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Test: Surface Area & Volumes- 2 - Class 9 MCQ


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25 Questions MCQ Test Mathematics (Maths) Class 9 - Test: Surface Area & Volumes- 2

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Test: Surface Area & Volumes- 2 - Question 1

The total surface area of a cube is 96 cm. The volume of the cube is

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 1

Surface area of a cube = 96 cm2
Surface area of a cube = 6 (Side)2 = 96 ⇒  (Side)2 = 16
⇒ (Side) = 4 cm
[taking positive square root because side is always a positive quantity]
Volume of cube = (Side)3 = (4)3 = 64 cm3
Hence, the volume of the cube is 64 cm3.

Test: Surface Area & Volumes- 2 - Question 2

If the outer diameter of a pipe 21 m long is 1 m, then its curved surface area is

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 2

To calculate the curved surface area of a pipe, use the formula:

  • Curved Surface Area = 2 × π × r × h

Where:

  • r is the radius of the pipe
  • h is the height or length of the pipe

Given:

  • Outer diameter = 1 m, thus radius (r) = 0.5 m
  • Length of the pipe (h) = 21 m

Now, substituting the values into the formula:

  • Curved Surface Area = 2 × π × 0.5 m × 21 m
  • ≈ 66 m2 (using π ≈ 3.14)

The curved surface area of the pipe is therefore approximately 66 m2.

Test: Surface Area & Volumes- 2 - Question 3

A cube of side 4 cm contains a sphere touching its sides. Find the approximate volume of the gap in between.

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 3

Let ABCD be cube which contain sphere touching its sides.The diameter of sphere is the sides of the cube.

Now,

The volume of the gap in between cube and sphere

= volume of cube - volume of sphere


\


= 30.48 cm3

Test: Surface Area & Volumes- 2 - Question 4

The cost of digging a pit of dimensions 4.5m×2.5m×2.5m at the rate of Rs 20 per cubic metre is

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 4

To calculate the cost of digging the pit, follow these steps:

  • Calculate the volume of the pit:
    • Dimensions: 4.5m × 2.5m × 2.5m
    • Volume = 4.5 × 2.5 × 2.5 = 28.125 cubic metres
  • Determine the cost per cubic metre:
    • Rate = Rs 20 per cubic metre
  • Calculate total cost:
    • Total cost = Volume × Rate
    • Total cost = 28.125 × 20 = Rs 562.50

The total cost of digging the pit is Rs 562.50.

Test: Surface Area & Volumes- 2 - Question 5

If the volume of a sphere is 4851 cm3, then its surface area is

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 5

The volume of a sphere is given as 4851 cm3.

To find the surface area, we can use the following formulas:

  • The volume of a sphere is calculated using the formula: V = (4/3)πr3, where r is the radius.
  • The surface area of a sphere is given by: A = 4πr2.

First, we need to find the radius:

  • Rearranging the volume formula to find r: r = ((3V)/(4π))1/3.
  • Substituting the volume: r = ((3 × 4851)/(4π))1/3.

Next, we calculate the surface area:

  • Using the radius in the surface area formula: A = 4πr2.
  • After calculations, the surface area is found to be: 1386 cm2.

Thus, the surface area of the sphere is 1386 cm2.

Test: Surface Area & Volumes- 2 - Question 6

The lateral surface area of a cube is 256 m2 The volume of the cube is

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 6

Test: Surface Area & Volumes- 2 - Question 7

The cost of cementing the inner curved surface of a 14 m deep well of radius 2 m at the rate of Rs 2 per m2 is

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 7

Depth of well(h) = 14m
radius of well(r) = 2m

Inner surface area of well(like a cylinder) = 2πrh 

surface area= 2 x 22/7 x 2 x 14 = 176 m2

cost of cementing = ₹2 per m2

Total Cost = 176 x 2 = ₹352

Test: Surface Area & Volumes- 2 - Question 8

The ratio of the radii of two spheres whose volumes are in the ratio 64 : 27 is

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 8

The ratio of the radii of two spheres, whose volumes are in the ratio 64 : 27, can be determined as follows:

  • The volume of a sphere is calculated using the formula: V = (4/3)πr³, where r is the radius.
  • If the volumes are in the ratio 64 : 27, we can express this as:
  • V1/V2 = 64/27
  • Substituting the volume formula gives us:
  • (4/3)πr1³ / (4/3)πr2³ = 64/27
  • This simplifies to:
  • r1³/r2³ = 64/27
  • Taking the cube root of both sides, we find:
  • r1/r2 = (64/27)^(1/3)
  • This results in:
  • r1/r2 = 4/3

Thus, the ratio of the radii of the two spheres is 4 : 3.

