Solution

(a) Because the shape of sharpened pencil is

Solution

(a) Because the shape of surahi is

Solution

(b)

Solution

(b) We know that, the shape of frustum of a cone is

So, the given figure is usually in the form of frustum of a cone.

Solution

(c)

Solution

(d) Because the shape of the shuttle cock is equal to sum of frustum of a cone and hemisphere.

Solution

(a)

A cone sliced by a plane parallel to base

The two parts separated

Frustum of a cone

[when we remove the upper portion of the cone cut off by plane, we get frustum of a cone]

Thinking Process

If we divide the total volume filled by marbles in a cube by volume of a marble, then we get the required number of marbles.

Solution

(a) Given, edge of the cube $=22 cm$

$\therefore \quad$ Volume of the cube $=(22)^{3}=10648 cm^{3} \quad[\because.$ volume of cube $.=(\text{ side })^{3}]$

Also, given diameter of marble $=0.5 cm$

$\therefore$ Radius of a marble, $r=\dfrac{0.5}{2}=0.25 cm \quad[\because$ diameter $=2 \times$ radius $]$

Volume of one marble $=\dfrac{4}{3} \pi r^{3}=\dfrac{4}{3} \times \dfrac{22}{7} \times(0.25)^{3}$

$ \begin{aligned} & \quad[\because \text{ volume of sphere }=\dfrac{4}{3} \times \pi \times(\text{ radius })^{3}] \\ & =\dfrac{1.375}{21}=0.0655 cm^{3} \\ & =\text{ Volume of the cube } \dfrac{1}{8} \times \text{ Volume of cube } \\ & =10648-10648 \times \dfrac{1}{8} \\ & =10648 \times \dfrac{7}{8}=9317 cm^{3} \end{aligned} $

Filled space of cube $=$ Volume of the cube $\dfrac{1}{8} \times$ Volume of cube $\therefore$ Required number of marbles $=\dfrac{\text{ Total space filled by marbles in a cube }}{\text{ Volume of one marble }}$

$ =\dfrac{9317}{0.0655}=142244 \text{ (approx) } $

Hence, the number of marbles that the cube can accomodate is 142244 .

Thinking Process

When a solid shape is melted and recast into the form other solid shape, then volume of both shapes are equal.

Solution

(b) Given, internal diameter of spherical shell $=4 cm$ and external diameter of shell $=8 cm$

$\therefore$ Internal radius of spherical shell, $r_1=\dfrac{4}{2} cm=2 cm \quad[\because$ diameter $=2$ xradius $]$ and external radius of shell, $r_2=\dfrac{8}{2}=4 cm \quad[\because$ diameter $=2$ xradius $]$

Now, volume of the spherical shell $=\dfrac{4}{3} \pi[r_2^{3}-r_1^{3}]$

$[\because.$ volume of the spherical shell $.=\dfrac{4}{3} \pi{(\text{ external radius })^{3}-(\text{ internal radius })^{3}}]$

$ \begin{aligned} & =\dfrac{4}{3} \pi(4^{3}-2^{3}) \\ & =\dfrac{4}{3} \pi(64-8) \\ & =\dfrac{224}{3} \pi cm^{3} \end{aligned} $

Let height of the cone $=h cm$

Diameter of the base of cone $=8 cm$

$\therefore$ Radius of the base of cone $=\dfrac{8}{2}=4 cm \quad[\because$ diameter $=2$ xradius $]$

According to the question,

Volume of cone $=$ Volume of spherical shell

$\Rightarrow \quad \dfrac{1}{3} \pi(4)^{2} h=\dfrac{224}{3} \pi \Rightarrow h=\dfrac{224}{16}=14 cm$

$[\because.$ volume of cone $=\dfrac{1}{3} \times \pi \times(\text{ radius })^{2} \times($ height $.)]$

Hence, the height of the cone is $14 cm$.

Solution

(a) Given, dimensions of the cuboid $=49 cm \times 33 cm \times 24 cm$

$\therefore$ Volume of the cuboid $=49 \times 33 \times 24=38808 cm^{3}$

$[\because$ volume of cuboid $=$ length $\times$ breadth $\times$ height $]$

Let the radius of the sphere is $r$, then

Volume of the sphere $=\dfrac{4}{3} \pi r^{3} \quad[\because.$ voulme of the sphere $.=\dfrac{4}{3} \pi \times(\text{ radius })^{3}]$

According to the question,

Volume of the sphere $=$ Volume of the cuboid

$ \begin{matrix} \Rightarrow & \dfrac{4}{3} \pi^{3} = 38808 \\ \Rightarrow & 4 \times \dfrac{22}{7} r^{3} = 38808 \times 3 \\ \Rightarrow & r^{3} = \dfrac{38808 \times 3 \times 7}{4 \times 22}=441 \times 21 \\ \Rightarrow & r^{3} = 21 \times 21 \times 21 \\ & & r = 21 cm \end{matrix} $

Hence, the radius of the sphere is $21 cm$.

Thinking Process

If we divide the volume of the wall except the volume of mortar are used on wall by the volume of one brick, then we get the required number of bricks used to construct the wall.

Solution

(b) Volume of the wall $=270 \times 300 \times 350=28350000 cm^{3}$

$[\because$ volume of cuboid $=$ length $\times$ breadth $\times$ height $]$

Since, $\dfrac{1}{8}$ space of wall is covered by mortar.

So, remaining space of wall $=$ Volume of wall - Volume of mortar

$ \begin{aligned} & =28350000-28350000 \times \dfrac{1}{8} \\ & =28350000-3543750=24806250 cm^{3} \end{aligned} $

Now, $\quad$ volume of one brick $=22.5 \times 11.25 \times 8.75=2214.844 cm^{3}$

$[\because$ volume of cuboid $=$ length $\times$ breadth $\times$ height $]$

$\therefore$ Required number of bricks $=\dfrac{24806250}{2214.844}=11200$ (approx)

Hence, the number of bricks used to construct the wall is 11200 .

Solution

(c) Given, diameter of the cylinder $=2 cm$

$\therefore$ Radius $=1 cm$ and height of the cylinder $=16 cm$

$\therefore$ Volume of the cylinder $=\pi \times(1)^{2} \times 16=16 \pi cm^{3}$

$[\because.$ volume of cylinder $=\pi \times(\text{ radius })^{2} \times$ height $]$

Now, let the radius of solid sphere $=r cm$

Then, its volume $=\dfrac{4}{3} \pi rm^{3} \quad[\because.$ volume of sphere $.=\dfrac{4}{3} \times \pi \times(\text{ radius })^{3}]$

According to the question,

Volume of the twelve solid sphere $=$ Volume of cylinder

$ \begin{aligned} \Rightarrow & 12 \times \dfrac{4}{3} \pi r^{3} = 16 \pi \\ \Rightarrow & r^{3} = 1 \Rightarrow r=1 cm \end{aligned} $

$\therefore \quad$ Diameter of each sphere, $d=2 r=2 \times 1=2 cm$

Hence, the required diameter of each sphere is $2 cm$.

