DMTCL 's recruitment exam was held on 29th JUNE 2018. Here is the solution of mechanical engineering question. Exam was held in Dhanmondi Govt. Boys High School, Dhaka. Hope this will be a great help to those who wants to do better in engineering job field of Bangladesh.

1. Page 1 of 5
Conducted by
Engr. Md. Al-amin, ME, BUET.
alaminbuet2008@gmail.com
Questions
MCQ:
1. Only Asian country of G8?
2. ধ্বনি নিসের প্রতীি ?
3. েনি নিসেদ িরঃ নিগ্রহ
4. নিনিয়ািা গ্যাে নিল্ড িসি আনিস্কার িরা হয় ?
5. িাাংলাসদসের োসে ভারসতর িতনি জেলা আসে ?
6. িাাংলাসদসের েযাসিলাইি িঙ্গিিু -১ িসি উৎসেনিত হয় ?
7. Passive form of “Let us write a letter” ?
Essay :
1. Metro rail
2. Banglai songkor jati
3.
Departmental Question:
1. A refrigerator makes 500 kg of water at 0 ֯C into ice at 0 ֯C by 30 min. Find the cooling
capacity in TR.
2. Find the power transmitted by a belt running over a pulley of o.6 m diameter at 200 rpm.
The coefficient of friction between the belt and pulley is 0.25, angle of lap 8π/9 radian and
maximum tension in the belt is 2500 N.
3. An engine working on Otto cycle has a cylinder bore 50 mm and stroke 450 mm. The
Clearance Volume is 1.23x105
mm3
. Find the compression ratio & air standard efficiency.
4. In a parallel flow heat exchanger hot fluid enter at 70 ֯C and leave at 40 ֯C while water is used
as cooler has got hot from 30 ֯C to 36 ֯C. Mass of the water is , specific heat is 4200 KJ/Kg ,
overall heat transfer is , then find out the area A required for those condition.
5.Determine an equation of PA-PB of bellow system in terms of γ1 and γ2 where manometric
height & flowing fluid heights are given h2 and h1.
Dhaka Mass Transit
Company Ltd. Project
Exam date: 29th
June
Exam Time: 3.00 PM ~ 5.00 PM ( 2 Hrs)
Exam venue: Dhanmondi Govt. Boys High
School, Dhaka
Question pattern:
GK MCQ + Essay = 15 + 10 = 25 Marks
Bangla MCQ + essay = 15+10 = 25 Marks
English MCQ + essay = 15+ 10 = 25 Marks
Dept. MCQ + Written = 20 + 80 = 100 Marks
-------------------------------------------------------
Total = 175 Marks

2. Page 2 of 5
Conducted by
Engr. Md. Al-amin, ME, BUET.
alaminbuet2008@gmail.com
6. The gearing of a machine tool is shown in fig. The motor shaft is connected to gear A and
rotates at 975 r.p.m. the gear wheel B & C are fixed to parallel shafts rotating together. Gear
wheel D & E are fixed to parallel shafts rotating together. The number of teeth on gear A, B,
C, D, E & F are 20, 50, 25, 75, 26 & 65 respectively. Find the speed of gear F that is fixed on
output shaft.
7. For the following pipe network find out the flow rate and velocity at pipe (3)
8. Calculate dynamic viscosity of an oil, which is used fo lubrication between a square plate
of size 0.8 m x 0.8 m and an inclinded plane with angle of inclination 30 ֯ as shown in the
Figure, The weight of the square plane is 300 N and it slides down the inclined plane with a
uniform velocity of 0.3 m/s. The thickness of the oil flim is 1.5 mm.
9. Draw SFD & BMD of the following load condition.
10. Drawing.

