$ kg and $d=1$ m gives $F=G=6.67\times10^{-11}$ N.

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2082 Chaitra · Lumbini · 2 marks

What is the gravitational force of attraction between these two heavenly bodies? In which condition gravitational force of attraction is equal to gravitational constant.
Figure: Two circles (bodies) of masses m1 and m2 joined by a line, separated by distance d.

Answer

The gravitational force of attraction between the bodies of masses $m_1$ and $m_2$ at distance $d$ is

F=\frac{Gm_1m_2}{d^2}

Condition for $F=G$ : when $m_1=m_2=1$ kg and $d=1$ m, then $F=G\times\frac{1\times1}{1^2}=G$ .

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Chaitra · Madhesh · 2 marks

Calculate the gravitational force of attraction between objects A and B as shown in the given figure?
Figure: Two circles A (mass 2,000 Kg) and B (mass 4,000 Kg), centres joined by a line labelled 80 Km.

Answer

Given: $m_1=2000$ kg, $m_2=4000$ kg, $d=80$ km $=8\times10^4$ m, $G=6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2}$ .

F=\frac{Gm_1m_2}{d^2}=\frac{6.67\times10^{-11}\times2000\times4000}{(8\times10^4)^2}

F=\frac{5.336\times10^{-4}}{6.4\times10^{9}}\approx8.34\times10^{-14}\ \text{N}

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2082 Chaitra · Bagmati · 2 marks

Calculate the force of gravitation between the heavenly bodies P and Q on the basis of information given in the figure.
Figure: Two circles: P of mass 1200 kg on the left and Q of mass 4000 kg on the right, with a line joining their centres labelled 4000 km.

Answer

Given: $m_P=1200$ kg, $m_Q=4000$ kg, $d=4000$ km $=4\times10^6$ m, $G=6.67\times10^{-11}$ .

F=\frac{Gm_Pm_Q}{d^2}=\frac{6.67\times10^{-11}\times1200\times4000}{(4\times10^6)^2}

F=\frac{3.2016\times10^{-4}}{1.6\times10^{13}}\approx2.0\times10^{-17}\ \text{N}

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2082 Asar · Grade increment · 2 marks

The mass of objects A and B as shown in the diagram are 2,500 kg and 4,000 kg respectively. Calculate gravitational force exerted between them.
Figure: Two circles A and B placed apart, with the distance between them marked as 600 m.

Answer

Given: $m_A=2500$ kg, $m_B=4000$ kg, $d=600$ m, $G=6.67\times10^{-11}$ .

F=\frac{Gm_Am_B}{d^2}=\frac{6.67\times10^{-11}\times2500\times4000}{600^2}=\frac{6.67\times10^{-4}}{3.6\times10^{5}}

F\approx1.85\times10^{-9}\ \text{N}

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Chaitra · Lumbini · 4 marks

Mass of the objects A and B as shown in the diagram are 2,500 kg and 4,000 kg respectively.
(i) Calculate the gravitational force between the objects A and B.
(ii) If acceleration due to gravity on body 'A' is 9.8 m/s $^2$ when it is on the surface of the earth? At which height from the surface of earth the body 'A' should be taken so that the acceleration due to gravity is 4 m/s $^2$ ?
Figure: Two spheres A and B drawn side by side with the distance between them marked 600 m.

Answer

(i) Given: $m_A=2500$ kg, $m_B=4000$ kg, $d=600$ m.

F=\frac{Gm_Am_B}{d^2}=\frac{6.67\times10^{-11}\times2500\times4000}{600^2}\approx1.85\times10^{-9}\ \text{N}

(ii) Given: $g=9.8\ \text{m/s}^2$ on the surface, $g'=4\ \text{m/s}^2$ , $R=6400$ km.

\frac{g'}{g}=\left(\frac{R}{R+h}\right)^2\Rightarrow\frac{4}{9.8}=\left(\frac{R}{R+h}\right)^2

\frac{R+h}{R}=\sqrt{2.45}=1.565

h=0.565\times6400\approx3617\ \text{km}

The body must be taken to about 3600 km above the Earth's surface.

Answer checked by Claude Sonnet · Oct 21, 2026

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2082 Chaitra · Koshi · 4 marks

Mass of the two heavenly bodies are $2\times10^{30}$ kg and $1.9\times10^{27}$ kg respectively. If the distance between their centers is $1.8\times10^{8}$ km. Calculate the gravitational force between them. What would be the force of attraction if the distance between the masses is made $3.6\times10^{8}$ km.

Answer

Given: $m_1=2\times10^{30}$ kg, $m_2=1.9\times10^{27}$ kg, $d=1.8\times10^{8}$ km $=1.8\times10^{11}$ m.

F=\frac{Gm_1m_2}{d^2}=\frac{6.67\times10^{-11}\times2\times10^{30}\times1.9\times10^{27}}{(1.8\times10^{11})^2}=\frac{2.535\times10^{47}}{3.24\times10^{22}}

F\approx7.8\times10^{24}\ \text{N}

When the distance is doubled ( $3.6\times10^{8}$ km), $F\propto\frac{1}{d^2}$ so the force becomes one-fourth:

F'=\frac{7.82\times10^{24}}{4}\approx1.96\times10^{24}\ \text{N}

Answer checked by Claude Sonnet · Oct 21, 2026

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2082 Chaitra · Madhesh · 1 mark

What is the gravitational force of attraction exerted between A and B in the given figure?
(a) 8.34×10 $^{-16}$ N (b) 8.34×10 $^{-17}$ N (c) 8.34×10 $^{-18}$ N (d) 8.34×10 $^{-19}$ N
Figure: Two circles A and B joined by a line labelled 800 km; A has mass 200 kg and B has mass 400 kg.

Answer

(c) $8.34\times10^{-18}$ N

$d=800$ km $=8\times10^5$ m:

F=\frac{6.67\times10^{-11}\times200\times400}{(8\times10^5)^2}=\frac{5.336\times10^{-6}}{6.4\times10^{11}}\approx8.34\times10^{-18}\ \text{N}

Answer checked by Claude Sonnet · Oct 21, 2026

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2082 Chaitra · Gandaki · 2+2 marks

(a) A stone is dropped from a tall tower that reaches to the ground after 4 s. Calculate the velocity with which the stone lands on the ground.
(b) Calculate the force of gravitational attraction between the two bodies shown in the diagram.
Figure: Two spheres separated by a distance of 500 km; the left sphere has mass 4,500 kg and the right sphere has mass 2,000 kg.

Answer

(a) Given: $u=0$ , $t=4$ s, $g=9.8\ \text{m/s}^2$ .

v=u+gt=0+9.8\times4=39.2\ \text{m/s}

(b) $m_1=4500$ kg, $m_2=2000$ kg, $d=500$ km $=5\times10^5$ m.

F=\frac{6.67\times10^{-11}\times4500\times2000}{(5\times10^5)^2}=\frac{6.003\times10^{-4}}{2.5\times10^{11}}\approx2.4\times10^{-15}\ \text{N}

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Chaitra · Gandaki · 1 mark

What could be the gravitational force produced between object A and B in the given hypothetical condition shown in the figure?
(i) $6.67\times10^{-10}$ N (ii) $6.67\times10^{-11}$ N (iii) $6.67\times10^{-12}$ N (iv) $6.67\times10^{-13}$ N
Figure: Two circles: object A = 4 kg on the left and object B = 10 kg on the right, joined by a line marked 2 m (distance between them).

