Let \(\Delta\)ABC be the right triangle, right angled at A whose sides are AB and AC measure 3 cm and 4 cm respectively
The length of the side BC (hypotenuse)
\(B C =\sqrt{3^{2}+4^{2}} \)
\(=\sqrt{25} \)
\(=5 \mathrm{~cm} \)
By revolving, \(\Delta\)ABC around its hypotenuse BC, the double Cone is formed
This solid consists two cones namely BAA' and, CA A'
AO or A'O is the common radius.
Height of the cone CAA' is co and slant height is 4 cm.
Height of the Cone BAA' is BO and slant height is 3 cm.
Now \(\Delta\)AOB is similar to \(\Delta\)CAB
Corresponding sides are proportional
\(\text { i.e. }, \frac{A O}{A C}=\frac{A B}{B C}=\frac{B O}{A B} \)
\(\Rightarrow \frac{A O}{4}=\frac{3}{5} \Rightarrow A O=\frac{12}{5} \mathrm{~cm} \)
\(\text { Similarly, } \frac{\mathrm{BO}}{3}=\frac{3}{5} \Rightarrow \mathrm{BO}=\frac{9}{5} \mathrm{~cm}\)
\(\mathrm{CO}=\mathrm{BC}-\mathrm{BO}\)
\(\therefore \text { Volume of double Cone }\)
\(v=\frac{12}{5}, h_{1}=B O= \frac{9}{5}, h_{2}=C O=\frac{16}{5} \)
\(=\left(\frac{1}{3} \pi r^{2} \times B O\right)+\left(\frac{1}{3} \pi r^{2} \times C O\right) \)
\(=\frac{1}{3} \pi r^{2}(B O+C O) \)
\(=\frac{22}{7 \times 3}\left(\frac{12}{5}\right)^{2} \cdot\left(\frac{9}{5}+\frac{16}{5}\right) \)
\(=\frac{22}{7 \times 3} \times \frac{12}{5} \times \frac{12}{5} \times 5 \)
\(=30.17 \mathrm{~cm}^{3}\)

radius of cone r = 5 cm, height h = 8 cm
\(\text { volume of cone }=\frac{1}{3} \pi r^{2} h \text { cubic units }\)
\(=\frac{1}{3} \pi(5)^{2}(8)=\frac{200 \pi}{3} \mathrm{~cm}^{3}\)
Given that the cone is filled to the brim. when read shots are dropped, one fourth of the water flown out. The volume of water flown out
\(=\frac{1}{4} \times \frac{200 \pi}{3}=\frac{50 \pi}{3} \mathrm{~cm}^{3}\)
\(\text { Volume of lead shot }=\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi\left(\frac{1}{2}\right)^{3}\)
\(=\frac{\pi}{6} \mathrm{~cm}^{3}\)
\(\therefore \text { Number of lead shots dropped into the vessel }\)
\(=\frac{50 \pi / 3}{\pi / 6}=100\)

\(\text { Radius of hemispherical bowl }=\frac{36}{2}=18 \mathrm{~cm}\)
\(\text { Volume of hemispherical bowl }=\frac{2}{3} \pi r^{3} \text { cubic units }\)
\(=\frac{2}{3} \pi(18)^{3} \mathrm{~cm}^{3}\)
Height of cylindrical bottle = 6 cm
Radius of cylindrical bottle = 3 cm
\(\text { Volume of cylindrical bottle }=\pi r^{2} h \text { cu. units }\)
\(=\pi(3)^{2}(6)\)
\(\therefore \text { Number of bottles required }=\)  \(\frac{Volume \ of \ hemispherical\ bowl }{Volume\ of\ a\ bottle}\)
\(\therefore \text { Number of bottles required }=\frac{\frac{2}{3} \pi(18)^{3}}{\pi(3)^{2}(6)}=72\)

radius of hemispherical ends
\(=\frac{1}{2} \times 0.7 \)
\(=\frac{0.7}{2} m=\frac{7}{20} m \)
Total length of Solid = 2.7 m
Volume of two hemispheres
\(=2\left(\frac{2}{3} \pi r^{3}\right) \mathrm{cu} \text {. units } \)
\(=\frac{4}{3} \times \frac{22}{7} \times\left(\frac{7}{20}\right)^{3} \)
\(=0.1797 \mathrm{~m}^{3} \)
\(\text { Volume of cylinder }=\pi r^{2} h\)
\(=\frac{22}{7} \times\left(\frac{7}{20}\right)^{2} \times 2 \)
\(=0.77 \mathrm{~m}^{3} \)
\(\therefore \text { Volume of solid }=0.1797+0.77\)
\(=0.95 \mathrm{~m}^{3}(\text { app })\) 

