MSW Sheet Metal Work: Mensuration, Marking, Riveting & Cylinder/Cone Fabrication: 17 MCQs with answers

All 17 multiple-choice questions from this kit, free and with no sign-up.

Warm-up · 6 questions

Q1. A cylinder has base radius $r$ and height $h$. Which formula gives its curved surface area?

Beginner
A $2\pi r(h+r)$ B $2\pi rh$ C $\pi r l$ D $\pi r^2 h$
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$2\pi rh$

The curved surface of a cylinder unrolls into a rectangle of width $2\pi r$ and height $h$, so CSA $= 2\pi rh$.

Q2. For a cone, which relationship correctly gives the slant height $l$ in terms of base radius $r$ and height $h$?

Beginner
A $l = \sqrt{r^2 + h^2}$ B $l = \sqrt{r^2 - h^2}$ C $l = r + h$ D $l = \frac{r h}{2}$
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$l = \sqrt{r^2 + h^2}$

The slant height, radius, and height form a right triangle, so Pythagoras gives $l = \sqrt{r^2 + h^2}$.

Q3. A cone and a cylinder have the same base radius and the same height. What is the ratio of their volumes (cone : cylinder)?

Beginner
A 3:1 B 1:3 C 1:1 D 1:2
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1:3

The cone’s volume is $\frac{1}{3}\pi r^2 h$ while the cylinder’s is $\pi r^2 h$, giving a ratio of 1:3.

Q4. Which tool is specifically designed to mark circles and equal intervals on sheet metal?

Beginner
A Divider B Try square C Scriber D Center punch
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Divider

A divider has two sharp points and is used to scribe circles and step off equal distances.

Q5. What is the main purpose of a center punch in sheet-metal work?

Beginner
A To cut sheet metal B To create a small indentation that guides the drill bit C To measure the thickness of the sheet D To mark 90° angles
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To create a small indentation that guides the drill bit

A center punch makes a conical dent (0.5–1 mm deep) so the drill bit does not wander on the smooth surface.

Q6. A cone’s volume is exactly one-third of the volume of a cylinder with the same base radius and height, giving a fixed 3:1 volume ratio.

Beginner
A True B False
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True

A cone’s volume is exactly one-third of the volume of a cylinder with the same base radius and height, giving a fixed 3:1 volume ratio.

Building up · 6 questions

Q7. A cylindrical pipe has outer radius $R$ and inner radius $r$. Which expression gives the volume of metal in the pipe of length $h$?

Intermediate
A $\pi(R^2 - r^2)h$ B $2\pi(R - r)h$ C $\pi(R - r)^2 h$ D $\pi(R^2 + r^2)h$
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$\pi(R^2 - r^2)h$

The metal volume is the difference between the outer and inner cylinder volumes: $\pi R^2 h - \pi r^2 h = \pi(R^2 - r^2)h$.

Q8. A cone has base radius $7\ \text{cm}$ and height $24\ \text{cm}$. What is its slant height?

Intermediate
A $17\ \text{cm}$ B $\sqrt{7^2 + 24^2}\ \text{cm}$ C $25\ \text{cm}$ D $31\ \text{cm}$
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$25\ \text{cm}$

$l = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25\ \text{cm}$.

Q9. A sheet-metal job has a total surface area of $500\ \text{cm}^2$, sheet thickness $0.2\ \text{cm}$, and density $7.8\ \text{g/cm}^3$. What is its weight?

Intermediate
A $390\ \text{g}$ B $7800\ \text{g}$ C $780\ \text{g}$ D $78\ \text{g}$
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$780\ \text{g}$

Volume $= 500 \times 0.2 = 100\ \text{cm}^3$; weight $= 100 \times 7.8 = 780\ \text{g}$.

Q10. In riveting, the hole diameter should be how much larger than the rivet diameter?

Intermediate
A $2\ \text{mm}$ larger B $0.5$–$1.0\ \text{mm}$ larger C $0.1$–$0.2\ \text{mm}$ larger D Exactly the same
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$0.1$–$0.2\ \text{mm}$ larger

A clearance of 0.1–0.2 mm allows the rivet to enter without force but still fills the hole when set.

