Questions 1-5 below

###Q1

A cube of side (4 , text{cm}) contains a sphere touching the sides.
Find the volume of the gap in between.

### Q2

The circumference of the base of a cone is (frac{220}{7} , text{cm})
and its slant height is (13 , text{cm}). Find the volume of the cone.

### Q3

A sphere and a cube have the same surface area. Show that the ratio of
the volume of the sphere to that of the cube is (sqrt{6}:sqrt{pi}).

### Q4

Two spheres made of the same metal have weights (5420 , text{g}) and
(740 , text{g}) respectively. Determine the radius of the larger
sphere if the diameter of the smaller one is (5 , text{cm}).

### Q5

A hemispherical tank full of water is emptied by a pipe at the rate of
(3 frac{4}{7} , text{lit/sec}). How much time will it take to empty
half the tank if it is (3 , text{m}) in diameter?

 

Solutions 1-5 below

Q1: Cube and Sphere

A cube of side (4 , text{cm}) encloses a sphere that touches its
sides. The volume of the gap in between is calculated as follows:

1. Volume of the cube:
[
V_{text{cube}} = s^3 = 4^3 = 64 , text{cm}^3.
]

2. Radius of the sphere (half of the cube’s side length):
[
r = 2 , text{cm}.
]

3. Volume of the sphere:
[
V_{text{sphere}} = frac{4}{3}pi r^3 = frac{32}{3}pi , text{cm}^3.
]

4. Volume of the gap:
[
V_{text{gap}} = V_{text{cube}} – V_{text{sphere}} = 64 –
frac{32}{3}pi , text{cm}^3.
]

### Q2: Cone

Given the circumference of the base of the cone as (220/7 , text{cm})
and slant height as (13 , text{cm}), the volume of the cone is found
as follows:

1. Radius of the base:
[
r = frac{220}{7 times 2pi} = 5 , text{cm}.
]

2. Height of the cone using the Pythagorean theorem:
[
h = sqrt{l^2 – r^2} = sqrt{13^2 – 5^2} = 12 , text{cm}.
]

3. Volume of the cone:
[
V_{text{cone}} = frac{1}{3}pi r^2 h = 100pi , text{cm}^3.
]

### Q3: Sphere and Cube

Given that a sphere and a cube have the same surface area, it can be
shown that the ratio of their volumes is (sqrt{6}:sqrt{pi}).

1. Equating the surface areas:
[
6s^2 = 4pi r^2.
]

2. Expressing the volume of both shapes:
[
frac{V_{text{sphere}}}{V_{text{cube}}} = frac{frac{4}{3}pi
r^3}{s^3}.
]

3. Substituting (r^2) in terms of (s^2):
[
frac{V_{text{sphere}}}{V_{text{cube}}} = frac{4sqrt{6}}{3sqrt{pi}}.
]

Thus, proving that the ratio of their volumes is (sqrt{6}:sqrt{pi}).

### Q4: Two Spheres

Two spheres made of the same material have weights in the ratio of
(5420:740). Given that the diameter of the smaller sphere is (5 ,
text{cm}), the radius of the larger sphere is:

[
R = sqrt[3]{frac{271}{37} times (2.5)^3} approx 7 , text{cm}.
]

### Q5: Hemispherical Tank

A hemispherical tank with a (3 , text{m}) diameter is emptied at a
rate of (3 frac{4}{7}) litres per second. The time to empty half the
tank is:

1. Volume of half the hemisphere in liters:
[
V_{text{liters}} = frac{1}{2} times frac{2}{3} pi (1.5)^3 times
1000.
]

2. Time to empty half the tank:
[
t = frac{V_{text{liters}}}{3 frac{4}{7}} , text{seconds}.
]

 

Questions 6-10 below

### Q6
(30) circular plates, each of radius (14 , text{cm}) and thickness
(3 , text{cm}), are placed one above the other to form a cylindrical
solid. Find:

a) The total surface area of the cylindrical solid.

b) The volume of the cylinder so formed.

