Let’s have a look at all the numerical problems from Chapter 9, Gravitation. We have selected only the numerical problems from the exercise section of the NCERT book. We also have 12 challenging questions for you at the end to practice further. Attempt these 12 questions after you complete the exercise questions from the book.
Formulas from Chapter 9 (Gravitation)
| Formula | Description |
|---|---|
| F = G (M × m) / d² | Universal Law of Gravitation: Force between two masses M and m separated by distance d, where G is the gravitational constant. |
| F = m × g | Gravitational force (Weight) on a body of mass m near Earth’s surface. |
| g = G M / R² | Acceleration due to gravity on Earth’s surface, where M is Earth’s mass and R is Earth’s radius. |
| v = u + at | Equation of motion under uniform acceleration (with a = g for free fall). |
| s = ut + ½ at² | Distance covered under uniform acceleration (replace a with g in free fall). |
| v² = u² + 2as | Relation between velocity, acceleration, and distance (a = g for free fall). |
| W = m × g | Weight of an object, directly proportional to its mass. |
| Wmoon = (1/6) Wearth | Weight of an object on the Moon is one-sixth of its weight on Earth. |
| Pressure = Thrust / Area | Pressure is thrust (force perpendicular) acting per unit area. SI unit: Pascal (Pa). |
| Density = Mass / Volume | Density of a substance. Determines floating or sinking in fluids. |
| Buoyant Force = Weight of displaced fluid | Archimedes’ Principle: “Any body completely or partially submerged in a fluid (gas or liquid) at rest is acted upon by an upward, or buoyant, force, the magnitude of which is equal to the weight of the fluid displaced by the body.” |
Class 9 Chapter 9 Physics Gravitation Solved Numerical Problems from NCERT Book
Q1. How does the force of gravitation between two objects change when the distance between them is reduced to half?
\[
F \propto \frac{1}{d^2}
\]
If \(d \to \frac{d}{2}\),
\[
F’ = \frac{1}{(d/2)^2} = 4 \cdot \frac{1}{d^2} = 4F
\]
\[
\text{Answer: Force becomes four times.}
\]
Q3. What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of earth = \(6 \times 10^{24}\,\text{kg}\), radius = \(6.4 \times 10^{6}\,\text{m}\))
\[
F = \frac{GMm}{R^2}
\]
\[
F = \frac{(6.7 \times 10^{-11})(6 \times 10^{24})(1)}{(6.4 \times 10^{6})^2}
\]
\[
F \approx 9.8 \, \text{N}
\]
Q6. What happens to the gravitational force if (i) one mass is doubled, (ii) distance is doubled/tripled, (iii) both masses are doubled?
\[
F \propto \frac{M m}{d^2}
\]
(i) \(M \to 2M \implies F’ = 2F\)
(ii) \(d \to 2d \implies F’ = \frac{F}{4}; \quad d \to 3d \implies F’ = \frac{F}{9}\)
(iii) \(M \to 2M, m \to 2m \implies F’ = 4F\)
Q12. Gravitational force on the Moon is \( \tfrac{1}{6} \) of that on Earth. What is the weight of a 10 kg object on Earth and on Moon?
On Earth:
\[
W_E = mg = 10 \times 9.8 = 98\, \text{N}
\]
On Moon:
\[
W_M = \frac{1}{6} W_E = \frac{98}{6} \approx 16.3 \, \text{N}
\]
Q13. A ball is thrown vertically upwards with a velocity of \(49\,\text{m/s}\). Calculate (i) maximum height, (ii) total time to return.
(i)
\[
H = \frac{u^2}{2g} = \frac{49^2}{2 \times 9.8} = 122.5 \, \text{m}
\]
(ii)
\[
T = \frac{2u}{g} = \frac{2 \times 49}{9.8} = 10 \, \text{s}
\]
Q14. A stone is released from the top of a tower of height \(19.6\,\text{m}\). Calculate its final velocity just before touching the ground.
\[
v^2 = u^2 + 2 g s = 0 + 2 \times 9.8 \times 19.6 = 384.16
\]
\[
v = 19.6 \, \text{m/s (downward)}
\]
Q15. A stone is thrown vertically upward with \(u = 40\,\text{m/s}\). Taking \(g = 10\,\text{m/s}^2\), find (i) maximum height, (ii) net displacement, (iii) total distance.
