Given: The volume of a right circular cone of height $9\ cm$ is $48 \pi\ cm^3$.To do:We have to find the diameter of its base.Solution:Height of the cone $h = 9\ cm$Volume of the right circular cone $=48 \pi  cm^{3}$This implies, $\frac{1}{3} \pi r^{2} h =48 \pi$ $\frac{1}{3} \pi r^{2} \times 9 ... Read More

 Given:A conical pit of top diameter $3.5\ m$ is $12\ m$ deep. To do:We have to find its capacity in kilolitres.Solution:Diameter of the top of the conical pit $= 3.5\ m$This implies, Radius of the top of the pit $(r)=\frac{3.5}{2}$$=1.75 \mathrm{~m}$Depth of the pit $(h)=12 \mathrm{~m}$Therefore, Volume of the pit $=\frac{1}{3} ... Read More

Given:A right triangle \( \mathrm{ABC} \) with sides \( 5 \mathrm{~cm}, 12 \mathrm{~cm} \) and \( 13 \mathrm{~cm} \) is revolved about the side \( 12 \mathrm{~cm} \).To do:We have to find the volume of the solid so obtained.Solution:Let in a triangle $ABC$, $AB=13\ cm$$BC=5\ cm$$CA=12\ cm$On revolving the right ... Read More

Given:A right triangle \( \mathrm{ABC} \) with sides \( 5 \mathrm{~cm}, 12 \mathrm{~cm} \) and \( 13 \mathrm{~cm} \) is revolved about the side \( 5 \mathrm{~cm} \).To do:We have to find the ratio of the volumes of the two solids obtained.Solution:Let in a triangle $ABC$, $AB=13\ cm$$BC=5\ cm$$CA=12\ cm$On ... Read More

 Given:The radius of a hemisphere is $10\ cm$.To do:We have to find the total surface area of the hemisphere.Solution:Radius of the hemisphere $(r)= 10\ cm$Therefore, The total surface area of the hemisphere $= 2\pi r^2$$= 2 \times 3.14 \times 10 \times 10$$= 628\ cm^2$Hence, the total surface area of the ... Read More

 Given:The radius of a spherical balloon increases from \( 7 \mathrm{~cm} \) to \( 14 \mathrm{~cm} \) as air is being pumped into it. To do:We have to find the ratio of surface areas of the balloon in the two cases.Solution:Let the initial radius of the spherical balloon be $r_1$ and ... Read More

 Given:A hemispherical bowl made of brass has an inner diameter of $10.5\ cm$.Rate of tin-plating is $Rs.\ 16$ per $100\ cm^2$.To do:We have to find the cost of tin-plating it on the inside.Solution:Inner diameter of the hemispherical bowl $= 10.5\ cm$This implies, Radius of the bowl $(r)=\frac{10.5}{2}$$=5.25$$=\frac{525}{100}$$=\frac{21}{4} \mathrm{~cm}$Therefore, The surface ... Read More

Given: The surface area of a sphere is $154\ cm^{2}$.To do: We have to find the radius of the sphere.Solution:Let $r$ be the radius of the sphere.Therefore, Surface area of the sphere$=4\pi r^2$$=154$This implies, $r^2=\frac{154}{4\pi}$$r^2=\frac{154}{4\times\frac{22}{7}}$$r^2=\frac{154\times7}{4\times22}$$r^2=\frac{49}{4}$$r^2=\frac{7^2}{2^2}$$r=\frac{7}{2}$$r=3.5\ cm$Therefore, the radius of the sphere is $3.5\ cm$.Read More

 Given:A hemispherical bowl is made of steel $0.25\ cm$ thick. The inside radius of the bowl is $5\ cm$.To do:We have to find the volume of steel used in making the bowl.Solution:The thickness of steel $= 0.25\ cm$$=\frac{1}{4}\ cm$Inside radius of the bowl $(r) = 5\ cm$This implies, Outside radius ... Read More

Given:A right circular cylinder just encloses a sphere of radius \( r \).To do:We have to find(i) surface area of the sphere, (ii) curved surface area of the cylinder, (iii) ratio of the areas obtained in (i) and (ii).Solution:(i) Surface area of a sphere of radius $r = 4\pi r^2$(ii) ... Read More