Take , unless stated otherwise.
Volume of the sphere
If h is theÂ height of a cylinder of radiusÂ 6 cm. Then its volume,
Since, the volume of metal in the form of sphereÂ and cylinder remains the same , we have
Â Â Â
Â Â Â h = 2.744
Q.2Â Â Â Â Â MetallicÂ spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. FindÂ the radius of the resulting sphere.
Sol.
Sum of the volumes of 3 gives spheres.
Let R beÂ the radius of the new spheres whose volume is the sum of theÂ volumes of 3 given spheres.
Therefore Â Â Â Â
Â Â Â
Â Â Â
Â Â Â R = 12
Hence, theÂ radius of theÂ resulting sphere is 12 cm.
Therefore 1728 = 2^{3} Ã— 6^{3} = (2 Ã— 6)^{3} = 12)^{3}
Q.3Â Â Â Â Â A 20 cm deep well withÂ diameter 7 cm is dug and theÂ earthÂ fromÂ digging is evenly spread out to form a platform 22 m by 14 m. Find theÂ height of the platform.
Sol.Â
Let h m be theÂ required heightÂ of the platform.
The shape of the platform will be like the shape of a cuboid 22 m Ã— 14 m Ã— h with a holeÂ in theÂ shape of cylinder of radiusÂ 3.5 m and depth h m.
The volumeÂ of the platform will be equal to theÂ volumeÂ of theÂ earthÂ dug out fromÂ the well.
Now, theÂ volumeÂ of the earth = VolumeÂ of the cylindrical well
Also, theÂ volume of the platform = 22 Ã— 14 Ã— h
But volume of the platform = Volume of theÂ well
i.e., Â Â Â Â 22 Ã— 14 Ã— h = 770
ThereforeÂ Height of the platform = 2.5 m
Q.4Â Â Â Â A well of diameter 3 m is dug 14 m deep. The earth taken out of it hasÂ been spread evenly all around it in the shape of a circularÂ ringÂ of width 4 m to form an embankment. Find the height of the embankment.
Sol.
Let h be the required height of the embankment.
The shape of the embankment will be like the shape of a cylinder of internal radius 1.5 m andÂ external radius (4 + 1.5)m = 5.5 m (see figure).
TheÂ volume of the embankment will be equalÂ to the volume of earthÂ dug out from theÂ well.
Now,Â the volume of the earth = VolumeÂ of the cylindrical well
AlsoÂ the volume of the embankment
Hence, we have
Â Â Â Â Â Â Â Â Â Â
Â Â Â
Hence, the required heightÂ of theÂ embankment = 1.125 m
Q.5Â Â Â Â A container shaped like a rightÂ circularÂ cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to beÂ filled into cones of height12 cm and diameter 6 cm, havingÂ hemispherical shape on the top. Find the number of suchÂ cones whichÂ can beÂ filled with ice cream.
Sol.
Volume of the cylinder
VolumeÂ of a cone havingÂ hemispherical shape on the top
Let the number of cone that can be filled with ice cream be h.
Then
Q.6 How many silverÂ coins, 1.75 cm in diameter and ofÂ thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm Ã— 10 cm Ã— 3.5 cm?
Sol. The shape of the coin will be like the shape of a cylinder of radius
= 0.875 cm and of height 2mm
Its volume Â
VolumeÂ of the cuboid
NumberÂ of coinsÂ required to form the cuboid
Hence, 400Â coins must be melted to form a cuboid.
The shape of the coin will be like the shape of a cylinder of radius
= 0.875 cm and of height 2mm
Its volume Â
VolumeÂ of the cuboid
NumberÂ of coinsÂ required to form the cuboid
Hence, 400Â coins must be melted to form a cuboid.
Q.7Â Â Â Â A cylindricalÂ bucket, 32 m highÂ and withÂ radius of base 18 cm, is filled with sand. The bucketÂ is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radiusÂ and slant heightÂ of the heap.
Sol.
Volume of theÂ sand = Volume of the cylindrical bucket
Volume of the conical heap
, whereÂ r = ? , h = 24 cm
The volumeÂ of the conicalÂ heap will be equalÂ to thatÂ of sand.
Therefore Â Â Â Â
Â Â Â
Â Â Â r = 18 Ã— 2 = 36
Here, slant height Â Â Â
Â Â Â
Â Â Â
Hence, the radius of the conical heap is 36 cm and its slant height is
Q.8Â Â Â Â Water in a canal 6 m widÂ and 1.5 m deep, is flowing with a speed of 10 km/h. HowÂ muchÂ area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Sol.
WidthÂ of theÂ canal = 6 m
Depth of the canal = 1.5 m
Length of water column per hour = 10 km
Length of waterÂ column in 30 minutes or hour
Volume of water flown in 30 minutes
= 1.5 Ã— 6 Ã— 5000
Since
i.e., 0.08 m standing water is desired
ThereforeÂ Area irrigated in 30 minutes
Q.9 A farmer connects a pipe of internal diameter 20 cm fromÂ a cannal into a cylindrical tank in his field, which is 10 m in diameter and 2m deep. If water flows through the pipe at theÂ rate of 3 km/h, in how much timeÂ will the tank be filled?
Sol.
Diameter of the pipe = 20 cm
Â Â Â Radius of the pipe = 10 cm
Length of water column per hour
= 3 km = 3 Ã— 1000 Ã— 100 cm
Volume of water flown in one hour
Tank to beÂ filled = VolumeÂ of cylinder (withÂ r = 5m = 500 cm and h = 2m = 200 cm)
Time required to fillÂ the tank
= 1 hour 40 minutes
= 60 + 40 minutes = 100 minutes.