Surface Areas and Volumes

Given: Volume of each cube = 64 cm³

Step 1: Find the side of each cube.
$$a^3 = 64 \implies a = 4 \text{ cm}$$

Step 2: Dimensions of the resulting cuboid.
When two cubes are joined end to end:
- Length $l = 4 + 4 = 8$ cm
- Breadth $b = 4$ cm
- Height $h = 4$ cm

Step 3: Surface area of the cuboid.
$$\text{SA} = 2(lb + bh + hl)$$
$$= 2(8 \times 4 + 4 \times 4 + 4 \times 8)$$
$$= 2(32 + 16 + 32)$$
$$= 2 \times 80 = 160 \text{ cm}^2$$

Answer: The surface area of the resulting cuboid is $\boxed{160 \text{ cm}^2}$.

Given:
- Diameter of hemisphere = 14 cm, so radius $r = 7$ cm
- Total height of vessel = 13 cm
- Height of cylindrical part $h = 13 - 7 = 6$ cm

Formula used:
$$\text{Inner surface area} = \text{CSA of cylinder} + \text{CSA of hemisphere}$$
$$= 2\pi r h + 2\pi r^2$$

Calculation:
$$= 2 \times \frac{22}{7} \times 7 \times 6 + 2 \times \frac{22}{7} \times 7^2$$
$$= 2 \times 22 \times 6 + 2 \times 22 \times 7$$
$$= 264 + 308 = 572 \text{ cm}^2$$

Answer: The inner surface area of the vessel is $\boxed{572 \text{ cm}^2}$.

Given:
- Radius of cone = radius of hemisphere $r = 3.5$ cm
- Total height of toy = 15.5 cm
- Height of cone $h = 15.5 - 3.5 = 12$ cm

Step 1: Find slant height of cone.
$$l = \sqrt{r^2 + h^2} = \sqrt{(3.5)^2 + (12)^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5 \text{ cm}$$

Step 2: Total surface area.
$$\text{TSA} = \text{CSA of cone} + \text{CSA of hemisphere}$$
$$= \pi r l + 2\pi r^2$$
$$= \frac{22}{7} \times 3.5 \times 12.5 + 2 \times \frac{22}{7} \times (3.5)^2$$
$$= \frac{22}{7} \times 3.5 \times 12.5 + 2 \times \frac{22}{7} \times 12.25$$
$$= 22 \times 0.5 \times 12.5 + 2 \times 22 \times 1.75$$
$$= 137.5 + 77 = 214.5 \text{ cm}^2$$

Answer: The total surface area of the toy is $\boxed{214.5 \text{ cm}^2}$.

Given: Side of cube = 7 cm

Step 1: Greatest diameter of hemisphere.
The greatest diameter the hemisphere can have equals the side of the cube.
$$\text{Diameter} = 7 \text{ cm}, \quad r = 3.5 \text{ cm}$$

Step 2: Surface area of the solid.
The total surface area = Total surface area of cube − Base area of hemisphere + Curved surface area of hemisphere
$$= 6a^2 - \pi r^2 + 2\pi r^2$$
$$= 6a^2 + \pi r^2$$
$$= 6 \times (7)^2 + \frac{22}{7} \times (3.5)^2$$
$$= 6 \times 49 + \frac{22}{7} \times 12.25$$
$$= 294 + 38.5 = 332.5 \text{ cm}^2$$

Answer: The greatest diameter is 7 cm and the surface area of the solid is $\boxed{332.5 \text{ cm}^2}$.

Given:
- Edge of cube = $l$
- Diameter of hemisphere = $l$, so radius $r = \dfrac{l}{2}$

Surface area of remaining solid:
= Total surface area of cube − Base area of hemisphere + Curved surface area of hemisphere
$$= 6l^2 - \pi r^2 + 2\pi r^2$$
$$= 6l^2 + \pi r^2$$
$$= 6l^2 + \pi \left(\frac{l}{2}\right)^2$$
$$= 6l^2 + \frac{\pi l^2}{4}$$
$$= \frac{l^2}{4}(24 + \pi)$$

Answer: The surface area of the remaining solid is $\boxed{\dfrac{l^2}{4}(24 + \pi)}$.

