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1.

Find the amount of water displaced by a solid spherical ball of diameter 4.2 cm , when it is completely immersed in water.

Answer» Given, diameter of a solid spherical ball = 42 cm
`therefore` Radius of a solid spherical ball, `r =(42)/(2) = 2.1 cm`
Now, volume of water displaced by a solid spherical ball when it is csompleley immersed in water= Volume of a solid spherical ball
`= (4)/(3) pir^(3) = (4)/(3) xx (22)/(7) xx (2.1)^(3)`
`= (4)/(3) xx (22)/(7) xx2.1 xx2.1 xx2.1`
`=(814.968)/(21) = 38.808 = 38.81 cm^(3)`
Hence, the volume of water displaced by a solid spherical ball, when it is
Discussion
2.

A colth having an area of `165 m^(2)`is shapped into the form of a conical tent of radius 5 m. (i) How many students can sit in the tent, if a student on an average occupies `(5)/(7)m^(2)` on the ground ? (ii) Find the volume of the cone.

Answer» (i) Given, radius of the base of conical tent = 5 m
and area needs to sit a student on the ground `= (5)/(7)m^(2)`
`therefore` Area of the base of a conical tent `= pir^(2)`.
`= (22)/(7) xx 5 xx 5m^(2)`
Now, number of students `= ("Area of the bases of a conical tent")/("Area needs to sit a student on the groud")`
`= ((22xx5xx5)/(7))/(5//7) = (22)/(7) xx 5 xx 5xx(7)/(5) = 110`
Hence, 110 students can sit in the conical tent .
(ii) Given, area of the form a conical tent `= 165m^(2)`
Radius of the base of a conical tent , r = 5 m
Curved surface area of the = Area of cloth to from a conical tent
`rArr" "pirl = 165`
`rArr" "(22)/(7)xx (5) xx l = 165`
`therefore" " l = (165 xx 7)/(22 xx5) = (33xx7)/(22) = 10.5m`
Now, height of a conical tent = `sqrt(l^(2) - r^(2)) = sqrt((10.5)^(5) - (5)^(2))`
` = sqrt(110.25 - 25 ) = sqrt(8528) = 9.23m`
Volume of a cone (conical tent) `=(1)/(3) pir^(2)h = (1)/(3) xx (22)/(7) xx 5 xx 5 xx 923`
` = (1)/(3) x (1550xx 923)/(7) = (50765)/(7xx3) = 241.7m^(3)`
Hence, the volume of the (conical tent ) is `241.7m^(3)`
Discussion
3.

How many square metres of canvas is required for a conical tent whose height is 3.5 m and the base is 12 m?

Answer» Given height of a conical tent, h = 3.5 m and radius of the base of a conical tent , r= 12m
Slant, height , `l = sqrt(h^(2) + r^(2)) = sqrt((3.5)+ (12)^(2))`
`" "=sqrt(12.25 + 144)`
`= sqrt(15625)= 12.5m`
`therefore` Canvas required = Curved surface area of the cone (conical tent)
`= pirl =(22)/(7) xx 12 xx12.5`
`= 471.42m^(2)`
Hence, the canvas required to makes a conical tent is `471.42m^(2)`
Discussion
4.

If the radius of a sphere is 2r , them its volume will beA. `(4)/(3)pir^(3)`B. `4pir^(3)`C. `(8pir^(3))/(3)`D. `(32)/(3) pir^(3)`

Answer» Correct Answer - D
Given radius of a sphere = 2r
Volume of a sphere ` = (4)/(3) pi ("Radius")^(3)`
`= (4)/(3) pi (2r)^(3) = (4)/(3)pi.8r^(3)`
`= (32pir^(3))/(3) "cu units"" "[because "radius =2r"]`
Hence, the volume of a sphere is `(32pir^(3))/( 3)` cu units.
Discussion
5.

A small village having a population of 5000, requires 75 L of water per head per day . The village has got an overhead tank of measurement `40 mxx 25 mxx 15 m`.For how days will the water of this tank last?

