Solution

Given:

$\text{Radius}(R)=2 r$

Now, the volume of sphere is:

$ \begin{aligned} \dfrac{4}{3} \pi R^{3} & =\dfrac{4}{3} \pi \times(2 r)^{3} \\ & =\dfrac{4}{3} \pi \times 8 r^{3} \\ & =\dfrac{24}{3} \pi r^{3} \end{aligned} $

Hence, the correct option is (D).

Solution

The formula of total surface area of cube is $6(\text{ edge })^{2}$.

So, $6(\text{ edge })^{2}=96$

$ \begin{aligned} (\text{ edge })^{2} & =\dfrac{96}{6} \\ (\text{ edge })^{2} & =16 \\ \text{ edge } & =\sqrt{16} \\ \text{ edge } & =4 ~cm \end{aligned} $

Therefore, the volume of cube is $=(\text{ edge })^{3}=(4 ~cm)^{3}=64 ~cm^{3}$.

Hence, the correct option is (C).

Solution

The formula of volume of the cone is $\dfrac{1}{3} \pi r^{2} h$.

So, $\dfrac{1}{3} \pi r^{2} h=\dfrac{1}{3} \pi(2.1)^{2} \times 8.4$

Now, volume of sphere $=\dfrac{4}{3} \pi r_1^{3}$

According to the question,

$\dfrac{4}{3} \pi r_1^{3}=\dfrac{1}{3} \pi(2.1)^{2} \times 8.4$

$4 r_1^{3}=(2.1)^{2} \times 8.4$

$r_1^{3}=\dfrac{(2.1)^{2} \times 8.4}{4}$

$r_1^{3}=(2.1)^{2} \times 2.1$

$r_1^{3}=(2.1)^{3}$

$r_1=2.1$

Hence, the correct option is (B).

Solution

The formula of curved surface area of cylinder is $2 \pi r h$.

According to the question, when radius is double and height is halved, then the curve surface are will be:

$=2 \pi \times(2 r) \times \dfrac{h}{2}$

$=2 \pi r h$

Since, the curved surface area will be same.

Hence, the correct option is (C).

Solution

The formula of Total surface area of cone $=$ Area of the base + Curved surface area of cone

$ \begin{aligned} & =\pi(\dfrac{r}{2})^{2}+\pi(\dfrac{r}{2}) \times 2 l \\ & =\dfrac{\pi r}{2}(\dfrac{r}{2}+2 l) \\ & =\dfrac{\pi r}{2}(r+4 l) \\ & =\pi r(l+\dfrac{r}{4}) \end{aligned} $

Hence, the correct option is (B).

Solution

Let the radii of two cylinders be $2 r$ and $3 r$ respectively and their heights are in the ratio $5 h$ and $3 h$. The volume of cylinders be $V_1$ and $V_2$. So,

$ \begin{aligned} \dfrac{V_1}{V_2} & =\dfrac{\pi(2 r)^{2}(5 h)}{\pi(3 r)^{2}(3 h)} \\ & =\dfrac{4 r^{2} \times 5 h}{3 r^{2} \times 3 h} \\ & =\dfrac{20}{27} \end{aligned} $

Hence, the correct option is (B).

Solution

The formula of the lateral surface area of a cube is $4(\text{ edge })^{2}$.

So,

$4(\text{ edge })^{2}=256$

$(\text{ edge })^{2}=\dfrac{256}{4}$

$(\text{ edge })^{2}=64$

edge $=\sqrt{64}=8 m$

Therefore, volume of cube $=(\text{ edge })^{3}=8^{3}=512 m^{3}$

Hence, the correct option is (A).

Solution

Volume of pit $=(16 \times 12 \times 4) m^{3}$

Volume of a plank $=(4 \times 0.5 \times 0.2) m^{3}$

Now, the required number of planks is calculated as follows:

Required number of planks $=\dfrac{\text{ Volume of pit }}{\text{ Volume of plank }}$

$ =\dfrac{16 \times 12 \times 4}{4 \times 0.5 \times 0.2} $

$ =1920 $

Therefore, the required number of planks is 1920 .

Hence, the correct option is (B).

Solution

The formula of the longest pole $=\sqrt{l^{2}+b^{2}+h^{2}}$

$ \begin{aligned} & =\sqrt{10^{2}+10^{2}+5^{2}} \\ & =\sqrt{100+100+25} \\ & =\sqrt{225} \\ & =15 ~cm \end{aligned} $

Hence, the correct option is (A).

