| Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4 |
Exercise 14.3
Question 1
Two cylindrical vessels are filled with oil. The radius of one vessel is 15 cm and its height is 25 cm. The radius and height of the other vessel are 10cm and 18 cm respectively. Find the radius of a cylindrical vessel 30 cm in height, which will just contain the oil of two given vessels.
Sol :Radius of 1st cylindrical vessel = 15cm
and height = 25cm
Radius of 2nd cylindrical vessel = 10cm
and height = 18cm
So, Volume of 1st cylindrical vessel = πr2h
= π × (15)2 × 25
= 5625π cm3
Volume of 2nd cylindrical vessel = π(r’)2h’
= π × (10)2 × 18
= 1800π cm3
Height of the third vessel = 30cm
and let its radius be R
So,
Volume of third cylindrical vessel = πR2H
= πR2×30
= 30πR2
Volume of 1st cylindrical vessel + Vol. of 2nd Cylindrical vessel
= Volume of the third cylindrical Vessel
⇒ 5625π + 1800π = 30πR2
⇒ 7425π = 30πR2
$\Rightarrow \frac{7425}{30}=\mathrm{R}^{2}$
⇒ R2 = 247.5
⇒ R = 15.73cm
Hence, radius of the required cylinder is 15.73cm
Question 2
A metal cube of edge 12 cm is melted and formed into three smaller cubes. If the edges of two smaller cubes are 6 cm and 8 cm, find the edge of the third smaller cube (Assume that there is no loss of metal during melting)Sol :
Let the edge of the third smaller cube be x cm.
Three small cubes are formed by melting the cube of edge 12 cm.
Edges of two small cubes are 6 cm and 8 cm.
Now, volume of a cube = (side)3
Volume of the big cube = sum of the volumes of the three small cubes
⇒ (12)3 = (6)3 + (8)3 + (x)
⇒ 1728 = 216 + 512 + x3
⇒ 1728 = 728 + x3
⇒ x3 = 1728 – 728
⇒ x3 = 1000
⇒ x = 10 cm
Therefore, the edge of the third cube is 10 cm
Question 3
A hemisphere of lead of radius 8 cm is cast into a right circular cone of base radius 6 cm. Determine the height of the cone, correct to two places of decimals.Sol :
Radius of hemisphere = 8cm
and Radius of right circular cone = 6cm
According to question,
Volume of the hemisphere = Volume of cone
$\Rightarrow \frac{2}{3} \pi r^{3}=\frac{1}{3} \pi R^{2} h$
$\Rightarrow \frac{2}{3} \times \pi \times 8 \times 8 \times 8=\frac{1}{3} \times \pi \times 6 \times 6 \times \mathrm{h}$
$\Rightarrow \frac{1024}{36}=\mathrm{h}$
⇒ h = 28.44cm
Question 4
A solid sphere of radius 3 cm is melted and then recast into small spherical balls each of diameter 0.6 cm. Find the number of small balls thus obtained.Sol :
Given that Radius of bigger sphere (R) = 3cm
and diameter of smaller spherical balls = 0.6cm
So, radius of small spherical balls (r) = 0.3cm
Let the number of small balls = n
According to the question,
n × volume of small balls = Volume of bigger sphere
$\Rightarrow \mathrm{n} \times \frac{4}{3} \pi \mathrm{r}^{3}=\frac{4}{3} \pi \mathrm{R}^{3}$
⇒ nr3 = R3
$\Rightarrow \mathrm{n}=\frac{\mathrm{R}^{3}}{\mathrm{r}^{3}}$
$\Rightarrow \mathrm{n}=\left(\frac{\mathrm{R}}{\mathrm{r}}\right)^{3}$
$\Rightarrow \mathrm{n}=\left(\frac{3}{0.3}\right)^{3}$
⇒ n = (10)3⇒ n = 1000
Hence, the number of small balls = 1000
Question 5
A solid metal cone with radius of base 12 cm and height 24 cm is melted to form solid spherical balls of diameter 6 cm each. Find the number of balls thus formed.