Test: Surface Area & Volumes- 2 - Question 9

The volume of resulting cuboid formed when two cubes each of side 6 cm are joined end to end is

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 9

The volume of the cuboid formed by joining two cubes with a side of 6 cm is calculated as follows:

  • Each cube has a volume of 216 cm3 (calculated as 6 cm x 6 cm x 6 cm).
  • When two cubes are joined end to end, the new dimensions of the cuboid are:
    • Length: 12 cm (6 cm + 6 cm)
    • Width: 6 cm
    • Height: 6 cm
  • The volume of the cuboid is calculated as:
    • Volume = Length x Width x Height
    • Volume = 12 cm x 6 cm x 6 cm = 432 cm3
Test: Surface Area & Volumes- 2 - Question 10

A hemispherical bowl is made of steel 0.25 cm thick. If the inner radius of the bowl is 3.25 cm, then the outer curved surface area of the bowl is

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 10

To find the outer curved surface area of a hemispherical bowl, follow these steps:

  • Identify the inner radius: The inner radius is given as 3.25 cm.
  • Calculate the outer radius: Add the thickness of the steel (0.25 cm) to the inner radius:
    • Outer radius = 3.25 cm + 0.25 cm = 3.5 cm.
  • Use the formula for curved surface area: The formula for the curved surface area of a hemisphere is:
    • Curved Surface Area = 2 * π * r².
  • Calculate the outer curved surface area: Substitute the outer radius into the formula:
    • Curved Surface Area = 2 * π * (3.5 cm)².
    • Curved Surface Area = 2 * π * 12.25 cm².
    • Curved Surface Area ≈ 2 * 3.14 * 12.25 cm².
    • Curved Surface Area ≈ 76.9 cm².

Rounding to the nearest whole number, the outer curved surface area of the bowl is approximately 77 cm².

Test: Surface Area & Volumes- 2 - Question 11

The difference between the total surface area of a cube of side 4 cm and its lateral surface area is

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 11

To find the difference between the total surface area and the lateral surface area of a cube with a side length of 4 cm, follow these steps:

  • The formula for the total surface area of a cube is: 6 × (side length)2.
  • For a side length of 4 cm, the total surface area is: 6 × (4 cm)2 = 6 × 16 cm2 = 96 cm2.
  • The formula for the lateral surface area of a cube is: 4 × (side length)2.
  • For a side length of 4 cm, the lateral surface area is: 4 × (4 cm)2 = 4 × 16 cm2 = 64 cm2.
  • Now, calculate the difference:
  • Difference = Total Surface Area - Lateral Surface Area = 96 cm2 - 64 cm2 = 32 cm2.
Test: Surface Area & Volumes- 2 - Question 12

The diameter of the base of a cylinder of curved surface area 88 cm2 and height 14 cm is

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 12

To find the diameter of the base of the cylinder:

  • The formula for the curved surface area (CSA) of a cylinder is given by:
  • CSA = 2πrh, where r is the radius and h is the height.
  • Given the CSA is 88 cm2 and the height is 14 cm, we can set up the equation:
  • 88 = 2πr × 14
  • This simplifies to:
  • 88 = 28πr
  • Next, solve for r:
  • r = 88 / (28π)
  • Calculating this gives:
  • r ≈ 1 cm
  • Since the diameter (d) is twice the radius:
  • d = 2r
  • This results in:
  • d ≈ 2 cm
Test: Surface Area & Volumes- 2 - Question 13

Given that the surface area of a spherical shot put is 616 cm2. its diameter is

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 13

To find the diameter of a spherical shot put given its surface area:

  • The formula for the surface area (A) of a sphere is: A = 4πr², where r is the radius.
  • Given that the surface area is 616 cm², we can set up the equation: 4πr² = 616.
  • To isolate , divide both sides by : r² = 616 / (4π).
  • Calculating this gives: r² ≈ 49.
  • Taking the square root of both sides, we find: r ≈ 7 cm.
  • The diameter (d) is twice the radius: d = 2r, which results in: d ≈ 14 cm.

The diameter of the shot put is 14 cm.

Test: Surface Area & Volumes- 2 - Question 14

The number of litres that a cuboidal water tank of dimensions 6m×5m×4.5m can hold is

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 14
The volume of a cuboidal water tank with dimensions 6m × 5m × 4.5m can be calculated using the formula for the volume of a cuboid:

Volume = Length × Width × Height

  • Substituting the dimensions:
  • Volume = 6m × 5m × 4.5m
  • Calculating the volume:
  • Volume = 30m² × 4.5m = 135m³

To convert cubic metres to litres, use the conversion factor:

  • 1m³ = 1000 litres
  • Therefore, 135m³ = 135 × 1000 litres = 135000 litres

The tank can hold 135000 litres of water.