### Q7
A semi-circular sheet of paper with a diameter of (42 , text{cm}) is
bent into an open conical cup. Find the depth and volume of the cup.

### Q8
The internal and external radii of a hollow hemispherical vessel are (12
, text{cm}) and (frac{25}{2} , text{cm}) respectively. The cost of
painting (1 , text{cm}^2) of the surface is Rs (0.07). Find the
total cost to paint the vessel.

### Q9
Solid spheres with a diameter of (4 , text{cm}) each are dropped into
a cylindrical beaker containing some water. If the diameter of the beaker
is (12 , text{cm}) and the water rises by (24 , text{cm}), find
the number of solid spheres dropped in the water.

### Q10
A hollow sphere with internal and external radii of (2 , text{cm}) and
(4 , text{cm}) respectively is melted into a cone with a base radius
of (4 , text{cm}). Find the height and the slant height of the cone.

Solutions (6-10 below)

### Q6
(30) circular plates, each of radius (14, text{cm}) and thickness
(3, text{cm}), are stacked to form a cylindrical solid.

a) Total surface area of the cylinder is calculated using the formula:
[
text{Total surface area} = 2pi r (r + h) = 2pi times 14 times (14 +
90) = 9280pi , text{cm}^2.
]

b) Volume of the cylinder:
[
text{Volume} = pi r^2 h = pi times 14^2 times 90 = 52920pi ,
text{cm}^3.
]

### Q7
A semi-circular sheet of paper with a diameter of (42, text{cm}) is
bent into an open conical cup.

Radius of the base of the cone:
[
r = 21, text{cm}.
]

Slant height of the cone (equal to the diameter of the semi-circular
sheet):
[
l = 42, text{cm}.
]

Using the Pythagorean theorem to find the depth of the cone:
[
h = sqrt{l^2 – r^2} = sqrt{42^2 – 21^2} = 21sqrt{3}, text{cm}.
]

Volume of the conical cup:
[
V = frac{1}{3}pi r^2 h = frac{1}{3}pi times 21^2 times 21sqrt{3} =
10206sqrt{3}pi , text{cm}^3.
]

### Q8
The internal and external radii of a hollow hemispherical vessel are
(12, text{cm}) and (frac{25}{2}, text{cm}) respectively.

Surface area to be painted:
[
text{Area} = 2pi left(frac{25}{2}right)^2 – 2pi (12)^2 = 170.5pi –
144pi = 26.5pi , text{cm}^2.
]

Total cost to paint the vessel:
[
text{Cost} = text{Area} times text{Rate} = 26.5pi times 0.07 =
1.855pi , text{Rs}.
]

### Q9
Solid spheres with a (4, text{cm}) diameter are dropped into a (12,
text{cm}) diameter beaker, causing the water to rise (24, text{cm}).

Volume of water displaced:
[
V = pi times 6^2 times 24 = 864pi , text{cm}^3.
]

Volume of a single sphere:
[
V_s = frac{4}{3}pi times 2^3 = frac{32}{3}pi , text{cm}^3.
]

Number of spheres:
[
n = frac{V}{V_s} = frac{864pi}{frac{32}{3}pi} = 81.
]

### Q10
A hollow sphere with internal and external radii (2, text{cm}) and
(4, text{cm}) respectively is melted into a cone with a base radius of
(4, text{cm}).

Volume of the hollow sphere:
[
V_{text{hollow sphere}} = frac{4}{3}pi (4^3 – 2^3) = frac{224}{3}pi
, text{cm}^3.
]

Volume of the cone:
[
V_{text{cone}} = V_{text{hollow sphere}} = frac{224}{3}pi ,
text{cm}^3.
]

Using the volume formula for a cone to find the height:
[
h = frac{3V}{pi r^2} = frac{3 times frac{224}{3}pi}{pi times 4^2}
= 14, text{cm}.
]

Using the Pythagorean theorem to find the slant height:
[
l = sqrt{r^2 + h^2} = sqrt{4^2 + 14^2} = 2sqrt{53}, text{cm}.
]

Questions 11-15

### Q11
If the number of square centimeters in the surface area of a sphere is equal
to the number of cubic centimeters in its volume, find the diameter of the
sphere.

