Q1: A closed box measures 66 cm, 36 cm and 21 cm from outside. If its walls are made of metal-sheet, 0.5 cm thick, then find the weight of the box, if 1 cm of metal weighs 3.6 g.
(a) 17985.6 g
(b) 15725.6 g
(c) 13625.6 g
(d) 15825.6 g
Ans: (d)
Given, The external dimensions of a closed wooden box are 66cm,36cm and 21cm respectively.
As the thickness is 0.5cm , the internal length, breadth and height of the closed box are 66−2(0.5) = 65cm, 36−2(0.5) = 35cm and 21−2(0.5) = 20cm, respectively.
Now, volume of wood used to make the box = Volume with the external dimensions − Volume with the internal dimensions ⇒ Volume of wood used to make the box (66 × 36 × 21) − (65 × 35 × 20) = 49896 − 45500 = 4396cucm
So, weight of the wood = 3.60 × 4396 = 15825.6g
Q2: What length of tarpaulin 3m wide will be required to make a conical tent of height 8 m and base radius 6 m? Assume that the extra length of material will be required for stitching margins and wastage in cutting is approximately 20 cm. (Use π = 3.14).
Ans: We have,
h = Height of tent = 8m, r = Base radius=6m, width of tarpaulin = 3m
∴ slant height = 1 =
Curved surface area of the cone=πrl = 3.14 × 6 × 10m2
We know that, curved surface area = 3.14 × 6 × 10m2 = Length of tarpaulin X width of tarpaulin
∴ Length of 3m wide tarpaulin required =
Extra length required = 0.2,
∴ Total length required = 63.00m.
Q3: A rectangular sheet of paper 44 cm × 18 cm is rolled along its length and a cylinder is formed. Find the radius of the cylinder.
Ans: When the rectangular sheet is rolled along its length, we find the length of the sheet forms the circumference of its base and breadth of the sheet becomes the height of the cylinder.
let r cm be the radius of the base and h cm can be height, then h = 18 cm
Now, circumference of the base = length of the sheet
⇒ 2πr = 44
hence radius of the cylinder is 7 cm.
Q4: A cylinder of radius 12 cm contains water upto the height of 20 cm. A spherical iron ball is dropped into the cylinder and thus water level is raised by 6.75 cm. what is the radius of the ball?
Ans: Given the radius of the cylinder, r = 12 cm.
It is also given that a spherical iron ball is dropped into the cylinder and the water level raised by 6.75 cm.
Hence volume of water displaced = volume of the iron ball
Height of the raised water level, h = 6.75 m
Volume of water displaced = πr2h
= π × 12 × 12 × 6.75 cm3
⇒ Volume of iron ball = π × 12 × 12 × 6.75 cm3 → (1)
But, volume of iron ball = 4/3 πr3 ---(2)
From (1) and (2) we get
4/3 πr3 = π × 12 × 12 × 6.75 r3
= π × 12 × 12 × 6.75 × 3/4
r3 = 729
r3 = 33
r = 9
Thus the radius of the iron ball is 9 cm.
Q5: A solid metallic sphere of diameter 21 cm is melted and recast into a number of smaller cones, each of diameter 3.5 cm and height 3 cm. Find the number of cones so formed.
Ans: Diameter of sphere = 21cm
Radius = 21/2 cm
Height of the cone = 3cm
Radius of the cone = 7/4 cm
Volume of the sphere =
Let n be the number of cone formed,
then
n=(volume of the sphere)/(volume of cone)
Number of cones formed = 504.
Q6: Water is pouring into a cubiodal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108m3, find the number of hours it will take to fill the reservoir.
Ans: Given that the reservoir is cuboidal in shape.
The volume of the reservoir = 108 m³ = 108 × 1000 litres = 108000 litres [Since 1 cubic meter = 1000 litre]
The volume of water flowing into the reservoir in 1 minute = 60 L
The volume of water pouring in the reservoir 1 hour = (60 × 60) Litres per hour = 3600 litres / hour
Thus, the required number of hours to fill the reservoir = Volume of the reservoir / Volume of water pouring in the reservoir 1 hour
108000 / 3600 hours = 30 hours
Thus, the number of hours it will take to fill the reservoir is 30 hours.
Q7: The internal and external radii of a metallic spherical shell are 4 cm and 8 cm, respectively. It is melted and recast into a solid right circular cylinder of height 9(1/3) cm. Find the diameter of the base of the cylinder.
(a) 16 cm
(b) 18 cm
(c) 12 cm
(d) 14 cm
Ans: (a)
Since the spherical shell is recasted into a cylinder,
Volume of the hollow sphere = Volume of the cylinder
Volume of a hollow sphere of outer Radius R and inner radius r =
Volume of a Cylinder of Radius "R" and height " "h" =πR2h
Hence, diameter of the base of the cylinder = 2 × 8 = 16 cm
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