Volume of Cylindrical Portion
\(=\pi r^{2} h \)
\(=\frac{22}{7} \times\left(\frac{21}{2}\right)^{2} \times 18 \)
\(=6237 \mathrm{~cm}^{3} \)
Volume of two Conical ends
\(=2\left(\frac{1}{3} \pi r^{2} h\right) \)
\(=\frac{2}{3} \times \frac{22}{7} \times\left(\frac{21}{2}\right)^{2} \times 9 \mathrm{~cm}^{3} \)
\(=\frac{174636}{84} \)
\(=2079 \mathrm{~cm}^{3} \)
capacity of the tank = Volume of Cylinder + Volume of 2 Cones
= 6237 + 2079
= 8316 cm³

\(\text { Radius of hemisphere ' } r \text { ' }=\frac{7}{2}=3.5 \mathrm{~cm}\)
\(\text { Radius of base of cone }=3.5 \mathrm{~cm}\)
\(\text { Total height of toy }=14.5 \mathrm{~cm}\)
\(\text { Height of conical part }=14.5-3.5=11 \mathrm{~cm}\)
\(\text { Slant height of cone } l=\sqrt{h^{2}+r^{2}}\)
\(=\sqrt{11^{2}+3.5^{2}} \)
\(=\sqrt{121+12.25} \)
\(=\sqrt{133.25}=11.54 \mathrm{~cm} \)
Volume of toy = Volume of cone + Volume of hemisphere
\(=\frac{1}{3} \pi r^{2} h+\frac{2}{3} \pi r^{3} \)
\(=\frac{\pi r^{2}}{3}(h+2 r) \)
\(=\frac{22}{3 \times 7} \times 12.25 \times 18 \)
\(=231 \mathrm{~cm}^{3} \)
Total surface area of toy = C.S.A of cone + C.S.A of hemisphere
\(=\pi r l+2 \pi r^{2}=\pi \mathrm{r}(l+2 \mathrm{r}) \)
\(=\frac{22}{7} \times 3.5(11.54+2 \times 3.5) \)
\(=22 \times 0.5 \times 18.54=203.94 \mathrm{~cm}^{2} \)

Volume of the original cone OAB
\(=\frac{1}{3} \pi R^{2} H \)
\(=\frac{1}{3} \pi\left(\mathrm{R}^{2}\right)(30)=10 \pi R^{2} \mathrm{~cm}^{3} \)
Volume of small cone OCD
\(=\frac{1}{3} \pi r^{2} h \text { cubic units. }\)
Given Volume of cone OCD
\(=\frac{1}{27}(\text { Volume of cone OAB) }\)
\(\frac{1}{3} \pi r^{2} h =\frac{1}{27}\left(10 \pi R^{2}\right) \)
\(h =\frac{10 \pi R^{2}}{27}\left(\frac{3}{\pi r^{2}}\right) \)
\(=\frac{10}{9}\left(\frac{R}{r}\right)^{2} \)
From similar triangles OQB and OPD
\(\text { We get } \frac{Q B}{P D}=\frac{O Q}{O P}=\frac{30}{h}\)
\(\Rightarrow \frac{R}{r}=\frac{30}{h}\)
\(\text { Substituting (2) in (1), }\)
\(h=\frac{10}{9}\left(\frac{30}{h}\right)^{2}\)
\(\mathrm{h}^{3}=1000\)
h = 10 cm
Hence, at (30 - 10) = 20 cm above the base, the section is made

Internal radius = 'r' m = 1 m
External radius = 'R' m = 1 + 0.01 = 1.01 m.
volume of iron used = External volume -Internal volume
\(=\frac{2}{3} \pi\left(R^{3}-r^{3}\right) \text { cubic units } \)
\(=\frac{2}{3} \pi\left((1.01)^{3}-1^{3}\right) \)
\(=\frac{2}{3} \times \frac{22}{7} \times 0.030301 \)
\(=0.06348 \mathrm{~m}^{3}(\mathrm{app}) \)

radius = r,
slant height = I
\(\text { and } l^{2}=\mathrm{r}^{2}+\mathrm{h}^{2}\)
\(=r^{2}+24^{2}=r^{2}+576\)
\(\text { C.S.A }=\pi r l\)
\(=\frac{22}{7} \times r \times \sqrt{r^{2}+576} \mathrm{~cm}^{2}\)
\(\frac{22}{7} \times r \times \sqrt{r^{2}+576} =550 \)
\(r \sqrt{r^{2}+576} =175 \)
\(\text { Squaring both sides }\)
\(r^{2}\left(r^{2}+576\right) =(175)^{2} \)
\(\Rightarrow r^{4}+576 r^{2}-(175)^{2} =0 \)
\(\Rightarrow \left(r^{2}-49\right)\left(r^{2}+625\right) =0 \)
\(r^{2}+625 \neq 0, \ \therefore r^{2}-49 =0 \)
\(r^{2} =49 \)
\(r =7 \mathrm{~cm} . \)
\(\therefore \text { Volume } =\frac{1}{3} \pi r^{2} h \text { cubic units. } \)
\(=\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24 \)
\(=1232 \mathrm{~cm}^{3} \)

