Test: Surface Area & Volumes- 1 - Class 9 MCQ

To calculate the volume of sand needed to fill the rectangular sandbox, we can use the formula:Volume = Length × Width × DepthGiven:
Length = 2 m
Width = 5 m
Depth = 10 cm = 0.1 mSubstituting the values into the formula:Volume = 2 m × 5 m × 0.1 m
Volume = 1 cubic meterTherefore, the correct answer is A: 1 cubic meter.

To find the weight of the iron beam, we need to calculate its volume and then multiply it by the weight of 1 cubic meter of iron, which is given as 50 kg.Given:
Length of beam (l) = 9 m
Breadth (b) = 40 cm = 0.4 m
Height (h) = 20 cm = 0.2 m
Weight of 1 cubic meter of iron = 50 kgVolume of the beam = l × b × h = 9 m × 0.4 m × 0.2 m = 0.72 cubic metersWeight of the beam = Volume × Weight of 1 cubic meter of iron = 0.72 cubic meters × 50 kg/cubic meter = 36 kg

Therefore, the weight of the iron beam is 36 kg, which corresponds to option B.

To find the maximum length of a pencil that can be kept in a rectangular box of dimensions 8 cm × 6 cm × 2 cm, we need to find the length of the diagonal of the box, which will be the maximum length of the pencil that can fit inside the box.Using the Pythagorean theorem, the length of the diagonal of the box can be found as:Diagonal = √(length² + breadth² + height²)
Diagonal = √(8² + 6² + 2²)
Diagonal = √(64 + 36 + 4)
Diagonal = √104
Diagonal = 2√26Therefore, the maximum length of a pencil that can be kept in the rectangular box is 2√26 cm, which corresponds to option C.

The surface area of a cube of side 27 cm can be found using the formula:Surface area of a cube = 6a^2, where a is the length of the side of the cube.Given:
Length of side of the cube = 27 cmSubstituting the value into the formula:Surface area of the cube = 6 × 27^2
Surface area of the cube = 6 × 729
Surface area of the cube = 4374 cm^2Therefore, the surface area of the cube of side 27 cm is 4374 cm^2, which corresponds to option C

Given that the perimeter of one face of a cube is 40 cm. Let each side of the cube be a cm. Each face of a cube is a square. Therefore, the perimeter of one face of the cube is 4a. We can write:4a = 40 cm
a = 10 cmTherefore, the length of each side of the cube is 10 cm. The volume of the cube is given by the formula V = a^3, where a is the length of each side of the cube. Substituting the value of a, we get:V = 10^3 = 1000 cm^3Therefore, the volume of the cube is 1000 cm^3, which corresponds to option B

The volume of a cube with a surface area of 384 sq. cm is 512 cm3, which corresponds to option D.

The length of the longest rod that can fit in a cubical vessel of side 10 cm is equal to the length of the diagonal of the cube. Using the Pythagorean theorem, the length of the diagonal of the cube can be found as:Diagonal = √(length² + breadth² + height²)
Diagonal = √(10² + 10² + 10²)
Diagonal = √300
Diagonal = 10√3 cmTherefore, the length of the longest rod that can fit in a cubical vessel of side 10 cm is 10√3 cm, which corresponds to option C.

Given:

Radius = 10.5 cm

Height = 6 m

Weight of 1 cm³ = 5 grams

Formula used:

Volume of cylinder = πr²h

1 m = 100 cm

⇒ 6 m = 600 cm

1 kg = 1000 gram

Calculation:

According to the question

Volume of cylinder = πr2h

⇒ (3.14 × 10.5 × 10.5 × 600) cm³

⇒ 207711 cm³

Now, Weight of 1 cm3​ = 5 grams

So, Weight of 207,711 cm³ = 5 × 207711

⇒ 1,038,555 grams

⇒ 1038.555 kg ~ 1038.5 kg

Given a cylinder with

Radius = 28 / 2 = 14 cm

Height = 20 cm

ii.) Total surface area = 2πr ( r + h )

= 2 x 22 / 7 x 14cm ( 14 + 20 )cm

= 2991.99cm2
=2992cm2
(b) 2992cm2 is the correct option.

The lateral surface area of a cone with height 3 cm and radius 4 cm can be found using the formula:
Lateral surface area of a cone = πrl, where r is the radius of the base of the cone and l is the slant height of the cone.
Given:
Height of the cone = 3 cm
Radius of the base of the cone = 4 cm
Using the Pythagorean theorem, the slant height of the cone can be found as:
l = √(r^2 + h^2) = √(4^2 + 3^2) = √(16 + 9) = √25 = 5 cm
Substituting the values into the formula:
Lateral surface area of the cone = πrl = π × 4 × 5 = 20π cm^2
Approximating π to 22/7, we get:
Lateral surface area of the cone = 20 × 22/7 = 440/7 cm^2
Therefore, the lateral surface area of the cone with height 3 cm and radius 4 cm is 440/7 cm^2, which corresponds to option A

Solution

Let r= radius of the base.
Given area = 346.5

π×r² =346.5

r² = 346.5×722 = 4414

r = 212

l = √r²+h²=√4414+142

l = 352 m

The area of canvas = surface area of the cone =πrl = π×212×352 m² = 33×352 m²

Given length of canvas = 1.1 m
Let width = x
Area of canvas = 1.1x
So, 1.1 x=33×352
x=33×352×1.1 = 525 m

Let three spheres are S₁, S₂ & S₃
having radii r₁ = 3cm, r₂ = 4cm & r₃ = 5cm respectively.
Let the radius of new Big sphere S is R.
A/Q,
Volume of new Sphere  S = Sum of volumes three Spheres S₁, S₂ & S₃