Solution

As we know that balloon is hemispherical in shape.

The formula of surface area of hemispherical balloon of radius is $2 \pi r^{2}$.

So, the ratio of the surface areas of two balloons $=1: 4$

Hence, the correct option is (A).

Solution

We consider the radius of the sphere be $r$.

Given in the question, that height and diameter of cylinder are equal to the diameter of sphere.

Therefore, the radius of cylinder is $r$ and its height be $2 r$.

As, volume of sphere $=\dfrac{2}{3}$ Volume of cylinder

$ \begin{aligned} \dfrac{4}{3} \pi r^{3} & =\dfrac{2}{3}(\pi r^{2} \times 2 r) \\ & =\dfrac{4}{3} \pi r^{3} \end{aligned} $

Hence, the given statement is true.

Solution

We consider the original radius of the cone be $r$ and height ne $h$.

The volume of cone $=\dfrac{1}{3} \pi r^{2} h$

As we know that the radius of a height circular cone is halved and height is doubled. So,

$V=\dfrac{1}{3} \pi(\dfrac{r}{2})^{2} \times 2 h$

$ \begin{aligned} & =\dfrac{1}{3} \pi \times \dfrac{r^{2}}{4} \times 2 h \\ & =\dfrac{1}{2}(\dfrac{1}{3} \pi r^{2} h) \end{aligned} $

Since, the volume become half of the original volume.

Hence, the given statement is false.

Solution

We consider that in a right circle cone, height (h), radius (r), and slant height(l) are always the sides of a right triangle that is $l^{2}=r^{2}+h^{2}$.

Hence, the given statement is false.

Solution

We consider that radius and height of the cylinder be $r$ and $h$ respectively.

So, the curved surface area of cylinder $=2 \pi r h$

Now, according to the question, when radius is doubled and the curved surface area is not changed that means the height must be halved. So,

The formula of curved surface area $=2 \pi \times(2 r) \times \dfrac{h}{2}=2 \pi r h$

Hence, the given statement is true.

Solution

Given in the question, edge of cube $=2 r$, then height of cube becomes $h=2 r$.

The formula of volume of a cone $=\dfrac{1}{3} \pi r^{2} h$

$=\dfrac{1}{3} \pi r^{2}(2 r)$

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

The formula of volume of a hemisphere $=\dfrac{2}{3} \pi r^{3}$

Therefore, the volume of a cone is equal to the volume of a hemisphere.

Hence, the given statement is true.

Solution

We consider that the radius of the base of a cylinder and a right circular cone be $r$ and height be h. So,

The formula of volume of a cylinder $=\pi r^{2} h$

The formula of volume of a cone $=\dfrac{1}{3} \pi r^{2} h$

Since, Volume of a cylinder $=3 \times$ Volume of a cone

Therefore, the volume of a cylinder is three times the volume of the right circular cone.

Hence, the given statement is true.

Solution

Let radius of hemisphere is $r$.

The formula of volume of a cone, $=V_1=\dfrac{1}{3} \pi r^{2} h$

$ \begin{aligned} V_1 & =\dfrac{1}{3} \pi r^{2}(r) \quad[\text{ As } h=r] \\ & =\dfrac{1}{3} \pi r^{3} \end{aligned} $

Volume of a hemisphere, $V_2=\dfrac{2}{3} \pi r^{3}$

volume of cylinder, $V_3=\pi r^{2} h=\pi r^{2} \times r=\pi r^{3}[$ As h $=r]$

$V_1: V_2: V_3=\dfrac{1}{2} \pi r^{3}: \dfrac{2}{3} \pi r^{3}: \pi r^{3}=1: 2: 3$

Therefore, the ratio of their volumes is $1: 2: 3$.

Hence, the given statement is true.

Solution

Given, the length of the diagonal of a cube $=6 \sqrt{3} ~cm$

we consider the edge (side) of a cube be a ~cm.

So, diagonal of a cube $=a \sqrt{3}$

$6 \sqrt{3}=a \sqrt{3}$

$a=6 ~cm$

Therefore, the edge of a cube is $6 ~cm$.

Hence, the given statement is false.

Solution

We consider a be the edge of the cube.

As the sphere is inscribed in a cube, the radius of the sphere is $\dfrac{a}{2}$.

$V_1=$ Volume of cube $=(\text{ edge })^{3}=a^{3}$ $V_2=$ Volume of sphere $=\dfrac{4}{3} \pi(\dfrac{a}{2})^{3}=\dfrac{1}{3} \pi a^{3}$

$V_1: V_2=a^{3}: \dfrac{1}{3} \pi a^{3}=1: \dfrac{\pi}{6}=6: \pi$

Hence, the given statement is true.