Solution

(a) Given, the radius of the top of the bucket, $R=28 cm$ and the radius of the bottom of the bucket, $r=7 cm$

Slant height of the bucket, $l=45 cm$

Since, bucket is in the form of frustum of a cone.

$\therefore$ Curved surface area of the bucket $=\pi l(R+r)=\pi \times 45(28+7)$

$ \begin{aligned} & {[\because \text{ curved surface area of frustum of a cone }=\pi(R+r) l]} \\ & \quad=\pi \times 45 \times 35=\dfrac{22}{7} \times 45 \times 35=4950 cm^{2} \end{aligned} $

Solution

(a) Given, diameter of cylinder $=$ Diameter of hemisphere $=0.5 cm$

[since, both hemispheres are attach with cylinder]

$\therefore$ Radius of cylinder $(r)=$ radius of hemisphere $(r)=\dfrac{0.5}{2}=0.25 cm$

$[\because$ diameter $=2$ xradius $]$

and total length of capsule $=2 cm$ $\therefore$ Length of cylindrical part of capsule,

$ \begin{aligned} h = \text{ Length of capsule }- \text{ Radius of both hemispheres } \\ & =2-(0.25+0.25)=1.5 cm \end{aligned} $

Now, capacity of capsule $=$ Volume of cylindrical part $+2 \times$ Volume of hemisphere

$ =\pi^{2} h+2 \times \dfrac{2}{3} \pi r^{3} $

$[\because.$ volume of cylinder $=\pi \times(\text{ radius })^{2} \times$ height and volume of hemispere $.=\dfrac{2}{3} \pi(\text{ radius })^{3}]$

$ \begin{aligned} & =\dfrac{22}{7}[(0.25)^{2} \times 1.5+\dfrac{4}{3} \times(0.25)^{3}] \\ & =\dfrac{22}{7}[0.09375+0.0208] \\ & =\dfrac{22}{7} \times 0.11455=0.36 cm^{3} \end{aligned} $

Hence, the capacity of capsule is $0.36 cm^{3}$.

Solution

(a) Because curved surface area of a hemisphere is $2 \pi^{2}$ and here, we join two solid hemispheres along their bases of radius $r$, from which we get a solid sphere. Hence, the curved surface area of new solid $=2 \pi^{2}+2 \pi r^{2}=4 \pi^{2}$

Solution

(b) Because the sphere encloses in the cylinder, therefore the diameter of sphere is equal to diameter of cylinder which is $2 r cm$.

Solution

(c) During conversion of a solid from one shape to another, the volume of the new shape will remain unaltered.

Solution

(a) Given, diameter of one end of the bucket,

$ 2 R=44 \Rightarrow R=22 cm \quad[\because \text{ diameter, } r=2 \text{ xradius }] $

and diameter of the other end,

$ 2 r=24 \Rightarrow r=12 cm \quad[\because \text{ diameter, } r=2 \text{ xradius }] $

Height of the bucket, $h=35 cm$

Since, the shape of bucket is look like as frustum of a cone. $\therefore \quad$ Capacity of the bucket $=$ Volume of the frustum of the cone

$ \begin{aligned} & =\dfrac{1}{3} \pi h[R^{2}+r^{2}+R r] \\ & =\dfrac{1}{3} \times \pi \times 35[(22)^{2}+(12)^{2}+22 \times 12] \\ & =\dfrac{35 \pi}{3}[484+144+264] \\ & =\dfrac{35 \pi \times 892}{3}=\dfrac{35 \times 22 \times 892}{3 \times 7} \end{aligned} $

$ =32706.6 cm^{3}=32.7 L \quad[\because 1000 cm^{3}=1 L] $

Hence, the capacity of bucket is $32.7 L$.

Solution

(b) We know that, if a cone is cut by a plane parallel to the base of the cone, then the portion between the plane and base is called the frustum of the cone.

Solution

(d) Let the radii of the two spheres are $r_1$ and $r_2$, respectively.

$\therefore \quad$ Volume of the sphere of radius, $r_1=V_1=\dfrac{4}{3} \pi r_1^{3}$

$$ \begin{equation*} [\because \text{ volume of sphere }=\dfrac{4}{3} \pi(\text{ radius })^{3}] \tag{i} \end{equation*} $$

and volume of the sphere of radius, $r_2=V_2=\dfrac{4}{3} \pi r_2^{3}$

Given, ratio of volumes $=V_1: V_2=64: 27 \Rightarrow \dfrac{\dfrac{4}{3} \pi r_1^{3}}{\dfrac{4}{3} \pi r_2^{3}}=\dfrac{64}{27} \quad$ [using Eqs. (i) and (ii)]

$\Rightarrow \quad \dfrac{r_1^{3}}{r_2^{3}}=\dfrac{64}{27} \Rightarrow \dfrac{r_1}{r_2}=\dfrac{4}{3}$

Now, ratio of surface area $=\dfrac{4 \pi_1^{2}}{4 \pi_2{ }^{2}} \quad[\because.$ surface area of a sphere $.=4 \pi(\text{ radius })^{2}]$

$ \begin{aligned} & =\dfrac{r_1^{2}}{r_2^{2}} \\ & ={\dfrac{r_1}{r_2}}^{2}=\dfrac{4}3^{2} \\ & =16: 9 \end{aligned} $

[using Eq. (iii)]

Hence, the required ratio of their surface area is $16: 9$.

Solution

False

Curved surface area of a hemisphere $=2 \pi^{2}$

Here, two identical soild hemispheres of equal radius are stuck together. So, base of both hemispheres is common.

$\therefore$ Total surface area of the combination

$ =2 \pi r^{2}+2 \pi r^{2}=4 \pi r^{2} $

Solution

False

Since, the total surface area of cylinder of radius, $r$ and height, $h=2 \pi h+2 \pi r^{2}$

When one cylinder is placed over the other cylinder of same height and radius, then height of the new cylinder $=2 h$

and radius of the new cylinder $=r$

$\therefore$ Total surface area of the new cylinder $=2 \pi(2 h)+2 \pi^{2}=4 \pi h+2 \pi r^{2}$

Solution

False

We know that, total surface area of a cone of radius, $r$

and

height, $h=$ Curved surface Area + area of base $=\pi l+\pi^{2}$

where,

$ l=\sqrt{h^{2}+r^{2}} $

and total surface area of a cylinder of base radius, $r$ and height, $h$

$ =\text{ Curved surface area }+ \text{ Area of both base }=2 \pi r h+2 \pi r^{2} $

Here, when we placed a cone over a cylinder, then one base is common for both.

So, total surface area of the combined solid

$ \begin{aligned} & =\pi r l+2 \pi h+\pi r^{2}=\pi r[l+2 h+r] \\ & =\pi r \sqrt{r^{2}+h^{2}}+2 h+r \end{aligned} $

Solution

False

Because solid ball is exactly fitted inside the cubical box of side $a$. So, $a$ is the diameter for the solid ball.