Sol :
Given that Radius of cone = 12cm
Height of the cone = 24cm
So,
Volume of cone $=\frac{1}{3} \pi \mathrm{R}^{2} \mathrm{h}$
$=\frac{1}{3} \pi \times 12 \times 12 \times 24$
= 1152π cm3
Height of the cone = 24cm
So,
Volume of cone $=\frac{1}{3} \pi \mathrm{R}^{2} \mathrm{h}$
$=\frac{1}{3} \pi \times 12 \times 12 \times 24$
= 1152π cm3
It is also given that diameter of each spherical balls = 6cm
so, Radius of each ball = 3cm
so, Radius of each ball = 3cm
Volume of each ball $=\frac{4}{3} \pi r^{3}$
$=\frac{4}{3} \pi \times 3 \times 3 \times 3$
= 36π cm3
$=\frac{4}{3} \pi \times 3 \times 3 \times 3$
= 36π cm3
Total number of balls formed by melting the cone
$=\frac{\text { Volume of cone }}{\text { Volume of a ball }}$
$=\frac{1152 \times \pi}{36 \times \pi}$
= 32
Hence, 32 balls are formed by melting the cone.
$=\frac{\text { Volume of cone }}{\text { Volume of a ball }}$
$=\frac{1152 \times \pi}{36 \times \pi}$
= 32
Hence, 32 balls are formed by melting the cone.
Question 6
A solid metallic sphere of diameter 21 cm is melted and recast into a number of smaller cones, each of diameter 3.5 cm and height 3cm. Find the number of cones so formed.Sol :
Let the number of cones formed be n
Diameter of a metallic sphere = 21cm
So, Radius $=\frac{21}{2} \mathrm{cm}$
Volume of the sphere $=\frac{4}{3} \pi r^{3}$
$=\frac{4}{3} \times \frac{22}{7} \times\left(\frac{21}{2}\right)^{3}$
= 11 × 21 × 21
= 4851cm3
Now, Diameter of cone = 3.5cm
So, Radius $=\frac{3.5}{2}=\frac{7}{4} \mathrm{cm}$
and height = 3cm
Volume of the cone $=\frac{1}{3} \pi r^{2} h$
$=\frac{1}{3} \times \frac{22}{7} \times\left(\frac{7}{4}\right)^{2} \times 3$
$=\frac{22 \times 7}{4 \times 4}$ cm3
According to question,
n × volume of cone = volume of sphere
$\Rightarrow \mathrm{n} \times \frac{22 \times 7}{4 \times 4}=4851$
$\Rightarrow \mathrm{n}=\frac{4851 \times 16}{22 \times 7}$
⇒ n = 504
Hence, the number of cones formed = 504
Question 7
Spherical ball of diameter 21cm is melted and recasted into cubes, each of side 1 cm. Find the number of cubes thus formed. [Use π=22/7]
Sol :Diameter of sphere = 21 cm
So, radius of the sphere $=\frac{21}{2} \mathrm{cm}$
Volume of the sphere $=\frac{4}{3} \pi r^{3}$
$=\frac{4}{3} \times \frac{22}{7} \times\left(\frac{21}{2}\right)^{3}$
= 11 × 21 × 21
= 4851cm3
Volume of cube = a3 = 13
Let number of cubes formed be n, then
Volume of sphere = n × (volume of cube)
⇒ 4851 =n×1
⇒ n = 4851
Hence, number of cubes is 4851.
Question 8
The internal and external diameters of a hollow hemispherical shell are 6cm and 10cm respectively. It is melted and recast into a solid cone of base diameter 14cm. Find the height of the cone so formed.
Sol :Given:
Internal Diameter of a hollow hemispherical shell = 6cm
So, internal radius, (r) = 3cm
External Diameter of a hollow hemispherical shell = 10cm
So, external radius (R) = 5cm
Volume of material in the shell $=\frac{2}{3} \pi\left[\mathrm{R}^{3}-\mathrm{r}^{3}\right]$
$=\frac{2}{3} \pi\left[5^{3}-3^{3}\right]$
$=\frac{2}{3} \times \frac{22}{7} \times 98$
So, radius = 7cm
Volume of cone $=\frac{1}{3} \pi\left(\mathrm{r}^{\prime}\right)^{2} \mathrm{h}$
$=\frac{1}{3} \times \frac{22}{7} \times(7)^{2} \mathrm{h}$
According to the question.
