Chapter 12 — Surface Areas and Volumes

Concept Used:

• This is a combination of two solids: cone + hemisphere.
• Total volume = sum of individual volumes.
• Volume formulas:

\[ \text{Volume of cone} = \frac{1}{3}\pi r^2 h \] \[ \text{Volume of hemisphere} = \frac{2}{3}\pi r^3 \]

Since both parts have the same radius, substitution becomes direct.

Solution Roadmap:

1. Identify given dimensions.
2. Write formula for each solid.
3. Add volumes.
4. Substitute \(r = 1\) and simplify carefully.

Solution:

Radius of cone \(r = 1 \, \text{cm}\)
Height of cone \(h = 1 \, \text{cm}\)
Radius of hemisphere \(r = 1 \, \text{cm}\)

Total volume \(V = V_{\text{cone}} + V_{\text{hemisphere}}\)

\[ V = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 \]

Substituting \(r = 1\), \(h = 1\):

\[ V = \frac{1}{3}\pi (1)^2 (1) + \frac{2}{3}\pi (1)^3 \]

Simplifying each term:

\[ V = \frac{1}{3}\pi + \frac{2}{3}\pi \]

Taking common factor:

\[ V = \frac{\pi}{3} (1 + 2) \] \[ V = \frac{\pi}{3} \times 3 \] \[ V = \pi \]

Final Answer: \( \pi \, \text{cm}^3 \)

Exam Significance:

• Very common CBSE board pattern question (3–4 marks).
• Tests ability to identify composite solids.
• Frequently asked in NTSE, polytechnic, and foundation exams.
• Common mistake: forgetting hemisphere volume or using full sphere formula.

Concept Used:

• The model is a combination of three solids: one cylinder + two identical cones.
• Total volume = volume of cylinder + 2 × volume of cone.
• Formulas used:

\[ \text{Volume of cylinder} = \pi r^2 h \] \[ \text{Volume of cone} = \frac{1}{3}\pi r^2 h \]

Solution Roadmap:

1. Convert diameter into radius.
2. Compute volume of cylinder.
3. Compute volume of one cone.
4. Multiply cone volume by 2.
5. Add all volumes carefully.

Solution:

Diameter = 3 cm
Radius \(r = \dfrac{3}{2} = 1.5\) cm
Height of cylinder \(h_1 = 12\) cm
Height of each cone \(h_2 = 2\) cm

Total volume \(V = V_{\text{cylinder}} + 2 \times V_{\text{cone}}\)

\[ V = \pi r^2 h_1 + 2 \left( \frac{1}{3}\pi r^2 h_2 \right) \]

Substituting values:

\[ V = \pi (1.5)^2 (12) + 2 \left( \frac{1}{3}\pi (1.5)^2 (2) \right) \]

First simplify \( (1.5)^2 \):

\[ (1.5)^2 = 2.25 \]

Now substitute:

\[ V = \pi (2.25)(12) + 2 \left( \frac{1}{3}\pi (2.25)(2) \right) \]

Simplify cylinder part:

\[ \pi \times 2.25 \times 12 = 27\pi \]

Simplify cone part:

\[ 2 \left( \frac{1}{3} \times 2.25 \times 2 \right)\pi \] \[ = 2 \left( \frac{4.5}{3} \right)\pi \] \[ = 2 \times 1.5\pi \] \[ = 3\pi \]

Total volume:

\[ V = 27\pi + 3\pi \] \[ V = 30\pi \]

Using \( \pi = \frac{22}{7} \):

\[ V = 30 \times \frac{22}{7} \] \[ V = \frac{660}{7} \] \[ V \approx 94.29 \ \text{cm}^3 \]

Final Answer: \( \dfrac{660}{7} \ \text{cm}^3 \approx 94.29 \ \text{cm}^3 \)

Exam Significance:

• Classic CBSE 4-mark question from combination of solids.
• Tests whether student correctly handles multiple components.
• Very common in JEE Foundation, NTSE for conceptual clarity.
• Frequent mistake: incorrect factorisation like combining terms wrongly.

Concept Used:

• The shape is a capsule: cylinder + two hemispheres.
• Two hemispheres = one complete sphere.
• Total volume = volume of cylinder + volume of sphere.

\[ \text{Volume of sphere} = \frac{4}{3}\pi r^3 \] \[ \text{Volume of cylinder} = \pi r^2 h \]

Solution Roadmap:

1. Find radius from diameter.
2. Determine cylinder height (subtract hemispherical parts).
3. Compute volume of one gulab jamun.
4. Multiply by 45.
5. Take 30% of total volume.

