1. Riser is designed so as to
(a) Freeze after the casting freezes
(b) Freezing before the casting freezes
(c) Freeze as the same as the casting
(d) Minimise the time of pouring
(1 Mark, 1987)

Ans: a
2. The contraction allowance provided on the pattern and core boxes compensates for the following type of contraction.
(a) Liquid contraction
(b) Solidification contraction
(c) Solid contraction
(d) All the above three type of contraction
(1 Mark, 1988)

Ans: c
Explanation:
Riser is given to compensate Liquid and Solidification Shrinkage.
Contraction allowance is provided for solid contraction.
3. Chills are used in moulds to
(a) Achieve directional solidification
(b) Reduce possibility of blow holes
(c) Reduce the freezing time
(d) Smoothen the metal for reducing the spatter.
(1 Mark, 1989)

Ans: a
Explanation:
Chills and Pads are provided to achieve directional solidification.
4. Increase in water content in the moulding sand causes
(a) Flowability to go through a maximum
(b) Permeability to go through a maximum
(c) Compressive strength to go through a maximum
(d) Strength to go through a maximum
(1 Mark, 1989)

Ans: a
5. Two cubical castings of the same metal and sizes of 2 cm side and 4cm side are moulded in green sand. If the smaller casting solidifies in 2mins, the expected time of solidifications of larger casting will be
(a) 16 min
(b) 2\sqrt { 8 } min
(c) 8 min
(d) 4 min
(2 Mark, 1989)

Ans: c
Explanation:
We know: { t }_{ s }\alpha { \left( \frac { V }{ A } \right) }^{ 2 }
=> { t }_{ s }\alpha { d }^{ 2 }
Now,\frac { { \left( { t }_{ s } \right) }_{ 1 } }{ { \left( { t }_{ s } \right) }_{ 2 } } =\frac { { 4 }^{ 2 } }{ { 2 }^{ 2 } }
=> \frac { { \left( { t }_{ s } \right) }_{ 1 } }{ 2 } =4
So, the time of solidification of larger casting is 2*4 = 8 min.
6. The pressure in the gate will be maximum with the gating system
(a) 4:8:3
(b) 1:3:3
(c) 1:2:4
(d) 1:2:1
(1 Mark, 1990)

Ans: a
Explanation:
Pressure is maximum if the gate area is minimum compare to sprue and runner area.
In gating ratio 4:8:3, the gate area is minimum.
7. When there is no room temperature change, the total shrinkage allowance on a pattern is independent of
(a) Pouring temperature of the liquid metal
(b) Freezing temperature of liquid metal
(c) The component size
(d) Coefficient of thermal contraction of solidified metal
(1 Mark, 1991)

Ans: a
Explanation:
Shrinkage allowance = L\alpha \left( { T }_{ r }{ T }_{ f } \right)
Where, L = Component length
\alpha = Coefficient of thermal contraction
{ T }_{ r } = Room temperature
{ T }_{ f } = Freezing temperature
So, the shrinkage allowance is independent of pouring temperature.
8. Converging passage is used for feeding the liquid molten metal into the mould to
(a) Increase the rate of feeding
(b) Quickly break off the protruding portion of the casting
(c) Decrease wastage of cast metal
(d) Avoid aspiration of air
(1 Mark, 1991)

Ans: d
9. In a greensand moulding process, uniform ramming leads to
(a) Less chance of gas porosity
(b) Uniform flow of molten metal into the mould cavity
(c) Greater dimensional stability of the casting
(d) Less sand expansion type of casting defect
(1 Mark, 1992)

Ans: c

Ans: A – S, B – R, C – Q, D – T
11. Centrifugally cast products have
(a) Large grain structure with high porosity
(b) Fine grain structure with high density
(c) Fine grain structure with low density
(d) Segregation of slug towards the outer skin of the casting
(1 Mark, 1993)

Ans: b
Explanation:
In centrifugal casting, the product has fine grain structure, high density, Low porosity and segregation of slug towards the inner skin of the casting.