Q11. A cone has base radius $3\ \text{cm}$ and slant height $5\ \text{cm}$. What is its total surface area? (Take $\pi = 3.14$.)

Intermediate
A $28.26\ \text{cm}^2$ B $47.1\ \text{cm}^2$ C $75.36\ \text{cm}^2$ D $150.72\ \text{cm}^2$
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$75.36\ \text{cm}^2$

TSA $= \pi r(l+r) = 3.14 \times 3 \times (5+3) = 3.14 \times 24 = 75.36\ \text{cm}^2$.

Q12. A cylinder has base area $154\ \text{cm}^2$ and height $10\ \text{cm}$. What is its total surface area? (Take $\pi = \frac{22}{7}$.)

Intermediate
A $440\ \text{cm}^2$ B $1540\ \text{cm}^2$ C $308\ \text{cm}^2$ D $748\ \text{cm}^2$
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$748\ \text{cm}^2$

From $\pi r^2 = 154$, $r = 7\ \text{cm}$. TSA $= 2 \times \frac{22}{7} \times 7 \times (10+7) = 44 \times 17 = 748\ \text{cm}^2$.

Topper level · 5 questions

Q13. A cone and a cylinder have the same volume and the same base radius. How does the cone’s height compare to the cylinder’s height?

Advanced
A The cone’s height is two times the cylinder’s height. B The cone’s height is three times the cylinder’s height. C The cone’s height is one-third of the cylinder’s height. D The cone’s height is equal to the cylinder’s height.
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The cone’s height is three times the cylinder’s height.

Set $\frac{1}{3}\pi r^2 h_{\text{cone}} = \pi r^2 h_{\text{cyl}}$. Cancelling gives $h_{\text{cone}} = 3 h_{\text{cyl}}$.

Q14. A cone has base radius $7\ \text{cm}$ and slant height $25\ \text{cm}$. What is the sector angle of its pattern? (Take $\pi = \frac{22}{7}$.)

Advanced
A $100.8^\circ$ B $50.4^\circ$ C $201.6^\circ$ D $360^\circ$
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$100.8^\circ$

$\theta = 360 \times \frac{r}{l} = 360 \times \frac{7}{25} = 360 \times 0.28 = 100.8^\circ$.

Q15. A double riveted lap joint is used instead of a single riveted lap joint. What is the main advantage?

Advanced
A It requires fewer rivets overall. B It is roughly twice as strong because it has two rows of rivets. C It is easier to align than a single row. D It eliminates the need for a center punch.
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It is roughly twice as strong because it has two rows of rivets.

Strength scales with the number of rivets in shear; two rows give approximately double the strength of a single row.

Q16. A cylindrical brass shell has base area $154\ \text{cm}^2$, height $10\ \text{cm}$, thickness $1\ \text{mm}$, and density $8.5\ \text{g/cm}^3$. What is its weight? (Take $\pi = \frac{22}{7}$.)

Advanced
A $4080\ \text{g}$ B $2040\ \text{g}$ C $40800\ \text{g}$ D $408\ \text{g}$
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$4080\ \text{g}$

TSA $= 748\ \text{cm}^2$; thickness $= 0.1\ \text{cm}$; volume $= 74.8\ \text{cm}^3$; weight $= 74.8 \times 8.5 = 4080\ \text{g}$.

Q17. A cylindrical pipe has an outer radius of 10 cm, an inner radius of 8 cm, and a length of 50 cm. Calculate the volume of metal in the pipe. Use the formula for the volume of metal in a cylindrical pipe: $V = \pi(R^2 - r^2)h$.

Advanced
A 3140 cm³ B 3150 cm³ C 3148 cm³ D 3144 cm³
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3144 cm³

To find the volume of metal in the pipe, we apply the formula: $V = \pi(R^2 - r^2)h$. Here, $R = 10\ ext{cm}$, $r = 8\ ext{cm}$, and $h = 50\ ext{cm}$. Substituting these values gives: $V = \pi(10^2 - 8^2)50 = \pi(100 - 64)50 = \pi(36)50 = 1800\pi$. Using $\pi \approx 3.14$, we find…

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