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H₃PO₄ → 3H⁺ + PO₄^3- (Phosphate ion)

Molecular weight of C₆H₆ = 6 × 12 + 6 × 1 = 78 g, 

Mass of 1 molecule = 78/6.002 x 10²³ = 12.94 × 10^-23 g = 1.294 × 10^-22 g 

Correct option: (C) Atomic number

Correct option: (C) 10.8

Atomic mass = \(\cfrac{10\times 19 + 81 \times11}{100}\) 

= \(\cfrac{1081}{100}\) = 10.81

Correct option: (A) 1.4 x 10^-21 g 

Molecular weight of C₆₀H₁₂₂

= 12 x 60 + 122 x 1

= 720 + 122

= 842

∵ 6 x 10²³ molecules = 842 g

1 molecule = \(\cfrac{842}{6\,\times 10^{23}}\) 

= 140.33 x 10^-23

= 1.4 x 10^-21 g

Correct option: (D) 78

Molecular formula of benzene is C₆H₆.

∴ Molecular mass = sum of atomic weight of all the atoms

∴ Molecular mass = 12 x 6 + 6 x 1 = 72 + 6 = 78

∴ According to Avogadro’s law, the gram molecule of benzene is equal to 78 g of C₆H₆.

Correct answer is (d) 4

Correct answer is (a) 4.00 mm

Correct answer is (a) 1.9 mm

percentage error in measurement of displacement = \(\frac{5}{200}\)x100

Percentage error in measurement of time = \(\frac{0.2}{20}\) x 100

∴ Maximum permissible error = 2.5 + 1 = 3.5%

(d) Coefficient of viscosity

Correct answer is (c) 3%

(a) Moment of inertia and moment of force

Correct answer is (c) [MLT^-1]

The energy possessed by a liquid by virtue of its pressure is called pressure energy.

By placing sleepers below the rails cross-sectional area is increased. This in turn reduces the pressure due to weight of the train on the rails as pressure is defined as the force per unit area.

Sudden decrease in atmospheric pressure indicates the occurrence of storm.

Newton-metre and radiam/s respectively.

Initial speed of the car, u = 35 ms^-1

Final speed of the car, v = 0

Let a be the retardation of the car.

Then using the relation,

v² = u² + 2as, we have 0 = (35)² + 2a × 200

a = – (35 x 35)/(2 x 200) = – 3.06 ms^-2

Let t be the time taken by the car to come to stop. Then,

v = u + at ;

0 = 35 – 3.06 × t or

t = 35/3.06 = 11.43 s.

1. Initial velocity of the car u = 0;

Final velocity v = 36 kmhr^-1

= (36 x 1000)/3600 = ms^-1

= 10 ms^-1

time taken t = 10 s

∴ acceleration a = (v - u)/t = (10 - 0)/10 = 1ms^-2

2. To find the retardation

During the motion with retardation, Initial velocity of the car u = 10 ms^-1

Final velocity v = 0, distance travelled, s = 50 m

Using the relation v² = u² + 2as,

0 = 10² – 2a × 50

a = (-100/(2 x 50)) = -1 ms^-2

Class 10 Maths MCQ Questions of Surface areas and volumes with Answers is recommended for the students. Students attempt to solve these questions first and afterward check with the appropriate answer. This practice will assist to gain problem-solving abilities and build their confidence level.

Students are encouraged to solve the MCQ Questions of Surface areas and volumes to know various ideas. Practicing the MCQ question for Class 1O Maths of Surface areas and volumes with answers will support your confidence subsequently assisting you with scoring great in the exam. Solve the underneath given Class 10 Maths MCQ Questions of Surface areas and volumes with Answers and check your answers with the detailed answers here.

Practice Class 10 MCQ Questions for Maths

1.  During conversion of a solid from one shape to another, the volume of the new shape will

(a) increase
(b) decrease
(c) remain unaltered
(d) be doubled

2.  A right circular cylinder of radius r cm and height h cm (h>2r) just encloses a sphere of diameter

(a) r cm
(b) 2r cm
(c) h cm
(d) 2h cm

3. If two solid hemispheres of same base radii r, are joined together along their bases, then curved surface area of this new solid is

(a) 4πr²
(b) 6πr²
(c) 3πr²
(d) 8πr²

4. The shape of an ice-cream cone is a combination of:

(a) Sphere + cylinder
(b) Sphere + cone
(c) Hemisphere + cylinder
(d) Hemisphere + cone

