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### Maths MCQ Class IX Ch-13 | Surface Area & Volume

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# **Mathematics MCQ | Class 09 | Chapter 13**

# SURFACE AREA & VOLUME

**MCQ Based on the different types of Solid Figures.****MCQ Based on the concept of Cube & Cuboid.****MCQ Based on the Right circular Cylinder and Cone.****MCQ Based on the Sphere & Hemi-Sphere.**

**MCQ Based on the different types of Solid Figures.****MCQ Based on the concept of Cube & Cuboid.****MCQ Based on the Right circular Cylinder and Cone.****MCQ Based on the Sphere & Hemi-Sphere.**__Features__

__Features__

**In the MCQ given below you find the important MCQ which are strictly according to the CBSE syllabus and are very useful for the CBSE Examinations.****Solution Hints are also given to some difficult problems.****Each MCQ contains four options from which one option is correct.**

__Action Plan__

__Action Plan__

**First of all students should Learn and write all basic points and Formulas related to the Chapter 13 Surface Area and Volume.****Start solving the NCERT Problems with examples.****Solve the important assignments on the Chapter 13 Class IX.****Then start solving the following MCQ.**

**MCQ BASED ON CUBE AND CUBOID**

**Cube and cuboid have following properties**

**Both have 6 faces, 8 vertices and 12 edges.**

**Faces of cube are squares and faces of cube are rectangles.**

**Q1) The formula to find the surface area of a cuboid of length (l), breadth (b) and height (h) is:**

**a)
lb + bh + hl**

**b)
2(lb + bh + hl)**

**c)
2(lbh)**

**d)
lbh/2**

**Answer: b**

**Q2) If the perimeter of one of the faces of a cube is 40 cm, then its
volume is:
a) 6000 cm³
b) 1600 cm³
c) 1000 cm³
d) 600 cm³**

**Answer: c**

**Q3) The surface area
of a cube whose edge equals to 3cm is:**

**a)
62 sq.cm**

**b)
30 sq.cm**

**c)
54 sq.cm**

**d)
90 sq.cm**

**Answer: c**

**Explanation: Given, a = 3 cm**

**Surface area of cube = 6a ^{2}**

**SA = 6 x 3 x 3 = 54 sq.cm**

**Q4) The
surface area of a cube whose edge equals to 3cm is:
a) 62 sq.cm
b) 30 sq.cm
c) 54 sq.cm
d) 90 sq.cm**

**Answer c**

**Q5) The surface area of
cuboid-shaped box having length = 80 cm, breadth = 40cm and height = 20cm is:**

**a)
11200 sq.cm**

**b)
13000 sq.cm**

**c)
13400 sq.cm**

**d)
12000 sq.cm**

**Answer: a**

**Explanation: surface area of the box = 2(lb + bh + hl)**

**S.A. = 2[(80 × 40) + (40 × 20) + (20 × 80)]**

**= 2[3200 + 800 + 1600]**

**= 2 × 5600 = 11200 sq.cm.**

**Q6) The
length of the longest pole that can be put in a room of dimension (10 m × 10 m
× 5 m) is
a) 15 m
b) 16 m
c) 10 m
d) 12 m**

**Answer: a**

**Q7) The total surface
area of a cube is 96 cm ^{2}. The volume
of the cube is:**

**a) 8 cm ^{3} **

**b) 512 cm ^{3}**

**c) 64 cm ^{3}**

**d) 27 cm ^{3}**

**Answer: c**

**Explanation:**

**We know
that the TSA of the cone = 6a ^{2}.**

**6a ^{2} = 96 cm^{2}**

**a ^{2} = 96/6 = 16**

**a = 4 cm**

**The volume
of cone = a ^{3} cubic units**

**V = 4 ^{3} = 64cm^{3}.**

**Q8) The
lateral surface area of a cube is 256 m³. The volume of the cube is**

**
a) 512 m³**

**
b) 64 m³**

**
c) 216 m³**

**
d) 256 m³**

**Answer: a**

**Q9)
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**

**Answer: a**

**Explanation:**

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

**The length
of the longest pole = √[10 ^{2 }+
10^{2 }+ 5^{2}]**

**= √(100 + 100
+ 25) = √225 = 15 m.**

**Q10)
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**

**Answer: b**

**Explanation:**

**Volume of
Plank = 400 cm × 50cm × 20cm = 400000cm ^{3}**

**Volume of
pits = 1600cm × 1200cm × 400cm = 768000000cm ^{3}**

**Number of
planks = Volume of planks/Volume of pits**

**=
768000000/400000**

**Hence, the
number of pits = 1920**

**Q11) 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**

**Answer: c**

**Q12)
The lateral surface area of a cube is 256 m ^{2}.
The volume of the cube is**

**a) 512 m ^{3}**

**b) 64 m ^{3}**

**c) 216 m ^{3}**

**d) 256 m ^{3}**

**Answer: a**

**Explanation:**

**The lateral
surface area of cube = 4a ^{2}**

**4a ^{2 }= 256**

**a ^{2} = 256/4 =64**

**a = 8 m**

**Hence, the
volume of cube = a ^{3} cube
units **

**V = 8 ^{3 } = 512 m^{3}.**

**MCQ BASED ON RIGHT CIRCULAR CYLLINDER**

**Cylinder is a solid figure which have three faces, two faces at the top and bottom are plan surface and one is curved surface or Lateral surface.**

