Q: Volume of a cylinder of height 3 cm is $48\pi$. Radius of the cylinder is:

Volume of cylinder $V = \pi r^2 h$. Given $h = 3\text{ cm}$ and $V = 48\pi$: $\pi r^2 (3) = 48\pi \implies 3r^2 = 48 \implies r^2 = 16 \implies r = 4\text{ cm}$. Correct Option: (c) [1 Mark]

Question 1

The total surface area of a solid sphere (S₁) and a solid hemisphere (S₂), as shown in the diagram, are equal. The ratio of radii R and r is:

Accepted Answer

Correct Option: (c) √3 : 2 Formula Equating Total Surface Area of solid sphere S₁ = 4πR². Total Surface Area of solid hemisphere S₂ = 3πr² (Curved 2πr² + Base flat πr²). Given: 4πR² = 3πr² ⇒ R² / r² = 3 / 4. Taking square root: R / r = √3 / 2 .

Question 2

An eye drop bottle is prepared consisting of a hemisphere, a cylinder and a conical cap, as shown in the given diagram. Height of the cylindrical and conical parts are each equal to the diameter (7 cm). Find the:
(a) minimum height of the cylindrical box required to pack this bottle.
(b) volume of the liquid medicine (shaded part) in the bottle. Give your answer to the nearest whole number. (Use π = 22/7)

Accepted Answer

Part (a) Height Diameter d = 7 cm ⇒ Radius r = 3.5 cm.
Height of cylinder h_cyl = 7 cm; Height of cone h_cone = 7 cm.
The bottom hemisphere is inverted into the cylinder base, so base of the bottle rests on the cylinder rim.
Total height of bottle = h_cyl + h_cone = 7 + 7 = 14 cm. [1.5 Marks]

Part (b) Liquid Volume Liquid is filled in the cylinder excluding the inverted hemisphere space:
Volume = Volume of cylinder − Volume of hemisphere = πr²h − (2/3)πr³ = πr² [h − (2/3)r]
= (22/7) × (3.5)² × [7 − (2/3 × 3.5)] = 38.5 × [7 − 2.333] = 38.5 × 4.667 ≈ 180 cm³. [2.5 Marks]

Question 3

A solid wooden toy is prepared by joining a cone, a cylinder and a sphere, as shown in the given diagram. The radius of each of the three solids is 7 cm and heights of each of the cone and the cylinder is 24 cm. Find:
(a) the total surface area of the given solid.
(b) the cost of painting the total surface at the rate of ₹0.50 per cm². (Use π = 22/7)

Accepted Answer

Slant Height of Cone l = √(r² + h²) = √(7² + 24²) = √(49 + 576) = √625 = 25 cm. [1 Mark]

Total Surface Area
The outer exposed surface consists of:
1. Surface of Sphere = 4πr²
2. Curved Surface of Cylinder = 2πrh
3. Curved Surface of Cone = πrl
TSA = πr [ 4r + 2h + l ] = (22/7) × 7 × [ 4(7) + 2(24) + 25 ]
= 22 × [ 28 + 48 + 25 ] = 22 × 101 = 2,222 cm². [2 Marks]

Cost of Painting Cost = TSA × Rate = 2,222 × 0.50 = ₹1,111. [1 Mark]

Question 4

If the volume of two spheres is in the ratio 27 : 64, then the ratio of their radii is:

Accepted Answer

V1 / V2 = (r1 / r2)³ = 27 / 64 ⇒ r1 / r2 = ³√(27/64) = 3 : 4 . [1 Mark]

Question 5

A solid wooden capsule is formed of a cylindrical block and two hemispheres. Find the sum of total surface area of the three parts as shown in Figure 2. Given radius = 3.5 cm and length of cylinder = 14 cm. (Use π = 22/7)