5.  If a cone is cut parallel to the base of it by a plane in two parts, then the shape of the top of the cone will be a:

(a) Sphere
(b) Cube
(c) Cone itself
(d) Cylinder

6. A cylindrical pencil sharpened at one edge is combination of:

(a) a cone and a cylinder
(b) frustum of a cone and cylinder
(c) a hemisphere and a cylinder
(d) four cylinders

7.  A shuttle cock used for playing badminton has the shape of a combination of:

(а) a cylinder and a sphere
(b) a cylinder and a hemisphere
(c) a cone and a sphere
(d) a frustum of a cone and a hemisphere

8.  The radius of a wire is decreased to one third. If volume remains the same, the length will become:

(a) 3 times
(b) 6 times
(c) 9 times
(d) 27 times

9. A cylinder and a cone area of same base radius and of same height. The ratio of the volume of cylinder to that of cone is:

(a) 3 : 1
(b) 1 : 3
(c) 2 : 3
(d) 1 : 1

10. If the surface areas of two spheres are in ratio 16 : 9, then their volumes will be in the ratio:

(a) 27 : 64
(b) 64 : 27
(c) 4 : 3
(d) 3 : 4

11. A solid cylinder of radius r and height h is placed over other cylinder of same height and radius. The total surface area of the shape so formed is

(a) 4πrh + 4πr²
(b) 4πrh − 4πr²
(c) 4πrh + 2πr²           
(d) 4πrh − 2πr²

12. The radii of the top and bottom of a bucket of slant height 45cm are 28cm and 7 cm respectively. The curved surface area of the bucket is:

(a) 4950 cm²
(b) 4951 cm²
(c) 4952 cm²
(d) 4953 cm²

13. If we cut a cone in two parts by a plane parallel to the base, then the bottom part left over is the:

(a) Cone
(b) Frustum of cone
(c) Sphere
(d) Cylinder

14. If r is the radius of the sphere, then the surface area of the sphere is given by;

(a) 4 π r²
(b) 2 π r²
(c) π r²
(d) 4/3 π r²

15. If we change the shape of an object from a sphere to a cylinder, then the volume of cylinder will

(a) Increase
(b) Decrease
(c) Remains unchanged
(d) Doubles

16. The radius of the top and bottom of a bucket of slant height 35 cm are 25 cm and 8 cm. The curved surface of the bucket is:

(a) 4000 sq.cm
(b) 3500 sq.cm
(c) 3630 sq.cm
(d) 3750 sq.cm

17. The total surface area of a hemispherical solid having radius 7 cm is

(a) 462 cm²
(b) 294 cm²
(c) 588 cm²
(d) 154 cm²

18. A solid formed on revolving a right angled triangle about its height is

(a) cylinder
(b) sphere
(c) right circular cone
(d) two cones

19. If we join two hemispheres of same radius along their bases, then we get a;

(a) Cone
(b) Cylinder
(c) Sphere
(d) Cuboid

20. Two identical solid hemispheres of equal base radius r cm are stuck together along their bases. The total surface area of the combination is 

(a) 6πr²
(b) 5πr²
(c) 4πr²
(d) 3πr²

Answer:

1. Answer:  (c) remain unaltered

Explanation: During conversion of one solid shape to another, the volume of the new shape will remain unaltered.

2. Answer: (b) 2r cm

Explanation: Because the sphere is enclosed inside the cylinder, therefore the diameter of sphere is equal to the diameter of cylinder which is 2r cm.

3. Answer: (a) 4πr²

Explanation: Because curved surface area of a hemisphere is and here we join two solid hemispheres along their bases of radii r, from which we get a solid sphere.

Hence the curved surface area of new solid = 2πr² + 2πr² = 4πr²  

4. Answer: (d) Hemisphere + cone

Explanation: The shape of an ice-cream cone is a combination of a hemisphere and a cone.

5. Answer: (c) Cone itself

Explanation: If we cut a cone into two parts parallel to the base, then the shape of the upper part remains the same.

6. Answer: (a) a cone and a cylinder

7. Answer: (d) a frustum of a cone and a hemisphere

8. Answer: (c) 9 times

9. Answer: (a) 3 : 1

10. Answer: (b) 64 : 27

11. Answer: (c) 4πrh + 2πr²           

Explanation: Since the total surface area of cylinder of radius r and height h = 2πrh + 2πr².