$ \begin{matrix} \therefore & \text{ Radius of the ball }=\dfrac{a}{2} \\ \text{ So, } \quad & \text{ volume of the ball }=\dfrac{4}{3} \pi \dfrac{a^{3}}{2}=\dfrac{1}{6} \pi a^{3} \end{matrix} $

Solution

False

Since, the volume of the frustum of a cone is $\dfrac{1}{3} \pi h[r_1^{2}+r_2^{2}+r_1 r_2]$, where $h$ is vertical height of the frustum and $r_1, r_2$ are the radii of the ends.

Solution

True

We know that, capacity of cylindrical vessel $=\pi r^{2} h cm^{3}$

and capacity of hemisphere $=\dfrac{2}{3} \pi r^{3} cm$

From the figure, capacity of the cylindrical vessel

$ =\pi r^{2} h-\dfrac{2}{3} \pi r^{3}=\dfrac{1}{3} \pi r^{2}[3 h-2 r] $

Solution

False

We know that, if $r_1$ and $r_2$ are the radii of the two ends of the frustum and $h$ is the vertical height, then curved surface area of a frustum is $\pi /(r_1+r_2)$, where $l=\sqrt{h^{2}+(r_1-r_2)^{2}}$.

Solution

True

Because the resulting figure is

Here, $A B C D$ is a frustum of a cone and $C D E F$ is a hollow cylinder.

Solution

Given, edges of three solid cubes are $3 cm, 4 cm$ and $5 cm$, respectively.

$\therefore \quad$ Volume of first cube $=(3)^{3}=27 cm^{3}$

Volume of second cube $=(4)^{3}=64 cm^{3}$

and

volume of third cube $=(5)^{3}=125 cm^{3}$

$\therefore \quad$ Sum of volume of three cubes $=(27+64+125)=216 cm^{3}$

Let the edge of the resulting cube $=R cm$

Then, volume of the resulting cube, $R^{3}=216 \Rightarrow R=6 cm$

Solution

Given, dimensions of cuboidal $=9 cm \times 11 cm \times 12 cm$

$\therefore$ Volume of cuboidal $=9 \times 11 \times 12=1188 cm^{3}$

and diameter of shot $=3 cm$

$ \begin{aligned} \therefore \quad \text{ Radius of shot, } r = \dfrac{3}{2}=1.5 cm \\ \text{ Volume of shot } = \dfrac{4}{3} \pi^{3}=\dfrac{4}{3} \times \dfrac{22}{7} \times(1.5)^{3} \\ & =\dfrac{297}{21}=14.143 cm^{3} \end{aligned} $

Solution

Given, volume of the frustum $=28.49 L=28.49 \times 1000 cm^{3}$

$ [\because 1 L=1000 cm^{3}] $

$ =28490 cm^{3} $

and radius of the top $(r_1)=28 cm$

radius of the bottom $(r_2)=21 cm$

Let height of the bucket $=h cm$

Now, volume of the bucket $=\dfrac{1}{3} \pi h(r_1^{2}+r_2^{2}+r_1 r_2)=28490$

[given]

$ \begin{matrix} \Rightarrow & \dfrac{1}{3} \times \dfrac{22}{7} \times h(28^{2}+21^{2}+28 \times 21)=28490 \\ \Rightarrow & h(784+441+588)=\dfrac{28490 \times 3 \times 7}{22} \\ \Rightarrow & 1813 h=1295 \times 21 \\ \therefore & h=\dfrac{1295 \times 21}{1813}=\dfrac{27195}{1813}=15 cm \end{matrix} $

Solution

Let ORN be the cone then given, radius of the base of the cone $r_1=8 cm$.

and height of the cone, (h) $O M=12 cm$

Let $P$ be the mid-point of $O M$, then

$ \begin{matrix} & O P = P M=\dfrac{12}{2}=6 cm \\ \text{ Now, } & \Delta O P D & \sim \triangle O M N \\ \therefore & \dfrac{O P}{O M} = \dfrac{P D}{M N} \\ \Rightarrow & \dfrac{6}{12} = \dfrac{P D}{8} \Rightarrow \dfrac{1}{2}=\dfrac{P D}{8} \\ \Rightarrow & P D = 4 cm \end{matrix} $

The plane along $C D$ divides the cone into two parts, namely

(i) a smaller cone of radius $4 cm$ and height $6 cm$ and (ii) frustum of a cone for which

Radius of the top of the frustum, $r_1=4 cm$

Radius of the bottom, $r_2=8 cm$

$ \begin{aligned} & \therefore \quad \begin{aligned} \text{ Volume of smaller cone } = \dfrac{1}{3} \pi \times 4 \times 4 \times 6=32 \pi cm^{3} \\ \text{ and volume of the frustum of cone } = \dfrac{1}{3} \times \pi \times 6[(8)^{2}+(4)^{2}+8 \times 4] \\ =2 \pi(64+16+32)=224 cm^{3} \end{aligned} \end{aligned} $

$\therefore$ Required ratio $=$ Volume of frustum : Volume of cone $=24 \pi: 32 \pi=1: 7$

Solution

Let the length of a side of a cube $=a cm$

Given, volume of the cube, $a^{3}=64 cm^{3} \Rightarrow a=4 cm$

On joining two cubes, we get a cuboid whose

$ \begin{aligned} \text{ length, } l = 2 acm \\ \text{ breadth, } b = a cm \\ \text{ height, } h = a cm \\ \text{ ting cuboid } = 2(l b+b h+h l) \\ & =2(2 a \cdot a+a \cdot a+a \cdot 2 a) \\ & =2(2 a^{2}+a^{2}+2 a^{2})=2(5 a^{2}) \\ & =10 a^{2}=10(4)^{2}=160 cm^{2} \end{aligned} $

$ \begin{aligned} & \text{ and height, } h=a cm \\ & \text{ Now, surface area of the resulting cuboid }=2(l b+b h+h l) \end{aligned} $

Solution

Given that, side of a solid cube $(a)=7 cm$

Height of conical cavity i.e., cone, $h=7 cm$

Since, the height of conical cavity and the side of cube is equal that means the conical cavity fit vertically in the cube.

Radius of conical cavity i.e., cone, $r=3 cm$ $\Rightarrow$ Diameter $=2 \times r=2 \times 3=6 cm$

Since, the diameter is less than the side of a cube that means the base of a conical cavity is not fit inhorizontal face of cube.

Now, volume of cube $=(\text{ side })^{3}=a^{3}=(7)^{3}=343 cm^{3}$

and volume of conical cavity i.e., cone $=\dfrac{1}{3} \pi \times r^{2} \times h$

$ \begin{aligned} & =\dfrac{1}{3} \times \dfrac{22}{7} \times 3 \times 3 \times 7 \\ & =66 cm^{3} \end{aligned} $

$\therefore$ Volume of remaining solid $=$ Volume of cube - Volume of conical cavity

$ =343-66=277 cm^{3} $

Hence, the required volume of solid is $277 cm^{3}$.

Solution

If two cones with same base and height are joined together along their bases, then the shape so formed is look like as figure shown.

