Question 1 |

Consider sand casting of a cube of edge length a. A
cylindrical riser is placed at the top of the casting.
Assume solidification time, t_s\propto V/A , where V isthe
volume and A is the total surface area dissipating
heat. If the top of the riser is insulated, which of the
following radius/radii of riser is/are acceptable?

**MSQ**\frac{a}{3} | |

\frac{a}{2} | |

\frac{a}{4} | |

\frac{a}{6} |

Question 1 Explanation:

Riser should take more time for solidification than
casting \left ( \frac{V}{A} \right )_r\geq\left ( \frac{V}{A} \right )_c

For top riser bottom area is not cooling surface and it is given that top cross section is also insulated. So only lateral area is cooling area.

\frac{\frac{\pi}{4}D^2H}{\pi DH}\geq \frac{a^3}{6a^2}\Rightarrow \frac{D}{4}\geq \frac{a}{6}\Rightarrow D\geq \frac{2a}{3}

Therefore, Radius R\geq \frac{a}{3} .

only answer is available \frac{a}{2}.

For top riser bottom area is not cooling surface and it is given that top cross section is also insulated. So only lateral area is cooling area.

\frac{\frac{\pi}{4}D^2H}{\pi DH}\geq \frac{a^3}{6a^2}\Rightarrow \frac{D}{4}\geq \frac{a}{6}\Rightarrow D\geq \frac{2a}{3}

Therefore, Radius R\geq \frac{a}{3} .

only answer is available \frac{a}{2}.

Question 2 |

A cast product of a particular material has dimensions 75 mm x 125 mm x 20 mm. The total solidification time for the cast product is found to be 2.0 minutes as calculated using Chvorinov's rule having the index, n = 2. If under the identical casting conditions, the cast product shape is changed to a cylinder having diameter = 50 mm and height = 50 mm, the total solidification time will be __________minutes (round off to two decimal places).

1.25 | |

2.83 | |

2.48 | |

3.12 |

Question 2 Explanation:

Casting dimensions =75 \mathrm{~mm} \times 125 \mathrm{~mm} \times 20 \mathrm{~mm}

Total solidification time \left(t_{s}\right)_{\mathrm{cu}}=2.0 \mathrm{~min}.

Cylindrical casting =\mathrm{H}=\mathrm{D}=50 \mathrm{~min}

Total solidification Time \left(t_{s}\right)_{c y}=? Chvorinov's rule,

\begin{aligned} t_{s} &\propto\left(\frac{V}{A}\right)^{n} \quad n=2 \\ \frac{\left(t_{s}\right)_{c u}}{\left(t_{s}\right)_{c y}}&=\frac{\left(\frac{V}{A}\right)_{c u}^{2}}{\left(\frac{V}{A}\right)_{c y}^{2}}\\ \left(\frac{V}{A}\right)_{c y} &=\frac{\frac{\pi}{4} d^{2} h}{2 \frac{\pi}{4} d^{2} h+\pi d h} \\ h &=d \\ \left(\frac{V}{A}\right)_{c y} &=\frac{d}{6} \\ \frac{\left(t_{s}\right)_{c U}}{\left(t_{s}\right)_{c_{y}}} &=\frac{\left[\frac{75 \times 125 \times 20}{2(75 \times 125+125 \times 20+20 \times 75)}\right]^{2}}{\left(\frac{50}{6}\right)^{2}} \\ \frac{2}{\left(t_{s}\right)_{c y}} &=\frac{49.131}{69.44}=0.7075 \\ \left(t_{s}\right)_{c y} &=2.83 \mathrm{~min} . \end{aligned}

Total solidification time \left(t_{s}\right)_{\mathrm{cu}}=2.0 \mathrm{~min}.

Cylindrical casting =\mathrm{H}=\mathrm{D}=50 \mathrm{~min}

Total solidification Time \left(t_{s}\right)_{c y}=? Chvorinov's rule,

\begin{aligned} t_{s} &\propto\left(\frac{V}{A}\right)^{n} \quad n=2 \\ \frac{\left(t_{s}\right)_{c u}}{\left(t_{s}\right)_{c y}}&=\frac{\left(\frac{V}{A}\right)_{c u}^{2}}{\left(\frac{V}{A}\right)_{c y}^{2}}\\ \left(\frac{V}{A}\right)_{c y} &=\frac{\frac{\pi}{4} d^{2} h}{2 \frac{\pi}{4} d^{2} h+\pi d h} \\ h &=d \\ \left(\frac{V}{A}\right)_{c y} &=\frac{d}{6} \\ \frac{\left(t_{s}\right)_{c U}}{\left(t_{s}\right)_{c_{y}}} &=\frac{\left[\frac{75 \times 125 \times 20}{2(75 \times 125+125 \times 20+20 \times 75)}\right]^{2}}{\left(\frac{50}{6}\right)^{2}} \\ \frac{2}{\left(t_{s}\right)_{c y}} &=\frac{49.131}{69.44}=0.7075 \\ \left(t_{s}\right)_{c y} &=2.83 \mathrm{~min} . \end{aligned}