Solution

Let the cylinder have radius $r$ and height $h$.

So, the volume of cylinder $(V_1)=\pi r^{2} h$

According to the question, when radius of cylinder is doubled and height is halved $(V_2)$. So,

$ \begin{aligned} V_2 & =\pi(2 r)^{2} \times \dfrac{h}{2} \\ & =\pi \times 4 r^{2} \times \dfrac{h}{2} \\ & =\pi \times 2 r^{2} \times h \\ & =2 \pi r^{2} h \\ & =2 V_1 \end{aligned} $

Hence, the given statement is true.

Solution

Given, radius of each metal sphere $=2 ~cm$

So, volume of a metallic sphere $=\dfrac{4}{3} \pi r^{3}$

$ \begin{aligned} & =\dfrac{4}{3} \times 3.14 \times(2 ~cm)^{3} \\ & =\dfrac{100.48}{3} ~cm^{3} \\ & =33.49 ~cm^{3} \end{aligned} $

Now, the volume of 16 such a sphere $=16 \times 33.49 ~cm^{3}=535.84 ~cm^{3}$

Dimensions of internal box is $16 ~cm \times 8 ~cm \times 8 ~cm$.

Now, internal volume of a rectangular box $=16 ~cm \times 8 ~cm \times 8 ~cm=1024 ~cm^{3}$

Volume of the preservative liquid $=1024 ~cm^{3}-535.84 ~cm^{3}=488.16 ~cm^{3}$

Solution

Suppose the edge of the cube is a.

So, the volume of cube $=a^{3}$

Now, the volume of water when the cube is full of water is $15.625 m^{3}$.

According to the question,

$ \begin{aligned} a^{3} & =15.625 m^{3} \\ a & =\sqrt[3]{15.625 m^{3}} \\ & =\sqrt[3]{(2.5)^{3} m^{3}} \\ & =2.5 m \end{aligned} $

So, edge of cube $=2.5 ~cm$

Now, present depth of water in the tank $=1.3 m \quad$ [Given]

So, remaining depth $=2.5 m-1.3 m=1.2 m$

Therefore, Volume of water already used in the tank $=2.5 m \times 2.5 m \times 1.2 m=7.5 m^{3}$

Solution

Given: Diameter of spherical ball $=4.2 ~cm$

Now, radius of spherical ball $(r)=\dfrac{4.2}{2} ~cm=2.1 ~cm$

Amount of water displaced by solid spherical ball $=$ Volume of solid spherical ball

So, volume of spherical ball $=\dfrac{4}{3} \pi r^{3}=\dfrac{4}{3} \times \dfrac{22}{7} \times(2.1)^{3}$

$ =\dfrac{88}{21} \times \dfrac{21}{10} \times \dfrac{21}{10} \times \dfrac{21}{10}=38.808 ~cm^{3} $

Therefore, the amount of water displaced by solid spherical ball when it completely immersed in water is $38.808 ~cm^{3}$.

Solution

Given: Height is $h=3.5 m$ and the radius of the base is $r=12 m$.

Now, Slant height $(l)$ will be:

$ \begin{aligned} \sqrt{h^{2}+r^{2}} & =\sqrt{(3.5 m)^{2}+(12 m)^{2}} \\ & =\sqrt{12.25 m^{2}+144 m^{2}} \\ & =\sqrt{156.25 m^{2}} \\ & =12.5 m \end{aligned} $

So, area of canvas required:

$=\pi r l$

$=\dfrac{22}{7} \times 12 m \times 12.5 m$

$=471.42 m^{2}$

Solution

Let weight of one solid sphere is $m_1=5920 g$ and its radius is $r_1$. Similarly, weight of another solid sphere is $m_2=740 g$ and its radius is $r_2$.

