Chapter-13, SURFACE AREAS AND VOLUMES

1. The radius of a sphere is 2r, then its volume will be 

A) 43πr3 B) 4πr3​​​​​​​C) 83πr3​​​​​​​ D) 323πr3

Explanation: 

The radius of a sphere is 2r, hence the response is D.

Sphere volume = 43π (radius) 3

=43π(2r)3=43π.8r3

Units of volume =32πr33 [∵radius=2r].

2. The total surface area of a cube is 96 cm2. The volume of the cube is: (A) 8 cm3 (B) 512 cm3 (C) 64 cm3 (D) 27 cm3

Explanation: 

A cube's total surface area is = 96 cm2.

A cube's total surface area is =6 (Side) 2 = 96 cm2.

⇒(16 cm2)/(Side)

⇒Size = 4 cm

(because the side is always a positive amount, we take the positive square root)

∴ Volume of a cube is equal to (Side)3=(4cm)3=64cm3.

3. A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. The radius of the sphere is : (A) 4.2 cm (B) 2.1 cm (C) 2.4 cm (D) 1.6 cm

Explanation:

Since the cone's height is 8.4 cm, the correct response is B.

The base's radius is 2.1 cm.

A cone's volume is equal to 13r2h=13(2.1).2×8.4 

=π×4.41×2.28 cm3

In order to create a spherical, the cone is melted.

Let R represent the sphere's radius.

So, the sphere's volume must equal the cone's volume.

⇒43πr3=π×4.41×2.8⇒r3=4.41×2.8×34

⇒r3=4.41×0.7×3

⇒r3=4.41×2.1

∴r=2.1

Consequently, the sphere has a 2.1 cm radius.

4. In a cylinder, the radius is doubled and height is halved, curved surface area will be (A) halved (B) doubled (C) the same (D) four times

Explanation:

The response is C.

Let a cylinder's height be h and radius be r.

∴The cylinder's curved surface area is equal to 2πrh.

We have: height = h2, radius = 2r.

=2π(2r)xh2=2πrh New curved surface area

The curved surface area will therefore be the same.

5. The total surface area of a cone whose radius is r/ 2  and slant height 2l is (A) 2πr (l + r) (B) πr (l + 4 ) (C) πr (l + r) (D) 2πrl

Explanation:

The right answer is B) 

We are aware that the cone's entire surface area equals πr(r+l).

Given are the values of radius (2r) and slant (2l).

The revised total surface area of the cone is therefore =πx 2r(2r+2l)

= πr(4r2+rl)

=π r(l+4r).

As a result, choice B is accurate.

6. The radii of two cylinders are in the ratio of 2:3 and their heights are in the ratio of 5:3. The ratio of their volumes is: (A) 10: 17 (B) 20: 27 (C) 17: 27 (D) 20: 37

Explanation:

The response is B.

Let r1 and r2 represent the cylinders' radii while h1 and h2 represent their heights.

Given that h1h2=53 and r1r2=23

∴Volume ratio =π r21h1 πr22h2 = (r1r2)2 (h1h2)

=(23)2(53)=49×53=20:27

As a result, their volumes are 20:27.

7. The lateral surface area of a cube is 256 m2. The volume of the cube is (A) 512 m3 (B) 64 m3 (C) 216 m3 (D) 256 m3

Explanation:

The right answer is A)

Given, a cube's lateral surface area is 256 m2.

As we know, the cube's lateral surface area is =4 (edge)2.

∴4(edge)2=256

⟹(edge)2=4256

⟹(edge)2=64

⟹ edge =8.

The volume of the cube also equals (edge)3.

∴ Cube volume = (8)3 = 512 m3.

As a result, choice A is accurate.

8. The number of planks of dimensions (4 m × 50 cm × 20 cm) that can be stored in a pit that is 16 m long, 12m wide and 4 m deep is (A) 1900 (B) 1920 (C) 1800 (D) 1840

Explanation:

Given that a pit is 16 metres long, 12 metres broad, and 40 metres deep.

The board measures 4 m, 50 cm, and 20 cm.

The maximum number of boards that can be stored in the pit must be determined.

The plank length is 4 metres as stated.

The plank's width is 50 cm or 0.5 metres.

Plank thickness: 20 cm = 0.2 metres

The plank's volume 

=4 x 0.5 x 0.2

= 4 m3.

Given that the pit is 16 metres long.

12 metres is the pit's width.

The pit is 40 metres high.

The pit's volume is 

= 16 × 12 × 40

= 16 × 480

= 7680 m³

Plank storage capacity in the pit is calculated as follows: volume of the pit/volume of

One board equals 7680/4, 

=1920.

