Q.1      2 cubes each of volume 64cm3 are joined end to end. Find the surface area of the resulting cuboid.
Sol.

 

Let  the  length  of each  edge of  the cube  be a cm.
Then, Volume  =64cm3
⇒  a3=64
⇒   a = 4
When two cubes of equal volumes (i.e., equal edges are joined end to end, we get a cuboid such that its.
l = Length = 4cm + 4cm = 8 cm
b = Breadth = 4 cm
and     h = Height = 4 cm
Therefore  Surface area of the cuboid
= 2(lb + bh + hl)
= 2(8 × 4 + 4 × 4 + 4 × 8) cm2
= 2(32 + 16 + 32) cm2
= (2 × 80) cm2 = 160 cm2

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Q.2  A vessel is in the form of a hollow  hemisphere mounted by a hollow cylinder. The diameter of the  hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Sol. 

Here r, the  radius of hemisphere  = 7 cm ,
h , the height  of cylinder = (13 – 7) cm = 6 cm
Clearly, radius of the base of cylindrical part is also r cm .

Surface area of the vessel = Curved surface area of the cylindrical part + Curved surface area of hemispherical part
=(2πrh+2πr2)cm2=2πr(h+r)cm2
=2×227×7×13cm2=572cm2

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Q.3  A toy is in the  form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total  height of the toy is 15.5 cm. Find  the total surface area of the toy.
Sol.

We have VO = 15.5 cm,
OA = OO’ = 3.5 cm

Let r be the  radius of the base of cone and h be the height of conical part of the by.
Then r = OA = 3.5 cm
h = VO = VO’ – OO = (15.5 – 3.5) cm = 12 cm
Also  radius  of the hemisphere = OA = r = 3.5 cm
Total surface area of the toy
= Curved surface area of cone + Curved surface area of hemisphere
=πrℓ+2πr2
where l = Slant height =OA2+OV2−−−−−−−−−−√=(3.5)2+122−−−−−−−−−−√
=12.25+144−−−−−−−−−√=156.25−−−−−√=12.5cm
=πr(ℓ+2r)
=227×3.5×(12.5+2×3.5)cm2
=227×3.5×19.5cm=214.5cm2

Q.4  A cubical block  of side 7 cm is surmounted by a hemisphere. What is the greatest diameter of the  hemisphere can have? Find the surface area of the  solid?
Sol.

 

The  greatest diameter that a hemisphere  can have = 7 cm .
Surface area of the solid  after surmounted hemisphere
=6ℓ2−πR2+2πR2
=6ℓ2+πR2
=6(7)2+227×(72)2
=6×49+11×72
= 294 + 38.5  = 332.5 cm2

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Q.5  A hemispherical depression is cut  of  one face of a cubical wooden block  such the  diameter l of the  hemisphere is equal to the edge of the cube. Determine th surface area of the  remaining solid.
Sol.

Edge of the cube  = ℓ
Diameter of the hemisphere  = ℓ
Therefore  Radius  of the hemisphere = ℓ2
Therefore  Area  of the remaining  solid  after cutting out the  hemispherical  depression.
=6ℓ2−π(ℓ2)2+2π(ℓ�2)2
=6ℓ2+π(ℓ2)2
=6ℓ2+π×ℓ24=ℓ24(24+π)

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Q.6  A medicine capsule is in the  shape of a cylinder with two hemispheres stuck to each of its ends (see figure). The length of the entire capsule is 14 mm and the  diameter of the  capsule is 5 mm. Find  its surface area.

Sol.

Let r cm  be the  radius and h cm  be the height  of the  cylinder. Then.

r=52mm=2.5mm
and     h=(14−2×52)mm
= (14 – 5)mm = 9mm
Also the  radius of hemisphere = 52cm=rcm
Now, surface area of the capsule = Curved surface of cylinder + Surface area of two hemispheres
=(2πrh+2×2πr2)mm2
=2πr(h+2r)mm2
=2×227×52×(9+2×52)mm2
=227×5×14mm2=220mm2

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Q.7 A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the  cylindrical part are 2.1 m and 4 m respectively, and the  slant height  of the top is 2.8 m, find  the area of the canvas used for making the tent. Also, find  the cost of the canvas of the tent at the rate of Rs 500 per m2. (Note that the base of the tent will not covered with  canvas).
Sol.

