S. Prime circle 1. Cam
So correct pairs are, P-4, Q-3, R-2, S-1
SOL 5.45 Option (D) is correct.
The ratio of the maximum fluctuation of speed to the mean speed is called the coefficient of fluctuation of speed ( )Cf .
Let, N1 & N2 = Maximum & Minimum speeds in r.p.m. during the cycle
SOL 5.46 Option (D) is correct.
In this question pair or mechanism is related to contact & machine related to it.
Column I Column II
P. Higher Kinematic Pair 2. Line Contact Q. Lower Kinematic Pair 6. Surface Contact R. Quick Return Mechanism 5. Shaper
S. Mobility of a Linkage 1. Grubler’s Equation So correct pairs are, P-2, Q-6, R-5, S-1
SOL 5.47 Option (C) is correct.
Given m=250 kg, k=100kN m/ , N=3600 rpm, e cc 0 15. Natural frequency of spring mass system,
wn = mk 10
SOL 5.48 Option (A) is correct.
GATE MCQ Mechanical Engineering (4-volumes)
Here P Q R, , , & S are the lengths of the links.
According to Grashof’s law : “For a four bar mechanism, the sum of the shortest and longest link lengths should not be greater than the sum of remaining two link lengths, if there is to be continuous relative motion between the two links
S+L GP+Q
SOL 5.49 Option (A) is correct.
The table of motions is given below. Take CW=+ve, CCW =-ve S.
No.
Condition of Motion Revolution of elements Gear 1
N1
Compound Gear 2-3, N2=N3
Gear 4 N4
Carrier N5
1. Carrier 5 is fixed
& Gear 1 rotates 1 rpm
+ (CW)
+1
ZZ
2
- 1 ZZ
ZZ
2 1
4
# 3 0
2. Gear 1 rotates through rpm
+x (CW)
+x
x ZZ
2
- 1 x Z ZZ Z
2 4
1 3 0
3. Add +y revolutions to all elements
+y +y +y +y
4. Total motion. x+y y x ZZ
2
- 1 y x Z ZZ Z
2 4 1 3
+ # +y
Note
(i) Speed ratio = Speed of drivenSpeed of driver ..
No of teeth on driver No of teeth on driven
=
i.e. NN
2
1 ZZ
1
= 2
CCW = Counter clock wise direction (-ve)
CW = Clock wise direction (+ve)
(ii) Gear 2 & Gear 3 mounted on the same shaft (Compound Gears)
So, N2 =N3
SOL 5.50 Option (D) is correct.
Given w1=60 rpm (CW), w4=-2#60 (CCW) =-120 rpm From the previous part,
4 5
Negative sign show the counter clock wise direction.
So, w5 =156 rpm, CCW
SOL 5.51 Option (D) is correct.
Given m=12.5 kg, k=1000 /N m, c=15Ns m/ Critical Damping,
cc =2m mk =2 km
On substituting the values, we get
cc =2 1000#12 5. =223.6Ns m/
SOL 5.52 None of these
We know logarithmic decrement,
d 1 Now, from equation (i), we get
d ( . )
SOL 5.53 Option (C) is correct.
Given l=8, j=9
We know that, Degree of freedom,
n =3(l-1)-2j =3 8( -1)-2#9 =3
SOL 5.54 Option (C) is correct.
The speed of sound in air =332 /m s
For frequency of instrument of 144 Hz, length of sound wave
GATE MCQ Mechanical Engineering (4-volumes)
LI =144332 =2.30 m For sample P of 64 Hz,
LP = 33264 =5.1875 m Q of 96 Hz LQ = 33296 =3.458 m
R of 128 Hz LR = 128332 =2.593 m S of 250 Hz LS = 256332 =1.2968 m
Here, the length of sound wave of sample R L( R=2.593 )m is most close to the length of sound wave of Instrument (LI =2.30 )m . Hence, sample R produce most perceptible induced vibration.
SOL 5.55 Option (B) is correct.
Given N=300 . .r p m Angular velocity of cam,
w N
2p60
= =10prad/ sec Time taken to move 30c is,
t 18010 30 sec 1061
601
# p p
= = =
Now, Cam moves 30c with a constant acceleration & then with a deceleration, so maximum speed of the follower is at the end of first 30c rotation of the cam and during this 30c rotation the distance covered is 10 mm, with initial velocity u=0.
