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In a right circular cone, the cross-section made by a plane parallel to the base is a(A) circle                                                         (B) frustum of a cone(C) sphere                                                       (D) hemisphere

According to the question if a right circular cone is cut by a plane parallel to its base the figure formed is

Here BECD is not a circle, not a sphere not a hemisphere but it is a frustum of a cone.

Hence in a right circular cone, the cross-section made by a plane parallel to the base is a frustum of a cone.

If two solid hemispheres of same base radius r are joined together along their bases, then curved surface area of this new solid is(A) $4 \pi r^{2}$                      (B) $6 \pi r^{2}$                  (C) $3 \pi r^{2}$                      (D) $8 \pi r^{2}$

The radius of hemisphere = r

$\text{ Curved surface area}$ $=2 \pi r^{2}$

The curved surface area of two solid hemisphere

$=2 \times 2 \pi r^{2}$

$=4 \pi r^{2}$

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A medicine-capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of entire capsule is 2 cm. The capacity of the capsule is(A) 0.36 cm3               (B) 0.35 cm3                (C) 0.34 cm3                (D) 0.33 cm3

Diameter of hemisphere = 0.5 cm

$Radius$ $=\frac{0.5}{2}=\frac{5}{20}=\frac{1}{4}=0.25cm$

$\text{ Volume of hemisphere }$$=\frac{2}{3}\pi r^{3}$

$= \frac{2}{3}\times\frac{22}{7}\times \frac{1}{4}\times\frac{1}{4}\times \frac{1}{4}=\frac{11}{336}$

$\text{ Volume of two hemispheres}$ $=\frac{2 \times11}{336} =\frac{11}{168}$

Similarly radius of cylinder = 0.25

Height  = 2- 0.25 -0.25

= 2- 0.5

=1.5 cm

$Volume$ $= \pi r ^{2}h$

$= \frac{22}{7}\times \frac{1}{4}\times\frac{1}{4}\times\frac{15}{10}\Rightarrow \frac{33}{112}$

The total volume of capsule = volume of two hemispheres + volume of the cylinder

$=\frac{11}{168}+\frac{33}{112}$

$=0.065 +0.294$

= 0.359cm3

=0.36cm3   (approximate)

A shuttle cock used for playing badminton has the shape of the combination of(A) a cylinder and a sphere(B) a cylinder and a hemisphere(C) a sphere and a cone(D) frustum of a cone and a hemisphere

Answer (D) frustum of a cone and a hemisphere

Solution

Cylinder – A cylinder is a three dimensional solid that has two parallel based joined by a curved surface at a fixed distance.

Sphere – The set of all points in thee dimensional space lying the same distance from a given point.

Similarly hemisphere

Cone – A cone is a three dimensional geometrical shape that tappers smoothly from a flat base to a point called the apex or vertex.

Similarly frustum of cone

Hence we conclude that shuttle cock which is used for playing badminton has the shape of combination of frustum of a cone and hemisphere.

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The shape of a gilli, in the gilli-danda game (See figure), is a combination of(A) two cylinders                                            (B) a cone and a cylinder(C) two cones and a cylinder                          (D) two cylinders and a cone

Answer(C) two cones and a cylinder

The shape of a glass (tumbler) (see figure) is usually in the form of(A) a cone                                                       (B) frustum of a cone(C) a cylinder                                                  (D) a sphere

Answer (B) frustum of a cone

Solution

A cone – A cone is a three-dimensional geometrical shape that tapers smoothly from a flat base to a point called the apex or vertex.

(B) Frustum of a cone – It is a portion of a solid that lies between one or two parallel planes cutting it.

(C) A cylinder – A cylinder is a three dimensional solid that has two parallel based joined by a curved surface at a fixed distance.

(D) A sphere – The set of all points in three-dimensional space lying the same distance from a given point.

Hence the shape of glass is usually in the form of the frustum of a cone.

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A plumbline (sahul) is the combination of (A) a cone and a cylinder                                (B) a hemisphere and a cone(C) frustum of a cone and a cylinder              (D) sphere and cylinder

Answer(B) a hemisphere and a cone

A Surahi is the combination of(A) a sphere and a cylinder                             (B) a hemisphere and a cylinder(C) two hemispheres                           (D) a cylinder and a cone.

Solution

(A) A Sphere – The set of all points in three dimensions space lying the same distance from a given point.

A cylinder – A cylinder is a three dimensional solid that holds two parallel bases joined by a curve surface at a fixed distance.

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A cylindrical pencil sharpened at one edge is the combination of(A) a cone and a cylinder                                (B) frustum of a cone and a cylinder(C) a hemisphere and a cylinder                     (D) two cylinders.

Answer(A) a cone and a cylinder

Solution

(A) A cone –A cone is a three-dimensional geometric shape that tapers smoothly from a flat base to a point called the apex or vertex.

A cylinder – A cylinder is a three dimensional solid that holds two parallel bases joined by a curve surface at a field distance.

(B) Frustum of a cone – It is the portion of a solid that lies between one or two parallel planes cutting it.

A cylinder – A cylinder is a three dimensional solid that holds two parallel bases joined by a curve surface at a fixed distance.

(C) Hemisphere – In geometry it is an exact half of a sphere and it is a three dimensional geometric.

A cylinder – A cylinder is a three dimensional solid that holds two parallel bases joined by a curve surface at a fixed distance.

(D) Cylinder – A cylinder is a three dimensional solid that holds two parallel bases joined by a curve surface at a fixed distance.

(Combination of two cylinders)

Hence form the above diagrams we conclude that option (A) is correct.

Therefore, A cylindrical pencil sharped at one edge is the combination of a cone and a cylinder.

The diameters of the two circular ends of the bucket are 44 cm and 24 cm. The height of the bucket is 35 cm. The capacity of the bucket is(A) 32.7 litres (B) 33.7 litres (C) 34.7 litres (D) 31.7 litres

$\text{Volume of frustum of cone }=\frac{\pi h}{3}(r_{1}^{2}+r_{2}{^2}+r_{1}r_{2})$
Diameter of first end = 44 cm

$(r_{1})=\frac{44}{2}=22cm$
Diameter of second end = 24 cm
$Radius%u200B%u200B%u200B%u200B%u200B%u200B%u200B$ $(r_{2})=\frac{24}{2}=12cm$
$\text{Height }(h)=35cm$
$\text{Volume}$ $=\frac{\pi h}{3}(r_{1}^{2}+r_{2}^{2}+r_{1}r_{2})$
$=\frac{22\times 35}{7\times 3}(22^{2}+12^{2}+22\times 12)$
$\Rightarrow \frac{110}{3}[484+144+264]$
$\frac{110}{3}[892]$

$=32706.67cm^{3}$
$\text{We know that 1 liter}= 1000 cm^{3}$???????
$\therefore 32706.67cm^{3}=32.7litres$