Exercise 13.1 9 Questions – Cubes & Cuboids
2 cubes each of volume 64 cm³ are joined end to end. Find surface area of resulting cuboid.
Volume 64 ' edge = 4 cm. Cuboid: l=8, b=4, h=4. SA = 2(32+16+32) = 2(80) = 160 cm²
Q2: A vessel is in form of hollow hemisphere mounted by hollow cylinder. Inner diameter=14cm, total height=13cm. Find inner SA. ' r=7. Cylinder h=13'7=6. CSA(cyl)=2π(7)(6)=264, CSA(hemi)=2π(49)=308. Total=572 cm²
Q5: A hemispherical depression is cut out from one face of a cubical wooden block of edge 21 cm. Find SA of remaining solid after cutting hemisphere of largest possible diameter. ' r=10.5. SA=6(21²)'π(10.5)²+2π(10.5)² = 2646+π(110.25) ≈ 2992.37 cm²
Q8: A hemispherical dome of a building needs painting. Circumference of base=17.6 m. Cost=₹5/100cm². ' r=17.6/2π=2.8m. CSA=2π(2.8)²=49.28m². Cost=₹24,640
Exercise 13.2 8 Questions – Cylinder + Cone Combinations
A solid is in shape of cone standing on hemisphere with both radii = 1 cm. Height of solid = 5 cm. Find volume (π=3.14).
Cone height = 5'1 = 4 cm. Vol(cone) = "π(1)²(4) = 4π/3. Vol(hemi) = (2/3)π(1)³ = 2π/3. Total = 2π = 6.28 cm³
Q3: A gulab jamun contains sugar syrup up to 30% of volume. Cylinder with two hemispherical ends, l=5cm, d=2.8cm. Find syrup in 45 gulab jamuns. ' r=1.4, cyl h=5'2.8=2.2. Vol(total)=πr²h+(4/3)πr³ = π(1.96—2.2+3.66) = 7.97π≈25.05cm³. Syrup=7.515cm³/unit. —45=338cm³
Q6: A solid iron pole consists of cylinder of height 220cm and base dia 24cm surmounted by another cylinder of height 60cm and radius 8cm. Find mass. ' Vol=π[12²—220+8²—60]=π—35520. Mass=35520π—8≈892.26kg
Exercise 13.3 9 Questions – Cone, Sphere & Hemisphere
A metallic sphere of radius 4.2 cm is melted and recast into shape of cylinder of radius 6 cm. Find height.
Vol(sphere) = (4/3)π(4.2)³ = (4/3)π(74.088). Vol(cyl) = π(6)²h = 36πh. Equate: 36πh = (4/3)π(74.088) ' h = 4—74.088/(3—36) = 2.744 cm
Q3: A 20 m deep well with diameter 7 m. Earth dug out is spread to form 22 m — 14 m platform. Find height. ' Vol(earth)=π(3.5)²—20=245π=770m³. Platform h=770/(22—14)=2.5 m
Q6: A cylindrical bucket 32 cm high, 18 cm radius, filled with sand. Emptied on ground forming conical heap. Sand heap height=24 cm. Find radius and slant height. ' Vol=π(18)²(32)=10368π. Cone=(1/3)πr²(24)=8πr². 8πr²=10368π ' r=36cm. l=√(36²+24²)=√1872=12√13 cm
Q8: Water in canal 6 m wide, 1.5 m deep flowing at 10 km/h. How much area will it irrigate in 30 min if 8 cm standing water needed? ' Vol in 30min=6—1.5—5000=45000m³. Area=45000/0.08=562500 m²
Exercise 13.4 5 Questions – Spheres & Hemispheres
A drinking glass is in shape of frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find capacity.
r₁=2, r₂=1, h=14. Vol(frustum)="πh(r₁²+r₂²+r₁r₂)="π(14)(4+1+2)=(14/3)π(7)=102.67 cm³
Q3: Shanti Sweets Stall was placing order for cardboard boxes. Two sizes: 25—20—5cm and 15—12—5cm. 5% extra for overlap. 1000 boxes total. Cardboard cost ₹4/1000cm². ' Total SA for both types with overlap = 1.05—[(1450)+(540)]—1000 = ₹8,358
Q5: A container shaped like right circular cylinder of diameter 12 cm, height 15 cm, full of ice cream. Ice cream to be filled in cones of height 12 cm and diameter 6 cm with hemispherical top. Find number of cones. ' Vol(cyl)=π(6)²(15)=540π. Vol(cone+hemi)="π(3)²(12)+(2/3)π(3)³=36π+18π=54π. Number=10
Exercise 13.5 (Optional) 7 Questions – Frustum
Q1: A copper rod of diameter 1 cm and length 8 cm is drawn into wire of length 18 m of uniform thickness. Find thickness. ' Vol=π(0.5)²(8)=2π. Wire: πr²—1800=2π ' r²=2/1800 ' r=1/30. Diameter=2/30=0.067 cm
Q5: A bucket is in form of frustum of cone of height 30 cm with radii 10 cm and 20 cm. Find capacity and cost of milk at ₹40/litre. ' Vol="π(30)(100+400+200)=10π(700)=21991cm³=21.99L. Cost=₹879.60
Q7: Derivation of frustum volume formula using similar triangles.
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