Ans: A – 2, B – 6, C – 5, D – 3, E – 4, F 1
13. Light impurities in the molten metal are prevented from reaching the mould cavity by providing a
(a) Strainer
(b) Button well
(c) Skim bob
(d) All of the above
(1 Mark, 1996)

Ans: c

Ans: A – 5, B – 6, C 1 , D – 3
15. Which of the following materials requires the largest shrinkage allowance, while making a pattern for casting?
(a) Aluminium
(b) Brass
(c) Cast iron
(d) Plain Carbon Steel
(1 Mark, 1999)

Ans: d
16. Disposable patterns are made of
(a) Wood
(b) Rubber
(c) Metal
(d) Polystyrene
(1 Mark, 2000)

Ans: d
17. Shrinkage allowance on pattern is provided to compensate for shrinkage when
(a) The temperature of liquid metal drops from pouring to freezing temperature
(b) The metal changes from liquid to solid state at freezing temperature
(c) The temperature of solid phase drops from freezing to room temperature
(d) The temperature of metal drops from pouring to room temperature
(1 Mark, 2001)

Ans: c
Explanation:
Shrinkage allowance = L\alpha \left( { T }_{ r }{ T }_{ f } \right)
Where, L = Component length
\alpha = Coefficient of thermal contraction
{ T }_{ r } = Room temperature
{ T }_{ f } = Freezing temperature
18. In centrifugal casting, the impurities are
(a) Uniformly distributed
(b) Forced towards the outer surface
(c) Trapped near the mean radius of the casting
(d) Collected at the center of the casting
(1 Mark, 2002)

Ans: d
19. The primary purpose of a sprue in a casting mould is to
(a) Feed the casting at a rate consistent with the rate of solidification
(b) Act as a reservoir for molten metal
(c) Feed molten metal from the pouring basin to the gate
(d) Help feed the casting until all solidification takes place
(1 Mark, 2002)

Ans: c
20. Hardness of green sand mould increases with
(a) Increase in moisture content beyond 6 percent
(b) Increase in permeability
(c) Decrease in permeability
(d) Increase in both moisture content and permeability
(1 Mark, 2003)

Ans: c
21. With a solidification factor of 0.97 \times10^{6} s/m^{2}, the solidification time (in seconds) for a spherical casting of 200 mm diameter is
(a) 539
(b) 1078
(c) 4311
(d) 3233
(2 Mark, 2003)

Ans: b
Explanation:
Given, d = 200 mm = 0.2 m
For sphere: { \left( \frac { V }{ A } \right) }=\frac { \frac { 1 }{ 6 } \pi { d }^{ 3 } }{ \pi { d }^{ 2 } } =\frac { d }{ 6 } = 0.1/3 m
Solidification time = K{ \left( \frac { V }{ A } \right) }^{ 2 } = 1078 s
22. Misrun is a casting defect which occurs due to
(a) Very high pouring temperature of the metal
(b) Insufficient fluidity of the molten metal
(c) Absorption of gases by the liquid metal
(d) Improper alignment of the mould flasks
(1 Mark, 2004)

Ans: b
23. Gray cast iron blocks 200*100*10 mm are to be cast in sand moulds. Shrinkage allowance for pattern making is 1%, The ratio of the volume of pattern to that of the casting will be
(a) 0.97
(b) 0.99
(c) 1.01
(d) 1.03
(2 Mark, 2004)

Ans: d
Explanation:
Ratio of volume of pattern to that of casting = \frac { (1.01\times 200)\times (1.01\times 100)\times (1.01\times 10) }{ 200\times 100\times 10 } = 1.03
24. A mould has a down sprue whose length is 20 cm and the cross sectional area at the base of the down sprue is 1 cm^{2}. The down sprue feeds a horizontal runner leading into the mould cavity of volume 1000 cm^{3}. The time required to fill the mould cavity will be
(a) 4.05 s
(b) 5.05 s
(c) 6.05 s
(d) 7.25 s
(2 Mark, 2005)