(i)
\[
H = \frac{u^2}{2g} = \frac{40^2}{20} = 80 \, \text{m}
\]
(ii) Net displacement = \(0 \, \text{m}\) (returns to start)
(iii) Total distance = \(80 + 80 = 160 \, \text{m}\)
Q16. Calculate the force of gravitation between Earth and Sun, given \(M_E = 6 \times 10^{24}\,\text{kg}\), \(M_S = 2 \times 10^{30}\,\text{kg}\), \(d = 1.5 \times 10^{11}\,\text{m}\).
\[
F = \frac{GM_E M_S}{d^2}
\]
\[
F = \frac{(6.7 \times 10^{-11})(6 \times 10^{24})(2 \times 10^{30})}{(1.5 \times 10^{11})^2}
\]
\[
F \approx 3.6 \times 10^{22} \, \text{N}
\]
Q17. A stone falls from a \(100\,\text{m}\) tower and another is projected upwards from the ground at \(25\,\text{m/s}\). When and where do they meet?
From top:
\[
y_A = 100 – \tfrac{1}{2} g t^2
\]
From ground:
\[
y_B = 25t – \tfrac{1}{2} g t^2
\]
Equating:
\[
100 – \tfrac{1}{2} g t^2 = 25t – \tfrac{1}{2} g t^2 \implies t = 4 \, \text{s}
\]
Position:
\[
y = 25(4) – 4.9(16) = 21.6 \, \text{m}
\]
Q18. A ball returns after \(6\,\text{s}\). Find (a) initial velocity, (b) max height, (c) position after \(4\,\text{s}\).
(a)
\[
T = \frac{2u}{g} \implies u = \frac{gT}{2} = \frac{9.8 \times 6}{2} = 29.4 \, \text{m/s}
\]
(b)
\[
H = \frac{u^2}{2g} = \frac{(29.4)^2}{19.6} = 44.1 \, \text{m}
\]
(c)
\[
y(4) = ut – \tfrac{1}{2} g t^2 = 29.4(4) – 4.9(16) = 39.2 \, \text{m}
\]
Q21. The volume of 50 g of a substance is 20 cm³. Will it float or sink in water?
\[
\rho = \frac{m}{V} = \frac{50}{20} = 2.5 \, \text{g/cm}^3
\]
Since \(2.5 > 1\), object is denser than water ⇒ **sinks**.
Q22. The volume of a 500 g packet is 350 cm³. Will it float or sink? What is the mass of water displaced?
\[
\rho = \frac{500}{350} \approx 1.43 \, \text{g/cm}^3 > 1
\]
⇒ Packet **sinks**.
Displaced water mass:
\[
m = \rho_{\text{water}} \times V = 1 \times 350 = 350 \, \text{g}
\]
Class 9 Chapter 9 Physics Gravitation Extra Challenging Questions
Q1. Two masses \(M_1=800\,\text{kg}\) and \(M_2=1200\,\text{kg}\) are \(d=0.50\,\text{m}\) apart.
(a) Find the gravitational force.
(b) If both masses are doubled and the distance is tripled, what is the new force and the ratio \(F’/F\)?
{Approach:} Use Newton’s law of gravitation
\[
F=\frac{GM_1M_2}{d^2}.
\]
{Step 1: Calculate original force.}
\[
F = \frac{(6.7\times10^{-11})(800)(1200)}{(0.50)^2} = 2.57\times10^{-4}\,\text{N}.
\]
{Step 2: Apply changes.}
\[
F’ = \frac{G(2M_1)(2M_2)}{(3d)^2} = \frac{4}{9}\,\frac{GM_1M_2}{d^2} = \frac{4}{9}F.
\]
{Step 3: Calculate new force.}
\[
F’ = \frac{4}{9}\times 2.57\times10^{-4} = 1.14\times10^{-4}\,\text{N}.
\]
{Final Answer:}
\[
F = 2.57\times10^{-4}\,\text{N},\quad F’ = 1.14\times10^{-4}\,\text{N},\quad \frac{F’}{F} = \frac{4}{9}.
\]
Q2. At what altitude \(h\) above Earth’s surface will a body weigh \(10\%\) less than at the surface? (Take \(R=6.4\times10^{6}\,\text{m}\)).
