Chapter 2:- Periodic Classification of Element

Chapter 3:- Chemical Reaction and Equation

Chapter 4:- Effect of Electric Current

Chapter 6 Refraction of Light

Chapter 7:- Lenses

Chapter 8 :- Metallurgy

Chapter 9 Carbon Compound

Chapter 10 :- Space Mission

## Exercise

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

I | II | III |
---|---|---|

Mass | kg | Measure of inertia |

Weight | N | Depends on height |

Acceleration due to gravity | m/s^{2} |
Zero at the centre |

Gravitational constant | Nm^{2}/kg^{2} |
Same in the entire universe |

**Solution:- **

I | II | III |
---|---|---|

Mass | kg | Measure of inertia |

Weight | N | Depends on height |

Acceleration due to gravity | m/s^{2} |
Zero at the centre |

Gravitational constant | Nm^{2}/kg^{2} |
Same in the entire universe |

**2. Answer the following questions.**

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

Answer:- Differential Between Mass and Weight

No. | Mass | Weight |
---|---|---|

1. | Mass is the amount of matter contained in a body. | Weight is the force exerted on a body due to the gravitational pull of another body such as Earth, the sun and the moon. |

2. | Mass is an intrinsic property of a body. | Weight is an extrinsic property of a body. |

3. | Mass is the measure of inertia. | Weight is the measure of force. |

4. | The mass of a body remains the same everywhere in the universe. | The weight of a body depends on the local acceleration due to gravity where it is placed. |

5. | The mass of a body cannot be zero. | The weight of a body can be zero. |

6. | The SI unit of mass is the kilogram (kg). | Since weight is a force, its SI unit is the newton (N). |

7. | The mass of a body can be measured using a beam balance and a pan balance. | The weight of a body can be measured using a spring balance and a weighing machine. |

**Will the mass and weight of an object on the earth be same as their values on Mars? Why?**

**Answer:-**

No, the mass of an object will remain the same wherever it is in the universe, but the weight of an object will vary depending on the strength of the gravitational field it is experiencing. The gravitational force on Mars is about 38% of the gravitational force on Earth, so an object with a mass of 10 kg on Earth would have a weight of 98 newtons (N) (10 kg x 9.8 m/s²) on Earth, but only a weight of 37.4 N (10 kg x 3.74 m/s²) on Mars.

### b What are (i) free fall, (ii) acceleration due to gravity (iii) escape velocity (iv) centripetal force ?

Answer:- **1) Freefall:-**

**A body is said to be under free fall when no other force except the force of gravity is acting on it**

**ii) Acceleration due to gravity:-**

Acceleration due to gravity, denoted as “g”, is the acceleration experienced by an object when it falls freely under the influence of gravity. It is the acceleration of an object towards the Earth or any other celestial body due to the force of gravity. On the surface of the Earth, the acceleration due to gravity is approximately 9.8 meters per second squared (m/s²), which means that an object in free fall near the surface of the Earth will accelerate at a rate of 9.8 m/s² towards the Earth’s center. The value of acceleration due to gravity can vary slightly depending on the location and altitude.

**iii) Escape Velocity:- **

Escape velocity is the minimum speed required for an object to escape from the gravitational pull of a celestial body such as a planet, star, or moon, and move away from it indefinitely. It is the speed at which the kinetic energy of an object equals the gravitational potential energy of the object at a given altitude.

The formula for escape velocity is given by:

v = √^{2GM}/_{r}

where:

- v is the escape velocity
- G is the universal gravitational constant
- M is the mass of the celestial body
- r is the distance from the center of the celestial body to the object.

This formula shows that the escape velocity depends on the mass of the celestial body and the distance from its center.

**iv) Centripetal Force:- **

The force required to keep an object under circular motion is known as centripetal force. This force always acts towards the centre of the circular path

**c. Write the three laws given by Kepler. How did they help Newton to arrive at the inverse square law of gravity**?

**Answer:-**

Three laws given by Kepler is as follows:**First Law:** The orbits of the planets are in the shape of an ellipse, having the Sun at one focus.**Second Law:** The area swept over per hour by the radius joining the Sun and the planet is the same in all parts of the planet’s orbit.**Third Law:** The squares of the periodic times of the planets are proportional to the cubes of their mean distances from the Sun.

