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 $=22cm$

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

Also, given diameter of marble $=0.5cm$

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

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

$\begin{array}{rl}& [\because \text{ volume of sphere }={\displaystyle \frac{4}{3}}\times \pi \times (\text{ radius }{)}^3]\\ & ={\displaystyle \frac{1.375}{21}}=0.0655c{m}^3\\ & =\text{ Volume of the cube }{\displaystyle \frac{1}{8}}\times \text{ Volume of cube }\\ & =10648-10648\times {\displaystyle \frac{1}{8}}\\ & =10648\times {\displaystyle \frac{7}{8}}=9317c{m}^3\end{array}$

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

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

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 $=4cm$ and external diameter of shell $=8cm$

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

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

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

$\begin{array}{rl}& ={\displaystyle \frac{4}{3}}\pi (4^3-2^3)\\ & ={\displaystyle \frac{4}{3}}\pi (64-8)\\ & ={\displaystyle \frac{224}{3}}\pi c{m}^3\end{array}$

Let height of the cone $=hcm$

Diameter of the base of cone $=8cm$

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

According to the question,

Volume of cone $=$ Volume of spherical shell

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

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

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

Solution

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

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

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

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

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

According to the question,

Volume of the sphere $=$ Volume of the cuboid

$\begin{array}{cc}\Rightarrow & {\displaystyle \frac{4}{3}}{\pi }^3=38808\\ \Rightarrow & 4\times {\displaystyle \frac{22}{7}}{r}^3=38808\times 3\\ \Rightarrow & r^3={\displaystyle \frac{38808\times 3\times 7}{4\times 22}}=441\times 21\\ \Rightarrow & r^3=21\times 21\times 21\\ & & r=21cm\end{array}$

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

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 $=270\times 300\times 350=28350000c{m}^3$

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

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

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

$\begin{array}{rl}& =28350000-28350000\times {\displaystyle \frac{1}{8}}\\ & =28350000-3543750=24806250c{m}^3\end{array}$

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

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

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

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

Solution

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

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

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

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

Now, let the radius of solid sphere $=rcm$

Then, its volume $={\displaystyle \frac{4}{3}}\pi r{m}^3[\because .$ volume of sphere $.={\displaystyle \frac{4}{3}}\times \pi \times (\text{ radius }{)}^3]$

According to the question,

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

$\begin{array}{rl}\Rightarrow & 12\times {\displaystyle \frac{4}{3}}\pi r^3=16\pi \\ \Rightarrow & r^3=1\Rightarrow r=1cm\end{array}$

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

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

Solution

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

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

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

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

$\begin{array}{rl}& [\because \text{ curved surface area of frustum of a cone }=\pi (R+r)l]\\ & =\pi \times 45\times 35={\displaystyle \frac{22}{7}}\times 45\times 35=4950c{m}^2\end{array}$

Solution

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

[since, both hemispheres are attach with cylinder]

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

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

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

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

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

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

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

$\begin{array}{rl}& ={\displaystyle \frac{22}{7}}[(0.25{)}^2\times 1.5+{\displaystyle \frac{4}{3}}\times (0.25{)}^3]\\ & ={\displaystyle \frac{22}{7}}[0.09375+0.0208]\\ & ={\displaystyle \frac{22}{7}}\times 0.11455=0.36c{m}^3\end{array}$

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

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 $2rcm$.

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,

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

and diameter of the other end,

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

Height of the bucket, $h=35cm$

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

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

$=32706.6c{m}^3=32.7L[\because 1000c{m}^3=1L]$

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

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.

