MSBSHSE Solutions For SSC (Class 10) Science Part 1 Chapter 1 Gravitation are provided here for students to practice thoroughly and prepare for their exam. We bring you Maharashtra Board Solutions for Class 10, designed by our subject experts, which facilitates smooth and accurate understanding of the concepts of the chapter.

This chapter mainly focuses on the concept of Gravitation and questions related to it. The chapter deals with Kepler’s laws, Newton’s universal law of gravitation, Free Fall, and other concepts related to Gravitation. Acceleration due to the gravitational force of the Earth as well as Escape velocity is also discussed in this chapter.

These solutions of MSBSHSE for Class 10 (SSC) come with detailed step-by-step explanations of the exercises asked in the Maharashtra Board Science Textbooks for SSC Part 1. The Maharashtra State Board Solutions for Chapter 1 Science Part 1 can be accessed here and students can use it as a reference tool to quickly review all the topics for the exam. Students can prepare for the exams by studying with the help of these solutions and then score high marks for the exam.

## Maharashtra Board SSC (Class 10) Science Part 1 Chapter 1- BYJU’S Important Questions & Answers

**1. Study the entries in the following table and rewrite them putting the connected items in a single row.**

I |
II |
III |

Mass |
m/s^{2} |
Zero at the centre |

Weight |
kg |
Measure of inertia |

Acceleration due to gravity |
Nm^{2}/kg^{2} |
Same in the entire universe |

G r a v i t ational constant |
N |
Depends on height |

**Answer:**

I |
II |
III |

Mass | kg | Measure of inertia |

Weight | N | Zero at the centre |

Acceleration due to gravity | m/s^{2} |
Same in the entire universe |

Gravitational constant | Nm^{2}/kg^{2} |
Depends on height |

**2. What is the difference between mass and weight of an object? Will the mass and weight of an object on the earth be the same as their values on Mars? Why?**

**Answer: Mass: **The amount of matter present in the object is Mass, a scalar quantity with SI unit kg. Its value is the same everywhere and does not change even when we go to another planet. According to Newton’s first law, it is the measure of the inertia of an Object. The higher the mass, higher is the inertia.

**Weight: **The weight of an object, on the other hand is the force with which the earth attracts the object. The force (F) on an object of mass m on the surface of the earth can be written using equation (4)

Weight (W) = F = m g …. ( g= GM/R^{2})

Since, Weight is a force, it is a vector quantity with the SI unit Newton and its direction is towards the centre of the earth. Given that the value of g is not the same everywhere, the weight of an object changes from place to place, even if its mass is constant everywhere. From this, it is seen that while the weight of an object on Earth and Mars may differ, its mass remains constant on Mars as well on Earth.

**3. What is a free fall? **

**Answer: **The force of gravity due to the earth acts on each and every object, hence, when we are holding a stone in our hand, the stone experiences this force. Nevertheless, it is balanced by another force that we apply to it in the opposite direction. The result of this is that the stone remains at rest. However, when we release the stone in our hands, the only force that acts on it is the gravitational force of the earth and the stone falls down by its influence. Any object that moves under just the influence of the force of gravity is falling freely. Thus, this released stone is in a state of free fall. The initial velocity of the object during the free fall is zero, which increases along with the acceleration resulting from the gravity of the earth. In free fall, the frictional force due to air opposes the motion of the object and a buoyant force is also applied to the object. Hence, free fall is only possible in a vacuum**. **

**4.** **What is acceleration due to gravity?**

**Answer: **The earth applies gravitational force on the objects near it. According to Newton’s second law of motion, a force applied to a body causes its acceleration. Hence, the gravitational force due to the earth on a body results in its acceleration. This is known as acceleration due to gravity denoted by ‘g’. Acceleration is a vector. As the gravitational force on any object due to the earth is directed towards the centre of the earth, the direction of the acceleration due to gravity is also directed towards the centre of the earth i.e. Vertically downwards.

**5. What is escape velocity? **

**Answer: **When a ball is thrown upwards, its velocity is decreased due to the gravitation of the earth, reaching zero after reaching a certain height. From there, the ball then starts to fall. Maximum height of the ball is dependent upon its initial velocity.

According to Newton’s third equation of motion,

v^{2} = u^{2}+2as, with v = the final velocity of the ball = 0 and a = – g

Therefore, 0 = u^{2} + 2 (-g) s and maximum height of the ball = s = -(u^{2}/ 2g)

Hence, when the initial velocity u is higher, the larger is the height reached by the ball.

For this reason the higher the initial velocity, the ball will oppose the gravity of the earth more and larger will be the height to which it can reach. The value of g keeps decreasing as we go higher above the surface of the earth. Thus, the force pulling the ball downward decreases as the ball goes up. Keep increasing the initial velocity of the ball, till it reaches larger and larger heights.

Above a particular value of initial velocity of the ball, the ball is able to overcome the downward pull of the earth and escape the earth forever without falling back to the earth. This velocity is called escape velocity. We can determine its value by using the law of conservation of energy.