Now, diameter of the smaller sphere: $r_1=5 m$

So, it’s radius $=r_2=\dfrac{5}{2}$

As we know that:

Density $(D)=\dfrac{\text{ Mass }}{\text{ Volume }}$ or Volume $=\dfrac{\text{ Mass }}{\text{ Density }}$

Then,

$V_1=\dfrac{5920}{D} ~cm^{3}$

And: $V_2=\dfrac{740}{D} ~cm^{3}$

Now, dividing equation, (I) and (II), get:

$\dfrac{V_1}{V_2}=\dfrac{\dfrac{5920}{D} ~cm^{3}}{\dfrac{740}{D} ~cm^{3}}$

$\dfrac{\dfrac{4}{3} \pi r_1^{3}}{\dfrac{4}{3} \pi r_2^{3}}=\dfrac{5920}{740}$ [Volume of a sphere is $\dfrac{4}{3} \pi r^{3}$ ]

$\dfrac{r_1^{3}}{r_2^{3}}=\dfrac{592}{74}$

$(\dfrac{r_1}{\dfrac{5}{2}})^{3}=\dfrac{592}{74}$

$r_1^{3}=\dfrac{592}{74} \times \dfrac{125}{8}$

$=\dfrac{74000}{592}$

$=125$

$r_1^{3}=125$

$r_1=5 ~cm$

Hence, the radius of larger sphere is $5 ~cm$.

Solution

Given: diameter of cylinder glass $=7 ~cm$

Glass is filled with milk upto an height of $12 ~cm$.

Radius of cylinder glass $=\dfrac{\text{ Diameter }}{2}=\dfrac{7}{2}=3.5 ~cm$

Volume of cylinder glass $(V)=\pi r^{2} h$

$ \begin{aligned} & =\dfrac{22}{7} \times(3.5)^{2} \times 12 \\ & =\dfrac{22}{7} \times 12.25 \times 12 \\ & =22 \times 1.75 \times 12 \\ & =462 ~cm^{2} \end{aligned} $

Quantity of milk needed for 1600 students $=(462 \times 1600) ~cm^{3}$

$ \begin{aligned} & =739200 ~cm^{3} \\ & =739.2 \text{ litres } \end{aligned} $

Solution

Given: Length of cylinder roller $(h)=2.5 m$

Radius of cylinder roller $(r)=1.75 m$

Total area on road covered by cylinder roller $=5500 m^{2}$

Now, area covered in one revolution $=$ lateral surface area of the cylinder

$ \begin{aligned} & =2 \pi r h \\ & =2 \times \dfrac{22}{7} \times 1.75 m \times 2.5 m \\ & =27.5 m^{2} \end{aligned} $

Since, the number of revolution (n) made by the roller is:

$=\dfrac{\text{Total area covered}}{\text{ Area covered in one revolution }}$

$=\dfrac{5500}{27.5}$

$=200$ revolutions.

Solution

Water contained in overhead tank $=40 m \times 25 m \times 15 m$.

$=(40 \times 25 \times 15) m^{3}$

$=(40 \times 25 \times 15 \times 1000)$ litres

Now, water needed for 5000 villages for one day $=(5000 \times 75)$ litres $=375000$ litres

So, total number of days the water of the tank last:

$ \begin{aligned} & =\dfrac{40 \times 25 \times 15 \times 1000}{375000} \\ & =\dfrac{15000}{375} \\ & =40 \text{ days } \end{aligned} $

Solution

Number of ladoos $=\dfrac{\text{ Volum of spherical ladoo of radius } 5 ~cm}{\text{ Volume of one spherical ladoo of radius } 2.5 ~cm}$

$ \begin{aligned} & =\dfrac{\dfrac{4}{3} \times \dfrac{22}{7} \times 5^{3}}{\dfrac{4}{3} \times \dfrac{22}{7} \times(\dfrac{5}{3})^{3}} \\ & =8 \end{aligned} $

Solution

According to the question, the solid formed is a cone whose height of a cone, $h=8 ~cm$ and radius of a cone, $r=6 ~cm$. Slant height of a cone, $1=10 ~cm$. So,

Volume of a cone $=\dfrac{1}{3} \pi r^{2} h=\dfrac{1}{3} \times \dfrac{22}{7} \times 6 \times 6 \times 8$

$ \dfrac{6336}{21}=301.7 ~cm^{3} $

Now, curved surface of the area of cone $=\pi rl$

$ \begin{aligned} & =\dfrac{22}{7} \times 6 \times 10 \\ & =\dfrac{1320}{7} \\ & =188.5 ~cm^{2} \end{aligned} $

Therefore, the volume and surface area of a cone are $301.7 ~cm^{3}$ and $188.5 ~cm^{2}$, respectively.