Test: Surface Area & Volumes- 2 - Question 15

If the surface area of a solid sphere is 1386 cm2. then the total surface area of the solid hemisphere of the same radius is

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 15

To find the total surface area of a solid hemisphere given the surface area of a sphere:

  • The formula for the surface area of a sphere is 4πr².
  • Given that the surface area is 1386 cm², we can set up the equation:
  • 4πr² = 1386
  • To find the radius (r), we first solve for :
  • r² = 1386 / (4π)
  • Next, we calculate the surface area of a hemisphere:
  • The total surface area of a hemisphere is given by the formula:
  • 2πr² + πr² = 3πr²
  • Now, substitute the value of :
  • Total surface area = 3 * (1386 / 4) = 1039.5 cm²

The total surface area of the solid hemisphere is therefore 1039.5 cm².

Test: Surface Area & Volumes- 2 - Question 16

The volume of a cube whose diagonal is 8√3 cm is

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 16

To find the volume of a cube when the diagonal is given:

  • The formula for the diagonal D of a cube in terms of its side length s is:
  • D = s√3
  • Given that the diagonal is 8√3 cm, we can set up the equation:
  • 8√3 = s√3
  • Dividing both sides by √3 gives:
  • s = 8 cm
  • To find the volume V of the cube, use:
  • V = s³
  • Calculating the volume:
  • V = 8³ = 512 cm³

The volume of the cube is 512 cm³.

Test: Surface Area & Volumes- 2 - Question 17

The total surface area of a right circular cylinder of height 4 cm and radius 3 cm is

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 17

To calculate the total surface area of a right circular cylinder:

  • Use the formula: Total Surface Area = 2πr(h + r)
  • Where:
    • r is the radius of the base
    • h is the height of the cylinder
  • Given:
    • Height (h) = 4 cm
    • Radius (r) = 3 cm
  • Substituting the values into the formula:
    • Total Surface Area = 2π(3)(4 + 3)
    • Total Surface Area = 2π(3)(7)
    • Total Surface Area = 42π cm2
  • Approximating π as 3.14:
    • Total Surface Area ≈ 42 × 3.14 ≈ 131.88 cm2

The total surface area of the cylinder is approximately 132 cm2.

Test: Surface Area & Volumes- 2 - Question 18

If a spherical balloon grows to twice its radius when inflated, then the ratio of the volume of the inflated balloon to the original balloon is

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 18

When a spherical balloon expands to twice its radius, the volume ratio between the inflated and original balloons can be calculated as follows:

  • The formula for the volume of a sphere is V = (4/3)πr³.
  • If the original radius is r, the original volume is V₁ = (4/3)πr³.
  • When the radius is doubled, the new radius becomes 2r.
  • The volume of the inflated balloon is V₂ = (4/3)π(2r)³.
  • Calculating this gives V₂ = (4/3)π(8r³) = (32/3)πr³.
  • The ratio of the volumes is V₂ / V₁ = ((32/3)πr³) / ((4/3)πr³).
  • Simplifying the ratio results in V₂ / V₁ = 8.

The final ratio of the volume of the inflated balloon to the original balloon is 8:1.

Test: Surface Area & Volumes- 2 - Question 19

The surface area of a cuboid whose length, breadth and height are 15 cm, 10 cm and 20 cm respectively is

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 19

The surface area of a cuboid can be calculated using the formula:

  • Surface Area = 2(lw + lh + wh)

Where:

  • l = length
  • w = breadth
  • h = height

For the given cuboid:

  • Length = 15 cm
  • Breadth = 10 cm
  • Height = 20 cm

Now, substituting the values into the formula:

  • Surface Area = 2(15 cm × 10 cm + 15 cm × 20 cm + 10 cm × 20 cm)
  • = 2(150 cm2 + 300 cm2 + 200 cm2)
  • = 2(650 cm2)
  • = 1300 cm2

Thus, the surface area of the cuboid is 1300 cm2.