Given:
- Total length of capsule = 14 mm
- Diameter = 5 mm, so radius $r = 2.5$ mm

Step 1: Find height of cylindrical part.
The two hemispheres together contribute $2r = 5$ mm to the total length.
$$h = 14 - 2 \times 2.5 = 14 - 5 = 9 \text{ mm}$$

Step 2: Total surface area.
$$\text{TSA} = \text{CSA of cylinder} + 2 \times \text{CSA of hemisphere}$$
$$= 2\pi r h + 2 \times 2\pi r^2$$
$$= 2\pi r h + 4\pi r^2$$
$$= 2\pi r(h + 2r)$$
$$= 2 \times \frac{22}{7} \times 2.5 \times (9 + 5)$$
$$= 2 \times \frac{22}{7} \times 2.5 \times 14$$
$$= 2 \times 22 \times 2.5 \times 2$$
$$= 220 \text{ mm}^2$$

Answer: The surface area of the capsule is $\boxed{220 \text{ mm}^2}$.

Given:
- Height of cylinder $h = 2.1$ m
- Diameter = 4 m, so radius $r = 2$ m
- Slant height of cone $l = 2.8$ m

Step 1: Area of canvas used.
$$\text{Area} = \text{CSA of cylinder} + \text{CSA of cone}$$
$$= 2\pi r h + \pi r l$$
$$= \pi r (2h + l)$$
$$= \frac{22}{7} \times 2 \times (2 \times 2.1 + 2.8)$$
$$= \frac{22}{7} \times 2 \times (4.2 + 2.8)$$
$$= \frac{22}{7} \times 2 \times 7$$
$$= 22 \times 2 = 44 \text{ m}^2$$

Step 2: Cost of canvas.
$$\text{Cost} = 44 \times 500 = ₹22000$$

Answer: Area of canvas = $44 \text{ m}^2$ and cost of canvas = $\boxed{₹22000}$.

Given:
- Height of cylinder = height of cone $h = 2.4$ cm
- Diameter = 1.4 cm, so radius $r = 0.7$ cm

Step 1: Slant height of cone.
$$l = \sqrt{r^2 + h^2} = \sqrt{(0.7)^2 + (2.4)^2} = \sqrt{0.49 + 5.76} = \sqrt{6.25} = 2.5 \text{ cm}$$

Step 2: Total surface area of remaining solid.
The remaining solid has:
- CSA of cylinder (outer curved surface)
- Base area of cylinder (bottom circle, solid)
- CSA of cone (inner hollow surface)

$$\text{TSA} = \text{CSA of cylinder} + \text{Base of cylinder} + \text{CSA of cone}$$
$$= 2\pi r h + \pi r^2 + \pi r l$$
$$= \pi r (2h + r + l)$$
$$= \frac{22}{7} \times 0.7 \times (2 \times 2.4 + 0.7 + 2.5)$$
$$= \frac{22}{7} \times 0.7 \times (4.8 + 0.7 + 2.5)$$
$$= \frac{22}{7} \times 0.7 \times 8$$
$$= 22 \times 0.1 \times 8 = 17.6 \text{ cm}^2$$

Answer: The total surface area of the remaining solid $\approx \boxed{18 \text{ cm}^2}$.

Given:
- Height of cylinder $h = 10$ cm
- Radius $r = 3.5$ cm

Total surface area of the article:
When hemispheres are scooped from each end, the flat circular ends of the cylinder are replaced by the curved surfaces of the two hemispheres.
$$\text{TSA} = \text{CSA of cylinder} + 2 \times \text{CSA of hemisphere}$$
$$= 2\pi r h + 2 \times 2\pi r^2$$
$$= 2\pi r h + 4\pi r^2$$
$$= 2\pi r(h + 2r)$$
$$= 2 \times \frac{22}{7} \times 3.5 \times (10 + 2 \times 3.5)$$
$$= 2 \times \frac{22}{7} \times 3.5 \times 17$$
$$= 2 \times 22 \times 0.5 \times 17$$
$$= 2 \times 11 \times 17 = 374 \text{ cm}^2$$

Answer: The total surface area of the article is $\boxed{374 \text{ cm}^2}$.

Given:
- Radius of cone = radius of hemisphere $r = 1$ cm
- Height of cone $h = r = 1$ cm

Volume of solid:
$$V = \text{Volume of cone} + \text{Volume of hemisphere}$$
$$= \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$$
$$= \frac{1}{3}\pi (1)^2 (1) + \frac{2}{3}\pi (1)^3$$
$$= \frac{\pi}{3} + \frac{2\pi}{3}$$
$$= \frac{3\pi}{3} = \pi \text{ cm}^3$$

Answer: The volume of the solid is $\boxed{\pi \text{ cm}^3}$.