Solution

Given:

Outer diameter of cylinder tube $(d)=16 ~cm$

Thickness of the iron sheet $=2 ~cm$

Height of the cylindrical tube $(h)=100 ~cm$

Outer radius of a cylindrical tube $(r_1)=\dfrac{d}{2}=\dfrac{16}{2} ~cm=8 ~cm$

Inner radius of a cylindrical tube $=(r_1-.$ thickness of the iron sheet $)$

$=8-2$

$=6 ~cm$

Now, volume of metal used in making cylindrical tube $=$ Outer volume of a cylindrical tube Inner volume of cylindrical tube

$ \begin{aligned} & =\pi r_1^{2} h-\pi r_2^{2} h \\ & =\pi h(r_1^{2}-r_2^{2}) \\ & =\dfrac{22}{7} \times 100(8^{2}-6^{2}) \\ & =\dfrac{22}{7} \times 100 \times(8-6) \times(8+6) \\ & =2200 \times 4 \\ & =8800 ~cm^{2} \end{aligned} $

Hence, $8800 ~cm^{2}$ of iron has been used in making the tube.

Solution

Given: Diameter of a semi-circular sheet is $28 ~cm$. So,

Radius $(r)=\dfrac{28}{2}=14 ~cm$

$=14 ~cm$

Suppose the radius of a conical cup be R.

So, Circumference of base of cone $=$ Circumference of semi-circle

$2 \pi R=\pi r$

$2 \pi R=\pi \times 14$

$R=7 ~cm$

Now,

$ \begin{aligned} h & =\sqrt{l^{2}-R^{2}} \\ & =\sqrt{14^{2}-7^{2}} \\ & =\sqrt{196-49} \\ & =\sqrt{147} \\ & =12.1243 ~cm \end{aligned} $

Volume of conical cup $=\dfrac{1}{3} \pi R^{2} h$

$ \begin{aligned} & =\dfrac{1}{3} \times \dfrac{22}{7} \times 7 \times 7 \times 12.1243 \\ & =622.38 ~cm^{3} \end{aligned} $

Hence, the capacity of an open conical cup is $622.38 ~cm^{3}$.

Solution

(i) Given:

Radius of the base of a conical tent $=5 ~cm$

Area needs to sit a student on the ground $=\dfrac{5}{7} m^{2}$

So, area of the base of a conical tent $=\pi r^{2}$

$ =\dfrac{22}{7} \times 5 \times 5 m^{2} $

Now, number of students $=\dfrac{\text{ Area of the base of a conical tent }}{\text{ Area needs to sit a student on the ground }}$

$ \begin{aligned} & =\dfrac{\dfrac{22 \times 5 \times 5}{7}}{\dfrac{5}{7}} \\ & =\dfrac{22}{7} \times 5 \times 5 \times \dfrac{7}{5} \\ & =110 \end{aligned} $

Hence, 110 students can sit in the conical tent.

(ii) Given: area of the cloth to form a conical tent $=165 m^{2}$

Radius of the base of a conical tent( $r)=5 m$

Now, Curved surface area of a conical tent $=$ Area of cloth to form a conical tent $\pi r l=165$

$\dfrac{22}{7} \times 5 \times l=165$

$ \begin{aligned} l & =\dfrac{165 \times 7}{22 \times 5} \\ & =\dfrac{33 \times 7}{22} \\ & =10.5 m \end{aligned} $

Now, height of the conical tent is calculated as:

$ \begin{aligned} h & =\sqrt{l^{2}-r^{2}} \\ & =\sqrt{(10.5)^{2}-(5)^{2}} \\ & =\sqrt{110.25-25} \\ & =\sqrt{85.25} \\ & =9.23 \end{aligned} $

Volume of a cone $=\dfrac{1}{3} \pi r^{2} h$

$ \begin{aligned} & =\dfrac{1}{3} \times \dfrac{22}{7} \times 5 \times 5 \times 923 \\ & =\dfrac{1}{3} \times \dfrac{1550 \times 923}{7} \\ & =\dfrac{50765}{7 \times 3} \\ & =241.7 m^{3} \\ & \approx 242 m^{3} \end{aligned} $

Hence, the volume of the cone is $242 m^{3}$.

Solution

Given: The tank contains 50 kilolitres of water.

Internal diameter of a hemispherical tank $=14 ~cm$

So, internal radius of hemispherical tank $=\dfrac{\text{ Diameter }}{2}=\dfrac{14}{2} m=7 m$

Now, Volume of hemisphere tank $=\dfrac{2}{3} \pi r^{3}$

$ \begin{aligned} & =\dfrac{2}{3} \times \dfrac{22}{7} \times 7^{3} \\ & =\dfrac{44 \times 49}{3} \\ & =718.66 m^{3} \end{aligned} $

The tank contains 50 kilolitres of water $=50,0001$ itres $=\dfrac{50,000}{1,000} m^{3}=50 m^{3}$

Volume of water pumped into the tank $=718.66 m^{3}-50 m^{3}=668.66 m^{3}$

Solution

Suppose $V_1$ and $V_2$ be the volume of two sphere.

