NCERT Solutions for class 10 Math Chapter – 13 Exercise – 13.5
Q1. A copper wire , 3 mm in diameter , is would 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 cm3.
Solution:-
Length of wire = 2πr
= 3.14 x 5
= 31.4 cm
The number of turns of the wire to cover 12 cm will be
= 12/3/10
= 40
The length of wire required to cover the whole surfaces = Length of wire required to complete 40 rounds
40 x 3.14 = 1256
Volume of wire = Area of cross section of wire x Length of wire
= π(0.15)2 x 1257.14
= 88.898 cm3
Mass = Volume x density
= 88.898 x 8.88
= 789.41 gm
Q2. A right triangle whose sides are 3 cm and 4 cm is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed.
Solution:-
Volume of double cone will be
= [1/3π x (12/5)2 9/5 + 1/3π x (12/5)2 x 16/5)
V = 30.14 cm3
The surface area of the double cones will be
= [π x 12/5 x3] + [π x 12/5 x 4]
= 52.75 cm2
Q3. A cistern , internally measuring 150 cm x 120 cm x 100 cm , has 129600 cm3 of water it. Porous bricks are placed in the water until the cistern is full to brim. Each brick absorbs one – seventeenth of its own volume of water. How many bricks can be put in without overflowing the water , each being 22.5 cm x 7.5 cm x 6.5 cm ?
Solution:-
Volume to be filled in cristen = 1980000 – 129600
= 1850400 cm3
Volume of n bricks will be = n x 22.5 x 7.5 x 6.5
Each brick absorbs one – seventeenth of its , the volume will be = n/(17) x (22.5 x 7.5 x 6.5)
= 1850400 + n/(17) x (22.5 x 7.5 x 6.5)
N = 1792.41
Q4. 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 97280 km2 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.
Solution:-
Total volume of 3 rivers = 3 x [(Surface area of river) x depth]
Surface area of river = [1072 x (75/1000)] km
Depth = (3/1000) km
Volume of 3 rivers = 3 x [ 1072 x (75/100)] x (3/1000)
= 0.72 km3
Volume of rainfall = Total surface area x Total height of rain
= 9780 x 10/100 x 1000
= 9.7 km3
Q5. An oil funnel made of thin 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
Solution:-
Area of tin sheet required = CSA of frustum part + CSA of cylindrical part
= π(r1 + r2)l + 2πr2h2
Area of tin sheet required = 782 x (4/7) cm2