CBSE MATH STUDY: Optional Exercise – 13.5 - 10th Mathematics

Question 1: 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 cm³.

Solution: 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= 2pr = 2p × 5 = 10p

Number of round = Height of cylinder/Demeter of wire = 12/0.3= 40 round

Length of wire in 40 rounds = 40 × 10p = 40x10 x3.14=1256cm=12.56m

Radius of wire = 0.3/2 = 0.15cm

Volume of wire = Area of cross-section of wire × Length of wire = p (0.15)² × 1256 = 3.14 x 0.0225 x 1256 = 88.736 cm³

Mass = Volume × Density = 88.736 × 8.88 = 787.979 gm

Question 2: 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 p as found appropriate.)

Solution: The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse is shown in the figure.

AC = √ (3² + 4²) = 5 cm

Area of DABC = ½ x AC x r =1/2 x BC x AB

Þ 5 x r = 3 x 4 Þ 12/5=2.4cm

Volume of double cone = Volume of cone 1 + Volume of cone 2

= 1/3 p r²H + /3 p r²h = 1/3 p r²(H + h)

= 1/3 x 3.14 x 2.4 x 2.4 x 5 = 30.14 cm³

Surface area of double cone = Surface area of cone 1 + Surface area of cone 2

= p r L + p r l

= p r ( L + l )

= 3.14 x 2.4 (4+3)

= 3.14 x 2.4 x7

= 52.75 cm2

Question 3: A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm3 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?

Solution: Volume of cistern = 150 × 120 × 110 = 1980000 cm3

Volume to be filled in cistern = 1980000 – 129600 = 1850400 cm3

Volume of n bricks = 22.5 × 7.5 × 6.5 = 1096.875

As each brick absorbs one-seventeenth of its volume,

Þ Volume absorbed by 1 bricks = 1/17 x 1096.875 =64.52

Actual volume of bricks without water = 1096.875 - 64.52 =1032.375

Number of bricks = 1850400/1032.375 =1792.37

Therefore, 1792 bricks were placed in the cistern.

Q 4. 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: Area of the valley, A = 97280 km2

Level in the rise of water in the valley, h = 10 cm = (10/100000) km = (1/10000) km

Thus, amount of rain fall in 14 day = Ah = 97280 km2 × (1/10000) km = 9.828 km3

Amount of rain fall in 1 day = 9.828 /14 = 0.702km3

Volume of water in 3 rivers = length × breadth × height

= 1072 km × 75 m × 3 m

= 1072 km × (75/1000) km × (3/1000) km = 0.2412 x3

= 0.7236 km3

This shows that the amount of rain fall is approximately equal to the amount of water in three rivers.

Question 5: 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 given figure).

Solution: Radius (R) of upper circular end of frustum

=18/2 = 9 cm

Radius (r) of lower circular end of frustum

= Radius of circular end of cylindrical part

= 8/2 = 4 cm

Height (H) of frustum

= 22 − 10 = 12 cm

Height (h) of cylindrical = 10 cm

Slant height (l) of frustum

= √(R-r)² + H² =√(9-4)² + 12² =13 cm

Area of tin sheet required = CSA of frustum t + CSA of cylindrical

= p (R+r)l + 2prh = p [{(9+4)x13} + {2x4x10}]=22/7{169+80}=782.57 cm2

Question 6: Derive the formula for the volume of the frustum of a cone

Solution: Let ABC be a cone. A frustum DECB is cut by a plane parallel to its base. Let R and r be the radii of the ends of the frustum of the cone and h be the height of the frustum of the cone.

In D ABG and D ADF, DF||BG

<A = <A and <AGB = <AFD
∴ D ABG ∼ D ADF (AA similarity)

DF/BG = AF/AG =AD/AB

Þ r/R = H-h/H=L-l/L

Þif r/R = H-h/HÞ H =(Rh/R-r)

Volume of frustum of cone = Volume of cone ABC − Volume of cone ADE

= 1/3xpR²H - 1/3xpr²(H-h)

= 1/3p [R²H - r²(H-h)]

= 1/3p [R²(Rh/R-r) - r²{(Rh/R-r)-h}]

= 1/3p [(R³h/R-r) - r²{(Rh –Rh+rh) /(R-r)}]

= 1/3p [(R³h - r³h) /(R-r)]

= 1/3ph [(R³ - r³) /(R-r)]