Ans: b
Explanation:
Velocity of molten metal = \sqrt { 2\times g\times h } =\quad \sqrt { 2\times 981\times 20 } =198.1cm/s
The time required to fill the mold cavity will be = \frac { Volume\quad of\quad mould\quad cavity }{ Area\times Velocity } =\frac { 1000 }{ 1\times 198.1 } =5.05s
26. An expendable pattern is used in
(a) Slush casting
(b) Squeeze casting
(c) Centrifugal casting
(d) Investment casting
(1 Mark, 2006)

Ans: d
Explanation:
Expendable mold casting is a generic classification that includes sand, plastic, shell, plaster, and investment (lostwax technique) molds.
All these methods use temporary, nonreusable molds. After the molten metal in the mold cavity solidifies, the mold is broken to take out the solidified cast.
Expendable mold casting processes are suitable for very complex shaped parts and materials with high melting point temperature.
However, the rate of production is often limited by the time to make mold rather than the casting itself.
27. In a sand casting operation, the total liquid head is maintained constant such that it is equal to the mould height. The time taken to fill the mould with a top gate is t_{A}. If the same mould is filed with a bottom gate, then the time taken is t_{B}. Ignore the time required to fill the runner and frictional effects. Assume atmospheric pressure at the top molten metal surfaces. The relation between t_{A }and t_{B} is:
(a) t_{B} = \sqrt { 2 }t_{A }
(b) t_{B} = 2t_{A }
(c) t_{B} =t_{A}/\sqrt { 2 }
(d) t_{B}=2\sqrt { 2 }t_{A}
(2 Mark, 2006)

Ans: b
Explanation:
Time taken to fill the mould with top gate is t_{A}.
{ t }_{ A }=\frac { AH }{ { A }_{ g }\sqrt { 2g{ h }_{ m } } } ………..(1)
Where, A = Area of mold
H = Height of mould
A_{g }= Area of gate
h_{m }= Total liquid head
Given that h_{m }= H, So equation 1 becomes:
{ t }_{ A }=\frac { A\sqrt { { h }_{ m } } }{ { A }_{ g }\sqrt { 2g } } ………….(2)
Time taken to fill the mold with bottom gate
{ t }_{ B }=\frac { 2A }{ { A }_{ g }\sqrt { 2g } } \left( \sqrt { { h }_{ m } } \sqrt { { h }_{ m }H } \right)though h_{m }= H,
{ t }_{ B }=\frac { 2A\sqrt { { h }_{ m } } }{ { A }_{ g }\sqrt { 2g } } …………..(3)
From equation 2 and 3 we get t_{B}= 2t_{A}
28. Which of the following engineering materials is the most suitable candidate for hot chamber die casting?
(a) Low carbon steel
(b) Titanium
(c) Copper
(d) Tin
(1 Mark, 2007)

Ans: d
Explanation:
Hot chamber die casting is used for low melting temperature alloys like lead, tin and zinc.
29. Volume of a cube of side ‘l ’ and volume of a sphere of radius ‘r ’ are equal. Both the cube and the sphere are solid and of same material. They are being cast. The ratio of the solidification time of cube to the same of the sphere is
(a){ \left( \frac { 4\pi }{ 6 } \right) }^{ 3 }{ \left( \frac { r }{ l } \right) }^{ 6 }
(b){ \left( \frac { 4\pi }{ 6 } \right) }{ \left( \frac { r }{ l } \right) }^{ 2 }
(c){ \left( \frac { 4\pi }{ 6 } \right) }^{ 2 }{ \left( \frac { r }{ l } \right) }^{ 3 }
(d){ \left( \frac { 4\pi }{ 6 } \right) }^{ 2 }{ \left( \frac { r }{ l } \right) }^{ 4 }
(1 Mark, 2007)