So, according to the question:

$\dfrac{V_1}{V_2}=\dfrac{64}{27}$

$$ \begin{matrix} \dfrac{\dfrac{4}{3} \pi r_1^{3}}{\dfrac{4}{3} \pi r_2^{3}}=\dfrac{64}{27} & \text{ [As volume of sphere is } \dfrac{4}{3} \pi r^{3} \text{ ] } \\ \dfrac{r_1^{3}}{r_2^{3}}=\dfrac{64}{27} & \\ \dfrac{r_1^{3}}{r_2^{3}}=\dfrac{4^{3}}{3^{3}} & \\ \dfrac{r_1}{r_2}=\dfrac{4}{3} & \ldots \text{ (I) } \tag{I} \end{matrix} $$

Let surface area of both spheres are $S A_1$ and $S A_2$ respectively. So,

$\dfrac{S A_1}{S A_2}=\dfrac{4 \pi r_1^{2}}{4 \pi r_2^{2}} \quad$ [As surface area of sphere is $4 \pi r^{2}$ ]

$\dfrac{S A_1}{S A_2}=\dfrac{r_1^{2}}{r_2^{2}}$

$\dfrac{S A_1}{S A_2}=(\dfrac{r_1}{r_2})^{2}$

$\dfrac{S A_1}{S A_2}=(\dfrac{4}{3})^{2} \quad[$ Using equation (I) $]$

$\dfrac{S A_1}{S A_2}=\dfrac{16}{9}$

Hence, the ratio of the surface area of the two sphere is $16: 9$.

Solution

Side of a cube $=4 ~cm$

As cube contains a sphere touching its sides. So, the diameter of the sphere $=4 ~cm$

Side of cube $=$ Diameter of sphere

4= Radius of sphere

Radius of sphere $=\dfrac{4}{2}=2$

Volume of the gap $=$ Volume of cube - Volume of sphere

$ \begin{aligned} & =(\text{ Side })^{3}-\dfrac{4}{3} \pi r^{3} \\ & =(4)^{3}-\dfrac{4}{3} \pi \times 2^{3} \quad[\text{ Since, side of cube }=\text{ diameter of sphere }] \end{aligned} $

$ \begin{aligned} & =(64-\dfrac{4}{3} \times \dfrac{22}{7} \times 8) \\ & =64-33.52 \\ & =30.48 ~cm^{3} \end{aligned} $

Hence, the volume of the gap between a cube and a sphere is $30.48 ~cm^{3}$.

Solution

Suppose the radius of sphere $=r=$ Radius of a right circular cylinder According to the question,

Volume of right circular cylinder $=$ Volume of a sphere [Given]

$ \begin{aligned} \pi r^{2} h & =\dfrac{4}{3} \pi r^{3} \\ h & =\dfrac{4}{3} r \end{aligned} $

As diameter of the cylinder $=2 r$

Increased diameter from height of the cylinder $=2 r-\dfrac{4 r}{3}=\dfrac{2 r}{3}$

Now, percentage increase in diameter of the cylinder $=\dfrac{\dfrac{2 r}{3} \times 100}{\dfrac{4}{3} r}$

$=50 %$

Hence, the diameter of the cylinder exceeds its height by $50 %$.

Solution

Given: radius of a circular plate( $r)=14 ~cm$

Thickness of the circular plates $=3 ~cm$

The height of the cylinder solid $(h)=$ Thickness of 30 circular plates $=30 \times 3=90 ~cm$

(i) Total surface area of the cylinder solid so formed $=2 \pi r(h+r)$

$=2 \times \dfrac{22}{7} \times 14 \times(90+14)$

$=44 \times 2 \times 104$

$=9152 ~cm^{2}$

Hence, the total surface area of the cylinder solid is $9152 ~cm^{2}$.

(ii) Volume of the cylinder so formed $=\pi r^{2} h$

$ \begin{aligned} & =\dfrac{22}{7} \times 14^{2} \times 90 \\ & =\dfrac{22}{7} \times 14 \times 14 \times 90 \\ & =22 \times 28 \times 90 \\ & =55440 ~cm^{3} \end{aligned} $

Hence, the volume of the cylinder so formed is $55440 ~cm^{3}$.