Given that, radius of cone, $r=8 cm$ and height of cone, $h=15 cm$

So, surface area of the shape so formed

$=$ Curved area of first cone + Curved surface area of second cone

$=2 \cdot$ Surface area of cone

$=2 \times \pi r l=2 \times \pi \times r \times \sqrt{r^{2}+h^{2}}$

[since, both cones are identical]

$=2 \times \dfrac{22}{7} \times 8 \times \sqrt{(8)^{2}+(15)^{2}}$

$=\dfrac{2 \times 22 \times 8 \times \sqrt{64+225}}{7}$

$=\dfrac{44 \times 8 \times \sqrt{289}}{7}$

$=\dfrac{44 \times 8 \times 17}{7}$

$=\dfrac{5984}{7}=854.85 cm^{2}$

$=855 cm^{2}$ (approx)

Hence, the surface area of shape so formed is $855 cm^{2}$.

Solution

Let volume of cone $A$ be $2 V$ and volume of cone $B$ be $V$. Again, let height of the cone $A=h_1$ $cm$, then height of cone $B=(21-h_1) cm$

Given, diameter of the cone $=6 cm$

$\therefore \quad$ Radius of the cone $=\dfrac{6}{2}=3 cm$

Now, volume of the cone, $A=2 V=\dfrac{1}{3} \pi r^{2} h=\dfrac{1}{3} \pi(3)^{2} h_1$

$\Rightarrow \quad V=\dfrac{1}{6} \pi 9 h_1=\dfrac{3}{2} h_1 \pi$

and volume of the cone, $B=V=\dfrac{1}{3} \pi(3)^{2}(21-h_1)=3 \pi(21-h_1)$

From Eqs. (i) and (ii),

$ \begin{aligned} & \Rightarrow \quad h_1=2(21-h_1) \\ & \Rightarrow \quad 3 h_1=42 \\ & \Rightarrow \quad h_1=\dfrac{42}{3}=14 cm \\ & \therefore \quad \text{ Height of cone, } B=21-h_1=21-14=7 cm \\ & \begin{matrix} \text{ Now, volume of the cone, } A=3 \times 14 \times \dfrac{22}{7}=132 cm^{3} \quad \text{ [using Eq. (i)] } \end{matrix} \\ & \text{ and volume of the cone, } B=\dfrac{1}{3} \times \dfrac{22}{7} \times 9 \times 7=66 cm^{3} \quad \text{ [using Eq. (ii)] } \\ & \text{ Now, } \quad \text{ volume of the cylinder }=\pi r^{2} h=\dfrac{22}{7}(3)^{2} \times 21=594 cm^{3} \end{aligned} $

$\therefore$ Required volume of the remaining portion $=$ Volume of the cylinder

$ \begin{aligned} & \quad-(\text{ Volume of cone } A+\text{ Volume of cone } B) \\ = & 594-(132+66) \\ = & 396 cm^{3} \end{aligned} $

Solution

Given, ice-cream cone is the combination of a hemisphere and a cone.

Also, radius of hemisphere $=5 cm$

$\therefore \quad$ Volume of hemisphere $=\dfrac{2}{3} \pi^{3}=\dfrac{2}{3} \times \dfrac{22}{7} \times(5)^{3}$

$ =\dfrac{5500}{21}=261.90 cm^{3} $

Now, $\quad$ radius of the cone $=5 cm$

and height of the cone $=10-5=5 cm$

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

$ \begin{aligned} & =\dfrac{1}{3} \times \dfrac{22}{7} \times(5)^{2} \times 5 \\ & =\dfrac{2750}{21}=130.95 cm^{3} \end{aligned} $

Now, total volume of ice-cream cone $=261.90+130.95=392.85 cm^{3}$

Since, $\dfrac{1}{6}$ part is left unfilled with ice-cream.

$\therefore$ Required volume of ice-cream $=392.85-392.85 \times \dfrac{1}{6}=392.85-65.475$

$ =327.4 cm^{3} $

Solution

Given, diameter of a marble $=1.4 cm$

$ \begin{aligned} \therefore \quad \text{ Radius of marble } = \dfrac{1.4}{2}=0.7 cm \\ \text{ So, volume of one marble } = \dfrac{4}{3} \pi(0.7)^{3} \\ & =\dfrac{4}{3} \pi \times 0.343=\dfrac{1.372}{3} \pi cm^{3} \end{aligned} $

Also, given diameter of beaker $=7 cm$

$\therefore$ Radius of beaker $=\dfrac{7}{2}=3.5 cm$

Height of water level raised $=5.6 cm$

$\therefore$ Volume of the raised water in beaker $=\pi(3.5)^{2} \times 5.6=68.6 \pi cm^{3}$

Now, required number of marbles $=\dfrac{\text{ Volume of the raised water in beaker }}{\text{ Volume of one spherical marble }}$

$ =\dfrac{68.6 \pi}{1.372 \pi} \times 3=150 $

Solution

Given that, lots of spherical lead shots made from a solid rectangular lead piece.

$\therefore$ Number of spherical lead shots

$$ \begin{equation*} =\dfrac{\text{ Volume of solid rectangular lead piece }}{\text{ Volume of a spherical lead shot }} \tag{i} \end{equation*} $$

Also, given that diameter of a spherical lead shot $i . e .$, sphere $=4.2 cm$

$\therefore$ Radius of a spherical lead shot, $r=\dfrac{4.2}{2}=2.1 cm \quad \because$ radius $=\dfrac{1}{2}$ diameter

So, volume of a spherical lead shot i.e., sphere

$ \begin{aligned} & =\dfrac{4}{3} \pi r^{3} \\ & =\dfrac{4}{3} \times \dfrac{22}{7} \times(2.1)^{3} \\ & =\dfrac{4}{3} \times \dfrac{22}{7} \times 2.1 \times 2.1 \times 2.1 \\ & =\dfrac{4 \times 22 \times 21 \times 21 \times 21}{3 \times 7 \times 1000} \end{aligned} $

Now, length of rectangular lead piece, $l=66 cm$

Breadth of rectangular lead piece, $b=42 cm$

Height of rectangular lead piece, $h=21 cm$

$\therefore$ Volume of a solid rectangular lead piece i.e., cuboid $=l \times b \times h=66 \times 42 \times 21$

From Eq. (i),

Number of spherical lead shots $=\dfrac{66 \times 42 \times 21}{4 \times 22 \times 21 \times 21 \times 21} \times 3 \times 7 \times 1000$

$ \begin{aligned} & =\dfrac{3 \times 22 \times 21 \times 2 \times 21 \times 21 \times 1000}{4 \times 22 \times 21 \times 21 \times 21} \\ & =3 \times 2 \times 250 \\ & =6 \times 250=1500 \end{aligned} $

Hence, the required number of spherical lead shots is 1500 .

Solution

Given that, lots of spherical lead shots made out of a solid cube of lead.