A total of 1920 planks can be kept as a result.

9. The length of the longest pole that can be put in a room of dimensions (10 m × 10 m × 5m) is (A) 15 m (B) 16 m (C) 10 m (D) 12 m

Explanation:

The diagonal of the room is equal to the length of the longest pole that can be placed there.

Diagonal length = (l2 + b2 + h2)

=√ (102 + 102 + 52)

= √(100 + 100 + 25) 

= √225

= 15.

10. The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is pumped into it. The ratios of the surface areas of the balloon in the two cases are (A) 1: 4 (B) 1 : 3 (C) 2 : 3 (D) 2: 1

Explanation:

The radius of a hemispherical balloon, r1, is 6 cm, 

hence the answer is A.

The balloon gets pumped full of air. Therefore, r2=12 cm is the radius of a hemispherical balloon.

The ratio of the balloon's surface areas in the two scenarios is 3.πr213πr22

⇒r21r22=(6)2(12)2=36144=14=1:4

Consequently, the surface area ratio between the balloons in the two scenarios is 1:4.

11. The volume of a sphere is equal to two-thirds of the volume of a cylinder whose height and diameter are equal to the diameter of the sphere.

Explanation:

Assume that both the sphere's and cylinder's radius is r.

Given that the cylinder's height equals its base's diameter, ⇒h=2r

Sphere radius equals Cylinder radius = r

Considering the stated circumstance,

Volume of sphere

= 23x cylinder volume

⇒43πr3=23×πr2×2r

⇒43πr3=43πr3

Therefore, the sphere's volume is equivalent to a cylinder's volume in two-thirds.

12. If the radius of a right circular cone is halved and the height is doubled, the volume will remain unchanged.

Explanation:

False.

Assume that a right circular cone has a radius of r and a height of h.

thus, right circular cone volume V=13πr2h

Rnew=r2, and hnew=2h for the new radius and height.

Right circular cone's new volume, Vnew=13πrnew2hnew=13π(r2)2x(2h)=13πxr242h=13πxr22h=12(13πr2h)=12V

Half of the original book makes up the new volume.

13. In a right circular cone, height, radius, and slant height do not always be sides of an aright triangle.

Explanation:

False.

It produces a cone by rotating a right-angled triangular lamina AOB around OA. The vertex of a cone is point A. A circle with an OB radius and O center serve as its base. The cone's height is measured by its length OA, and its slant height is determined by its length AB.

Obviously, "AOB=90"

Let height = h units and base radius = r units.

L units = slant height

Then,

l2=h2+r2⇒l=√h2+r2

As a result, the three sides of a right triangle are always the height, radius, and slant height of a right circular cone.

14. If the radius of a cylinder is doubled and its curved surface area is not changed, the height must be halved.

Explanation:

True.

Let r be the cylinder's radius and h be its height.

Cylinder's curved surface area = 2π rh

Let H be the height and r1=2r be the new radius.

The cylinder's curved surface is left unchanged, therefore

2πrh=2π(2r)×H

⇒2πrh=4πrH

⇒H=2πrh4πr⇒H=h2

Thus, the height must be cut in half.

15. The volume of the largest right circular cone that can be fitted in a cube whose edge is 2r equals the volume of a hemisphere of radius r.

Explanation:

True.

Given,

The edge of the cube equals 2r

H = 2r determines the cone's height.

Cone volume =13 πr2h=13π r2(2 r)=23π r3.

Hemisphere volume =23 πr3

As a result, the cone's and sphere's volumes are equal.

16. A cylinder and a right circular cone are having the same base and same height. The volume of the cylinder is three times the volume of the cone.

Explanation:

True.

Let r represent the base radius of a cylinder and h represent the height of a right circular cone.

The volume of the cylinder therefore =π r2h.

Cone volume = 13 πr 2 h.

The volume of the cone 3 times the volume of the cylinder.

17. A cone, a hemisphere, and a cylinder stand on equal bases and have the same height. The ratio of their volumes is 1: 2 : 3.

Explanation:

The cone, hemisphere, and cylinder bases are all same. Let each object's radius be r and its height be r. Cone volume = 13πr2 + 13πr3

The volume of the hemisphere =23πr3 and the volume of the cylinder =πr2(r)=πr3 have a ratio of =13πr3:23πr3=13:23:1=1:2:3.

18. If the length of the diagonal of a cube is 6 3 cm, then the length of the edge of the cube is 3 cm.

Explanation:

False.

Given that a cube's diagonal is= 6 √3 cm in length.

