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Q1. |
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Ans. |
 \begin{align*} \text{Given that:}
& \text {Volume of each cube }= 64 \text { cm}^{3}\\
\text {Solution :}
& \text {Side of cube } = a \text { cm}\\
& \text {We know that -}\\
& \text {Volume of cube }= (Side)^{3}\\
& 64 = a^{3}\\
& \text {or, } \sqrt[3]{64} = a\\
& \text {or, } a = 4 \text { cm}\\
& \text {When we join 2 cube it become a cuboid, then}\\
& \text {Length of Cuboid }=l = 8 \text { cm}\\
& \text {Height of Cuboid }=h = 4 \text { cm}\\
& \text {Breadth of Cuboid }=b = 4 \text { cm}\\
& \text {As we now TSA of cuboid} = 2(lb + bl + bh)\\
& \text {Putting value}\\
& \Rightarrow 2(4\times 4 +4\times 8 + 8 \times 4)\\
& \Rightarrow 2 (16+32+32)\\
& \Rightarrow 2 \times 80 \\
& \Rightarrow 160 \text { cm}^{2} \\
\end{align*}
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Q2. | A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel. |
Ans. |  \begin{align*} \text{Given that:} & \text {Diameter of Hemisphere }= 14 \text { cm}\\ & \text {Height of Hemisphere}=\text {Radious of Hemisphere}\\ & \Rightarrow \frac {Diameter}{2}=\frac {14}{2}=7 \text { cm}\\ & \text {Height of vessel }= 13 \text { cm}\\ & \therefore \text {Height of cylinder (h)}=\\ & \text {Height of vessel}-\text {Height of hemisphere}\\ & \Rightarrow h = 13-7 = 6 \text { cm}\\ & \text {Inner surface are of vessel}=\\ & \text {CSA of Hemisphere } + \text { CSA of cylinder}\\ & \Rightarrow 2\pi r^{2}+2\pi rh\\ & \Rightarrow 2\pi r (r+h)\\ & \text {Putting value -}\\ & \Rightarrow 2\times \frac {22}{7}\times 7 (7+6)\\ & \Rightarrow 2\times \frac {22}{7}\times 7 (13)\\ & \Rightarrow 2\times 22\times 13\\ & \Rightarrow 572 \text { cm}^{2}\\ \end{align*} |
Q3. | A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy. |
Ans. |  \begin{align*} \text{Given that:} &\text {Radius of Cone }=3.5 \text { cm}\text {Let, }r\\ & \text {Total height of Toy }=15.5 \text { cm}\text {Let, }H\\ & \text {Radius of Cone }=\text {Radius of Hemisphere }=3.5 \text { cm}\\ & \text {Let, } h = \text {Height of cone}\\ & \text {Let, Slant height of cone } = l \text { cm}\\ \text {Solution: } & \text {As per construction -}\\ & \Rightarrow AO = h \text {and}\\ & \Rightarrow h = \text {Total height of toy}-\text {Radius of Hemisphere}\\ & \Rightarrow 15.5-3.5 = 12 \text { cm}\\ & \text {In }\triangle AOB,\\ & \text {Using Pythagoras Theorem}\\ & \Rightarrow (AB)^{2}=(OA)^{2}+(OB)^{2}\\ & \text {or, } l^{2}=r^{2}+h^{2}\\ & \Rightarrow l = \sqrt {r^{2}+h^{2}}\\ & \Rightarrow l = \sqrt {(3.5)^{2}+(12)^{2}}\\ & \Rightarrow l = \sqrt {12.25+144}\\ & \Rightarrow l = \sqrt {156.25}\\ & \Rightarrow l = 12.5 \text { cm}\\ & \text {TSA of toy}=\text {SA of cone}+\text {SA of Hemisphere}\\ & \Rightarrow \pi r l + 2 \pi r^{2}\\ & \Rightarrow \pi r (l + 2 r)\\ & \Rightarrow \frac {22}{7}\times 3.5 \left(12.5+2\times 3.5\right)\\ & \Rightarrow \frac {22}{7}\times 3.5 \left(12.5+7\right)\\ & \Rightarrow \frac {22}{7}\times 3.5 \left(19.5\right)\\ & \Rightarrow \frac {22}{7}\times 68.25\\ & \Rightarrow 214.5 \text { cm}^{2}\\ \end{align*} |