Ans: d
Explanation:
Given, Volume of cube = Volume of sphere
The ratio of the solidification time = \frac { { { { \left( \frac { V }{ A } \right) }_{ C }^{ 2 } } } }{ { \left( \frac { V }{ A } \right) }_{ S }^{ 2 } } ={ \left( \frac { { A }_{ S } }{ { A }_{ C } } \right) }^{ 2 }={ \left( \frac { 4\pi }{ 6 } \right) }^{ 2 }{ \left( \frac { r }{ l } \right) }^{ 4 }
30. A 200 mm long down sprue has an area of crosssection of 650 mm^{2} where the pouring basin meets the down sprue (i.e at the beginning of the down sprue). A constant head of molten metal is maintained by the pouring basin. The molten metal flow rate is 6.5*10^{5} mm^{3}/s. Considering the end of down sprue to be open to atmosphere and an acceleration due to gravity of 10^{4} mm/s^{2}, the area of the down sprue in mm^{2} at its end (avoiding aspiration effect) should be
(a) 650.0
(b) 350.0
(c) 290.7
(d) 190.0
(2 Mark, 2007)

Ans: c
Explanation:
The velocity of molten metal at the beginning of sprue
= \frac { Q }{ A } =\frac { 6.5\times { 10 }^{ 5 } }{ 650 } =1000{ mm/s}
Velocity at the exit of sprue = \sqrt { { 1000 }^{ 2 }+2\times 10000\times 200 } =2236.07mm/s
Area of the own sprue = \frac { Q }{ V } =\frac { 6.5\times { 10 }^{ 5 } }{ 2236.067 } =290.68{ mm }^{ 2 }
31. While cooling, a cubical casting of side 40 mm undergoes 3%, 4% and 5% volume shrinkage during the liquid state, phase transition and solid state, respectively. The volume of metal compensated from the riser is
(a) 2%
(b) 7%
(c) 8%
(d) 9%
(2 Mark, 2008)

Ans: b
Explanation:
The volume of metal is compensated from the riser = 3 + 4 = 7%
Note:
The liquid and solidification shrinkage is compensated by riser.
The solid shrinkage is compensated by shrinkage allowance.
32. Match the items in Column I and Column II.
(b) P1, Q4, R2, S3
(c) P3, Q4, R2, S1
(d) P4, Q1, R2, S3
(2 Mark, 2009)

Ans: d
33. Two streams of liquid metal, which are not hot enough to fuse properly, result into a casting defect known as
(a) Cold shut
(b) Swell
(c) Sand wash
(d) Scab
(1 Mark, 2009)

Ans: a
34. In a gating system, the ratio 1:2:4 represents
(a) Sprue base area: runner area: ingate area
(b) Pouring basin area: ingate area: runner area
(c) Sprue base area: ingate area: casting area
(d) Runner area: ingate area: casting area
(1 Mark, 2010)

Ans: a
35. Green sand mould indicates that
(a) Polymeric mould has been cured
(b) Mould has been totally dried
(c) Mould is green in colour
(d) Mould contains moisture
(1 Mark, 2011)

Ans: d
36. A cubic casting of 50mm side undergoes volumetric solidification shrinkage and volumetric solid contraction of 4% and 6% respectively. No. riser is used. Assume uniform cooling in all directions. The side of the cube after solidification and contraction is
(a) 48.32mm
(b) 49.90mm
(c) 49.94mm
(d) 49.96mm
(2 Mark, 2011)

Ans: a
Explanation:
Volume of the cubic casting (a^{3}) = 50^{3} mm^{3}
Volume of the cube after solidification
= 0.96\times 0.94\times 50 = 112800 mm^{3}
Now, side of the cube after solidification
= \sqrt [ 3 ]{ 112800 } = 48.317 mm
37. A cube shaped casting solidifies in 5 minutes. The solidification time in minutes for a cube of the same material, which is 8 times heavier than the original casting will be
(a) 10
(b) 20
(c) 24
(d) 40
(1 Mark, 2013)

Ans: b
Explanation:
Given m_{2} = 8m_{1}
=> V_{2} = 8V_{1} [Because density is constant]
=> \left( { a }_{ 2 } \right) ^{ 3 }=8\left( { a }_{ 1 } \right) ^{ 3 }
=>{ a }_{ 2 }=2{ a }_{ 1 }
We know, t\alpha { \left( \frac { V }{ A } \right) }^{ 2 }
=>t\alpha { \left( \frac { a }{ 6 } \right) }^{ 2 }
Now \frac { { t }_{ 2 } }{ { t }_{ 1 } } ={ \left( \frac { { a }_{ 2 } }{ { a }_{ 1 } } \right) }^{ 2 }
=>\frac { { t }_{ 2 } }{ 5 } ={ \left( \frac { { 2a }_{ 1 } }{ { a }_{ 1 } } \right) }^{ 2 }
=> t_{2 }= 20 min
38. Match the casting defects (Group A) with the probable causes (Group B):
(a) P1, Q3, R2, S4
(b) P4, Q3, R2, S1
(c) P3, Q4, R2, S1
(d) P1, Q2, R4, S3
(1 Mark, 2014[3])