Let "a" cm represent the cube's side (edge) length.

So, the cube's diagonal equals a3.

⇒6√3=a√3

⇒a=6 cm

The edge length is 6 cm as a result.

19. If a sphere is inscribed in a cube, then the ratio of the volume of the cube to the volume of the sphere will be 6: π.

Explanation:

If a cube's edge equals a, 

then its volume equals a3

 The cube has an inscribed sphere with a radius of (a2) and a diameter of (a2)∴Volume of sphere=43πr3

=43π(a2)2

=43π×a38

=πa36

Now, the ratio of their volumes is as follows:

=a3:πa36

= 6a3:πa3.

=6:π.

20. If the radius of a cylinder is doubled and the height is halved, the volume will be doubled.

Explanation:

Assume that the cylinder's height is h and its radius is r.

The cylinder's volume is then given by V1=πr2h.

Radius change, R=2r

The cylinder's new height is H=h2.

∴ The volume of a cylinder is given by V2=πR2H=π(2r)2h2

=πx4r2xh2 

=2πr2h

=2V1.

Therefore, if a cylinder's height is halved and its radius is doubled.

21. Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm × 8 cm × 8 cm. When 16 spheres are packed the box is filled with preservative liquid. Find the volume of this liquid. Give your answer to the nearest integer. [Use 3.14 π=]

Explanation:

Given that each metal sphere's radius is 2 cm

A rectangular box's inside measurements are 

1=16 cm, b=8 cm, and h=8 cm.

∴ A metallic sphere's volume is equal to 43πr3=43π3.14x(2).3

 =43×3.14×8

The volume of 16 metallic spheres is now equal to 4x3.14x8x163=100.48x163

=1607.683=163 = 535.89 cm3

The rectangular box's internal volume is equal to

=l×b×h=16×8×8

=1024 cm3 

 ∴Preservative liquid volume is =1024 - 535.89

= 488.11 cm3≈ 488 cm3.

Thus, when 16 metallic spheres are put in the rectangular box, the amount of preservative liquid is 488 cm3.

22. A storage tank is in the form of a cube. When it is full of water, the volume of water is 15.625 m3. If the present depth of water is 1.3 m, find the volume of water already used from the tank.

Explanation:

The formula for the cube's volume is a3=15.625m3a=2.5m.

The water is 1.3 meters deep.

⟹ The used water is 1.2 meters tall.

Thus, the amount of water utilized is x2.5 x 2.5 x 1.2 = 7.5 m3.

23. Find the amount of water displaced by a solid spherical ball of diameter 4.2 cm, when it is completely immersed in water.

Explanation:

Given that the solid spherical ball has a diameter of 4.2 cm

∴ The solid spherical ball's radius is r=4.22=2.1 cm.

Now, the Volume of a solid spherical ball when fully submerged in water = Volume of the solid spherical ball 

= 43πr3

=43x227x(2.1)3

=43x227x2.1x2.1x2.1

=814.96821 

=38.808≈38.81 cm3

The volume (amount) of water that a solid, spherical ball displaces when submerged in water is therefore 38.81 cm3.

24. How many square meters of the canvas is required for a conical tent whose height is 3.5 m and the radius of the base is 12 m?

Explanation:

Given, A conical tent's height is 3.5 meters, and its base's radius is 12 meters.

Slant height, 

l=√h2+r2=√(3.5)2+(12)2

=√12.25+144

=√156.25=12.5 m

∴ The conical tent's curved surface area divided by the necessary canvas area is 

=227x 12x 1 2.5,

= 471.42 m2.

Consequently, 471.42 m2 of fabric is needed to construct the conical tent.

25. Two solid spheres made of the same metal have weights of 5920 g and 740 g, respectively. Determine the radius of the larger sphere, if the diameter of the smaller one is 5 cm.

Explanation:

For the same metal, density, mass, and volume are directly related.

Let M1 stand for the Mass of Solid 1, V1 for Volume 1, M2 for the Mass of Solid 2, and V2 for Volume.

Now, M2/M1=V2/V1.

R3 and the volume of a sphere are intimately related.

M2/M1=V2/V1=R23/R13,

7405920=R23/R13,

R23/R13=8,

R2/R1=38, 

and R2/R1=2 as a result.

So, R1/R2=2

So, R1=2.5×2=5.

26. A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. If the glass is filled with milk upto an height of 12 cm, find how many litres of milk is needed to serve 1600 students.

Explanation:

Given that the glass's diameter is 7 cm

Glass's radius, r, is 72 cm. 

