Surface Area and Volume Class 10 Maths Important Questions

Original Volume=$\pi R^2 H$
When diameter is decreased by 5%, radius also is decreased by 5%.
New Radius = R - 5R/100 = 95R/100= 19R/20

New Volume=$\pi (\frac {19R}{20})^2 H = \frac {\pi \times H \times 361R^2}{400}$

But you want to maintain the volume at a constant, so Length H is to be multiplied by 400/361 to keep it constant.
So New Length= $ \frac {400H}{361}$
Increase in Length=$\frac {400H}{361 -H = \frac {39H}{361}$
% Increase in Length = $ \frac {39}{361} \times 100 = 10.8%$

Let number of small balls = x
Volume of x small balls = Volume of sphere
$x \times \frac{4}{3}\pi (.3)^3 = \frac{4}{3} \pi (3)^3$
x=1000

Edge of the solid cube = 44 cm
Volume of the cube =L³= 44³ = 85184 cm³
Diameter of the bullet = 4 cm
Radius = 2 cm
Volume of each bullet = $\frac {4}{3}\pi R^3$
= 33.5 cm³
Number of bullets = 85184 / 33.5 = 2542.8=2542(approx)

Width 300cm=3m
depth 120cm=1.2m
speed of canal 20km/h= 20 * 1000m/h
vol. of water flow in 1 hour = width of canal �depth of canal * speef of canal water
=3 *1.2 * 20 * 1000=72000m³
So vol. of water flow in 20 min = 72000 * 20/60=24000m³
Area irrigated in 20 min =240000/0.08=300000m²

Given
diameter of the sphere (d) = 6cm �
radius of the sphere ( r ) = d/2 = 6/2 = 3cm
diameter of the cylindrical vessel ( D ) = 12 cm
radius of the cylindrical vessel ( R ) = 12/2 = 6 cm
Let level of water raised = h cm
Now according to the Question
volume of the water raised in the vessel = volume of the sphere
$ \pi R^2h = \frac {4}{3}�\pi r^3$
Substituting the values
h=1 cm

For initial right circular core
h = 3.6 cm, r = 1.6 cm
Volume of the given cone =$ \frac {1}{3} \pi r^2 h = \frac {1}{3} \pi (1.6)^2 \times 3.6 cm^2$
For Final recast right circular cone

Radius of the second cone = 1.2 cm. Let its height be h
Volume of the second cone = $ \frac {1}{3} \pi r^2 h = \frac {1}{3} \pi (1.2)^2 \times h cm^2$
Now Volume of initial right circular core= Volume of Final recast right circular cone
$ \frac {1}{3} pi (1.6)^2 \times 3.6 =\frac {1}{3} pi (1.2)^2 \times h$
h=6.4 cm

Volume of water in conical vessel= $ \frac {1}{3} \pi r^2 h = \frac {1}{3} \pi (5)^2 24$

Let H be the height of water in cylinderical vessel.
Volume of water in cylinderical vessel=$ \pi r^2 h= \pi (10)^2 h$
Now Volume of water in conical vessel=Volume of water in cylinderical vessel
$\frac {1}{3} \pi (5)^2 24=\pi (10)^2 h$
h=2 cm

$l= (3^3 + 4^3 + 5^3)^{1/3} = 6$

Each one of those coins is also a small cylinder , and its volume is�
$ V_1 = \pi r^2h = \pi (.75)^2(.2)$
� The right circular cylinder has volume $
$V_2 = \pi r^2 h=\pi (2.25)^2(10)$�
Number of Coins =$\frac {V_2}{V_1} = 450$ coins�

Let h be the height water rise in cylinder
So Volume of the portion of increase in height = $\pi (10)^2 h$
Volume of cube=$ (8)^3$
Now Volume of cube= Volume of the portion of increase in height
$(8)^3=\pi (10)^2 h$
h=1.62 cm

Let x be the width
Volume of well=$\pi r^2 h= \pi \times (3.5)^2 \times 22.5 $
Volume of embankment=$ \pi R^2 h - \pi r^2 h=\pi \times (3.5 +x )^2 \times 1.5 - \pi \times (3.5)^2 \times 1.5$
Volume of well = Volume of embankment
So $\pi \times (3.5)^2 \times 22.5 = \pi \times (3.5 +x )^2 \times 1.5 - \pi \times (3.5)^2 \times 1.5$
On solving we get x = 10.5m

Volume of deep well= $\pi r^2 h = \pi \times (3.5)^2 \times 10$
Let h be the rise in level
Volume of earth spread out = $20 \times 14 \times h$
Now,
Volume of earth spread out=Volume of deep well
$\pi \times (3.5)^2 \times 10=20 \times 14 \times h$
h=11/8 m