### Q12
A sphere, a cylinder, and a cone have the same radius and the same height.
Find the ratio of their curved surface areas.

### Q13
A hemispherical dome, open at the base, is made from a fibre sheet. If the
diameter of the hemispherical dome is (80 , text{cm}) and
(frac{13}{170}) of the sheet actually used was wasted in making the dome,
find the cost of the dome at the rate of Rs (0.35/text{cm}^2). (Take
(pi = 3.14))

### Q14
A solid ball (3 , text{cm}) in radius, of a certain material, weighs
(339 frac{3}{7} , text{gms}). Find the weight of a spherical shell of
that material, the internal and external diameters of which are respectively
(10 , text{cm}) and (12 , text{cm}).

### Q15
A solid sphere of radius (r) is bisected. Find the surface area of each
piece.

Solutions 11-15

### Q11
If the surface area of a sphere in square centimeters is equal to its volume
in cubic centimeters, find the diameter of the sphere.

Given that
[
4pi r^2 = frac{4}{3}pi r^3,
]
we find
[
r = 3 , text{cm},
]
so the diameter is
[
d = 6 , text{cm}.

### Q12
A sphere, cylinder, and cone have the same radius and height. We need to
find the ratio of their curved surface areas.

For a sphere with radius ( r ):
[
A_{text{sphere}} = 4pi r^2.
]

For a cylinder of height ( h ) and radius ( r ):
[
A_{text{cylinder}} = 2pi rh.
]

For a cone with slant height ( l ) and radius ( r ):
[
A_{text{cone}} = pi rl.
]

Since the height and radius are equal,
[
l = sqrt{r^2 + h^2} = sqrt{2}r,
]
so
[
frac{A_{text{sphere}}}{A_{text{cylinder}}} :
frac{A_{text{cylinder}}}{A_{text{cone}}} = frac{2r}{h} :
frac{h}{sqrt{2}r} = 2 : sqrt{2}.

### Q13
A hemispherical dome with a ( 80 , text{cm} ) diameter is made from a
fibre sheet. ( frac{13}{170} ) of the sheet was wasted. We need to find
the cost of the dome at ( text{Rs} , 0.35/text{cm}^2 ) with ( pi =
3.14 ).

The surface area of the hemisphere is
[
A = 2pi r^2 = 2 times 3.14 times 40^2 , text{cm}^2.
]

The wasted material amounts to
[
A_{text{wasted}} = frac{13}{170}A,
]
so the useful material is
[
A_{text{useful}} = A – A_{text{wasted}}.
]

The cost of the dome is
[
text{Cost} = A_{text{useful}} times 0.35.

### Q14
A solid ball with a ( 3 , text{cm} ) radius weighs ( 339 frac{3}{7} ,
text{g} ). We need to find the weight of a spherical shell with internal
and external diameters of ( 10 , text{cm} ) and ( 12 , text{cm} ),
respectively.

The volume of the solid ball is
[
V_{text{ball}} = frac{4}{3}pi r^3,
]
and its density is
[
rho = frac{339 frac{3}{7}}{V_{text{ball}}}.
]

The volume of the spherical shell is
[
V_{text{shell}} = frac{4}{3}pi R^3 – frac{4}{3}pi r^3,
]
so its weight is
[
text{Weight} = rho V_{text{shell}}.

### Q15
A solid sphere of radius ( r ) is bisected. We need to find the surface
area of each piece.

The curved surface area of one half is the same as the other:
[
A_{text{half}} = 2pi r^2.
]

Adding the flat circular base of area ( pi r^2 ) gives
[
A_{text{total}} = 3pi r^2.
]