Surface Areas and Volumes: Advanced 5-Mark Problem Solving Guide

In the CBSE Class 10 Mathematics board examination, Section D features long-answer questions carrying 5 marks each. Among them, questions from Chapter 12 (Surface Areas and Volumes) are universally considered the most calculation-heavy and time-consuming problems on the paper.

A single arithmetic slip—such as confusing cone height with slant height, mixing up internal and external radii, or dividing recurring decimals prematurely—can destroy an entire 5-mark solution. High-scoring students succeed because they approach composite mensuration with an algebraic factorization-first mindset: factoring out common terms like $\pi$ and $r$ before performing any numerical arithmetic.

This advanced guide walks through three heavyweight 5-mark board exam problems step-by-step.

What You Will Learn

• The Factor-First Strategy: Eliminating multi-step multiplications
• Master Problem 1: The Solid Wooden Cylinder with Hemispherical Scoops (Total Surface Area & Volume)
• Master Problem 2: Water Flow Through a Cylindrical Pipe into a Conical Tank
• Master Problem 3: Melting Solid Spheres into a Hollow Spherical Shell
• Presentation templates to guarantee full step marks

1. The Factor-First Strategy

    SLOPPY APPROACH:
    Calculate πr²h = (22/7) × 14 × 14 × 30 = 18480
    Calculate (4/3)πr³ = (4/3) × (22/7) × 14 × 14 × 14 = 11498.67
    Add 18480 + 11498.67 = 29978.67  (Messy decimals, highly error-prone!)

    EXAMINER-PREFERRED FACTOR-FIRST APPROACH:
    V = πr²h + (4/3)πr³ = πr² [ h + (4/3)r ]
    Substitute values ONCE at the end!

Important: <u>Never substitute $\pi = 22/7$ or calculate numerical values in piecemeal intermediate steps! Combine formulas algebraically, factor out common multiples of $\pi$ and $r$, and evaluate numbers in a single final step.</u>

2. Advanced Solved 5-Mark Board Problems

Problem 1: Solid Cylinder with Hemispherical Scoops (NCERT Classic)

Problem: A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is $10\text{ cm}$, and its base is of radius $3.5\text{ cm}$, find: (a) the total surface area of the article, and (b) the volume of wood left in the article. (Use $\pi = 22/7$).

                     Hemispherical Scoop 1 (Radius 3.5 cm)
                             (               )
                             |               |
                             |   CYLINDER    | Height = 10 cm
                             |               |
                             (               )
                     Hemispherical Scoop 2 (Radius 3.5 cm)

Solution:

1. Analyze Dimensions:

   • Radius of cylinder and hemispheres: $r = 3.5\text{ cm} = \frac{7}{2}\text{ cm}$.
   • Height of cylinder: $h = 10\text{ cm}$.

2. Part (a): Total Surface Area of the Article: When hemispherical cavities are scooped out, the flat circular ends disappear, and two exposed curved hemispherical surfaces are created: $\mathbf{\text{TSA} = \text{CSA of Cylinder} + 2 \times (\text{CSA of Hemisphere})}$ $\text{TSA} = 2\pi r h + 2(2\pi r^2) = 2\pi r h + 4\pi r^2 = \mathbf{2\pi r (h + 2r)}$ Substitute values: $\text{TSA} = 2 \times \frac{22}{7} \times \frac{7}{2} \times [10 + 2(3.5)]$ $\text{TSA} = 22 \times [10 + 7] = 22 \times 17 = \mathbf{374\text{ cm}^2}$

3. Part (b): Volume of Wood Left in the Article: Here, material was carved out, so volumes subtract: $\mathbf{V_{\text{left}} = \text{Volume of Cylinder} - 2 \times (\text{Volume of Hemisphere})}$ $\mathbf{V_{\text{left}} = \pi r^2 h - \frac{4}{3}\pi r^3 = \pi r^2 \left(h - \frac{4}{3}r\right)}$ Substitute values: $V_{\text{left}} = \frac{22}{7} \times \left(\frac{7}{2}\right)^2 \times \left[10 - \frac{4}{3}\left(\frac{7}{2}\right)\right]$ $V_{\text{left}} = \frac{22}{7} \times \frac{49}{4} \times \left[10 - \frac{14}{3}\right] = \frac{77}{2} \times \left[\frac{30 - 14}{3}\right]$ $V_{\text{left}} = \frac{77}{2} \times \frac{16}{3} = \frac{77 \times 8}{3} = \frac{616}{3} = \mathbf{205.33\text{ cm}^3}$

4. Therefore:

   • <u>(a) The total surface area of the article is $374 ext{ cm}^2$</u>.
   • <u>(b) The volume of wood left is $205.33 ext{ cm}^3$</u>.

Problem 2: Water Flow Through a Pipe Filling a Tank (Rate of Flow)

Problem: A farmer connects a pipe of internal diameter $20\text{ cm}$ from a canal into a cylindrical tank in her field, which is $10\text{ m}$ in diameter and $2\text{ m}$ deep. If water flows through the pipe at the rate of $3\text{ km/h}$, in how much time will the tank be filled?

Solution:

1. Harmonize All Units to METRES:

   • Pipe internal diameter $= 20\text{ cm} \implies$ Radius $r = 10\text{ cm} = \frac{10}{100} = \mathbf{0.1\text{ m} = \frac{1}{10}\text{ m}}$.
   • Speed of water flow: $v = 3\text{ km/h} = 3 \times 1000 = 3000\text{ m/h}$ In metres per minute: $v = \frac{3000\text{ m}}{60\text{ min}} = \mathbf{50\text{ m/min}}$
   • Cylindrical tank dimensions: Diameter $= 10\text{ m} \implies$ Radius $R = 5\text{ m}$; Depth $H = 2\text{ m}$.

2. Calculate Volume of the Cylindrical Tank: $V_{\text{tank}} = \pi R^2 H = \pi \times 5^2 \times 2 = \mathbf{50\pi\text{ m}^3}$

3. Calculate Volume of Water Delivered by Pipe in 1 Minute: In $1\text{ minute}$, the length of the water column is $L = 50\text{ m}$: $V_{\text{1 min}} = \pi r^2 L = \pi \times \left(\frac{1}{10}\right)^2 \times 50 = \pi \times \frac{1}{100} \times 50 = \mathbf{\frac{\pi}{2}\text{ m}^3}$

4. Calculate Time Required to Fill the Tank ($t$): $t = \frac{\text{Total Volume of Tank}}{\text{Volume Delivered per Minute}} = \frac{50\pi}{\frac{\pi}{2}} = 50 \times 2 = \mathbf{100\text{ minutes}}$

5. Therefore, <u>the tank will be completely filled in $100\text{ minutes}$ (or $1\text{ hour } 40\text{ minutes}$)</u>.

3. Summary and Examination Tips

Exam Tip: In pipe flow problems, always convert water speed to metres per minute ($ext{m/min}$). This eliminates massive numbers and yields the time directly in clean integer minutes!

Common Mistake: Forgetting that scooping cavities INCREASES surface area. Scooping removes volume, but exposes new interior curved surfaces!