Accepted Answer

For the 3 separated components:
TSA of cylinder = 2πrh + 2πr².
TSA of 2 solid hemispheres = 2 × (3πr²) = 6πr².
Sum of TSAs = 2πrh + 8πr² = 2πr(h + 4r). [2 Marks]
With r = 3.5 = 7/2 cm, h = 14 cm:
2πr = 2 × (22/7) × 3.5 = 22 cm. h + 4r = 14 + 14 = 28 cm.
Total Area = 22 × 28 = 616 cm². [2 Marks]

Question 6

A hollow sphere of external diameter 10 cm and internal diameter 6 cm is melted and made into a solid right circular cone of height 8 cm. Find radius of cone so formed. [Use π = 22/7]

Accepted Answer

External R = 5 cm, internal r = 3 cm.
Volume of hollow sphere = 4/3 π (5³ − 3³) = 4/3 π (125 − 27) = (392/3)π cm³. [1.5 Marks]
Volume of cone = 1/3 π r_cone² (8) = (8/3)π r_cone².
Equating: (8/3)π r_cone² = (392/3)π ⇒ r_cone² = 49 ⇒ radius of cone = 7 cm. [1.5 Marks]

Question 7

A rectangular sheet of paper of size 11 cm × 7 cm is first rotated about side 11 cm and then about side 7 cm to form a cylinder. The ratio of their curved surface areas is:

Accepted Answer

Correct Option: (a) 1 : 1 CSA formed by rectangular sheet = area of sheet = 11 × 7 = 77 cm 2 in both cases. Ratio is 1 : 1 . [1 Mark]

Question 8

A solid metallic cylinder is cut into two identical halves along its height. Diameter is 7 cm and height is 10 cm. Find:
(a) The total surface area (both the halves).
(b) The total cost of painting the two halves at ₹30 per cm². (Take π = ²²/₇)

Accepted Answer

r = 3.5 cm, h = 10 cm. (a) Original cylinder TSA = 2π r ( h + r ) = 22 × 13.5 = 297 cm 2 . [1.5 Marks] Two new cut rectangular faces = 2 × (7 × 10) = 140 cm 2 . Total surface area = 297 + 140 = 437 cm 2 . [1.5 Marks] (b) Cost = 437 × ₹30 = ₹13,110 . [1 Mark]

Question 9

Oil is stored in a spherical vessel occupying 3 / 4 of its capacity (radius = 28 cm). This oil is poured into a cylindrical vessel with radius 21 cm. Find the height of the oil in the cylinder to the nearest cm.

Accepted Answer

Volume of oil = 3 / 4 × ( 4 / 3 π R 3 ) = π R 3 = π(28) 3 . [1.5 Marks] In cylinder: π r 2 h = π(21) 2 h = π(28) 3 [1 Mark] h = (28 × 28 × 28) / (21 × 21) = 448/9 ≈ 49.78 cm. To nearest cm: h = 50 cm . [1.5 Marks]

Question 10

Volume of a cylinder of height 3 cm is $48\pi$. Radius of the cylinder is:

Accepted Answer

Volume of cylinder $V = \pi r^2 h$. Given $h = 3	ext{ cm}$ and $V = 48\pi$: $\pi r^2 (3) = 48\pi \implies 3r^2 = 48 \implies r^2 = 16 \implies r = 4	ext{ cm}$. Correct Option: (c) [1 Mark]

Question 11

A solid is in the shape of a hemisphere of radius 7 cm, surmounted by a cone of height 4 cm. The solid is immersed completely in a cylindrical container filled with water to a certain height. If the radius of the cylinder is 14 cm, find the rise in the water level.

Accepted Answer

Step 1: Volume of Solid
Solid consists of hemisphere ($r = 7\text{ cm}$) and cone ($r = 7\text{ cm}, h = 4\text{ cm}$):
$V_{\text{hemisphere}} = \frac{2}{3}\pi r^3 = \frac{2}{3}\pi (7)^3 = \frac{686\pi}{3}\text{ cm}^3$.1 Mark
$V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (7^2)(4) = \frac{196\pi}{3}\text{ cm}^3$.1 Mark
Total Volume of Solid $V_{\text{solid}} = \frac{686\pi + 196\pi}{3} = \frac{882\pi}{3} = 294\pi\text{ cm}^3$.

