3 Marks Question - Maths STD 10 Questions - Vidyadip

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33 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks

150 spherical marbles, each of diameter 14cm, are dropped in a cylindrical vessel of diameter 7cm containing some water, which are completely immersed in water. Find the rise in the level of water in the vessel.

Answer

We have,
The radius of spherical marble, $\text{r}=\frac{1.4}{2}=7\text{cm}$ and
The radius of the cylindrical vessel, $\text{R}=\frac{7}{2}\text{cm}=3.5\text{cm}$
Let the rise in the level of water in the vessel be H.
Now,
Volume of water rised in the cylindrical vessel= Volume of 150 spherical marbles
$\Rightarrow\pi\text{R}^2\text{H}=150\times\frac{4}{3}\pi\text{r}^3$
$\Rightarrow\text{R}^2\text{H}=200\text{r}^3$
$\Rightarrow3.5\times3.5\times\text{H}=200\times7\times7\times7$
$\Rightarrow\text{H}=\frac{200\times7\times7\times7}{3.5\times3.5}$
$\therefore\text{H}=5600\text{cm}$
So, the rise in the level of water in the vessel is 5.6cm.
Disclaimer: The diameter of the spherical marbles should be 1.4cm instead 14cm. The has been corrected above.

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Question 23 Marks

The volume of a sphere is $4851\ cm^3$. Find its curved surface area.

Answer

Let the radius of the sphere be r. As, Volume of the sphere $= 4851\ cm^3$
$\Rightarrow\frac{4}{3}\pi\text{r}^3=4851$
$\Rightarrow\frac{4}{3}\times\frac{22}{7}\times\text{r}^3=4851$
$\Rightarrow\text{r}^3=4851\times\frac{3\times7}{4\times22}$
$\Rightarrow\text{r}^3=\frac{9261}{8}$
$\Rightarrow\text{r}=\sqrt[3]{\frac{9261}{8}}$
$\Rightarrow\text{r}=\frac{21}{2}\text{cm}$
Now, Curved surface area of the sphere $=4\pi\text{r}^2$ $=4\times\frac{22}{7}\times\frac{21}{2}\times\frac{21}{2}$ $=1386\text{cm}^2$

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Question 33 Marks

In a corner of a rectangular field with dimensions 35m × 22m, a well with 14m inside diameter is dug 8m deep. The earth dug out is spread evenly over the remaining part of the field. Find the rise in the level of the field.

Answer

We have,
Length of the fie!, l = 35m,
Width of the field, b = 22m,
Depth of the well, H = 8m and
Radius of the well, $\text{R}=\frac{14}{2}=7\text{m},$
Let the rise in the level of the field be h.
Now,
Volume of the earth on remaining part of the field= Volume of earth dug out
⇒ Area of the remaining field × h = Volume of the well
⇒ (Area of the field-Area of base of the well) $\times\text{h}\pi\text{R}^2\text{H}$
$\Rightarrow(\text{lb}-\pi\text{R}^2)\times\text{h}=\pi\text{R}^2\text{H}$
$\Rightarrow(35\times22-\frac{22}{7}\times7\times7)\times\text{h}=\frac{22}{7}\times7\times7\times8$
$=(770-154)\times\text{h}=1232$
$\Rightarrow616\times\text{h}=1232$
$\Rightarrow\text{h}=\frac{1232}{616}$
$\therefore\text{h}=2\text{m}$
So, the rise in the level of the field is 2m.

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Question 43 Marks

The surface areas of two spheres are in the ratio of 4 : 25. Find the ratio of their volumes.

