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:
\[
r = \frac{220}{7 \times 2\pi} = 5 \, \text{cm}.
\]

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

### 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:
\[
6s^2 = 4\pi r^2.
\]

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 \(s^2\):
\[
\frac{V_{\text{sphere}}}{V_{\text{cube}}} = \frac{4\sqrt{6}}{3\sqrt{\pi}}.
\]

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:
\[
\text{Total surface area} = 2\pi r (r + h) = 2\pi \times 14 \times (14 +
90) = 9280\pi \, \text{cm}^2.
\]

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

### 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:
\[
h = \sqrt{l^2 – r^2} = \sqrt{42^2 – 21^2} = 21\sqrt{3}\, \text{cm}.
\]

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

### 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:
\[
\text{Area} = 2\pi \left(\frac{25}{2}\right)^2 – 2\pi (12)^2 = 170.5\pi –
144\pi = 26.5\pi \, \text{cm}^2.
\]

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

### 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:
\[
V = \pi \times 6^2 \times 24 = 864\pi \, \text{cm}^3.
\]

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

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

### 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:
\[
l = \sqrt{r^2 + h^2} = \sqrt{4^2 + 14^2} = 2\sqrt{53}\, \text{cm}.
\]

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
\[
4\pi r^2 = \frac{4}{3}\pi r^3,
\]
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 \( r \):
\[
A_{\text{sphere}} = 4\pi r^2.
\]

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

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
\[
A = 2\pi r^2 = 2 \times 3.14 \times 40^2 \, \text{cm}^2.
\]

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:
\[
A_{\text{half}} = 2\pi r^2.
\]

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