Question 3 |

A true centrifugal casting operation needs to be performed horizontally to make copper tube sections with outer diameter of 250 mm and inner diameter of 230 mm. The value of acceleration due to gravity, g = 10 \; m/s^2. If a G-factor (ratio of centrifugal force to weight) of 60 is used for casting the tube, the rotational speed required is _____rpm

(round off to the nearest integer).

(round off to the nearest integer).

124 | |

284 | |

662 | |

847 |

Question 3 Explanation:

\begin{aligned} \text { Given: } D_{0}&=250 \mathrm{~mm} ; \\ D_{i}&=230 \mathrm{~mm} ; \\ g&=10 \mathrm{~m} / \mathrm{s}^{2} ; \\ \text { G-factor }&=60 \\ F_{c} &=m r \omega^{2} \\ F_{g} &=m g \\ \text { G-factor } &=\frac{F_{c}}{F_{g}}=60 \\ \text { G-factor } &=\frac{F_{c}}{F_{g}}=\frac{m r \omega^{2}}{m g}=\frac{D_{0}}{2 g}\left(\frac{2 \pi N}{60}\right)^{2} \\ N &=\sqrt{\frac{2 \times G-\text { Factor } \times g}{D_{0}}} \times\left(\frac{60}{2 \pi}\right) \\ N &=\sqrt{\frac{2 \times 60 \times 10}{0.25}} \times\left(\frac{60}{2 \pi}\right) \\ N &=661.93 \\ N &=662 \mathrm{rpm} \end{aligned}

Question 4 |

A mould cavity of 1200 cm^3 volume has to be filled through a sprue of 10 cm length
feeding a horizontal runner. Cross-sectional area at the base of the sprue is 2 cm^2.
Consider acceleration due to gravity as 9.81 m/s^2. Neglecting frictional losses due to
molten metal flow, the time taken to fill the mould cavity is _______ seconds (round off
to 2 decimal places).

10.6 | |

2.24 | |

4.28 | |

15.3 |

Question 4 Explanation:

Volume of mould cavity (V)=1200 \mathrm{cm}^{3}

Height of sprue \left(h_{s}\right)=10 \mathrm{cm}

Area of sprue at the bottom \left(A_{s}\right)=2 \mathrm{cm}^{2}

g=9.81 \mathrm{m} / \mathrm{s}^{2}

Sprue is feed a horizontal runner: Filling time required \left(t_{t}\right)=?

By assuming top gate, A_{g}=A_{s}=A_{3}=2 \mathrm{cm}^{2}

\begin{aligned} \text { Filling time }\left(t_{t}\right) &=\frac{V}{A_{g} v_{g}}=\frac{1200}{2 \sqrt{2 \times 981}}=4.28 \mathrm{s} \\ t_{f} &=4.28 \mathrm{s} \end{aligned}

Height of sprue \left(h_{s}\right)=10 \mathrm{cm}

Area of sprue at the bottom \left(A_{s}\right)=2 \mathrm{cm}^{2}

g=9.81 \mathrm{m} / \mathrm{s}^{2}

Sprue is feed a horizontal runner: Filling time required \left(t_{t}\right)=?

By assuming top gate, A_{g}=A_{s}=A_{3}=2 \mathrm{cm}^{2}

\begin{aligned} \text { Filling time }\left(t_{t}\right) &=\frac{V}{A_{g} v_{g}}=\frac{1200}{2 \sqrt{2 \times 981}}=4.28 \mathrm{s} \\ t_{f} &=4.28 \mathrm{s} \end{aligned}

Question 5 |

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 x 30 cm x 15 cm) is___

12.55 | |

45.28 | |

28.92 | |

36.28 |

Question 5 Explanation:

\frac{\text { Volume }}{\mathrm{t}}=\mathrm{A}_{\mathrm{c}} \sqrt{2 \mathrm{gh}_{\mathrm{c}}}

At point (3) is open to atmosphere so (3) is the choke.

