Surface Areas and Volumes: 9 Math
NCERT Exemplar Exercise 13.1 q6-10: 9th math
NCERT Exemplar Exercise 13.1 Surface Areas And Volumes Class nine Math Question (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 volume is:
(A) 10:17
(B) 20:27
(C) 17:27
(D) 20:37
Answer (B) 20:27
Solution
Given, ratio of radii of two cylinders = 2 : 3
And, ratio of height of the given two cylinders = 5 : 3
Thus, ratio of their volumes = ?
For Cylinder 1
Let radius = 2r
Let height = 5h
Now, we know that, Volume of a Cylinder
= ℼ r2 h
Thus, volume of cylinder 1
= ℼ × (2r)2 × 5 h
= ℼ × 4 r2 × 5 h
Thus, volume of cylinder1 = ℼ 20 r2 h
And for Cylinder 2
Let radius = 3 r
And, Let height = 3 h
Volume of cylinder 2 = ℼ r2 h
= ℼ × (3 r)2 × 3 h
= ℼ × 9 r2 × 3 h
Thus, volume of cylinder2 = 27 r2 h
Now, Ratio of volumes of cylinder 1 and cylinder 2 = volume1/volume2
`=(pi\ 20r^2xx h)/(pi\ 27r^2xxh)`
= 20/27
Thus, ratio of volumes of given cylinders = 20:27 Answer
NCERT Exemplar Exercise 13.1 Surface Areas And Volumes Class nine Math Question (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
Answer (A) 512 m3
Explanation
Given, the Lateral surface Area of a Cube
= 256 m2
Thus, volume of the given cube = ?
We know that, Lateral surface area of a cube
= 4 a2
Where a = edge of the cube
Thus, lateral surface area of the given cube
= 256 m2 = 4 a2
⇒ 256/4 m2 = a2
⇒ a2 = 64 m2
⇒ a = 8 m
Now, we know that Volume of a Cube = a3
Thus, volume of the given cube = (8 m)3
= 8 m × 8 m × 8 m ×
= 512 m3
Thus, volume of the given cube = 512 m3 Answer
NCERT Exemplar Exercise 13.1 Surface Areas And Volumes Class nine Math Question (8) The number of planks of dimensions (4m × 50cm × 20cm) that can be stored in a pit which is 16 m long, 12 m wide and 4m deep is
(A) 1900
(B) 1920
(C) 1800
(D) 1840
Answer (B) 1920
Explanation
Given, Length of plank = 4 m
And, Breadth of the plank = 50 cm
= 50/100 = 0.5 m
And, Height of the plank = 20 cm
= 20/100 = 0.2 m
And, length of the pit = 16 m
And, width of the pit = 12 m
And, depth of the pit = 4 m
Thus, total number of planks that can be stored in the given pit = ?
Now, volume of planks = `l` × b × h
= 0.5 m × 0.2 m × 4 m
= 0.4 m3
And, volume of the pit = `l`b h
= 16 m × 12 m × 4 m
= 768 m3
Now, the total no of planks can be stored in the pit
= volume of the pit/volume of the plank
= 768/0.4
= 1920 planks
Thus, total number of planks that can be stored in the given pit = 1920 Answer
NCERT Exemplar Exercise 13.1 Surface Areas And Volumes Class nine Math Question (9) The length of the longest pole that can be put in a room of dimensions (10 m × 10 m × 5 m ) is
(A) 15m
(B) 16m
(C) 10m
(D) 12m
Answer (A) 15m
Explanation
The longest pole which can be put in the given room = diagonal of the given cuboid
Here, given, Length of room = 10 m
And, breadth (b) of the room = 10 m
And, height (h) of the room = 5 m
We know that, diagonal of a cuboid
Thus, diagonal of the given room = 15 m
Thus, longest pole which can be stored in the given room = 15 m Answer
NCERT Exemplar Exercise 13.1 Surface Areas And Volumes Class nine Math Question (10) 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) 1:4
Explanation
Given, Radius of the hemispherical balloon in first case (r) = 6 cm
And, the radius of the hemispherical balloon after pumping of more air in second case (R) = 12 cm
Thus, ratio of the surface areas of the given two hemispherical balloons = ?
Now, we know that, surface area of a hemisphere = 3 ℼ r2
Case-1
Thus, surface area of hemispherical balloon in 1st case
= 3 × ℼ × (6 cm)2
= 3 ℼ × 36 cm2
Case-2
Radius of the hemispherical balloon = 12 cm
Therefore, surface area of the hemispherical in given case-2
= 3 ℼ r2
= 3 × ℼ × (12 cm)2
= 3 ℼ × (12 cm)2
= 3 ℼ× 144 cm2
Now, ratio of surface areas in the given two cases
`=(3\ pi\ 36cm^2)/(3\ pi\ 144cm^2)`
= 1/4 = 1 : 4
Thus, ratio of the surface areas of the given balloon in two cases = 1:4 Answer
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