Volume of a shell = Volume of a cone
$\frac{2}{3} \times \frac{22}{7} \times 98=\frac{1}{3} \times \frac{22}{7} \times(7)^{2} \mathrm{h}$
⇒ $\frac{2 \times 98}{7 \times 7}=h$
⇒ h = 4cm
Let inner radius of the cylinder be x cm
Given that radius of sphere = 6cm
$\therefore$ Volume of the sphere $=\frac{4}{3} \pi\left(\mathrm{r}^{\prime}\right)^{3}$
= $\frac{4}{3} \times \pi \times(6)^{3}$
= 4 × π × 2 × 36 cm3
It is also given that external radius of base of cylinder (R) = 5cm
and height = 32cm
∴ Volume of hollow cylinder = πh[R2 – r2]
= π × 32 × [(5)2 – x2]
= 32π [25 – x2]
Solid sphere is melted and casted into hollow cylinder
∴ Volume of sphere = Volume of hollow cylinder
⇒ 4 × π × 2 × 36 = 32π [25 – x2]
$\Rightarrow \frac{8 \times 36}{32}=25-\mathrm{x}^{2}$
⇒ 9 – 25 = -x2
⇒ x2 = 16
⇒ x = ±4
⇒ x = 4cm
Hence, the inner radius of the cylinder is 4cm
So, the thickness of the cylinder = external radius – inner radius
= 5 – 4
= 1cm
Diameter of a copper sphere = 6 cm
So, Radius of copper sphere = 3cm
Volume of the sphere $=\frac{4}{3} \pi r^{3}$
$=\frac{4}{3} \times \pi \times(3)^{3}$
= 36π cm3
Length of wire = 36cm
Let the radius of wire be R cm
Volume of wire = πR2h
=πR2×36
But the volume of wire = volume of sphere
⇒ 36πR2 = 36π
⇒ R2 = 1
⇒ R = 1cm [taking positive root, because radius can’t be negative]
Hence, the radius of wire is 1cm
Sol :
Let the radius of the base of conical ice cream = x cm
Then, height of the conical ice cream = 2 × diameter
= 2 × (2x)
= 4x cm
Volume of ice- cream cone
= Volume of conical portion + Volume of hemispherical portion
$=\frac{1}{3} \pi r^{2} h+\frac{2}{3} \pi r^{3}$
$=\frac{1}{3} \pi x^{2}(4 x)+\frac{2}{3} \pi x^{3}$
$=\frac{4 \pi x^{3}+2 \pi x^{3}}{3}$
= 2πx3 cm3
Now,
Diameter of cylindrical container = 12cm
So, radius = 6cm
and height = 15cm
∴ Volume of cylindrical container = πr2h
= π × (6)2 × (15)
= 540π cm3
Number of children $=\frac{\text { Volume of cylindrical container }}{\text { Volume of one ice }-\text { cream cone }}$
⇒ x = 3
So, the radius of the base of conical ice cream is 3cm
Hence, the diameter of the base of conical ice-cream = 2×3 = 6cm
$=\frac{1}{3} \times \frac{22}{7} \times(7)^{2} \mathrm{h}$
According to the question.
Volume of a shell = Volume of a cone
$\frac{2}{3} \times \frac{22}{7} \times 98=\frac{1}{3} \times \frac{22}{7} \times(7)^{2} \mathrm{h}$
⇒ $\frac{2 \times 98}{7 \times 7}=h$
⇒ h = 4cm
Question 9
A solid sphere of radius 6cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5cm and its height is 32 cm, find the uniform thickness of the cylinder.
