SURFACE AREAS AND VOLUMES — NCERT Solutions | Class 9 Mathematics | Academia Aeternum
Ch 11  ·  Q–
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Class 9 Mathematics Exercise-11.2 NCERT Solutions Olympiad Board Exam
Chapter 11

SURFACE AREAS AND VOLUMES

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

9 Questions
20–30 min Ideal time
Q1 Now at
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Q1
NUMERIC3 marks
Find the surface area of a sphere of radius:
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm
📘 Concept & Theory Concept Used ›

A sphere is a perfectly round three-dimensional solid where every point on its surface is at the same distance from the centre.

The total outer area covered by a sphere is called its surface area.

Formula for the surface area of a sphere:

\[\small \text{Surface Area of Sphere} = 4\pi r^2 \]

where,

  • \[\small r = \text{radius of the sphere} \]
  • \[\small \pi = \frac{22}{7} \text{ (used when radius is multiple of 7)} \]
🗺️ Solution Roadmap Step-by-step Plan ›

  1. Write the formula for surface area of a sphere.

  2. Substitute the given radius value.

  3. Square the radius carefully.

  4. Multiply with \(\small 4\pi\).

  5. Write the final answer with correct unit \(\small \text{cm}^2\).

📊 Graph / Figure Graph / Figure ›
r Surface Area = 4πr²
Surface Area of a Sphere
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  9 steps
  1. Surface area of a sphere is given by:\[\small SA = 4\pi r^2\]
  2. (i) - Given: Radius \(\small = 10.5\text{ cm}\)
  3. Substituting the value of radius in the formula: \[\small \begin{aligned} SA &= 4\pi r^2 \\ &= 4 \times \frac{22}{7} \times (10.5)^2\\ &= 4 \times \frac{22}{7} \times 110.25\\ &= 4 \times 22 \times \frac{110.25}{7} \\ &= 4 \times 22 \times 15.75 \\ &= 88 \times 15.75 \\ &= 1386 \end{aligned}\]
  4. Hence,\[\small \boxed{SA = 1386\ \text{cm}^2}\]
  5. (ii) - Given: Radius \(\small = 5.6\text{ cm}\)
  6. Using the formula:\[\small SA = 4\pi r^2 \]
  7. Substituting the value of radius in the formula \[\small \begin{aligned} SA &= 4\pi r^2 \\ &= 4 \times \frac{22}{7} \times (5.6)^2\\ &= 4 \times \frac{22}{7} \times 31.36 \\ &= 4 \times 22 \times \frac{31.36}{7} \\ &= 4 \times 22 \times 4.48 \\ &= 88 \times 4.48 \\ &= 394.24 \end{aligned} \]
  8. Hence,\[\small \boxed{SA = 394.24\ \text{cm}^2}\]
  9. (iii) - Given Radius \(\small = 14\text{ cm}\)
  10. Using the formula:\[\small SA = 4\pi r^2\]
  11. Substituting the value of radius in the formula \[\small \begin{aligned} SA &= 4\pi r^2 \\ &= 4 \times \frac{22}{7} \times (14)^2\\ &= 4 \times \frac{22}{7} \times 196 \\ &= 4 \times 22 \times \frac{196}{7} \\ &= 4 \times 22 \times 28 \\ &= 88 \times 28 \\ &= 2464 \end{aligned} \]
  12. Hence,\[\small \boxed{SA = 2464\ \text{cm}^2}\]
💡 Answer Final Answer ›
Answers:
Surface Areas: \(\small \text{(i)}\ 1386\ \text{cm}^2\) \(\small \text{(ii)}\ 394.24\ \text{cm}^2\) \(\small \text{(iii)}\ 2464\ \text{cm}^2\)
🎯 Exam Significance Exam Significance ›

Questions based on surface area of spheres are very important for CBSE board examinations. Students are frequently asked to apply the formula directly or solve application-based problems involving hemispheres, balls, tanks, globes, and solid combinations.

This concept is also useful for competitive entrance examinations such as NTSE, Olympiads, Polytechnic entrance exams, and foundation-level preparation for JEE and NEET, where formula application and unit handling are tested.