Answer

(i) $6.67\times10^{-10}$ N

F=\frac{6.67\times10^{-11}\times4\times10}{2^2}=\frac{6.67\times10^{-11}\times40}{4}=6.67\times10^{-10}\ \text{N}

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Grade increment · 4 marks

In the given condition what will be the gravitational force produced between P and Q and between Q and R? In which case gravitational force is more by how much? Calculate.
Figure: Diagram: three spheres in a line. P (mass 500 kg) is 600 m from Q (mass 1100 kg); Q is 300 m from R (mass 900 kg). A radius of 100 m is marked on R.

Answer

Taking the distances between the centres as given: $d_{PQ}=600$ m, $d_{QR}=300$ m; $m_P=500$ kg, $m_Q=1100$ kg, $m_R=900$ kg.

Force between P and Q:

F_{PQ}=\frac{6.67\times10^{-11}\times500\times1100}{600^2}=\frac{3.67\times10^{-5}}{3.6\times10^{5}}\approx1.02\times10^{-10}\ \text{N}

Force between Q and R:

F_{QR}=\frac{6.67\times10^{-11}\times1100\times900}{300^2}=\frac{6.60\times10^{-5}}{9\times10^{4}}\approx7.34\times10^{-10}\ \text{N}

The force is greater between Q and R, by

7.34\times10^{-10}-1.02\times10^{-10}\approx6.32\times10^{-10}\ \text{N}

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Chaitra · Bagmati · 4 marks

In the given figure, mass of A and B are $6\times10^{24}$ kg and 400 kg respectively, and the radius of A is 6400 km.
(i) Calculate the gravitational force between the objects A and B.
(ii) To what height the body B should be taken from the surface of the body 'A' so that the value of acceleration due to gravity becomes 6 m/s $^2$ ?
Figure: Circle A (large body) with small body B on its surface, joined by a radius line.

Answer

Given: $M_A=6\times10^{24}$ kg, $m_B=400$ kg, $R=6400$ km $=6.4\times10^6$ m.

(i) B is on the surface of A, so $d=R$ :

F=\frac{6.67\times10^{-11}\times6\times10^{24}\times400}{(6.4\times10^6)^2}=\frac{1.6\times10^{17}}{4.096\times10^{13}}\approx3.9\times10^{3}\ \text{N}

(ii) $g'=6\ \text{m/s}^2=\frac{GM}{(R+h)^2}$ :

(R+h)^2=\frac{6.67\times10^{-11}\times6\times10^{24}}{6}=6.67\times10^{13}

R+h=8.17\times10^{6}\ \text{m}

h=8.17\times10^{6}-6.4\times10^{6}\approx1.77\times10^{6}\ \text{m}\approx1770\ \text{km}

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Shrawan · Grade increment · 2 marks

Mass and radius of a heavenly body are $7.342 \times 10^{22}$ kg and 1,737 km respectively. Calculate the acceleration due to gravity on its surface.

Answer

Given: $M=7.342\times10^{22}$ kg, $R=1737$ km $=1.737\times10^{6}$ m, $G=6.67\times10^{-11}$ .

g=\frac{GM}{R^2}=\frac{6.67\times10^{-11}\times7.342\times10^{22}}{(1.737\times10^{6})^2}=\frac{4.897\times10^{12}}{3.017\times10^{12}}

g\approx1.62\ \text{m/s}^2

Answer checked by Claude Sonnet · Oct 21, 2026

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2082 Chaitra · Karnali · 2 marks

Mass of a planet is $1.9 \times 10^{27}$ kg and its radius is 71,000 km, calculate its acceleration due to gravity of the planet.

Answer

Given: $M=1.9\times10^{27}$ kg, $R=71000$ km $=7.1\times10^{7}$ m.

g=\frac{GM}{R^2}=\frac{6.67\times10^{-11}\times1.9\times10^{27}}{(7.1\times10^{7})^2}=\frac{1.267\times10^{17}}{5.041\times10^{15}}

g\approx25.1\ \text{m/s}^2

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Chaitra · Madhesh · 1 mark

If the mass and radius of heavenly object are four times greater than that of the earth. What is the value of acceleration due to gravity of that object. (a) 1.22 m/s $^2$ (b) 2.45 m/s $^2$ (c) 4.9 m/s $^2$ (d) 9.8 m/s $^2$

Answer

(b) 2.45 m/s $^2$

$g'=\frac{G(4M)}{(4R)^2}=\frac{4}{16}\cdot\frac{GM}{R^2}=\frac{g}{4}=\frac{9.8}{4}=2.45\ \text{m/s}^2$ .

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Chaitra · Sudurpashchim · 1 mark

If the radius of the Earth is halved keeping the mass constant, how will the value of acceleration due to gravity 'g' change?
(i) becomes $\frac{1}{2}$ times of the original value (ii) becomes $\frac{1}{3}$ of the original value (iii) becomes 4 times of the original value (iv) becomes 6 times of the original value

Answer

(iii) becomes 4 times of the original value

$g'=\frac{GM}{(R/2)^2}=4\frac{GM}{R^2}=4g$ .

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2081 Chaitra · Gandaki · 2 marks

What is free fall? Write the value of acceleration due to gravity when a body falls freely towards the surface of the Earth.

Answer

Free fall: The fall of a body towards the Earth under the action of gravity alone (with no air resistance or other force) is called free fall.

The acceleration of a freely falling body is the acceleration due to gravity, $g=9.8\ \text{m/s}^2$ (about $9.8\ \text{m/s}^2$ directed towards the Earth's centre).

Answer checked by Claude Sonnet · Oct 21, 2026

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2082 Chaitra · Sudurpashchim · 1 mark

Define free fall.

Answer

The motion of a body falling towards the Earth only under the influence of gravity, without any other force such as air resistance, is called free fall.

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2081 Chaitra · Madhesh · 2 marks

Write any two conditions for an object to be weightlessness?

Answer

Two conditions for weightlessness:

When an object is in free fall (e.g. falling in a freely falling lift or orbiting satellite) so that it has no reaction from its support.
When an object is at a point very far from any heavenly body where gravitational force is negligible (or where the gravitational forces cancel, e.g. at the centre of the Earth).

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Chaitra · Madhesh · 1 mark

What is a dependent variable in a process of throwing a stone with a stretched rubber band?

Answer

The dependent variable is the distance travelled by the stone (how far the stone goes), which depends on how much the rubber band is stretched.

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2081 Chaitra · Lumbini · 2 marks

What are the dependent variable and independent variable included when a piece of paper is thrown with stretched rubber.

Answer

Independent variable: the amount by which the rubber band is stretched (the force applied), which is changed by the experimenter.
Dependent variable: the distance travelled by the piece of paper, which depends on the stretch.

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Chaitra · Sudurpashchim · 2 marks

A student has to test the change in effort applied and the number of pulleys used. In this experiment, identify a controlled variable and a dependent variable from number of pulleys, load of object, effort applied, friction produced in the pulleys and type of pulley.

Answer

Controlled variable: the load of the object (also the type of pulley and friction are kept same).
Dependent variable: the effort applied (it changes as the number of pulleys, the independent variable, is changed).

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2080 Chaitra · Bagmati · 4 marks

Acceleration due to gravity of three heavenly bodies are given in the table alongside. On the basis of it write the answer of the following.

Heavenly body A B C
Acceleration due to gravity (g) 25 m/s² 9.8 m/s² 1.67 m/s²
i. In which heavenly body it is easier to lift object? Why?
ii. If an object is dropped from the same height in all three heavenly bodies, in which heavenly body does the object fall faster? Why?