We have, Total  canvas used  =  Curved surface area of cylinder + Curved surface area  of cone
=(2πrh+πrℓ)m2
=πr(2h+ℓ)m2
=227×2×(2×2.1+2.8)m2
=227×2×7m2
=44m2

Now,  cost of 1m2 the canvas for the  tent = Rs 500
So,  cost of 44m2 the  canvas for the  tent = Rs 44 × 500 = Rs 22000

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Q.8 From  a solid  cylinder whose height  is 2.4 cm and diameter 1.4 cm, a conical  cavity of the same height  and same diameter is  hollowed out. Find the  total surface area of the remaining  solid to the  nearest cm2.
Sol.

Radius  of the cylinder =1.42cm=7cm
Height of the cylinder = 2.4 cm
Radius of the cone = .7 cm
Height  of the cone = 2.4 cm
Slant  height  of the cone
=(.7)2+(2.4)2−−−−−−−−−−−√cm
=0.49+5.76−−−−−−−−−√cm
=6.25−−−−√cm=2.5cm

Surface area of the remaining  solid = Curved surface area of cylinder + Curved surface of the  cone + Area of upper circular base of cylinder
=2πrh+πrℓ+πr2=πr(2h+ℓ+r)
=227×.7×(2×2.4+2.5+.7)cm2
=2×.1×(4.8+2.5+.7)cm2
2.2×8.0cm2=17.6cm2=18cm2

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Q.9  A wooden article  was made by scooping out  a hemisphere  from each  end of a solid cylinder, as shown in figure. If the height  of the cylinder is 10 cm,and its base is of radius  3.5 m, find the  total  surface area of the article.

Sol.

Surface area of the article when it is ready = Curved surface area of cylinder + 2 × Curved surface area of hemisphere.
=2πrh+2×2πr2
=2πr(h+2r)

where  r = 3.5 m and h = 10 cm
=2×227×3.5×(10+2×3.5)cm2
=2×227×3.5×17cm2=374cm2

 

Q.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.
Sol.

 

Volume of the  solid = Volume  of the cone  + Volume  of the  hemisphere

=13πr2h+23πR3
=13πr2×r+23πr3        [Since h = r and R = r]
=π3r3(1+2)=πz3r3×3
=πr3=π(1)3=π×1=πcm3    [Since r = 1 cm]

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Q.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 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 the Rachel made. (Assume the outer and inner dimensions of the model be nearly the same).
Sol.

Volume  of the air  contained in the model =  Volume of the  cylindrical portion of the model + Volume  of its two  conical ends.

=πr2h1+2×13πr2h2
=πr2(h1+23h2)
where r=32cm
h1=8cmandh2=2cm
=227×(32)2×(8+23×2)cm3
=227×94×24+43cm3
=(227×94×283)cm3
=66cm3(approx)

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Q.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 hemsipherical ends with length 5 cm and diameter is 2.8 cm (see figure).

Sol.

Volume of the gulab jamun = Volume of the cylindrical portion + Volume of the hemispherical ends
=πr2h+2(23πr3)
=πr2(h+43r), where  r  = 1.4 cm, h = 2.2 cm
=227×(1.4)2×(2.2+43×1.4)cm3
=227×1.96×(6.6+5.63)cm3
=227×1.96×12.23cm3
Volume  of 45 gulab jamuns
=45×227×1.96×12.23cm3

Quantity of syrup in gulab jamuns  = 30% of their  volume
=30100×45×227×1.96×12.23cm3
9×11×1.96×12.27cm3=338.184cm3
=338cm3(approx)

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Q.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 m and the  depth is 1.4 cm. Find the volume  of wood  in the  entire  stand (see figure).

Sol.

Volume  of wood  in the entire  stand   = Volume  of the cuboid – 4 × Volume  of a depression (i.e., cone)
=15×10×3.5cm3−4×13×227×0.5×0.5×1.4cm3
=(525−1.47)cm3=523.53cm3

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Q.5       A vessel 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.
Sol.