Answer» Given, total population of a small village = 5000
Water required per head per day `= 5000 xx 75 = 375000L`
`= (375000L)/(1000)m^(3) = 375m^(3)" "[because 1m^(3) = 1000L]`
Total capacity of water in overhead tank = Volume of overhead tank
`= 40 xx 25 xx 15 = 15000m^(3)`
`therefore` Number of days `=("Total capacity of water in over spead tank")/("Volume of water required for a small village per day")`
`= (15000)/(375) = 40` days
Hence, water of this tank will be last in 40 days.
Discussion
6.

A school provides milk to the studenst daily in a cylinder glasses of diameter 7 cm. If the glass is filled with milk upto an height of 12 cm, find how many liters of milk is needed to server 1600 students.

Answer» Given, diameter of glass = 7 cm
Radius of glass, `r = (7)/(2) cm`
`therefore` Milk contained in the cyllindrical glass = volume of cylindrical glass
` pir^(2)h = (22)/(7) xx (7)/(2) xx (7)/(2) xx 12 = 462 cm^(3)`
Now, milk required for 1600 students `= 462 xx 1600 = 739200 cm^(3) `
`" "(739200)/(1000) = 7392 L" "[because 1 cm^(3) =(1)/(1000)L]`
Hence, 739.2 L milk is needed to serve 1600 students.
Discussion
7.

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

Answer» Given,side of a cube = 4cm
Side of cube = Diameter of sphere
4 = Diameter of sphere
Radius of sphere = `(4)/(2)= 2 cm `
Volume of the gap = Volume of cube - Volume of sphere
`= ("side")^(3) - (4)/(3)pir^(3)`
`= (4)^(3) - (4)/(3) pi(2)^(2)" " [because "side of cube" = "diameter of sphere"]`
` = (64 - (4)/(3) xx (22)/(7) xx 8) = 64- 33.52 = 30.48cm^(3)`
Hence, the volume of the gap in between a cube and a sphere is `30.48 cm^(3)`.
Discussion
8.

A sphere and a right circule cylinder of the same radius have equal volume. By what percentage does the diameter of the cylinder exceed its height ?

Answer» Let the radius of sphere= r= Radius of a right circular cylinder
According to the question,
Volume of cylinder= volume of a sphere
`rArr " "pir^(2) h = (4)/(3) pir^(3) rArr h = (4)/(3) r`
`because` Diameter of the cylinder = 2r
`therefore`Inreased diameter from height of the cylinder `= 2r-(4r)/(3) = (2r)/(3)`
Now,percentage increase in diameter of the cylinder `= (((2r)/(3) xx100))/((4)/(3)r) = 50%`
Hence, the diameter of the cylinder exceeds its height by 50%.
Discussion
9.

If theradii of two cylinders are in the ratio 2:3 and their heights are in theratio 5:3, then find the ratio of their volumes.A. `10:17`B. `20:27`C. `17: 27`D. `20: 37`

Answer» Correct Answer - B
Let the radii of two cylinders be `r_(1)` and `r_(2)` and height of the cylinder be `h_(1)` and `h_(2)`
Given , `" "(r_(1))/(r_(2)) =(2)/(3) and (h_(1))/(h_(2)) = (5)/(3)`
`therefore`Ratio of volumes `(pir_(2)^(1) h_(1))/(pir_(2)^(2) h_(2)) =((r_(1))/(r_(2)))^(2)((h_(1))/(h_2))`
`= ((2)/(3))^(2)((5)/(3)) =(4)/(9) xx (5)/(3) = 20: 27`
Hence, the radius of their volumes is 20: 27
Discussion
10.

If the radius of a cylinder is doubled and its curved surface area is not changed, the height must be halved.

Answer» True
Let the radius of a cylinder be r and height be h.
Curbved surface area of a cylinder `= 2pirh`
Again, let it the radius of a cylinder be r = 2r and height be H.
Curved surface of a cylinder is not charged, therefore .
` 2 pi rh = 2 pi (2r) xxH`
`rArr " "2pirh = 4pirh `
`rArr " " H= (2pirh)/(4pir)rArr H =(h)/(2)`
Hence, the height must be halved.
Discussion