From Newton’s second law of motion, S =ut+21at2
0.01 0 21 a
601 2
# #
= + b l
a =0 01. #2#( )60 2 =72 / secm 2 Maximum velocity,
vmax = +u at 72 601
= # =1.2 / secm
SOL 5.56 Option (C) is correct.
Given m1=m2=0.5 kg, r1=0.05 m, r2=0.06 m
Balancing mass m =0.1 kg
Let disc rotates with uniform angular velocity w and x & y is the position of balancing mass along X & Y axis.
Resolving the forces in the x-direction, we get Fx
S =0
0.56-0.06cos30c+0.05cos0c w@ 2 =0 1. #x# w2
. ( . )
0 5# -0 00196 =0 1. x Fc=mrw2
x =-0.0098 m =-9.8 mm Similarly in y-direction, SFy =0
. ( . sin . sin )
0 5 0 06# 30c+0 05# 0 w2 =0 1. #y# w2
. .
0 5#0 03 =0 1 #. y
y =0.15 m =150 mm
Position of balancing mass is given by, r = x2+y2 = (-9 8. )2+(150)2 150.31mmb150
= mm
SOL 5.57 Option (C) is correct.
Given m=0.1 kg, k=1kN m/
Let, wd be the frequency of damped vibration & wn be the natural frequency of spring mass system.
Hence, wd =90% of wn=0 9. wn (Given) ...(i) Frequency of damped vibration
wd = (1-e w2) n ...(ii)
From equation (i) and equation (ii), we get (1-e w2) n =0 9. wn
On squaring both the sides, we get 1-e2 =( . )0 9 2 =0 81.
e2 = -1 0 81. =0 19. e = 0 19. =0 436. And Damping ratio is given by,
e cc
= c
km c
= 2
c =2 km # e =2 1000#0 1. #0 436. =8.72Ns m/ b8.7Ns m/
SOL 5.58 Option (B) is correct.
GATE MCQ Mechanical Engineering (4-volumes)
From Triangle ABC,
AB = (100)2+(240)2 = 67600 =260 mm Length of shortest link l1 =60 mm
Length of longest link l3 =260 mm From the Grashof’s law,
l1+l3 $ l2+l4
60+260$ 160+240 320 $ 400 So, l1+l3 <l2+l4
Also, when the shortest link O A2 will make a complete revolution relative to other three links, if it satisfies the Grashof’s law. Such a link is known as crank.
The link O B4 which makes a partial rotation or oscillates is known as rocker. So, crank rocker mechanism is obtained.
Here, O A2 = =l1 60 mm is crank (fixed link) Adjacent link, O O2 4 =240 mm is fixed
So, crank rocker mechanism will be obtained.
SOL 5.59 Option (B) is correct.
Let, w4 is the angular velocity of link O B4
From the triangle ABC, tan q 100240
125
= = ...(i)
q =tan-1b l125 =22 62c. Also from the triangle O O A1 2 ,
tan q O OO A
1 2
= 2
O O1 2 tanO A2
= q 5
12
= 60 =144 mm
From the angular velocity ratio theorem.
V24 =w4#I I24 14 =w#I I24 12
w4 I II I
24 14 24 12
= # w =(240144+144)#8 = 384144#8 =3rad/ sec
SOL 5.60 Option (D) is correct.
From the given data the component of force at joint A along AO2 is necessary to find the joint reaction at O2. So, it is not possible to find the magnitude of the joint reaction at O2.
SOL 5.61 Option (D) is correct.
Mechanical advantage in the form of torque is given by, . .
M A TT
input output
output input
ww
= =
Here output link is a slider, So, woutput=0 Therefore, M A. . =3
SOL 5.62 Option (C) is correct.
Given
wwn= =r 0 5.
And due to isolation damping ratio, e cc 0
= c = For isolation c=0
We know the transmissibility ratio of isolation is given by, . .
T R 1 2
1 2
n n
n
2 2 2
2
ww
e ww e ww
= - +
+ a
a k
k
: D 9 C 1 ( . )0 5 0
1 0
= 2 2
- +
+
6 @ 0 75.
1 34
= =
SOL 5.63 Option ( D) is correct.
Given planar mechanism has degree of freedom, N=1 and two infinite parallel lines meet at infinity. So, the instantaneous centre I24 will be at N, but for single degree of freedom, system moves only in one direction.
Hence, I24 is located at infinity( )3 .