Height  of the conical  vessel, h = 8 cm.
Its  radius r = 5 cm
Volume of cone = Volume of water in cone
=13πr2h
=13×227×5×5×8cm3
=440021cm3
Volume  of water flows out = Volume of  lead shots

=14 of the  volume of water in the cone
=14×440021cm3=110021cm3
Radius  of the lead shot  = 0.5 cm
Volume of one  spherical  lead shot =43πr3
=43×227×510×510×510cm3=1121cm3
Therefore   Number of lead  shots dropped into the vessel
=VolumeofwaterflowsoutVolumeofoneleadshot
=110021÷1121=110021×2111=100

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Q.6       A solid iron  pole  consists of a cylindrical 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 cm3 of iron  has approximately 8 g mass. (Use π=3.14)
Sol.

Volume  of the solid iron  pole = Volume  of the cylindrical  portion + Volume  of the other  cylindrical portion
=πr12h1+πr22h2
=(3.14×(12)2×220+3.14×(8)2×60)cm3
=(3.14×144×220+3.14×64×60)cm3
=(99475.2+12057.6)cm3
=111532.8cm3

Hence, the mass of the pole = (111532.8 × 8) grams
=(111532.8×81000)kg=892.26kg

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Q.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 circular 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.
Sol.

Volume  of the cylinder =πr2h
=227×(60)2×180cm3
=22×3600×1807cm3
=142560007cm3
Volume of the  solid = Volume of cone + Volume of hemisphere

=(13×227×602×120+23×227×603)cm3
=(31680007+31680007)cm3
=63360007cm3
Volume of water left in the cylinder = Volume  of the cylinder – Volume  of the solid
=(142560007−63360007)cm3
=(79200007)cm3
=1131428.57142cm3
=1.131m3(approx)

 

Q.1      A metallic sphere  of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height  of the cylinder.
Sol.

 

Volume of the sphere
=43πr3=43×π×(4.2)3cm3
If h is the  height of a cylinder of radius  6 cm. Then its volume,
=π(6)2hcm3=36πhcm3
Since, the volume of metal in the form of sphere  and cylinder remains the same , we have
36πh=43×π×4.2×4.2×4.2
⇒    h=136×43×4.2×4.2×4.2
⇒    h = 2.744

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Q.2      Metallic  spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find  the radius of the resulting sphere.
Sol.

Sum of the volumes of 3 gives spheres.
=43π(r13+r23+r33)
=43π(63+83+103)cm3
=43π(216+512+1000)cm3
=43π(1728)cm3
Let R be  the radius of the new spheres whose volume is the sum of the  volumes of 3 given spheres.
Therefore      =43πR3=43π(1728)
⇒    R3=1728
⇒    R3=(12)3
⇒    R = 12
Hence, the  radius of the  resulting sphere is 12 cm.

Therefore 1728 = 2³ × 6³ = (2 × 6)³ = 12)³

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Q.3      A 20 cm deep well with  diameter 7 cm is dug and the  earth  from  digging is evenly spread out to form a platform 22 m by 14 m. Find the  height of the platform.
Sol.

Let h m be the  required height  of the platform.
The shape of the platform will be like the shape of a cuboid 22 m × 14 m × h with a hole  in the  shape of cylinder of radius  3.5 m and depth h m.
The volume  of the platform will be equal to the  volume  of the  earth  dug out from  the well.
Now, the  volume  of the earth = Volume  of the cylindrical well
=πr2h
=227×12.25×20m3
=770m3
Also, the  volume of the platform = 22 × 14 × h m3
But volume of the platform = Volume of the  well
i.e.,      22 × 14 × h = 770
h=77022×14=2.5
Therefore  Height of the platform = 2.5 m

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Q.4     A well of diameter 3 m is dug 14 m deep. The earth taken out of it has  been spread evenly all around it in the shape of a circular  ring  of width 4 m to form an embankment. Find the height of the embankment.
Sol.

Let h be the required height of the embankment.
The shape of the embankment will be like the shape of a cylinder of internal radius 1.5 m and  external radius (4 + 1.5)m = 5.5 m (see figure).