Solution:

Number of gulab jamuns = 45
Diameter = 2.8 cm
Radius \(r = \dfrac{2.8}{2} = 1.4\) cm
Total length = 5 cm

Height of cylindrical part:

\[ h = 5 - 2r \] \[ h = 5 - 2(1.4) \] \[ h = 5 - 2.8 = 2.2 \text{ cm} \]

Volume of one gulab jamun:

\[ V = \text{Volume of cylinder} + \text{Volume of sphere} \] \[ V = \pi r^2 h + \frac{4}{3}\pi r^3 \]

Substituting values:

\[ V = \pi (1.4)^2 (2.2) + \frac{4}{3}\pi (1.4)^3 \]

Compute powers:

\[ (1.4)^2 = 1.96, \quad (1.4)^3 = 2.744 \] \[ V = \pi (1.96)(2.2) + \frac{4}{3}\pi (2.744) \]

Simplify:

\[ V = \pi (4.312) + \frac{4}{3}\pi (2.744) \] \[ V = 4.312\pi + \frac{10.976}{3}\pi \]

Convert to common form:

\[ V = 4.312\pi + 3.6587\pi \] \[ V = 7.9707\pi \]

Using \( \pi = \frac{22}{7} \):

\[ V = 7.9707 \times \frac{22}{7} \] \[ V \approx 25.05 \ \text{cm}^3 \]

Volume of 45 gulab jamuns:

\[ V_{\text{total}} = 45 \times 25.05 \] \[ V_{\text{total}} = 1127.25 \ \text{cm}^3 \]

Sugar syrup = 30% of total volume:

\[ = \frac{30}{100} \times 1127.25 \] \[ = 338.175 \] \[ \approx 338 \ \text{cm}^3 \]

Final Answer: \( 338 \ \text{cm}^3 \) of syrup

Exam Significance:

• Very important real-life application problem in CBSE exams.
• Tests understanding of capsule geometry.
• Frequently appears in NTSE and Olympiad foundation.
• Common mistake: not converting two hemispheres into one sphere.

Concept Used:

• This is a subtractive solid problem.
• Volume of wood = volume of cuboid − volume removed (cones).
• Total removed volume = 4 × volume of one cone.

\[ \text{Volume of cuboid} = l \times b \times h \] \[ \text{Volume of cone} = \frac{1}{3}\pi r^2 h \]

Solution Roadmap:

1. Find volume of cuboid.
2. Find volume of one conical depression.
3. Multiply by 4.
4. Subtract from cuboid volume.

Solution:

Length \(l = 15\) cm
Breadth \(b = 10\) cm
Height \(h = 3.5\) cm
Radius of cone \(r = 0.5\) cm
Height of cone \(h_c = 1.4\) cm

Volume of cuboid:

\[ V_{\text{cuboid}} = 15 \times 10 \times 3.5 \] \[ V_{\text{cuboid}} = 525 \ \text{cm}^3 \]

Volume of one cone:

\[ V_{\text{cone}} = \frac{1}{3}\pi r^2 h \] \[ V_{\text{cone}} = \frac{1}{3} \times \pi \times (0.5)^2 \times 1.4 \]

Compute powers:

\[ (0.5)^2 = 0.25 \] \[ V_{\text{cone}} = \frac{1}{3} \times \pi \times 0.25 \times 1.4 \]

Multiply:

\[ 0.25 \times 1.4 = 0.35 \] \[ V_{\text{cone}} = \frac{1}{3} \times \pi \times 0.35 \]

Using \( \pi = \frac{22}{7} \):

\[ V_{\text{cone}} = \frac{1}{3} \times \frac{22}{7} \times 0.35 \]

Simplify \(0.35 = \frac{35}{100} = \frac{7}{20}\):

\[ V_{\text{cone}} = \frac{1}{3} \times \frac{22}{7} \times \frac{7}{20} \] \[ = \frac{1}{3} \times \frac{22}{20} \] \[ = \frac{22}{60} \] \[ = 0.3667 \ \text{cm}^3 \]

Volume of 4 cones:

\[ 4 \times 0.3667 = 1.4668 \ \text{cm}^3 \]

Volume of wood:

\[ V = 525 - 1.4668 \] \[ V = 523.5332 \ \text{cm}^3 \] \[ V \approx 523.53 \ \text{cm}^3 \]

Final Answer: \( 523.53 \ \text{cm}^3 \)

Exam Significance:

• Standard CBSE 3–4 mark question from subtractive solids.
• Tests clarity in removal of volume concept.
• Important for JEE foundation & NTSE.
• Common mistake: forgetting to multiply by 4 or incorrect decimal handling.