Learning this problem carefully improves:

  • Formula application skills
  • Accuracy in decimal calculations
  • Understanding of three-dimensional geometry
  • Speed and confidence in mensuration problems
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  5 points

  1. Surface area of a sphere is given by: \[\small SA = 4\pi r^2 \]

  2. Radius must always be squared before multiplication.

  3. Final answers of surface area are written in square units.

  4. Use \[\small \pi = \frac{22}{7} \] whenever convenient for simplification.

  5. Careful step-by-step calculation reduces mistakes in board examinations.
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1 / 9  ·  11%
Q2 →
Q2
NUMERIC3 marks
Find the surface area of a sphere of diameter:
(i) 14 cm
(ii) 21 cm
(iii) 3.5 m
📘 Concept & Theory Concept Used ›

The surface area of a sphere is the total curved area covering its outer surface.

Formula for surface area of a sphere:

\[\small SA = 4\pi r^2 \]

Here,

  • \[\small SA = \text{Surface Area} \]
  • \[\small r = \text{radius of the sphere} \]
  • \[\small \pi = \frac{22}{7} \]

Since the diameter is given in the question, we first find the radius using:

\[\small r = \frac{d}{2} \]

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Write the given diameter.

  2. Convert diameter into radius using \[\small r = \frac{d}{2} \]

  3. Use the formula \[\small SA = 4\pi r^2 \]

  4. Perform calculations carefully step-by-step.

  5. Write the final answer with correct square units.

📊 Graph / Figure Graph / Figure ›
d r Surface Area = 4πr²
Fig. 1 — Free body diagram
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  9 steps
  1. (i) Given: Diameter \(\small = 14\text{ cm}\)
  2. Radius of the sphere: \[\small \begin{aligned} r &= \frac{d}{2} \\ &= \frac{14}{2} \\ &= 7\text{ cm} \end{aligned} \]
  3. Surface area of sphere: \[\small \require{cancel} \begin{aligned} SA &= 4\pi r^2 \\ &= 4 \times \frac{22}{\cancel{7}} \times 7 \times \cancel{7}\\ &=4 \times 22\times 7\\ &=88\times7\\ &=616 \end{aligned} \]
  4. Hence,\[\small \boxed{SA = 616\ \text{cm}^2}\]
  5. (ii) Given: Diameter \(\small = 21\text{ cm}\)
  6. Radius of the sphere: \[\small \begin{aligned} r &= \frac{d}{2} \\ &= \frac{21}{2} \\ &= 10.5\text{ cm} \end{aligned} \]
  7. Surface area of sphere: \[\small \begin{aligned} SA &= 4\pi r^2 \\ &= 4 \times \frac{22}{7} \times (10.5)^2\\ &= 4 \times \frac{22}{7} \times 110.25 \\ &= 4 \times 22 \times \frac{110.25}{7} \\ &= 4 \times 22 \times 15.75 \\ &= 88 \times 15.75 \\ &= 1386 \end{aligned} \]
  8. Hence,\[\small \boxed{SA = 1386\ \text{cm}^2}\]
  9. (iii) Given: Diameter \(\small = 3.5\text{ m}\)
  10. Radius of the sphere: \[\small \begin{aligned} r &= \frac{d}{2} \\ &= \frac{3.5}{2} \\ &= 1.75\text{ m} \end{aligned} \]
  11. Surface area of sphere: \[\small \begin{aligned} SA &= 4\pi r^2 \\ &= 4 \times \frac{22}{7} \times (1.75)^2\\ &= 4 \times \frac{22}{7} \times 3.0625 \\ &= 4 \times 22 \times \frac{3.0625}{7} \\ &= 4 \times 22 \times 0.4375 \\ &= 88 \times 0.4375 \\ &= 38.5 \end{aligned} \]
  12. Hence,\[\small \boxed{SA = 38.5\ \text{m}^2}\]
💡 Answer Final Answer ›
Answers: \(\small \text{(i)}\ 616\ \text{cm}^2\), \(\small \text{(ii)}\ 1386\ \text{cm}^2\), \(\small \text{(iii)}\ 38.5\ \text{m}^2\)
🎯 Exam Significance Exam Significance ›

This problem is important because students must understand the relationship between diameter and radius before applying mensuration formulas.

