Gravitation — NCERT Solutions

Given:

• Mass of object, $m = 10\,\text{kg}$
• $g_{\text{Earth}} = 9.8\,\text{m/s}^2$
• $g_{\text{Moon}} = \dfrac{1}{6} \times g_{\text{Earth}} = \dfrac{9.8}{6} \approx 1.633\,\text{m/s}^2$

Formula: $W = mg$

Weight on Earth:
$$W_{\text{Earth}} = m \times g_{\text{Earth}} = 10 \times 9.8 = \mathbf{98\,N}$$

Weight on Moon:
$$W_{\text{Moon}} = m \times g_{\text{Moon}} = 10 \times \frac{9.8}{6} = \frac{98}{6} \approx \mathbf{16.3\,N}$$

Answer:

• Weight of the object on Earth $= 98\,\text{N}$
• Weight of the object on Moon $\approx 16.3\,\text{N}$

Given:

• Initial velocity, $u = 49\,\text{m/s}$ (upward)
• At maximum height, final velocity, $v = 0$
• Acceleration due to gravity, $g = 9.8\,\text{m/s}^2$ (acting downward, so $a = -9.8\,\text{m/s}^2$)

(i) Maximum height:

Using the equation: $v^2 = u^2 + 2as$
$$0 = (49)^2 + 2 \times (-9.8) \times h$$
$$0 = 2401 - 19.6h$$
$$h = \frac{2401}{19.6} = \mathbf{122.5\,m}$$

(ii) Total time to return to the surface:

Time to reach maximum height using $v = u + at$:
$$0 = 49 + (-9.8) \times t$$
$$t = \frac{49}{9.8} = 5\,\text{s}$$

By symmetry, the time to fall back from maximum height to the ground is also $5\,\text{s}$.

$$\text{Total time} = 5 + 5 = \mathbf{10\,s}$$

Answers:

• (i) Maximum height $= 122.5\,\text{m}$
• (ii) Total time $= 10\,\text{s}$

Given:

• Initial velocity, $u = 0$ (released from rest)
• Height of tower, $s = 19.6\,\text{m}$
• Acceleration due to gravity, $g = 9.8\,\text{m/s}^2$

Formula: $v^2 = u^2 + 2gs$

Calculation:
$$v^2 = 0 + 2 \times 9.8 \times 19.6$$
$$v^2 = 2 \times 9.8 \times 19.6 = 384.16$$
$$v = \sqrt{384.16} = 19.6\,\text{m/s}$$

Answer: The final velocity of the stone just before touching the ground is $\mathbf{19.6\,m/s}$.

Given:

• Initial velocity, $u = 40\,\text{m/s}$ (upward)
• $g = 10\,\text{m/s}^2$
• At maximum height, $v = 0$

Maximum Height:

Using $v^2 = u^2 - 2gh$:
$$0 = (40)^2 - 2 \times 10 \times h$$
$$h = \frac{1600}{20} = \mathbf{80\,m}$$

Net Displacement:

The stone is thrown upward and returns to the same point (the thrower's hand/ground level).
$$\text{Net displacement} = \mathbf{0\,m}$$

Total Distance Covered:

The stone travels $80\,\text{m}$ upward and then $80\,\text{m}$ downward.
$$\text{Total distance} = 80 + 80 = \mathbf{160\,m}$$

Summary:

• Maximum height $= 80\,\text{m}$
• Net displacement $= 0\,\text{m}$
• Total distance covered $= 160\,\text{m}$

Given:

• Mass of Earth, $M_E = 6 \times 10^{24}\,\text{kg}$
• Mass of Sun, $M_S = 2 \times 10^{30}\,\text{kg}$
• Distance, $d = 1.5 \times 10^{11}\,\text{m}$
• $G = 6.67 \times 10^{-11}\,\text{N m}^2\,\text{kg}^{-2}$

Formula:
$$F = G\frac{M_E \times M_S}{d^2}$$

Calculation:
$$F = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 2 \times 10^{30}}{(1.5 \times 10^{11})^2}$$

Numerator:
$$6.67 \times 6 \times 2 \times 10^{-11+24+30} = 80.04 \times 10^{43}$$

Denominator:
$$(1.5)^2 \times 10^{22} = 2.25 \times 10^{22}$$

$$F = \frac{80.04 \times 10^{43}}{2.25 \times 10^{22}} = \frac{80.04}{2.25} \times 10^{21} = 35.57 \times 10^{21}$$

$$F \approx 3.56 \times 10^{22}\,\text{N}$$

Answer: The gravitational force between the Earth and the Sun is approximately $\mathbf{3.56 \times 10^{22}\,N}$.

Given:

• Height of tower, $H = 100\,\text{m}$
• Stone 1: falls from top, initial velocity $u_1 = 0$
• Stone 2: projected upward from ground, initial velocity $u_2 = 25\,\text{m/s}$
• $g = 10\,\text{m/s}^2$

Let the two stones meet after time $t$ seconds.

Let the meeting point be at height $x$ from the ground.