Answer

Let the radii of the two spheres be r and R. As, $\frac{\text{Surface area of the first sphere}}{\text{Surface area of the second sphere}}=\frac{4}{25}$ $\Rightarrow\frac{4\pi\text{r}^2}{4\pi\text{R}^2}=\frac{4}{25}$ $\Rightarrow\Big(\frac{\text{r}}{\text{R}}\Big)^2=\frac{4}{25}$ $\Rightarrow\frac{\text{r}}{\text{R}}=\sqrt{\frac{4}{25}}$ $\Rightarrow\frac{\text{r}}{\text{R}}=\frac{2}{5}\ ...(\text{i})$The ratio of the volumes of the two spheres $=\frac{\text{Volume of the first sphere}}{\text{Volume of the second sphere}}$
$=\frac{\Big(\frac{4}{3}\pi\text{r}^3\Big)}{\Big(\frac{4}{3}\pi\text{R}^3\Big)}$ $=\Big(\frac{\text{r}}{\text{R}}\Big)^3$ $=\Big(\frac{2}{5}\Big)^3$ [Using (i)] $=8:25$ So, the ratio of the volumes of the given spheres is 8 : 125

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Question 53 Marks

A 5-m-wide cloth is used to make a conical tent of base diameter 14m and height 24m. Find th cost of cloth used at the rate of Rs. 25 per metre.

Answer

Let the length of the cloth be l m
Given radius of cone = 7m and heigth = 24m
Now,
Slant height $=\sqrt{\text{h}^2+\text{r}^2}$
$\Rightarrow\text{l}=\sqrt{\text{h}^2+\text{r}^2}$
$\Rightarrow\text{l}\sqrt{(24)^2+(7)^2}$
$\Rightarrow\sqrt{576+49}$
$\Rightarrow\text{l}=\sqrt{625}$
$\Rightarrow\text{l}=25\text{m}$
Area of the doth = curved surface area of cone
$=\pi\text{r}\text{l}$
$=\frac{22}{7}\times7\times25$
$=22\times25$
$=550\text{m}^2$
Now,
Area of the doth = length × breadth
⇒ 550 = length × 5
⇒ length $=\frac{550}{5}$
⇒ length 110m
Given cost of 1m doth = Rs. 25
⇒ cost od 110m cloth =110 × 25
= Rs. 2750

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Question 63 Marks

Two cubes each of volume $125\ cm^3$ are joined end to end to form a solid. Find the surface area of the resulting cuboid.

Answer

Let the edge of the cube be a.
As,
Volwne of the cube $= 125\ cm^3$
$\Rightarrow\text{a}^3=125$
$\Rightarrow\text{a}=\sqrt[3]{125}$
$\Rightarrow\text{a}=5\text{cm}$
So,
Length of the resulting cuboid, $l = 2 × 5 = 10\ cm$,
Breadth of the resulting cuboid, $b = 5\ cm$ and
Height of the resulting cuboid, $h = 5\ cm$
Now,
Surface area of the resulting cuboid $= 2 (lb + bh + hl)$
$= 2 \times (10 \times 5 + 5 \times 5 + 5 \times 10)$
$= 2 \times (50 + 25 + 50)$
$= 2 \times 125$
$= 250\ cm^2$
So, the surface area of the resulting cuboid is $250\ cm^2$.

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Question 73 Marks

Five identical cubes, each of edge $5\ cm$, are placed adjacent to each other. Find the volume of the resulting cuboid.

Answer

We have,
Length of the resulting cuboid, $l = 5 \times 5 = 25\ cm$,
Breadth of the resulting cuboid, $b = 5\ cm$ and
Height of the resulting cuboid, $h = 5\ cm$
Now,
Volume of the resulting cuboid = lbh
$= 25 \times 5 \times 5$
$= 625\ cm^3$

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Question 83 Marks

A spherical cannon ball 28cm in diameter is melted and recast into a right circular conical mould, base of which is 35cm in diameter. Find the height of the cone.