Heavenly body	A	B	C
Acceleration due to gravity (g)	25 m/s²	9.8 m/s²	1.67 m/s²

Answer

i. It is easiest to lift an object on heavenly body C, because it has the least $g$ (1.67 m/s $^2$ ), so the weight $W=mg$ of the object is least there.

ii. The object falls fastest on heavenly body A, because it has the greatest acceleration due to gravity (25 m/s $^2$ ), so the object's speed increases fastest and it takes the shortest time to reach the ground ( $t=\sqrt{2h/g}$ ).

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Chaitra · Madhesh · 1 mark

Which formula will you use to find out height from the earth surface from where a stone is dropped down and time taken by the stone to reach the ground is known?

Answer

$h=ut+\frac12gt^2$ , which for a dropped stone ( $u=0$ ) becomes $h=\frac12gt^2$ .

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Chaitra · Madhesh · 4 marks

Rita and Pema dropped stone from the top of two different towers standing nearby. Rita dropped the stone of mass 1 Kg. After 1s time Pema dropped the stone of mass 1.5 Kg which reached the ground 0.46s later than the Rita's stone. If the tower from where Rita dropped the stone is 60 m high, what is the height of tower from where Pema dropped the stone?

Answer

Given: Rita's tower $h_1=60$ m, $g=9.8\ \text{m/s}^2$ , $u=0$ .

Time of fall of Rita's stone:

t_1=\sqrt{\frac{2h_1}{g}}=\sqrt{\frac{120}{9.8}}=3.50\ \text{s}

Pema dropped her stone 1 s later and it landed 0.46 s after Rita's, so Pema's stone fell for

t_2=3.50+0.46-1=2.96\ \text{s}

h_2=\frac12gt_2^2=\frac12\times9.8\times(2.96)^2\approx42.9\ \text{m}

The height of Pema's tower is about 43 m. (The masses do not affect the time of fall.)

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Shrawan · Grade increment · 1 mark

Weight of object is 800 N on the surface of earth whose radius is 6,371 km. What is the weight of the object at the height 6,371 km from the surface of the earth?
(a) 200 N (b) 800 N (c) 1600 N (d) 3200 N

Answer

(a) 200 N

At a height equal to $R$ the distance from the Earth's centre is $2R$ . Since $W\propto\frac{1}{d^2}$ , the weight becomes $\frac{1}{4}$ : $W'=\frac{800}{4}=200$ N.

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Chaitra · Koshi · 2 marks

How does this fall differ from object falling on the moon? Write any two.
Figure: A tall coconut palm tree with a point X near the top among the fruits and a point Y on the ground below, joined by a vertical dotted line (object falls from X to Y).

Answer

Differences between a fall on the Earth and a fall on the Moon:

On the Moon $g\approx1.63\ \text{m/s}^2$ (about $\frac16$ of the Earth's), so the object falls more slowly and takes longer to reach the ground.
The Moon has no atmosphere, so there is no air resistance and a fruit and a feather fall together; on the Earth air resistance affects light objects.

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Chaitra · Koshi · 2 marks

If a cricket ball is thrown vertically upward into the sky with velocity of 18 m/s. Calculate the maximum height attained by the ball. [Note: Neglect the air resistance.]

Answer

Given: $u=18$ m/s, $v=0$ at the top, $g=-9.8\ \text{m/s}^2$ .

v^2=u^2-2gh

0=18^2-2\times9.8\times h

h=\frac{324}{19.6}\approx16.5\ \text{m}

The maximum height attained is about 16.5 m.

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2081 Chaitra · Lumbini · 1 mark

For how long does a stone remain in air when it is thrown upward with an initial velocity of 20 m/s?
(a) 2 s (b) 4 s (c) 10 s (d) 20 s

Answer

(b) 4 s

Time of flight $T=\frac{2u}{g}=\frac{2\times20}{9.8}\approx4$ s (about 4 s; 2 s up and 2 s down with $g\approx10$ ).

Answer checked by Claude Sonnet · Oct 21, 2026

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2082 Chaitra · Madhesh · 1 mark

In the figure height of four different mountains are given. Among them which mountain has least acceleration due to gravity?
Figure: Four mountain peaks A, B, C, D drawn on a baseline; B is the tallest, A and C are shorter, D is of medium-high height.

Answer

Mountain B, because it is the tallest. The value of $g$ decreases with height, so it is least at the highest peak (farthest from the Earth's centre).

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2082 Chaitra · Madhesh · 1 mark

Write a difference between mass and weight.

Answer

Mass is the quantity of matter in a body and is constant everywhere (unit kg), whereas weight is the gravitational force on a body ( $W=mg$ ), which changes from place to place (unit N).

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2082 Chaitra · Sudurpashchim · 1 mark

Which of the following statements is correct? (i) Value of acceleration due to gravity decreases when we go higher from the surface of the Earth. (ii) Value of acceleration due to gravity decreases when the height increases from the surface of the Earth. (iii) Value of acceleration due to gravity is less at the polar regions than the equatorial region of the Earth. (iv) Value of acceleration due to gravity is the most at the highest peak of the Earth.

Answer

(ii) Value of acceleration due to gravity decreases when the height increases from the surface of the Earth.

$g=\frac{GM}{(R+h)^2}$ falls as $h$ increases (statement (i) says the same thing). Statements (iii) and (iv) are wrong, since $g$ is greater at the poles and least at the highest peak.

Answer checked by Claude Sonnet · Oct 21, 2026

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2082 Chaitra · Karnali · 1 mark

If a stone is dropped from the top of the tower having height 22.05 m, how long does it take to reach the ground?
(a) 1 second (b) 1.5 second (c) 2 second (d) 2.5 second

Answer

(c) 2 second

h=\frac12gt^2\Rightarrow t=\sqrt{\frac{2\times22.05}{9.8}}=\sqrt{4.5}\approx2.1\ \text{s}

which is closest to 2 s.

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2079 Chaitra · Koshi · 1 mark

What is the value of $g$ on the surface of the moon?

Answer

The value of $g$ on the moon is about $1.63\ \text{m/s}^2$ (roughly $\tfrac{1}{6}$ of the earth's).

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2079 Chaitra · Karnali · 1 mark

What is the acceleration due to gravity on the surface of Jupiter?

Answer

The acceleration due to gravity on Jupiter is about $24.8\ \text{m/s}^2$ (often taken as about $25\ \text{m/s}^2$ ).

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2079 Chaitra · Lumbini · 1 mark

What is the value of acceleration due to gravity of the earth on the equator?

Similar questions: Define g, value at pole

Answer

The value of $g$ at the equator is about $9.78\ \text{m/s}^2$ , which is the lowest value on the earth's surface.

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2079 Chaitra · Sudurpashchim · 1 mark

What is the unit of the universal gravitational constant ( $G$ )?

Answer

The SI unit of $G$ is $\text{N m}^2\,\text{kg}^{-2}$ (that is, $\text{N m}^2/\text{kg}^2$ ).

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2079 Chaitra · Bagmati · 1 mark

If 'm' is mass and 'w' is weight, write down the relation between 'm' and 'w'.

Answer

$w = mg$

where $g$ is the acceleration due to gravity, so weight is directly proportional to mass.

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2079 Chaitra · Gandaki · 1 mark

State Newton's universal law of gravitation.

Answer

Every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

$F = \frac{G m_1 m_2}{d^2}$

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2079 Chaitra · Koshi · 3 marks

In between the objects shown in the diagram, $6\times10^{-10}$ N gravitational force is exerted. If the distance is halved and the mass of 40 kg is doubled, what will be the gravitational force between them?
Figure: Two spheres of masses 40 kg and 90 kg placed 20 m apart.

