Category : 10th Class
Volume and Surface Area of Solids
|
S. no. |
SOLID |
Lateral/ Curved Surface area |
Total Surface Area |
Volume |
|
1. |
Cube \[Each\text{ }side\text{ }=\text{ }a\] |
\[4{{a}^{2}}\] |
\[6{{a}^{2}}\] |
\[{{A}^{3}}\] |
|
2. |
Cuboid \[Length\text{ }=\text{ }l\] \[Breadth\text{ }=\text{ }b\] \[Height\text{ }=\text{ }h\] |
\[2\left( l+b \right)\times h\] |
\[2(lb+bh+lh)\] |
\[(l\times b\times h)\] |
|
3. |
Right Circular Cylinder \[Radius\text{ }of\text{ }base\text{ }=\text{ }r\] \[Height\text{ }=\text{ }h\] |
\[2\pi rh\] |
\[2\pi r\,(h\,+\,r)\] |
\[\pi {{r}^{2}}h\] |
|
4. |
Cone \[Radius\text{ }of\text{ }base\text{ }=\text{ }r\] \[Height\text{ }=\text{ }h\] \[Slant\text{ }height\text{ }=\] \[l\,\,=\,\,\sqrt{{{r}^{2}}\,\,+\,\,{{h}^{2}}}\] |
\[\pi rl\] |
\[\pi r(l+r)\] |
\[\frac{1}{3}\pi {{r}^{2}}h\] |
|
5. |
Sphere \[Radius\text{ }=\text{ }r\] |
__________ |
\[4\pi {{r}^{2}}\] |
\[\frac{4}{3}\pi {{r}^{3}}\]
|
|
6. |
Hemisphere \[Radius\text{ }=\text{ }r\] |
\[2\pi {{r}^{2}}\] |
\[3\pi {{r}^{2}}\] |
\[\frac{2}{3}\pi {{r}^{3}}\]
|
|
7. |
Hollow Cylinder \[Inner\text{ }radius\text{ }=~r\] \[Outer\text{ }radius\text{ }=\text{ }R\] \[Height\text{ }=\text{ }h\] |
\[2\pi h\,(R\,+\,r)\] |
\[2\pi h\,(R\,+\,r)\,+\,\,2\pi ({{R}^{2}}-{{r}^{2}})\] \[Or\,\,\,2\pi \,(R\,+\,r)\,(R+h-r)\] |
\[\pi h\,({{R}^{2}}\,\,+\,\,{{r}^{2}})\] |
|
8. |
Spherical Shell \[Inner\text{ }radius\text{ }=~r\]\[Outer\text{ }radius\text{ }=\text{ }R\] |
\[2\pi ({{R}^{2}}+{{r}^{2}})\] |
\[\pi (3{{R}^{2}}\,\,+\,\,{{r}^{2}})\] |
\[\frac{4}{3}\pi ({{R}^{3}}-{{r}^{3}})\] |
|
9. |
Frustum of a cone \[Radius\text{ }of\text{ }base\text{ }=\text{ }R\] \[Radius\text{ }of\text{ }top\text{ }=\text{ }r\] \[Height\text{ }=\text{ }h\] \[Slant\text{ }height\text{ }=\text{ }l\] \[{{L}^{2}}=\text{ }{{h}^{2}}+\text{ }{{\left( R\text{ }-\text{ }r \right)}^{2}}\] |
\[\pi l\,(R\,\,+\,\,r)\] |
\[\pi [{{R}^{2}}+{{r}^{2}}+l\,(R+r)]\] |
\[\frac{\pi h}{3}[{{R}^{2}}+{{r}^{2}}+Rr]\] |
Snap Test
(a) \[350\text{ }c{{m}^{2}}\]
(b) \[355\text{ }c{{m}^{2}}\]
(c) \[340\text{ }c{{m}^{2}}\]
(d) \[345\text{ }c{{m}^{2}}\]
(e) None of these
Ans. (a)
Explanation: \[Length\text{ }of\text{ }the\text{ }cuboid\text{ }=\text{ }15\text{ }cm,\text{ }breadth\text{ }=\text{ }5\text{ }cm,\text{ }height\text{ }=\text{ }5\text{ }cm\]
\[\therefore \] Surface area of the cuboid \[=\text{ }2\text{ }\left( Ib\text{ }+\text{ }bh\text{ }+\text{ }Ih \right)\]
= \[2\text{ }\left[ 15\text{ }\times \text{ }5\text{ }+\text{ }5\text{ }\times \text{ }5\text{ }+\text{ }1\text{ }5\text{ }\times \text{ }5 \right]\text{ }c{{m}^{2}}\]
\[=\,\,\,350\text{ }c{{m}^{2}}.\]
(a) \[2\text{ }:\text{ }2\] (b) \[3\text{ }:\text{ }1\]
(c) \[2\text{ }:\text{ }1\] (d) \[4\text{ }:\text{ }5\]
(e) None of these
Ans. (c)
Explanation: Area of base of the cone = radius of base of the hemisphere
\[\Rightarrow \,\,p{{r}^{2}}\,\,=\,p{{R}^{2}}\,\,\,\,\,\,\,\,\,\Rightarrow \,\,r=R\]
Thus, the radius of base of the cone = radius of base of the hemisphere = r.