$\therefore$ Number of spherical lead shots

$$ \begin{equation*} =\dfrac{\text{ Volume of a solid cube of lead }}{\text{ Volume of a spherical lead shot }} \tag{i} \end{equation*} $$

Given that, diameter of a spherical lead shot i.e., sphere $=4 cm$

$\Rightarrow \quad$ Radius of a spherical lead shot $(r)=\dfrac{4}{2}$

$ r=2 cm \quad[\because \text{ diameter }=2 \times \text{ radius }] $

So, volume of a spherical lead shot i.e., sphere

$ \begin{aligned} & =\dfrac{4}{3} \pi r^{3} \\ & =\dfrac{4}{3} \times \dfrac{22}{7} \times(2)^{3} \\ & =\dfrac{4 \times 22 \times 8}{21} cm^{3} \end{aligned} $

Now, since edge of a solid cube $(a)=44 cm$

So, volume of a solid cube $=(a)^{3}=(44)^{3}=44 \times 44 \times 44 cm^{3}$

From Eq. (i),

Number of spherical lead shots $=\dfrac{44 \times 44 \times 44}{4 \times 22 \times 8} \times 21$

$ \begin{aligned} & =11 \times 21 \times 11=121 \times 21 \\ & =2541 \end{aligned} $

Hence, the required number of spherical lead shots is 2541.

Solution

Given that, a wall is constructed with the help of bricks and mortar.

$ \begin{aligned} & \therefore \quad \text{ Number of bricks }=\dfrac{\text{ (Volume of wall) }-\dfrac{1}{10} \text{ th volume of wall }}{\text{ Volume of a brick }} \\ & =24 \times 0.4 \times 6=\dfrac{24 \times 4 \times 6}{10} m^{3} \\ & \text{ Breadth of a brick }(b_1)=16 cm=\dfrac{16}{100} m \\ & \text{ Height of a brick }(h_1)=10 cm=\dfrac{10}{100} m \\ & \text{ volume of a brick }=l_1 \times b_1 \times h_1 \\ & =\dfrac{25}{100} \times \dfrac{16}{100} \times \dfrac{10}{100}=\dfrac{25 \times 16}{10^{5}} m^{3} \end{aligned} $

From Eq. (i),

$ \begin{aligned} \text{ Number of bricks } = \dfrac{\dfrac{24 \times 4 \times 6}{10}-\dfrac{24 \times 4 \times 6}{100}}{\dfrac{25 \times 16}{10^{5}}} \\ & =\dfrac{24 \times 4 \times 6}{100} \times 9 \times \dfrac{10^{5}}{25 \times 16} \\ & =\dfrac{24 \times 4 \times 6 \times 9 \times 1000}{25 \times 16} \\ & =24 \times 6 \times 9 \times 10=12960 \end{aligned} $

Hence, the required number of bricks used in constructing the wall is 12960 .

Solution

Given that, lots of metallic circular disc to be melted to form a right circular cylinder. Here, a circular disc work as a circular cylinder.

Base diameter of metallic circular disc $=1.5 cm$

$\therefore$ Radius of metallic circular disc $=\dfrac{1.5}{2} cm \quad[\because$ diameter $=2 \times$ radius $]$ and height of metallic circular disc i.e., $=0.2 cm$

$\therefore \quad$ Volume of a circular disc $=\pi \times(\text{ Radius })^{2} \times$ Height

$ \begin{aligned} & =\pi \times \dfrac{1.5}2^{2} \times 0.2 \\ & =\dfrac{\pi}{4} \times 1.5 \times 1.5 \times 0.2 \end{aligned} $

Now, height of a right circular cylinder $(h)=10 cm$

and diameter of a right circular cylinder $=4.5 cm$

$\Rightarrow$ Radius of a right circular cylinder $(r)=\dfrac{4.5}{2} cm$

$\therefore \quad$ Volume of right circular cylinder $=\pi r^{2} h$

$ =\pi \dfrac{4.5}2^{2} \times 10=\dfrac{\pi}{4} \times 4.5 \times 4.5 \times 10 $

$\therefore$ Number of metallic circular disc $=\dfrac{\text{ Volume of a right circular cylinder }}{\text{ Volume of a metallic circular disc }}$

$ \begin{aligned} & =\dfrac{\dfrac{\pi}{4} \times 4.5 \times 4.5 \times 10}{\dfrac{\pi}{4} \times 1.5 \times 1.5 \times 0.2} \\ & =\dfrac{3 \times 3 \times 10}{0.2}=\dfrac{900}{2}=450 \end{aligned} $

Hence, the required number of metallic circular disc is 450 .

Solution

Let height of the cone be $h$.

Given, radius of the base of the cone $=6 cm$

$\therefore \quad$ Volume of circular cone $=\dfrac{1}{3} \pi r^{2} h=\dfrac{1}{3} \pi(6)^{2} h=\dfrac{36 \pi h}{3}=12 \pi h cm^{3}$

Also, given radius of the hemisphere $=8 cm$

$\therefore \quad$ Volume of the hemisphere $=\dfrac{2}{3} \pi r^{3}=\dfrac{2}{3} \pi(8)^{3}=\dfrac{512 \times 2 \pi}{3} cm^{3}$

According to the question,

Volume of the cone $=$ Volume of the hemisphere

$ \begin{aligned} \Rightarrow & 12 \pi h = \dfrac{512 \times 2 \pi}{3} \\ \therefore & h = \dfrac{512 \times 2 \pi}{12 \times 3 \pi} \\ & = \dfrac{256}{9}=28.44 cm \end{aligned} $

Solution

Given, dimensions of base of rectangular tank $=11 m \times 6 m$ and height of water $=5 m$

Volume of the water in rectangular tank $=11 \times 6 \times 5=330 m^{3}$

Also, given radius of the cylindrical tank $=3.5 m$

Let height of water level in cylindrical tank be $h$.

Then, volume of the water in cylindrical tank $=\pi r^{2} h=\pi(3.5)^{2} \times h$

$ \begin{aligned} & =\dfrac{22}{7} \times 3.5 \times 3.5 \times h \\ & =11.0 \times 3.5 \times h=38.5 hm^{3} \end{aligned} $

According to the question,

$ \begin{aligned} & 330 = 38.5 h \quad \text{ [since, volume of water is same in both tanks] } \\ \therefore \quad & h = \dfrac{330}{38.5}=\dfrac{3300}{385} \\ \therefore \quad & = 8.57 m \text{ or } 8.6 m \end{aligned} $

Hence, the height of water level in cylindrical tank is $8.6 m$.

Solution

Let the length $(l)$, breadth (b) and height ( $h$ ) be the external dimension of an open box and thickness be $x$.

Given that,

external length of an open box $(l)=36 cm$

external breadth of an open box $(b)=25 cm$

and external height of an open box $(h)=16.5 cm$

$\therefore$ External volume of an open box $=l$ bh

$ \begin{aligned} & =36 \times 25 \times 16.5 \\ & =14850 cm^{3} \end{aligned} $

Since, the thickness of the iron $(x)=1.5 cm$

So, internal length of an open box $(l_1)=l-2 x$

$ \begin{aligned} & =36 \times 2 \times 1.5 \\ & =36-3=33 cm \end{aligned} $

Therefore, internal breadth of an open box $(b_2)=b-2 x$

$ =25-2 \times 1.5=25-3=22 cm $

and internal height of an open box $(h_2)=(h-x)$

$ =16.5-1.5=15 cm $

So, internal volume of an open box $=(l-2 x) \cdot(b-2 x) \cdot(h-x)$

$ =33 \times 22 \times 15=10890 cm^{3} $

Therefore, required iron to construct an open box

$ \begin{aligned} & =\text{ External volume of an open box }- \text{ Internal volume of an open box } \\ & =14850-10890=3960 cm^{3} \end{aligned} $

Hence, required iron to construct an open box is $3960 cm^{3}$.

