Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Filter By

Clear Filter

All Questions

#12.3

A wall 24 m long, 0.4 m thick and 6 m high is constructed with the bricks each of dimensions $25 cm \times 16 cm \times 10 cm.$ If the mortar occupies1/10 th of the volume of the wall, then find the number of bricks used in constructing the wall.

Answer 12960

Solution

Length of wall $=24 m =24 \times 100 =2400 cm$ ( because 1m = 100cm )

Breadth of wall $=0.4m =0.4 \times 100 =40 cm$

Height of wall $= 6m = 6 \times 100 = 600cm$

The volume of wall = length × breadth × height

$=2400 \times 40 \times 600$

$=5760000cm^3$

$\text{ Mortar occupied}=\frac{5760000}{10} =5760000cm^3$

Remaining volume = 57600000 – 5760000 = 51840000 cm³

Length of brick = 25 cm

Breadth of brick = 16cm

Height of brick = 10cm

Volume of brick = length × breadth × height

= 25 × 16 × 10 = 4000 cm³

$\text{Number of bricks}=\left [ \frac{\text{remaining volume} }{\text {volume of brick}} \right ]=\frac{5184000}{4000}=12960$

Hence the number of bricks is 12960

View Full Answer(1)

Posted by

infoexpert21

#12.3

How many spherical lead shots of diameter 4 cm can be made out of a solid cube of lead whose edge measures 44 cm.

Answer 2542

Solution

Given : Diameter of spherical lead shot = 4 cm

Edge of cube = 44 cm

Volume of cube =a³

=44³ (=44 cm)

=85184cm³

Radius of spherical lead shot =2cm

$\text{ Volume of lead shot}=\frac{4}{3}\pi r^3$

$=\frac{4}{3}\times 3.14 \times (2)^3=33.5$

$\text{ Number of lead shots}=\frac{\text {volume of cube}}{\text{volume of lead shot}}$

$=\frac{85184}{33.5}=2542$

Hence 2542 lead shots can be made out of a cube of lead whose edge measures 44 cm.

View Full Answer(1)

Posted by

infoexpert21

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

#12.3

How many spherical lead shots each of diameter 4.2 cm can be obtained from a solid rectangular lead piece with dimensions 66 cm, 42 cm and 21 cm.

Answer 1501

Solution

It is given that the length, breadth and height of rectangular solid is 66cm, 42cm and 21cm respectively.

Volume of solid rectangular lead piece = $l \times b \times h$

$66 \times 42 \times 21$

$=5821 cm^3$

The diameter of spherical lead shot =4.2

The radius of spherical lead shot

$=\frac{4.2}{2}=2.1 cm$

$\text{ The volume of lead shot }=\frac{4}{3}\pi r^3$

$=\frac{4}{3}\times 3.14 \times (2.1)^3=38.77 cm^3$

$\text{ Number of lead shot can be obtained}=\frac{\text {volume of lead piece}}{\text{volume of lead shot}}$

$=\frac{58212}{38.77}=1501$

Hence 1501 lead shot can be obtained from the lead piece of dimensions 66cm, 42cm and 21cm

View Full Answer(1)

Posted by

infoexpert21

#12.3

Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6 cm.

Answer 150

Solution

Given:- Diameter of marble = 1.4 cm

Diameter of beaker = 7cm

Diameter of marble = 1.4 cm

$\text{ Radius of marble}=\frac{1.4}{2}=0.7 cm$

The volume of 1 marble

$=\frac{4}{3} \pi r^3=\frac{4}{3}\pi (0.7)^3$

$=\frac{4}{3} \times 3.14 \times 0.343=1.43cm^3$

Diameter of beaker = 7 cm

$\text{ Radius of beaker}=\frac{7}{2}=3.5 cm$

Water level rises(h) = 5.6cm

Volume of water

$=\pi r^2 h=(3.14)(3.5)^2(5.6)$

$=215.40$

$\text{ Number of marbles required}=\frac{\text {volume of water }}{\text {volume of 1 marble }}$

$=\frac{215.40}{1.43}=150$

Hence 150 marbles should be dropped into the beaker.

So that the water level rises by 5.6 cm.

View Full Answer(1)

Posted by

infoexpert21

JEE Main high-scoring chapters and topics

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

#12.3

An ice cream cone full of ice cream having radius 5 cm and height 10 cm as shown in the Figure. Calculate the volume of ice cream, provided that its 1/8 part is left unfilled with ice cream.

In this figure, there is a hemisphere of radius of 5 cm

And a cone of radius 5 cm and of height $10-5 = 5 cm$

Volume of cone

$=\frac{1}{3}\pi r^2 h=\frac{1}{3}\times 3.14 \times (5)^2\times 5=130.83cm^3$

Volume of hemisphere

$=\frac{2}{3}\pi r^3$

$=\frac{2}{3}\times 3.14 \times (5)^3=261.66cm^2$

The volume of complete figure = volume of cone + volume of the hemisphere

$= 130.83 +261.66 = 392.49 cm^3$

$\text{ The volume of the unfilled part }=\frac{392.49}{6}=65.41cm^3$

Volume of ice cream = volume of complete figure - volume of unfilled part

$392.49- 65.41=327.08 cm^3$

Hence the volume of ice cream is 327.08 cm³

View Full Answer(1)

Posted by

infoexpert21

#12.3

Two solid cones A and B are placed in a cylindrical tube as shown in the Figure. The ratio of their capacities are 2:1. Find the heights and capacities of cones. Also, find the volume of the remaining portion of the cylinder.