Concept Used:

• This is based on volume displacement principle.
• When solids are immersed, they displace equal volume of liquid.
• Thus, volume of water overflow = total volume of lead shots.

\[ \text{Volume of cone} = \frac{1}{3}\pi r^2 h \] \[ \text{Volume of sphere} = \frac{4}{3}\pi r^3 \]

Solution Roadmap:

1. Find volume of conical vessel.
2. Find volume of one spherical shot.
3. Compute overflow water (1/4 of cone volume).
4. Equate displaced volume with total volume of spheres.
5. Solve for number of spheres.

Solution:

Radius of cone \(r = 5\) cm
Height of cone \(h = 8\) cm
Radius of sphere \(r_s = 0.5\) cm

Volume of conical vessel:

\[ V_{\text{cone}} = \frac{1}{3}\pi r^2 h \] \[ V_{\text{cone}} = \frac{1}{3}\pi (5)^2 (8) \] \[ V_{\text{cone}} = \frac{1}{3}\pi \times 25 \times 8 \] \[ V_{\text{cone}} = \frac{200}{3}\pi \]

Water overflow = \( \frac{1}{4} \) of vessel volume:

\[ V_{\text{overflow}} = \frac{1}{4} \times \frac{200}{3}\pi \] \[ V_{\text{overflow}} = \frac{200}{12}\pi \] \[ V_{\text{overflow}} = \frac{50}{3}\pi \]

Volume of one spherical lead shot:

\[ V_s = \frac{4}{3}\pi r^3 \] \[ V_s = \frac{4}{3}\pi (0.5)^3 \] \[ (0.5)^3 = 0.125 \] \[ V_s = \frac{4}{3}\pi \times 0.125 \] \[ V_s = \frac{0.5}{3}\pi \] \[ V_s = \frac{\pi}{6} \]

Let number of lead shots = \(n\)

Total volume of lead shots = \(n \times \frac{\pi}{6}\)

Using displacement principle:

\[ n \times \frac{\pi}{6} = \frac{50}{3}\pi \]

Cancel \( \pi \):

\[ n \times \frac{1}{6} = \frac{50}{3} \]

Multiply both sides by 6:

\[ n = \frac{50}{3} \times 6 \] \[ n = 50 \times 2 \] \[ n = 100 \]

Final Answer: Number of lead shots = 100

Exam Significance:

• Highly important CBSE application-based question.
• Tests volume displacement concept (frequent exam trap).
• Common in NTSE, Olympiad, JEE foundation.
• Common mistake: not equating overflow volume with volume of spheres.

Concept Used:

• The pole is a combination of two cylinders.
• Total volume = sum of volumes of both cylinders.
• Mass = Volume × Density.

\[ \text{Volume of cylinder} = \pi r^2 h \] \[ \text{Mass} = \text{Volume} \times \text{Density} \]

Solution Roadmap:

1. Find volume of each cylinder.
2. Add to get total volume.
3. Multiply by density.
4. Convert grams to kilograms.

Solution:

First cylinder:
Radius \(r_1 = 12\) cm, Height \(h_1 = 220\) cm

Second cylinder:
Radius \(r_2 = 8\) cm, Height \(h_2 = 60\) cm

Total volume:

\[ V = \pi r_1^2 h_1 + \pi r_2^2 h_2 \] \[ V = \pi (12)^2 (220) + \pi (8)^2 (60) \] \[ V = \pi (144 \times 220) + \pi (64 \times 60) \] \[ V = \pi (31680 + 3840) \] \[ V = \pi (35520) \]

Using \( \pi = 3.14 \):

\[ V = 3.14 \times 35520 \] \[ V = 111532.8 \ \text{cm}^3 \]

Given: 1 cm³ of iron has mass = 8 g

Total mass in grams:

\[ \text{Mass} = 111532.8 \times 8 \] \[ \text{Mass} = 892262.4 \ \text{g} \]

Convert into kilograms:

\[ \text{Mass} = \frac{892262.4}{1000} \] \[ \text{Mass} = 892.2624 \ \text{kg} \] \[ \approx 892.26 \ \text{kg} \]

Final Answer: \( 892.26 \ \text{kg} \)

Exam Significance:

• Classic CBSE 4-mark question involving volume + density.
• Tests unit conversion accuracy (g → kg).
• Frequently appears in NTSE and engineering foundation exams.
• Common mistake: incorrect multiplication or missing unit conversion.

Concept Used:

• This is based on volume displacement principle.
• Volume of water displaced = volume of immersed solid.
• Water left = volume of cylinder − volume of solid.

\[ \text{Volume of cylinder} = \pi r^2 h \] \[ \text{Volume of cone} = \frac{1}{3}\pi r^2 h \] \[ \text{Volume of hemisphere} = \frac{2}{3}\pi r^3 \]

Solution Roadmap:

1. Find volume of cylinder.
2. Find volume of cone.
3. Find volume of hemisphere.
4. Add solid volume.
5. Subtract from cylinder volume.