Given:
- Diameter = 3 cm, so radius $r = 1.5$ cm
- Total length = 12 cm
- Height of each cone $h_c = 2$ cm

Step 1: Height of cylindrical part.
$$h = 12 - 2 - 2 = 8 \text{ cm}$$

Step 2: Volume of air in the model.
$$V = \text{Volume of cylinder} + 2 \times \text{Volume of cone}$$
$$= \pi r^2 h + 2 \times \frac{1}{3}\pi r^2 h_c$$
$$= \pi r^2 \left(h + \frac{2h_c}{3}\right)$$
$$= \frac{22}{7} \times (1.5)^2 \times \left(8 + \frac{2 \times 2}{3}\right)$$
$$= \frac{22}{7} \times 2.25 \times \left(8 + \frac{4}{3}\right)$$
$$= \frac{22}{7} \times 2.25 \times \frac{28}{3}$$
$$= \frac{22 \times 2.25 \times 28}{21}$$
$$= \frac{22 \times 63}{21} = \frac{1386}{21} = 66 \text{ cm}^3$$

Answer: The volume of air contained in the model is $\boxed{66 \text{ cm}^3}$.

Given:
- Total length of each gulab jamun = 5 cm
- Diameter = 2.8 cm, so radius $r = 1.4$ cm
- Syrup = 30% of total volume

Step 1: Height of cylindrical part.
$$h = 5 - 2r = 5 - 2 \times 1.4 = 5 - 2.8 = 2.2 \text{ cm}$$

Step 2: Volume of one gulab jamun.
$$V = \pi r^2 h + \frac{4}{3}\pi r^3$$
$$= \pi r^2 h + \frac{4}{3}\pi r^3$$
$$= \frac{22}{7} \times (1.4)^2 \times 2.2 + \frac{4}{3} \times \frac{22}{7} \times (1.4)^3$$
$$= \frac{22}{7} \times 1.96 \times 2.2 + \frac{4}{3} \times \frac{22}{7} \times 2.744$$
$$= \frac{22 \times 1.96 \times 2.2}{7} + \frac{4 \times 22 \times 2.744}{21}$$
$$= \frac{94.864}{7} + \frac{241.472}{21}$$
$$= 13.552 + 11.498 = 25.05 \text{ cm}^3 \approx 25.05 \text{ cm}^3$$

Step 3: Volume of syrup in 45 gulab jamuns.
$$\text{Syrup} = 45 \times 30\% \times 25.05$$
$$= 45 \times \frac{30}{100} \times 25.05$$
$$= 45 \times 0.3 \times 25.05$$
$$= 338.18 \approx 338 \text{ cm}^3$$

Answer: The volume of syrup in 45 gulab jamuns is approximately $\boxed{338 \text{ cm}^3}$.

Given:
- Dimensions of cuboid: $l = 15$ cm, $b = 10$ cm, $h = 3.5$ cm
- Radius of each conical depression $r = 0.5$ cm
- Depth of each depression $h_c = 1.4$ cm
- Number of depressions = 4

Step 1: Volume of cuboid.
$$V_{\text{cuboid}} = l \times b \times h = 15 \times 10 \times 3.5 = 525 \text{ cm}^3$$

Step 2: Volume of one conical depression.
$$V_{\text{cone}} = \frac{1}{3}\pi r^2 h_c = \frac{1}{3} \times \frac{22}{7} \times (0.5)^2 \times 1.4$$
$$= \frac{1}{3} \times \frac{22}{7} \times 0.25 \times 1.4 = \frac{1}{3} \times \frac{22 \times 0.35}{7} = \frac{1}{3} \times 1.1 = \frac{1.1}{3} \approx 0.3667 \text{ cm}^3$$

Step 3: Volume of wood.
$$V_{\text{wood}} = V_{\text{cuboid}} - 4 \times V_{\text{cone}}$$
$$= 525 - 4 \times \frac{11}{30}$$
$$= 525 - \frac{44}{30}$$
$$= 525 - 1.4\overline{6} \approx 523.53 \text{ cm}^3$$

Answer: The volume of wood in the entire stand is approximately $\boxed{523.53 \text{ cm}^3}$.

Given:
- Height of cone $h = 8$ cm, radius $r = 5$ cm
- Radius of each lead shot $r_s = 0.5$ cm
- Water that flows out = $\dfrac{1}{4}$ of total volume

Step 1: Volume of water in cone.
$$V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 25 \times 8 = \frac{4400}{21} \text{ cm}^3$$

Step 2: Volume of water that flows out.
$$V_{\text{out}} = \frac{1}{4} \times \frac{4400}{21} = \frac{1100}{21} \text{ cm}^3$$

Step 3: Volume of one lead shot.
$$V_{\text{shot}} = \frac{4}{3}\pi r_s^3 = \frac{4}{3} \times \frac{22}{7} \times (0.5)^3 = \frac{4}{3} \times \frac{22}{7} \times 0.125 = \frac{11}{21} \text{ cm}^3$$

Step 4: Number of lead shots.
$$n = \frac{V_{\text{out}}}{V_{\text{shot}}} = \frac{\dfrac{1100}{21}}{\dfrac{11}{21}} = \frac{1100}{11} = 100$$

Answer: The number of lead shots dropped in the vessel is $\boxed{100}$.

