Let BPC be the hemisphere and ABC be the cone standing on the base of the hemisphere (see Figure). The radius BO of the hemisphere (as well as of the cone) \(=\frac{1}{2} \times 4 \mathrm{~cm}=2 \mathrm{~cm} .\)
So, volume of the toy = \(\begin{aligned} & =\frac{2}{3} \pi r^3+\frac{1}{3} \pi r^2 h \end{aligned}\)
\(\begin{aligned} & =\left[\frac{2}{3} \times 3.14 \times(2)^3+\frac{1}{3} \times 3.14 \times(2)^2 \times 2\right] \mathrm{cm}^3=25.12 \mathrm{~cm}^3 \end{aligned}\)
Now, let the right circular cylinder EFGH circumscribe the given solid. The radius of the base of the right circular cylinder = HP = BO = 2 cm, and its height is EH = AO + OP = (2 + 2) cm = 4 cm
So, the colume required = volume of the right circular cylinder - volume of the toy
= 3.14 \(\times\)2² \(\times\)4 - 25.12) cm³
= 25.12 cm³
hence, the required difference of the two volumes = 25.12 cm³.

The volume of air inside the shed (when there are no people or machinery) is given by the volume of air inside the cuboid and inside the half cylinder, taken together.
Now, the length, breadth and height of the cuboid are 15 m, 7 m and 8 m, respectively. Also, the diameter of the half cylinder is 7 m and its height is 15 m.
So, the required volume = volume of the cuboid \(+\frac{1}{2}\) volume of the cylinder 
\(=\left[15 \times 7 \times 8+\frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 15\right] \mathrm{m}^{3}=1128.75 \mathrm{~m}^{3}\)
Next, the total space occupied by the machinery = 300 m³
And the total space occupied by the workers = 20 × 0.08 m³ = 1.6 m³
Therefore, the volume of the air, when there are machinery and workers
= 1128.75 – (300.00 + 1.60) = 827.15 m³

The total surface area of the cube = 6 × (edge)² = 6 × 5 × 5 cm² = 150 cm².
Note that the part of the cube where the hemisphere is attached is not included in the surface area.
So, the surface area of the block = TSA of cube – base area of hemisphere + CSA of hemisphere
\(=150-\pi r^{2}+2 \pi r^{2}=\left(150+\pi r^{2}\right) \mathrm{cm}^{2} \)
\(=150 \mathrm{~cm}^{2}+\left(\frac{22}{7} \times \frac{4.2}{2} \times \frac{4.2}{2}\right) \mathrm{cm}^{2} \)
\(=(150+13.86) \mathrm{cm}^{2}=163.86 \mathrm{~cm}^{2}\)

(i) 8 cm
(ii) 1386 cm²
(iii) Social justice, caring

Increase in water level when sphere is dropped into cylindrical vessel = \(\frac{32}{9}\)cm
Volume of water level raised = \(\pi r^2h\) \(=\frac{22}{7}=r^2=\frac{32}{9}\)
Volume of water level raised = volume of sphere
\(\frac{22}{7}=r^2=\frac{32}{9}\)= \(\frac{4}{3}\times\frac{22}{7}\times(6)^3\)
\(\Rightarrow\)  r² = 81
\(\Rightarrow\)  r = 9 cm
Diameter,  d =  18 cm

Given, one gulabjamun is a combination of a cylinder and two hemispheres. Here, total length of one gulabjamun = 5 cm and diameter = 2.8 cm