{Approach:} At altitude, \(g’ = g\left(\frac{R}{R+h}\right)^2\). Set \(g’/g=0.9\).
{Step 1: Set ratio.}
\[
\left(\frac{R}{R+h}\right)^2 = 0.9.
\]
{Step 2: Simplify.}
\[
R+h = \frac{R}{\sqrt{0.9}}.
\]
{Step 3: Solve for \(h\).}
\[
h = R\left(\frac{1}{\sqrt{0.9}}-1\right) = 6.4\times10^6(1.05409-1) = 3.46\times10^5\,\text{m}.
\]
{Final Answer:}
\[
h = 3.46\times10^{5}\,\text{m} \ (346\,\text{km}).
\]
Q3. A ball is projected upward at \(u=30\,\text{m s}^{-1}\) from the edge of a \(45\,\text{m}\) high cliff. Find (i) time to hit the ground, (ii) impact speed, (iii) total path length.
{Approach:} Use vertical motion equation \(y=45+ut-\tfrac12gt^2\). Solve \(y=0\).
{Step 1: Solve for time.}
\[
0 = 45+30t-4.9t^2 \Rightarrow t = \frac{30+\sqrt{900+882}}{9.8} = 7.37\,\text{s}.
\]
{Step 2: Final speed.}
\[
v = u-gt = 30-9.8(7.37) = -42.2\,\text{m s}^{-1}.
\]
{Step 3: Max rise.}
\[
H = \frac{u^2}{2g} = \frac{900}{19.6} = 45.9\,\text{m}.
\]
{Step 4: Total distance.}
\[
\text{Upward} = 45.9, \quad \text{Downward} = 45+45.9=90.9,\quad \text{Total}=136.8\,\text{m}.
\]
{Final Answer:}
\[
t = 7.37\,\text{s},\quad v=42.2\,\text{m s}^{-1}\ \text{down},\quad \text{distance}=136.8\,\text{m}.
\]
Q4. Stone A is thrown upward at \(u=40\,\text{m s}^{-1}\). After \(1.5\,\text{s}\), stone B is dropped from \(60\,\text{m}\). When and where do they meet?
{Approach:} Write positions:
Stone A: \(y_A=40t-4.9t^2\).
Stone B: \(y_B=60-4.9(t-1.5)^2\).
{Step 1: Set equal.}
\[
40t-4.9t^2 = 60-4.9(t-1.5)^2.
\]
{Step 2: Solve.}
\[
t = 1.94\,\text{s}.
\]
{Step 3: Height.}
\[
y=40(1.94)-4.9(1.94)^2 = 59.1\,\text{m}.
\]
{Final Answer:}
\[
t=1.94\,\text{s},\quad y=59.1\,\text{m above ground}.
\]
Q5. A \(2.5\,\text{kg}\) copper block (\(\rho=8900\,\text{kg m}^{-3}\)) is fully immersed in water. Find the apparent weight.
{Approach:} Apparent weight \(=mg-F_b\), with \(F_b=\rho_w g V\).
{Step 1: Volume.}
\[
V=\frac{m}{\rho}=\frac{2.5}{8900}=2.81\times10^{-4}\,\text{m}^3.
\]
{Step 2: Buoyant force.}
\[
F_b=1000\times9.8\times2.81\times10^{-4}=2.75\,\text{N}.
\]
{Step 3: Apparent weight.}
\[
W_{\text{app}}=mg-F_b=24.5-2.75=21.8\,\text{N}.
\]
{Final Answer:}
\[
W_{\text{app}}=21.8\,\text{N}.
\]
Q6. A \(3.0\,\text{kg}\) wood block (\(\rho=600\)) floats in water. (a) What fraction is submerged? (b) How much mass must be added to just submerge it?
{Step 1: Volume.}
\[
V=\frac{3}{600}=5.0\times10^{-3}\,\text{m}^3.
\]
{Step 2: Submerged fraction.}
\[
\frac{V_{\text{sub}}}{V}=\frac{\rho}{\rho_w}=\frac{600}{1000}=0.6.
\]
{Step 3: Added mass.}
\[
m_{\text{add}}=(\rho_w-\rho)V=(1000-600)\times5.0\times10^{-3}=2.0\,\text{kg}.