\(\text { radius }=\mathrm{r} \text {, height }=\mathrm{h}\)
\(\text { T.S.A }=2 \pi r(h+r) \text { sq. units }\)
\(\text { Given, } 2 \pi r(h+r)=462\)
\(r(h+r)=\frac{462 \times 7}{2 \times 22}=\frac{147}{2}\)
\(\text { C.S.A }=2 \pi r h \text { sq. units }\)
\(\text { Given, } 2 \pi r h=\frac{1}{3}(2 \pi r(h+r))\)
\(\Rightarrow \mathrm{h}=\frac{h+r}{3} \Rightarrow \mathrm{r}=2 \mathrm{~h}\)
\(\text { Substituting (2) in (1) }\)
\(2 h(h+2 h) =\frac{147}{2} \)
\(6 h^{2} =\frac{147}{2} \)
\(h^{2} =\frac{49}{4} \Rightarrow h=\frac{7}{2} \)
\(\therefore \mathrm{r} =2 \mathrm{~h}=2\left(\frac{7}{2}\right)=7 \mathrm{~cm} \)
\(\therefore \text { Volume } =\pi r^{2} h=\frac{22}{7} \times(7)^{2}\left(\frac{7}{2}\right) \)
\(=539 \mathrm{~cm}^{3} . \)

Let the diameter be 'x' units
\(\text { radius }=\frac{x}{2} \text { units }\)
\(\therefore \text { C.S.A of sphere }=4 \pi r^{2} \text { sq. units. }\)
\(=4 \pi\left(\frac{x}{2}\right)^{2} \)
\(=4 \pi\left(\frac{x^{2}}{4}\right)=\pi x^{2} \)
Given that diameter is decreased by 25%.
New diameter = x - 25 % of x
\(=x\left(1-\frac{25}{100}\right)=\frac{3}{4} x\)
\(\text { New Radius }=\frac{3 x}{8}\)
\(\text { C.S.A of new sphere }=4 \pi\left(\frac{3 x}{8}\right)^{2}=\frac{9 \pi x^{2}}{16}\)
\(\text { Decrease in C.S.A }=\pi x^{2}-\frac{9 \pi x^{2}}{16}=\frac{7 \pi x^{2}}{16}\)
\(\text { Hence, percentage in decrease }\)
\(= \frac{7 \pi x^{2}}{16} \times 100 \% \)
\(= 43.75 \% \)

Let r₁, be the radius and I₁, be the slant height of first cone and r₂, be the radius and l₂ be the slant height of the second cone.
\(\text { C.S.A of I cone } =\pi r_{1} l_{1} \)
\(\text { C.S.A of II cone } =\pi r_{2} l_{2} \)
\(\text { Given } \pi r_{1} l_{1} =2 \pi r_{2} l_{2} \)
\(\Rightarrow r_{1} l_{1} =2 r_{2} l_{2} \)
\(\text { and } r_{1} l_{1} =2 r_{2}\left(2 l_{1}\right) \)
\(r_{1} l_{1} =4 r_{2} l_{1} \)
\(r_{1} =4 r_{2} \)
\(\frac{r_{1}}{r_{2}} =\frac{4}{1} \)
\(\text { Hence, the ratio of their radii is } 4: 1 \text {. }\)

Let the radius of the base be 'r' cm.
\(\text { Circumference } =2 \pi r \)
\(2 \pi r =1.5 \mathrm{~cm} \)
\(r =\frac{1.5}{2 \pi}=\frac{10.5}{44} \mathrm{~cm} \)
\(\text { Curved surface area of pencil }\)
\(=2 \pi r h \)
\(=2 \times \frac{22}{7} \times \frac{10.5}{44} \times 25 \)
\(\text { Cost of colouring }100 \mathrm{~cm}^{2}=\text { र. } 0.05\)
\([\because 1 \text {. sq. } \mathrm{dm}=100 \mathrm{sq} \cdot \mathrm{cm}]\)
\(\text { Cost of colouring }=\frac{0.05}{100} \times 2 \times \frac{22}{7} \times \frac{10.5}{44} \times 25\)
\(=Rs \frac{3}{160}\)
\(\text { Hence, cost of colouring } 1,20,000 \text { pencils }\)
\(=\frac{3}{160} \times 120000=\text { Rs } 2250\)