TOPICS

Exercise - 13.5

Surface Area and Volume

**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 cm^{3}.

Given that:- Diameter of wire = 3 mm = 0.3 cm Radius(r) of wire = 0.3/2 = 0.15 cm Lenght(h) of wire(Height of cylinder) = 12 cm It can be observed that 1 round of wire will cover 3 mm height of cylinder. No. of rounds = Height of cylinder/Diameter of wire = 12/0.3 = 40 rounds Length of wire required in 1 round = Circumference of base of cylinder = 2πr = 2π × 5 = 10π Length of wire in 40 rounds = 40 × 10π = 400 x 22/7 = 8800/7 = 1257.14 cm = 12.57 m Therefore, Volume of wire = Area of cross-section of wire × Length of wire = πr^{2}x 1257.14 = 22/7 x 0.15 x 0.15 × 1257.14 = 88.898 cm^{3}Now, Mass of the wire = Volume × Density = 88.898 × 8.88 = 789.41 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 π as found appropriate.)

Given that:- The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse is shown in the figure. So, AB = 4 cm and BC = 3 cm Hypotenuse(AC) = √AB² + BC² = √4² + 3² = √16 + 9 = √25 = 5 cm Area of ΔABC = 1/2 x AB x BC 1/2 x AC x OB = 1/2 x 4 x 5 1/2 x 5 x OB = 6 OB = 12/5 cm OB = 2.4 cm i.e., radius(r) = 2.4 cm Therefore, Volume of double cone = Volume of cone 1 + Volume of cone 2 = 1/3 x πr^{2}h_{1}+ 1/3 x πr^{2}h_{2}= 1/3 x πr^{2 x}x (h_{1}+ h_{2}) = 1/3 x πr^{2 x}x (OA + OC) = 1/3 x 3.14 x 2.4 x 2.4 x 5 (Here, OA + OC = AC = 5 cm) = 3.14 x 0.8 x 2.4 x 5 = 30.14 cm^{3}Now, Surface area of double cone = Surface area of cone 1 + Surface area of cone 2 = πrl_{1}+ πrl_{2}= πr x (l_{1}+ l_{2}) = 3.14 x 2.4 x (4 + 3) (Here, l_{1}= AB = AD = 4 cm, l_{2}= BC = CD = 3 cm) = 3.14 x 2.4 x 7 = 52.75 cm^{2}

**Question-3 :-** A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm^{3} 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?

Given that:- Volume of cistern = 150 × 120 × 110 = 1980000 cm^{3}Volume to be filled in cistern = 1980000 − 129600 = 1850400 cm^{3}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/17 x (1096.875) 1850400 + n/17 x (1096.875) = 1096.875n 1850400 = 16n/17 x 1096.875 n = (1850400 x 17)/(1096.875 x 16) n = 1792.41 Therefore, 1792 bricks were placed in the cistern.

**Question-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 km^{2}, 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.

Given that:- Area of the valley = 97280 km^{2}Level in the rise of water in the valley, h = 10 cm = 10/100000 Km = 1/10000 Km Thus, amount of rainfall in 14 days = Area of the valley x h = 97280 Km^{2}x 1/10000 Km = 9.828 Km^{3}Therefore, Amount of rainfall in 1 day = 9.828/14 = 0.702 Km^{3}Dimensions single river = 1072 km x 75 m x 3 m = 1072 km x 75/1000 km x 3/1000 km Therefore, Volume of water in three rivers = 3 x (l x b x h) = 3 x (1072 x 75/1000 x 3/1000) = 3 x 1072 x 0.075 x 0.003 = 0.7236 Km^{3}Hence, this shows that the amount of rainfall 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).