Sol :Let inner radius of the cylinder be x cm
Given that radius of sphere = 6cm
$\therefore$ Volume of the sphere $=\frac{4}{3} \pi\left(\mathrm{r}^{\prime}\right)^{3}$
= $\frac{4}{3} \times \pi \times(6)^{3}$
= 4 × π × 2 × 36 cm3
It is also given that external radius of base of cylinder (R) = 5cm
and height = 32cm
∴ Volume of hollow cylinder = πh[R2 – r2]
= π × 32 × [(5)2 – x2]
= 32π [25 – x2]
Solid sphere is melted and casted into hollow cylinder
∴ Volume of sphere = Volume of hollow cylinder
⇒ 4 × π × 2 × 36 = 32π [25 – x2]
$\Rightarrow \frac{8 \times 36}{32}=25-\mathrm{x}^{2}$
⇒ 9 – 25 = -x2
⇒ x2 = 16
⇒ x = ±4
⇒ x = 4cm
Hence, the inner radius of the cylinder is 4cm
So, the thickness of the cylinder = external radius – inner radius
= 5 – 4
= 1cm
Question 10
The diameter of a copper sphere is 6cm. The sphere is melted and is drawn into a long wire of uniform circular cross-section. If the length of the wire is 36cm, find its radius [Take π=2.14]
Sol :Diameter of a copper sphere = 6 cm
So, Radius of copper sphere = 3cm
Volume of the sphere $=\frac{4}{3} \pi r^{3}$
$=\frac{4}{3} \times \pi \times(3)^{3}$
= 36π cm3
Length of wire = 36cm
Let the radius of wire be R cm
Volume of wire = πR2h
=πR2×36
But the volume of wire = volume of sphere
⇒ 36πR2 = 36π
⇒ R2 = 1
⇒ R = 1cm [taking positive root, because radius can’t be negative]
Hence, the radius of wire is 1cm
Question 11
A cylindrical container is filled with ice-cream. Its diameter is 12cm and height is 15cm. The whole ice-cream is distributed among 10 children in equal cones having hemispherical tops. If the height of the conical portion is twice the diameter of its base, find the diameter of the ice-cream cone.Sol :
Let the radius of the base of conical ice cream = x cm
Then, height of the conical ice cream = 2 × diameter
= 2 × (2x)
= 4x cm
Volume of ice- cream cone
= Volume of conical portion + Volume of hemispherical portion
$=\frac{1}{3} \pi r^{2} h+\frac{2}{3} \pi r^{3}$
$=\frac{1}{3} \pi x^{2}(4 x)+\frac{2}{3} \pi x^{3}$
$=\frac{4 \pi x^{3}+2 \pi x^{3}}{3}$
= 2πx3 cm3
Now,
Diameter of cylindrical container = 12cm
So, radius = 6cm
and height = 15cm
∴ Volume of cylindrical container = πr2h
= π × (6)2 × (15)
= 540π cm3
Number of children $=\frac{\text { Volume of cylindrical container }}{\text { Volume of one ice }-\text { cream cone }}$
$\Rightarrow 10=\frac{540 \times \pi}{2 \times \pi \times \mathrm{x}^{3}}$
$\Rightarrow \mathrm{x}^{3}=\frac{540}{2 \times 10}$
⇒ x3 = 27⇒ x = 3
So, the radius of the base of conical ice cream is 3cm
Hence, the diameter of the base of conical ice-cream = 2×3 = 6cm
| S.no | Chapters | Links |
|---|---|---|
| 1 | Real numbers | Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 |
| 2 | Polynomials | Exercise 2.1 Exercise 2.2 Exercise 2.3 |
| 3 | Pairs of Linear Equations in Two Variables | Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 |
| 4 | Trigonometric Ratios and Identities | Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4 |
| 5 | Triangles | Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 |
| 6 | Statistics | Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4 |
| 7 | Quadratic Equations | Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5 |
| 8 | Arithmetic Progressions (AP) | Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4 |
| 9 | Some Applications of Trigonometry: Height and Distances | Exercise 9.1 |
| 10 | Coordinates Geometry | Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 |
| 11 | Circles | Exercise 11.1 Exercise 11.2 |
| 12 | Constructions | Exercise 12.1 |
| 13 | Area related to Circles | Exercise 13.1 |
| 14 | Surface Area and Volumes | Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4 |
| 15 | Probability | Exercise 15.1 |
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