Test: Surface Area & Volumes- 2 - Question 20

The number of spherical bullets each 5 dm in diameter which can be cast from a rectangular block of lead 11 m long, 10 m broad and 5 high is

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 20

To find out how many spherical bullets can be cast from a rectangular block of lead, follow these steps:

  • Calculate the volume of the rectangular block:
    • Length = 11 m = 1100 dm
    • Breadth = 10 m = 1000 dm
    • Height = 5 m = 500 dm
    • Volume = Length × Breadth × Height = 1100 × 1000 × 500 = 550000000 dm³
  • Calculate the volume of one spherical bullet:
    • Diameter = 5 dm, so radius = 2.5 dm
    • Volume = (4/3) × π × (radius)³ = (4/3) × π × (2.5)³
    • Volume ≈ 65.45 dm³
  • Determine the number of bullets that can be made:
    • Number of bullets = Volume of block ÷ Volume of one bullet
    • Number of bullets = 550000000 ÷ 65.45 ≈ 8400

The result is that approximately 8400 spherical bullets can be cast from the block of lead.

Test: Surface Area & Volumes- 2 - Question 21

The number of planks of dimensions (5m×25cm×10cm) that can be placed in a pit which is 20 m long, 6 m wide and 80 cm deep is

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 21

To determine the number of planks that can fit in the pit, follow these steps:

  • Calculate the volume of the pit:
    • Length: 20 m
    • Width: 6 m
    • Depth: 80 cm (0.8 m)
    • Volume = Length × Width × Depth = 20 m × 6 m × 0.8 m = 96 m³
  • Calculate the volume of one plank:
    • Dimensions: 5 m × 25 cm (0.25 m) × 10 cm (0.1 m)
    • Volume = 5 m × 0.25 m × 0.1 m = 0.125 m³
  • Determine the number of planks:
    • Number of planks = Volume of pit / Volume of one plank
    • Number of planks = 96 m³ / 0.125 m³ = 768

The pit can accommodate 768 planks.

Test: Surface Area & Volumes- 2 - Question 22

If the lateral surface area of a cylinder is 132 cm2 and its height is 7 cm, then its base diameter is

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 22

To find the base diameter of the cylinder:

We start with the formula for the lateral surface area of a cylinder:

  • Lateral Surface Area = 2 × π × r × h

Where:

  • r = radius of the base
  • h = height of the cylinder

Given:

  • Lateral Surface Area = 132 cm2
  • Height (h) = 7 cm

Substituting the values into the formula:

  • 132 = 2 × π × r × 7

To isolate r:

  • 132 = 14πr
  • r = 132 / (14π)

Now, calculate the radius:

  • Using π ≈ 3.14, r ≈ 132 / 43.96 ≈ 3 cm

To find the diameter:

  • Diameter = 2 × r = 2 × 3 cm = 6 cm

Therefore, the base diameter of the cylinder is 6 cm.

Test: Surface Area & Volumes- 2 - Question 23

If the total surface area of a hemisphere is 1848 cm2, then the diameter is

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 23

Test: Surface Area & Volumes- 2 - Question 24

The volume of a cylinder whose circumference of the base is 132 cm and height 25 cm is

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 24

To find the volume of a cylinder:

  • First, we need to determine the radius of the base using the circumference.
  • The formula for circumference (C) is: C = 2πr, where r is the radius.
  • Given the circumference is 132 cm, we can rearrange the formula to find r:
  • r = C / (2π)
  • Substituting the values: r = 132 / (2π) ≈ 21 cm.
  • Next, we calculate the volume (V) of the cylinder using the formula:
  • V = πr²h, where h is the height.
  • We know that the height (h) is 25 cm. Now, substitute the values:
  • V ≈ π(21)²(25).
  • Calculating further:
  • V ≈ π(441)(25) ≈ π(11025) ≈ 34650 cm³.

The volume of the cylinder is approximately 34650 cm³.

Test: Surface Area & Volumes- 2 - Question 25

The number of solid sphere each 6 cm in diameter, which can be moulded from a solid cylinder of height 45 cm and diameter 4 cm without any loss is

Detailed Solution for Test: Surface Area & Volumes- 2 - Question 25

To determine how many solid spheres can be moulded from a solid cylinder, follow these steps:

  • Calculate the volume of the cylinder:
    • Diameter of the cylinder = 4 cm, so radius = 2 cm.
    • Height of the cylinder = 45 cm.
    • Volume = πr²h = π(2)²(45) = 180π cm³.
  • Calculate the volume of one sphere:
    • Diameter of the sphere = 6 cm, so radius = 3 cm.
    • Volume = (4/3)πr³ = (4/3)π(3)³ = 36π cm³.
  • Determine the number of spheres:
    • Number of spheres = Volume of cylinder / Volume of sphere.
    • Number of spheres = 180π / 36π = 5.

The number of solid spheres that can be moulded from the cylinder is 5.

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