Fill in the blanks by suitable conversion of units

(a) 1 kg m2 s–2 = ….g cm2 s–2

(b) 1 m = ….. ly

(c) 3.0 m s–2 = …. km h–2

(d) G = 6.67 × 10–11 N m2 (kg)–2 = …. (cm)3s–2 g–1

Asked by Abhisek | 1 year ago |  98

##### Solution :-

(a) 1 kg m2 s–2 = ….g cm2 s–2

1 kg m2 s-2 = 1kg x 1m2 x 1s -2

We know that,

1kg = 103

1m = 100cm = 102cm

When the values are put together, we get:

1kg x 1m2 x 1s-2 = 103g x (102cm)2 x 1s-2

= 103g x 104 cm2 x 1s-2  = 107 gcm2s-2

1kg m2 s-2 = 107 gcm2s-2

(b) 1 m = ….. ly

Using the formula,

Distance = speed x time

Speed of light = 3 x 108 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 108 m/s) x (365 x 24 x 60 x 60)

= 9.46×1015m

9.46 x 1015 m = 1ly

So that, 1m = 1/9.46 x 1015ly

1.06 x 10-16ly

1 meter = 1.06 x 10-16ly

(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)-2h-2)

= 3 x 10-3km x (3600)2 hr-2 = 3.88 x 104 km h-2

=> 3.0 m s-2 = 3.88 x 104 km h­-2

(d) G = 6.67 × 10–11 N m2 (kg)–2 = …. (cm)3s–2 g–1

G = 6.67 x 10-11 N m2 (kg)-2

We know that,

1N = 1kg m s-2

1 kg = 103 g

1m = 100cm= 102 cm

Put the values together, we get:

6.67 x 10-11 Nm2 kg-2

= 6.67 x 10-11 x (1kg m s -2) (1m2) (1kg-2)

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

6.67 x 10-11 x (1kg -1 x 1m3 x 1s-2)

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

6.67 x 10-11 x (103g)-1 x (102 cm)3 x (1s-2)

6.67 x 10-8 cm3 s-2 -1

G = 6.67 x 10-11 Nm2(kg)-2

= 6.67 x 10-8  (cm)3 s-2 g -1

Answered by Pragya Singh | 1 year ago

### Related Questions

#### It is a well-known fact that during a total solar eclipse-the disk of the moon almost completely

It is a well-known fact that during a total solar eclipse-the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.

#### The farthest objects in our Universe discovered by modern astronomers are so distant that light

The farthest objects in our Universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us?

#### A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects

A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects underwater. In a submarine equipped with a SONAR, the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine? (Speed of sound in water = 1450 m s–1).