**6. What is centripetal force?**

**Answer: **Imagine an object moving in a circle with constant speed. Such a motion is possible only if there is a force directed towards the centre of the circle, acting on it constantly. This force is known as the centripetal force. Suppose, m is the mass of the object with v as its speed and r as the radius of the circle, then this force is equivalent to F = m (v^{2}/r).

**7. Write the three laws given by Kepler. **

**Answer: **Johannes Keppler, studied the data available about planetary positions and motion and noticed that the motion of planets follows certain laws. According to him, there are three laws describing planetary motion known as Kepler’s laws which are given here.

Kepler’s first law states that the orbit of a planet is an ellipse with the Sun at one of the foci. Take the image given below:

This image shows the elliptical orbit of a planet revolving around the sun with the position of the Sun indicated by S.

Kepler’s second law states The line joining the planet and the Sun sweeps equal areas in equal intervals of time. Distances covered by the planet in equal time, are AB and CD, i.e. After equal intervals of time, the positions of the planet starting from A and C are shown by B and D, respectively.

Then there is Kepler’s third law that states the square of its period of revolution around the Sun is directly proportional to the cube of the mean distance of a planet from the Sun.

Hence,

if r is the average distance of the planet from the Sun and

T is its period of revolution then,

T^{2} a r^{3} i.e. T^{2}/T^{3}= constant = K.

**8. A stone thrown vertically upwards with initial velocity u reaches a height ‘h’ before coming down. Show that the time taken to go up is the same as the time taken to come down.**

**Answer**: During vertical upward motion of the stone

Given that S=h, u=u, v=0 and a= -g

And if ‘t’ is the time taken by the ball to reach the height ‘h’, then by applying a second equation of motion, you get – 2gh = v^{2}-u^{2}.

Hence, u= **√**2gh

Applying the first equation of motion you get v = u-gt

Therefore, t = u/g= (**√**2h/g)

Now, for vertical downward motion of the stone

S=h, u=u and a= g

Given that “v” is the velocity of the ball with which it hits the ground and ‘t” is the time taken for the ball to reach the ground.

Here, applying second motion of the equation gives you

2gh = v^{2}

So, v = **√**2gh

Now, applying second motion of the equation

v= u + gt

Hence, t= v/g = (**√**2h/g)

This proves that the time taken by the stone to go up is the same as the time taken to come down.

**9. Explain why the value of g is zero at the centre of the earth.**

**Answer: **At the centre of the earth, the force driven by the upper half of the earth is cancelled out due to the force of the lower half of the earth. In this manner, the force caused by any portion of the earth will be nullified at its centre by the force due to the opposite side. Thus, the gravitational force at the centre will be 0. Now, by applying Newton’s Law F= mg. Given that mass of any object cannot be zero, then if F=0, then g has to be 0. So, the value of g is 0 at the centre of the earth.

**10. The radius of planet A is half the radius of planet B. If the mass of A is M _{A}, what must be the mass of B so that the value of g on B is half that of its value of A?**

**Answer: **Suppose the radius of planet A= R_{A} And radius of planet B = R_{B} And

Mass of planet A = M_{A}, then find mass of planet B = M_{B}.

Here, R_{A}= R_{B}/2, g_{B}= ½ g_{A}

Then, g= GM / R_{2} And g_{A} = GM_{A} /R^{2}_{A}

Therefore, g_{B} = G M_{B}/ R^{2}_{B}

Now, M_{B}/ R^{2}-B = ½ [GM_{A} /(R_{B}/2)^{2}]

M_{B}/ R^{2}_{B}= ½ [4 (GM_{A} /(R_{B})^{2})]

So, M_{B}= 2M_{A}

**11. Why is force necessary?**

**Answer: **A force is required in order to change the speed as well as the direction of

motion of an object.

**12. What is Newton’s universal law of gravitation? **

**Answer: **According to Newton’s Universal Law of Gravitation, every object in the Universe attracts every other object using a definite force. This force is directly proportional to the product of the masses of the two objects. It is also inversely proportional to the square of the distance between them. Now, if you take two objects of masses M and m kept at a distance from each other, then the gravitational force of attraction between these two bodies is written mathematically, as F ∝ Mm/ d^{2} Or F= G (Mm/d^{2}). In this, G known as the constant of proportionality is also the Universal gravitational constant.

**13. What is the value of G?**

**Answer: **The value of G is the gravitational force acting between two unit masses kept at a unit distance away from each other. Thus, in SI units, the value of G is equal to the gravitational force between two masses of 1 kg kept 1 m apart. Meanwhile, in SI units, the unit of G is Newton m^{2} kg^{-2}. The value of G was first experimentally measured by Henry Cavendish. As per SI units, its value is 6.673 x 10-11 N m^{2} kg^{-2}.

