**Question 1:** If the diagonal of cube is √18 cm, then its volume is

**Solution :** Let the side of the cube is x cm.

We know diagonal of cube = a√3 cm

Put equal both

a√3 = √18

Squaring both sides

a^{2}(3) = 18

a^{2} = 6

a = √6

Volume of the cube = a^{3}

= (√6)^{3}

= 6√6 cm^{3}

**Question 2:** The areas of three consecutive faces of a cuboid are 27 cm^{2}, then the volume of the cuboid is

**Solution :** Let the three sides of the cuboid is l, b and h.

=>lb = bh = hl = 27

=>l^{2}b^{2}h^{2} = 27 x 27 x 27

=>lbh = 27√27

=>lbh = 81√3

Hence, volume of the cuboid is 81√3 cm^{3}.

**Question 3:** 4cm of rain has fallen on a square km of land. Assuming that 60% of the raindrops could have been collected and contained in a pool having a 200m x 20m base, by what level would the water level in the pool have increased?

**Solution :** Area of square land = 1000 x 1000

Volume of rain in square land = 1000 x 1000 x (4/100)

Only 60% of water is collected and contained in a box.

(1000 x 1000 x 4/100)x(60/100)

Let a is the water level which increased.

200 x 20 x a = (1000 x 1000 x 4/100)x(60/100)

4000a = 40000 x 6/10

4000a = 24000

a = 6 m

Hence, the water level is increased by **6 m** in the container.

**Question 4:** If two adjacent sides of a rectangular parallelopiped are 4cm and 6cm and the total surface area of parallelopiped is 88 cm^{2}, then the diagonal of the parallelopiped is

**Solution :**Let length = 4cm

Breadth = 6cm

Height = h cm

Total surface area

2(lb + bh + hl) = 88

24 + 6h + 4h = 44

10h = 20

h = 2cm

Diagonal = √(4^{2} + 6^{2} + 2^{2})

= 2√14 cm

**Question 5:** A largest possible rod of length 70√3 cm can be placed in a cubical room.The surface area of the largest possible sphere that fit within the cubical room is (π =22/7) is

**Solution :** Diagonal of the cube = side√3

side√3 = 70√3

side= 70 cm

For largest sphere, diameter of the sphere = side of the cube

2 x radius = 70

radius = 35

Surface area of the sphere = 4 π r^{2}

=> 4 x 22/7 x 35 x 35

=> 15400 cm^{3}

**Question 6:** A metallic hemisphere is melted and recast in the shape of cone with the same base radius 3 cm as that of the hemisphere. If H is the height of the cone is:

**Solution :** When we change shape volume remains constant.

Volume of the hemisphere = 2/3 π r^{3}

Volume of the cone = 1/3 π r^{2} h

So, 2/3 π r^{3} = 1/3 π r^{2} h

2 r = h

h = 6 cm

Hence, height of the cone is **6cm**.

**Question 7:** If the radius of the sphere is increased by 4 cm, its surface area increased by 704 cm^{2}. The radius of the sphere before change is :

**Solution :** Let the radius of the sphere before change is a cm.

Acc. to question

4π(a+4)^{2} – 4πa^{2} = 704

4π{(a+4)^{2} – a^{2}} = 704

4π{(a^{2} + 8a + 16 – a^{2}} = 704

4π{8a + 16} = 704

π(a + 2) = 22

22/7(a + 2) = 22

a + 2 = 7

a = 5 cm

Hence, the radius of sphere before change is **5 cm**.

**Question 8:** The height of a conical tank is 12cm and the diameter of its base is 32cm. The cost of painting if from outside at the rate of Rs 21 per sq. m. is

**Solution :**

We have to find slant height(l) of the cone.

Radius of cone = 32/2 = 16 cm

l = √(r^{2} + h^{2})

l = √(16^{2} + 12^{2})

l = √400 = 20 cm

Cost of painting = Surface area of cone x 21

= πrl x 21

= 22/7 x 16 x 20 x 21

= 21120

Hence, the cost of painting is **Rs 21120**.

**Question 9:** A cylindrical tank of radius 21 cm is full of water. If 13.86 litres of water is drawn off, the water level in the tank will drop by:

**Solution :** Initial height = H

Final height = h

Volume of cylinder = π(r)^{2}h

Acc. to question

π(21)^{2}x H – π(21)^{2}x h = 13860 cm^{3}

π(21)^{2}x (H – h) = 13860

22/7 x 21 x 21 x (H – h) = 13860

H – h = (13860 x 7) / (21 x 21 x 22)

H – h = 10 cm

Hence, water level is drop by **10 cm**.