\mathrm{g}=9.81 \mathrm{m} / \mathrm{sec}^{2}=981 \mathrm{cm} / \mathrm{sec}^{2} \\ \frac{40 \times 30 \times 15}{\mathrm{t}}=\frac{\pi}{4} \times 2^{2} \sqrt{2 \times 981 \times 20} \Rightarrow \frac{18000}{\mathrm{t}}=622.32088

\Rightarrow \mathrm{t}=28.923 \mathrm{sec}

At point (3) is open to atmosphere so (3) is the choke.

\mathrm{g}=9.81 \mathrm{m} / \mathrm{sec}^{2}=981 \mathrm{cm} / \mathrm{sec}^{2} \\ \frac{40 \times 30 \times 15}{\mathrm{t}}=\frac{\pi}{4} \times 2^{2} \sqrt{2 \times 981 \times 20} \Rightarrow \frac{18000}{\mathrm{t}}=622.32088

\Rightarrow \mathrm{t}=28.923 \mathrm{sec}

Question 6 |

The fluidity of molten metal of cast alloys (without any addition of fluxes) increases with increase in

viscosity | |

surface tension | |

freezing range | |

degree of superheat |

Question 6 Explanation:

Fluidity increases with increase of degree of super heat.

Question 7 |

Match the following sand mold casting defects with their respective causes.

P-4, Q-3, R-1, S-2 | |

P-3, Q-4, R-2, S-1 | |

P-2, Q-4, R-1, S-3 | |

P-3, Q-4, R-1, S-2 |

Question 7 Explanation:

Blow hole \Rightarrow Poor Permeability

Misrun \Rightarrow Insufficient Fluidity

Hot tearing \Rightarrow Poor Collapsibility

Wash \Rightarrow Mold erosion

Misrun \Rightarrow Insufficient Fluidity

Hot tearing \Rightarrow Poor Collapsibility

Wash \Rightarrow Mold erosion

Question 8 |

In a casting process, a vertical channel through which molten metal flows downward from pouring basin to runner for reaching the mold cavity is called

blister | |

sprue | |

riser | |

pin hole |

Question 8 Explanation:

Vertical channel is called sprue and the horizontal channel is called runner

Question 9 |

For sand-casting a steel rectangular plate with dimensions 80 mm \times 120 mm \times 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).

51.84 | |

41.28 | |

85.96 | |

74.87 |

Question 9 Explanation:

\begin{aligned} \text { casting } \operatorname{size} &=80 \times 120 \times 20 \mathrm{mm} \\ \left(t_{s}\right)_{\text {casting }} &=2 \mathrm{min} \\ \left(t_{s}\right)_{\text {riser }} &=3 \mathrm{min} \\ 2 &=k\left(\frac{80 \times 120 \times 20}{2(80 \times 120+120 \times 20+20 \times 80)}\right)^{2}\\ k &=0.040138 \mathrm{min} / \mathrm{mm}^{2} \\ \text { Riser : } \left(t_{S}\right)_{\text {riser }}&=k\left(\frac{V}{A}\right)^{2} \\ 3 &=0.040138\left(\frac{V}{A}\right)^{2} \\ \left(\frac{V}{A}\right) &=8.64 \\ \frac{d}{6} &=8.64\\ \text{Riser if } h&=d \text{ then }\frac{V}{A}=\frac{d}{6} \\ \frac{V}{A}&=\frac{\frac{\pi}{4} d^{2} h}{2 \frac{\pi}{4} d^{2}+\pi d h}\\ \text{If } h=d, \text{ then }\left(\frac{V}{A}\right)&=\frac{d}{6} \\ h=d&=51.84 \mathrm{mm} \end{aligned}

Question 10 |

Match the following products with the suitable manufacturing process

P-4, Q-3, R-1, S-2 | |

P-2, Q-1, R-3, S-4 | |

P-4, Q-1, R-2, S-3 | |

P-1, Q-3, R-4, S-2 |

Question 10 Explanation:

Tooth paste tube is made by impact extrusion process

Centrifugal casting is used to manufacture metallic pipes.

Plastic bottles are produced by blow moulding

Rolling process is used producing threaded bolts.

Centrifugal casting is used to manufacture metallic pipes.

Plastic bottles are produced by blow moulding

Rolling process is used producing threaded bolts.

There are 10 questions to complete.

Question 11solution is wrong

Thank you Rahul for your suggestions.

Sir,I think question no 3 not put height of the sprue plz Correct other wise thank you for your effort❤️

Sir,plz check the answer of the question no 12 because cavity of the sphere neglected

This is very useful for every gate student, thanks

Solution for 38 question is wrong ..it should be 0.97

Paper is insufficient

In question no. 17. (Neglect heat loss of cylinder top and bottom) this line is missing.Please add this it.

Dear Swapnil Sonawane,

This statement is not mentioned in official GATE 2015 Paper.