Solution:

Radius of cylinder \(r = 60\) cm
Height of cylinder \(h = 180\) cm

Cone: \(r = 60\) cm, \(h = 120\) cm
Hemisphere: \(r = 60\) cm

Volume of cylinder:

\[ V_{\text{cyl}} = \pi r^2 h \] \[ V_{\text{cyl}} = \pi (60)^2 (180) \] \[ (60)^2 = 3600 \] \[ V_{\text{cyl}} = \pi \times 3600 \times 180 \] \[ V_{\text{cyl}} = 648000\pi \]

Volume of cone:

\[ V_{\text{cone}} = \frac{1}{3}\pi r^2 h \] \[ V_{\text{cone}} = \frac{1}{3}\pi (60)^2 (120) \] \[ V_{\text{cone}} = \frac{1}{3}\pi \times 3600 \times 120 \] \[ V_{\text{cone}} = \pi \times 1200 \times 120 \] \[ V_{\text{cone}} = 144000\pi \]

Volume of hemisphere:

\[ V_{\text{hemisphere}} = \frac{2}{3}\pi r^3 \] \[ V_{\text{hemisphere}} = \frac{2}{3}\pi (60)^3 \] \[ (60)^3 = 216000 \] \[ V_{\text{hemisphere}} = \frac{2}{3}\pi \times 216000 \] \[ V_{\text{hemisphere}} = 144000\pi \]

Total volume of solid:

\[ V_{\text{solid}} = 144000\pi + 144000\pi \] \[ V_{\text{solid}} = 288000\pi \]

Volume of water left:

\[ V = 648000\pi - 288000\pi \] \[ V = 360000\pi \]

Using \( \pi = 3.14 \):

\[ V = 360000 \times 3.14 \] \[ V = 1130400 \ \text{cm}^3 \]

Convert to \( \text{m}^3 \):

\[ 1 \ \text{m}^3 = 1000000 \ \text{cm}^3 \] \[ V = \frac{1130400}{1000000} \] \[ V = 1.1304 \ \text{m}^3 \] \[ \approx 1.131 \ \text{m}^3 \]

Final Answer: \( \approx 1.131 \ \text{m}^3 \)

Exam Significance:

• High-weight CBSE 4–5 mark question.
• Combines composite solids + displacement.
• Frequently appears in NTSE, Olympiad, JEE foundation.
• Common mistake: forgetting subtraction or miscalculating hemisphere volume.

Concept Used:

• This is a composite solid: sphere + cylinder.
• Total capacity = sum of volumes.
• We verify by comparing calculated volume with observed value.

\[ \text{Volume of sphere} = \frac{4}{3}\pi r^3 \] \[ \text{Volume of cylinder} = \pi r^2 h \]

Solution Roadmap:

1. Find radius of sphere and cylinder.
2. Compute volume of sphere.
3. Compute volume of cylinder.
4. Add both volumes.
5. Compare with given value.

Solution:

Cylindrical neck:
Radius \(r_1 = 1\) cm, Height \(h = 8\) cm

Spherical part:
Radius \(r_2 = \dfrac{8.5}{2} = 4.25\) cm

Total volume:

\[ V = \frac{4}{3}\pi r_2^3 + \pi r_1^2 h \]

Compute \( (4.25)^3 \):

\[ (4.25)^2 = 18.0625 \] \[ (4.25)^3 = 18.0625 \times 4.25 = 76.765625 \]

Now substitute:

\[ V = \frac{4}{3}\pi (76.765625) + \pi (1)^2 (8) \]

Using \( \pi = 3.14 \):

\[ V = 3.14 \left( \frac{4}{3} \times 76.765625 + 8 \right) \]

Simplify inside bracket:

\[ \frac{4}{3} \times 76.765625 = 102.354167 \] \[ V = 3.14 (102.354167 + 8) \] \[ V = 3.14 (110.354167) \] \[ V \approx 346.51 \ \text{cm}^3 \]

Measured volume = 345 cm³

Difference:

\[ 346.51 - 345 = 1.51 \ \text{cm}^3 \]

Since the difference is small (due to approximation), the measured value is approximately correct.

Final Conclusion: The child's measurement is approximately correct.

Exam Significance:

• Important verification-type CBSE question.
• Tests precision + approximation judgement.
• Common in NTSE and Olympiad reasoning.
• Common mistake: wrong cube calculation or ignoring approximation.