Filter By

## All Questions

#### The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen is used upon writing 3300 words on average. How many words can be written in a bottle of ink containing one-fifth of a litre?

$\\\text{Volume of the barrel}=\pi r^2h=\frac{22}{7}\times(0.25)^2\times7\\=1.375cm^3=\frac{1.375}{1000}L$

3300 words can be written with 0.001375 L of ink

So with 1L of ink 3300/0.001375 words can be written=2400000

So with 1/5th of a litter 2400000/5 words can be written=48000

#### A pen stand made of wood is in the shape of a cuboid with four conical depressions and a cubical depression to hold the pens and pins, respectively. The dimension of the cuboid are 10 cm, 5 cm and 4 cm. The radius of each of the conical depressions is 0.5 cm and the depth is 2.1 cm. The edge of the cubical depression is 3 cm. Find the volume of the wood in the entire stand.

Radius of conical depression = 0.5 cm

Depth = 2.1 cm

$Volume=\frac{1}{3}\pi r^2 h$

$=\frac{1}{3}\times \frac{22}{7} \times (0.5)^2 \times 2.1$

$=0.55cm^3$

$\therefore$the volume of 4 cones $=4 \times 0.55=2.2cm^3$

Edge of cube = 3

The volume of cube     =33    (  Because the volume of cube = a3 )

Length of cuboid = 10 cm

Height = 4 cm

$Volume$ $=l \times b \times h$

$=10 \times 5 \times 4=200 cm^3$

Volume of wood = volume of cuboid – volume of cube - volume of 4 cones

=200 - 2.2 -27 = 170.8 cm3

## Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support

#### The rain water from a roof of dimensions 22 m × 20 m drains into a cylindrical vessel having diameter of base 2 m and height 3.5 m. If the rain water collected from the roof just fill the cylindrical vessel, then find the rainfall in cm.

The radius of the cylindrical vessel

$=\frac{2}{2}=1cm$

Height = 3.5 cm

$\text{ Volume}$ $=\pi r^2 h=\frac{22}{7}\times 1 \times 1 \times \frac{35}{10}=11m^3$

Let the height of rainfall = x

Length = 22m

$Volume= l \times b \times h$

$= 22 \times 20 \times x$

Rainfall = volume of water = volume of the cylindrical vessel

$22 \times 20 \times x=11$

$x=\frac{11}{22 \times 20}=\frac{1}{40}=0.025 m$

x=0.025 m

Or        x= 2.5cm        [Q 1 m = 100 cm]

Hence the rainfall is in 2.5 cm

#### Water flows through a cylindrical pipe, whose inner radius is 1 cm, at the rate of 80 cm/sec in an empty cylindrical tank, the radius of whose base is 40 cm. What is the rise of water level in tank in half an hour?

Radius of cylindrical pipe = 1 cm

Height = 80 cm

$Volume=\pi r ^2 h$

$=\frac{22}{7} \times 1 \times 1 \times 80 = 251.4285 cm^3 /sec$

In half an hour volume of water is

$=251.4285 \times30 \times60 = 452571.5 cm^3$

Radius of cylindrical tank = 40 cm

Let height = h

Volume $=\pi r ^2 h$

$\frac{22}{7}\times (40)^2 h= 5828.5714 cm^3$

According to question

The volume of cylindrical pipe = volume of the cylindrical tank

$= 452571.5 = 5028.5714h$

$\frac{452571.5}{5028.5714} = h$

h =89.99

h = 90 cm (approximate)

## JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

#### A solid right circular cone of height 120 cm and radius 60 cm is placed in a right circular cylinder full of water of height 180 cm such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is equal to the radius of the cone.

Height of cone = 120cm

$Volume =\frac{1}{3}\pi r^2 h$

$=\frac{1}{3}\pi( 60)^2 120$

$=452571.43cm^3$

Height of cylinder = 180

Radius = 60 cm           (given that radius of the cylinder is equal to the radius of the cone)

$Volume=\pi r^2 h$

$=pi( 60)^2 180=2036571.43cm^3$

Volume of water left in the cylinder = volume of cylinder – volume of cone

2036571.43 – 452571.43

= 1584000 cm3

Or        1.584 m3          [Q 1m = 100cm]

2m3 (approximate)

#### A hemispherical bowl of internal radius 9 cm is full of liquid. The liquid is to be filled into cylindrical shaped bottles each of radius 1.5 cm and height 4 cm. How many bottles are needed to empty the bowl?