Step 2: Water Level Rise
Let the rise in water level in the cylinder be $H$.
Volume of water displaced = Volume of cylinder of radius $R = 14\text{ cm}$ and height $H$:
$\pi R^2 H = V_{\text{solid}}$1 Mark
$\pi (14)^2 H = 294\pi \implies 196 H = 294 \implies H = \frac{294}{196} = \frac{3}{2} = \mathbf{1.5\text{ cm}}$.
Rise in water level = 1.5 cm.1 Mark

Question 12

A solid spherical ball of radius 6 cm is melted and recast into 64 identical spherical marbles. Find the radius of each marble.

Accepted Answer

Step 1 Radius of big spherical ball $R = 6\text{ cm}$.
Volume of big sphere $V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (6)^3\text{ cm}^3$. [1 Mark]
Step 2 Let $r$ be the radius of each small spherical marble.
Total volume of 64 marbles $= 64 \times \left(\frac{4}{3} \pi r^3\right)$.
By conservation of volume during recasting:
$$\frac{4}{3} \pi (6)^3 = 64 \times \frac{4}{3} \pi r^3 \implies 216 = 64 r^3$$ [1 Mark]
Step 3 Solving for $r$:
$$r^3 = \frac{216}{64} = \frac{27}{8} \implies r = \sqrt[3]{\frac{27}{8}} = \frac{3}{2} = \mathbf{1.5\text{ cm}}$$
Answer: The radius of each marble is $1.5\text{ cm}$. [1 Mark]

Question 13

From a solid wooden cylinder of height 28 cm and diameter 6 cm, two conical cavities are hollowed out. The diameters of the cones are also of 6 cm and height 10.5 cm. Taking $\pi = \frac{22}{7}$, find the volume of the remaining solid.

Accepted Answer

Radius of cylinder and cones $r = \frac{6}{2} = 3\text{ cm}$.
Height of cylinder $H = 28\text{ cm}$. Height of each cone $h = 10.5\text{ cm} = \frac{21}{2}\text{ cm}$.
Step 1 Volume of cylinder:
$$V_{\text{cylinder}} = \pi r^2 H = \frac{22}{7} \times 3^2 \times 28 = \frac{22}{7} \times 9 \times 28 = 22 \times 9 \times 4 = \mathbf{792\text{ cm}^3}$$ [1 Mark]
Step 2 Volume of two identical conical cavities:
$$2 \times V_{\text{cone}} = 2 \times \left(\frac{1}{3} \pi r^2 h\right) = \frac{2}{3} \times \frac{22}{7} \times 3^2 \times \frac{21}{2} = \frac{2}{3} \times \frac{22}{7} \times 9 \times \frac{21}{2} = 22 \times 9 = \mathbf{198\text{ cm}^3}$$ [1 Mark]
Step 3 Volume of remaining solid:
$$V_{\text{remaining}} = V_{\text{cylinder}} - 2 \times V_{\text{cone}} = 792 - 198 = \mathbf{594\text{ cm}^3}$$
Answer: Volume of remaining solid = $594\text{ cm}^3$. [1 Mark]

Question 14

A solid metallic sphere of radius $6	ext{ cm}$ is melted and made into a solid cylinder of height $32	ext{ cm}$. Find the: radius of the cylinder curved surface area of the cylinder Take $\pi = 3.1$.

