RD Sharma Class 9 Solutions Chapter 18 MCQS (Updated 2024)

RD Sharma Class 9 Solutions Chapter 18 MCQS: MCQs are one of the important parts of class 9 exams. You can get the RD Sharma Class 9 Solutions Chapter 18 MCQS from the link provided below. 

• RD Sharma Class 9 Maths Solutions

Access RD Sharma Class 9 Solutions Chapter 18 MCQS

Question 1.
The length of the longest rod that can be fitted in a cubical vessel of edge 10 cm long, is
(a) 10 cm
(b) 102–√ cm
(c) 103–√ cm
(d) 20 cm
Solution:
Edge of cuboid (a) = 10 cm
∴ Longest edge = 3–√ a cm
= 3–√ x 10 = 103–√ cm (c)

Question 2.
Three equal cubes are placed adjacent in a row. The ratio of the total surface area of the resulting cuboid to that of the sum of the surface areas of three cubes is
(a) 7: 9
(b) 49: 81
(c) 9: 7
(d) 27: 23
Solution:
Let a be the side of three equal cubes
∴ The surface area of 3 cubes
= 3 x 6a² = 18a²
and length of so formed cuboid = 3a
Breadth = a
and height = a
∴ Surface area = 2(lb + bh + hl)
= 2[3a x a + a x a+a x 3a] = 2[3a² + a² + 3a²] = 2 x 7a² = 14a²
∴ The ratio in the surface areas of the cuboid and three cubes = 14a²: 18a²= 7:9 (a)

Question 3.
The length, width, and height of a rectangular solid are in the ratio of 3: 2 : 1. If the volume of the box is 48 cm³, the total surface area of the box is
(a) 27 cm²
(b) 32 cm²
(c) 44 cm²
(d) 88 cm²
Solution:
The ratio in the dimensions of a cuboid =3: 2 : 1
Let length = 3x
Breadth = 2x
and height = x
Then volume = lbh = 3x x 2x x x = 6×3
∴ 6x³ = 48 ⇒ x³= 486 = 8 = (2)³
∴ x = 2
∴ Length (l) = 3 x 2 = 6 cm
Breadth (b) = 2 x 2 = 4 cm
Height (h) = 1 x 2 = 2 cm
Now surface area = 2[lb + bh + hl]
= 2[6 x 4 + 4 x 2 + 2 x 6] cm²
= 2[24 + 8-+ 12] = 2 x 44 cm²
= 88 cm² (d)

Question 4.
If each edge of a cube, of volume V, is doubled, then the volume of the new cube is
(a) 2V
(b) 4V
(c) 6V
(d) 8V
Solution:
Let a be the edge of a cube whose Volume = V
∴ a3 = V
By doubling the edge, we get 2a
Then volume = (2a)3 = 8a³
∴ Volume of new cube = 8a³ = 8V (d)

Question 5.
If each edge of a cuboid of surface area S is doubled, then the surface area of the new cuboid is
(a) 2S
(b) 4S
(c) 6S
(d) 8S
Solution:
Let each edge of a cube = a
The surface area = 6a²
∴ S = 6a²
Now doubling the edge, we get
The new edge of a new cube = 2a
∴ Surface area = 6(2a)²
= 6 x 4a² = 24a²
= 4 x 6a² = 4S (b)

Question 6.
The area of the floor of a room is 15 m2. If its height is 4 m, then the volume of the air contained in the room is
(a) 60 dm³
(b) 600 dm³
(c) 6000 dm³
(d) 60000 dm³
Solution:
Area of a floor of a room = 15 m²
Height (h) = 4 m
∴ The volume of air in the room = Floor area x Height
= 15 m² x 4 m = 60 m³
= 60 x 10 x 10 x 10 dm² = 60000 dm² (d)

Question 7.
Volume of a cuboid is 12 cm³. The volume (in cm³) of a cuboid whose sides are double of the above cuboid is
(a) 24
(b) 48
(c) 72
(d) 96
Solution:
Volume of cuboid = 12 cm³
By doubling the sides of the cuboid the
volume will be = 12 cm³ x 2 x 2 x 2
= 96 cm³ (d)

Question 8.
If the sum of all the edges of a cube is 36 cm, then the volume (in cm3) of that cube is
(a) 9
(b) 27
(c) 219
(d) 729
Solution:
The sum of all edges of a cube = 36 cm
No. of the edge of a cube is 12
∴ Length of its one edge = 3612 = 3 cm
Then volume = (edge)³ = (3)³ cm³
= 27 cm³ (b)

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