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A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per .
It can be observed that 1 round of wire will cover 3 mm height of cylinder.
Length of wire required in 1 round = Circumference of base of cylinder
= 2πr = 2π × 5 = 10π
Length of wire in 40 rounds = 40 × 10π
= 1257.14 cm = 12.57 m
Radius of wire 
Volume of wire = Area of cross-section of wire × Length of wire
= π(0.15)2 × 1257.14
= 88.898 cm3
Mass = Volume × Density
= 88.898 × 8.88
= 789.41 gm
A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of as found appropriate.)
The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse is shown in the figure.
Hypotenuse
= 5 cm
Area of ΔABC
Volume of double cone = Volume of cone 1 + Volume of cone 2
= 30.14 cm3
Surface area of double cone = Surface area of cone 1 + Surface area of cone 2
= πrl1 + πrl2
= 52.75 cm2
A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm * 7.5 cm * 6.5 cm?
Volume of cistern = 150 × 120 × 110
= 198000 cm3
Volume to be filled in cistern = 198000 − 129600
= 185040 cm3
Let n numbers of porous bricks were placed in the cistern.
Volume of n bricks = n × 22.5 × 7.5 × 6.5
= 1096.875n
As each brick absorbs one-seventeenth of its volume, therefore, volume absorbed by these bricks 
n = 1792.41
Therefore, 1792 bricks were placed in the cistern.
In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 7280 , show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.
Area of the valley = 7280 km2
If there was a rainfall of 10 cm in the valley then amount of rainfall in the valley = Area of the valley × 10 cm
Amount of rainfall in the valley = 7280 km2 × 10 cm
Length of each river, l = 1072 km = 1072 × 1000 m = 107200 m
Breadth of each river, b = 75 m
Depth of each river, h = 3 m
Volume of each river = l × b × h
= 107200 × 75 × 3 m3
= 2.412 × 108 m3
Volume of three such rivers = 3 × Volume of each river
= 3 × 2.412 × 108 m3
= 7.236 × 108 m3
Thus, the total rainfall is approximately same as the volume of the three rivers.
An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see the figure).
Radius (r1) of upper circular end of frustum part 
Radius (r2) of lower circular end of frustum part = Radius of circular end of cylindrical
part
Height (h1) of frustum part = 22 − 10 = 12 cm
Height (h2) of cylindrical part = 10 cm
Slant height (l) of frustum part 
Area of tin sheet required = CSA of frustum part + CSA of cylindrical part
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