3. Page 3 of 5
Conducted by
Engr. Md. Al-amin, ME, BUET.
alaminbuet2008@gmail.com
Solution
Departmental Question:
1. A refrigerator makes 500 kg of water at 0 ֯C into ice at 0 ֯C by 30 min. Find the cooling
capacity in TR.
Ans:
Cooling capacity ,
TR =
𝑚𝑙𝑓
𝑡
=
500 𝑥 335
30
KJ/ min
= 5583.33 KJ/min =
5583.33
210
= 26.587 TR
[ as 1 TR = 210 KJ/min]
Given Data:
m= mass = 500 Kg
t= time = 30 min
TR = Ton of Refrigeration= ?
lf = 335 KJ/Kg
2. Find the power transmitted by a belt running over a pulley of o.6 m diameter at 200 rpm.
The coefficient of friction between the belt and pulley is 0.25, angle of lap 8π/9 radian and
maximum tension in the belt is 2500 N.
Ans:
We know,
𝑇1
𝑇2
= 𝑒µ𝛳
⇒
2500
𝑇2
= 𝑒0.25 𝑥 8𝜋/9
= 2.0099
⇒ T2 =
2500
2.0099
= 1245.7N
Again, Power = (T1-T2)v= (T1-T2)x
𝜋𝑑𝑁
60
= (2500-1245.7)x
𝜋 𝑥 0.6 𝑥 200
60
= 7880.99 W
Given Data:
d=pulley dia = 0.6 m
N = speed = 200 rpm
µ = coefficient of friction = 0.25
ϴ= angle of lap = 8π/9
T1 = maximum tension = 2500N
P = power =?
3. An engine working on Otto cycle has a cylinder bore 50 mm and stroke 450 mm. The
Clearance Volume is 1.23x105
mm3
. Find the compression ratio & air standard efficiency.
Ans:
We know, swept volume=Vs = πB2
S/4
= π x 502
x 450 / 4= 8.83 X 105
mm3
Now, rc =
𝑉𝑠+𝑉𝑐
𝑉𝑐
=
8.83 𝑋 10^5 +1.23 𝑋 10^5
1.23 𝑋 10^5
=8.18
Again,air standard efficiency,
ηotto = 1 -
1
𝑟𝑐ϒ−1
⇒ ηotto = 1 -
1
8.181.4−1
= 56.85%
Given Data:
B = bore = 50 mm
S = stroke = 450 mm
Vc = Clearance volume = 1.23 X 105
mm3
rc = Compression ratio = ?
ηotto = air standard efficiency=?
ϒ = 1.4
4. In a parallel flow heat exchanger hot fluid enter at 70 ֯C and leave at 40 ֯C while water is used
as cooler has got hot from 30 ֯C to 36 ֯C. Mass of the water is , specific heat is 4200 KJ/Kg ,
overall heat transfer is , then find out the area A required for those condition.
Ans:
we know,
LMTD =
∆𝑇1− ∆𝑇2
ln
∆𝑇1
∆𝑇2
=
40− 4
ln
40
4
=15.63
Again,
Q = mS ΔT = UA(LMTD)
A=mS ΔT/U(LMTD)
= □ * 4200*(36-30)/ □*15.63
= □ m2 (ans:)
Given Data:
Th1=70 ֯C, Th2=40 ֯C
Tc1=30 ֯C, Tc2=36 ֯C
ΔT1= Th1- Tc1=70-30=40
ΔT2= Th2- Tc2=40-36=4
m = mass of water= □ Kg
S = 4200 KJ/Kg
U = overall heat transfer
=□
A = required area = ?

4. Page 4 of 5
Conducted by
Engr. Md. Al-amin, ME, BUET.
alaminbuet2008@gmail.com
5. Determine an equation of PA-PB of bellow system in terms of γ1 and γ2 where manometric
height & flowing fluid heights are given h2 and h1.
Ans:
Difference of pressure head between A & B ,
H = h2 (
ϒ2
ϒ1
-1)
Again,
PA-PB = ϒ1 H = ϒ1 h2 (
ϒ2
ϒ1
-1)= h2 ( ϒ2- ϒ1 )
6. The gearing of a machine tool is shown in fig. The motor shaft is connected to gear A and
rotates at 975 r.p.m. the gear wheel B & C are fixed to parallel shafts rotating together. Gear
wheel D & E are fixed to parallel shafts rotating together. The number of teeth on gear A, B,
C, D, E & F are 20, 50, 25, 75, 26 & 65 respectively. Find the speed of gear F that is fixed on
output shaft.
Ans:
We have,
NF/ NA = TA TCTE/ TB TDTF
NF = NA. TA TCTE/ TB TDTF
= 975*20*25*26
/(50*75*65)
=52 rpm
Given data:
NA= 975 rpm
TA = 20, TB = 50
TA = 25, TD = 75
TE = 26, TF = 65
NF = ?
7. For the following pipe network find out the flow rate and velocity at pipe (3)
According to fig.
Q1 = A1V1 = πD1
2
/4 . V1 = π(0.06)2
/4 . 2 = 0.00565 m3
/s
Q2 = 30% of Q1 = 0.3 Q1
Q3 = Q1 – Q2 = Q1 – 0.3Q1 = .7 Q1 = .7 X 0.00565
= 0.003955 m3/s
Again, Q3 = A3V3 ⇒ V3 = Q3 / A3 = Q3 /( πD3
2
/4)
= 0.003955 /{ π(.04)2
/4}
= 3.1472 m/s

5. Page 5 of 5
Conducted by
Engr. Md. Al-amin, ME, BUET.
alaminbuet2008@gmail.com
8. Calculate dynamic viscosity of an oil, which is used fo lubrication between a square plate
of size 0.8 m x 0.8 m and an inclinded plane with angle of inclination 30 ֯ as shown in the
Figure, The weight of the square plane is 300 N and it slides down the inclined plane with a
uniform velocity of 0.3 m/s. The thickness of the oil flim is 1.5 mm.
Ans:
We know,
Newton’s laws of viscoity
τ =F/A= µ du/dy
150/0.64=
µ* (0.3/0.0015)
or,
µ = 1.17 Ns/m2
Given data:
A=area =0 .8*0.8 = 0.64 m2
ϴ = inclination angle = 30֯
W = weight = 300N
F = force along plane
= W sin 30֯ = 300 * sin 30֯ =
150 N
du = 0.3 m/s
dy = 1.5 mm = 0.0015 m
µ= dynamic viscosity = ?
9. Draw SFD & BMD of the following load condition.
10. Drawing.