Ans: b
39. The hot tearing in a metal casting is due to
(a) High fluidity
(b) High melt temperature
(c) Wide range of solidification temperature
(d) Low coefficient of thermal expansion
(1 Mark, 2014[3])

Ans: c
40. A cylindrical blind riser with diameter d and height h, is placed on the top of the mold cavity of a closed type sand mold as shown in the figure. If the riser is of constant volume, then the rate of solidification in the riser is the least when the ratio h/d is
(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1
(2 Mark, 2014[3])

Ans: a
Explanation:
Volume of the riser (V) = \frac { \pi }{ 4 } { d }^{ 2 }h
=> h = \frac { 4V }{ { \pi d }^{ 2 } }
Surface area of the riser (A)
= \pi dh+\frac { \pi }{ 4 } { d }^{ 2 }=\frac { 4V }{ { d } } +\frac { \pi }{ 4 } { d }^{ 2 }
For least solidification time, Surface area should be maximum.
So, \frac { dA }{ dd } =0
=> \frac { 4V }{ { d }^{ 2 } } +\frac { \pi d }{ 2 } =0
=> { d }^{ 3 }=\frac { 8 }{ \pi } V=\frac { 8 }{ \pi } \times \frac { \pi }{ 4 } { d }^{ 2 }h
=> \frac { h }{ d } =\frac { 1 }{ 2 }
41. A cylindrical riser of 6 cm diameter and 6 cm height has to be designed for a sand casting mould for producing a steel rectangular plate casting of 7cm × 10cm×2cm dimensions having the total solidification time of 1.36 minute. The total solidification time (in minute) of the riser is ___.
(2 Mark, 2014[4])

Ans: 3 min. (Range 2.5 – 4.5)
Explanation:
For casting:
{ \left( \frac { V }{ A } \right) }_{ C }=\frac { 7\times 10\times 2 }{ 2(7\times 2+10\times 2+7\times 10) } =0.673 cmFor riser: (d = h)
{ \left( \frac { V }{ A } \right) }_{ R }=\frac { \frac { \pi }{ 4 } { d }^{ 3 } }{ \frac { \pi }{ 4 } { d }^{ 2 }+\pi { d }^{ 2 } } =1.2\quad cmNow, \frac { { t }_{ R } }{ { t }_{ C } } =\frac { { \left( \frac { V }{ A } \right) }_{ R }^{ 2 } }{ { \left( \frac { V }{ A } \right) }_{ C }^{ 2 } } ={ \left( \frac { 1.2 }{ 0.673 } \right) }^{ 2 }
=> \frac { { t }_{ R } }{ 1.36 } =3.17
=> Total solidification time for the riser = 3.17*1.36 =4.32 min
42. The solidification time of a casting is proportional to { \left( \frac { V }{ A } \right) }^{ 2 }, where V is the volume of the casting and A is the total casting surface area losing heat. Two cubes of same material and size are cast using sand casting process. The top face of one of the cubes is completely insulated. The ratio of the solidification time for the cube with top face insulated to that of the other cube is
(a) \frac { 25 }{ 36 }
(b) \frac { 36 }{ 25 }
(c) 1
(d) \frac { 6 }{ 5 }
(2 Mark, 2015[1])