SOL 5.64 Option (A) is correct.
Given N2=120 rpm, v1=12 / secm
So, coriolis component of the acceleration of link 1 is, ac12 =2 vw2 1 2#2 #60120 12
p #
= =301.44 /m s2 -302 /m s2
SOL 5.65 Option (C) is correct.
GATE MCQ Mechanical Engineering (4-volumes)
Given l=300mm=0.3m, W=300 N
Let, rod is twisted to the left, through an angle q. From the similar triangle OCD & OAB,
tan q .y .x 0 15 0 30
= =
If q is very very small, then tan q .y .x 0 15 0 30 -q = =
x=0.30q and y=0.15q ..(i)
On taking moment about the hinged point O 300
kx# +W#y =0 k Wyx
=-300 xy
300300
=- #a k =- =-12 0.5 /N mm From equation (i) xy
.. 0 300 15
= =-500 /N m
Negative sign shows that the spring tends to move to the point B.
In magnitude, k =500 /N m
SOL 5.66 Option (C) is correct.
Types of Mechanisms Motion Achieved P. Scott-Russel Mechanism 4. Straight Line Motion Q. Geneva Mechanism 1. Intermittent Motion R. Off-set slider-crank Mechanism 2. Quick Return Mechanism S. Scotch Yoke Mechanism 3. Simple Harmonic Motion So, correct pairs are, P-4, Q-1, R-2, S-3
SOL 5.67 Option (C) is correct.
Types of Joint Degree of constraints
P. Revolute 2. Five
Q. Cylindrical 3. Four
R. Spherical 1. Three
So, correct pairs are P-2, Q-3, R-1
SOL 5.68 Option (A) is correct.
Given M=20 kg, l=1000mm=1m, A=25#25 mm2 Esteel=200 GPa =200#10 Pa9
Mass moment of inertia of a square section is given by,
I b 12
= 4 ( )
12 25#10 3 4
= - =3.25#10 m-8 4
Deflection of a cantilever, Loaded with a point load placed at the free end is, d WlEI
3
= 3 mglEI 3
= 3
. . ( ) 3 200 10 3 25 10
20 9 81 1
9 8
3
# # # #
# #
= - =19500196 2. =0.01 m wn g
= d ..
0 019 81
= =31.32rad/ sec Therefore, critical damping constant
cc =2 Mwn =2#20#31 32. 1252.8Ns m/
= -1250Ns m/
SOL 5.69 Option (B) is correct.
Let, Z is the number of teeth and motor rotates with an angular velocity w1 in clockwise direction & develops a torque T1.
Due to the rotation of motor, the gear 2 rotates in anti-clockwise direction & gear 3 rotates in clock wise direction with the same angular speed.
Let, T2 is the torque developed by gear.
Now, for two equal size big gears,
Module m
( . )
( )
No of teeths Pitch circle diameter DZ
= =
D =mZ =2#80=160 mm
(Due to rotation of gear 2 & gear 3 an equal force (F) is generated in the downward direction because teeth are same for both the gears)
For equilibrium condition, we have Downward force = upward force
F+F =1000
GATE MCQ Mechanical Engineering (4-volumes)
F =500 N
And h = Power OutputPower Input 2 TT
1 1 2 2
#w w
=
Output power is generated by the two gears
T
F D
2 2
1 1
# # 2
w w
= b l ...(i)
We know velocity ratio is given by NN
2
1 ZZ
2 1
1
ww 2
= = N
260 w= p
From equation (i), h T
F D
ZZ
2 2
1 2
1
# #
= b l#
T1 F D ZZ
2
# 1
h #
= b l . .
500 0 80 160 8020
# #
= =25 N m
-SOL 5.70 Option (C) is correct.
Given pressure angle f =20c, FT =500 N from previous question.
From the given figure we easily see that force action along the line of action is F. From the triangle ABC,
cos f = FFT
F cosFT
= f
cos 20500
= c =532 N
SOL 5.71 Option (D) is correct.
A single slider crank chain is a modification of the basic four bar chain. It is find, that four inversions of a single slider crank chain are possible. From these four inversions, crank and slotted lever quick return motion mechanism is used in shaping machines, slotting machines and in rotary internal combustion engines.
SOL 5.72 Option (A) is correct.
Given p< < <q r s
“Double crank” mechanism occurs, when the shortest link is fixed. From the given pairs p is the shortest link. So, link of length p should be fixed.