Volume of a frustum of a cone =$\frac {1}{3} \pi h(r_1^2 + r_2^2 + r_1r_2)$
Here h=11 cm , $r_2= \frac {48}{2 \pi}=\frac {24}{\pi}$ , $r_1 =frac {36}{2 \pi}=\frac {18}{\pi}$
Substituting these values
V=1554 cm³

Radius of cylindrical pipe = 7/2 cm =3.5 cm
Volume of water flows in 1 min = 192.5 L
Therefore,Volume of water flows in 1 hour = $192.5 \times 60 \times 1000 cm^3$

Let l be the distance covered by water in 1 hour. Volume of water that covers the diameter of 1 hour =volume of cylinder per hour=$ \pi \times r^2 \times l=\pi \times (3.5)^2 \times l cm^3$

Now
$ \pi \times (3.5)^2 \times l=192.5 \times 60 \times 1000 $
l=300000 cm= 3km
So rate of flow is 3km/hr

Radius of circular pipe (r) = 7/2 =3.5cm
Now Length of water per sec = 72cm
Therefore, volume of water per second = volume of cylinder per second =$ \pi r^2 h = \frac {22}{7} \times (3.5)^2 \times 72 = 2772 cm^3$
Now $ 1 L = 1000cm^3 $
So, volume of water per sec = 2.772 litres
Hence volume of water in hr= $2.772 \times 60=9979.2 litres$

) = 20/2 cm = 10 cm = 1/10 m
Height of water in 1 hour( h)= 3 km = 3000 m
Volume of water passing through the pipe in a hour = Volume of the cylinder formed
$V = \pi r^2 h= \pi \times (1/10)^2 \times 3000 = 30 \pi$

Volume of cistern
$V = \pi r^2 h=\pi \times (5)^2 \times 2=50 \pi$

Therefore time needed to fill the cistern =$ \frac {50 \pi}{30 \pi} = \frac {5}{3}$ hr

Volume of water in 1/2 hours= $ \pi \times (1)^2 \times 700 \times 30 \times 60$
Let H be the increase in the water level
Then
$\pi \times (40)^2 \times H=\pi \times (1)^2 \times 700 \times 30 \times 60$
H=787.5cm

The situation is shown in below figure

By similarity of triangle ABC and AXY
$\frac {AB}{AX} = \frac {BC}{XY}$
$\frac {12}{4}=\frac {9}{r}$
r=3 cm

The area of the internal surface of the cone not in contact with water is the area of the frustum of the cone which is given as
$A =\pi L(r_1+ r_2)$
Here $r_1=3 cm $ ,$r_2= 9 cm$
Also height of the frustum = 12 -4 =8 cm
Now L is given by
$L= \sqrt {h^2 + (r_2 -r_1)^2} =10$ cm
Therefore Area
$A= \frac {22}{7} \times 10 ( 3 + 9) = 376.8 cm^2$

Answer 2.5cm

Answer 225000m²

Answer 51 minutes 12 seconds

Answer 26.73 minutes

Radius of the barrel=5/2 = 2.5mm= .25 cm
Height of bareel= 7 cm
Volume of the barrel= $\pi r^2 H= $ \frac {22}{7} \times (.25)^2 \times 7=1.375 cm^3$

Now 1/5 litre means = $200 cm^3$
Now as per the question
$1.375 cm^3$ of ink can write 3300 words
$1 cm^3$ of ink will write $\frac {3300}{1.375}$ words
$200 cm^3$ will write $\frac {3300}{1.375} \times 200=480000$ words

Answer 509.14dm³

Volume of the solid iron rectangular block= $4.4 \times 2.6 \times 1=11.44 m^3$
Internal Radius(r)=30 cm
External radius(R) = 30 +5=35 cm
Let L be the lenght of the pipe,Then volume of the pipe will be
$V=\pi R^2 L - \pi r^2 L = \pi L[(.35)^2 - (.30)^2]$
Now As per the question
Volume of the solid iron rectangular block=volume of the pipe
$11.44=\pi L[(.35)^2 - (.30)^2]$
L=112m

Let the radius of the hemispherical dome be r metres and the total height of the building be h metres.
Since the base diameter of the dome is equal to 2/3 of the total height, therefore
$2r = \frac {2}{3} h$. This implies $r = \frac {h}{3}$
Let H metres be the height of the cylindrical portion.
Therefore, $H =h � \frac {h}{3} = \frac {2h}{3}$ m

Volume of the air inside the building = Volume of air inside the dome + Volume of the air inside the cylinder
$V= \frac {2}{3} \pi r^3 + \pi r^2 H$
Substituting the value of r and H, we get
$=\frac {8}{81} \pi h^3$
Now as per the question
$\frac {8}{81} \pi h^3=67 \frac {1}{27}$
or h=6 m