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

$$\begin{array}{}\text{(i)}& [\because \text{ volume of sphere }={\displaystyle \frac{4}{3}}\pi (\text{ radius }{)}^3]\end{array}$$

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

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

$\Rightarrow {\displaystyle \frac{r_1^3}{r_2^3}}={\displaystyle \frac{64}{27}}\Rightarrow {\displaystyle \frac{r_1}{r_2}}={\displaystyle \frac{4}{3}}$

Now, ratio of surface area $={\displaystyle \frac{4{\pi }_1^2}{4{\pi }_2{}^2}}[\because .$ surface area of a sphere $.=4\pi (\text{ radius }{)}^2]$

$\begin{array}{rl}& ={\displaystyle \frac{r_1^2}{r_2^2}}\\ & ={{\displaystyle \frac{r_1}{r_2}}}^2={{\displaystyle \frac{4}{3}}}^2\\ & =16:9\end{array}$

[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 $=2h$

and radius of the new cylinder $=r$

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

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, $h$

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

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

So, total surface area of the combined solid

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

Solution

False

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

$\begin{array}{cc}\therefore & \text{ Radius of the ball }={\displaystyle \frac{a}{2}}\\ \text{ So, }& \text{ volume of the ball }={\displaystyle \frac{4}{3}}\pi {\displaystyle \frac{a^3}{2}}={\displaystyle \frac{1}{6}}\pi a^3\end{array}$

Solution

False

Since, the volume of the frustum of a cone is ${\displaystyle \frac{1}{3}}\pi h[r_1^2+r_2^2+r_1{r}_2]$, 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 $=\pi r^{2}hc{m}^3$

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

From the figure, capacity of the cylindrical vessel

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

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, $ABCD$ is a frustum of a cone and $CDEF$ is a hollow cylinder.

Solution

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

$\therefore$ Volume of first cube $=(3{)}^3=27c{m}^3$

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

and

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

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

Let the edge of the resulting cube $=Rcm$

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

Solution

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

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

and diameter of shot $=3cm$

$\begin{array}{r}\therefore \text{ Radius of shot, }r={\displaystyle \frac{3}{2}}=1.5cm\\ \text{ Volume of shot }={\displaystyle \frac{4}{3}}{\pi }^3={\displaystyle \frac{4}{3}}\times {\displaystyle \frac{22}{7}}\times (1.5{)}^3\\ & ={\displaystyle \frac{297}{21}}=14.143c{m}^3\end{array}$

Solution

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

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

$=28490c{m}^3$

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

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

Let height of the bucket $=hcm$

Now, volume of the bucket $={\displaystyle \frac{1}{3}}\pi h(r_1^2+r_2^2+r_1{r}_2)=28490$

[given]

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

Solution

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

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

Let $P$ be the mid-point of $OM$, then

$\begin{array}{cc}& OP=PM={\displaystyle \frac{12}{2}}=6cm\\ \text{ Now, }& \mathrm{\Delta }OPD& \sim \mathrm{△}OMN\\ \therefore & {\displaystyle \frac{OP}{OM}}={\displaystyle \frac{PD}{MN}}\\ \Rightarrow & {\displaystyle \frac{6}{12}}={\displaystyle \frac{PD}{8}}\Rightarrow {\displaystyle \frac{1}{2}}={\displaystyle \frac{PD}{8}}\\ \Rightarrow & PD=4cm\end{array}$

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

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

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

Radius of the bottom, $r_2=8cm$

$\begin{array}{rl}& \therefore \begin{array}{r}\text{ Volume of smaller cone }={\displaystyle \frac{1}{3}}\pi \times 4\times 4\times 6=32\pi c{m}^3\\ \text{ and volume of the frustum of cone }={\displaystyle \frac{1}{3}}\times \pi \times 6[(8{)}^2+(4{)}^2+8\times 4]\\ =2\pi (64+16+32)=224c{m}^3\end{array}\end{array}$

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

Solution

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

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

On joining two cubes, we get a cuboid whose

$\begin{array}{r}\text{ length, }l=2acm\\ \text{ breadth, }b=acm\\ \text{ height, }h=acm\\ \text{ ting cuboid }=2(lb+bh+hl)\\ & =2(2a\cdot a+a\cdot a+a\cdot 2a)\\ & =2(2a^2+a^2+2a^2)=2(5a^2)\\ & =10a^2=10(4{)}^2=160c{m}^2\end{array}$