Answer

Given: $m_1 = 40$ kg, $m_2 = 90$ kg, $d = 20$ m and $F = 6\times10^{-10}$ N.

New values: $m_1' = 80$ kg and $d' = 10$ m.

Since $F \propto \dfrac{m_1 m_2}{d^2}$ , we compare the two cases.

$\frac{F'}{F} = \frac{m_1'}{m_1}\times\frac{d^2}{d'^2} = \frac{80}{40}\times\frac{20^2}{10^2} = 2\times 4 = 8$

$F' = 8 \times 6\times10^{-10} = 4.8\times10^{-9}\ \text{N}$

Check with $G = 6.67\times10^{-11}$ : $F' = \dfrac{6.67\times10^{-11}\times 80\times 90}{10^2} = 4.8\times10^{-9}$ N.

The new gravitational force is $4.8\times10^{-9}$ N.

Answer checked by Claude Sonnet · Oct 21, 2026

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2079 Chaitra · Karnali · 2 marks

The mass of the large object is 50 kg and that of the small object is 20 kg as shown in the diagram. The gravitational force between them is $5\times10^{-5}$ N. If they are separated by 14 m between their surfaces, what force will be exerted between them?
Figure: Two spheres with radii 10 m and 8 m touching each other.

Answer

When the spheres touch, the distance between their centres is the sum of the radii:

$d_1 = 10 + 8 = 18\ \text{m},\qquad F_1 = 5\times10^{-5}\ \text{N}$

When their surfaces are 14 m apart, the distance between centres is

$d_2 = 14 + 10 + 8 = 32\ \text{m}$

The masses stay the same, so $F \propto \dfrac{1}{d^2}$ :

$\frac{F_2}{F_1}=\frac{d_1^2}{d_2^2}\;\Rightarrow\; F_2 = 5\times10^{-5}\times\frac{18^2}{32^2}=5\times10^{-5}\times\frac{324}{1024}\approx 1.58\times10^{-5}\ \text{N}$

The force between them is about $1.58\times10^{-5}$ N.

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2079 Chaitra · Karnali · 3 marks

An object is at a height of $6.6\times10^{7}$ m from the centre of the earth. The radius of the earth and its mass are $6\times10^{6}$ m and $6\times10^{24}$ kg respectively. Calculate the value of acceleration due to gravity and the weight of a 12 kg mass at that height.

Answer

Given: $r = 6.6\times10^{7}$ m (distance from the centre of the earth), $M = 6\times10^{24}$ kg, $m = 12$ kg, $G = 6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2}$ .

Acceleration due to gravity at that height

$g' = \frac{GM}{r^2}=\frac{6.67\times10^{-11}\times 6\times10^{24}}{(6.6\times10^{7})^2}=\frac{4.002\times10^{14}}{4.356\times10^{15}}\approx 0.092\ \text{m/s}^2$

Weight of the 12 kg mass

$W = mg' = 12\times 0.0919 \approx 1.10\ \text{N}$

The radius of the earth is not needed because the height is already given from the centre.

Answer checked by Claude Sonnet · Oct 21, 2026

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2079 Chaitra · Madhesh · 4 marks

The gravitational force between two masses when they are separated by a certain distance is shown in the given table.

S.N. Gravitational force Distance Mass first Mass second
1 $F$ $d$ $m_1$ $m_2$
2 $F/2$ $2d$ $2m_1$ $2m_2$
3 $4F$ $2d$ $2m_1$ $\tfrac{1}{2}m_2$
4 $F/2$ $2d$ $2m_1$ $m_2$
(a) With reference to serial 1, which of them are true? (b) If the value of $F$ is 400 N in serial 1, show by calculation the value of $F$ when the distance between the bodies is doubled keeping their masses constant.

S.N.	Gravitational force	Distance	Mass first	Mass second
1	$F$	$d$	$m_1$	$m_2$
2	$F/2$	$2d$	$2m_1$	$2m_2$
3	$4F$	$2d$	$2m_1$	$\tfrac{1}{2}m_2$
4	$F/2$	$2d$	$2m_1$	$m_2$

Answer

From Newton's law of gravitation, $F = \dfrac{G m_1 m_2}{d^2}$ .

(a) Compare each row with serial 1:

Serial 2: $\dfrac{G(2m_1)(2m_2)}{(2d)^2}=\dfrac{4Gm_1m_2}{4d^2}=F$ , not $F/2$ , so it is false.
Serial 3: $\dfrac{G(2m_1)(m_2/2)}{(2d)^2}=\dfrac{Gm_1m_2}{4d^2}=\dfrac{F}{4}$ , not $4F$ , so it is false.
Serial 4: $\dfrac{G(2m_1)(m_2)}{(2d)^2}=\dfrac{2Gm_1m_2}{4d^2}=\dfrac{F}{2}$ , so it is true.

Only serial 4 is true.

(b) The masses are constant, so $F\propto \dfrac{1}{d^2}$ . When $d$ is doubled:

$F' = \frac{F}{(2)^2}=\frac{400}{4}=100\ \text{N}$

The force becomes 100 N.

Answer checked by Claude Sonnet · Oct 21, 2026

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SEE 2080 model question, Set 3, Science and Technology · 2 marks

What is acceleration due to gravity? What is the value of acceleration due to gravity on the pole of the earth?

Similar questions: g at equator (2079 Chaitra · Lumbini)

Answer

Acceleration due to gravity ( $g$ ): the acceleration produced in a freely falling body due to the Earth's gravitational pull. Its value is $g=\frac{GM}{R^2}$ .

At the poles of the Earth the value of $g$ is maximum, about $9.83\ \text{m/s}^2$ .

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SEE 2080 model question, Set 1, Science and Technology · 1 mark

Calculate the mass that a person can lift on moon if he can lift a mass of 50 kg on the earth. The value of g on the surface of the Earth is 9.8 m/s $^2$ and that of moon is 6 times less.
(a) 200.6 kg (b) 300.6 kg (c) 400.6 kg (d) 500.6 kg

Answer

(b) 300.6 kg (about 300 kg)

The force the person can apply is fixed: $F=50\times9.8=490$ N. On the Moon $g_m=\frac{9.8}{6}=1.63\ \text{m/s}^2$ , so $m=\frac{490}{1.63}\approx300$ kg, i.e. 6 times more.

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SEE 2080 model question, Set 2, Science and Technology · 1 mark

The acceleration due to gravity on the surface of the earth is $g$ . If mass and radius both are made halved, what is the new value of acceleration due to gravity on the earth surface?
(a) $g$ (b) $2g$ (c) $3g$ (d) $4g$

Answer

(b) $2g$

$g'=\frac{G(M/2)}{(R/2)^2}=\frac{GM/2}{R^2/4}=2\frac{GM}{R^2}=2g$ .

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SEE 2080 model question, Set 2, Science and Technology · 2 marks

Give any two examples of free fall.

Answer

Two examples of free fall:

A stone or fruit (e.g. a coconut) dropped from a height falls freely towards the ground (neglecting air resistance).
A satellite or astronaut in an orbiting spacecraft is in continuous free fall around the Earth (also a skydiver before the parachute opens, approximately).