In CBSE board examinations, direct formula-based questions and application-based questions from spheres are frequently asked. Such questions test calculation accuracy, unit handling, and conceptual clarity.

Similar concepts are also used in NTSE, Olympiads, Polytechnic entrance exams, and foundation-level preparation for engineering and medical entrance examinations.

Practising such questions improves:

  • Understanding of 3-dimensional geometry
  • Formula application speed
  • Decimal computation accuracy
  • Confidence in mensuration problems
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  5 points

  1. Surface area of a sphere is: \[\small SA = 4\pi r^2 \]

  2. If diameter is given, first convert it into radius using: \[\small r = \frac{d}{2} \]

  3. Radius must always be squared before multiplication.

  4. Surface area is always written in square units.

  5. Step-by-step simplification helps avoid calculation mistakes.
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2 / 9  ·  22%
Q3 →
Q3
NUMERIC3 marks
Find the total surface area of a hemisphere of radius 10 cm. (Use \(\small \pi = 3.14\))
📘 Concept & Theory Concept Used ›

A hemisphere is exactly half of a sphere obtained by cutting a sphere through its centre.

The total surface area of a hemisphere includes:

  • Curved surface area of the hemisphere
  • Area of the circular base

Curved surface area of a hemisphere:

\[\small 2\pi r^2 \]

Area of circular base:

\[\small \pi r^2 \]

Therefore,

\[\small \begin{aligned} \text{Total Surface Area} &= 2\pi r^2 + \pi r^2 \\ &= 3\pi r^2 \end{aligned} \]

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Identify the radius of the hemisphere.

  2. Write the formula for total surface area: \[\small TSA = 3\pi r^2 \]

  3. Substitute the given values carefully.

  4. Perform multiplication step-by-step.

  5. Write the final answer in square units.

📊 Graph / Figure Graph / Figure ›
r TSA = 3πr²
Total Surface Area of a Hemisphere
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  4 steps
  2. Formula for total surface area of a hemisphere:\[\small TSA = 3\pi r^2\]
  3. Substituting the values: \[\small \begin{aligned} TSA &= 3 \times 3.14 \times (10)^2\\ &= 3 \times 3.14 \times 100 \\ &= 3 \times 314 \\ &= 942 \end{aligned} \]
  4. Hence,\[\small \boxed{TSA = 942\ \text{cm}^2}\]
💡 Answer Final Answer ›
Answer: \(\small 942\ \text{cm}^2\)
🎯 Exam Significance Exam Significance ›

Questions based on hemispheres are frequently asked in CBSE board examinations. Students are often required to distinguish between curved surface area and total surface area.

This question is conceptually important because many students forget to include the circular base while calculating total surface area.

Similar concepts are also useful in NTSE, Olympiads, Polytechnic entrance examinations, and foundation-level preparation for JEE and other competitive exams.

Practising such problems improves:

  • Understanding of solid geometry
  • Formula selection accuracy
  • Speed in mensuration calculations
  • Confidence in 3-dimensional geometry problems
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  5 points

  1. Total surface area of a hemisphere is: \[\small TSA = 3\pi r^2 \]

  2. Curved surface area and total surface area are different quantities.

  3. Total surface area includes the circular base.

  4. Surface area is always written in square units.

  5. Step-by-step calculations help reduce mistakes in board examinations.
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3 / 9  ·  33%
Q4 →
Q4
NUMERIC3 marks
The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
📘 Concept & Theory Concept Used ›

A balloon in spherical shape follows the surface area formula of a sphere.

Surface area of a sphere is:

\[\small SA = 4\pi r^2 \]

This formula shows that the surface area depends on the square of the radius.

Therefore,

\[\small SA \propto r^2 \]

So, when the radius changes, the ratio of surface areas can be found by squaring the ratio of radii.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Identify the initial and final radii.

