NIOS Class 12 Physics Chapter 5 Gravitation

NIOS Class 12 Physics Chapter 5 Gravitation Solutions English Medium As Per New Syllabus to each chapter is provided in the list so that you can easily browse throughout different chapters NIOS Class 12 Physics Chapter 5 Gravitation Notes in English and select need one. NIOS Class 12 Physics Solutions English Medium Download PDF. NIOS Study Material of Class 12 Physics Notes Paper Code: 312.

NIOS Class 12 Physics Chapter 5 Gravitation

Also, you can read the NIOS book online in these sections Solutions by Expert Teachers as per National Institute of Open Schooling (NIOS) Book guidelines. These solutions are part of NIOS All Subject Solutions. Here we have given NIOS Class 12 Physics Notes, NIOS Senior Secondary Course Physics Solutions in English for All Chapter, You can practice these here.

Gravitation

Chapter: 5

Module – I: Motion, Force and Energy

INTEXT QUESTIONS 5.1

1. The period of revolution of the moon around the earth is 27.3 days. Remember that this is the period with respect to the fixed stars (the period of revolution with respect to the moving earth is about 29.5 days; it is this period that is used to fix the duration of a month in some calendars). The radius of moon’s orbit is 3.84 × 10⁸ m (60 times the earth’s radius). Calculate the centripetal acceleration of the moon and show that it is very close to the value given by 9.8 m s⁻² divided by 3600, to take account of the variation of the gravity as 1/r².

Ans: Given: T = 27.3 days

= 27.3 × 24 × 3600 s

R = 3.84 × 10⁸ m

a = 4π²R / T²

a = 4π²(3.84 × 10⁸) / (27.3 × 24 × 3600)²

T = 2.358 × 10⁶ s

T² = 5.56 × 10¹² s²

4π²R = 3.80 × 10¹⁰ m

a = 3.80 × 10¹⁰ / 5.56 × 10¹² 

= 6.8 × 10⁻³ m s⁻²

9.8 m s⁻² / 3600

= 2.72 × 10⁻³ m s⁻²

close to the calculated value when the factor 1/r² is included.

2. From Eqn. (5.1), deduce dimensions of G.

Ans: Given:

F = G m₁m₂ / r²

Formula (dimensional):

3. Using Eqn. (5.1), show that G may be defined as the magnitude of force between two masses of 1 kg each separated by a distance of 1 m.

Ans: Given: m₁ = 1kg

m₂ = 1 kg, 

r = 1 m

F = G m₁m₂ / r²

Substitution:

F = G · 1 · 1 / 1² 

= G

Thus the numerical value of G equals the force between two 1 kg masses 1 m apart.

4. The magnitude of force between two masses placed at a certain distance is F. What happens to F if (i) the distance is doubled without any change in masses, (ii) the distance remains the same but each mass is doubled, (iii) the distance is doubled and each mass is also doubled?

Ans: Formula: F ∝ m₁m₂ / r²

(i) r→2r 

→ F′

= F/4

(ii) m₁→2m₁,

m₂→2m₂ 

→ F′

= 4F

(iii) m₁,m₂ doubled and r doubled 

→ F′ = 4F/4 

= F.

5. Two bodies having masses 50 kg and 60 kg are separated by a distance of 1 m. Calculate the gravitational force between them.

Ans: Given:

m₁ = 50 kg, 

m₂ = 60 kg,

r = 1 m, 

G = 6.67 × 10⁻¹¹ N m² kg⁻²

F = G m₁m₂ / r²

F = 6.67 × 10⁻¹¹ × 50 × 60 / 1²

Calculation:

F = 2.0 × 10⁻⁷ N.

INTEXT QUESTIONS 5.2

1. The mass of the earth is 5.97 × 10²⁴kg and its mean radius is 6.371 × 10⁶m. Calculate the value of g at the surface of the earth.

Ans: Given:

M = 5.97 × 10²⁴ kg, 

R = 6.371 × 10⁶ m, 

G = 6.67 × 10⁻¹¹ N m² kg⁻²

g = G M / R²

g = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / (6.371 × 10⁶)²

g = 9.81 m s⁻².

2. Careful measurements show that the radius of the earth at the equator is 6378 km while at the poles it is 6357 km. Compare values of g at the poles and at the equator.

Ans: Given:

Rₑ = 6.378 × 10⁶ 

m, Rₚ = 6.357 × 10⁶ m

g ∝ 1/R²

gₑ/gₚ = (Rₚ/Rₑ)² 

= (6.357/6.378)²

gₑ/gₚ = 0.9979 

⇒ gₚ ≈ 1.00 gₑ or gₚ is about 0.2% larger than gₑ.

3. A particle is thrown up. What is the direction of g when (i) the particle is going up, (ii) when it is at the top of its journey, (iii) when it is coming down, and (iv) when it has come back to the ground?

Ans: In all four cases the acceleration due to gravity g is vertically downward toward the earth’s centre.

4. The mass of the moon is 7.3 × 10²² kg and its radius is 1.74 × 10⁶ m. Calculate the gravitational acceleration at its surface.

Ans: Given:

M = 7.3 × 10²² kg, 

R = 1.74 × 10⁶ m

g = G M / R²

g = 6.67 × 10⁻¹¹ × 7.3 × 10²² / (1.74 × 10⁶)²

g = 1.62 m s⁻².