Newton used Kepler’s third law of planetary motion to arrive at the inverse-square rule. He assumed that the orbits of the planets around the Sun are circular, and not elliptical, and so derived the inverse-square rule for gravitational force using the formula for centripetal force. This is given as:

F = mv^{2}/ r^{ }...(i) where, m is the mass of the particle, r is the radius of the circular path of the particle and v is the velocity of the particle. Newton used this formula to determine the force acting on a planet revolving around the Sun. Since the mass m of a planet is constant, equation (i) can be written as:

F ∝ v^{2}/ r^{ }...(ii)

Now, if the planet takes time T to complete one revolution around the Sun, then its velocity v is given as:

v = 2πr/ T^{ }...(iii) where r is the radius of the circular orbit of the planet

or, v ∝ r/ T ...(iv) [as the factor 2π is a constant]

On squaring both sides of this equation, we get:

v^{2} ∝ r^{2}/ T^{2}...(v)

On multiplying and dividing the right-hand side of this relation by r, we get:

`v^2alpha 1 \\r `...(vi)

According to Kepler’s third law of planetary motion, the factor r^{3}/ T^{2} is a constant. Hence, equation (vi) becomes:

v^{2} ∝ 1/ r...(vii)

On using equation (vii) in equation (ii), we get:

`F alpha 1/r^2`

Hence, the gravitational force between the sun and a planet is inversely proportional to the square of the distance between them.

**d. 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 same as the time taken to come down.**

**Answer:-**

We have v = u + at

and s = ut + 1/2 “at”

^{2}∴ s = (“v” - “at”)t + 1/2“at”

^{2}+ 1/2“at”^{2}s = ut - 1/2 “at”

^{2}As the stone moves upward from A → B

S = AB = h, t = t

_{1}, a = - g (retardation)u = u and v = 0

h = 0 - 1/2 (- “g”)t

_{1}^{2}h = 1/2“gt”

_{1}^{2}As the stone moves downward from B → A

t = t

_{2}, u = 0, s = h and a = gh = 1/2“gt”

_{2}^{2}“t”

_{1}^{2}= “t”_{2}^{2}“t”

_{1}= “t”_{2}

#### e. If the value of g suddenly becomes twice its value, it will become two times more difficult to pull a heavy object along the floor. Why?

**Answer:- **

**W = F = mg.**

### Que:- 3Explain why the value of g is zero at the centre of the earth.

Answer:_

This is because all of the forces on you that result from the **Earth’s gravity** are balanced. Mass attracts mass. There is the same amount of earthly mass in every direction. So you would be pulled equally in each direction and the net **force** on **your** body would be **zero**

**4. Let the period of revolution of a planet at a distance R from a star be T. Prove that if it was at a distance of 2R from the star, its period of revolution will be 8 T**

**Answer:-**

From Kepler's third law of planetary motion, we have

or

$$ T_1^2 \propto 8 R^3 . . . . . $$(iii)

$$\Rightarrow T_1 = \sqrt{8}T$$

**5. Solve the following examples**.

**a. An object takes 5 s to reach the ground from a height of 5 m on a planet. What is the value of g on the planet?**

**Answer:- **

**Given:** Time (t) = 5 s, height (s) = 5 m

**To find:** Gravitational acceleration (g)

**Formula: **s = ut + `1/2 "gt"^2`

**Calculation:** From formula,

5 = `0 xx 5 + 1/2 "g"(5)^2`

∴ `5 = 1/2 "g" xx 25`

∴ g = `2/5`

∴ g = 0.4 m/s^{2 }

The gravitational acceleration of the planet is **0.4 m/s ^{2}**.

**b. The radius of planet A is half the radius of planet B. If the mass of A is MA, what must be the mass of B so that the value of g on B is half that of its value on A?**

The acceleration due to gravity of a planet is given as

\[\text{g} = \frac{\text{GM}}{\text{r}^2}\]

For planet A:

For planet B:

\[\text{g}_{B} = \frac{\text{GM}_\text{B}}{\text{r}_\text{B}^2}\]

\[\text{g}_\text{B} = \frac{1}{2} \text{g}_\text{A}\] ...(Given) or,

\[\frac{\text{GM}_\text{B}}{\text{r}_\text{B}^2} = \frac{\text{G M}_\text{A}}{2 \text{r}_\text{A}^2}\]

\[\Rightarrow \text{M}_\text{B} = \frac{\text{M}_\text{A} \text{r}_\text{B}^2}{2 \text{r}_\text{A}^2}\]

\[\Rightarrow \text{M}_\text{B} = \frac{\text{M}_\text{A} \text{r}_\text{B}^2}{2(\frac{1}{2} \text{r}_\text{B})^2} = 2 \text{M}_\text{A}\]

**c. The mass and weight of an object on earth are 5 kg and 49 N respectively. What will be their values on the moon? Assume that the acceleration due to gravity on the moon is 1/6th of that on the earth.**

**Given:** Mass on earth (me) = 5 kg, weight on earth (W_{e}) = 49 N,

acceleration due to gravity on moon (g_{m}) = `9.8/6` m/s^{2} = 1.63 m/s^{2}

**To find:** Mass (m_{m}), weight (W_{m}) on moon

**Formula:** W_{m} = m_{m}g_{m}

**Calculation:** The mass of the object is independent of gravity and remains unchanged i.e., **5 kg**.

From formula,

W_{m} = 5 × 1.63

∴ W_{m} = **8.15 N**

On moon, the mass of the object is **5 kg** and the weight is **8.15 N**.