  2. Write the formula for surface area of a sphere.

  3. Form the ratio of the two surface areas.

  4. Cancel common terms carefully.

  5. Simplify the ratio completely.

📊 Graph / Figure Graph / Figure ›
7 cm 14 cm Surface Area ∝ r²
Increase in Surface Area with Radius
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  10 steps
  1. Given

  2. Initial radius of the balloon:

    \[\small r_1 = 7\text{ cm} \]

    Final radius of the balloon:

    \[\small r_2 = 14\text{ cm} \]

  3. Step-by-step Solution
  4. Surface area of a sphere is given by:\[\small SA = 4\pi r^2\]
  5. Let, \[\small SA_1 = \text{Initial Surface Area}\] and \[\small SA_2 = \text{Final Surface Area}\]
  6. Then, \[\small \begin{aligned} SA_1 &= 4\pi r_1^2 \end{aligned} \] and \[\small \begin{aligned} SA_2 &= 4\pi r_2^2 \end{aligned} \]
  7. Ratio of the two surface areas: \[\small \begin{aligned} \frac{SA_1}{SA_2} &= \frac{4\pi r_1^2}{4\pi r_2^2} \end{aligned} \]
  8. Cancelling common terms: \[\small \begin{aligned} \frac{SA_1}{SA_2} &= \frac{r_1^2}{r_2^2} \end{aligned} \]
  9. Substituting the values: \[\small \begin{aligned} \frac{SA_1}{SA_2} &= \left(\frac{7}{14}\right)^2 \end{aligned} \]

  10. Simplifying: \[\small \begin{aligned} \frac{7}{14} &= \frac{1}{2} \end{aligned} \]
  11. Therefore, \[\small \begin{aligned} \frac{SA_1}{SA_2} &= \left(\frac{1}{2}\right)^2 \\ &= \frac{1}{4} \end{aligned} \]
  12. Hence, the required ratio is:\[\small 1 : 4\]
💡 Answer Final Answer ›
Answer: The required ratio is:\(\small 1 : 4\)
🎯 Exam Significance Exam Significance ›

Ratio-based mensuration problems are frequently asked in CBSE board examinations. Such questions test conceptual understanding rather than direct substitution in formulas.

This problem is especially important because it demonstrates how surface area changes with the square of the radius.

Similar concepts are used in NTSE, Olympiads, Scholarship examinations, and foundation-level preparation for engineering entrance examinations.

Practising such problems improves:

  • Understanding of proportional relationships
  • Algebraic simplification skills
  • Speed in solving ratio-based questions
  • Conceptual clarity in surface area formulas
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  5 points

  1. Surface area of a sphere is: \[\small SA = 4\pi r^2 \]

  2. Surface area is proportional to the square of the radius: \[\small SA \propto r^2 \]

  3. Common terms cancel while finding ratios.

  4. Radius doubling causes surface area to become four times.

  5. Ratio questions are often solved faster using proportional reasoning.
← Q3
4 / 9  ·  44%
Q5 →
Q5
NUMERIC3 marks
A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹16 per 100 \(\small \text{cm}^2\).
📘 Concept & Theory Concpet Used ›

A hemispherical bowl is shaped like half of a sphere.

Since tin-plating is done only on the inside surface of the bowl, we calculate only the curved surface area of the hemisphere.

Curved surface area of a hemisphere:

\[\small CSA = 2\pi r^2 \]

where,

  • \[\small r = \text{radius of the hemisphere} \]
  • \[\small \pi = \frac{22}{7} \]

After finding the area, we multiply it by the given rate to calculate the total cost.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Find the radius from the given diameter.

  2. Use the curved surface area formula of a hemisphere.

  3. Calculate the inner surface area carefully.

  4. Use the given rate of tin-plating.

  5. Compute the final cost with correct currency unit.

📊 Graph / Figure Graph / Figure ›
Diameter = 10.5 cm r= 5.25 cm CSA of Hemisphere = 2πr²
Tin-plating the Inner Surface of a Hemisphere
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. Given:

    Inner diameter of the hemispherical bowl:

    \[\small d = 10.5\text{ cm} \]

    Radius of the bowl:

    \[\small \begin{aligned} r &= \frac{d}{2} \\ &= \frac{10.5}{2} \\ &= 5.25\text{ cm} \end{aligned} \]