$\begin{array}{rl}& \text{ and height, }h=acm\\ & \text{ Now, surface area of the resulting cuboid }=2(lb+bh+hl)\end{array}$

Solution

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

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

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, $r=3cm$ $\Rightarrow$ Diameter $=2\times r=2\times 3=6cm$

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 $=(\text{ side }{)}^3=a^3=(7{)}^3=343c{m}^3$

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

$\begin{array}{rl}& ={\displaystyle \frac{1}{3}}\times {\displaystyle \frac{22}{7}}\times 3\times 3\times 7\\ & =66c{m}^3\end{array}$

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

$=343-66=277c{m}^3$

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

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

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

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

[since, both cones are identical]

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

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

$={\displaystyle \frac{44\times 8\times \sqrt{289}}{7}}$

$={\displaystyle \frac{44\times 8\times 17}{7}}$

$={\displaystyle \frac{5984}{7}}=854.85c{m}^2$

$=855c{m}^2$ (approx)

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

Solution

Let volume of cone $A$ be $2V$ 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 $=6cm$

$\therefore$ Radius of the cone $={\displaystyle \frac{6}{2}}=3cm$

Now, volume of the cone, $A=2V={\displaystyle \frac{1}{3}}\pi r^{2}h={\displaystyle \frac{1}{3}}\pi (3{)}^2{h}_1$

$\Rightarrow V={\displaystyle \frac{1}{6}}\pi 9h_1={\displaystyle \frac{3}{2}}{h}_1\pi$

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

From Eqs. (i) and (ii),

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

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

$\begin{array}{rl}& -(\text{ Volume of cone }A+\text{ Volume of cone }B)\\ =& 594-(132+66)\\ =& 396c{m}^3\end{array}$

Solution

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

Also, radius of hemisphere $=5cm$

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

$={\displaystyle \frac{5500}{21}}=261.90c{m}^3$

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

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

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

$\begin{array}{rl}& ={\displaystyle \frac{1}{3}}\times {\displaystyle \frac{22}{7}}\times (5{)}^2\times 5\\ & ={\displaystyle \frac{2750}{21}}=130.95c{m}^3\end{array}$

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

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

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

$=327.4c{m}^3$

Solution

Given, diameter of a marble $=1.4cm$

$\begin{array}{r}\therefore \text{ Radius of marble }={\displaystyle \frac{1.4}{2}}=0.7cm\\ \text{ So, volume of one marble }={\displaystyle \frac{4}{3}}\pi (0.7{)}^3\\ & ={\displaystyle \frac{4}{3}}\pi \times 0.343={\displaystyle \frac{1.372}{3}}\pi c{m}^3\end{array}$

Also, given diameter of beaker $=7cm$

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

Height of water level raised $=5.6cm$

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

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

$={\displaystyle \frac{68.6\pi }{1.372\pi }}\times 3=150$

Solution

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

$\therefore$ Number of spherical lead shots

$$\begin{array}{}\text{(i)}& ={\displaystyle \frac{\text{ Volume of solid rectangular lead piece }}{\text{ Volume of a spherical lead shot }}}\end{array}$$

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

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

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

$\begin{array}{rl}& ={\displaystyle \frac{4}{3}}\pi r^3\\ & ={\displaystyle \frac{4}{3}}\times {\displaystyle \frac{22}{7}}\times (2.1{)}^3\\ & ={\displaystyle \frac{4}{3}}\times {\displaystyle \frac{22}{7}}\times 2.1\times 2.1\times 2.1\\ & ={\displaystyle \frac{4\times 22\times 21\times 21\times 21}{3\times 7\times 1000}}\end{array}$

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

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

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

$\therefore$ 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 $={\displaystyle \frac{66\times 42\times 21}{4\times 22\times 21\times 21\times 21}}\times 3\times 7\times 1000$