Accepted Answer

Step 1 Equate volumes of metallic solids:
Radius of sphere $R = 6\text{ cm}$.
$\text{Volume of sphere} = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (6)^3 = \frac{4}{3}\pi (216) = 288\pi\text{ cm}^3$. [1 Mark]
Step 2 (i) Determine radius of cylinder ($r$):
Height of cylinder $h = 32\text{ cm}$.
$\text{Volume of cylinder} = \pi r^2 h = \pi r^2 (32) = 32\pi r^2$.
Since the sphere is recast into the cylinder:
$32\pi r^2 = 288\pi \implies r^2 = \frac{288}{32} = 9 \implies r = \sqrt{9} = \mathbf{3\text{ cm}}$. [1 Mark]
Step 3 (ii) Calculate Curved Surface Area ($CSA$):
$CSA = 2\pi r h = 2 \times (3.1) \times 3 \times 32$
$CSA = 6 \times 3.1 \times 32 = 18.6 \times 32 = \mathbf{595.2\text{ cm}^2}$. [2 Marks]

Question 15

A hemispherical and a conical hole is scooped out of a solid wooden cylinder. Find the volume of the remaining solid where the measurements are as follows:
The height of the solid cylinder is $7\text{ cm}$, radius of each of hemisphere, cone and cylinder is $3\text{ cm}$. Height of cone is $3\text{ cm}$.
Give your answer correct to the nearest whole number. Take $\pi = \frac{22}{7}$.

Accepted Answer

3 cm 7 cm Cone h = 3 cm Hemisphere r = 3 cm Vector Diagram: Cylinder with conical hole scooped at top and hemispherical hole scooped at bottom Step 1 Calculate individual geometric volumes in terms of $\pi$: Volume of Cylinder: $V_{	ext{cyl}} = \pi R^2 H = \pi (3)^2 (7) = 63\pi	ext{ cm}^3$. [1 Mark] Volume of Conical Hole: $V_{	ext{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (3)^2 (3) = 9\pi	ext{ cm}^3$. Volume of Hemispherical Hole: $V_{	ext{hemi}} = \frac{2}{3}\pi r^3 = \frac{2}{3}\pi (3)^3 = \frac{2}{3}\pi(27) = 18\pi	ext{ cm}^3$. [1 Mark] Step 2 Subtract the scooped volumes from the cylinder: $V_{	ext{remaining}} = V_{	ext{cyl}} - (V_{	ext{cone}} + V_{	ext{hemi}})$ $V_{	ext{remaining}} = 63\pi - (9\pi + 18\pi) = 63\pi - 27\pi = 36\pi	ext{ cm}^3$. [1 Mark] Step 3 Evaluate numerically using $\pi = \frac{22}{7}$: $V = 36 	imes \frac{22}{7} = \frac{792}{7} = 113.1428\dots	ext{ cm}^3$. Rounding to nearest whole number: $\mathbf{V \approx 113	ext{ cm}^3}$. [1 Mark]

Question 16

Two right circular cones, $C_1$ and $C_2$, have their heights in the ratio $2 : 1$ and radii in the ratio $1 : 2$. The ratio of their volumes is:

Accepted Answer

Solution Given: $\frac{h_1}{h_2} = \frac{2}{1}$ and $\frac{r_1}{r_2} = \frac{1}{2}$.
$\text{Ratio of volumes } \frac{V_1}{V_2} = \frac{\frac{1}{3}\pi r_1^2 h_1}{\frac{1}{3}\pi r_2^2 h_2} = \left(\frac{r_1}{r_2}\right)^2 \times \left(\frac{h_1}{h_2}\right) = \left(\frac{1}{2}\right)^2 \times \left(\frac{2}{1}\right) = \frac{1}{4} \times 2 = \mathbf{\frac{1}{2}}$.
Correct Answer: (b) 1 : 2 [1 Mark]

Question 17

A semi-circular sheet of paper having radius $7	ext{ cm}$ is folded to form a right circular cone, as shown in the diagram. The slant height of the cone so formed is:

Accepted Answer

Solution When a semi-circular sheet of radius $R$ is folded into a cone, the radius of the semicircle becomes the slant height ($l$) of the cone.
$\therefore \mathbf{l = 7\text{ cm}}$.
Correct Answer: (b) 7 cm [1 Mark]