For Stone 1 (falling from top):

Distance fallen from top $= H - x = 100 - x$
$$100 - x = \frac{1}{2}g t^2 = \frac{1}{2} \times 10 \times t^2 = 5t^2 \quad \cdots (1)$$

For Stone 2 (projected upward from ground):
$$x = u_2 t - \frac{1}{2}g t^2 = 25t - 5t^2 \quad \cdots (2)$$

Adding equations (1) and (2):
$$(100 - x) + x = 5t^2 + 25t - 5t^2$$
$$100 = 25t$$
$$t = 4\,\text{s}$$

Finding the height (from equation 2):
$$x = 25 \times 4 - 5 \times (4)^2 = 100 - 80 = 20\,\text{m}$$

Answer: The two stones will meet after $\mathbf{t = 4\,s}$ at a height of $\mathbf{20\,m}$ from the ground.

Given:

• Total time of flight $= 6\,\text{s}$
• $g = 9.8\,\text{m/s}^2$

Since the motion is symmetric, time to reach maximum height $= \dfrac{6}{2} = 3\,\text{s}$.

(a) Initial velocity:

At maximum height, $v = 0$. Using $v = u - gt$:
$$0 = u - 9.8 \times 3$$
$$u = 29.4\,\text{m/s}$$

The ball was thrown up with a velocity of $\mathbf{29.4\,m/s}$.

(b) Maximum height:

Using $s = ut - \dfrac{1}{2}gt^2$ for upward journey ($t = 3\,\text{s}$):
$$h = 29.4 \times 3 - \frac{1}{2} \times 9.8 \times (3)^2$$
$$h = 88.2 - \frac{1}{2} \times 9.8 \times 9$$
$$h = 88.2 - 44.1 = 44.1\,\text{m}$$

Maximum height reached $= \mathbf{44.1\,m}$.

(c) Position after 4 s:

The ball reaches maximum height at $t = 3\,\text{s}$ and then starts falling. At $t = 4\,\text{s}$, it has been falling for $4 - 3 = 1\,\text{s}$ from the maximum height.

Distance fallen in 1 s from maximum height:
$$d = \frac{1}{2}g(1)^2 = \frac{1}{2} \times 9.8 \times 1 = 4.9\,\text{m}$$

Position from ground $= 44.1 - 4.9 = 39.2\,\text{m}$

After 4 s, the ball is at a height of $\mathbf{39.2\,m}$ from the ground (on its way down).

The buoyant force (upthrust) on an object immersed in a liquid acts in the vertically upward direction.

Explanation: When an object is immersed in a liquid, the liquid exerts pressure on all surfaces of the object. The pressure on the lower surface of the object is greater than the pressure on the upper surface (since pressure increases with depth). This difference in pressure results in a net upward force called the buoyant force or upthrust, which always acts vertically upward — opposite to the direction of gravity.

Reason:

When a block of plastic is released under water, two forces act on it:

1. Weight ($W = mg$) — acting vertically downward.
2. Buoyant force (upthrust) — acting vertically upward.

The density of plastic is less than the density of water. According to Archimedes' Principle, the buoyant force equals the weight of water displaced by the plastic block.

Since the plastic has lower density than water:
$$\text{Buoyant force} > \text{Weight of plastic block}$$

The net force on the plastic block is directed upward, so it accelerates upward and rises to the surface of the water.

Conclusion: A block of plastic comes up to the surface because the buoyant force acting on it is greater than its weight.

Given:

• Mass of substance, $m = 50\,\text{g}$
• Volume of substance, $V = 20\,\text{cm}^3$
• Density of water, $\rho_{\text{water}} = 1\,\text{g cm}^{-3}$

Density of substance:
$$\rho_{\text{substance}} = \frac{m}{V} = \frac{50}{20} = 2.5\,\text{g cm}^{-3}$$

Comparison:
$$\rho_{\text{substance}} = 2.5\,\text{g cm}^{-3} > \rho_{\text{water}} = 1\,\text{g cm}^{-3}$$

Conclusion: Since the density of the substance ($2.5\,\text{g cm}^{-3}$) is greater than the density of water ($1\,\text{g cm}^{-3}$), the substance will sink in water.

Given:

• Mass of packet, $m = 500\,\text{g}$
• Volume of packet, $V = 350\,\text{cm}^3$
• Density of water, $\rho_{\text{water}} = 1\,\text{g cm}^{-3}$

Step 1: Find density of packet:
$$\rho_{\text{packet}} = \frac{m}{V} = \frac{500}{350} \approx 1.43\,\text{g cm}^{-3}$$

Step 2: Compare with density of water:
$$\rho_{\text{packet}} = 1.43\,\text{g cm}^{-3} > \rho_{\text{water}} = 1\,\text{g cm}^{-3}$$

Conclusion: Since the density of the packet is greater than the density of water, the packet will sink.

Step 3: Mass of water displaced:

When the packet sinks, it displaces water equal to its own volume ($350\,\text{cm}^3$).
$$\text{Mass of water displaced} = \rho_{\text{water}} \times V = 1 \times 350 = \mathbf{350\,g}$$

Answers:

• The packet will sink in water.
• Mass of water displaced $= 350\,\text{g}$.