SOL 5.73 Option (B) is correct.
We clearly see from the figure that cylinder can either revolve about x-axis or slide along x-axis & all the motions are restricted.
Hence, Number of degrees of freedom =2 & movability includes the six degrees of freedom of the device as a whole, as the ground link were not fixed. So, 4 degrees of freedom are constrained or arrested.
SOL 5.74 Option (B) is correct.
Given N=1200 rpm, DE=2kJ=2000J, D=1 m, Cs=0 02. Mean angular speed of engine,
w N
2p60
= .
2#3 1460#1200
= =125.66rad/ sec Fluctuation of energy of the flywheel is given by,
DE =I Cw2 s 21mR C
2w2 s
= For solid disc I mR
2
= 2
m R C
E 2
2Dw2 s
= (125 66. ) 0 02. 2 2000
21 2 2
# #
= #
^ h (125 664 . )2 20000 02.
2#
# #
= =50.66kg-51kg
SOL 5.75 Option (B) is correct.
Given m =10 kg, d =30mm=0.03m, l =500mm=0.5m, 2.1 10 Pa
Eshaft 11
= #
We know that, static deflection due to 10 kg of Mass at the centre is given by, d WlEI
48
= 3 mglEI 48
= 3 ...(i)
The moment of inertia of the shaft,
I = 64p d4 = 64 0 03p ( . )4 =3.974#10 m-8 4 ...(ii) Substitute values in equation (i), we get
d . .
. ( . ) 48 2 1 10 3 974 10
10 9 81 0 5
11 8
3
# # # #
# #
=
-. . 400 5812 262510
# 3
= =3.06#10 m-5 If wc is the critical or whirling speed in r.p.s. then,
wc g
= d & 2pfc g
= d
GATE MCQ Mechanical Engineering (4-volumes)
fc g
21 p d
= . .
2 13 14 .
3 06 10 9 81 5
# #
=
-. . .
6 281
30 69 8110
# 6
= - =90.16Hz-90Hz
SOL 5.76 Option (C) is correct.
Given, the circular disc rotates about the point O at a uniform angular velocity w.
Let vA is the linear velocity of point A & vB is the linear velocity of point B.
vA=wrA and vB=wrB.
Velocity of point B with respect to point A is given by, vBA =vB-vA =wrB-wrA =w(rB-rA) From the given figure,
rB >rA
So, wrB >wrA
vB >vA
Therefore, relative velocity w(rB-rA) in the direction of point B.
SOL 5.77 Option (D) is correct.
Acceleration of point B with respect to point A is given by,
aBA =wvBA =w#w(rB-rA) =w2(rB-rA) ...(i) This equation (i) gives the value of centripetal acceleration which acts always towards the centre of rotation.
So, aBA acts towards to O i.e. its direction from Z to O
SOL 5.78 Option (A) is correct.
Given m =10 kg, k =2kN m/ , c =500Ns m/ , k =q 1kN m rad/ / 0.5 m
l1= , l2=0.4 m
Let, the rigid slender bar twist downward at the angle q. Now spring & damper exert a force kx1 & cx2 on the rigid bar in the upward direction.
From similar triangle OAB & OCD, tan q x. x.
0 42 0 51
= =
Let q be very very small, then tan -q q, q x. x.
0 42 0 51
= =
x2 =0 4q. or x1=0 5. q ...(i) On differentiating the above equation, we get
xo2 =0 4q. o or xo1=0 5. qo ...(ii) We know, the moment of inertia of the bar hinged at the one end is,
I ml 31
= 2 ( . )
3 10# 0 5 2
= =0.833 kg m- 2
As no external force acting on the system. So, governing equation of motion from the Newton’s law of motion is,
Iqp+cx lo2 2+kx l1 1+kqq =0 0.833qp+500#0.4xo2+2000#(0. )5 x1+1000q =0
. x x
0 833qp+200o2+1000 1+1000q =0 ...(iii)
. . .
0 833qp+200#0 4qo+1000#0 5q+1000q =0 .
0 833qp+80qo+1500q =0 ...(iv) On comparing equation (iv) with its general equation,
Iqp+cqo+kq =0 We get, I=0 833. , c=80, k=1500
So, undamped natural frequency of oscillations is given by
wn = Ik = 0 8331500. = 1800 72. =42.43rad/ sec
SOL 5.79 Option (C) is correct.