Given:
- Lower cylinder: height $h_1 = 220$ cm, diameter = 24 cm → radius $r_1 = 12$ cm
- Upper cylinder: height $h_2 = 60$ cm, radius $r_2 = 8$ cm
- Density = 8 g/cm³

Step 1: Volume of lower cylinder.
$$V_1 = \pi r_1^2 h_1 = 3.14 \times 144 \times 220 = 3.14 \times 31680 = 99475.2 \text{ cm}^3$$

Step 2: Volume of upper cylinder.
$$V_2 = \pi r_2^2 h_2 = 3.14 \times 64 \times 60 = 3.14 \times 3840 = 12057.6 \text{ cm}^3$$

Step 3: Total volume.
$$V = V_1 + V_2 = 99475.2 + 12057.6 = 111532.8 \text{ cm}^3$$

Step 4: Mass of the pole.
$$\text{Mass} = V \times 8 = 111532.8 \times 8 = 892262.4 \text{ g} \approx 892.26 \text{ kg}$$

Answer: The mass of the pole is approximately $\boxed{892.26 \text{ kg}}$.

Given:
- Cone: height $h_c = 120$ cm, radius $r = 60$ cm
- Hemisphere: radius $r = 60$ cm
- Cylinder: height $H = 180$ cm, radius $R = 60$ cm

Step 1: Volume of cylinder.
$$V_{\text{cyl}} = \pi R^2 H = \frac{22}{7} \times 3600 \times 180 = \frac{22 \times 3600 \times 180}{7} = \frac{14256000}{7} \text{ cm}^3$$

Step 2: Volume of solid (cone + hemisphere).
$$V_{\text{cone}} = \frac{1}{3}\pi r^2 h_c = \frac{1}{3} \times \frac{22}{7} \times 3600 \times 120 = \frac{22 \times 3600 \times 40}{7} = \frac{3168000}{7} \text{ cm}^3$$
$$V_{\text{hemi}} = \frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 216000 = \frac{2 \times 22 \times 216000}{21} = \frac{9504000}{21} = \frac{3168000}{7} \text{ cm}^3$$
$$V_{\text{solid}} = \frac{3168000}{7} + \frac{3168000}{7} = \frac{6336000}{7} \text{ cm}^3$$

Step 3: Volume of water left.
$$V_{\text{water}} = V_{\text{cyl}} - V_{\text{solid}} = \frac{14256000}{7} - \frac{6336000}{7} = \frac{7920000}{7} \approx 1131428.57 \text{ cm}^3$$

$$= \frac{7920000}{7} \text{ cm}^3 \approx 1.131 \text{ m}^3$$

Answer: The volume of water left in the cylinder is $\dfrac{7920000}{7} \text{ cm}^3 \approx \boxed{1131428.57 \text{ cm}^3}$.

Given:
- Cylindrical neck: length $h = 8$ cm, diameter = 2 cm → radius $r_c = 1$ cm
- Spherical part: diameter = 8.5 cm → radius $r_s = 4.25$ cm
- $\pi = 3.14$

Step 1: Volume of cylindrical neck.
$$V_{\text{cyl}} = \pi r_c^2 h = 3.14 \times 1^2 \times 8 = 25.12 \text{ cm}^3$$

Step 2: Volume of spherical part.
$$V_{\text{sphere}} = \frac{4}{3}\pi r_s^3 = \frac{4}{3} \times 3.14 \times (4.25)^3$$
$$= \frac{4}{3} \times 3.14 \times 76.765625$$
$$= \frac{4 \times 3.14 \times 76.765625}{3}$$
$$= \frac{964.33}{3} \approx 321.39 \text{ cm}^3$$

Step 3: Total volume.
$$V = 25.12 + 321.39 = 346.51 \text{ cm}^3$$

Conclusion: The actual volume ≈ 346.51 cm³, whereas the child measured 345 cm³. The child's answer is not correct (the correct volume is approximately 346.51 cm³).

Answer: The child's answer is incorrect. The correct volume is approximately $\boxed{346.51 \text{ cm}^3}$.