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Volume and Surface Area
1.A right triangle with sides 3 cm, 4 cm and 5 cm is rotated the side of 3 cm to form a cone. The volume of the cone so formed is:

Explanation:

Solution:

2.In a shower, 5 cm of rain falls. The volume of water that falls on 1.5 hectares of ground is:

Explanation:

Solution:
$\begin{array}{rl}& \text{1}\phantom{\rule{thinmathspace}{0ex}}\text{hectare}=10000\phantom{\rule{thinmathspace}{0ex}}{m}^{2}\\ & \text{So,}\phantom{\rule{thinmathspace}{0ex}}\text{Area}=\left(1.5×10000\right){m}^{2}\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=15000\phantom{\rule{thinmathspace}{0ex}}{m}^{2}\\ & \text{Depth}=\frac{5}{100}m=\frac{1}{20}m\\ & \therefore \text{Volume}=\left(\text{Area}\phantom{\rule{thinmathspace}{0ex}}×\phantom{\rule{thinmathspace}{0ex}}\text{Depth}\right)\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\left(15000×\frac{1}{20}\right){m}^{3}\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=750\phantom{\rule{thinmathspace}{0ex}}{m}^{3}\end{array}$

3.A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is:

Explanation:

Solution:
$\begin{array}{rl}& 2\left(15+12\right)×h=2\left(15×12\right)\\ & ⇒h=\frac{180}{27}m=\frac{20}{3}m\\ & \therefore \text{Volume}=\left(15×12×\frac{20}{3}\right){m}^{3}\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=1200\phantom{\rule{thinmathspace}{0ex}}{m}^{3}\end{array}$

4.A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weighs 8 g/cm3, then the weight of the pipe is:

Explanation:

Solution:
$\begin{array}{rl}& \text{External}\phantom{\rule{thinmathspace}{0ex}}\text{radius}=4\phantom{\rule{thinmathspace}{0ex}}cm\\ & \text{Internal}\phantom{\rule{thinmathspace}{0ex}}\text{radius}=3\phantom{\rule{thinmathspace}{0ex}}cm\\ & \text{Volume}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{iron}\\ & =\left(\frac{22}{7}×\left[{\left(4\right)}^{2}-{\left(3\right)}^{2}\right]×21\right)c{m}^{3}\\ & =\left(\frac{22}{7}×7×1×21\right)c{m}^{3}\\ & =462\phantom{\rule{thinmathspace}{0ex}}c{m}^{3}\\ & \therefore \text{Weight}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{iron}=\left(462×8\right)gm\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=3696\phantom{\rule{thinmathspace}{0ex}}gm\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=3.696\phantom{\rule{thinmathspace}{0ex}}kg\end{array}$

5.A boat having a length 3 m and breadth 2 m is floating on a lake. The boat sinks by 1 cm when a man gets on it. The mass of the man is:

Explanation:

Solution:
Volume of water displaced
= (3 x 2 x 0.01) m3
= 0.06 m3.
∴ Mass of man = Volume of water displaced x Density of water
= (0.06 x 1000) kg
= 60 kg.

6.50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the average displacement of water by a man is 4 m3, then the rise in the water level in the tank will be:

Explanation:

Solution:
$\begin{array}{rl}& \text{Total}\phantom{\rule{thinmathspace}{0ex}}\text{Volume}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{water}\phantom{\rule{thinmathspace}{0ex}}\text{displaced}\\ & =\left(4×50\right){m}^{3}=200\phantom{\rule{thinmathspace}{0ex}}{m}^{3}\\ & \therefore \text{Rise}\phantom{\rule{thinmathspace}{0ex}}\text{in}\phantom{\rule{thinmathspace}{0ex}}\text{water}\phantom{\rule{thinmathspace}{0ex}}\text{level}\\ & =\left(\frac{200}{40×20}\right)m\\ & =0.25\phantom{\rule{thinmathspace}{0ex}}m\\ & =25\phantom{\rule{thinmathspace}{0ex}}cm\end{array}$

7.The slant height of a right circular cone is 10 m and its height is 8 m. Find the area of its curved surface.

Explanation:

Solution:
$\begin{array}{rl}& l=10\phantom{\rule{thinmathspace}{0ex}}m\\ & h=8\phantom{\rule{thinmathspace}{0ex}}m\\ & So,\phantom{\rule{thinmathspace}{0ex}}r=\sqrt{{l}^{2}-{h}^{2}}=\sqrt{{\left(10\right)}^{2}-{8}^{2}}=6\phantom{\rule{thinmathspace}{0ex}}m\\ & \therefore \text{Curved}\phantom{\rule{thinmathspace}{0ex}}\text{surface}\phantom{\rule{thinmathspace}{0ex}}\text{area}\\ & \pi \phantom{\rule{thinmathspace}{0ex}}rl=\left(\pi ×6×10\right){m}^{2}\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=60\pi \phantom{\rule{thinmathspace}{0ex}}{m}^{2}\end{array}$

8.A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. From each of its corners, a square is cut off so as to make an open box. If the length of the square is 8 m, the volume of the box (in m3) is:

Explanation:

Solution:
Clearly, l = (48 - 16)m = 32 m,
b = (36 -16)m = 20 m,
h = 8 m.
∴ Volume of the box = (32 x 20 x 8) m3 = 5120 m3.

9.The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3. Find the ratio of its diameter to its height.

Explanation:

Solution:
$\begin{array}{rl}& \frac{\pi \phantom{\rule{thinmathspace}{0ex}}{r}^{2}h}{2\pi \phantom{\rule{thinmathspace}{0ex}}rh}=\frac{924}{264}\phantom{\rule{thinmathspace}{0ex}}\\ & ⇒r=\left(\frac{924}{264}×2\right)=7m\\ & \text{And},\phantom{\rule{thinmathspace}{0ex}}2\pi \phantom{\rule{thinmathspace}{0ex}}fh=264\\ & ⇒h=\left(264×\frac{7}{22}×\frac{1}{2}×\frac{1}{7}\right)=6m\\ & \therefore \text{Required}\phantom{\rule{thinmathspace}{0ex}}\text{ratio}\\ & =\frac{2r}{h}=\frac{14}{6}=7:3\end{array}$

10.A cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.1 m and its walls are 5 cm thick. The thickness of the bottom is:

$\begin{array}{rl}& \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{thickness}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{bottom}\phantom{\rule{thinmathspace}{0ex}}\text{be}\phantom{\rule{thinmathspace}{0ex}}x\phantom{\rule{thinmathspace}{0ex}}cm\\ & \text{Then},\phantom{\rule{thinmathspace}{0ex}}\left[\left(330-10\right)×\left(260-10\right)×\left(110-x\right)\right]=8000×1000\\ & ⇒320×250×\left(110-x\right)=8000×1000\\ & ⇒\left(110-x\right)=\frac{8000×1000}{320×250}=100\\ & ⇒x=10\phantom{\rule{thinmathspace}{0ex}}cm=1\phantom{\rule{thinmathspace}{0ex}}dm\end{array}$