  2. Step-by-step Solution
  3. Since tin-plating is required only on the inside, we use the curved surface area of hemisphere:\[\small CSA = 2\pi r^2 \]
  4. Substituting the values: \[\small \begin{aligned} CSA &= 2 \times \frac{22}{7} \times (5.25)^2\\ &= 2 \times \frac{22}{7} \times 27.5625 \\ &= 2 \times 22 \times \frac{27.5625}{7} \\ &= 2 \times 22 \times 3.9375 \\ &= 44 \times 3.9375 \\ &= 173.25 \end{aligned} \]
  5. Hence, the inner surface area of the bowl is:\(\small 173.25\ \text{cm}^2\)
  6. Rate of tin-plating:\[\small \text{₹ }16 \text{ per } 100\ \text{cm}^2\]
  7. Cost of tin-plating: \[\small \begin{aligned} \text{Cost} &= \frac{16}{100} \times 173.25\\ &= 0.16 \times 173.25 \\ &= 27.72 \end{aligned} \]
💡 Answer Final Answer ›
Answer: Cost of Tinning is ₹27.72
🎯 Exam Significance Exam Significance ›

Application-based mensuration questions are frequently asked in CBSE board examinations. This problem is important because students must identify which surface area formula should be used according to the situation.

In this question, only the inside curved surface is plated, so curved surface area is used instead of total surface area.

Similar commercial mathematics and mensuration concepts are used in NTSE, Olympiads, Scholarship examinations, and foundation-level engineering entrance preparation.

Practising such problems improves:

  • Understanding of practical geometry applications
  • Ability to choose correct formulas
  • Unit conversion and cost calculation skills
  • Accuracy in decimal computations
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  5 points

  1. Curved surface area of a hemisphere is: \[\small CSA = 2\pi r^2 \]

  2. Radius is half of the diameter: \[\small r = \frac{d}{2} \]

  3. For inner plating, only the curved surface is considered.

  4. Commercial mensuration problems involve area and rate calculations together.

  5. Step-by-step simplification helps avoid mistakes in calculations.
← Q4
5 / 9  ·  56%
Q6 →
Q6
NUMERIC3 marks
Find the radius of a sphere whose surface area is \(\small 154\ \text{cm}^2\).
📘 Concept & Theory concept Used ›

The surface area of a sphere depends on the square of its radius.

Formula for surface area of a sphere:

\[\small SA = 4\pi r^2 \]

where,

  • \[\small SA = \text{Surface Area of Sphere} \]
  • \[\small r = \text{radius of the sphere} \]
  • \[\small \pi = \frac{22}{7} \]

When the surface area is given, we rearrange the formula to find the radius.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Write the given surface area.

  2. Step 2: Use the formula \[\small SA = 4\pi r^2 \]

  3. Substitute the given value of surface area.

  4. Solve for \(\small r^2\).

  5. Take the square root to find the radius.

📊 Graph / Figure Graph / Figure ›
r SA = 154 cm² SA = 4πr²
Finding Radius from Surface Area
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. Given:
  2. SA = \(\small 154\ \text{cm}^2\)
  3. Step-by-step Solution
  4. Let the radius of the sphere be:\[\small r\text{ cm}\]
  5. Formula for surface area of a sphere:\[\small SA = 4\pi r^2\]
  6. Substituting the given value: \[\small \begin{aligned} 154 &= 4\pi r^2\\ &= 4 \times \frac{22}{7} \times r^2\\ \Rightarrow r^2&= \frac{154}{4 \times \frac{22}{7}}\\ &= \frac{154 \times 7}{4 \times 22}\\ &= \frac{1078}{88} \\ &= 12.25\\ \Rightarrow r &= \sqrt{12.25} \\ &= 3.5 \end{aligned} \]
  7. Hence, the radius of the sphere is:\[\small \boxed{3.5\ \text{cm}}\]
💡 Answer Final Answer ›
Answer: Radius of the sphere is:\(\small \boxed{3.5\ \text{cm}}\)
🎯 Exam Significance Exam Significance ›

Reverse mensuration problems, where students are required to find dimensions from area or volume, are very important for CBSE board examinations.