Answer

Diameter of cannon ball = 28cm
Radius of the cannon ball = 14cm
Volume of ball $=\frac{4}{3}\pi\text{r}^3=\frac{4}{3}\pi\times(14)^3\text{cm}^3$
Diameter of base of cone = 35cm
Radius of base of cone $=\frac{35}{2}\text{cm}$
Let the height of the cone be h cm.
Volume of cone $=\frac{1}{3}\pi\text{r}^2\text{h}=\frac{1}{3}\pi\Big(\frac{35}{2}\Big)^2\times\text{h}\ \text{cm}^3$
From the above results and from the given conditions,
Volume of ball = Volume of cone
Or, $\frac{4}{3}\pi\times(14)^3=\frac{1}{3}\pi\times\Big(\frac{35}{2}\Big)^2\times\text{h}$
$\Rightarrow\text{h}=\frac{\frac{4}{3}\times(14)^3}{\frac{1}{3}\pi\times\Big(\frac{3\pi}{2}\Big)^2}=\frac{4\times14\times14\times14\times2\times2}{35\times35}=35.84\text{cm}$

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Question 93 Marks

A spherical ball of radius 3cm is melted and recast into three spherical balls. The radii of two of these balls 1.5cm and 2cm. Find t radius of the third ball

Answer

Radius of sphere = 3cm
Radius of first ball = 1.5cm
Radius of second ball = 2cm
Let radius of the third ball be rem.
Volume of third ball = Volume of original sphere - Sum of volumes of other two balls
$=\frac{4}{3}\pi\times3^3=\big(\frac{4}{3}\pi\times\frac{3}{2}^3+\frac{4}{3}\pi\times2^3\big)$
$=\frac{4}{3}\pi\times3\times3\times3-\Big(\frac{4}{3}\pi\times\frac{3}{2}\times\frac{3}{2}\times\frac{3}{2}+\frac{4}{3}\pi\times2\times2\times2\Big)$
$=4\pi\times3\times3-\Big(\pi\frac{3\times2}{2}+\frac{4}{3}\pi\times2\times2\times2\Big)$
$=36\pi-\Big(\pi\frac{9}{2}+\frac{32}{3}\pi\Big)$
$=\Big(\frac{36\times6-9\times3-32\times2}{6}\Big)\pi$
$=\Big(\frac{216-27-64}{6}\Big)\pi=\frac{125\pi}{6}$
Therefore,
$\frac{4}{3}\pi\text{r}^3=\frac{125\pi}{6}$
Or, $\text{r}=\sqrt[3]{\frac{125\times3}{4\times6}}=\sqrt[3]{\frac{125}{8}}=\frac{5}{2}=2.5\text{cm}$

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Question 103 Marks

The volume of a hemisphere is $2425\frac{1}{2}\text{cm}^3.$ find its curved surface area.

Answer

Let r be the radius the hemisphere.
now, volume of hemisphere $2425\frac{1}{2}\text{cm}^3$
$\Rightarrow\frac{2}{3}\pi\text{r}^3=\frac{4851}{2}$
$\Rightarrow\frac{2}{3}\times\frac{22}{7}\times\text{r}^3=\frac{4851}{2}$
$\Rightarrow\frac{44}{21}\times\text{r}^3=\frac{4851}{2}$
$\Rightarrow\text{r}^3=\frac{4851\times21}{88}$
$\Rightarrow\text{r}=\sqrt[3]{1157.625}$
$\Rightarrow\text{r}=10.5\text{cm}$
Now,
Cueved surface area $2\pi\text{r}^2$
$=2\times\frac{22}{7}\times(10.5)^2$
$=\frac{44}{7}\times110.25$
$=693\text{cm}^2$

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Question 113 Marks

A solid metallic sphere of diameter 21cm is melted and recast into a number of smaller cones, each of diameter 3.5cm and height 3cm Find the number of cones so formed.