When one cylinder is placed over the other cylinder of same height and radius,

Then height of new cylinder = 2h

And radius of the new cylinder = r

Therefore total surface area of new cylinder

= 2πr (2h) + 2πr²

= 4πrh + 2πr²  

12. Answer: (a) 4950 cm²

Explanation: Curved Surface area of the bucket = π (R + r) l

⇒ Curved surface area of the bucket = π (28 + 7) X 45

⇒ Curved surface area of the bucket = 4950 cm²  

13. Answer: (b) Frustum of cone

14. Answer: (a) 4 π r²

Explanation: If r is the radius of the sphere, then the surface area of the sphere is given by 4 π r².

15. Answer: (c) Remains unchanged

Explanation: If we change the shape of a three-dimensional object, the volume of the new shape will be same.

16. Answer: (c) 3630 sq.cm

Explanation: Curved surface of bucket = π(R₁ + R₂) x slant height (l)

Curved Surface = (22/7) x (25 + 8) x 35

CSA = 22 x 33 x 5 = 3630 sq.cm.

17. Answer: (a) 462 cm²
Explanation: Reason: Total surface area of hemisphere = 3πr² 

= 3 × 22/7 × 7 × 7 = 462 cm²  

18. Answer: (c) right circular cone

19. Answer: (c) Sphere

Explanation: If we join two hemispheres of same radius along their bases, then we get a Sphere.

20. Answer: (c) 4πr²

Explanation: When two hemispheres are joined together along their bases, a sphere of the same base radius is formed.

Curved Surface Area of a sphere = 4πr²

Click here for practice more MCQ questions from Chapter Surface Areas and Volumes Class 10 Maths

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The Class 9 MCQ Questions of Surface Areas and Volumes with Answer will teach students about the connection between the length and expansiveness concerning various shapes and the incorporation of stature any place it is required.

Practice MCQ Questions for Class 9 Maths

1. In a cylinder, radius is doubled and height is halved, curved surface area will be

(a) halved
(b) doubled
(c) same
(d) four time

2. During conversion of a solid from one shape to another, the volume of the new shape will

(A) increase
(B) decrease
(C) remain unaltered
(D) be doubled

3. 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

4.  The total surface area of a cone whose radius is r/2 and slant height 2l is

(a) 2πr(l+r)
(b) πr(l+(r/4))
(c) πr(l+r)
(d) 2πrl

5. 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

6. A cuboid having surface areas of 3 adjacent faces as a, b and c has the volume:

(a) 3\(\sqrt{abc}\)
(b) \(\sqrt{abc}\)
(c) abc
(d) (abc)²

7. 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

8. 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

9. The length of the longest pole that can be put in a room of dimensions (10 m × 10 m × 5m) is

(a) 15m
(b) 16m
(c) 10m
(d) 12m

10. 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

11. Volume of spherical shell is

(a) 2/3 πr³
(b) 3/4 πr³
(c) 4/3 π(R³ – r³)
(d) None of these

12. Volume of hollow cylinder

(a) π(R² – r²)h
(b) πR²h
(c) πr²h
(d) πr²(h₁ – h₁)

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

(a) 4/3 πr³
(b) 4πr³
(c) 8/3 πr³
(d) 32/3 πr³

14. The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratio of the surface areas of the balloon in the two cases is

(a) 1:4
(b) 1:3
(c) 2:3
(d) 2:1

15. The surface area of a sphere of radius 14 cm is:

(a) 1386 sq.cm
(b) 1400 sq.cm
(c) 2464 sq.cm
(d) 2000 sq.cm

16. If slant height of the cone is 21cm and the diameter of the base is 24 cm. The total surface area of a cone is:

(a) 1200.77 sq.cm
(b) 1177 sq.cm
(c) 1222.77 sq.cm
(d) 1244.57 sq.cm

17. The diameter of the base of a cone is 10.5 cm, and its slant height is 10 cm. The curved surface area is:

(a) 150 sq.cm
(b) 165 sq.cm
(c) 177 sq.cm
(d) 180 sq.cm

18. The Curved surface area of a right circular cylinder is 4.4 sq.cm. The radius of the base is 0.7 cm. The height of the cylinder will be:

(a) 2 cm
(b) 3 cm
(c) 1 cm
(d) 1.5 cm

19. The curved surface area of a right circular cylinder of height 14 cm is 88 cm². The diameter of the base is:

(a) 2 cm
(b) 3cm
(c) 4cm
(d) 6cm

20. The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratios of the surface areas of the balloon in the two cases is

(a) 1 : 4
(b) 1 : 3
(c) 2 : 3
(d) 2 : 1

Answer: 

1. Answer: (c) same

2. Answer:  (C) remain unaltered

Explanation: During conversion of one solid shape to another, the volume of the new shape will remain unaltered.