Given that, $1 cm^{3}$ of iron weights $=7.5 g=\dfrac{7.5}{1000} kg=0.0075 kg$

$\therefore \quad 3960 cm^{3}$ of iron weights $=3960 \times 0.0075=29.7 kg$

Solution

Given, length of the barrel of a fountain pen $=7 cm$ and diameter $=5 mm=\dfrac{5}{10} cm=\dfrac{1}{2} cm$

$\therefore$ Radius of the barrel $=\dfrac{1}{2 \times 2}=0.25 cm$

Volume of the barrel $=\pi^{2} h \quad$ [since, its shape is cylindrical]

$ \begin{aligned} & =\dfrac{22}{7} \times(0.25)^{2} \times 7 \\ & =22 \times 0.0625=1.375 cm^{3} \end{aligned} $

Also, given volume of ink in the bottle $=\dfrac{1}{5}$ of litre $=\dfrac{1}{5} \times 1000 cm^{3}=200 cm^{3}$

Now, $1.375 cm^{3}$ ink is used for writing number of words $=3300$

$\therefore 1 cm^{3}$ ink is used for writing number of words $=\dfrac{3300}{1.375}$

$\therefore 200 cm^{3}$ ink is used for writing number of words $=\dfrac{3300}{1.375} \times 200=480000$

Solution

Given, speed of water flow $=10 m min^{-1}=1000 cm / min$

and diameter of the pipe $=5 mm=\dfrac{5}{10} cm$

$\therefore \quad$ Radius of the pipe $=\dfrac{5}{10 \times 2}=0.25 cm$

$\therefore$ Area of the face of pipe $=\pi^{2}=\dfrac{22}{7} \times(0.25)^{2}=0.1964 cm^{2}$

Also, given diameter of the conical vessel $=40 cm$

$\therefore$ Radius of the conical vessel $=\dfrac{40}{2}=20 cm$

and depth of the conical vessel $=24 cm$

$\therefore \quad$ Volume of conical vessel $=\dfrac{1}{3} \pi^{2} h=\dfrac{1}{3} \times \dfrac{22}{7} \times(20)^{2} \times 24$

$ \begin{aligned} =\dfrac{211200}{21}=10057.14 cm^{3} \\ \therefore \quad \text{ Required time } = \dfrac{\text{ Volume of the conical vessel }}{\text{ Area of the face of pipe } \times \text{ Speed of water }} \\ =\dfrac{10057.14}{0.1964 \times 10 \times 100} \\ =51.20 min=51 min \dfrac{20}{100} \times 60 s=51 min 12 s \end{aligned} $

Solution

Given that, a heap of rice is in the form of a cone.

Height of a heap of rice i.e., cone $(h)=3.5 m$

and diameter of a heap of rice i.e., cone $=9 m$

Radius of a heap of rice i.e., cone $(r)=\dfrac{9}{2} m$

So,

$ \begin{aligned} \text{ volume of rice } = \dfrac{1}{3} \pi \times r^{2} h \\ & =\dfrac{1}{3} \times \dfrac{22}{7} \times \dfrac{9}{2} \times \dfrac{9}{2} \times 3.5 \\ & =\dfrac{6237}{84}=74.25 m^{3} \end{aligned} $

Now, canvas cloth required to just cover heap of rice

$=$ Surface area of a heap of rice

$=\pi r l$

$=\dfrac{22}{7} \times r \times \sqrt{r^{2}+h^{2}}$

$=\dfrac{22}{7} \times \dfrac{9}{2} \times \sqrt{\dfrac{9}2^{2}+(3.5)^{2}}$

$=\dfrac{11 \times 9}{7} \times \sqrt{\dfrac{81}{4}+12.25}$

$=\dfrac{99}{7} \times \sqrt{\dfrac{130}{4}}=\dfrac{99}{7} \times \sqrt{32.5}$

$=14.142 \times 5.7$

$=80.61 m^{2}$

Hence, $80.61 m^{2}$ canvas cloth is required to just cover heap.

Solution

Given, pencils are cylindrical in shape.

Length of one pencil $=25 cm$

and circumference of base $=1.5 cm$

$\Rightarrow$ $ r=\dfrac{1.5 \times 7}{22 \times 2}=0.2386 cm $

Now, curved surface area of one pencil $=2 \pi h$

$ \begin{matrix} =2 \times \dfrac{22}{7} \times 0.2386 \times 25 & \\ =\dfrac{262.46}{7}=37.49 cm^{2} \\ =\dfrac{37.49}{100} dm^{2} & \because 1 cm=\dfrac{1}{10} dm \\ =0.375 dm^{2} & \end{matrix} $

$\therefore$ Curved surface area of 120000 pencils $=0.375 \times 120000=45000 dm^{2}$

Now, cost of colouring $1 dm^{2}$ curved surface of the pencils manufactured in one day

$=\text{₹} 0.05$

$\therefore$ Cost of colouring $45000 dm^{2}$ curved surface $=\text{₹} 2250$

Solution

Given, length of the pond $=50 m$ and width of the pond $=44 m$

$ \text{ Depth required of water }=21 cm=\dfrac{21}{100} m $

$\therefore$ Volume of water in the pond $=50 \times 44 \times \dfrac{21}{100}{ }^{3}=462 m^{3}$

Also, given radius of the pipe $=7 cm=\dfrac{7}{100} m$

and speed of water flowing through the pipe $=(15 \times 1000)=15000 mh^{-1}$

Now, volume of water flow in $1 h=\pi R^{2} H$

$ \begin{aligned} & =\dfrac{22}{7} \times \dfrac{7}{100} \times \dfrac{7}{100} \times 15000 \\ & =231 m^{3} \end{aligned} $

Since, $231 m^{3}$ of water falls in the pond in $1 h$.

So, $1 m^{3}$ water falls in the pond in $\dfrac{1}{231} h$.

Also, $462 m^{3}$ of water falls in the pond in $\dfrac{1}{231} \times 462 h=2 h$

Hence, the required time is $2 h$.

Solution

Given that, a solid iron cuboidal block is recast into a hollow cylindrical pipe.

Length of cuboidal pipe $(l)=4.4 m$

Breadth of cuboidal pipe $(b)=2.6 m$ and height of cuboidal pipe $(h)=1 m$

So, volume of a solid iron cuboidal block $=l \cdot b \cdot h$

$ =4.4 \times 2.6 \times 1=11.44 m^{3} $

Also, internal radius of hollow cylindrical pipe $(r_i)=30 cm=0.3 m$

and thickness of hollow cylindrical pipe $=5 cm=0.05 m$

So, external radius of hollow cylindrical pipe $(r_e)=r_i+$ Thickness

$ \begin{aligned} & =0.3+0.05 \\ & =0.35 m \end{aligned} $

$\therefore$ Volume of hollow cylindrical pipe $=$ Volume of cylindrical pipe with external radius

-Volume of cylindrical pipe with internal radius

$ \begin{aligned} & =\pi r_e^{2} h_1-\pi r_i^{2} h_1=\pi(r_e^{2}-r_i^{2}) h_1 \\ & =\dfrac{22}{7}[(0.35)^{2}-(0.3)^{2}] \cdot h_1 \\ & =\dfrac{22}{7} \times 0.65 \times 0.05 \times h_1=0.715 \times h_1 / 7 \end{aligned} $

where, $h_1$ be the length of the hollow cylindrical pipe.