Answer

Diameter of sphere = 21cm
Radius of sphere $=\frac{21}{2}\text{cm}$
Volume of sphere $=\frac{4}{3}\pi\text{r}^3=\frac{4\times21\times21\times21\pi}{3\times2\times2\times2}=\frac{21\times21\times21\pi}{3\times2}\text{cm}^3$
Diameter of the cone= 3.5cm
Radius of the cone $=\frac{3.5}{2}=\frac{7}{4}\text{cm}$
Height = 3cm
Volume of each cone $=\frac{1}{3}\pi\text{r}^2\text{h}=\frac{1}{3}\pi\times3\Big(\frac{7}{4}\Big)^2=\Big(\frac{7}{4}\Big)^2\pi\text{cm}^3$
Total number of cones $=\frac{\text{Volume of sphere}}{\text{Volume of a cone}}=\frac{\frac{21\times21\times21\times\pi}{3\times2}}{\Big(\frac{7}{4}\Big)^2\pi}=\frac{21\times21\times21\times\pi\times4\times4}{3\times2\times\pi\times7\times7}=504$

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Question 123 Marks

A copper sphere of diameter 18cm is drawn into a wire of diameter 4mm. Find the length of the wire.

Answer

We have, Radius of the sphere, $\text{R}=\frac{18}{2}=9\text{cm}$ and Radius of the wire, $\text{r}=\frac{4}{2}=2\text{mm}=0.2\text{cm}$ Let the length of the wire be l. Now, Volume of the wire = Volume of the copper sphere $\Rightarrow\pi\text{r}^2\text{l}=\frac{4}{3}\pi\text{R}^3$$\Rightarrow\text{l}=\frac{4\text{R}^3}{3\text{r}^2}$
$\Rightarrow\text{l}=\frac{4\times9\times9\times9}{3\times0.2\times0.2}$
$\therefore\text{l}=24300\text{cm}=243\text{m}$
So. the length of the wire is 243m.

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Question 133 Marks

The diameter of a sphere is 42cm. It is melted and drawn into a cylindrical wire of diameter 2.8cm. Find the length of the wire.

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Question 143 Marks

The radii of the circular ends of a frustum of height 6cm are 1cm and 6cm, respectively. Find the slant height of the frustum.

Answer

We have,
Height of the frustum, h = 6cm,
Radii of tbe circular ends, R = 14cm and r = 6cm
Let the slant height of the frustum be l.
Now,
$\text{l}=\sqrt{(\text{R}-\text{r})^2+\text{h}^2}$
$=\sqrt{(14-6)^2+6^2}$
$=\sqrt{8^2+6^2}$
$=\sqrt{64+36}$
$=\sqrt{100}$
$=10\text{cm}$
So, the slant heiglt of the frustum is 10cm.

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Question 153 Marks

The given figure represents a solid consisting of a cylinder surmounted by a cone at one end and a hemisphere at the other. Find the volurne of the solid.

Answer

Height of cylinder $= 6.5\ cm$
height of cone $= h_2$ =$ (12.8 - 6.5)cm = 6.3cm$
Radius of cylinder = radius of cone
= radius of hemisphere
$=\Big(\frac{7}{2}\Big)\text{cm}$
Volume of solid = volume of cylinder + volume of cone + volume of hemisphere
$=\pi\text{r}^2\text{h}_1+\frac{1}{3}\pi\text{r}^2\text{h}_2+\frac{2}{3}\pi\text{r}^2\Big(\text{h}_1+\frac{1}{3}\text{h}_2+\frac{2}{3}\text{r}\Big)$
$=\Big[\frac{22}{7}\times3.5\times3.5\times\Big(6.5+6.3\times\frac{1}{3}+\frac{2}{3}\times3.5\Big)\Big]$
$=[(38.5)\times(6.5+2.1+2.33)]\text{cm}^3$
$=(38.5\times10.93)\text{cm}^3=420.80\text{cm}^3$

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Question 163 Marks

Water running in a cylindrical pipe of inneer diameter $7\ cm$, is collected in a container at the rate of $192.5$ litres per minute. Find the rate o flow of water in the pipe in km/hr.