**Q13) If the radius of
a cylinder is 4cm and height is 10cm, then the total surface area of a cylinder
is:**

**a)
440 sq.cm**

**b)
352 sq.cm.**

**c)
400 sq.cm**

**d)
412 sq.cm**

**Answer: b**

**Explanation: Total Surface Area of a Cylinder = 2πr(r + h)**

**TSA = 2 x 22/7 x 4(4 + 10)**

**= (2 x 22 x 4 x 14)/7**

**= (2 x 22x 4 x 2)**

**= 352 sq.cm**

**Q14) The radius of a cylinder is doubled and the height remains the
same. The ratio between the volumes of the new cylinder and the original
cylinder is
a) 1 : 2
b) 3 : 1
c) 4 : 1
d) 1 : 8**

**Answer: c**

**Q15) In a cylinder, the
radius is doubled and height is halved, the curved surface area will be**

**a) Halved**

**b) Doubled**

**c) Same**

**d) Four times**

**Answer: c**

**Explanation:**

**We know
that the curved surface area of a cylinder is 2πrh**

**Given that,
r = 2R, h = H/2**

**Hence, the
CSA of new cylinder = 2π(2R)(H/2) = 2πRH **

**Therefore,
the answer is “Same”.**

**Q16) The curved
surface area of a right circular cylinder of height 14 cm is 88 cm ^{2}. The diameter of the base is:**

**a)
2 cm**

**b)
3cm**

**c)
4cm**

**d)
6cm**

**Answer: a**

**Q17)
The curved surface area of a right circular cylinder of height 14 cm is 88 cm ^{2}. The diameter of the base is:
a) 2 cm
b) 3cm
c) 4cm
d) 6cm**

**Answer: a**

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

**Height = 14 cm**

**2πrh = 88**

**r = 88/2πh**

**r =1 cm**

**Diameter = 2r = 2cm**

**Q18) Volume of hollow cylinder is**

**
a) π(R² – r²)h**

**
b) πR²h**

**
c) πr²h**

**
d) πr²(h _{1} – h_{1})**

**Answer: a**

**Q19) 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**

**Answer: c**

**Explanation: Curved surface area of cylinder = 2πrh**

**2πrh = 4.4**

**h = 4.4/(2π x 0.7)**

**h = 1 cm**

**Q20) 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**

**Answer: b**

**Q 21) 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**

**Answer: b**

**Explanation:**

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

**Hence, r _{1}= 2r, r_{2} = 3r**

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

**Hence, h _{1} = 5h, h_{2} = 3h**

**The ratio
of the volume of two cylinders = V _{1}/V_{2}**

**= πr _{1}^{2}h_{1}/πr_{2}^{2}h_{2}**

**= [(2r) ^{2}(5h)]/[(3r)^{2}(3h)]**

**Ratio of
their volumes = (20r ^{2}h)/(27r^{2}h) = 20/27 = 20 : 27.**

**Q 22)
If the radius of a cylinder is 4cm and height is 10cm, then total surface area
of cylinder is:**

**
a) 440 sq.cm**

**
b) 352 sq.cm.**

**
c) 400 sq.cm**

**
d) 412 sq.cm**

**Answer: b**

**MCQ BASED ON RIGHT CIRCULAR CONE**

**Cone is a solid figure which has two faces one is plane surface at the bottom and another is a curved surface.**

**Q 23) 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**

**Answer: b**

**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**

**Q 24) A cone
is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast
into a shape. The radius of the sphere is**

**
a) 4.2 cm**

**
b) 2.1 cm**

**
c) 2.4 cm**

**
d) 1.6 cm**

**Answer b**

**Q 25) 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**

**Answer: d**

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

**r = 24/2 = 12 cm**

*l* = 21 cm

**TSA = π(12)(21 + 12) = 1244.57 sq.cm**

**Q
26) 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**

**Answer: b**

**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**

**(⅓)πr ^{2}h = (4/3)πr^{3}**

**(⅓)π(2.1) ^{2}(8.4) = (4/3)πr^{3}**

**37.044 = 4r ^{3}**

**r ^{3}= 37.044/4**

**r ^{3}= 9.261**

**r = 2.1**

**Therefore,
the radius of the sphere is 2.1 cm.**

**Q27)
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**

**Answer: b**

**Q28)
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) 2πr l**

**Answer: b**

**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[(r l)
+ (r2/4)]**

**TSA of new
cone =πr[ l + (r/4)]**

**MCQ BASED ON SPHERE AND HEMI-SPHERE**

**Sphere is solid figure which has only one curved surface.**

**Hemi sphere is a solid figure which has two faces, one is plane surface at the top and another is curved surface.**

**Q29) The volume of a hemisphere whose radius is r is:**

**a)
4/3 πr ^{3}**

**b)
4πr ^{3}**

**c)
2πr ^{3}**

**d) 2/3 π r ^{3}**

**Answer: d**

**Q30) 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**

**Answer: c**

**Explanation: Radius of sphere, r = 14 cm**

**Surface area = 4πr ^{2}**

**= 4 x 22/7 x (14) ^{2} =
2464 sq.cm.**

**Q 31)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: a**

**Q32)
The radius of a sphere is 2r, then its volume will be**

**a) (4/3) πr ^{3}**

**b) 4πr ^{3}**

**c) (8/3) πr ^{3}**

**d) (32/3) πr ^{3}**

**Answer: d**

**Explanation:**

**Given : r =
2r**

**The volume
of sphere = (4/3)πr ^{3 }=
(4/3)π(2r)^{3}**

**V =
(4/3)π(8r ^{3}) = (32/3)πr^{3}.**

**Q
33) 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**

**Answer: a**

**Explanation:**

**We know
that the total surface area of the hemisphere = 3πr ^{2} 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 = ¼,
which is equal to 1:4.**

**Download Math MCQ Worksheet For Class IX**

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