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Mensuration 3D | Set-2

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  • Difficulty Level : Hard
  • Last Updated : 04 Mar, 2022

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 
a2(3) = 18 
a2 = 6 
a = √6 
Volume of the cube = a3 
= (√6)3 
= 6√6 cm3 

Question 2: The areas of three consecutive faces of a cuboid are 27 cm2, then the volume of the cuboid is 
Solution : Let the three sides of the cuboid is l, b and h. 
=>lb = bh = hl = 27 
=>l2b2h2 = 27 x 27 x 27 
=>lbh = 27√27 
=>lbh = 81√3 
Hence, volume of the cuboid is 81√3 cm3

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 cm2, 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 = √(42 + 62 + 22
= 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 r2 
=> 4 x 22/7 x 35 x 35 
=> 15400 cm3 

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 r3 
Volume of the cone = 1/3 π r2
So, 2/3 π r3 = 1/3 π r2
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 cm2. 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 – 4πa2 = 704 
4π{(a+4)2 – a2} = 704 
4π{(a2 + 8a + 16 – a2} = 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 = √(r2 + h2
l = √(162 + 122
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
Acc. to question 
π(21)2x H – π(21)2x h = 13860 cm3 
π(21)2x (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 cm3. What will be the product of their radii? 
Solution : Let the radii are R and r. 
R + r = 20 
(R + r)2 = 400 
R2 + r2 + 2Rr = 400 
R2 + r2 = 400 – 2Rr 
Sum of volumes 
4/3 π R3 + 4/3 π r3 = 1760 
4/3 π (R3 + r3) = 1760 
(R + r)(R2 + r2 – 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 cm2. 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 πr2 
3 πr2 = 462 
r2 = 49 
r = 7 cm 
Acc. to question 
2/3 πr3= 1/3 π r2
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)2x4.9 
=> 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 transferred through pipe in a hours is equal to Volume of water in the rectangular tank. 
π r2 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 r3 
Acc . to question 
4/3 π 63 = 4/3π{(3a)2 + (4a)2 + (5a)2
216 = 27a3 + 64a3 + 125a3 
216 = 216a3 
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 r3 = π r2
=> 4/3 r = h 
Curved surface area 
 

  Cylinder         Sphere
  2πrh        :    4πr2
  2πr4r/3     :    4πr2
  8/3         :     4
   2          :     3

 


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