From the previous part of the question Damping coefficient, c =80Nms rad/
SOL 5.80 Option (C) is correct.
From the Kutzbach criterion the degree of freedom, n =3(l-1)-2j-h
For single degree of Freedom (n=1), 1 =3(l-1)-2j-h
l j h
3 -2 - -4 =0 ...(i)
The simplest possible mechanisms of single degree of freedom is four-bar mechanism. For this mechanism j=4, h=0
From equation (i), we have l
3 -2#4- -4 0 =0 & l =4
SOL 5.81 Option (B) is correct.
When a point on one link is sliding along another rotating link, such as in quick return motion mechanism, then the coriolis component of the acceleration must be calculated. Quick return motion mechanism is used in shaping machines, slotting machines and in rotary internal combustion engines.
GATE MCQ Mechanical Engineering (4-volumes)
SOL 5.82 Option (C) is correct.
The deflection of a cantilever beam loaded at the free end is given by, d MgLEI
3
= 3
And natural frequency, wn g
= d
MLEI 3 3
= ...(i)
If the length of the cantilever beam is halved, then w l n
M EIL
MLEI 2
3 3 8 3 3
#
= =
b b
l l
From equation (i)
w ln = 8wn
So, natural frequency is increased by a factor 8.
SOL 5.83 Option (C) is correct.
For a spring loaded roller follower driven with a disc cam, the pressure angle should be large during rise as well as during return for ease of transmitting motion.
If pressure angle is large, then side thrust will be minimum. Pressure angles of up to about 30c to 35c are about the largest that can be used without causing difficulties.
SOL 5.84 Option (B) is correct.
Let initial length of the spring =L Potential energy at A,
PEA =mg L( -d)
and at B, PEB =mg L6 -(d+x)@+21kx2
So, change in potential energy from position A to position B is PEAB
T =PEB-PEA
mgL mgd mgx 21kx2 mgL mgd
= - - + - +
PEAB
T = 21kx2-mgx
SOL 5.85 Option (A) is correct.
The mean speed of the engine is controlled by the governor. If load increases then fluid supply increases by the governor and vice-versa.
Flywheel stores the extra energy and delivers it when needed. So, Flywheel reduces speed fluctuations.
Flywheel reduce speed fluctuations during a cycle for a constant load, but Flywheel does not control the mean speed N N N
1 2 2
= +
b l of the engine.
SOL 5.86 Option (B) is correct.
First make the table for the motion of the gears.
Take CW =+ve, CCW=-ve
S.
No.
Condition of Motion Arm Sun Gear NS
Planet Gear NP
Ring Gear NG
(i) Arm is fixed & sun gear rotates +1 rpm
(CW)
0 +1
ZZ
P
- S ZZ
R
- S
(ii) Sun Gear rotates through +x rpm
(CW)
0 +x
x ZZ
P
- S x ZZ
R
- S
(iii) Add +y revolution to all elements
+y +y +y +y
(iv) Total Motion +y x+y y x ZZ
P
- S y x ZZ
R
- S
Let Teethes and speed of the sum gear, planet gear and ring gear is represented by ZG, ZP, ZR and NG, NP, NR respectively.
Given sun gear is driven clockwise at 100 rpm. So, From the table
x+y =100 ...(i)
Ring gear is held stationary. From the table y x ZZ
P
- S =0 y x 8020
= # y x
= 4 & x =4y ...(ii) From equation on (i) and (ii)
y y
4 + =100 y =20 rpm
SOL 5.87 Option (D) is correct.
Give a small displacement q to the assembly. So assembly oscillates about its mean position.
GATE MCQ Mechanical Engineering (4-volumes)
From this a restoring torque is acts along the line of oscillation.
Net restoring torque,
T =mgsin(a+q)#L-mgsin(a-q)#L
T =mgL6sin cosa q+cos sina q-sin cosa q+cos sina q@ T =2mgLcos sina q
For very small deflection q, sin q , q
T =2mgLqcosa Now from newton’s law,
Iq +p T =0 cos Iqp+2mgLq a =0
( cos ) mL dt
d mgL
2 2 2q2 + 2 a q =0 I=mL2+mL2
cos dt
d g L
2
2q + a q =0
On comparing with qp+w qn2 =0, we get
n2
w cos L g a
=
wn cos g L a
=
*********