Q4. | A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid. |
Ans. |  \begin{align*} \text{Given that : }
&\text {Side of Cubical block }=7 \text { cm}\text { Let, } a\\
\text {Solution : }
& \text {As per construction-}\\
& \text {Greatest Diameter (D) of Hemisphere}=\text {Side of Cube (a)} = 7 \text { cm}\\
& \text {So, Radius } = \frac {Diameter}{2}=\frac {7}{2}=3.5 \text { cm}\\
& \text {SA of Solid so formed}\\
& \Rightarrow TSA_{\text {Cubical Block}} +CSA_{Hemisphere}-AREA_{\text {Hemisphere Base}}\\
& \Rightarrow 6a^{2}+\left(2\pi r^{2}-\pi r^{2}\right)\\
& \Rightarrow 6a^{2}+\pi r^{2}\\
& \text {Putting value}\\
& \Rightarrow \left(6\times{7}^{2}\right)+\left(\frac {22}{7}\times(3.5)^{2}\right)\\
& \Rightarrow 294+\left(\frac {22}{7}\times12.25\right)\\
& \Rightarrow 294+38.5\\
& \Rightarrow 332.5 \text { cm}^{2}\\
\end{align*} |
Q5. | A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter \(l\) of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid. |
Ans. | \begin{align*} \text{Given that : }
&\text {Because, Hemisphere cut from a cubical box}\\
&\text {which edge }= l\\
&\text {So, we can say that diameter of hemisphere}=l\\
\text {Solution : }
& \text {We know that- }\\
& \text {SA of a cube =} 6(side)^{2} = 6l^{2}\\
& \text {Let, Radius of hemisphere }= r\\
& \text {Radius }= \frac {l}{2}\\
& \text {SA of Remaining Solid}=\\
& \text {SA of Cubical box}-\text {SA of Hemisphere}\\
& \Rightarrow 6l^{2}-\pi r^{2}+2 \pi r^{2}\\
& \Rightarrow 6l^{2}-\pi \left(\frac {l}{2}\right)^{2}+2\pi \left(\frac {l}{2}\right)^{2}\\
& \Rightarrow 6l^{2}- \frac {\pi l^{2}}{4}+\frac {\pi l^{2}}{2}\\
& \Rightarrow \frac {24l^{2}-\pi l^{2}+2\pi l^{2}}{4}\\
& \Rightarrow \frac {24l^{2}+\pi l^{2}}{4}\\
& \frac {l^{2}}{4}\left(24+\pi\right) \text {Square Units}
\end{align*} |
Q6. | A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig. 12.10). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.  |
Ans. |  \begin{align*} \text{Given that : }
& \text {Length of Capsule, Let } l=14 \text { mm}\\
& \text {Diameter of Capsule } = \text {Diameter of cylinder}= 5 \text { mm} \\
& \text {Radius of Hemisphere}=\text {Radius of Cylinder}\\
& =\frac {Diameter}{2}=\frac {5}{2}= 2.5 \text { mm}\\
& \text {Length of cylinder }=MN \\
& = \text {Total length of capsule}-\text {Radius of Right Hemisphere}\\
& -\text {Radius of left Hemisphere}\\
& \Rightarrow 14-2.5-2.5 = 9 \text { mm}\\
& \text {TSA of Capsule}=\\
& =\text {CSA of Cylinder}+\text {SA of Right Hemisphere}\\
& +\text {SA of Left Hemisphere}\\
& \Rightarrow 2\pi rl +2\pi r^{2} + 2\pi r^{2}\\
& \Rightarrow 2\pi rl +4\pi r^{2}\\
& \Rightarrow 2\pi r \left( l + 2r\right) \\
& \text {Putting value}\\
& \Rightarrow 2\times \frac {22}{7}\times 2.5 \left(9+2\times 2.5\right) \\
& \Rightarrow 2\times \frac {22}{7}\times 2.5 \times 14 \\