This question tests algebraic manipulation, formula application, and simplification skills together.

Similar concepts are frequently used in NTSE, Olympiads, Scholarship examinations, and foundation-level engineering entrance preparation.

Practising such problems improves:

  • Algebraic equation-solving ability
  • Formula rearrangement skills
  • Numerical simplification accuracy
  • Confidence in mensuration problems
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  5 points

  1. Surface area of a sphere is: \[\small SA = 4\pi r^2 \]

  2. When surface area is given, rearrange the formula to find radius.

  3. Square root is taken after finding \(\small r^2\).

  4. Step-by-step simplification reduces calculation mistakes.

  5. Reverse-formula problems are commonly asked in examinations.
← Q5
6 / 9  ·  67%
Q7 →
Q7
NUMERIC3 marks
The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
📘 Concept & Theory Concept Used ›

The surface area of a sphere depends on the square of its radius.

Formula for surface area of a sphere:

\[\small SA = 4\pi r^2 \]

Therefore,

\[\small SA \propto r^2 \]

Since the radius is half of the diameter, the ratio of diameters is the same as the ratio of radii.

If the radius ratio is known, then the surface area ratio is obtained by squaring the radius ratio.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Write the given ratio of diameters.

  2. Relate the ratio of diameters to the ratio of radii.

  3. Use the surface area formula of a sphere.

  4. Form the ratio of surface areas.

  5. Square the ratio of radii to get the final answer.

📊 Graph / Figure Graph / Figure ›
Moon Earth Diameter Ratio = 1 : 4 Surface Area Ratio = 1 : 16
Comparison of Surface Areas of Moon and Earth
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  11 steps
  1. Let,\[\small r_m = \text{radius of the moon}\] and \[\small r_e = \text{radius of the earth}\]
  2. Given that the diameter of the moon is one fourth of the diameter of the earth.
    Therefore, the ratio of their radii is also: \[\small \begin{aligned} r_m &= \frac{1}{4}r_e \end{aligned} \]
  3. Surface area of a sphere is:\[\small SA = 4\pi r^2\]
  4. Let,\[\small SA_m = \text{surface area of the moon}\] and SA_e = \text{surface area of the earth}
  5. Then, \[\small \begin{aligned} SA_m &= 4\pi r_m^2 \end{aligned} \] and \[\small \begin{aligned} SA_e &= 4\pi r_e^2 \end{aligned} \]
  6. Ratio of their surface areas: \[\small \begin{aligned} \frac{SA_m}{SA_e} &= \frac{4\pi r_m^2}{4\pi r_e^2} \end{aligned} \]
  7. Cancelling common terms: \[\small \begin{aligned} \frac{SA_m}{SA_e} &= \frac{r_m^2}{r_e^2} \end{aligned} \]
  8. Substituting \[\small r_m = \frac{1}{4}r_e \]
  9. \[\small \begin{aligned} \frac{SA_m}{SA_e} &= \left(\frac{r_m}{r_e}\right)^2 \\ &= \left(\frac{1}{4}\right)^2 \end{aligned} \]
  10. Squaring the fraction: \[\small \begin{aligned} \left(\frac{1}{4}\right)^2 &= \frac{1^2}{4^2} \\ &= \frac{1}{16} \end{aligned} \]
  11. Therefore, the ratio of the surface areas of the moon and the earth is: \boxed{1 : 16}
🎯 Exam Significance Exam Significance ›

Ratio and proportionality problems involving surface areas are frequently asked in CBSE board examinations.

This question is important because it develops conceptual understanding that surface area changes according to the square of dimensions.

Such concepts are also widely used in NTSE, Olympiads, Scholarship examinations, and foundation-level competitive examinations.