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CDC model paper 2080, Science and Technology · 1 mark

The mass of the Earth is $6 \times 10^{24}$ kg. What would be the acceleration due to gravity of the meteor falling from 10000 km height from the centre. (a) 4.2 m/s $^2$ (b) 4.002 m/s $^2$ (c) 4 m/s $^2$ (d) 4.02 m/s $^2$

Answer

(b) 4.002 m/s $^2$

g=\frac{GM}{r^2}=\frac{6.67\times10^{-11}\times6\times10^{24}}{(10^{7})^2}=\frac{4.002\times10^{14}}{10^{14}}=4.002\ \text{m/s}^2

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CDC model paper 2080, Science and Technology · 2 marks

Kabin dropped a stone from a tower of 49 m high. One second later, he threw another stone. They both hit the ground at the same time. Find out the velocity with which he threw the second stone.

Answer

Given: $h=49$ m, $g=9.8\ \text{m/s}^2$ . Find $u$ of the second stone.

Time taken by the first stone (dropped, $u=0$ ):

h=\frac12gt^2\Rightarrow49=4.9t^2\Rightarrow t=\sqrt{10}=3.16\ \text{s}

The second stone is thrown 1 s later and lands at the same time, so its time $t'=3.16-1=2.16$ s. For it:

h=ut'+\frac12gt'^2

49=u(2.16)+4.9(2.16)^2=2.16u+22.9

u=\frac{49-22.9}{2.16}\approx12.1\ \text{m/s}

The second stone was thrown downwards with a velocity of about 12 m/s.

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SEE 2080 model question, Set 1, Science and Technology · 2 marks

Write any two factors on which gravity on the surface of a heavenly body depends.

Answer

Two factors on which gravity on the surface of a heavenly body depends ( $g=\frac{GM}{R^2}$ ):

The mass of the heavenly body (greater mass gives greater $g$ ).
The radius of the heavenly body (greater radius gives smaller $g$ ).

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SEE 2080 model question, Set 1, Science and Technology · 2 marks

Prove that if the earth attracts two bodies placed at the same distance from the centre of the earth with the same force, then their masses are equal.

Answer

Let the mass of the Earth be $M$ , and the two bodies of masses $m_1$ and $m_2$ be at the same distance $d$ from its centre.

Force on the first body: $F_1=\frac{GMm_1}{d^2}$

Force on the second body: $F_2=\frac{GMm_2}{d^2}$

Given $F_1=F_2$ :

\frac{GMm_1}{d^2}=\frac{GMm_2}{d^2}

m_1=m_2

Hence their masses are equal.

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SEE 2080 model question, Set 2, Science and Technology · 2 marks

A body is dropped from top of the tower of height 'h' and at the same time next body is thrown up from the bottom of tower with velocity 20 m/s. If these bodies meet after 2 sec. at a point of the tower, calculate the height of the tower. ( $g = 10\ \text{m/s}^2$ )

Answer

Given: $g=10\ \text{m/s}^2$ , $u_2=20$ m/s (upward), $t=2$ s.

Distance fallen by the dropped body:

s_1=\frac12gt^2=\frac12\times10\times2^2=20\ \text{m}

Distance risen by the thrown body:

s_2=u t-\frac12gt^2=20\times2-\frac12\times10\times4=40-20=20\ \text{m}

They meet, so the height of the tower is

h=s_1+s_2=20+20=40\ \text{m}

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SEE 2080 model question, Set 3, Science and Technology · 1 mark

What would be the weight of a body of mass 20 kg at distance 12000 km from centre of the earth of mass $6 \times 10^{24}$ kg?
(a) 35 N (b) 45 N (c) 55 N (d) 25 N

Answer

(c) 55 N

W=\frac{GMm}{d^2}=\frac{6.67\times10^{-11}\times6\times10^{24}\times20}{(1.2\times10^{7})^2}=\frac{8.0\times10^{15}}{1.44\times10^{14}}\approx55.6\ \text{N}

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SEE 2080 model question, Set 3, Science and Technology · 2 marks

If the given gravitational force between A and B is F. What is the gravitational force between B and C in terms of F?
Figure: Diagram of three spheres: A of mass M, B of mass 2M, C of mass 3M. Distance between A and B is d; distance between B and C is 2d.

Answer

Given: $F_{AB}=G\frac{M\times2M}{d^2}=\frac{2GM^2}{d^2}=F$ .

F_{BC}=G\frac{2M\times3M}{(2d)^2}=\frac{6GM^2}{4d^2}=\frac{3GM^2}{2d^2}

F_{BC}=\frac{3}{2}\times\frac{GM^2}{d^2}=\frac{3}{2}\times\frac F2=\frac{3F}{4}

So the force between B and C is $\frac{3}{4}F$ .

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Questions from SEE 2082 Science and Technology, 1041 Koshi (Board exam, Chaitra 2082 (scanned PDF, bilingual)), SEE 2082 Science and Technology, 1041 Madhesh (Board exam, Chaitra 2082 (scanned PDF, bilingual)), SEE 2082 Science and Technology, 1041 Bagmati (Board exam, Chaitra 2082 (scanned PDF, bilingual)), SEE 2082 Science and Technology, 1041 Gandaki (Board exam, Chaitra 2082 (scanned PDF, bilingual)), SEE 2082 Science and Technology, 1041 Lumbini (Board exam, Chaitra 2082 (scanned PDF, bilingual)), SEE 2082 Science and Technology, 1041 Karnali (Board exam, Chaitra 2082 (scanned PDF, bilingual)), SEE 2082 Science and Technology, 1041 Sudurpashchim (Board exam, Chaitra 2082 (scanned PDF, bilingual)), SEE 2081 Science and Technology, 1041 Koshi (Board exam, Chaitra 2081 (PDF; official answer keys exist on the bidhyamandir page)), SEE 2081 Science and Technology, 1041 Madhesh (Board exam, Chaitra 2081 (PDF; official answer keys exist on the bidhyamandir page)), SEE 2081 Science and Technology, 1041 Bagmati (Board exam, Chaitra 2081 (PDF; official answer keys exist on the bidhyamandir page)), SEE 2081 Science and Technology, 1041 Gandaki (Board exam, Chaitra 2081 (PDF; official answer keys exist on the bidhyamandir page)), SEE 2081 Science and Technology, 1041 Lumbini (Board exam, Chaitra 2081 (PDF; official answer keys exist on the bidhyamandir page)), SEE 2081 Science and Technology, 1041 Karnali (Board exam, Chaitra 2081 (PDF; official answer keys exist on the bidhyamandir page)), SEE 2081 Science and Technology, 1041 Sudurpashchim (Board exam, Chaitra 2081 (PDF; official answer keys exist on the bidhyamandir page)), SEE 2080 Science and Technology, 1041 Koshi (Board exam, 2080 (scanned PDF, bilingual)), SEE 2080 Science and Technology, 1041 Madhesh (Board exam, 2080 (scanned PDF, bilingual)), SEE 2080 Science and Technology, 1041 Bagmati (Board exam, 2080 (scanned PDF, bilingual)), SEE 2080 Science and Technology, 1041 Gandaki (Board exam, 2080 (scanned PDF, bilingual)), SEE 2080 Science and Technology, 1041 Lumbini (Board exam, 2080 (scanned PDF, bilingual)), SEE 2080 Science and Technology, 1041 Karnali (Board exam, 2080 (scanned PDF, bilingual)), SEE 2080 Science and Technology, 1041 Sudurpashchim (Board exam, 2080 (scanned PDF, bilingual)), SEE 2082 Science and Technology, GI-1041 (grade increment) (Grade increment, Asar 2083 (PDF)), SEE 2081 Science and Technology, GI-1041 (grade increment) (Grade increment 2081 (8 page images); month taken from the August 2025 post, check), SEE 2080 Science and Technology, GI-1041 (grade increment) (Grade increment 2080 (PDF)), CDC model paper 2080, Science and Technology (with answer key) (Official CDC model paper, pages 2-9; answer key pages 10-15), SEE 2080 model question, Set 1, Science and Technology (Model paper Set 1 (typed PDF, publisher not named, no answers)), SEE 2080 model question, Set 2, Science and Technology (Model paper Set 2 (typed PDF, publisher not named, no answers)), SEE 2080 model question, Set 3, Science and Technology (Model paper Set 3 (typed PDF, publisher not named, no answers)), SEE 2079 Compulsory Science (RE-111) Koshi (Board exam, Chaitra 2079 (2023), last year of the old course (scanned PDF, bilingual, 75 marks); only questions that fit the current outline were taken), SEE 2079 Compulsory Science (RE-111) Madhesh (Board exam, Chaitra 2079 (2023), last year of the old course (scanned PDF, bilingual, 75 marks); only questions that fit the current outline were taken), SEE 2079 Compulsory Science (RE-111) Bagmati (Board exam, Chaitra 2079 (2023), last year of the old course (scanned PDF, bilingual, 75 marks); only questions that fit the current outline were taken), SEE 2079 Compulsory Science (RE-111) Gandaki (Board exam, Chaitra 2079 (2023), last year of the old course (scanned PDF, bilingual, 75 marks); only questions that fit the current outline were taken), SEE 2079 Compulsory Science (RE-111) Lumbini (Board exam, Chaitra 2079 (2023), last year of the old course (scanned PDF, bilingual, 75 marks); only questions that fit the current outline were taken), SEE 2079 Compulsory Science (RE-111) Karnali (Board exam, Chaitra 2079 (2023), last year of the old course (scanned PDF, bilingual, 75 marks); only questions that fit the current outline were taken) and SEE 2079 Compulsory Science (RE-111) Sudurpashchim (Board exam, Chaitra 2079 (2023), last year of the old course (scanned PDF, bilingual, 75 marks); only questions that fit the current outline were taken). Answers are written for this site; check them against your class notes.

