# Mensuration 3D | Set-2

Last Updated : 17 Apr, 2024

Question 1: If the diagonal of cube is âˆš18 cm, then its volume isÂ
Solution : Let the side of the cube is x cm.Â
We know diagonal of cube = aâˆš3 cmÂ
Put equal bothÂ
aâˆš3 = âˆš18Â
Squaring both sidesÂ
a2(3) = 18Â
a2 = 6Â
a = âˆš6Â
Volume of the cube = a3Â
= (âˆš6)3Â
= 6âˆš6 cm3Â

Question 2: The areas of three consecutive faces of a cuboid are 27 cm2, then the volume of the cuboid isÂ
Solution : Let the three sides of the cuboid is l, b and h.Â
=>lb = bh = hl = 27Â
=>l2b2h2 = 27 x 27 x 27Â
=>lbh = 27âˆš27Â
=>lbh = 81âˆš3Â
Hence, volume of the cuboid is 81âˆš3 cm3.Â

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?Â
Solution : Area of square land = 1000 x 1000Â
Volume of rain in square land = 1000 x 1000 x (4/100)Â
Only 60% of water is collected and contained in a box.Â
(1000 x 1000 x 4/100)x(60/100)Â
Let a is the water level which increased.Â
200 x 20 x a = (1000 x 1000 x 4/100)x(60/100)Â
4000a = 40000 x 6/10Â
4000a = 24000Â
a = 6 mÂ
Hence, the water level is increased by 6 m in the container.Â

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 isÂ
Solution :Let length = 4cmÂ
Height = h cmÂ
Total surface areaÂ
2(lb + bh + hl) = 88Â
24 + 6h + 4h = 44Â
10h = 20Â
h = 2cmÂ
Diagonal = âˆš(42 + 62 + 22)Â
= 2âˆš14 cmÂ

Question 5: A largest possible rod of length 70âˆš3 cm can be placed in a cubical room.The surface area of the largest possible sphere that fit within the cubical room is (Ï€ =22/7) isÂ
Solution : Diagonal of the cube = sideâˆš3Â
sideâˆš3 = 70âˆš3Â
side= 70 cmÂ
For largest sphere, diameter of the sphere = side of the cubeÂ
Surface area of the sphere = 4 Ï€ r2Â
=> 4 x 22/7 x 35 x 35Â
=> 15400 cm3Â

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:Â
Solution : When we change shape volume remains constant.Â
Volume of the hemisphere = 2/3 Ï€ r3Â
Volume of the cone = 1/3 Ï€ r2 hÂ
So, 2/3 Ï€ r3 = 1/3 Ï€ r2 hÂ
2 r = hÂ
h = 6 cmÂ
Hence, height of the cone is 6cm.Â

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 :Â
Solution : Let the radius of the sphere before change is a cm.Â
Acc. to questionÂ
4Ï€(a+4)2 – 4Ï€a2 = 704Â
4Ï€{(a+4)2 – a2} = 704Â
4Ï€{(a2 + 8a + 16 – a2} = 704Â
4Ï€{8a + 16} = 704Â
Ï€(a + 2) = 22Â
22/7(a + 2) = 22Â
a + 2 = 7Â
a = 5 cmÂ
Hence, the radius of sphere before change is 5 cm.Â

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. cm. isÂ
Solution :Â
Â

We have to find slant height(l) of the cone.Â
Radius of cone = 32/2 = 16 cmÂ
l = âˆš(r2 + h2)Â
l = âˆš(162 + 122)Â
l = âˆš400 = 20 cmÂ
Cost of painting = Surface area of cone x 21Â
= Ï€rl x 21Â
= 22/7 x 16 x 20 x 21Â
= 21120
Hence, the cost of painting is Rs 21120.Â

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:Â
Solution : Initial height = HÂ
Final height = hÂ
Volume of cylinder = Ï€(r)2hÂ
Acc. to questionÂ
Ï€(21)2x H – Ï€(21)2x h = 13860 cm3Â
Ï€(21)2x (H – h) = 13860Â
22/7 x 21 x 21 x (H – h) = 13860Â
H – h = (13860 x 7) / (21 x 21 x 22)Â
H – h = 10 cmÂ
Hence, water level is drop by 10 cm.Â

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?Â
Solution : Let the radii are R and r.Â
R + r = 20Â
(R + r)2 = 400Â
R2 + r2 + 2Rr = 400Â
R2 + r2 = 400 – 2RrÂ
Sum of volumesÂ
4/3 Ï€ R3 + 4/3 Ï€ r3 = 1760Â
4/3 Ï€ (R3 + r3) = 1760Â
(R + r)(R2 + r2 – Rr) = 1760 x 7/22 x 3/4Â
20(400 – 2Rr – Rr) = 420Â
400 – 3Rr = 21Â
3Rr = 379Â
Rr = 379/3Â
Hence, product of the radii is 379/3.Â

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 isÂ
Solution :Â
Total surface area of the hemisphere = 3 Ï€r2Â
3 Ï€r2 = 462Â
r2 = 49Â
r = 7 cmÂ
Acc. to questionÂ
2/3 Ï€r3= 1/3 Ï€ r2hÂ
2r = hÂ
h = 2 x 7 = 14Â
Hence, the height of the cone is 14 cm.Â

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?Â
Solution :Diameter of the beaker = 14 cmÂ
Radius of the beaker = 7 cmÂ
Level of water rises by 4.9 cmÂ
Diameter of a marble = 1.4 cmÂ
Let n marbles dropped.Â
So, volume of n marbles dropped = n x 4/3 Ï€ (0.7)3Â
=> n x 4/3 Ï€ (0.7)3 = Ï€ (7)2x4.9Â
=> n x 4/3 x 7/10 x 7/10 x 7/10 = 7 x 7 x 7 x 7/10Â
=> n = 2100/4Â
=> n = 525Â
Hence, 525 marbles are dropped in water.Â

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 isÂ
Solution : Let the time taken to fill is a hours.Â
Volume of water transferred through pipe in a hours is equal to Volume of water in the rectangular tank.Â
Ï€ r2 h x a = 100 x 44 x 14/100Â
22/7 x 14/100 x 14/100 x 10000 x a = 100 x 44 x 14/100Â
a = (100 x 44 x 14 x 7 x 100 x 100) / (22 x 14 x 14 x 10000 x 100)Â
a = 1 hour
Hence, time taken is 1 hour.Â

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 isÂ
Solution : Let the radius of small spheres is 3a, 4a and 5a.Â
Volume of sphere =4/3 Ï€ r3Â
Acc . to questionÂ
4/3 Ï€ 63 = 4/3Ï€{(3a)2 + (4a)2 + (5a)2}Â
216 = 27a3 + 64a3 + 125a3Â
216 = 216a3Â
a = 1Â
Hence, Radius of largest sphere is 5×1 = 5 cm.Â

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 isÂ
Solution : Let the radius of sphere is r.Â
Let the height of the cylinder is h.Â
Given volume of sphere = Volume of cylinderÂ
4/3 Ï€ r3 = Ï€ r2 hÂ
=> 4/3 r = hÂ
Curved surface areaÂ
Â

`  Cylinder         Sphere  2Ï€rh        :    4Ï€r2  2Ï€r4r/3     :    4Ï€r2  8/3         :     4   2          :     3`

Â

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