Ans: b
Explanation:
Solidification time (t) \alpha { \left( \frac { V }{ A } \right) }^{ 2 }
Ratio of solidification time for the cube with top face insulated to that of the other cube is: \frac { { t }_{ 1 } }{ { t }_{ 2 } } =\frac { { \left( \frac { { V }_{ 1 } }{ { A }_{ 1 } } \right) }^{ 2 } }{ { \left( \frac { { V }_{ 2 } }{ { A }_{ 2 } } \right) }^{ 2 } }
=> \frac { { t }_{ 1 } }{ { t }_{ 2 } } =\frac { { { A }_{ 2 } }^{ 2 } }{ { { A }_{ 1 } }^{ 2 } } [Given: V_{1} = V_{2}]
=> \frac { { t }_{ 1 } }{ { t }_{ 2 } } =\frac { { \left( 6{ a }^{ 2 } \right) }^{ 2 } }{ { \left( 5{ a }^{ 2 } \right) }^{ 2 } }
=> \frac { { t }_{ 1 } }{ { t }_{ 2 } } =\frac { 36 }{ 25 }
43. A cube and a sphere made of cast iron (each of volume 1000 cm^{3}) were cast under identical conditions. The time taken for solidifying the cube was 4 s. The solidification time (in s) for the sphere is _____.
(2 Mark, 2015[2])

Ans: 6.16 (Range 6 – 6.3)
Explanation:
According to Chvorinov’s rule,
Solidification time (t) \alpha { \left( \frac { V }{ A } \right) }^{ 2 }
Ratio of solidification time for the sphere and cube is: \frac { { t }_{ s } }{ { t }_{ c } } =\frac { { \left( \frac { { V }_{ s } }{ { A }_{ s } } \right) }^{ 2 } }{ { \left( \frac { { V }_{ c } }{ { A }_{ c } } \right) }^{ 2 } }
=> \frac { { t }_{ s } }{ { t }_{ c } } =\frac { { { A }_{ c } }^{ 2 } }{ { { A }_{ s } }^{ 2 } } [Given: V_{s} = V_{c}]
\frac { { t }_{ s } }{ { t }_{ c } } ={ \left( \frac { 6{ a }^{ 2 } }{ 4\pi { r }^{ 2 } } \right) }^{ 2 }Volume of cube (a^{3 })= 1000 cm^{3 }
=> a^{ }= 10 cm
Volume of sphere (\frac { 4 }{ 3 } \pi { r }^{ 3 })= 1000 cm^{3 }
=> r = 6.203 cm
Putting the values of a and r:
\frac { { t }_{ s } }{ { t }_{ c } } ={ \left( \frac { 6{ a }^{ 2 } }{ 4\pi { r }^{ 2 } } \right) }^{ 2 }=> \frac { { t }_{ s } }{ 4 } ={ \left( \frac { 6{ \times 10 }^{ 2 } }{ 4\pi \times { 6.203 }^{ 2 } } \right) }^{ 2 }
=> t_{s }= 6.157 s
44. Ratio of solidification time of a cylindrical casting (height = radius) to that of a cubic casting of side two times the height of cylindrical casting is_____.
(2 Mark, 2015[3])

Ans: 0.5625 (Range 0.5 – 0.6)
Explanation:
For cylindrical casting:
>{ V }_{ cy }=\pi { R }^{ 2 }H=\pi { R }^{ 3 }
>{ A }_{ cy }=2\pi { R }^{ 2 }+2\pi RH=4\pi { H }^{ 2 }
We know: { t }_{ cy }\alpha { \left( \frac { { V }_{ cy } }{ { A }_{ cy } } \right) }^{ 2 }
=> { t }_{ cy }\alpha \frac { 1 }{ 16 } H^{ 2 }
For cubic casting:
>{ V }_{ cu }={ a }^{ 3 }=8{ H }^{ 3 }
>{ A }_{ cu }={ 6a }^{ 2 }={ 24H }^{ 2 }
Again, { t }_{ cu }\alpha { \left( \frac { { V }_{ cu } }{ { A }_{ cu } } \right) }^{ 2 }
=>{ t }_{ cu }\alpha \frac { 1 }{ 9 } { H }^{ 2 }
Now ratio of solidification time
= \frac { { t }_{ cy } }{ t_{ cu } } =\frac { 9 }{ 16 } =0.5625
45. In full mould(cavityless) casting process, the pattern is made of
(a) Expanded polystyrene
(b) Wax
(c) Epoxy
(d) Plaster of paris
(2 Mark, 2015[3])