Now, by given condition,

Volume of solid iron cuboidal block $=$ Volume of hollow cylindrical pipe

$\begin{matrix} \Rightarrow & 11.44 = 0.715 \times h / 7 \\ & \therefore & h = \dfrac{11.44 \times 7}{0.715}=112 m\end{matrix} $

Hence, required length of pipe is $112 m$.

Solution

Let the rise of water level in the pond be $h m$, when 500 persons are taking a dip into a cuboidal pond.

Given that,

Length of the cuboidal pond $=80 m$

Breadth of the cuboidal pond $=50 m$

Now, volume for the rise of water level in the pond

$ \begin{aligned} & =\text{ Length } \times \text{ Breadth } \times \text{ Height } \\ & =80 \times 50 \times h=4000 hm^{3} \end{aligned} $

and the average displacement of the water by a person $=0.04$

$m^{3}$

So, the average displacement of the water by 500 persons $=500 \times 0.04 m^{3}$

Now, by given condition,

Volume for the rise of water level in the pond =Average displacement of the water by

500 persons

$ \begin{aligned} \Rightarrow & 4000 h = 500 \times 0.04 \\ \therefore & h = \dfrac{500 \times 0.04}{4000}=\dfrac{20}{4000}=\dfrac{1}{200} m \\ & = 0.005 m \\ & = 0.005 \times 100 cm \\ & = 0.5 cm \end{aligned} $

Hence, the required rise of water level in the pond is $0.5 cm$.

Solution

Given, dimensions of the cuboidal $=16 cm \times 8 cm \times 8 cm$

$\therefore$ Volume of the cuboidal $=16 \times 8 \times 8=1024 cm^{3}$

Also, given radius of one glass sphere $=2 cm$

$\therefore \quad$ Volume of one glass sphere $=\dfrac{4}{3} \pi r^{3}=\dfrac{4}{3} \times \dfrac{22}{7} \times(2)^{3}$

$ =\dfrac{704}{21}=33.523 cm^{3} $

Now, volume of 16 glass spheres $=16 \times 33.523=536.37 cm^{3}$

$\therefore \quad$ Required volume of water $=$ Volume of cuboidal - Volume of 16 glass spheres

$ =1024-536.37=487.6 cm^{3} $

Solution

Given that, height of milk container $(h)=16 cm$,

Radius of lower end of milk container $(r)=8 cm$

and radius of upper end of milk container $(R)=20 cm$

$\therefore$ Volume of the milk container made of metal sheet in the form of a frustum of a cone

$ \begin{aligned} & =\dfrac{\pi h}{3}(R^{2}+r^{2}+R r) \\ & =\dfrac{22}{7} \times \dfrac{16}{3}[(20)^{2}+(8)^{2}+20 \times 8] \\ & =\dfrac{22 \times 16}{21}(400+64+160) \\ & =\dfrac{22 \times 16 \times 624}{21}=\dfrac{219648}{21} \\ & =10459.42 cm^{3}=10.45942 L \end{aligned} $

$ =\dfrac{22 \times 16 \times 624}{21}=\dfrac{219648}{21} \quad[\because 1 L=1000 cm^{3}] $

So, volume of the milk container is $10459.42 cm^{3}$.

$\because$ Cost of $1 L$ milk $=$ ₹ 22

$\therefore$ Cost of $10.45942 L$ milk $=22 \times 10.45942=\text{₹} 230.12$

Hence, the required cost of milk is ₹ 230.12

Solution

Given, radius of the base of the bucket $=18 cm$

Height of the bucket $=32 cm$

So, volume of the sand in cylindrical bucket $=\pi r^{2} h=\pi(18)^{2} \times 32=10368 \pi$

Also, given height of the conical heap $(h)=24 cm$

Let radius of heap be $r cm$.

Then, volume of the sand in the heap $=\dfrac{1}{3} \pi r^{2} h$

$ =\dfrac{1}{3} \pi r^{2} \times 24=8 \pi r^{2} $

According to the question,

Volume of the sand in cylindrical bucket $=$ Volume of the sand in conical heap

$\Rightarrow$ $10368 \pi=8 \pi^{2}$

$\Rightarrow$ $10368=8 r^{2}$

$\Rightarrow$ $r^{2}=\dfrac{10368}{8}=1296$

$\Rightarrow \quad r=36 cm$ Again, let the slant height of the conical heap $=l$

Now, $$ \begin{aligned} l^2 & =h^2+r^2=(24)^2+(36)^2 \\ & =576+1296=1872 \\ l & =43.267 \mathrm{~cm} \end{aligned} $$ $$ \therefore \quad l=43.267 \mathrm{~cm} $$

Hence, radius of conical heap of sand $=36 cm$

and slant height of conical heap $=43.267 cm$

Solution

Since, rocket is the combination of a right circular cylinder and a cone.

Given, diameter of the cylinder $=6 cm$

$\therefore \quad$ Radius of the cylinder $=\dfrac{6}{2}=3 cm$

and height of the cylinder $=12 cm$

$\therefore$ Volume of the cylinder $=\pi^{2} h=3.14 \times(3)^{2} \times 12$

$ =339.12 cm^{3} $

and curved surface area $=2 \pi r h$

$ =2 \times 3.14 \times 3 \times 12=226.08 $

Now, in right angled $\triangle A O C$,

$\therefore$ Height of the cone, $h=4 cm$

$ h=\sqrt{5^{2}-3^{2}}=\sqrt{25-9}=\sqrt{16}=4 $

Radius of the cone, $r=3 cm$

Now, volume of the cone

$ \begin{aligned} & =\dfrac{1}{3} \pi r^{2} h=\dfrac{1}{3} \times 3.14 \times(3)^{2} \times 4 \\ & =\dfrac{113.04}{3}=37.68 cm^{3} \end{aligned} $

and curved surface area $=\pi r l=3.14 \times 3 \times 5=47.1$

Hence, total volume of the rocket

$ =339.12+37.68=376.8 cm^{3} $

and total surface area of the rocket $=$ CSA of cone + CSA of cylinder

$ \begin{aligned} & =47.1+226.08+28.26 \\ & =301.44 cm^{2} \end{aligned} $

Solution

Let total height of the building $=$ Internal diameter of the dome $=2 rm$

$\therefore \quad$ Radius of building (or dome) $=\dfrac{2 r}{2}=r m$

Height of cylinder $=2 r-r=r m$

$\therefore \quad$ Volume of the cylinder $=\pi r^{2}(r)=\pi r^{3} m^{3}$

and volume of hemispherical dome cylinder $=\dfrac{2}{3} \pi r^{3} m^{3}$

$\therefore$ Total volume of the building $=$ Volume of the cylinder + Volume of hemispherical dome