\(\therefore\) Radius of cylindrical part = Radius of hemispherical part
\(=r=\frac{2.8}{2}=1.4 \mathrm{~cm}\)
Height of cylindrical part,
h = PQ - (PR + SQ) = 5 - (1.4 + 1.4)
= 5 - 2.8 = 2.2 cm
\(\therefore\) Volume of one gulabjamun = 2 \(\times\) Volume of hemispherical part + Volume of cylindrical part
\(\begin{aligned} & =2 \times\left[\frac{2}{3} \pi r^{3}\right]+\pi r^2 h=\frac{4}{3} \pi r^{3}+\pi r^2 h \\ \end{aligned}\)
\(\begin{aligned} & =\pi r^2\left[\frac{4 r}{3}+h\right]=\frac{22}{7} \times 1.4 \times 1.4\left[\frac{4}{3} \times 1.4+2.2\right] \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{7} \times \frac{14}{10} \times \frac{14}{10}\left[\frac{4}{3} \times \frac{14}{10}+\frac{22}{10}\right] \\ \end{aligned}\)
\(\begin{aligned} & =22 \times \frac{1}{5} \times \frac{7}{5}\left[\frac{28}{15}+\frac{11}{5}\right]=\frac{154}{25} \times \frac{61}{15}=\frac{9394}{375} \mathrm{~cm}^3 \end{aligned}\)
Now, volume of 45 gulabjamuns
\(=45 \times \frac{9394}{375}=\frac{3 \times 9394}{25}=\frac{28182}{25}=1127.28 \mathrm{~cm}^3\)
Since, one gulabjamun contains sugar syrup upto about 30% of its volume.
Hence, quantity of syrup found in 45 gulabjamuns
\(=1127.28 \times \frac{30}{100}=1127.28 \times \frac{3}{10}=338.184 \approx 338 \mathrm{~cm}^3\)

(i) (a) \(2 \pi r(r+h)\) 
(ii) (a)  Area of canvas used = Cuurved Surface Area of conical part + Curved Surface Area of cylindrical part
\(=\pi r l+2 \pi r h\) 
(iii) (d)  Volume of tent = Volume of cylinder + Volume of cone
= \(\frac{1}{3} \pi R^{2} H+\pi R^{2} h\) 
(iv) (d) \(\text { Ratio of volume of two cylinders }=\frac{\pi R_{1}^{2} H_{1}}{\pi R_{2}^{2} \mathrm{H}_{2}}\) 
= \(\frac{2 \times 2 \times 5}{3 \times 3 \times 5}=\frac{4}{9}\) 
(v)(b) \(\text { Area of canvas used }=\pi r l+2 \pi r h\) 
= \(\pi r(2 h+l)\) 
= \(\frac{22}{7} \times 2(2 \times(2.1)+2.8)\) 

(i) (b) 8m
(ii) (a) 14m
(iii) (c) 26m
(iv) (a) 22m
(v) (d) 6m

(a) (i) Volume of cuboid = lbh
= 15 X 10 X 3.5
= 15 X 35 = 525cm³
(b) (ii) Volume of cone =  \(\frac{1}{3} \pi \mathrm{R}^{2} \mathrm{H}=\frac{1}{3} \times \frac{22}{7} \times 0.5 \times 0.5 \times 1.4\) 
\(=\frac{1}{3} \times \frac{22}{7} \times \frac{1}{2} \times \frac{1}{2} \times \frac{14}{10}=\frac{11}{30} \mathrm{~cm}^{3} .\) 
(c) (iii) \(\text { Volume of } 4 \text { cones }=4 \times \frac{11}{30} \mathrm{~cm}^{3}=\frac{22}{15} \mathrm{~cm}^{3}=1.47 \mathrm{~cm}^{3}\) 
(d) (iii) Volume of wood = 525 cm³ – 1.47 cm³
= 523.53 cm³
(e) (ii) )Surface Areas & Volumes

(a) (iii) Height of the cone = height of the top – height (radius) of the hemispherical part
= 5 – (3.5/2) = 5 – 1.75
= 3.25 cm
(b) (ii) Slant height of the cone \((l)=\sqrt{r^{2}+h^{2}}\) 
\(=\sqrt{\left(\frac{3.5}{2}\right)^{2}+(3.25)^{2}} \mathrm{~cm}\) 
= 3.7 cm(approx)
(c) (iii) \(\text { CSA of cone }=\pi r l=\left(\frac{22}{7} \times \frac{3.5}{2} \times 3.7\right) \mathrm{cm}^{2}\) 
= 20.35 cm²
(d) (iv) Curved surface area of the hemisphere =\(2 \pi r^{2}\)
=\(\left(2 \times \frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2}\right) \mathrm{cm}^{2} \\\) 
=19.25 cm²
(e) (i)  Total Surface Area of the toy = CSA of hemisphere + CSA of cone
= 19.25 + 20.35 cm²
= 39.6 cm²