Question
The diameter of a cylinder is 7 cm and its height is 16 cm. Using the value of $ \pi = \dfrac {22} {7}$, the lateral surface area of the cylinder is
Solution
Correct
option
is 352 cm $ ^ 2 $
$\Rightarrow$ Lateral surface area of the cylinder = $2\pi rh$ $ = 2 \times \dfrac{22}{7} \times \dfrac72 \times 16 = 352$ sq.cm
$\Rightarrow$ Lateral surface area of the cylinder = $2\pi rh$ $ = 2 \times \dfrac{22}{7} \times \dfrac72 \times 16 = 352$ sq.cm
Question
From a right circular cylinder of radius 10 cm and height 21 cm, a right circular cone of same base radius is removed. If the volume of the remaining part is 4400 cm $^3 $, then the height of the removed cone is$\left( \space \pi = \dfrac{22}{7}\right)$?
Solution
Correct
option
is 21 cm
$\Rightarrow$ Volume of the cylinder = $\pi r^2 h $ $ = \dfrac{22}{7} \times 10 \times 10 \times 21 = 6600$ cu. cm Volume of cone = $6600 - 4400 = 2200$cu.cm $\therefore 2200 = \dfrac13 \pi r \times 10^2 \times h $ $\Rightarrow 2200 = \dfrac{2200}{21} \times h \Rightarrow h = 21$cm .
$\Rightarrow$ Volume of the cylinder = $\pi r^2 h $ $ = \dfrac{22}{7} \times 10 \times 10 \times 21 = 6600$ cu. cm Volume of cone = $6600 - 4400 = 2200$cu.cm $\therefore 2200 = \dfrac13 \pi r \times 10^2 \times h $ $\Rightarrow 2200 = \dfrac{2200}{21} \times h \Rightarrow h = 21$cm .
Question
The area of an equilateral triangle is $ \dfrac {\sqrt {243}}{4} cm^2 $. The length of each side of the triangle will be.
Solution
Correct
option
is $3$ cm
$\Rightarrow$ Side = $\sqrt{\dfrac{4}{\sqrt3}\times \dfrac{\sqrt{243}}{4}}$ $ = \sqrt{\sqrt{\dfrac{243}{3 }}} = \sqrt9$ $= 3$cm
$\Rightarrow$ Side = $\sqrt{\dfrac{4}{\sqrt3}\times \dfrac{\sqrt{243}}{4}}$ $ = \sqrt{\sqrt{\dfrac{243}{3 }}} = \sqrt9$ $= 3$cm
Question
The diameters of two cylinders, whose volumes are equal, are in the ratio 3 : 2. Their height will be in the ratio?
Solution
Correct
option
is $4 : 9$
$\Rightarrow \dfrac{\pi {R}^2_1H}{\pi {R}^2_2h} =1 $ $\Rightarrow \dfrac{3^2 \times H}{2^2 \times h} =1 $ $\Rightarrow \dfrac{H}{h} = \dfrac49$
$\Rightarrow \dfrac{\pi {R}^2_1H}{\pi {R}^2_2h} =1 $ $\Rightarrow \dfrac{3^2 \times H}{2^2 \times h} =1 $ $\Rightarrow \dfrac{H}{h} = \dfrac49$
Question
Find the length of the largest rod which can be placed in a room 16 m long, 12 m wide and $10\dfrac23$ m high.
Solution
Correct
option
is $ 22 \dfrac23$ m
$\Rightarrow$ Since the room is cuboid shape Length of largest rod = Diagonal of cuboid $= \sqrt{L^2 + b^2 + h^2}$ $= \sqrt{16^2 + 12^2 + \dfrac{32^2}{3^2}}$ $ = \sqrt{256 + 144 + \dfrac{1024}{9}}$ $ = \sqrt{\dfrac{2304 + 1296 + 1024}{9}}$ $ = \sqrt{\dfrac{4624}{9}} = \dfrac{68}{3} = 22 \dfrac23$ m
$\Rightarrow$ Since the room is cuboid shape Length of largest rod = Diagonal of cuboid $= \sqrt{L^2 + b^2 + h^2}$ $= \sqrt{16^2 + 12^2 + \dfrac{32^2}{3^2}}$ $ = \sqrt{256 + 144 + \dfrac{1024}{9}}$ $ = \sqrt{\dfrac{2304 + 1296 + 1024}{9}}$ $ = \sqrt{\dfrac{4624}{9}} = \dfrac{68}{3} = 22 \dfrac23$ m
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