$ =\pi r^{3}+\dfrac{2}{3} \pi r^{3} m^{3}=\dfrac{5}{3} \pi r^{3} m^{3} $

According to the question,

$ \begin{matrix} \Rightarrow & \text{ Volume of the building } = \text{ Volume of the air } \\ \Rightarrow & \dfrac{5}{3} \pi r^{3} = 41 \dfrac{19}{21} \\ \Rightarrow \quad & \dfrac{5}{3} \pi r^{3} = \dfrac{880}{21} \\ \Rightarrow \quad & r^{3} = \dfrac{880 \times 7 \times 3}{21 \times 22 \times 5}=\dfrac{40 \times 21}{21 \times 5}=8 \\ \Rightarrow \quad & r^{3} = 8 \Rightarrow r=2 \\ & \text{ Height of the building } = 2 r=2 \times 2=4 m \end{matrix} $

Solution

Given, radius of hemispherical bowl, $r=9 cm$ and radius of cylindrical bottles, $R=1.5 cm$ and height, $h=4 cm$

$\therefore$ Number of required cylindrical bottles $=\dfrac{\text{ Volume of hemispherical bowl }}{\text{ Volume of one cylindrical bottle }}$

$ =\dfrac{\dfrac{2}{3} \pi r^{3}}{\pi R^{2} h}=\dfrac{\dfrac{2}{3} \times \pi \times 9 \times 9 \times 9}{\pi \times 1.5 \times 1.5 \times 4}=54 $

Solution

(i) Whenever we placed a solid right circular cone in a right circular cylinder with full of water, then volume of a solid right circular cone is equal to the volume of water falled from the cylinder.

(ii) Total volume of water in a cylinder is equal to the volume of the cylinder.

(iii) Volume of water left in the cylinder $=$ Volume of the right circular cylinder -volume of a right circular cone.

Now, given that

Height of a right circular cone $=120 cm$

Radius of a right circular cone $=60 cm$

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

$ \begin{aligned} & =\dfrac{1}{3} \times \dfrac{22}{7} \times 60 \times 60 \times 120 \\ & =\dfrac{22}{7} \times 20 \times 60 \times 120 \\ & =144000 \pi cm^{3} \end{aligned} $

$\therefore$ Volume of a right circular cone $=$ Volume of water falled from the cylinder

$ =144000 \pi cm^{3} $

[from point (i)]

Given that, height of a right circular cylinder $=180 cm$

and radius of a right circular cylinder $=$ Radius of a right circular cone

$ =60 cm $

$\therefore$ Volume of a right circular cylinder $=\pi r^{2} \times h$

$ \begin{aligned} & =\pi \times 60 \times 60 \times 180 \\ & =648000 \pi cm^{3} \end{aligned} $

So, volume of a right circular cylinder $=$ Total volume of water in a cylinder

$ =648000 \pi cm^{3} \quad[\text{ from point (ii)] } $

From point (iii), Volume of water left in the cylinder

$$ \begin{aligned} & =\text { Total volume of water in a cylinder }- \text { Volume of water falled from } \\ & =648000 \pi-144000 \pi \quad \text { the cylinder when solid cone is placed } \\ & =504000 \pi=504000 \times \dfrac{22}{7}=1584000 \mathrm{~cm}^3 \\ & =\dfrac{1584000}{(10)^6} \mathrm{~m}^3=1.584 \mathrm{~m}^3 \end{aligned} $$

the cylinder when solid cone is placed in it

Hence, the required volume of water left in the cylinder is $1.584 m^{3}$.

Solution

Given, radius of tank, $r_1=40 cm$

Let height of water level in tank in half an hour $=h_1$

Also, given internal radius of cylindrical pipe, $r_2=1 cm$

and speed of water $=80 cm / s$ i.e., in 1 water flow $=80 cm$

$\therefore$ In $30(min)$ water flow $=80 \times 60 \times 30=144000 cm$

According to the question,

Volume of water in cylindrical tank =Volume of water flow from the circular pipe

$ \begin{matrix} \Rightarrow & r_1^{2} h_1=\pi r_2^{2} h_2 \\ \Rightarrow & 40 \times 40 \times h_1=1 \times 1 \times 144000 \\ \therefore & h_1=\dfrac{144000}{40 \times 40}=90 cm \end{matrix} $

in half an hour

Hence, the level of water in cylindrical tank rises $90 cm$ in half an hour.

Solution

Given, length of roof $=22 m$ and breadth of roof $=20 m$

Let the rainfall be $acm$.

$\therefore \quad$ Volume of water on the roof $=22 \times 20 \times \dfrac{a}{100}=\dfrac{22 a}{5} m^{3}$

Also, we have radius of base of the cylindrical vessel $=1 m$

and height of the cylindrical vessel $=3.5 m$

$\therefore$ Volume of water in the cylindrical vessel when it is just full

$ =\dfrac{22}{7} \times 1 \times 1 \times \dfrac{7}{2}=11 m^{3} $

Now, volume of water on the roof $=$ Volume of water in the vessel

$ \begin{aligned} & \Rightarrow \quad \dfrac{22 a}{5}=11 \\ & \therefore \quad a=\dfrac{11 \times 5}{22}=2.5 \quad[\because \text{ volume of cylinder }=\pi \times(\text{ radius })^{2} \times \text{ height }] \end{aligned} $

Hence, the rainfall is $2.5 cm$.

Solution

Given that, length of cuboid pen stand $(l)=10 cm$

Breadth of cubiod pen stand $(b)=5 cm$

and height of cuboid pen stand $(h)=4 cm$

(Pen)

with conical base

(Pin)

Cuboid

with cubical base

$\therefore$ Volume of cuboid pen stand $=l \times b \times h=10 \times 5 \times 4=200 cm^{3}$

Also, radius of conical depression $(r)=0.5 cm$

and height (depth) of a conical depression $(h_1)=2.1 cm$

$\therefore$ Volume of a conical depression $=\dfrac{1}{3} \pi r^{2} h_1$

$ \begin{aligned} & =\dfrac{1}{3} \times \dfrac{22}{7} \times 0.5 \times 0.5 \times 2.1 \\ & =\dfrac{22 \times 5 \times 5}{1000}=\dfrac{22}{40}=\dfrac{11}{20}=0.55 cm^{3} \end{aligned} $

Also, given

Edge of cubical depression (a) $=3 cm$

$\therefore$ Volume of cubical depression $=(a)^{3}=(3)^{3}=27 cm^{3}$

So, volume of 4 conical depressions

$=4 \times$ Volume of a conical depression

$ =4 \times \dfrac{11}{20}=\dfrac{11}{5} cm^{3} $

Hence, the volume of wood in the entire pen stand

$=$ Volume of cuboid pen stand - Volume of 4 conical

depressions - volume of a cubical depressions

$=200-\dfrac{11}{5}-27=200-\dfrac{146}{5}$

$=200-29.2=170.8 cm^{3}$

So, the required volume of the wood in the entire stand is

Conical depression