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## NCERT Class 11 Physics Textbook Chapter 2 With Answer PDF Free Download

### Chapter 2: Units and Measurements

#### 2.1 Introduction

Measurement of any physical quantity involves comparison with a certain basic, arbitrarily chosen, internationally accepted reference standard called unit.

The result of a measurement of a physical quantity is expressed by a number (or numerical measure) accompanied by a unit.

Although the number of physical quantities appears to be very large, we need only a limited number of units for expressing all the physical quantities, since they are interrelated with one another.

The units for the fundamental or base quantities are called fundamental or base units.

The units of all other physical quantities can be expressed as combinations of the base units. Such units obtained for the derived quantities are called derived units.

A complete set of these units, both the base units and derived units is known as the system of units.

#### 2.2 The International System of Units

In earlier times scientists of different countries were using different systems of units for measurement.

Three such systems, the CGS, the FPS (or British) system, and the MKS the system were in use extensively till recently.

The base units for length, mass, and time in these systems were as follows :

• In the CGS system, they were centimeter, gram, and second respectively.

• In the FPS system, they were foot, pound, and second respectively.

• In the MKS system, they were a meter, kilogram, and second respectively.

The system of units that is at present internationally accepted for measurement is the Système Internationale d’ Unites (French for International System of Units).

#### 2.3.1 Measurement of Large Distances

Large distances such as the distance of a planet or a star from the earth cannot be measured directly with a meter scale. An important method in such cases is the parallax method.

When you hold a pencil in front of you against some specific point on the background (a wall) and look at the pencil first through your left eye A (closing the right eye) and then look at the pencil through your right eye B (closing the left eye), you would notice that the position of the pencil seems to change with respect to the point on the wall.

This is called parallax. The distance between the two points of observation is called the basis. In this example, the basis is the distance between the eyes.

Author | NCERT |

Language | English |

No. of Pages | 23 |

PDF Size | 697 KB |

Category | Physics |

Source/Credits | ncert.nic.in |

### NCERT Solutions Class 11 Physics Chapter 2 Units and Measurements

**2.1 Fill in the blanks**

**(a) The volume of a cube of side 1 cm is equal to …..m ^{3}**

**(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to**

**…(mm)**

^{2}**(c) A vehicle moving with a speed of 18 km h**

^{–1}covers….m in 1 s**(d) The relative density of lead is 11.3. Its density is ….g cm ^{–3} or ….kg m^{–3}.**

Answer:

(a) Volume of cube, V = (1 cm)^{3} = (10^{-2} m)^{3} = 10^{-6} m^{3}.

(b) Surface area = curved area + area on top /base = 2πrh + 2πr^{2} = 2πr (h + r)

r = 2 cm = 20 mm

h = 10 cm = 100 mm

Surface area = 2πr (h + r) = 2 x 3.14 x 20 (100 + 20) = 15072 mm^{2}

Hence, answer is 15072 mm^{2}

(c) Speed of vehicle = 18 km/h

1 km = 1000 m

1 hr = 60 x 60 = 3600 s

1 km/hr = 1000 m/3600 s = 5/18 m/s

18 km/h = = (18 x 1000)/3600

= 5 m/s

Distance travelled by the vehicle in 1 s = 5 m

(d) The Relative density of lead is 11.3 g cm^{-3}

=> 11.3 x 10^{3 }kg m^{-3} [1 kilogram = 10^{3}g, 1 meter = 10^{2 }cm]

=> 11.3 x 10^{3} kg m^{-4}

**2.2 Fill in the blanks by suitable conversion of units**

(a) 1 kg m^{2} s^{–2} = ….g cm^{2} s^{–2}

(b) 1 m = ….. ly

(c) 3.0 m s^{–2} = …. km h^{–2}

(d) G = 6.67 × 10^{–11} N m^{2} (kg)^{–2} = …. (cm)3s^{–2} g^{–1}

Answer:

(a) 1 kg m^{2} s^{–2} = ….g cm^{2} s^{–2}

1 kg m^{2} s^{-2 }= 1kg x 1m^{2} x 1s ^{-2}

We know that,

1kg = 10^{3}

1m = 100cm = 10^{2}cm

When the values are put together, we get:

1kg x 1m^{2} x 1s^{-2 }= 10^{3}g x (10^{2}cm)^{2} x 1s^{-2 } = 10^{3}g x 10^{4} cm^{2} x 1s^{-2 } = 10^{7} gcm^{2}s^{-2}

=>**1kg m ^{2} s^{-2} = 10^{7} gcm^{2}s^{-2}**

(b) 1 m = ….. ly

Using the formula,

Distance = speed x time

Speed of light = 3 x 10^{8} m/s

Time = 1 yr = 365 days = 365 x 24 hr = 365 x 24 x 60 x 60 sec

Put these values in the formula mentioned above, we get:

One light year distance = (3 x 10^{8} m/s) x (365 x 24 x 60 x 60) = 9.46×10^{15}m

9.46 x 10^{15} m = 1ly

So that, 1m = 1/9.46 x 10^{15}ly

=> 1.06 x 10^{-16}ly

=>**1 meter = 1.06 x 10 ^{-16}ly**

(c) 3.0 m s^{–2} = …. km h^{–2}

1 km = 1000m so that 1m = 1/1000 km

3.0 m s^{-2} = 3.0 (1/1000 km) (1/3600 hour) ^{-2} = 3.0 x 10^{-3} km x ((1/3600)^{-2}h^{-2})

= 3 x 10^{-3}km x (3600)^{2} hr^{-2 }= 3.88 x 10^{4} km h^{-2}

**=> 3.0 m s ^{-2} = 3.88 x 10^{4} km h^{-2}**

(d) G = 6.67 × 10^{–11} N m^{2} (kg)^{–2} = …. (cm)3s^{–2} g^{–1}

G = 6.67 x 10^{-11} N m^{2} (kg)^{-2}

We know that,

1N = 1kg m s^{-2}

1 kg = 10^{3} g

1m = 100cm= 10^{2} cm

Put the values together, we get:

=> 6.67 x 10^{-11} Nm^{2} kg^{-2} = 6.67 x 10^{-11} x (1kg m s ^{-2}) (1m^{2}) (1kg^{-2})

Solve the following and cancelling out the units, we get:

=> 6.67 x 10^{-11} x (1kg ^{-1} x 1m^{3} x 1s^{-2})

Put the above values together to convert kg to g and m to cm

=> 6.67 x 10^{-11} x (10^{3}g)^{-1} x (10^{2} cm)^{3} x (1s^{-2})

=> 6.67 x 10^{-8 }cm^{3} s^{-2 }g ^{-1}

=>**G = 6.67 x 10 ^{-11 }Nm^{2}(kg)^{-2} = 6.67 x 10^{-8 } (cm)^{3} s^{-2} g ^{-1}**

Units and Measurements Textbook With Solutions PDF Free Download