Chapter titles and hours from the official syllabus ↗

Past questions and answers

1.Where does an object have more weight between the place lying in lower altitude from the sea level like Janakpur and places lying higher altitude like Kathmandu?

2.Check the correctness of the equation s=ut+12at2s = ut + \frac{1}{2}at^2s=ut+21​at2 by using unit-wise analysis.

3.Check the validity of equation S=ut+12atS = ut + \frac{1}{2}atS=ut+21​at, by unit wise analysis.

4.Check the validity of the equation v=u+at2v = u + at^2v=u+at2 by its unit wise analysis.

5.Check the correctness of the equation v2=u2+2as2v^2 = u^2 + 2as^2v2=u2+2as2 by using unit wise analysis.

6.Write any two effects of gravity.

7.What is the weight of body on the moon surface if its weight is 150 N on the earth surface? (a) 20 N (b) 25 N (c) 30 N (d) 35 N

8.If the Earth could be squeezed to the size of the Moon, what would be the effect on the value of acceleration due to gravity of the Earth? (a) increases (b) decreases (c) becomes zero (d) remains the same

9.Which of the following group of fundamental units represent the derived unit of energy Joule (J)? (a) kg ms−2\text{kg ms}^{-2}kg ms−2 (b) kg ms−1\text{kg ms}^{-1}kg ms−1 (c) kg m2s−2\text{kg m}^2\text{s}^{-2}kg m2s−2 (d) kg ms−3\text{kg ms}^{-3}kg ms−3

10.What is the relationship between weight of object (w) and acceleration due to gravity (g)? (i) W = g (ii) W ∝ g (iii) W ∝ 1/g (iv) W = 2g

11.State Newton's Universal law of gravitation and prove that F=Gm1m2d2F = \frac{Gm_1m_2}{d^2}F=d2Gm1​m2​​ with the help of appropriate diagram.

12.In which condition gravitational constant and gravitational force are equal?

13.Write any two differences between free fall and weightlessness.

15.Write any two differences between acceleration due to gravity (g) and gravitational constant (G).

16.The value of acceleration due to gravity is more in the polar region than the equatorial region, why?

18.What is the weight of a body having 50 kg mass on the moon? (i) 8.35 N (ii) 83.5 N (iii) 835 N (iv) 8355 N

19.What could be the weight of a body on the Earth if its weight on the Moon is 5 N? (a) 56\frac{5}{6}65​ N (b) 5 N (c) 30 N (d) 50 N

20.What will be the weight of stone on the surface of the Earth if it weighs 78.4 N on the surface of the Moon? (i) 380 N (ii) 490 N (iii) 590 N (iv) 690 N

21.If the weight of an object on the surface of the Moon is 66 N, what might be the weight of the object on Earth? (Acceleration due to gravity of Moon is 1.62 m/s2^22) (a) 396 N (b) 66 N (c) 11 N (d) 6.6 N

22.If the weight of a person on the surface of the Earth is 294 N, what might be the weight of the person on the moon? (a) 6 N (b) 9.8 N (c) 49 N (d) 60 N

23.What will be the weight of a man on the Moon, if his weight on the Earth is 750 N? (Acceleration due to gravity of the Moon = 1.63 m/s2^22) (a) 124.7 N (b) 125 N (c) 126.8 N (d) 127.8 N

24.What will be the weight of an object on the planet having acceleration due to gravity 2.5 m/s2^22 if the object weighs 700 N on the Earth? (a) 240 N (b) 1750 N (c) 175 N (d) 700 N

25.What mass can be lifted by a person on the Moon if he can lift 100 kg mass on the Earth? Calculate.

26.Test the validity of power (P) = Wt\frac{W}{t}tW​ by unit wise analysis.

27.Prove that 'Watt' is derived unit showing the proper relation among the fundamental units involved in it. Check unit wise homogeneity in s=ut2+12at2s = ut^2 + \tfrac{1}{2}at^2s=ut2+21​at2

28.Which fundamental units of following represents the proper relation for Newton? (i) kg m−2s−2kg\,m^{-2}s^{-2}kgm−2s−2 (ii) kg m2s−1kg\,m^{2}s^{-1}kgm2s−1 (iii) kg m2s2kg\,m^{2}s^{2}kgm2s2 (iv) kg m s−2kg\,m\,s^{-2}kgms−2

29.Write the fundamental units included in the unit of force.

30.Why is watt called derived unit?

31.Prove that unit of the work is (J) = kg m2/s2kg\,m^2/s^2kgm2/s2.

32.Prove that, fundamental units of power is kgm2/s3kgm^2/s^3kgm2/s3.

33.Prove that unit of gravitational constant is derived unit.

35.What happens to the gravitational force when masses of two objects kept constant and the distance between them is made half? Explain with suitable equation.

36.What change occurs in the gravitational force between two bodies when their masses are kept constant and the distance between their centres is reduced by 3 times?

38.What is the gravitational force of attraction between these two heavenly bodies? In which condition gravitational force of attraction is equal to gravitational constant. Figure: Two circles (bodies) of masses m1 and m2 joined by a line, separated by distance d.

39.Calculate the gravitational force of attraction between objects A and B as shown in the given figure? Figure: Two circles A (mass 2,000 Kg) and B (mass 4,000 Kg), centres joined by a line labelled 80 Km.

40.Calculate the force of gravitation between the heavenly bodies P and Q on the basis of information given in the figure. Figure: Two circles: P of mass 1200 kg on the left and Q of mass 4000 kg on the right, with a line joining their centres labelled 4000 km.

41.The mass of objects A and B as shown in the diagram are 2,500 kg and 4,000 kg respectively. Calculate gravitational force exerted between them. Figure: Two circles A and B placed apart, with the distance between them marked as 600 m.