Question 18

A solid sphere completely fits into a cylindrical container of radius $3.5\text{ cm}$ and height $7\text{ cm}$ as shown in the figure below. Find the volume of water required to fill the remaining (unshaded) space of the container to the nearest of a whole number.
(Take $\pi = \frac{22}{7}$)

Accepted Answer

Step 1 Dimensions of Cylinder:
Radius $r = 3.5\text{ cm} = \frac{7}{2}\text{ cm}$, height $h = 7\text{ cm}$.
$\text{Volume of Cylinder } V_{\text{cyl}} = \pi r^2 h = \frac{22}{7} \times \left(\frac{7}{2}\right)^2 \times 7 = \frac{22}{7} \times \frac{49}{4} \times 7 = \frac{1078}{4} = 269.5\text{ cm}^3$. [1.5 Marks]
Step 2 Dimensions of Inscribed Sphere:
Since the sphere fits completely into the container, its diameter $= h = 7\text{ cm} \implies$ radius $R = 3.5\text{ cm} = \frac{7}{2}\text{ cm}$.
$\text{Volume of Sphere } V_{\text{sph}} = \frac{4}{3}\pi R^3 = \frac{4}{3} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^3 = \frac{4}{3} \times \frac{22}{7} \times \frac{343}{8} = \frac{539}{3} \approx 179.67\text{ cm}^3$. [1.5 Marks]
Step 3 Volume of water to fill remaining space:
$V_{\text{water}} = V_{\text{cyl}} - V_{\text{sph}} = 269.5 - 179.67 = 89.83\text{ cm}^3$.
Rounding to nearest whole number: $\mathbf{V_{\text{water}} \approx 90\text{ cm}^3}$. [1 Mark]

Question 19

The surface area of a solid metallic sphere is $616	ext{ cm}^2$. It is melted and recast into solid right circular cones of diameter $3.5	ext{ cm}$ and height $3.5	ext{ cm}$. Find the: radius of the sphere. number of cones formed. (Take $\pi = \frac{22}{7}$)

Accepted Answer

Step 1 (a) Find radius of sphere ($R$):
$\text{Surface Area} = 4\pi R^2 = 616 \implies 4 \times \frac{22}{7} \times R^2 = 616$
$R^2 = \frac{616 \times 7}{88} = 7 \times 7 = 49 \implies \mathbf{R = 7\text{ cm}}$. [1.5 Marks]
Step 2 Calculate Volume of Sphere:
$V_{\text{sph}} = \frac{4}{3}\pi R^3 = \frac{4}{3} \times \frac{22}{7} \times (7)^3 = \frac{4312}{3}\text{ cm}^3$. [1 Mark]
Step 3 Calculate Volume of One Cone ($r = \frac{3.5}{2} = \frac{7}{4}\text{ cm}, h = \frac{7}{2}\text{ cm}$):
$V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times \left(\frac{7}{4}\right)^2 \times \frac{7}{2} = \frac{539}{48}\text{ cm}^3$. [1 Mark]
Step 4 (b) Calculate number of cones formed:
$\text{Number of cones} = \frac{V_{\text{sph}}}{V_{\text{cone}}} = \frac{4312/3}{539/48} = \frac{4312}{3} \times \frac{48}{539} = 8 \times 16 = \mathbf{128\text{ cones}}$. [0.5 Mark]

Question 20

The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. Find the: (i) radius of the cylinder (ii) volume of cylinder. (use $\pi = \frac{22}{7}$)

Accepted Answer

Step 1 Base circumference $= 2\pi r = 132\text{ cm}$:
$$2 \times \frac{22}{7} \times r = 132 \implies \frac{44}{7} r = 132$$ $$r = \frac{132 \times 7}{44} = 3 \times 7 = \mathbf{21\text{ cm}}$$ [1.5 Marks]
Step 2 Volume of cylinder $= \pi r^2 h$ with $r = 21\text{ cm}, h = 25\text{ cm}$:
$$V = \frac{22}{7} \times 21 \times 21 \times 25 = 22 \times 3 \times 21 \times 25$$ $$V = 66 \times 525 = \mathbf{34,650\text{ cm}^3}$$ [1.5 Marks]
Answers: (i) Radius $= 21\text{ cm}$, (ii) Volume $= 34,650\text{ cm}^3$.