∴ The volume of the cylindrical glass =π r2h=227x72x72x12=462 cm3 for milk contained in the glass.

Now, the amount of milk required for 1600 students is equal to the amount of milk required for one student, 

= 462 x 1600 x 739 200 cm3 

= 739 200 1000 l = 739 9.2 L. [∵1 cm3=11000 L]

Hence, 739. It takes 2 L of milk to serve 1600 pupils.

27. A cylindrical roller 2.5 m in length, and 1.75 m in radius when rolled on a road was found to cover the area of 5500 m2 . How many revolutions did it make?

Explanation:

One revolution of a cylindrical roller equals the area of its curved surface.

The cylinder's curved surface is equal to 2πrxl=2πx(1.75)(2.5)=27.5 m2.

Area covered in one revolution during a revolutionWhole Area =27.55500=200.

28. A small village, having a population of 5000, requires 75 liters of water per head per day. The village has got an overhead tank of measurement 40 m × 25 m × 15 m. For how many days will the water in this tank last?

Explanation:

Given that a small community has 5000 residents

Each person in the village needs 75 liters of water each day.

Dimensions of the above tank are 40 m, 25 m, and 15 m.

The number of days the tank's water supply will last must be determined.

The daily water needed is (75 x 5000) liters or 375000 liters.

Water volume is equal to 375000/1000 or 375 m3.

The volume of an overhead tank equals its length, width, and depth

Given: length = 40 m, width = 25 m, and depth = 15 m.

Tank volume is 40(25)(15)

= 1000(15)

= 15000 m3.

The number of days equals the sum of the overhead tank's capacity and the amount of water used each day (15000/375 = 40). The water in the tank will therefore remain fresh for 40 days.

29. A shopkeeper has one spherical laddoo of radius 5cm. With the same amount of material, how many laddoos of radius 2.5 cm can be made?

Explanation:

The ideal decision is

The bigger laddoo's (R) radius is 5 cm.

∴Volume=43π(5)3cm3=5003πcm3

Small Laddoo Radius (r) = 2.5 cm 

∴Small laddoo volume = 43 + (2.5)3cm3=62.5π3cm3=625π30cm3

Laddoos that can be manufactured are 500π3÷625π30=500π3×30625π=8.

30. A right triangle with sides 6 cm, 8 cm, and 10 cm is revolved about the side 8 cm. Find the volume and the curved surface of the solid so formed.

Explanation:

When the side of 8 cm of a right triangle with sides 6 cm, 8 cm, and 10 cm is rotated, a solid cone with a height of 8 cm results.

The cone's radius, r, is 6 cm.

Cone's slant height, l = 10 cm

∴Cone volume 

=13πr2h=13x227x6x6x8

=633621=301.7 cm3.

Cone's curved surface area =πrl

=227x 6x 10=13 207 = 188.5 cm2.

As a result, the cone has a volume and surface area of 301.7 cm3 and 188.5 cm2, respectively.

31. A cylindrical tube opened at both ends is made of an iron sheet that is 2 cm thick. If the outer diameter is 16 cm and its length is 100 cm, find how many cubic centimeters of iron have been used in making the tube.

Explanation:

The cylindrical tube's outside diameter, d, is 16 cm,

and its outer radius, r1, is 8 cm.

=>(r1 - thickness of the iron sheet)/

(r2 = inner radius of the cylindrical tube) 

=82=6 cm

The cylindrical tube's height is 100 cm.

The volume of the metal used to create cylindrical tubes 

is equal to the inner volume of a cylindrical tube divided by its outer volume. For example, 

=πr21h−πr22h=πh(r21−r22)

=227×100(82−62)

=227×100(8+6)(8−6)

=227×100×14×2

=2200×4=8800 cm3

As a result, 8800 cm3 of iron was used to create the cylindrical tube.

32. A semi-circular sheet of metal of diameter 28cm is bent to form an open conical cup. Find the capacity of the cup.

Explanation:

Assume that a semi-circular metal sheet is bent into an open conical cup.

The metal sheet diameter is 28 cm.

We must determine the cup's volume.

Cone base circumference equals semicircle circumference

Let R be the cup's radius.

The base of cone circumference = 2πR

Semicircle's diameter is 28 cm.

Radius = 28/2, or 14 cm, therefore.

Semicircle circumference is equal to πr.

Where r is the semicircle's radius, which is equal to π14

Now, 2πR = 14π

2R = 14 

R = 14/2

 R = 7 cm

Cone volume = 1/3π r2h

where r is the cone's radius and h is its height.

Cone slant height equals semicircle radius, which is 14 cm.