The  volume of the embankment will be equal  to the volume of earth  dug out from the  well.
Now,  the volume of the earth = Volume  of the cylindrical well
=π×(1.5)2×14m3=31.5πm3
Also  the volume of the embankment
=π(5.52−1.52)hm3
=π(5.5+1.5)(5.5−1.5)hm3
=π×7×4hm3=28πhm3
Hence, we have
28πh=31.5π
⇒    h=31.528=1.125
Hence, the required height  of the  embankment = 1.125 m

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Q.5     A container shaped like a right  circular  cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be  filled into cones of height12 cm and diameter 6 cm, having  hemispherical shape on the top. Find the number of such  cones which  can be  filled with ice cream.
Sol.

Volume of the cylinder
=πr2h
=π(122)2×15
=π×62×15

Volume  of a cone having  hemispherical shape on the top
=13πr2h+23πr3=13πr2(h+2r)
=13π(62)2(12+2×62)
=13π×32×18
Let the number of cone that can be filled with ice cream be h.
Then 13π×32×18×n=π×62×15
n=π×6×6×15π×3×3×18×3=10
Q.6 How many silver  coins, 1.75 cm in diameter and of  thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?
Sol. The shape of the coin will be like the shape of a cylinder of radius 1.752cm
= 0.875 cm and of height 2mm =210cm=.2cm
Its volume
=πr2h=227×0.875×0.875×.2cm3
=0.48125cm3
Volume  of the cuboid =5.5×10×3.5cm3=192.5cm3
Number  of coins  required to form the cuboid
=VolumeofthecuboidVolumeofthecoin=192.50.48125=400
Hence, 400  coins must be melted to form a cuboid.

 

Q.1     A drinking glass is in the  shape of a frustum of a cone of height  14 cm. The  diameters of its two circular ends are 4 cm and 2 cm. Find  the capacity of the glass.
Sol.

 

Capacity of the glass,
V=π×h3×(R2+r2+Rr)
Here, R = 2 cm , r = 1 cm , h = 14 cm
Therefore     V=2273×(22+12+2×1)
=443×(4+1+2)
=443×7=308310223cm3

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Q.2     The slant height  of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find  the curved surface area  of the frustum.
Sol.

Slant height ,l = 4 cm (Given)
2πr1=6        ⇒    πr1=3
and   2πr2=18       ⇒    πr2=9

Curved surface of the  frustum
=(πr1+πr2)ℓ
=(3+9)×4cm2
=12×4cm2
=48cm2

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Q.3      A fez, the  (cap) used by the Turks, is  shaped like the frustum of a cone (see figure). If its radius on the open side is 10 cm, radius at the upper base is 4 cmand its slant height is 15 cm, find the area of material used for  making it.

Sol.

Here  R = 10 cm, r = 4 cm and l = 15 cm
Area of the material used for making the feza = Surface area of frustum
+  The  surface of top circular section

=π(R+r)ℓ+πr2
=227(10+4)15+227×4×4
=(227×14×15+3527)cm2
=4620+3527cm2=49727cm2
=71027cm2

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Q.4      A container  opened from  the top and made up of a metal  sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of milk which can completely fill the container, at the rate of Rs 20 per litre. Also  find the  cost of metal sheet used to make the container, if it costs Rs 8 per 100 cm2. (Take π=3.14).
Sol.

Here R = 20 cm, r = 8 cm and h = 16 cm Capacity of the container
= Volume of the frustum

=13πh(R2+r2+Rr)
=13×3.14×16(202+82+20×8)cm3
=50.243×(400+64+160)cm3
=50.243×624cm3=50.24×208cm3
=10449.921000litres
Cost of milk @ Rs 20 per litre
=Rs(20×10449.921000)
=Rs208.99≈Rs209
To find the slant height

ℓ=162+122−−−−−−−−√
=256+144−−−−−−−−√
400−−−√=20cm
=π(R+r)ℓ
=227×(20+8)×20cm2
=227×28×20cm2=1758.4cm2
and area of the bottom
=πr2=227×(8)2=227×64=200.96cm2
Therefore  Total area of metal required
= 1758.4 cm2+200.96cm2=1959.36cm2
Cost of metal sheet used to manufacture the container @ Rs 8 per 100cm2
=Rs(8100×1959.36)=Rs156.75