Answer

We have,
Radius of cylindrical pipe, $\text{r}=\frac{7}{2}\text{cm}$ and
The rate of flow of water $= 192.5\ L/min$
$=\frac{192.5\text{L}}{1\text{min}}$
$=\frac{192.5\times1000\text{cm}^3}{1\text{min}}$ (As, $1 L = 1000\ cm^3$)
$= 192500\ cm^3/min$
$\Rightarrow $ The volume of water flowing out from the cylindrical pipe in $1 min = 192500\ cm^3$
Now, the rate of flow of water in the pipe $=\frac{\text{The volume or water flowing out from the cylindrical pipe}}{\pi\text{r}^2}$
$=\frac{192500}{\Big(\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\Big)}$
$=\frac{192500}{77}$
$=5000\text{cm}/\text{min}$
$=\frac{5000\times60}{1\times100000}\text{km}/\text{hr}$
$=3\text{km}/\text{hr}$
So, the rate of flow of water in the pipe is $3\ km/hr$.

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Question 173 Marks

If the total surface area of a solid hemisphere is 462cm, find its volume.

Answer

Total surface area of solid hemisphere $=3\pi\text{r}^2$
$\Rightarrow462=3\pi\text{r}^2$
$\Rightarrow462=3\times\frac{22}{7}\times\text{r}^2$
$\Rightarrow462=\frac{66}{7}\times\text{r}^2$
$\Rightarrow\text{r}^2=\frac{3234}{66}$
$\Rightarrow\text{r}^2=49$
$\Rightarrow\text{r}=7\text{cm}$
Now,
Volume of a solid hemisphere $=\frac{2}{3}\pi\text{r}^3$
$=\frac{2}{3}\times\frac{22}{7}\times7\times7\times7$
$=\frac{2156}{3}$
$=718.67\text{cm}^3$

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Question 183 Marks

How many lead shots each 3mm in diameter can be made from a cuboid of dimensions 9cm × 11cm × 12cm?

Answer

We have, Radius of a lead shot, $\text{r}=\frac{3}{2}=1.5\text{mm}=0.15\text{cm}$ and Dimensions of the cuboid are 9cm × 11cm × 12cm Now,$=\frac{9\times11\times12}{\Big(\frac{4}{3}\pi\text{r}^3\Big)}$
$=\frac{9\times11\times12}{\Big(\frac{4}{3}\times\frac{22}{7}\times0.15\times0.15\times0.15\Big)}$
$=84000$
So, the number of lead shots that can be made from the cuboid is 84000.

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Question 193 Marks

A solid metallic sphere of radius 8cm is melted and recast into spherical balls each of radius 2cm. Find the number of spherical balls obtained.

Answer

We have, Radius of the solid metallic sphere, R = 8cm and Radius of the spherical ball, r = 2cm Now, The number spherical balls obtained $=\frac{\text{Volume of the solid metallic spher}}{\text{Volume of a spherical ball}}$$=\frac{\Big(\frac{4}{3}\pi\text{R}^3\Big)}{\Big(\frac{4}{3}\pi\text{r}^3\Big)}$
$=\Big(\frac{\text{R}}{\text{r}}\Big)^3$
$=\Big(\frac{8}{2}\Big)^3$
$=4^3$
$=64$
So, the number of spherical balls obtained is 64.

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Question 203 Marks

Three metallic cubes whose edges are $3\ cm, 4\ cm$ and $5\ cm$, are melted and recast into a single large cube. Find the edge of the new cube formed.

Answer

We have,
Edges of the cubes : $a_1 = 3cm, a_2 = 4cm$ nnd $a_3 = 5cm$
Let the edge of the new cube be a.
Now
Volume of the new cube $=\text{a}_1^3+\text{a}_2^3+\text{a}_3^2$
$\Rightarrow\text{a}^3=3^3+4^3+5^3$
$\Rightarrow\text{a}^3=27+64+125$
$\Rightarrow\text{a}^3={216}$
$\Rightarrow\text{a}=\sqrt[3]{216}$
$\therefore\text{a}=6\text{cm}$
so, the edge of the new cube so formed is $6\ cm$.