We know, l² = r² + h²

(14)²= (7)² + h²

196 = 49 + h²

h² = 196 - 49

h² = 147

Using square roots, h equals 7√3 cm.

Cup volume =  1/3 (22/7)(7)²(7√3)

= 22(49)(√3)/3

= 1867.096/3

= 622.37 cm³

Consequently, the cup's volume is 622.37 cm3.

33. A cloth having an area of 165 m2 is shaped into the form of a conical tent of radius 5cm (i) How many students can sit in the tent if a student, on average, occupies 5/7 m on the ground?

Explanation: 

Given: 165 m2 of fabric will be needed to create a conical tent.

Based on the conical tent's radius, r = 5 m

Conical tent surface area curved = area of fabric used to create a conical tent

⇒πrl=165m2

⇒22/7×5×l =165

∴l=165×7/22×5=33×7/22=10.5 m

Currently, the conical tent's height 

=√−r=√−(5)=√110.25−25=√85.25=9.23 m

Conical tent's cone's volume 

=1/3πr2h=1/3×22/7×5×5×9.23
 =1/3×1550×9.23/7=5076.5/7×3=241.7 m3

Consequently, the cone's (conical tent's) volume is 241.7 m3.

34. The water for a factory is stored in a hemispherical tank whose internal diameter is 14 m. The tank contains 50 kilolitres of water. Water is pumped into the tank to fill it to its capacity. Calculate the volume of water pumped into the tank.

Explanation:

Hemispherical tank's radius, r=7 m

Thus, the volume of the hemispherical tank is equal to the volume of the water to be pumped: 50 kl = 1 m3; 32 r3 = 50 m3; 722 (7) 3 = 50 m3; 718.67 m3; 50 m3 = 668.67 m3

∵1 kl=1 m3

⇒3/2πr3−50 m3

⇒3/2×7/22×(7)3−50 m3

⇒718.67−50 m3

⇒668.67 m3.

35. The volumes of the two spheres are in the ratio 64: 27. Find the ratio of their surface areas.

Explanation:

Let r1 and r2 be the spheres' respective radii.

Given that the volume of the two spheres has a ratio of 64:27

V1V2=6427

 ⇒4/3πr31/43πr32=6427 [ spherical volume = 43 r3]

⇒(r1/r2)3=(4/3)3

 ⇒r1/r2=4/3

Let S1 and S2 represent the two spheres' surface areas.

S1/S2=4πr2/4πr22=(r1/r2)2 [Sphere surface area =4r2]

⇒S1:S2=(4/3)2=169 ⇒S1:S2=16/9

Consequently, their surface area ratio is 16:9.

36. A cube of side 4 cm contains a sphere touching its sides. Find the volume of the gap in between.

Explanation:

Volume = (side)3=444=64cm3 for a cube with a side of 4 cm.

The largest sphere contacting its sides has a diameter of 4 cm.

Radius=4/2=2cm

∴Volume=4/3πr3=4/3×22/7×2×2×2cm3=704/21cm3

Now, the difference in volume is 64 704/21 cm3,

1344 704/21 cm3, 

640/21 cm3, and 30.48 cm3.

37. A sphere and a right circular cylinder of the same radius have equal volumes. By what percentage does the diameter of the cylinder exceed its height?

Explanation:

a sphere and a right circular cylinder with the same radius are both given.

Sphere Volume = 43 r3

Right circular cylinder volume = πr2h

Given Sphere Volume = Right Circular Cylinder Volume.

4/3πr3=πr2h

4/3r=h.

2r=3h/2

∴d=3h/2

Let h now be 100%.

D =3h/2=3/2×100% = 150%

Therefore, the needed difference is 150%100%50%.

As a result, the cylinder's diameter is 50% larger than its height.

38. 30 circular plates, each of radius 14 cm and thickness 3cm are placed one above the other to form a cylindrical solid. Find : (i) the total surface area (ii) volume of the cylinder so formed.

Explanation:

Given, a cylindrical solid is created by stacking 30 circular plates, each with a radius of 14 cm and a thickness of 3 cm.

The whole surface area needs to be determined.

Given that one circular plate's radius is 14 cm

One round plate has a thickness of 3 cm.

Now, the cylinder's base's radius is 14 cm.

Cylinder height = 30; circular plate thickness = 30(3); height = 90 cm

r2h = volume of the cylinder

where r is the cylinder's radius and h is the cylinder's height

The volume is equal to (22/7)(14)2(90) = 22(14)(2)(90) = 44(14)(90) = 55440 cm3.

Consequently, the cylinder's volume is 55440 cm3.