& \Rightarrow 2\times 22\times 2.5 \times 2\\
& \Rightarrow 220 \text { mm}^{2}
\end{align*} |
Q7. | A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs. \(500 \text {per m}^{2}\). (Note that the base of the tent will not be covered with canvas.) |
Ans. |  \begin{align*} \text{Given that : }
& \text {Height of Cylinderical part, Let } h=2.1 \text { m}\\
& \text {Diameter of Cylinderical part } = 4 \text { m}\\
& \text {So, Radius of Cylinderical part } = \frac {4}{2}=2 \text { m}\\
& \text {Slant height of Conical part, Let } l = 2.8 \text { m}\\
& \text {Cost of } 1 \text {m}^{2} \text {Canvas}= Rs. 500\\
\text {Solution : }& \text {So, Total area of canvas used }\\
& = \text {CSA of Conical part}+\text {CSA of Cylinderical part}\\
& \Rightarrow \pi rl + 2\pi rh\\
& \Rightarrow \pi r (l+2h)\\
& \text {Putting value -}\\
& \Rightarrow \frac {22}{7}\times 2 \left(2.8+2\times 2.1\right)\\
& \Rightarrow \frac {22}{7}\times 2\times 7\\
& \Rightarrow 44 \text { m}^{2}\\
& \text {So, Cost of } 44 \text { m}^{2} \text {Canvas}\\
& = 44 \times 500 = Rs. 22000 \\
\end{align*} |
Q8. | From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest \(\text {cm}^{2}\). |
Ans. |  \begin{align*} \text{Given that : }
& \text {Height of cylinder } Let, h = 2.4 \text { cm}\\
& \text {Diameter of cylinder }= 1.4 \text { cm}\\
& \text {Radius of Cylinder } r = \frac {1.4}{2}=0.7 \text { cm}\\
& \text {Radius of cylinder = Radius of Cone}\\
& \text {Let, Slant height of Cone } = l \text { cm}\\
& \text {Using Pythagoras Theorem -}\\
& l^{2} = h^{2}+r^{2}\\
& l = \sqrt {h^{2}+r^{2}}\\
& l = \sqrt {(2.4)^{2}+(0.7)^{2}}\\
& l = \sqrt {5.76+0.49}\\
& l = \sqrt {6.25} = 2.5 \text { cm}\\
& \text {SA of Remaining Sold}\\
& = \text {SA of Cylinder}+\text {Inner SA of Hollow cone}\\
& \Rightarrow \left(2 \pi rh + \pi r^{2}\right)+\pi rl\\
& \Rightarrow \pi r \left(2h+r+l\right)\\
& \Rightarrow \frac {22}{7}\times 0.7 \left(2\times 2.4 + 0.7+2.5\right)\\
& \Rightarrow \frac {22}{7}\times 0.7 \left(8\right)\\
& \Rightarrow \frac {22}{7}\times 5.6\\
& \Rightarrow 22\times 0.8\\
& \Rightarrow 17.6 \text { cm}^{2}\\
\\ \end{align*} |
Q9. | A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 12.11. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.  |
Ans. | \begin{align*} \text{Given that : }
& \text {Height of cylinder }= 10 \text { cm}\\
& \text {Radius of base } = 3.5 \text { cm}\\
& \text {As per construction, TSA of the article}\\
& = \text {CSA of the cylinder} + \text {2 SA of a hemisphere}\\
& \Rightarrow 2 \pi rh +2\pi r^{2}+2\pi r^{2}\\
& \Rightarrow 2 \pi r \left(h+r+r\right)\\
& \Rightarrow 2 \pi r \left(h+2r\right)\\
& \Rightarrow 2\times \frac {22}{7}\times 3.5 \left(10+2\times 3.5\right)\\
& \Rightarrow 2\times \frac {22}{7}\times 3.5 \times 17\\
& \Rightarrow 2\times \frac {22}{7}\times 59.5\\
& \Rightarrow 2\times 22 \times 8.5\\
& \Rightarrow 374 \text { cm}^{2}
\end{align*} |
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