Practising such problems improves:

  • Understanding of proportional relationships
  • Ratio simplification skills
  • Conceptual clarity in mensuration
  • Speed in solving geometry-based reasoning questions
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  5 points

  1. Surface area of a sphere is: \[\small SA = 4\pi r^2 \]

  2. Surface area is proportional to the square of the radius: \[\small SA \propto r^2 \]

  3. Ratio of diameters and ratio of radii are the same.

  4. Squaring the ratio of dimensions gives the ratio of surface areas.

  5. Ratio-based mensuration problems are important for board and competitive exams.
← Q6
7 / 9  ·  78%
Q8 →
Q8
NUMERIC3 marks
A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
📘 Concept & Theory Concept Used ›

A hemispherical bowl has two radii:

  • Inner radius
  • Outer radius

Since the bowl has thickness, the outer radius becomes greater than the inner radius.

Relationship:

\[\small \text{Outer Radius} = \text{Inner Radius} + \text{Thickness} \]

The curved surface area of a hemisphere is:

\[\small CSA = 2\pi r^2 \]

Since we need the outer curved surface area, we use the outer radius in the formula.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Write the inner radius and thickness.

  2. Find the outer radius.

  3. Use the curved surface area formula of hemisphere.

  4. Substitute the outer radius carefully.

  5. Simplify step-by-step and write the answer in square units.

📊 Graph / Figure Graph / Figure ›
3D Hemisphere Diagram A geometric diagram of a hemisphere showing inner and outer radii, thickness, and the formula for Curved Surface Area. Outer Radius (R) = 5.25 cm Inner Radius (r) = 5 cm Thickness = R - r = 0.25 cm CSA of Hemisphere = 2πr² (Where r is the inner radius for internal CSA)
Outer Curved Surface of a Hemispherical Bowl
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  8 steps
  1. Inner radius of the hemispherical bowl:\[\small r = 5\text{ cm}\]
  2. Thickness of steel:\[\small 0.25\text{ cm}\]
  3. Therefore, outer radius of the bowl: \[\small \begin{aligned} R &= 5 + 0.25 \\ &= 5.25\text{ cm} \end{aligned} \]
  4. Formula for curved surface area of hemisphere:\[\small CSA = 2\pi R^2\]
  5. Substituting the value of outer radius: \[\small \begin{aligned} CSA &= 2 \times \frac{22}{7} \times (5.25)^2 \end{aligned} \]
  6. Squaring the radius: \[\small \begin{aligned} (5.25)^2 &= 5.25 \times 5.25 \\ &= 27.5625 \end{aligned} \]
  7. Therefore, \[\small \begin{aligned} CSA &= 2 \times \frac{22}{7} \times 27.5625 \\ &= 2 \times 22 \times \frac{27.5625}{7} \\ &= 2 \times 22 \times 3.9375 \\ &= 44 \times 3.9375 \\ &= 173.25 \end{aligned} \]
  8. Hence, the outer curved surface area of the bowl is:\[\small \boxed{173.25\ \text{cm}^2}\]
🎯 Exam Significance Exam Significance ›
Answer: The outer curved surface area of the bowl is: \(\small \boxed{173.25\ \text{cm}^2}\)
🎯 Exam Significance Exam Significance ›

Questions involving thickness are very important in CBSE board examinations because they test conceptual understanding of inner and outer dimensions.

Students often make mistakes by directly using the inner radius instead of calculating the outer radius first.

Similar geometry and mensuration concepts are widely used in NTSE, Olympiads, Polytechnic entrance exams, and foundation-level engineering entrance preparation.

Practising such problems improves:

  • Understanding of practical mensuration
  • Ability to interpret thickness-based geometry
  • Formula application accuracy
  • Confidence in surface area calculations
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  5 points

  1. Curved surface area of a hemisphere is: \[\small CSA = 2\pi r^2 \]

  2. Outer radius is obtained by adding thickness to inner radius.

  3. Always identify whether inner or outer surface area is required.

  4. Thickness-based questions are common in examinations.

  5. Step-by-step calculations reduce mistakes in decimal simplification.
← Q7
8 / 9  ·  89%
Q9 →
Q9
NUMERIC3 marks
A right circular cylinder just encloses a sphere of radius \(\small r\) (see Fig. 11.10). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).
📘 Concept & Theory Concept Used ›

A sphere is said to be just enclosed by a cylinder when:

  • The radius of the cylinder equals the radius of the sphere.
  • The height of the cylinder equals the diameter of the sphere.