**Question 10:**The sum of the radii of two spheres is 20 cm and the sum of their volume is 1760 cm^{3}. What will be the product of their radii?

**Solution :** Let the radii are R and r.

R + r = 20

(R + r)^{2} = 400

R^{2} + r^{2} + 2Rr = 400

R^{2} + r^{2} = 400 – 2Rr

Sum of volumes

4/3 π R^{3} + 4/3 π r^{3} = 1760

4/3 π (R^{3} + r^{3}) = 1760

(R + r)(R^{2} + r^{2} – Rr) = 1760 x 7/22 x 3/4

20(400 – 2Rr – Rr) = 420

400 – 3Rr = 21

3Rr = 379

Rr = 379/3

Hence, product of the radii is **379/3**.

**Question 11:** The total surface area of a metallic hemisphere is 462 cm^{2}. A solid right circular cone is formed after melting the hemisphere. If the radius of the base of the cone is same as the radius of the hemisphere its height is

**Solution :**

Total surface area of the hemisphere = 3 πr^{2}

3 πr^{2} = 462

r^{2} = 49

r = 7 cm

Acc. to question

2/3 πr^{3}= 1/3 π r^{2}h

2r = h

h = 2 x 7 = 14

Hence, the height of the cone is **14 cm**.

**Question 12:** Balls of the marbles of diameter 1.4 cm diameter are dropped into a cylindrical beaker containing some water and fully submerged. The diameter of the beaker is 14cm. Find how many marbles have been dropped in it if the water rises by 4.9 cm?

**Solution :**Diameter of the beaker = 14 cm

Radius of the beaker = 7 cm

Level of water rises by 4.9 cm

Diameter of a marble = 1.4 cm

Radius of marble= 0.7 cm

Let n marbles dropped.

So, volume of n marbles dropped = n x 4/3 π (0.7)^{3}

=> n x 4/3 π (0.7)^{3} = π (7)^{2}x4.9

=> n x 4/3 x 7/10 x 7/10 x 7/10 = 7 x 7 x 7 x 7/10

=> n = 2100/4

=> n = 525

Hence, **525 marbles** are dropped in water.

**Question 13:** Water is flowing at the rate of 10km/h through a pipe of diameter 28 cm into a rectangular tank which is 100m long 44m wide. The time taken for the rise in the level of water in the tank be 14 cm is

**Solution :** Let the time taken to fill is a hours.

**Volume of water transfered through pipe in a hours is equal to Volume of water in the rectangular tank.**

π r^{2} h x a = 100 x 44 x 14/100

22/7 x 14/100 x 14/100 x 5000 x a = 100 x 44 x 14/100

a = (100 x 44 x 14 x 7 x 100 x 100) / (22 x 14 x 14 x 5000 x 100)

a = 2 hours

Hence, time taken is **2 hours**.

**Question 14:** By melting a solid lead sphere of radius 6 cm, three small spheres are made whose radii are in ratio 3:4:5. The radius of the largest sphere is

**Solution :** Let the radius of small spheres is 3a, 4a and 5a.

Volume of sphere =4/3 π r^{3}

Acc . to question

4/3 π 6^{3} = 4/3π{(3a)^{2} + (4a)^{2} + (5a)^{2}}

216 = 27a^{3} + 64a^{3} + 125a^{3}

216 = 216a^{3}

a = 1

Hence, Radius of largest sphere is 5×1 = 5 cm.

**Question 15:** A sphere and a cylinder have equal volume and equal radius. The ratio of the curved surface area of the cylinder to that of the sphere is

**Solution :** Let the radius of sphere is r.

Let the height of the cylinder is h.

Given volume of sphere = Volume of cylinder

4/3 π r^{3} = π r^{2} h

=> 4/3 r = h

Curved surface area

Cylinder Sphere 2πrh : 4πr^{2}2πr4r/3 : 4πr^{2}8/3 : 4 2 : 3

## Recommended Posts:

- Mensuration 2D
- Mensuration 3D
- Mensuration 2D | Set 2
- Profit and loss | Set-2
- Permutation and Combination | Set-2
- Probability | Set-2
- Progressions (AP,GP, HP) | Set-2
- Compound Interest | Set-2
- Simple Interest | Set-2
- Algebra | Set-2
- Trigonometry & Height and Distances | Set-2
- Mixture and Alligation | Set 2
- Mensuration 2D | Set 2
- Ratio proportion and partnership | Set-2