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Question 213 Marks

A cylinder with base radius 8cm and height 2cm is melted to form a cone of height 6cm. Calculate the radius of the base of the cone.

Answer

We have,
Bose radius of the cylinder, r = 8cm,
Heigbt of tbe cylinder, h = 2cm and
Height of the cone, H = 6cm
Let the base radius of the cone be R.
Now,
Volume of the coue = Volume of tbe cyliuder
$\Rightarrow\frac{1}{3}\pi\text{R}^2\text{H}=\pi\text{r}^2\text{h}$
$\Rightarrow\text{R}^2=\frac{3\text{r}^2\text{h}}{\text{H}}$
$\Rightarrow\text{R}^2=\frac{3\times8\times8\times2}{6}$
$\Rightarrow\text{R}^2=64$
$\Rightarrow\text{R}=\sqrt{64}$
$\therefore\text{R}=8\text{cm}$
so, the radius of the base of the cone is 8cm

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Question 223 Marks

If the area of the base of a right circular cone is $3850\ cm^2$ and its height is $84\ cm$, then find the slant height of the cone.

Answer

We have,
Height $= 84\ cm$
Let the radius and the slant height of the cone be r and l, respectively.
As,
Arca of the base of the cone $= 385cm2$
$\Rightarrow\pi\text{r}^2=3850$
$\Rightarrow\frac{22}{7}\times\text{r}^2=3850$
$\Rightarrow\text{r}^2=3850\times\frac{7}{22}$
$\Rightarrow\text{r}^2=1225$
$\Rightarrow\text{r}^2=\sqrt{1225}$
$\therefore\text{r}=35\text{cm}$
Now,
$\text{l}=\sqrt{\text{h}^2+\text{r}^2}$
$=\sqrt{84^2+35^2}$
$=\sqrt{7056+1225}$
$=\sqrt{8281}$
$=91\text{cm}$
So, the slant height of the given cone is $91\ cm$.

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Question 233 Marks

Find the ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height?

Answer

Let the radius of the sphere be r.
We have,
The radius of the cone = The radius of the cylinder = The radius of the sphere= r and
The height of the cylinder = The height of the cone = The height of the sphere = 2r
Now,
Volume of the cylinder $=\pi\text{r}^2(2\text{r})=2\pi\text{r}^3,$
Volume of the ocne $=\frac{1}{3}\pi\text{r}^2(2\text{r})=\frac{2}{3}\pi\text{r}^3$ and
Volume of the sphere $=\frac{4}{3}\pi\text{r}^3$
So,
The ratio of the volumes of the cylinder, the cone and the sphere $=2\pi\text{r}^3:\frac{2}{3}\pi\text{r}^3:\frac{4}{3}\pi\text{r}^3$
$=1:\frac{1}{3}:\frac{2}{3}$
$=3:1:2$
So, the ratio of the volumes of the cylinder, the cone and the sphere is 3 : 1 : 2.

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Question 243 Marks

A solid metallic sphere of diameter 8cm is melted and drawn into a cylindrical wire of uniform width. If the length of the wire is 12m, then find its width.

Answer

We have,
Radius of the metallic sphere, $\text{R}=\frac{8}{2}=4\text{cm}$ and
Height of the cylindrical wire, h = 12m = 1200cm
Let tbe radius of the base be r.
Now,
Volume of the cylindrical wire = Volume of the metallic sphere
$\Rightarrow\pi\text{r}^2\text{h}=\frac{4}{3}\pi\text{R}^3$
$\Rightarrow\text{r}^2=\frac{4\text{R}^3}{3\text{h}}$
$\Rightarrow\text{r}^2=\frac{4\times4\times4\times4}{3\times1200}$
$\Rightarrow\text{r}^2=\frac{16}{225}$
$\Rightarrow\text{r}=\sqrt{\frac{16}{225}}$
$\therefore$ The width of the wire = 2r
$=2\times\frac{4}{15}$
$=\frac{8}{15}\text{cm}$
So. the width of the wire is $\frac{8}{15}\text{cm}$

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Question 253 Marks

A right cylindrical vessel is full of water. How many right cones having the same radius and height as those of the right cylinder will be needed to store that water?