**14. Mahendra and Virat are sitting at a distance of 1 metre from each other. Their masses are 75 kg and 80 kg, respectively. What is the gravitational force between them? Given : r = 1 m, m _{1} = 75 kg, m_{2} = 80 kg and G = 6.67 x 10 ^{-11} Nm^{2}/kg^{2.}. **

**Answer: **As per the Newton’s law F = Gm_{1}m_{2} /r^{2}

Replace the equation with values

Then, F= 6.67 x 10 ^{-11}x 75 x 80/ 1^{2} N = 4.002 x 10^{-7} N

Hence, the gravitational force between Mahendra and Virat is 4.002 x 10^{-7} N

**15. Why does velocity increase with time? **

**Answer:** The velocity will increase with time because of the acceleration. Acceleration between two moving objects will not remain constant. When the distance between the objects decreases this will result in increased gravitational force leading to increased acceleration.

**16. What is the cause of high tide? **

**Answer:** High and low tides take place at different times at different places. The level of water in the sea changes due to the gravitational force exerted by the moon. Water directly under the moon gets pulled towards the moon and the level of water, there goes up causing a high tide at that place.

**17. The moon and the artificial satellites orbit the earth. The earth attracts them towards ****itself but unlike the falling apple, they do not fall on the earth, why?**

**Answer: **The moon and the artificial satellites that orbit the earth does not fall on the earth, despite the earth attracting them towards it. Reason for this is the velocity of the moon and the satellites along their orbits. If this velocity was not there, they would have fallen on the earth.

**18. Will the value of g be the same everywhere on the surface of the earth?**

**Answer: **No, the value of g is not the same everywhere on the surface of the earth as the shape of the earth is not exactly spherical.The distance of a point on the surface of the earth from its centre varies somewhat from place to place. Due to its rotation, the earth bulges at the equator and is flatter at the poles. Its radius is largest at the equator and smaller at the poles. The value of g is thus the highest (9.832 m/s^{2}) at the poles and decreases slowly with decreasing latitude. It is lowest (9.78 m/s^{2}) at the equator.

**19. Why does the value of g decrease as we go deep inside the earth?**

**Answer: **The value of g changes with height above the earth’s surface and it also goes inside the earth. When the value of r decreases, then the value of g is also expected to go down, as per the formula = f= Gm_{1}m_{2}/r^{2}_{. }Meanwhile, the part of the earth, which contributes towards the gravitational force felt by the object also decreases. This means that the value of M to be used in the equation also decreases. As a combined result of change in r and M, the value of g decreases as we go deep inside the earth.

**20. Any two objects, irrespective of their masses or any other properties, when dropped from the same height and falling freely will reach the earth at the same time. How? Give an example. **

**Answer: **Value of g is constant for all the objects at any given pace on the earth. Hence, if any two objects (irrespective of their masses or other properties) are dropped from the same height, then these objects will fall freely and touch the ground at the same time. To support this, Galileo performed an experiment around 1590 in the Italian city of Pisa. He dropped two spheres of different masses from the leaning tower of Pisa and demonstrated that both the spheres reached the ground at the same time.

**21. When we drop a feather and a heavy stone at the same time from a height, they do ****not reach the earth at the same time. Why?**

**Answer:** A buoyant and frictional force was applied on the feather due to air and so it floats and reaches the ground slowly, much after the heavy stone. Meanwhile, the buoyant and frictional forces on the stone is also much less than the weight of the stone and so does not affect the speed of the stone mulch. Scientists had performed this experiment in vacuum and showed that the feather and stone indeed reach the earth at the same time.

**22. A tennis ball is thrown up and reaches a height of 4.05 m before coming down. What was its initial velocity? (Assume g = 10 m/s ^{2}). **

**Answer: **For the upward motion of the ball, the final velocity of the ball (v) is given as 0. Now, if Distance travelled by the ball is 4.05 m, and acceleration a= – g = -10 m/s^{2}

Apply Newton’s third equation of motion and get

v^{2} = u^{2} + 2 a s

0 = u^{2} + 2 (-10) x 4.05

u^{2} = 81

Then, u = 9 m/s T

Hence, the initial velocity of the ball is 9 m/s.

**23. Write a short note on Gravitational potential energy. **

**Answer: **Potential energy is the energy stored in an object because of its position or state and it is relative. Hence, it increases as we go to greater heights from the surface of the earth. We had assumed that the potential energy of an object of mass m, at a height h from the ground is mgh and on the ground it is zero. When h is small compared to the radius R of the earth, let g be constant and let us use the above formula (mgh). However, for h with larger values, the value of g decreases with increase in height (h). For an object placed at infinite distance from the earth, the value of g is zero and earth’s gravitational force does not act on the object. So it is more appropriate to assume the value of potential energy to be zero there. Thus, for smaller distances, i.e. Heights, the potential energy is less than zero, i.e. It is negative.

When an object is at a height h from the surface of the earth, its potential energy is -GMm/(R+h), with M and R as the earth’s mass and radius, respectively.

Passionately studying from the textbooks and answering all the questions will do wonders when it comes to acing the exams. Students are recommended to go through all the topics thoroughly and prepare more effectively for the Maharashtra State Board exam.