$\begin{array}{rl}& ={\displaystyle \frac{3\times 22\times 21\times 2\times 21\times 21\times 1000}{4\times 22\times 21\times 21\times 21}}\\ & =3\times 2\times 250\\ & =6\times 250=1500\end{array}$

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

$$\begin{array}{}\text{(i)}& ={\displaystyle \frac{\text{ Volume of a solid cube of lead }}{\text{ Volume of a spherical lead shot }}}\end{array}$$

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

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

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

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

$\begin{array}{rl}& ={\displaystyle \frac{4}{3}}\pi r^3\\ & ={\displaystyle \frac{4}{3}}\times {\displaystyle \frac{22}{7}}\times (2{)}^3\\ & ={\displaystyle \frac{4\times 22\times 8}{21}}c{m}^3\end{array}$

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

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

From Eq. (i),

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

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

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

Solution

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

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

From Eq. (i),

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

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 $=1.5cm$

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

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

$\begin{array}{rl}& =\pi \times {{\displaystyle \frac{1.5}{2}}}^2\times 0.2\\ & ={\displaystyle \frac{\pi }{4}}\times 1.5\times 1.5\times 0.2\end{array}$

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

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

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

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

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

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

$\begin{array}{rl}& ={\displaystyle \frac{{\displaystyle \frac{\pi }{4}}\times 4.5\times 4.5\times 10}{{\displaystyle \frac{\pi }{4}}\times 1.5\times 1.5\times 0.2}}\\ & ={\displaystyle \frac{3\times 3\times 10}{0.2}}={\displaystyle \frac{900}{2}}=450\end{array}$

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 $=6cm$

$\therefore$ Volume of circular cone $={\displaystyle \frac{1}{3}}\pi r^{2}h={\displaystyle \frac{1}{3}}\pi (6{)}^{2}h={\displaystyle \frac{36\pi h}{3}}=12\pi hc{m}^3$

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

$\therefore$ Volume of the hemisphere $={\displaystyle \frac{2}{3}}\pi r^3={\displaystyle \frac{2}{3}}\pi (8{)}^3={\displaystyle \frac{512\times 2\pi }{3}}c{m}^3$

According to the question,

Volume of the cone $=$ Volume of the hemisphere

$\begin{array}{rl}\Rightarrow & 12\pi h={\displaystyle \frac{512\times 2\pi }{3}}\\ \therefore & h={\displaystyle \frac{512\times 2\pi }{12\times 3\pi }}\\ & ={\displaystyle \frac{256}{9}}=28.44cm\end{array}$

Solution

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

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

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

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

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

$\begin{array}{rl}& ={\displaystyle \frac{22}{7}}\times 3.5\times 3.5\times h\\ & =11.0\times 3.5\times h=38.5h{m}^3\end{array}$

According to the question,

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

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

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 $(l)=36cm$

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

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

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

$\begin{array}{rl}& =36\times 25\times 16.5\\ & =14850c{m}^3\end{array}$

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

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

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

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

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

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

$=16.5-1.5=15cm$

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

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

Therefore, required iron to construct an open box

$\begin{array}{rl}& =\text{ External volume of an open box }-\text{ Internal volume of an open box }\\ & =14850-10890=3960c{m}^3\end{array}$

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

Given that, $1c{m}^3$ of iron weights $=7.5g={\displaystyle \frac{7.5}{1000}}kg=0.0075kg$

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

Solution

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

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

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

$\begin{array}{rl}& ={\displaystyle \frac{22}{7}}\times (0.25{)}^2\times 7\\ & =22\times 0.0625=1.375c{m}^3\end{array}$

Also, given volume of ink in the bottle $={\displaystyle \frac{1}{5}}$ of litre $={\displaystyle \frac{1}{5}}\times 1000c{m}^3=200c{m}^3$

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

$\therefore 1c{m}^3$ ink is used for writing number of words $={\displaystyle \frac{3300}{1.375}}$

$\therefore 200c{m}^3$ ink is used for writing number of words $={\displaystyle \frac{3300}{1.375}}\times 200=480000$