Therefore,

\[\small h = 2r \]

Formula for surface area of a sphere:

\[\small SA = 4\pi r^2 \]

Formula for curved surface area of a cylinder:

\[\small CSA = 2\pi rh \]

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Write the formula for surface area of a sphere.

  2. Identify the dimensions of the enclosing cylinder.

  3. Calculate the curved surface area of the cylinder.

  4. Compare both areas.

  5. Simplify the ratio completely.

📊 Graph / Figure Graph / Figure ›
r h = 2r Archimedes' Cylinder & Sphere CSA of Cylinder = 2πr(2r) = 4πr²
Sphere Just Enclosed by a Cylinder
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  13 steps
  1. Let the radius of the sphere be:\[\small r\]
  2. (i) Surface Area of the Sphere
  3. Formula for surface area of a sphere:\[\small SA_{\text{sphere}} = 4\pi r^2\]
  4. Therefore,\[\small \boxed{SA_{\text{sphere}} = 4\pi r^2}\]
  5. Curved Surface Area of the Cylinder
  6. Since the cylinder just encloses the sphere:
  7. Radius of cylinder:\[\small r\]
  8. Height of cylinder:\[\small h = 2r\]
  9. Formula for curved surface area of cylinder:\[\small CSA_{\text{cylinder}} = 2\pi rh\]
  10. Substituting \(\small h = 2r\): \[\small \begin{aligned} CSA_{\text{cylinder}} &= 2\pi r \times 2r \end{aligned} \]
  11. Multiplying:\[\small \begin{aligned} CSA_{\text{cylinder}} &= 4\pi r^2 \end{aligned} \]
  12. Therefore,\[\small CSA_{\text{cylinder}} = 4\pi r^2\]
  13. (iii) Ratio of the Areas
  14. Ratio of surface area of sphere to curved surface area of cylinder: \[\small \begin{aligned} \frac{SA_{\text{sphere}}}{CSA_{\text{cylinder}}} &= \frac{4\pi r^2}{4\pi r^2} \end{aligned} \]
  15. Cancelling common terms: \[\small \begin{aligned} \frac{SA_{\text{sphere}}}{CSA_{\text{cylinder}}} &= 1 \end{aligned} \]
  16. Therefore, the ratio is:\[\small \boxed{1 : 1}\]
💡 Answer Final Answer ›
Answer: \(\small \text{(i)}\ 4\pi r^2\), \(\small \text{(ii)}\ 4\pi r^2\), \(\small \text{(iii)}\ 1 : 1\)
🎯 Exam Significance Exam Significance ›

This is an important conceptual mensuration problem frequently asked in CBSE board examinations.

The question develops understanding of relationships between different solids, especially spheres and cylinders.

Such geometry relationships are also useful in NTSE, Olympiads, Scholarship examinations, and foundation-level competitive exams.

Practising such problems improves:

  • Spatial visualization skills
  • Formula application accuracy
  • Ability to connect dimensions of different solids
  • Conceptual understanding of mensuration
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  5 points

  1. Surface area of a sphere is: \[\small SA = 4\pi r^2 \]

  2. Curved surface area of a cylinder is: \[\small CSA = 2\pi rh \]

  3. For a cylinder enclosing a sphere: \[\small h = 2r \]

  4. The curved surface area of the cylinder equals the surface area of the sphere.

  5. Visual geometry relationships are important in competitive examinations.
← Q8
9 / 9  ·  100%
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Chapter Complete!

All 9 solutions for SURFACE AREAS AND VOLUMES covered.

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NCERT Class 9 Maths Exercise 11.2 Solutions
NCERT Class 9 Maths Exercise 11.2 Solutions — Complete Notes & Solutions · academia-aeternum.com
Explore expertly solved textbook exercises from NCERT Mathematics Class 9 Chapter 11, “Surface Areas and Volumes.” These clear, stepwise solutions help you master concepts of three-dimensional geometry, applying key formulas for cubes, cuboids, cones, cylinders, spheres, and hemispheres. Perfect for CBSE exam preparation, revision, and building strong problem-solving skills.
🎓 Class 9 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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