Model Test – Solved ഉത്തരങ്ങൾ

G (ഗുരുത്വാകർഷണസ്ഥിരാങ്കം) is the force between two 1 kg masses kept 1 m apart. Henry Cavendish measured it in 1798. It has the same value everywhere in the universe.

Weight (ഭാരം) is a force, W = mg, so it is measured in newton. “kg” is the unit of mass; kgwt is a practical unit of weight.

F ∝ 1/d². With 2d the force is multiplied by 1/2² = ¼ (Table 4.2: 12.5 G → 3.125 G).

At the centre (ഭൂകേന്ദ്രം) a body is pulled equally from all sides, so the pulls cancel: g = 0 and the weight there is 0 N (the mass is unchanged).

1 kgwt is the force with which the Earth attracts a 1 kg mass: F = mg = 1 kg × 9.8 m/s² = 9.8 N.

The Earth’s gravity, always directed towards the Earth’s centre, itself acts as the centripetal force (അഭികേന്ദ്രബലം) on the satellite (കൃത്രിമ ഉപഗ്രഹം).

Centripetal force (അഭികേന്ദ്രബലം). For a stone on a string the hand supplies it through the string; without it the body moves off along the tangent.

Mass (മാസ്) is the quantity of matter – it is the same everywhere. Only the weight changes: 12 × 1.62 = 19.44 N on the Moon, 12 × 9.8 = 117.6 N on the Earth.

Law (സാർവിക ഗുരുത്വാകർഷണനിയമം): every object in the universe attracts every other object. The force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Equation: F = G m₁m₂/d², where F = gravitational force (N), m₁, m₂ = masses (kg), d = distance between their centres (m), G = universal gravitational constant, 6.67 × 10⁻¹¹ N m²/kg².

Given: m = 60 kg; g on the Earth = 9.8 m/s²; g on the Moon = 1.62 m/s²

Formula: W = mg

Substitution: (a) W = 60 × 9.8   (b) W = 60 × 1.62

Answer: (a) 588 N   (b) 97.2 N – about 1/6 of the Earth value; the mass is 60 kg in both places.

F = G m₁m₂/d².

(a) F′ = G(3m₁)m₂/d² = 3 × Gm₁m₂/d² = 3F (F ∝ product of masses).

(b) F″ = Gm₁m₂/(d/2)² = Gm₁m₂/(d²/4) = 4 × Gm₁m₂/d² = 4F (F ∝ 1/d²).

(a) No. The speed is constant, but velocity also has a direction, and the direction of motion changes at every instant on a circle. So the velocity is not uniform (it is accelerated motion, സമവർത്തുളചലനം).

(b) The string supplied the centripetal force. When it breaks there is no force towards the centre, so the stone moves in a straight line along the tangent (തൊടുവര) to the circle at the point where it was released.

Given: u = 0 (it falls from rest), t = 1.5 s, a = g = 9.8 m/s²

Formula: (a) s = ut + ½gt² = ½gt²   (b) v = u + gt = gt

Substitution: (a) h = ½ × 9.8 × 1.5² = 4.9 × 2.25   (b) v = 9.8 × 1.5

Answer: (a) h = 11.025 m ≈ 11 m   (b) v = 14.7 m/s

Any three correct pairs.

Given: m₁ = m₂ = 5 × 10⁴ kg, d = 10 m, G = 6.67 × 10⁻¹¹ N m²/kg²

Formula: F = G m₁m₂/d²

Substitution: F = 6.67 × 10⁻¹¹ × 5 × 10⁴ × 5 × 10⁴ / 10² = 6.67 × 25 × 10^{−11 + 8 − 2} = 166.75 × 10⁻⁵

Answer: (a) F = 1.6675 × 10⁻³ N ≈ 1.67 × 10⁻³ N

(b) This force is tiny – about the weight of a 0.17 g speck on the Earth. The resistance of the water and the pull of the anchor ropes are far larger, so it cannot move the boats (it would give each boat an acceleration of only F/m ≈ 3.3 × 10⁻⁸ m/s²). Gravitational forces between everyday objects are too small to notice.

(a) Force of gravity on a mass m at the surface: F = GMm/R². By the second law, F = ma = mg. So mg = GMm/R²; cancelling m on both sides, g = GM/R².

Given: G = 6.67 × 10⁻¹¹ N m²/kg², M = 6 × 10²⁴ kg, R = 6.4 × 10⁶ m

Formula: g = GM/R²

Substitution: g = 6.67 × 10⁻¹¹ × 6 × 10²⁴ / (6.4 × 10⁶)² = 40.02 × 10¹³ / 40.96 × 10¹²

Answer: (b) g = 9.77 m/s² ≈ 9.8 m/s²

(c) The mass m of the falling body is not in g = GM/R², so every body falls with the same acceleration. In air the feather is slowed much more by air resistance; with the air removed nothing else acts, so both fall together (feather and coin experiment, തൂവലും നാണയവും പരീക്ഷണം).

(a) The reading increases in England (the textbook example: 20 N more).

(b) The Earth is not a perfect sphere – it is flattened at the poles, so the distance to the centre is smaller near the poles (ധ്രുവപ്രദേശം) than at the equator (ഭൂമധ്യരേഖാപ്രദേശം). By g = GM/R² a smaller R gives a bigger g (≈ 9.78 m/s² at the equator, ≈ 9.83 m/s² at the poles). The mass is unchanged, but W = mg is bigger, so the spring stretches more.

(c) No. A common balance compares the goods with standard masses; both pans are pulled more by the same factor, so it still balances – it measures mass, which does not change.

Given: u = 8.1 m/s (upwards), g = 1.62 m/s² on the Moon, v = 0 at the highest point

Formula: (a) v = u − gt   (b) v² = u² − 2gh, so h = u²/2g   (c) time up = time down

Substitution: (a) 0 = 8.1 − 1.62 t → t = 8.1/1.62   (b) h = 8.1²/(2 × 1.62) = 65.61/3.24 (check: ½gt² = ½ × 1.62 × 5²)   (c) 2 × 5   (d) h = 65.61/(2 × 9.8) = 65.61/19.6

Answer: (a) 5 s   (b) 20.25 m   (c) 10 s   (d) 3.35 m; 20.25/3.35 ≈ 6, so the ball goes about 6 times higher on the Moon, because g there is about 1/6 of the Earth’s.

(a) At rest the spring holds up the whole weight: W = mg = 5 × 9.8 = 49 N (5 kgwt).

(b) 0 N. In free fall (നിർബാധ പതനം) the balance and the object both fall with the same acceleration g, so the object does not pull on the spring and the spring is not stretched. The whole of gravity is used to accelerate the object – weightlessness (ഭാരമില്ലായ്മ). Its true weight, 49 N, is still acting.

Given: R = 6400 km, h = 400 km, R + h = 6800 km, g = 9.8 m/s² at the surface

Formula: g = GM/R² and g′ = GM/(R + h)², so g′ = g × R²/(R + h)²

Substitution: g′ = 9.8 × (6400/6800)² = 9.8 × 0.886

Answer: (c) g′ ≈ 8.68 m/s² ≈ 8.7 m/s² – about 89 % of the surface value, not zero.

The station and the astronauts are falling freely round the Earth together, with the same acceleration (gravity acts as the centripetal force). Nothing presses on the floor or a balance, so they feel weightless.