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Practice each question, then open the answer or worked solution.
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.
(A) a cone and a cylinder
The Nib of a sharpened pencil = conical shape
The rest of the part of a sharpened pencil = cylindrical
Therefore, a pencil is a combination of cylinder 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.
(A) a sphere and a cylinder
The top part of surahi = cylindrical shape
Bottom part of surahi = spherical shape
Therefore, surahi is a combination of Sphere and a cylinder.
A plumbline (sahul) is the combination of (see Fig. 12.2)
(A) a cone and a cylinder (B) a hemisphere and a cone
(C) frustum of a cone and a cylinder (D) sphere and cylinder
(B) a hemisphere and a cone
The upper part of plumbline = hemispherical,
The bottom part of plumbline = conical
Therefore, it is a combination of hemisphere and cone.
The shape of a glass (tumbler) (see Fig. 12.3) is usually in the form of
(A) a cone (B) frustum of a cone
(C) a cylinder (D) a sphere
(B) frustum of a cone
The shape of glass is a frustum or specifically, an inverted frustum.
The shape of a gilli, in the gilli-danda game (see Fig. 12.4), 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
(C) two cones and a cylinder
The left and right part of a gilli = conical
The central part of a gilli = cylindrical
Therefore, it is a combination of a cylinder and two cones.
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
(D) frustum of a cone and a hemisphere
The cork of a shuttle = hemispherical shapes
The upper part of a shuttle = shape of frustum of a cone.
Therefore, it is a combination of frustum of a cone and a hemisphere.
A cone is cut through a plane parallel to its base and then the cone that is formed on one side of that plane is removed. The new part that is left over on the other side of the plane is called
(A) a frustum of a cone (B) cone
(C) cylinder (D) sphere
(A) a frustum of a cone
When a cone is divided into two parts by a plane through any point on its axis parallel to its base, the upper and lower parts obtained are cone and a frustum, respectively.
A hollow cube of internal edge 22cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that 1/8 space of the cube remains unfilled. Then the number of marbles that the cube can accommodate is
(A) 142296 (B) 142396
(C) 142496 (D) 142596
(A) 142296
According to the question,
Volume of cube =223=10648cm3
Volume of cube that remains unfilled =1/8×10648=1331cm3
volume occupied by spherical marbles =10648−1331=9317cm3
Radius of the spherical marble = 0.5/2=0.25cm=1/4cm
Volume of 1 spherical marble = 4/3×22/7 × (1/4)3=11/168cm3
Numbers of spherical marbles, n = 9317 × (11/168) =142296
A metallic spherical shell of internal and external diameters 4 cm and 8 cm, respectively is melted and recast into the form a cone of base diameter 8cm. The height of the cone is
(A) 12cm (B) 14cm (C) 15cm (D) 18cm
(B) 14cm
Volume of spherical shell = Volume of cone recast by melting
For Spherical Shell,
Internal diameter, d1 = 4 cm
Internal radius, r1 = 2 cm
External diameter, d2 = 8 cm
External radius, r2 = 4 cm
Now,
As volume of spherical shell= 4/3 π (r23 – r13)
where r1 and r2 are internal and external radii respectively.
volume of given shell = 4/3 π (43 – 23)
= 4/3 π (56)
= (224/3) π
We know that,
Volume of cone = 224π /3 cm3
For cone,
Base diameter = 8 cm
Base radius, r = 4 cm
Let Height of cone = ‘h’.
We know,
Volume of cone = (1/3) π r2h,
Where r = Base radius and h = height of cone
Volume of given cone = (1/3) π 42h
⇒ 224π /3 = 16πh /3
⇒ 16h = 224
h = 14 cm
So, Height of cone is 14 cm.
A solid piece of iron in the form of a cuboid of dimensions 49cm × 33cm × 24cm, is moulded to form a solid sphere. The radius of the sphere is
(A) 21cm (B) 23cm (C) 25cm (D) 19cm
(A) 21cm
As we know,
Volume of cuboid = lbh
Where, l = length, b = breadth and h = height
For given cuboid,
Length, l = 49 cm
Breadth, b = 33 cm
Height, h = 24 cm
Volume of cube = 49× (33) × (24) cm3
Now,
Let the radius of cube be r.
As volume of sphere = 4/3 πr3
Where r = radius of sphere
Also,
Volume of cuboid = volume of sphere moulded
So,
49(33)(24) = 4/3 πr3
⇒ πr3 = 29106
⇒ r3 = 29106 × 22/7
⇒ r3 = 9261
⇒ r = ∛ 9261 cm = 21 cm
Hence, radius of sphere is 21 cm