Now, let h be the height of the cone.
Then, volume of the cone = volume of the hemisphere
\[\Rightarrow \,\,\,\frac{1}{3}\pi {{r}^{2}}h\,\,\,=\,\,\,\frac{2}{3}\pi {{r}^{3}}\,\,\,\Rightarrow \,\,\,\frac{h}{2}\,\,=\,\,\frac{2}{1}\,\,\,\Rightarrow \,\,\,\frac{height\,of\,the\,cone}{height\,of\,the\,hemisphere}\,\,\,=\,\,\,\frac{2}{1}\]
\[\therefore \] \[Ratio\text{ }of\text{ }their\text{ }height\text{ }=\text{ }2\text{ }:\text{ }1.\]
(a) \[450\text{ }c{{m}^{2~~~~~~~~~~~~~~~~~~~~~~~~}}\]
(b) \[440\text{ }c{{m}^{2}}\]
(c) \[420\text{ }c{{m}^{2}}\]
(d) \[430\text{ }c{{m}^{2}}\]
(e) None of these
Ans. (b)
Explanation: Radius of the base of the cylinder \[r\text{ }=\text{ }5\text{ }cm\]. Let its height be h cm.
\[Then,\text{ }volume\text{ }of\text{ }the\text{ }cylinder\text{ }=\text{ }1100\text{ }c{{m}^{3}}\]
\[\Rightarrow {{r}^{2}}h\,\,=\,1100\,\,\Rightarrow \,\,\frac{22}{7}\times 5\times 5\times h\,\,=\,\,1100\,\,\Rightarrow \,\,h\,\,=\,\,\frac{1100\times 7}{22\times 5\times 5}\,\,=\,\,14\,cm.\]
\[\therefore \] Curved surface area of the cylinder \[=\,2\pi rh\,\,=\,\,\left( 2\times \frac{22}{7}\times 5\times 14 \right)\,c{{m}^{2}}\,=\,\,440\,c{{m}^{2}}\]
(a) \[154\text{ }c{{m}^{2}}\]
(b) \[270\text{ }c{{m}^{2}}\]
(c) \[550\text{ }c{{m}^{2}}\]
(d) \[740\text{ }c{{m}^{2}}\]
(e) None of these
Ans. (c)
Explanation: Height of the cone \[h\text{ }=\text{ }24\text{ }cm.\]
Let radius of the base of the cone be r cm
Then, volume of the cone = \[\frac{1}{3}\pi {{r}^{2}}h\,\,=\,\,1232\,c{{m}^{2}}\]
\[\Rightarrow \,\,\,\,\,\frac{1}{3}\times \frac{22}{7}\times {{r}^{2}}\times 24\,\,=\,\,1232\,c{{m}^{2}}\]
\[\Rightarrow \,\,\,\,\,{{r}^{2}}\,\,=\,\,\frac{1232\,\,\times \,3\,\,\times \,\,7}{22\,\,\times \,\,24}\,\,\,\Rightarrow \,\,r=7\,cm\]
So, Slant height \[l\,\,=\,\,\sqrt{{{h}^{2}}+{{r}^{2}}}\,\,=\,\,\sqrt{{{24}^{2}}+{{7}^{2}}}\,\,=\,\,25\,cm\]
Therefore curved surface area \[=\,\,\pi rl\,\,~=\text{ }550\text{ }c{{m}^{2}}.\]
(a) \[\frac{4}{3}\] (b) \[\frac{r}{3}\]
(c) \[2r\] (d) \[\frac{2}{3}\,r\]
(e) None of these
Ans. (b)
Explanation: Volume of sphere \[=\,\,\frac{4}{3}\pi r3\,\,=\,\,\frac{r}{3}(4\pi r2)\,\,\frac{r}{3}\times (surface\,\,area)\]
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