49.Mass and radius of a heavenly body are 7.342×10227.342 \times 10^{22}7.342×1022 kg and 1,737 km respectively. Calculate the acceleration due to gravity on its surface.

50.Mass of a planet is 1.9×10271.9 \times 10^{27}1.9×1027 kg and its radius is 71,000 km, calculate its acceleration due to gravity of the planet.

51.If the mass and radius of heavenly object are four times greater than that of the earth. What is the value of acceleration due to gravity of that object. (a) 1.22 m/s2^22 (b) 2.45 m/s2^22 (c) 4.9 m/s2^22 (d) 9.8 m/s2^22

53.What is free fall? Write the value of acceleration due to gravity when a body falls freely towards the surface of the Earth.

54.Define free fall.

55.Write any two conditions for an object to be weightlessness?

56.What is a dependent variable in a process of throwing a stone with a stretched rubber band?

57.What are the dependent variable and independent variable included when a piece of paper is thrown with stretched rubber.

58.A student has to test the change in effort applied and the number of pulleys used. In this experiment, identify a controlled variable and a dependent variable from number of pulleys, load of object, effort applied, friction produced in the pulleys and type of pulley.

60.Which formula will you use to find out height from the earth surface from where a stone is dropped down and time taken by the stone to reach the ground is known?

62.Weight of object is 800 N on the surface of earth whose radius is 6,371 km. What is the weight of the object at the height 6,371 km from the surface of the earth? (a) 200 N (b) 800 N (c) 1600 N (d) 3200 N

63.How does this fall differ from object falling on the moon? Write any two. Figure: A tall coconut palm tree with a point X near the top among the fruits and a point Y on the ground below, joined by a vertical dotted line (object falls from X to Y).

64.If a cricket ball is thrown vertically upward into the sky with velocity of 18 m/s. Calculate the maximum height attained by the ball. [Note: Neglect the air resistance.]

65.For how long does a stone remain in air when it is thrown upward with an initial velocity of 20 m/s? (a) 2 s (b) 4 s (c) 10 s (d) 20 s

66.In the figure height of four different mountains are given. Among them which mountain has least acceleration due to gravity? Figure: Four mountain peaks A, B, C, D drawn on a baseline; B is the tallest, A and C are shorter, D is of medium-high height.

67.Write a difference between mass and weight.

69.If a stone is dropped from the top of the tower having height 22.05 m, how long does it take to reach the ground? (a) 1 second (b) 1.5 second (c) 2 second (d) 2.5 second

70.What is the value of ggg on the surface of the moon?

71.What is the acceleration due to gravity on the surface of Jupiter?

72.What is the value of acceleration due to gravity of the earth on the equator?

73.What is the unit of the universal gravitational constant (GGG)?

74.If 'm' is mass and 'w' is weight, write down the relation between 'm' and 'w'.

75.State Newton's universal law of gravitation.

76.In between the objects shown in the diagram, 6×10−106\times10^{-10}6×10−10 N gravitational force is exerted. If the distance is halved and the mass of 40 kg is doubled, what will be the gravitational force between them? Figure: Two spheres of masses 40 kg and 90 kg placed 20 m apart.

80.What is acceleration due to gravity? What is the value of acceleration due to gravity on the pole of the earth?

81.Calculate the mass that a person can lift on moon if he can lift a mass of 50 kg on the earth. The value of g on the surface of the Earth is 9.8 m/s2^22 and that of moon is 6 times less. (a) 200.6 kg (b) 300.6 kg (c) 400.6 kg (d) 500.6 kg

82.The acceleration due to gravity on the surface of the earth is ggg. If mass and radius both are made halved, what is the new value of acceleration due to gravity on the earth surface? (a) ggg (b) 2g2g2g (c) 3g3g3g (d) 4g4g4g

83.Give any two examples of free fall.

84.The mass of the Earth is 6×10246 \times 10^{24}6×1024 kg. What would be the acceleration due to gravity of the meteor falling from 10000 km height from the centre. (a) 4.2 m/s2^22 (b) 4.002 m/s2^22 (c) 4 m/s2^22 (d) 4.02 m/s2^22

85.Kabin dropped a stone from a tower of 49 m high. One second later, he threw another stone. They both hit the ground at the same time. Find out the velocity with which he threw the second stone.

86.Write any two factors on which gravity on the surface of a heavenly body depends.

87.Prove that if the earth attracts two bodies placed at the same distance from the centre of the earth with the same force, then their masses are equal.

88.A body is dropped from top of the tower of height 'h' and at the same time next body is thrown up from the bottom of tower with velocity 20 m/s. If these bodies meet after 2 sec. at a point of the tower, calculate the height of the tower. (g=10 m/s2g = 10\ \text{m/s}^2g=10 m/s2)

89.What would be the weight of a body of mass 20 kg at distance 12000 km from centre of the earth of mass 6×10246 \times 10^{24}6×1024 kg? (a) 35 N (b) 45 N (c) 55 N (d) 25 N

90.If the given gravitational force between A and B is F. What is the gravitational force between B and C in terms of F? Figure: Diagram of three spheres: A of mass M, B of mass 2M, C of mass 3M. Distance between A and B is d; distance between B and C is 2d.

Where does an object have more weight between the place lying in lower altitude from the sea level like Janakpur and places lying higher altitude like Kathmandu?

Check the correctness of the equation $s = ut + \frac{1}{2}at^2$ by using unit-wise analysis.

Check the validity of equation $S = ut + \frac{1}{2}at$ , by unit wise analysis.

Check the validity of the equation $v = u + at^2$ by its unit wise analysis.

Check the correctness of the equation $v^2 = u^2 + 2as^2$ by using unit wise analysis.

Write any two effects of gravity.

What is the weight of body on the moon surface if its weight is 150 N on the earth surface?
(a) 20 N (b) 25 N (c) 30 N (d) 35 N

If the Earth could be squeezed to the size of the Moon, what would be the effect on the value of acceleration due to gravity of the Earth?
(a) increases (b) decreases (c) becomes zero (d) remains the same

Which of the following group of fundamental units represent the derived unit of energy Joule (J)?
(a) $\text{kg ms}^{-2}$ (b) $\text{kg ms}^{-1}$ (c) $\text{kg m}^2\text{s}^{-2}$ (d) $\text{kg ms}^{-3}$

What is the relationship between weight of object (w) and acceleration due to gravity (g)?
(i) W = g (ii) W ∝ g (iii) W ∝ 1/g (iv) W = 2g

State Newton's Universal law of gravitation and prove that $F = \frac{Gm_1m_2}{d^2}$ with the help of appropriate diagram.

In which condition gravitational constant and gravitational force are equal?

Write any two differences between free fall and weightlessness.

Write any two differences between acceleration due to gravity (g) and gravitational constant (G).

The value of acceleration due to gravity is more in the polar region than the equatorial region, why?

