Tag: maths

Let R be the radius and H be the height of the cylinder and let rand h be the radius and height of the cone respectively. Then,
$3r=2R$
and $H:h=4:3$ ......(i)
$\Rightarrow \dfrac {H}{h}=\dfrac {4}{3}$
$\Rightarrow 3H=4h$ .......(ii)
Let n be the required number of cones which can be made from the materials of the cylinder. Then, the volume of the cylinder will be equal to the sum of the volumes of n cones. Hence, we have
$\pi R^2H=\dfrac {n}{3}\pi r^2h$
$\Rightarrow 3R^2H=nr^2h$
$\Rightarrow n=\dfrac {3R^2H}{r^2H}=\dfrac {3\times \dfrac {9r^2}{4}\times \dfrac {4h}{3}}{r^2h}$ [$\because$ From (i) and (ii), $R=\dfrac {3r}{2}$ and $H=\dfrac {4h}{3}$]
$\Rightarrow n=\dfrac {3\times 9\times 4}{3\times 4}$
$\Rightarrow n=9$
Hence, the required number of cones is 9.

Total surface area of the cube after hemispherical depression
$=$ T.S.A. of cube -Base area of hemisphere + C.S.A of hemisphere
$=6(edge)^2-\pi r^2+2\pi r^2$

Since the thickness of wall and brick is same, we only need to figure out the area of the wall, which on dividing with the area of brick will give us the number of bricks.

Radius of the bowl $ = \dfrac {36}{2} = 18  cm $

Volume of the bowl $ = \dfrac { 2 }{ 3 }

\pi { r }^{ 3 } = \dfrac {2}{3} \times \pi \times 18 \times 18 \times

18 {cm}^{3} $

Volume

of a Cylinder of Radius "R" and height "h" $ = \pi { R }^{

2 }h $





Hence, Volume of one cylindrical bottle, $ = \pi \times 3 \times 3 \times  6 $

Hence, number of bottled required $

= \dfrac {Volume  of  bowl} {Volume  of  each  bottle} = \dfrac{\dfrac {2}{3} \times \pi \times 18 \times 18 \times

18}{ \pi \times 3 \times 3 \times 

6} = 72 $

Volume of sphere = (4/3) * pi * r^3 = (4/3) * pi * 2^3 = (32/3) * pi. Volume of water rise in cylinder = pi * R^2 * h = pi * 4^2 * h = 16 * pi * h. Equating: (32/3) * pi = 16 * pi * h, so h = 32 / (3 * 16) = 2/3 cm.

Assuming '210m³' of sand, volume = 210 m³. Tank base area = 12 × 5 = 60 m². Rise in water level = 210/60 = 3.5 m. This matches option A. Options B and C are incorrect calculations.