Ans: a
46. The dimensions of a cylindrical side riser (height = diameter) for a 25cm \times 15cm \times 5cm steel casting are to be determined. For the tabulated shape factor values given below, the diameter of the riser (in cm) is __.
(2 Mark, 2015[3])

Ans: 10.6
Explanation:
Given, Length (L) = 25 cm,
Width (w) = 15 cm, Thickness (t) = 5 cm
Shape factor (Z) = \frac { L+w }{ t } =\frac { 25+15 }{ 5 } =8
From the table in the question, Corresponding value of 8 is 0.5.
So, \frac { Riser\quad volume }{ Casting\quad volume } = 0.5
=> \frac { \frac { \pi }{ 4 } { d }^{ 2 }h }{ L\times w\times t } =\frac { \frac { \pi }{ 4 } { d }^{ 3 } }{ 25\times 15\times 5 } =0.5
Solving, We get d = 10.6 cm
47. The part of a gating system which regulates the rate of pouring of molten metal is
(a) Pouring basin
(b) Runner
(c) Choke
(d) Ingate
(1 Mark, 2016[1])

Ans: c
48. A cylindrical job with diameter of 200 mm and height of 100 mm is to be cast using modulus method of riser design. Assume that the bottom surface of cylindrical riser does not contribute as cooling surface. If the diameter of the riser is equal to its height, then the height of the riser (in mm) is
(a) 150
(b) 200
(c) 100
(d) 125
(2 Mark, 2016[1])

Ans: a
Explanation:
Given, for riser d = h.
So, { \left( \frac { V }{ A } \right) }_{ R }=\frac { \frac { \pi }{ 4 } { d }^{ 2 }h }{ \pi dh+\frac { \pi }{ 4 } { d }^{ 2 } } =\frac { d }{ 5 }
For casting: D = 200 mm, H = 100 mm
Now, { \left( \frac { V }{ A } \right) }_{ C }=\frac { \frac { \pi }{ 4 } { D }^{ 2 }H }{ \pi DH+2\times \frac { \pi }{ 4 } { D }^{ 2 } } =25
According to modulus method:
{ M }_{ R }=1.2\times { M }_{ C }
=> { \left( \frac { V }{ A } \right) }_{ R }=1.2\times { \left( \frac { V }{ A } \right) }_{ c }=> \frac { d }{ 5 } =1.2\times 25
=> d = h = 150 mm
49. Heat is removed from a molten metal of mass 2 kg at a constant rate of 10 kW till it is completely solidified. The cooling curve is shown in the figure.
Assuming uniform temperature throughout the volume of the metal during solidification, the latent heat of fusion of the metal (in kJ/kg) is _______.
(2 Mark, 2016[1])

Ans: 50
Explanation:
Total heat removed at 873K is = \dot { Q } \times \Delta t=10kW\times 10s=100kJ
The latent heat of fusion (in kJ/kg) = 100/2 = 50kJ/kg
50. Gray cast iron of size 100mm \times 50mm \times 10mm with a central cavity of diameter 4 mm are sand cast. The shrinkage allowance for the pattern is 3%. The ratio of the volume of the pattern to volume of the casting is______.
(2 Mark, 2016[2])

Ans: 1.093
Explanation:
The size of pattern must be larger then the casting because of the the solid shrinkage of cast iron.
The solid shrinkage occurs linearly and it is given 3%.
Now, Ratio of volume of pattern to volume of casting is given by:
\frac { (1.03\times 100)\times (1.03\times 50)\times (1.03\times 10) }{ 100\times 50\times 10 } = 1.093
Note: Volume of cavity is assumed to be small and negligible.
51. A sprue in a sand mould has a top diameter of 20 mm and height of 200 mm. The velocity of the molten metal at the entry of the sprue is 0.5 m/s. Assume accelaration due to gravity at 9.8 m/s^{2} and neglect all losses. If the mould is well ventilated, the velocity (upto 3 decimal points accuracy) of the molten metal at the bottom of the sprue is _____ m/s.
(2 Mark, 2017[1])