What is the weight of a body having 50 kg mass on the moon? (i) 8.35 N (ii) 83.5 N (iii) 835 N (iv) 8355 N

What could be the weight of a body on the Earth if its weight on the Moon is 5 N?
(a) $\frac{5}{6}$ N (b) 5 N (c) 30 N (d) 50 N

What will be the weight of stone on the surface of the Earth if it weighs 78.4 N on the surface of the Moon? (i) 380 N (ii) 490 N (iii) 590 N (iv) 690 N

If the weight of an object on the surface of the Moon is 66 N, what might be the weight of the object on Earth? (Acceleration due to gravity of Moon is 1.62 m/s $^2$ )
(a) 396 N (b) 66 N (c) 11 N (d) 6.6 N

If the weight of a person on the surface of the Earth is 294 N, what might be the weight of the person on the moon? (a) 6 N (b) 9.8 N (c) 49 N (d) 60 N

What will be the weight of a man on the Moon, if his weight on the Earth is 750 N? (Acceleration due to gravity of the Moon = 1.63 m/s $^2$ )
(a) 124.7 N (b) 125 N (c) 126.8 N (d) 127.8 N

What will be the weight of an object on the planet having acceleration due to gravity 2.5 m/s $^2$ if the object weighs 700 N on the Earth?
(a) 240 N (b) 1750 N (c) 175 N (d) 700 N

What mass can be lifted by a person on the Moon if he can lift 100 kg mass on the Earth? Calculate.

Test the validity of power (P) = $\frac{W}{t}$ by unit wise analysis.

Prove that 'Watt' is derived unit showing the proper relation among the fundamental units involved in it. Check unit wise homogeneity in
$s = ut^2 + \tfrac{1}{2}at^2$

Which fundamental units of following represents the proper relation for Newton? (i) $kg\,m^{-2}s^{-2}$ (ii) $kg\,m^{2}s^{-1}$ (iii) $kg\,m^{2}s^{2}$ (iv) $kg\,m\,s^{-2}$

Write the fundamental units included in the unit of force.

Why is watt called derived unit?

Prove that unit of the work is (J) = $kg\,m^2/s^2$ .

Prove that, fundamental units of power is $kgm^2/s^3$ .

Prove that unit of gravitational constant is derived unit.

What happens to the gravitational force when masses of two objects kept constant and the distance between them is made half? Explain with suitable equation.

What change occurs in the gravitational force between two bodies when their masses are kept constant and the distance between their centres is reduced by 3 times?

What is the gravitational force of attraction between these two heavenly bodies? In which condition gravitational force of attraction is equal to gravitational constant.
Figure: Two circles (bodies) of masses m1 and m2 joined by a line, separated by distance d.

Calculate the gravitational force of attraction between objects A and B as shown in the given figure?
Figure: Two circles A (mass 2,000 Kg) and B (mass 4,000 Kg), centres joined by a line labelled 80 Km.

Calculate the force of gravitation between the heavenly bodies P and Q on the basis of information given in the figure.
Figure: Two circles: P of mass 1200 kg on the left and Q of mass 4000 kg on the right, with a line joining their centres labelled 4000 km.

The mass of objects A and B as shown in the diagram are 2,500 kg and 4,000 kg respectively. Calculate gravitational force exerted between them.
Figure: Two circles A and B placed apart, with the distance between them marked as 600 m.

Mass and radius of a heavenly body are $7.342 \times 10^{22}$ kg and 1,737 km respectively. Calculate the acceleration due to gravity on its surface.

Mass of a planet is $1.9 \times 10^{27}$ kg and its radius is 71,000 km, calculate its acceleration due to gravity of the planet.

If the mass and radius of heavenly object are four times greater than that of the earth. What is the value of acceleration due to gravity of that object. (a) 1.22 m/s $^2$ (b) 2.45 m/s $^2$ (c) 4.9 m/s $^2$ (d) 9.8 m/s $^2$

What is free fall? Write the value of acceleration due to gravity when a body falls freely towards the surface of the Earth.

Define free fall.

Write any two conditions for an object to be weightlessness?

What is a dependent variable in a process of throwing a stone with a stretched rubber band?

What are the dependent variable and independent variable included when a piece of paper is thrown with stretched rubber.

A student has to test the change in effort applied and the number of pulleys used. In this experiment, identify a controlled variable and a dependent variable from number of pulleys, load of object, effort applied, friction produced in the pulleys and type of pulley.

Which formula will you use to find out height from the earth surface from where a stone is dropped down and time taken by the stone to reach the ground is known?

Weight of object is 800 N on the surface of earth whose radius is 6,371 km. What is the weight of the object at the height 6,371 km from the surface of the earth?
(a) 200 N (b) 800 N (c) 1600 N (d) 3200 N

How does this fall differ from object falling on the moon? Write any two.
Figure: A tall coconut palm tree with a point X near the top among the fruits and a point Y on the ground below, joined by a vertical dotted line (object falls from X to Y).

If a cricket ball is thrown vertically upward into the sky with velocity of 18 m/s. Calculate the maximum height attained by the ball. [Note: Neglect the air resistance.]

For how long does a stone remain in air when it is thrown upward with an initial velocity of 20 m/s?
(a) 2 s (b) 4 s (c) 10 s (d) 20 s

In the figure height of four different mountains are given. Among them which mountain has least acceleration due to gravity?
Figure: Four mountain peaks A, B, C, D drawn on a baseline; B is the tallest, A and C are shorter, D is of medium-high height.

Write a difference between mass and weight.

If a stone is dropped from the top of the tower having height 22.05 m, how long does it take to reach the ground?
(a) 1 second (b) 1.5 second (c) 2 second (d) 2.5 second

What is the value of $g$ on the surface of the moon?

What is the acceleration due to gravity on the surface of Jupiter?

What is the value of acceleration due to gravity of the earth on the equator?

What is the unit of the universal gravitational constant ( $G$ )?

If 'm' is mass and 'w' is weight, write down the relation between 'm' and 'w'.

State Newton's universal law of gravitation.

In between the objects shown in the diagram, $6\times10^{-10}$ N gravitational force is exerted. If the distance is halved and the mass of 40 kg is doubled, what will be the gravitational force between them?
Figure: Two spheres of masses 40 kg and 90 kg placed 20 m apart.

What is acceleration due to gravity? What is the value of acceleration due to gravity on the pole of the earth?

Calculate the mass that a person can lift on moon if he can lift a mass of 50 kg on the earth. The value of g on the surface of the Earth is 9.8 m/s $^2$ and that of moon is 6 times less.
(a) 200.6 kg (b) 300.6 kg (c) 400.6 kg (d) 500.6 kg

The acceleration due to gravity on the surface of the earth is $g$ . If mass and radius both are made halved, what is the new value of acceleration due to gravity on the earth surface?
(a) $g$ (b) $2g$ (c) $3g$ (d) $4g$

Give any two examples of free fall.

The mass of the Earth is $6 \times 10^{24}$ kg. What would be the acceleration due to gravity of the meteor falling from 10000 km height from the centre. (a) 4.2 m/s $^2$ (b) 4.002 m/s $^2$ (c) 4 m/s $^2$ (d) 4.02 m/s $^2$

Kabin dropped a stone from a tower of 49 m high. One second later, he threw another stone. They both hit the ground at the same time. Find out the velocity with which he threw the second stone.

Write any two factors on which gravity on the surface of a heavenly body depends.

Prove that if the earth attracts two bodies placed at the same distance from the centre of the earth with the same force, then their masses are equal.

A body is dropped from top of the tower of height 'h' and at the same time next body is thrown up from the bottom of tower with velocity 20 m/s. If these bodies meet after 2 sec. at a point of the tower, calculate the height of the tower. ( $g = 10\ \text{m/s}^2$ )

What would be the weight of a body of mass 20 kg at distance 12000 km from centre of the earth of mass $6 \times 10^{24}$ kg?
(a) 35 N (b) 45 N (c) 55 N (d) 25 N

If the given gravitational force between A and B is F. What is the gravitational force between B and C in terms of F?
Figure: Diagram of three spheres: A of mass M, B of mass 2M, C of mass 3M. Distance between A and B is d; distance between B and C is 2d.