Answer

Let the radius and height of the cone be r and h, respectively. Then,Radius of the cylindrical vessel = r and
Height of the cylindrical vessel = h
Now,
The number of cones $=\frac{\text{Volume of the cylindrical vessel}}{\text{Volume of a cone}}$
$=\frac{\pi\text{r}^2\text{h}}{\Big(\frac{1}{3}\pi\text{r}^2\text{h}\Big)}$
$=3$
So, the number of cones that will be needed to store the water is 3.

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Question 263 Marks

A vessel is in the form of a hemispherical bowl surmounted by a hollow cylinder. The diameter of the hemisphere is 21cm and the total height of the vesel is 14.5cm. find its capacity.

Answer

Radius of hemisphere = 10.5cm
Height of cylinder = (14.5 10.5)cm = 4cm
Radius of cylinder = 10.5cm
Capacity = volume of cylinder + volume of hemisphere
$=\Big(\pi\text{r}^2\text{h}+\frac{2}{3}\pi\text{r}^2\Big)\text{cm}^3=\pi\text{r}^2\Big(\text{h}+\frac{2}{3}\text{r}\Big)\text{cm}^3$
$=\Big[\frac{22}{7}\times10.5\times10.5\Big(4+\frac{2}{3}\times10.5\Big)\Big]\text{cm}^3$
$=(346.5\times11)\text{cm}^2=3811.5\text{cm}^2$

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Question 273 Marks

The surface area of a sphere is $2464\ cm^2$ lf its radius be doubled, what will be the surface area of the new sphere?

Answer

Let the original radius be r.
⇒ original surface area $=4\pi\text{r}^2=2464\text{cm}^2\ ...(1)$
Given new radius $=2\text{r}$
⇒ New surface area $4\pi(2\text{r})^2$
$=4\times4\pi\text{r}^2$
$=4\times4\pi\text{r}^2$
$=4\times2464\ ...(\text{From}(1))$
$=9856\text{cm}^2$

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Question 283 Marks

A spherical ball of diameter $21\ cm$ is melted and recast into cubes. each of side $1\ cm$. Find the number of cubes so formed.

Answer

Diameter of the spherical ball $= 21\ cm$
Radius of the ball $\frac{21}{2}\text{cm}$
Volume of spherical ball $=\frac{4}{3}\pi\text{r}^3=\frac{4}{3}\times\frac{22}{7}\times\frac{21}{2}\times\frac{21}{2}\times\frac{21}{2}$
$=11\times21\times21=4851\text{cm}^3$
Volume of each cube $= 1^3 = 1\ cm^3$
Number of cubes $=\frac{\text{Volume of spherical ball}}{\text{Volume of each cube}}=\frac{4851}{1}=4851$

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Question 293 Marks

Two cubes each of volume $27cm^3$ are joined end to end to form a solid. Find the surface area of the resulting cuboid.

Answer

Let the length of each side of each cube $=5 cm$.
Now,
Volume of each cuboid $=27 cm^3$
$\Rightarrow s^3=27$
$\Rightarrow s=3 cm$
When two cubes of each side, 3 cm is joined end to end, then a cuboid is formed.
Now, length of cubiod $( l )=6 cm$,
breadth of cubiod (b) $=3 cm$ and
height of cubiod (h) $=3 cm$
$\therefore \text { Total surface area }=(lb+bh+lh)$
$=2[(6 \times 3)+(3 \times 3)(3 \times 6)]$
$=2[18+9+18]$
$=2 \times 45$
$=90 cm^2$

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Question 303 Marks

The slant height of a conical mountain is $2.5\ km$ and the area of its base is $1.54 km^2$. Find the height of the untain.