\]
{Final Answer:}
\[
60\%\ \text{submerged},\quad m_{\text{add}}=2.0\,\text{kg}.
\]
Q7. A block \(30\times20\times10\) cm, mass \(6\,\text{kg}\), rests on table in 3 ways. Find pressures.
{Step 1: Weight.}
\[
W=6\times9.8=58.8\,\text{N}.
\]
{Step 2: Areas and pressures.}
\[
A_1=0.30\times0.20=0.060,\quad P_1=58.8/0.060=980\,\text{Pa}.
\]
\[
A_2=0.30\times0.10=0.030,\quad P_2=58.8/0.030=1960\,\text{Pa}.
\]
\[
A_3=0.20\times0.10=0.020,\quad P_3=58.8/0.020=2940\,\text{Pa}.
\]
{Final Answer:}
\[
P_{\max}=2940\,\text{Pa}.
\]
Q8. A \(60\,\text{kg}\) person stands (a) on two heels of area \(1.5\,\text{cm}^2\) each, (b) on flats of area \(150\,\text{cm}^2\) each. Find pressures.
{Step 1: Weight.}
\[
W=60\times9.8=588\,\text{N}.
\]
{Step 2: Heels.}
\[
A=3\times10^{-4}\,\text{m}^2,\quad P=588/3\times10^{-4}=1.96\times10^6\,\text{Pa}.
\]
{Step 3: Flats.}
\[
A=3.0\times10^{-2}\,\text{m}^2,\quad P=588/3.0\times10^{-2}=1.96\times10^4\,\text{Pa}.
\]
{Final Answer:}
\[
P_{\text{heels}}=1.96\times10^6,\quad P_{\text{flats}}=1.96\times10^4.
\]
Q9. An asteroid has mass \(4.0\times10^{20}\) kg, radius \(5.0\times10^{5}\) m. Find its gravity and the weight of 2 kg.
{Step 1: Gravity.}
\[
g=\frac{GM}{R^2}=\frac{6.7\times10^{-11}\times4.0\times10^{20}}{(5.0\times10^5)^2}=0.107.
\]
{Step 2: Weight.}
\[
W=2.0\times0.107=0.214\,\text{N}.
\]
{Final Answer:}
\[
g=0.107,\quad W=0.214\,\text{N}.
\]
Q10. A body weighs 5.0 N in air and 3.8 N in water. Find (a) volume, (b) density.
{Step 1: Buoyant force.}
\[
F_b=5.0-3.8=1.2\,\text{N}.
\]
{Step 2: Volume.}
\[
V=\frac{F_b}{\rho_w g}=\frac{1.2}{1000\times9.8}=1.22\times10^{-4}\,\text{m}^3.
\]
{Step 3: Mass and density.}
\[
m=5.0/9.8=0.510,\quad \rho=0.510/1.22\times10^{-4}=4.17\times10^3.
\]
{Final Answer:}
\[
V=1.22\times10^{-4},\quad \rho=4.17\times10^3.
\]
Q11. A hollow sphere (volume \(1.20\times10^{-3}\)) of mass 0.50 kg floats in oil \(\rho=800\). Find submerged volume and fraction.
{Step 1: Submerged volume.}
\[
V_{\text{sub}}=m/\rho=0.50/800=6.25\times10^{-4}.
\]
{Step 2: Fraction.}
\[
\frac{V_{\text{sub}}}{V}=6.25\times10^{-4}/1.20\times10^{-3}=0.521.
\]
{Final Answer:}
\[
V_{\text{sub}}=6.25\times10^{-4},\quad 52.1\%\ \text{submerged}.
\]
Q12. A 1.8 kg body (volume \(1.5\times10^{-3}\)) is immersed in a liquid. Balance reads 12 N. Find density of liquid.
{Step 1: True weight.}
\[
W=1.8\times9.8=17.64.
\]
{Step 2: Buoyant force.}
\[
F_b=W-12=5.64.
\]
{Step 3: Density.}
\[
\rho=\frac{F_b}{gV}=\frac{5.64}{9.8\times1.5\times10^{-3}}=384.
\]
{Final Answer:}
\[
\rho=3.84\times10^2.
\]