Given that:- Diameter of the top of the funnel = 18 cm Radius(r_{1}) of upper circular end of frustum part = 18/2 = 9 cm Diameter of cylinderical portion = 8 cm Radius(r_{2}) of lower circular end of frustum part = Radius of circular end of cylindrical part = 8/2 = 4 cm Height(h_{1}) of frustum part = 22 − 10 = 12 cm Height(h_{2}) of cylindrical part = 10 cm Slant height(l) of frustum part = √(r_{1}- r_{2})² + h_{1}² = √(9 - 4)² + 12² = √25 + 144 = √169 = 13 cm Therefore, Area of tin sheet required = CSA of frustum part + CSA of cylindrical part = π(r_{1}+ r_{2})l + 2πr_{2}h_{2}= π[(r_{1}+ r_{2})l + 2r_{2}h_{2}] = 22/7 x [(9 + 4) x 13 + (2 x 4 x 10)] = 22/7 x [169 + 80] = 22/7 x 249 = 5478/7 = 782.57 cm^{2}

**Question-6 :-** Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

Let ABC be a cone. A frustum DECB is cut by a plane parallel to its base. Let r_{1}and r_{2}be the radii of the ends of the frustum of the cone and h be the height of the frustum of the cone. In ΔABG and ΔADF, DF||BG ∴ ΔABG ∼ ΔADF DF/BG = AF/AG = AD/AB r_{2}/r_{1}= (h_{1}- h)/h_{1}= (l_{1}- l)/l_{1}r_{2}/r_{1}= 1 - h/h_{1}= 1 - l/l_{1}1 - l/l_{1}= r_{2}/r_{1}l/l_{1}= l - r_{2}/r_{1}= (r_{1}- r_{2})/r_{1}l_{1}/l = r_{1}/(r_{1}- r_{2}) l_{1}= r_{1}l/(r_{1}- r_{2}) Therefore, CSA of frustum DECB = CSA of cone ABC − CSA cone ADE = πr_{1}l_{1}- πr_{2}(l_{1}- l) = πr_{1}[r_{1}l/(r_{1}- r_{2})] - πr_{2}[r_{1}l/(r_{1}- r_{2}) - l)] = πr_{1}^{2}l/(r_{1}- r_{2}) - πr_{2}[(r_{1}l - r_{1}l + r_{2}l)/(r_{1}- r_{2})] = πr_{1}^{2}l/(r_{1}- r_{2}) - πr_{2}^{2}l/(r_{1}- r_{2}) = πl[(r_{1}^{2}- r_{2}^{2})/(r_{1}- r_{2})] = π(r_{1}+ r_{2})l CSA of frustum = π(r_{1}+ r_{2})l TSA of frustum = CSA of frustum + Area of upper circular end + Area of lowercircular end = π(r_{1}+ r_{2})l + πr_{2}^{2}+ πr_{1}^{2}= π[(r_{1}+ r_{2})l + r_{1}^{2}+ r_{2}^{2}]

**Question-7 :-** Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

Given that:- Let ABC be a cone. A frustum DECB is cut by a plane parallel to its base. Let r_{1}and r_{2}be the radii of the ends of the frustum of the cone and h be the height of the frustum of the cone. In ΔABG and ΔADF, DF||BG ∴ ΔABG ∼ ΔADF DF/BG = AF/AG = AD/AB r_{2}/r_{1}= (h_{1}- h)/h_{1}= (l_{1}- l)/l_{1}r_{2}/r_{1}= 1 - h/h_{1}= 1 - l/l_{1}1 - h/h_{1}= r_{2}/r_{1}h/h_{1}= 1 - r_{2}/r_{1}= (r_{1}- r_{2})/r_{1}h_{1}/h = r_{1}/(r_{1}- r_{2}) h_{1}= r_{1}h/(r_{1}- r_{2}) Therefore, Volume of frustum of cone = Volume of cone ABC − Volume of cone ADE = 1/3 x πr_{1}^{2}h_{1}- 1/3 x πr_{2}^{2}(h_{1}- h) = π/3 x [r_{1}^{2}h_{1}- r_{2}^{2}(h_{1}- h)] = π/3 x [r_{1}^{2}{hr_{1}/(r_{1}- r_{2})} - r_{2}^{2}{hr_{1}/(r_{1}- r_{2}) - h}] = π/3 x [hr_{1}^{3}/(r_{1}- r_{2}) - r_{2}^{2}(hr_{1}- hr_{1}+ hr_{2})/(r_{1}- r_{2})] = π/3 x h x [(r_{1}^{3}- r_{2}^{3})/(r_{1}- r_{2})] = π/3 x h x [(r_{1}- r_{2})(r_{1}^{2}+ r_{2}^{2}+ r_{1}r_{2})] = 1/3 x πh[r_{1}^{2}+ r_{2}^{2}+ r_{1}r_{2}]

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