3. Answer: (b) 20: 27

Explanation: Given that, the radii of two cylinders are in the ratio of 2:3

Hence, r₁= 2r, r₂ = 3r

Also, given that, the height of two cylinders are in the ratio 5:3.

Hence, h₁ = 5h, h₂ = 3h

The ratio of the volume of two cylinders = V₁/V₂

= πr₁²h₁/πr₂²h₂

= [(2r)²(5h)]/[(3r)²(3h)]

Ratio of their volumes =(20r²h)/(27r²h) = 20/27 = 20: 27.

4. Answer: (b) πr(l+(r/4))

Explanation: The total surface area of cone = πr(l+r) square units.

If r = r/2 and l= 2l, then the TSA of cone becomes,

TSA of cone = π(r/2)[(2l+(2/r)]

=π[(rl)+(r2/4)]

TSA of new cone =πr[l+(r/4)]

5. Answer: (a) 512 m³

Explanation: The lateral surface area of cube = 4a²

4a²= 256

a² = 256/4 =64

a = 8 m

Hence, the volume of cube = a³ cube units 

V = 8³  = 512 m³.

6.  Answer:  (b) \(\sqrt{abc}\)

7. Answer: (b) 1920

Explanation: Volume of Plank = 400 cm×50cm×20cm=400000cm³

Volume of pits = 1600cm×1200cm×400cm = 768000000cm³

Number of planks = Volume of planks/Volume of pits

= 768000000/400000

Hence, the number of pits = 1920

8. Answer: (b) 2.1 cm

Explanation: Given that the height of cone = 8.4 cm

Radius of cone = 2.1 cm

Also, given that the volume of cone = volume of a sphere

(1/3)πr²h = (4/3)πr³

(1/3)π(2.1)2(8.4) = (4/3)πr³

37.044= 4r³

r³= 37.044/4

r³= 9.261

r = 2.1

Therefore, the radius of the sphere is 2.1 cm.

9. Answer: (a) 15m

Explanation: Given: l=10m, b= 10m, h= 5m

The length of the longest pole = √[10²+10²+5²]

= √(100+100+25) = √225 = 15 m.

10. Answer: (c) 64 cm³

Explanation: We know that the TSA of the cone = 6a².

6a² = 96 cm²

a² = 96/6 = 16

a =4 cm

The volume of cone = a³ cubic units

V = 4³ = 64cm³.

11. Answer: (c) 4/3 π(R³ – r³)

12. Answer: (a) π(R² – r²)h

13. Answer: (d) 32/3 πr³ 

14. Answer: (a) 1:4

Explanation: We know that the total surface area of the hemisphere = 3πr² square units.

If r= 6cm, then TSA = 3π(6)² = 108π

If r = 12 cm, then TSA = 3π(12)²= 432π

Then the ratio = (108π)/(432π)

Ratio = 1/4, which is equal to 1:4.

15. Answer: (c) 2464 sq.cm

Explanation: Radius of sphere, r = 14 cm

Surface area = 4πr²

= 4 x 22/7 x (14)² = 2464 sq.cm.

16. Answer: (d) 1244.57 sq.cm

Explanation: Total surface area = πr(l + r)

r = 24/2 = 12 cm

l = 21 cm

TSA = π(12)(21 + 12) = 1244.57 sq.cm

17. Answer: (b) 165 sq.cm

Explanation: Diameter = 10.5, Radius = 10.5/2

Slant height, l = 10cm

Curved surface area of cone = πrl = π(5.25)(10)

CSA = 165 sq.cm

18. Answer: (c) 1 cm

Explanation: Curved surface area of cylinder = 2πrh

2πrh = 4.4

h = 4.4/(2π x 0.7)

h = 1 cm

19. Answer: (a) 2 cm

Explanation: Curved surface area of cylinder = 88 sq.cm

Height = 14 cm

2πrh = 88

r = 88/2πh

r=1 cm

Diameter = 2r = 2cm

20. Answer: (a) 1 : 4

Click here to practice: – Surface Areas and Volumes MCQ Question for Class 9 Maths

(a) \(92\frac23\)%

Volume of cone = \(\frac13\)πr²h = \(\frac13\times\frac{22}{7}\times1\times7=\frac{22}{3}\) cu. cm 

Volume of cubical block = (10 × 5 × 2) cm³ = 100 cm³

∴ Wastage of wood = \(\frac{\big(100-\frac{22}{7}\big)}{100}\times100\)

= \(\frac{278}{3}\)% = \(92\frac23\)%

1) Rice: It is a kharif crop which requires high temperature and high humidity with annual rainfall above 100 cm.

