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Exercise - 13.2

Surface Area and Volume

**Question-1 :-** A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π. (Take π = 22/7)

Given that:- A solid is in the shape of a cone standing on a hemisphere. radius(r) of the hemispherical part = 1 cm = radius(r) of the conical part height(h) of the conical part = 1cm = radius(r) of the conical part Therefore, Volume of solid = Volume of conical part + Volume of hemispherical part = 1/3 x πr^{2}h + 2/3 x πr^{3}= 1/3 x πr^{2}(h + 2r) = 1/3 x π x 1 x 1 x [1 + (2 x 1)] = 1/3 x π x 3 = π cm^{3}

**Question-2 :-** Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.) (Take π = 22/7)

Given that:- Diameter of the model = 3 cm Radius (r) of cylindrical part = Radius of conical part = 3/2 cm Length of the model = 12 cm Height (h_{1}) of each conical part = 2 cm Height (h_{2}) of cylindrical part = 12 − 2 × Height of conical part = 12 − 2 × 2 = 8 cm Therefore, Volume of air present in the model = Volume of cylinder + 2 × Volume of cones = πr^{2}h_{2}+ 2 x 1/3 x π x r^{2}x h_{1}= πr^{2}[h_{2}+ 2/3 x h_{1}] = 22/7 x 3/2 x 3/2 x [8 + (2/3 x 2)] = 22/7 x 9/4 x 28/3 = 22 x 3 = 66 cm_{2}

**Question-3 :-** A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see Fig.). (Take π = 22/7)

Give that:- Length of the gulab jamun = 5 cm Diameter = 2.8 cm Radius (r) of cylindrical part = Radius (r) of hemispherical part = 2.8/2 = 1.4 cm Length of each hemispherical part = Radius of hemispherical part = 1.4 cm Length (h) of cylindrical part = 5 − 2 × Length of hemispherical part = 5 - (2 x 1.4) = 5 - 2.8 = 2.2 cm Therefore, volume of one gulab jamun = vol. of cylindrical part + 2 × vol. of hemispherical part = πr^{2}h + 2 x 2/3 x πr^{3}= πr^{2}[h + (4/3 x r)] = 22/7 x 1.4 x 1.4 x [2.2 + (4/3 x 1.4)] = 22 x 0.2 x 1.4 x [2.2 + 5.6/3] = 6.16 x 12.2/3 = 25.05 cm^{3}Volume of 45 gulab jamuns = 45 x 25.05 = 1,127.25 cm^{3}Volume of sugar syrup = 30% of volume = 30/100 x 1127.25 = 338.17 cm^{3}= 338 cm^{3}(Approx.)

**Question-4 :-** A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see Fig.). (Take π = 22/7)

Gien that:- Dimensions of the cuboid = 15 cm x 10 cm x 3.5 cm Radius(r) of the depressions(conical part) = 0.5 cm Depth(h) of the depressions(conical part) = 1.4 cm Therefore, Volume of wood = Volume of cuboid − 4 × Volume of cones = lbh - 4 x 1/3 x πr^{2}h = (15 x 10 x 3.5) - (4/3 x 22/7 x 0.5 x 0.5 x 1.4) = 525 - 30.8/21 = 10994.2/21 = 523.53 cm^{3}

**Question-5 :-** A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel. (Take π = 22/7)

Given that:- Height(h) of the conical vessel = 8 cm radius(r_{1}) of the conical vessel = 5 cm radius(r_{2}) of lead shots = 0.5 cm Let n number of lead shots were dropped in the vessel. Volume of water spilled = Volume of dropped lead shots 1/4 x volume of conical vessel = n x 4/3 x πr_{2}^{3}1/4 x 1/3 x πr_{1}^{2}h = n x 4/3 x πr_{2}^{3}5 x 5 x 8 = 16 x n x 0.5 x 0.5 x 0.5 200 = 16 x n x 0.125 200/(16 x 0.125) = n n = 200/2 n = 100 Hence, the number of lead shots dropped in the vessel is 100.

**Question-6 :-** A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm^{3} of iron has approximately 8g mass. (Use π = 3.14)

Given that:- Height(h_{1}) of the large cylinder = 220 cm Base diameter of large cylinder = 24 cm radius(r_{1}) of the large cylinder = 24/2 = 12 cm Height(h_{2}) of the smaller cylinder = 60 cm radius(r_{2}) of the smaller cylinder = 8 cm Therefore, total volume of the pole = volume of the large cylinder + volume of the smaller cylinder = πr_{1}^{2}h_{1}+ πr_{2}^{2}h_{2}= π(r_{1}^{2}h_{1}+ r_{2}^{2}h_{2}) = 3.14 x [(12 x 12 x 220) + (8 x 8 x 60)] = 3.14 x [31680 + 3840] = 3.14 x 35520 = 111532.8 cm^{3}Now, Mass of 1 cm^{3}iron = 8 g Mass of 111532.8 cm^{3}iron = 111532.8 × 8 = 892262.4 g = 892.262 kg

**Question-7 :-** A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm. (Take π = 22/7)

Given that:- Radius (r) of hemispherical part = Radius (r) of conical part = 60 cm Height (h_{2}) of conical part of solid = 120 cm Height (h_{1}) of cylinder = 180 cm Radius (r) of cylinder = 60 cm Therefore, Volume of water left = Volume of cylinder − Volume of solid = Volume of cylinder − [Volume of cone + volume of hemisphere] = πr^{2}h_{1}- [(1/3 x πr^{2}h_{2}) + (2/3 x πr^{3})] = (π x 60 x 60 x 180) - (π/3 x 60 x 60) x [120 + 2 x 60] = (π x 60 x 60 x 180) - (π/3 x 60 x 60 x 240) = (π x 60 x 60) x (180 - 80) = 3600 x π x 100 = 360000 x 22/7 = 1131428.57 cm^{3}= 1.131 m^{3}

**Question-8 :-** A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm^{3}. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.

Given that:- Height(h) of the cylindrical part = 8 cm Diameter of the cylindrical part = 2 cm radius(r_{2}) of the cylinderical part = 2/2 = 1 cm Diameter of the spherical part = 8.5 cm radius(r_{1}) of the spherical part = 8.5/2 cm Therefore, Volume of vessel = Volume of sphere + Volume of cylinder = 4/3 x πr_{1}^{3}+ πr_{2}^{2}h = π x [(4/3 x r_{1}^{3}) + (r_{2}^{2}h)] = 3.14 x [(4/3 x 8.5/2 x 8.5/2 x 8.5/2) + (1 x 1 x 8)] = 3.14 x [102.35 + 8] = 3.14 x 110.35 = 346.50 cm^{3}Hence, she is wrong.

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