Answer

Let the radius of the base of the conical mountain be $r$ km.
$\Rightarrow$ Area of the base of the conical mounrain $=\pi\text{r}^2$
$\Rightarrow1.54=\frac{22}{7}\times\text{r}^2$
$\Rightarrow\text{r}^2=\frac{10.78}{22}$
$\Rightarrow\text{r}^2=0..49$
$\Rightarrow\text{r}=0.7\text{ Km}$
Slant height of the conical mountain (l) $= 2.5\ km$
Let the height of the mountain be h km.
Now,
$\Rightarrow\text{l}^2=\text{h}^2+\text{r}^2$
$\Rightarrow(2.5)^2=\text{h}^2+(0.7)^2$
$\Rightarrow\text{h}=\sqrt{(2.5)^2-(0.7)^2}$
$\Rightarrow\text{h}-\sqrt{6.25-0.49}$
$\Rightarrow\text{h}=\sqrt{5.76}$
$\Rightarrow\text{h}=2.4\text{ Km}$
Thus, the height of the moutain is $2.4\ km$.

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Question 313 Marks

The radius and height of a solid right-circular cone are in the ratio of $5 : 12$. If its volume is $314\ cm^3$, find its total surface area. [Take $\pi$ = $3.14$.]

Answer

Let the Radius be $5n$
And the heigth be $12n$
Volume $= 314\ cm^3$
$\frac{1}{3}\pi\text{r}^2\text{h}=314$
$\frac{1}{3}\times3.14\times(5)^2\times12=314$
$\frac{1}{3}\times3.14\times300\text{n}=314$
$3.14\times100=314$
$314\text{n}=314$
$\text{n}=\frac{314}{314}$
$\text{n}=1$
$\pi\text{r}(\text{l}+\text{r})$
$\text{l}^2=\sqrt{(12)^2+(5)^2}$
$\text{l}^2=\sqrt{144+25}$
$\text{l}^2=\sqrt{169}$
$\text{l}=13$
$\pi\text{r}(1+\text{r})\Rightarrow3.14\times5(13+5)$
$\Rightarrow15.70\times18$
$\Rightarrow282.6\text{cm}^3$

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Question 323 Marks

The curved surface area of a sphere is $5544 cm^2$. Find its volume.

Answer

Let the radius of the sphere be $r$.
As,
Curved surface area or the sphere $= 5544 cm^2$
$\Rightarrow4\pi\text{r}^2=5544$
$\Rightarrow4\times\frac{22}{7}\times\text{r}^2=5544$
$\Rightarrow\text{r}^2=441$
$\Rightarrow\text{r}=21\text{cm}$
Now,
Volume of the sphere $=\frac{4}{3}\pi\text{r}^3$
$=\frac{4}{3}\times\frac{22}{7}\times21\times21\times21$
$= 38808cm^3$
So, the volume of the spllere is $38808 cm^3$.

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Question 333 Marks

Find the ratio of the volume of a cube to that of a sphere which will fit inside it.

Answer

Let the radius of the shere be R and the edge of the cube be a. As, the sphere is fit iuside the cube. So, diameter of the sphere = edge of the cube $\Rightarrow2\text{R}=\text{a}\ ...(\text{i})$ Now, The ratio of the volume of the cube to that of the sphere $=\frac{\text{Volume of the cube}}{\text{Volume of the sphere }}$ $=\frac{\text{a}^3}{\Big(\frac{4}{3}\pi\text{R}^3\Big)}$ $=\frac{(2\text{R})^3}{\Big(\frac{4}{3}\pi\text{R}^3\Big)}$ [Using (i)] $=\frac{3\times8\text{R}^3}{4\pi\text{R}^3}$ $=\frac{6}{\pi}$ $=6:\pi$ So, the ratio of the volume of the cube to that of the sphere is 6 : $\pi$

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