The radius of hemispherical bowl = 9 cm

$Volume =\frac{2}{3}\pi r^3$ $=\frac{2}{3} \times \frac{22}{7}\times 9\times 9\times 9$

Radius of cylindrical bottle = 1.5 cm

Height = 4cm

$Volume$$=\pi r^2h$

$=\frac{22}{7}\times \frac{15}{10}\times \frac{15}{10}\times 4$

$\text{ Number of bottles needed}=\frac{\text {volume of hemispherical bowl}}{\text {volume of cylindrical bottle}}$

$=\frac{2}{3} \times \frac{22}{7}\times 9\times 9\times 9$

_________________

$=\frac{22}{7}\times \frac{15}{10}\times \frac{15}{10}\times 4$

$=\frac{2 \times 9 \times 9 \times \times 10 \times 10}{3 \times 15 \times 15 \times 4} =54 bottles$

## NEET 2024 Most scoring concepts

Just Study 32% of the NEET syllabus and Score up to 100% marks

#### A building is in the form of a cylinder surmounted by a hemispherical vaulted dome and contains $41\frac{19}{21}m^3$ of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building?

Give: Volume of building

$41\frac{19}{21}=\frac{880}{21}$

Let total height above the floor = h

Hemisphere‘s diameter = h (given)

$Radius =\frac{h}{2}$

$Volume=\frac{2}{3}\pi r^3=\frac{2}{3}\pi \left ( \frac{h}{2} \right )^3$

Height of cylinder = total height – the height of hemisphere

$h-\frac{h}{2}=\frac{h}{2}$

Volume $\pi r^2 h=\pi \times \left (\frac{h}{2} \right )^2 \times \frac{h}{2} =\pi \left (\frac{h}{2} \right )^3$

According to question

The volume of building = Volume of cylinder + volume of the hemisphere

$\frac{880}{21}=\left ( \frac{h}{2} \right )^3\left [ \pi + \frac{2}{3} \pi \right ]$

$\frac{880}{21}=\left ( \frac{h}{2} \right )^3\left [ \frac{3\pi +2\pi }{3} \right ]$

$\frac{880 \times 2^3}{21}=h^3 \left [ \frac{5\pi }{3} \right ]$

$\frac{880 \times 2\times 2\times 2\times 7\times 3}{21\times 5\times 22}=h^3$

$h^3 = {2 \times 2 \times 2\times 2\times 2\times 2$

$h = \sqrt[3]{2 \times 2 \times 2\times 2\times 2\times 2}$

$h = 2 \times 2$

h=4m

#### A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and height of the cylinder are 6 cm and 12 cm, respectively. If the slant height of the conical portion is 5 cm, find the total surface area and volume of the rocket [Use $\pi = 3.14$].

Diameter of cylinder = 6 cm

Height = 12 cm

$\text{ Surface are}$ $=2 \pi r h = 2\pi \times 3 \times 12 = 72 \pi cm^2$

Similarly radius of circle = 3 cm

$\text{ Area}$ $=\pi r^2=\pi 3 \times 3 = 9 \pi cm^2$

Slant height of cone (l)=5cm

$\text{ Surface area}=\pi r l =\pi \times 5 \times 3 = 15\pi cm^2$

Total surface area = area of cylinder + area of circle + area of the cone

$=72\pi + 9 \pi + 15 \pi$

$=96\pi$

$=96\times 3.14=301.7 cm^2$

Slant height (l) = 5cm

We know that,

$h^2+r^2=l^2$

$h^2+3^2=5^2$

$\\h^2= 25 -9 \\ h^2=16 \\\ h = 4$

$\text{ Volume of cylinder= }\pi r^2 h \\ = \pi \times 3 \times 3 \times 12 = 108 \pi cm^3$

$\\\text{ Volume of cone}=\frac{1}{3}\pi r^2 h \\ = \frac {1}{3} \pi \times 3 \times 3 \times 12 = 12 \pi cm^3$

$\text{ Volume of rocket }$$=180 \pi + 12 \pi$

$=120 \pi$

$=120 \times 3.14 =377.14 cm^3$

## Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support

#### A milk container of height 16 cm is made of metal sheet in the form of a frustum of a cone with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk at the rate of Rs. 22 per litre which the container can hold.

Solution

Upper radius of frustum of cone (R) = 20cm

Lower radius of frustum of cone r = 8cm

Height (h)= 16cm

Volume $\frac{\pi h}{3}[R^2+r^2+Rr]$

$\frac{22 \times 16}{7 \times 3}(20^2+8^2+20 \times 8)$

$\frac{22 \times 16}{21}(400 + 64+160)$

$\frac{22 \times 16}{21}(624)$

$\frac{22 \times 16\times 208}{7}=10459.428 m^3$

= 10.459 liter

Cost of 1 liter milk = 22 Rs.

Cost of 10.459 liter milk $22 \times 10.459$

=   2301.10 Rs

#### 500 persons are taking a dip into a cuboidal pond which is 80 m long and 50 m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is 0.04m3?

Length of cuboidal pond = 80m

Let, Height = h

$\text{ Volume}\ l \times b \times h$

$80 \times 50 \times h=400 h m^3$

Average water of one person =0.04m3

Average water of 500 persons = $0.04 \times 500$

According to question

The volume of cuboidal pond = Average water of 500 persons

$400 h =0.04 \times 500$

$h = 4 \times \frac{500 }{100\times 400}=\frac{5}{100}=0.05m$