Question 21

The following figure represents a solid consisting of a right circular cylinder with a hemisphere at one end and a cone at the other. Their common radius is 7 cm. The height of the cylinder and cone are each of 4 cm. Find the volume of the solid.

Accepted Answer

Given Common radius $r = 7	ext{ cm}$, Cylinder height $h_{	ext{cyl}} = 4	ext{ cm}$, Cone height $h_{	ext{cone}} = 4	ext{ cm}$. 1. Volume of Cone $$V_1 = \frac{1}{3} \pi r^2 h_{	ext{cone}} = \frac{1}{3} 	imes \frac{22}{7} 	imes 7^2 	imes 4 = \frac{616}{3}	ext{ cm}^3$$ [1 Mark] 2. Volume of Cylinder $$V_2 = \pi r^2 h_{	ext{cyl}} = \frac{22}{7} 	imes 7^2 	imes 4 = 616	ext{ cm}^3$$ [1 Mark] 3. Volume of Hemisphere $$V_3 = \frac{2}{3} \pi r^3 = \frac{2}{3} 	imes \frac{22}{7} 	imes 7^3 = \frac{2156}{3}	ext{ cm}^3$$ [1 Mark] Total Volume $$V = V_1 + V_2 + V_3 = \frac{616}{3} + 616 + \frac{2156}{3} = \frac{2772}{3} + 616 = 924 + 616 = \mathbf{1540	ext{ cm}^3}$$ [1 Mark] Answer: Volume of the solid is $1540	ext{ cm}^3$ .

Question 22

In the given figure $ABCD$ is a rectangle. It consists of a circle and two semi circles each of which are of radius 5 cm. Find the area of the shaded region. Give your answer correct to three significant figures.

Accepted Answer

Step 1: Rectangle Dimensions Radius $r = 5\text{ cm}$.
Width $AB = 2r = 2 \times 5 = 10\text{ cm}$.
Length $BC = r + 2r + r = 4r = 4 \times 5 = 20\text{ cm}$.
$$\text{Area of Rectangle } ABCD = 20 \times 10 = 200\text{ cm}^2$$ [1.5 Marks]
Step 2: Unshaded Circular Area
One full circle + two semicircles of radius $5\text{ cm}$:
$$\text{Total circular area} = \pi r^2 + 2\left(\frac{1}{2} \pi r^2\right) = 2\pi r^2 = 2 \times 3.1416 \times 25 = 157.08\text{ cm}^2$$ [1.5 Marks]
Step 3: Shaded Region Area
$$\text{Area} = 200 - 157.08 = 42.92\text{ cm}^2$$ To three significant figures: $\mathbf{42.9\text{ cm}^2}$ [1 Mark]
Answer: Area of shaded region $= 42.9\text{ cm}^2$.

Question 23

A conical tent is to accommodate 77 persons. Each person must have $16	ext{ m}^3$ of air to breathe. Given the radius of the tent as $7	ext{ m}$ find the height of the tent and also its curved surface area.

Accepted Answer

Step 1: Volume Total volume of air $= 77 \times 16 = 1232\text{ m}^3$.
Radius $r = 7\text{ m}$.
$$V = \frac{1}{3} \pi r^2 h \implies 1232 = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times h$$ $$1232 = \frac{154}{3} h \implies h = \frac{1232 \times 3}{154} = \mathbf{24\text{ m}}$$ [1.5 Marks]
Step 2: Curved Surface Area
Slant height $l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25\text{ m}$.
$$\text{Curved Surface Area} = \pi r l = \frac{22}{7} \times 7 \times 25 = 22 \times 25 = \mathbf{550\text{ m}^2}$$ [1.5 Marks]
Answers: Height of tent $= 24\text{ m}$; Curved surface area $= 550\text{ m}^2$.

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