Mensuration - Volume and Surface Areas

Sponsor Area

Question
SSCCGLENQA12042729

A ball of lead 4 cm in diameter is covered with gold. If the volume of the gold and lead are equal, then the thickness of gold [given cube root of 2 space equals space 1.259] is approximately.

  • 5.038 m

  • 5.190 cm

  • 1.038 cm

  • 0.518 cm

Solution

D.

0.518 cm

Diameter of lead = 4 cm
Radius of lead = 2 cm
Volume of lead = 4 over 3 πr cubed space space equals space 4 over 3 straight pi cross times 2 cubed
Let the thickness of gold be x cm, then
        Volume of gold = 4 over 3 straight pi open parentheses left parenthesis 2 plus straight x right parenthesis cubed minus 2 cubed close parentheses space cu. space cm
      therefore space space 4 over 3 straight pi open parentheses open parentheses 2 plus straight x close parentheses cubed minus 2 cubed close parentheses space equals space 4 over 3 straight pi space cross times space 2 cubed
         rightwards double arrow space space left parenthesis 2 plus straight x right parenthesis cubed space minus space 2 cubed space equals space 2 cubed
rightwards double arrow space space left parenthesis 2 plus straight x right parenthesis cubed space space equals space 8 space plus space 8 space equals space 16
rightwards double arrow space space left parenthesis 2 plus straight x right parenthesis cubed space equals space 2 cubed. space 2
          rightwards double arrow space space 2 space plus space straight x space equals space 2 space cross times space cube root of 2
rightwards double arrow space space 2 space space plus space straight x space equals space 2 space cross times space 1.259 space equals space 2.518
therefore space space space space straight x space equals space 2.518 space minus space 2 space equals space 0.518 space cm
  

Sponsor Area

Question
SSCCGLENQA12042197

A circular swimming pool is surrounded by a concrete wall 4m wide. If the area of the concrete wall surrounding the pool is 11 over 25 that of the pool, then the radius (in m) of the pool is

  • 20

  • 30

  • 16

  • 8

Solution

A.

20


Area of pool  = A
therefore    Area of wall = fraction numerator 11 straight A over denominator 25 end fraction
Area of (pool + wall) = straight pi left parenthesis straight r plus 4 right parenthesis squared comma Area of pool = πr squared
Area of wall = Area of (pool + wall) - Area of pool
  fraction numerator 11 space cross times space πr squared over denominator 25 end fraction space equals space straight pi left parenthesis straight r plus 4 right parenthesis squared space minus space πr squared
11 over 25 cross times πr squared space equals space straight pi left parenthesis straight r squared plus 16 plus 8 straight r right parenthesis minus πr squared space equals space πr squared plus 16 straight pi plus 8 πr minus πr squared
11 over 25 πr squared space equals space 8 straight pi left parenthesis 2 plus straight r right parenthesis
   fraction numerator 11 straight r squared over denominator 25 end fraction space equals space 8 left parenthesis 2 plus straight r right parenthesis space rightwards double arrow space space space 11 straight r squared space equals space 200 space cross times space 2 space plus space 200 straight r
rightwards double arrow space space 11 straight r squared minus 200 straight r minus 400 space equals space 0 space space rightwards double arrow space space 11 straight r squared minus 220 straight r plus 20 straight r minus 400 space equals space 0
rightwards double arrow space space 11 straight r left parenthesis straight r minus 20 right parenthesis space space plus 20 left parenthesis straight r minus 20 right parenthesis space equals space 0
rightwards double arrow space space space space space space left parenthesis straight r minus 20 right parenthesis thin space left parenthesis 11 straight r plus 20 right parenthesis space equals space 0 space space rightwards double arrow space space straight r space equals space 20 space space equals space 0 space space space therefore space space straight r space equals space 20 space straight m

Question
SSCCGLENQA12042130

A cone, a cylinder and a hemisphere stand on equal bases and have equal heights. The ratio of their volumes is

  • 2 : 3 : 1

  • 2 : 1 : 3

  • 1 : 3 : 2

  • 1 : 2  :3

Solution

C.

1 : 3 : 2

Ratio of their volumes
 = cone : cylinder : hemi-sphere
 equals space 1 third πr squared straight h space colon thin space πr squared straight h space colon space 2 over 3 πr cubed
equals space 1 third πr cubed space colon space πr cubed space colon space 2 over 3 πr cubed space space space space space space left square bracket as space straight r space equals space straight h right square bracket
equals space 1 third space colon space 1 space colon space 2 over 3 space equals space 1 space colon space 3 space thin space colon 2

Question
SSCCGLENQA12042725

A conical cup is filled with ice cream. The ice-cream forms a hemispherical shape on its open top. The height of the hemispherical part is 7 cm. The radius of the hemispherical part equals the height of the cone. Then the volume of the ice-cream is open square brackets straight pi space equals space 22 over 7 close square brackets

  • 1078 cubic cm

  • 1708 cubic cm

  • 7108 cubic cm

  • 7180 cubic cm

Solution

A.

1078 cubic cm


Volume of the ice-cream = Volume of hemisphere part + Volume of conical part.
Radius (r) = 7 cm.
Volume of hemisphere = 2 over 3 πr cubed
equals space open parentheses 2 over 3 cross times 22 over 7 cross times 7 cross times 7 close parentheses space cu. space cm.
Volume of conical part = 1 third πr squared straight h
            open square brackets because space straight r space equals space straight h close square brackets
equals space open parentheses 1 third cross times 22 over 7 cross times 7 cross times 7 cross times 7 close parentheses space cu. space cm.
∴  Volume of ice-cream
equals space 2 over 3 cross times 22 over 7 cross times 7 cubed space plus space 1 third cross times 22 over 7 cross times 7 cubed
equals space 22 over 7 cross times 7 cubed space equals space 22 space cross times space 7 squared
equals space 1078 space cu. space cm.

Question
SSCCGLENQA12042203

A conical iron piece having diameter 28 cm and height 30cm is totally immersed into the water of a cylindrical vessel, resulting in the rise of water level by 6.4 cm. The diameter (in cm) of the vessel is

  • 3.5 

  • 35 over 2
  • 35

  • 32

Solution

C.

35

Let the diameter is 2r cm.
  According to the question,
    1 third straight pi open parentheses 28 over 2 close parentheses squared space cross times space 30 space equals space straight pi open parentheses fraction numerator 2 straight r over denominator 2 end fraction close parentheses squared space cross times space 64
   rightwards double arrow space space space space 10 cross times 14 cross times 14 space equals space straight r squared space cross times space 64 space space
rightwards double arrow space space space space straight r squared space equals space fraction numerator 10 space cross times space 14 space cross times space 14 space cross times space 10 over denominator 64 end fraction
therefore                 straight r space equals space fraction numerator 10 space cross times space 14 over denominator 8 end fraction space equals space fraction numerator 5 space cross times space 7 over denominator 2 end fraction cm
Diameter  = 2 cross times fraction numerator 5 cross times 7 over denominator 2 end fraction space equals space 35 cm