**d. An object thrown vertically upwards reaches a height of 500 m. What was its initial velocity? How long will the object take to come back to the earth? Assume g = 10 m/s2**

From newton’s third equation of the motion;

V^{2} = u^{2} + 2as

Where;

V = Final velocity;

U = initial velocity;

T = time taken;

S = distance travelled;

A = acceleration;

According to our question;

The figure below illustrates the situation given in the question

V = 0 (Velocity at maximum height is zero);

S = 500m ;

A = -10 m/s^{2} (because when object will be going up the acceleration due to gravity will be acting downwards to make object to fall. Hence by sign convention direction of motion and acceleration is opposite therefore a is negative);

Putting the above values we get

0^{2} = u^{2} + (2 x (-10) x 500)

0 = u^{2} - 10000

u^{2} = 10000

⇒ u = √10000 =100

**Therefore initial velocity is 100m/s.**

From the Newton’s first law of motion;

v = u + at

Where symbols have there usual meanings as above;

v = 0 (velocity at maximum height is zero);

u = initial velocity = 100m/s;

a = - 10m/s^{2} (because when object will be going up the acceleration due to gravity will be acting downwards to make object to fall. Hence by sign convention direction of motion and acceleration is opposite therefore a is negative);

Putting the values we get;

0 = 100 + ( -10t )

= 100 = 10t

t = `100/10` = 10s

Now we know that time required by an object to go up is same as time required to come down.

Therefore;

Total time = time of ascent + time of descent

= 10 + 10 = 20 s

**Hence total time to come back to earth is 20 seconds.**

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

**e. A ball falls off a table and reaches the ground in 1 s. Assuming g = 10 m/s2, calculate its speed on reaching the ground and the height of the table.**

**Answer:-**

Here, t =1 s

g = 10 m/s^{2}

u = 0

Let v be the velocity of the ball on reaching the ground.

Thus, from first equation of motion, we have

v = u + gt

⇒ v = 10 x 1 = 10 m/s

**Hence, the speed of the object on reaching the ground is 10 m/s.**

Let h be the height of the table. Thus, from the second equation of motion, we have

`"S" = "ut" + 1/2 "gt"^2`

`⇒ "h" = 0 + 1/2 xx 10xx1^2`

⇒ h = 5 m

**Hence, the height of the table is 5 m.**

**f. The masses of the earth and moon are 6 x 1024 kg and 7.4×1022 kg, respectively. The distance between them is 3.84 x 105 km. Calculate the gravitational force of attraction between the two?**

**Answer:-**

the mass of the earth, M = 6 × 10^{24} kg

the mass of the moon,

m = 7.4 × 10^{22}

the distance between the earth and the moon,

d = 3.84 × 10^{5} km

= 3.84 × 10^{5} × 1000 m

= 3.84 × 10^{8} m

G = 6.7 × 10^{-11} Nm^{2}kg^{-2}

the force exerted by the earth on the moon is

F = `("G" "M" xx "m")/"d"^2`

`= (6.7 xx 10^-11 "Nm"^2"kg"^-2 xx 6 xx 10^24 "kg" xx 7.4 xx 10^22 "kg")/(3.84 xx 10^8 "m")^2`

`= ((6.7 xx 10^-11) xx (6 xx 10^24) xx (7.4 xx 10^22))/(3.84 xx 10^8)^2`

`= (6.7 xx 6 xx 7.4)/(3.84 xx 3.84) xx 10^19`

= **2 × 10 ^{20}^{ }N.**

Thus the force exerted by the earth on the moon is 2 × 10^{20}^{ }N.

**g. The mass of the earth is 6 x 1024 kg. The distance between the earth and the Sun is 1.5x 1011 m. If the gravitational force between the two is 3.5 x 1022 N, what is the mass of****the Sun? Use G = 6.7 x 10-11 N m2 kg-2**

**Answer:-**

**Given:** Mass of the earth (M_{e}) = 6 × 10^{24} kg,

Gravitational force (F) = 3.5 × 10^{22} N,

Distance (R) = 1.5 × 10^{11} m,

Gravitational constant (G) = 6.7 × 10^{-11} Nm^{2}/kg^{2}

The gravitational force between the Sun and the Earth can be found out using the formula,

where, Me and Ms are the masses of the Earth and the Sun, respectively. Using all the given values, we have

\[3 . 5 \times {10}^{22} = (6 . 7 \times {10}^{- 11} )\frac{(6 \times {10}^{24} ) \times M_s}{(1 . 5 \times {10}^{11} )^2}\]

\[ \Rightarrow M_s = \frac{(3 . 5 \times {10}^{22} ) \times (1 . 5 \times {10}^{11} )^2}{(6 . 7 \times {10}^{- 11} ) \times (6 \times {10}^{24} )}\]

`= (7.88 xx 10^44)/(40.2 xx 10^13)`

**= 1.96 × 10 ^{30} kg**

The mass of the Sun is **1.96 × 10 ^{30} kg**.