Answer 396.18 cm³

Solution

Height of the tube = 21 cm

Base radius of the tube =3cm

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

$=\frac{22}{7}\times 3\times 3\times21=594cm^3$

Let the height of cone A is h cm

Height of cone $B=21-h cm$

Base radius of both A and B =3 cm

Volume of cone

$A=\frac{1}{3}\pi r^2h=\frac{1}{}3\pi (3)^2h$

$=3 \pi h$

Volume of cone

$B=\frac{1}{3}\pi r^2h=\frac{1}{}3\pi (3)^2(21-h)$

$=3 \pi (21-h)$

It is given that the ratio of the volume is 2: 1

$=\frac{3\pi h}{3 \pi (21-h)}=\frac{2}{1}$

$h=2(21-h)$

$h=42-2h$

$3h=42$

$h=14$

Height of cone A = 14 cm

Height of cone B =21-4=7 cm

Volume of cone $A=3 \pi h=3(3.14)(14)$

$=131.88cm^3$

Volume of cone $B = 3\pi (21-h)=2(3.14)(7)$

$=65.94cm^3$

The volume of remaining portion = Volume of the tube – the volume of cone A – the volume of cone B

$594-131.88 -65.94$

= 396.18 cm³

Volume of remaining portion = 396.18 cm³

View Full Answer(1)

Posted by

infoexpert21

NEET 2024 Most scoring concepts

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

#12.3

Two cones with same base radius 8 cm and height 15 cm are joined together along their bases. Find the surface area of the shape so formed.

Answer 854 cm²

Solution

According to question

Here are two cones joined together along their bases

Height of both cone = 15 cm

Base radius of both cone = 8 cm

Surface area of combination = 2(surface area of one cone)

(Q both cones are same)

$=2(\pi r l)$

$=2\pi r \sqrt{r^2+h^2}$ $\left (Q1= \sqrt{r^2+h^2} \right )$

$= 2 \times 3.14 \times 8 \times \sqrt{64+225}$

$= 2 \times 3.14 \times 8 \times 17$

$=854cm^2$

Hence the surface area of the combination is 854 cm²

View Full Answer(1)

Posted by

infoexpert21

#12.3

From a solid cube of side 7 cm, a conical cavity of height 7 cm and radius 3 cm is hollowed out. Find the volume of the remaining solid.

Answer 277 cm³

Solution

The figure formed when a conical cavity is cut out from a cube.

Volume of cube $=a^3$

$=(7)^3=343 cm^3$

The volume of the conical cavity

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

$=\frac{1}{3}\times 3.14 \times (3)^2 \times 7$ $\left ( \because r=3cm ,h=7cm \right )$

$=21\times 3.14= 66cm^3$

The volume of remaining solid = volume of the cube – the volume of the conical cavity

$=343-66= 277cm^3$

View Full Answer(1)

Posted by

infoexpert21

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

#12.3

Two identical cubes each of volume 64 cm³ are joined together end to end. What is the surface area of the resulting cuboid?

It is given that volume of cube = 64 cm³

$a^3=64cm^3$ ( Because the volume of cube = a³ )

$a^3=4^3$

$a=4$

So the side of the two cubes are 4 cm

The cuboid formed by joining two cubes.

The surface area of the cuboid $=2(lb +bh+hl)$

$=2(8 \times 4 + 4 \times + 4 \times 8)$

$=2(8 \times 4 +4 \times 4 + 4 \times 8)$

$=2(32+16 +32)$

$=2(80)=160cm^2$

View Full Answer(1)

Posted by

infoexpert21

#12.3

A cone of radius 8 cm and height 12 cm is divided into two parts by a plane through the mid-point of its axis parallel to its base. Find the ratio of the volumes of two parts.

Answer 1:7

Solution

When a cone is divided into two parts by a plane through the mid-point the image formed is

In figure $\Delta AGE:\Delta EFC$ $Q\angle E$ is common angle $\angle F=\angle G=90^{\circ}$

So the corresponding sides are in equal ratio.

$\frac{EF}{FG}=\frac{FC}{GA}$

$\frac{ 6}{12} =\frac{ FC}{}8$

FC = 4 cm

Volume of cone $EDC= \frac{1}{3}\pi r_{2}^2h$

$= \frac{1}{3}\times 3.14 \times (4)^2 \times 6=100.48cm^3$

Volume of frustum of cone

$ABCD =\frac{1}{3}\pi h[r_{1}^2+r_{2}^2 +r_{1}r_{2}]$

$=\frac{1}{3}\times 3.14 \times 6[(8)^2+(4)^2 +8 \times 4]$

$=6.28 [64+16+32]$

$= 6.28[112] = 703.36$

Volume of cone EDC : volume of ABCD

100.48 : 703.36

1 : 7

View Full Answer(1)

Posted by

infoexpert21

JEE Main high-scoring chapters and topics

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

1
2
3
4
5
6
7
8
Next