2) Wheat: It requires a cool growing season and a bright sunshine at the time of ripening.it requires 50 to 75 cm of annual rainfall evenly distributed over the growing season.

3) Sugar cane: It is a tropical as well as sub tropical crop .it grows well in hot and humid climate with a temperature of 21 to 27 degrees and an annual rainfall between 75 to 100 cm.

4)  Cotton: It grows well on black soil.it requires high temperature,light rainfall or irrigation,210 frost free days for its growth.it is a kharif crop and requires 6 to 8 months to mature.

5) Jute: It is also known as golden fibre. grows well on well-drained fertile soils in the flood plains where soils are renewed every year.high temperature is required.

6) Tea: It grows well in tropical and sub-tropical climates endowed with deep and fertile well-drained soil,rich in humus and organic matter.tea bushes require warm and moist frost free climate all through the year.

The emperor became restless due to Bajirao’s attack.

He called the Nizam to protect Delhi.

The Nizam attacked Bajirao with his large army.

Bajirao defeated him at Bhopal.

The Nizam agreed to secure the sanad of Malwa subhedari to Marathas from the Badshah.

Bajirao was a great warrior, a powerful ruler who established the Maratha dominance in North.

Bajirao defeated the Nizam at Palkhed near Aurangabad.

vector v = -{vector(mv)}/{M + m}

1. The centripetal force necessary for the particle to move in a circular path is provided by the tension in the string. Hence net force on the particle is nothing but tension T in the string.

The direction of acceleration can be opposite to that of velocity.- True

(i) is correct. The particle moving along the circular path is constantly accelerated and hence it must be under the action of an external force. The net force on the particle is T.

(a) Action: Backward push on the water

Reaction: Forward push (by water) on the man.

(b) Action: Pull on the cart.

Reaction: Backward tension in the rope.

Work is a vector quantity.- False

a) Zero net force.

The rain drop is falling with a constant speed.

Hence, it acceleration is zero. As per Newton’s second law of motion, the net force acting on the rain drop is zero.

b) Zero net force

The weight of the cork is acting downward. It is balanced by the buoyant force exerted by the water in the upward direction. 

Hence, no net force is acting on the floating cork.

c) Zero net force

The kite is stationary in the sky, i.e., it is not moving at all. 

Hence, as per Newton’s first law of motion, no net force is acting on the kite.

d) Zero net force

The car is moving on a rough road with a constant velocity. 

Hence, its acceleration is zero. As per Newton’s second law of motion, no net force is acting on the car.

e) Zero net force

The high speed electron is free from the influence of all fields. 

Hence, no net force is acting on the electron.

As the drop of rain is falling down with a constant speed, in accordance with first law of motion, the net force on the drop of rain is zero.

(1) According to first law of motion F = 0 as a = 0 (particle moves with constant speed) 

(2) Since kite is stationary net force on the kite is also zero. 

Displacement is always greater than distance.- False

Here, M = 20 kg, F = 50 (retarding force)

As F = mg ⇒ a = F/m = -50/20 = -2.5 ms^-2

Also, v = u + at, Here u = 15 ms^-1, v = 0

⇒ 0 = 15 + (-2.5) x t ⇒ t = 6s

Here, u = 2 ms^-1, v = 3.5 ms^-1, t = 25 s

From v = u + at ⇒ 3.5 = 2 + a x 25

⇒ a = {3.5 - 2}/{25} = 0.06 ms^-2

Mass of the body = m = 3 kg

Force acting on the body

F = ma = 3 x 0.06 = 0.18 N

Since the applied force increases the speed of the body, it acts in the direction of the motion.

The distance and displacement are equal only if, motion is along a straight path.- True

(1) Isaac Newton

(2) Uniform circular motion

(3) Recoil

(4) Uniform motion

(5) m/s²

(6) g cm/s

(1) True

(2) False

(3) False

(4) True

(5) True

(6) True

(b) the force acting on the scale is zero and the torque acting about centre of mass of the scale is also zero.

The difference in the true value and the measured value of a quantity is called error of measurement.

Precision is a measure of how consistently a device records nearly identical values i.e., reproducible results.

Correct answer is (B) 3/2