Ans: 2.042 m/s (Range 2.04 – 2.07)
Explanation:
Applying Bernoulli’s theorem,
\frac { { P }_{ 1 } }{ \rho g } +\frac { { V }_{ 1 }^{ 2 } }{ 2g } +{ h }_{ 1 }=\frac { { P }_{ 2 } }{ \rho g } +\frac { { V }_{ 2 }^{ 2 } }{ 2g } +{ h }_{ 2 }=> \frac { 0.5^{ 2 } }{ 2g } +0.2=\frac { { P }_{ 2 } }{ \rho g } +\frac { { V }_{ 2 }^{ 2 } }{ 2g } +0
=> V_{2 }= 2.042 m/s
52. For sandcasting a steel rectangular plate with dimensions 80 mm × 120 mm × 20 mm, a cylindrical riser has to be designed. The height of the riser is equal to its diameter. The total solidification time for the casting is 2 minutes. In Chvorinov’s law for the estimation of the total solidification time, exponent is to be taken as 2. For a solidification time of 3 minutes in the riser, the diameter (in mm) of the riser is __________ (correct to two decimal places).
(2 Mark, 2018[2])

Ans: 51.87 (Range 42 – 52)
Explanation:
Using Chvorinov’s principle,
For the casting:
{ t }_{ c }=k{ \left( \frac { V }{ A } \right) }^{ 2 }=> 2 = k{ \left( \frac { 80\times 120\times 20 }{ 2(80\times 120+120\times 20+20\times 80) } \right) }^{ 2 }
=> k = 0.040138 min/mm^{2}
For the riser:
{ t }_{ r }=k{ \left( \frac { V }{ A } \right) }^{ 2 }Where, \frac { V }{ A } =\frac { (\pi /4){ d }^{ 2 }h }{ 2(\pi /4){ d }^{ 2 }+\pi dh }
Since, d = h
\frac { V }{ A } =\frac { d }{ 6 }Now, 3 = 0.40138{ \left( \frac { d }{ 6 } \right) }^{ 2 }
=> d = 51.87 mm
53. In a casting process, a vertical channel through which molten metal flows downward from pouring basin to runner for reaching the mould cavity is called
(a) Blister
(b) Sprue
(c) Riser
(d) Pin hole
(1 Mark, 2019[1])

Ans: b
54. Match the following sand mold casting defects with their respective causes.
(a) P – 4, Q – 3, R – 1, S – 2
(b) P – 3, Q – 4, R – 2, S – 1
(c) P – 2, Q – 4, R – 1, S – 3
(d) P – 3, Q – 4, R – 1, S – 2
(2 Mark, 2019[1])

Ans: d
55. The fluidity of molten metal of cast alloys (without any addition of fluxes) increases with increase in
(a) Viscosity
(b) Surface tension
(c) Freezing range
(d) Degree of superheat
(1 Mark, 2019[2])

Ans: d
Explanation:
The fluidity of molten metal is inversely proportional to Viscosity, Surface tension and freezing range.
Where as, It is directly proportional to degree of super heat.
56. The figure shows a pouring arrangement for casting of a metal block. Frictional losses are negligible. The acceleration due to gravity is 9.81 m/s^{2} . The time (in s, round off to two decimal places) to fill up the mold cavity (of size 40 cm 30 cm 15 cm) is_____.
(2 Mark, 2019[2])

Ans: 28.92
Explanation:
V = Volume of mold = 40\times 30\times 15 = 18000 cm^{3}
A = Area at the bottom of sprue = \frac { \pi }{ 4 } { d }^{ 2 } = \frac { \pi }{ 4 } \times { 2 }^{ 2 } = 3.142 cm^{2}
v = Velocity at the bottom of sprue = \sqrt { 2\times g\times h } = \sqrt { 2\times 981\times 20 } = 198.1 cm/s
Mold filling time = \frac { V }{ A\times v\quad } = \frac { 18000 }{ 3.142\times 198.1 } = 28.92 s