CE-GATE-2015 PAPER|
www.gateforum.com
GATE 2015 –CE on 8th February, 2015 – (Forenoon Session)
General Aptitude Questions
Q.No-1-5 Carry One Mark Each
1.
Select the pair that best expresses a relationship similar to that expressed in the pair:
Children: Pediatrician
(A) Adult: Orthopaedist
(B) Females: Gynaecologist
(C) Kidney: Nephrologist
(D) Skin: Dermatologist
Answer:
(B)
Exp:
Community of people: Doctor
2.
Extreme focus on syllabus and studying for test has become such a dominant concern of
Indian students that this has closed their minds to anything ________ to the requirements of
the exam
(A) related
(B) extraneous
(C) outside
(D) useful
Answer:
(B)
Exp:
extraneous -irrelevant or unrelated to the subject being dealt with.
3.
If ROAD is written as URDG, then SWAN should be written as:
(A) VXDQ
(B) VZDQ
(C) VZDP
(D) UXDQ
Answer: (B)
Exp: R+3=U, O+3=R, A+3=D, D+3=G;
S+3=V, W+3=Z, A+3=D, N+3=Q
4.
The Tamil version of __________ John Abraham-starrer Madras Café _____ cleared by the
censor board with no cuts last week, but the film’s distributors _______ no takers among the
exhibitors for a release in Tamil Nadu ________ this Friday.
(A) Mr., was, found, on
(B) a, was found, at
(C) the, was, found, on
Answer:
(D) a, being, find, at
(C)
Exp:
John-Abraham starrer Madras Café talks about the movie not the person, so Mr. is ruled out.
‘Find no takers’ is not the correct phrase. At this Friday is incorrect. So, option C is correct.
5.
A function f(x) is linear and has a value of 29 at x = -2 and 39 at x = 3. Find its value at x = 5.
(A) 59
Answer:
Exp:
(B) 45
(C) 43
(D) 35
(C)
f(x)=2x+33
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
1
CE-GATE-2015 PAPER|
www.gateforum.com
Q.No-6-10 Carry Two Marks Each
6.
The head of a newly formed government desires to appoint five of the six selected members
P,Q,R,S,T and U to portfolios of Home, Power, Defense, Telecom and Finance. U does not
want any portfolio if S gets one of the five. R wants either Home or Finance or no portfolio.
Q says that if S gets either Power or Telecom, then she must get the other one. T insists on a
portfolio if P gets one.
Which is the valid distribution of portfolio?
(A) P-Home, Q-Power, R-Defense, S-Telecom, T-Finance
(B) R-Home, S-Power, P-Defense, Q-Telecom, T-Finance
(C) P-Home, Q-Power, T-Defense, S-Telecom, U-Finance
(D) Q-Home, U-Power, T-Defense, R-Telecom, P-Finance
Answer:
(B)
Exp:
Since U does not want any portfolio, (C) and (D) are ruled out. R wants Home, or Finance or
No portfolio, (A) is not valid. Hence option (B) is correct
7.
The exports and imports (in crores of Rs.) of a country from the year 2000 to 2007 are given
in the following bar chart. In which year is the combined percentage increase in imports and
exports the highest?
120
Exports
110
Imports
100
90
80
70
60
50
40
30
20
10
0
2000
Answer:
Exp:
2001
2002
2003
2004
2005
2006
2007
2006
Increase in exports in 2006=
100 − 70
= 42.8%
70
Increase in imports in 2006=
120 − 90
= 33.3%
90
which is more than any other year
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
2
CE-GATE-2015 PAPER|
8.
www.gateforum.com
Most experts feel that in spite of possessing all the technical skills required to be a batsman of
the highest order., he is unlikely to be so due to lack of requisite temperament. He was guilty
of throwing away his wicket several times after working hard to lay a strong foundation. His
critics pointed out that until he addressed to this problem, success at the highest level will
continue to elude him.
Which of the statement (s) below is/are logically valid and can be inferred from the above
passage?
(i) He was already a successful batsman at the highest level
(ii) He has to improve his temperament in order to become a great batsman
(iii) He failed to make many of his good starts count
(iv) Improving his technical skills will guarantee success
(A) (iii) and (iv)
Answer:
9.
(B) (ii) and (iii)
(C) (i), (ii) and (iii)
(D) (ii) only
(B)
Choose the most appropriate equation for the function drawn as a thick line, in the plot below.
y
( 2,0)
π
( 0, −1)
(A) x = y-|y|
Answer:
10.
(B) x = -(y-|y|)
(C) x = y+|y|
(D) x = -(y+|y|)
(B)
Alexander turned his attention towards India, since he had conquered Persia.
Which one of the statements below is logically valid and can be inferred from the above
sentence?
(A) Alexander would not have turned his attention towards India had he not conquered
Persia.
(B) Alexander was not ready to rest on his laurels, and wanted to march to India
(C) Alexander was completely in control of his army and could command it to move towards
India.
(D) Since Alexander’s kingdom extended to Indian borders after the conquest of Persia, he
was keen to move further.
Answer:
(A)
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
3
CE-GATE-2015 PAPER|
www.gateforum.com
Section Name: Civil Engineering
Q.No-1-25 Carry One Mark Each
1.
Consider the following statements for air-entrained concrete:
(i) Air-entrainment reduces the water demand for a given level of workability
(ii) Use of air-entrained concrete is required in environments where cyclic freezing and
thawing is expected.
Which of the following is TRUE?
(A) Both (i) and (ii) are True
(B) Both (i) and (ii) are False
(C) (i) is True and (ii) is False
(D) (i) is False and (ii) is True
Answer:
Exp:
(A)
(i) Air-entrainment reduces the water demand for a given level of workability-True
(ii) Use of air-entrained concrete is required in environments where cyclic freezing and
thawing is expected. -True
2.
Which of the following statements is TRUE for the relation between discharge velocity and
seepage velocity?
(A) Seepage velocity is always smaller than discharge velocity
(B) Seepage velocity can never be smaller than discharge velocity
(C) Seepage velocity is equal to the discharge velocity
(D) No relation between seepage velocity and discharge velocity can be established
Answer:
(B)
Vs =
Exp:
V
n
∵ n <1
So, VS > V always
So, seepage velocity (VS) can never by smaller than discharge velocity
3.
The integral
∫
x2
x1
x 2 dx with x 2 > x1 > 0 is evaluated analytically as well as numerically using a
single application of the trapezoidal rule. If I is the exact value of the integral obtained
analytically and J is the approximate value obtained using the trapezoidal rule, which of the
following statements is correct about their relationship?
(A) J > 1
(B) J<1
(C) J = 1
(D) Insufficient data to determine the relationship
Answer:
(A)
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
4
CE-GATE-2015 PAPER|
Exp:
We know that the approximated value of
∫
b
a
www.gateforum.com
f (x)dx obtained by trapezoidal rule is always
greater than the analytical value.
∴ J > I where J=approximate value
I=analytical value
4.
A circular pipe has a diameter of 1m, bed slope of 1 in 1000, and Manning’s roughness
coefficient equal to 0.01. It may be treated as an open channel flow when it is flowing just
full, i.e., the water level just touches the crest. The discharge in this condition is denoted by
Qfull. Similarly, the discharge when the pipe is flowing half-full, i.e., with a flow depth of
0.5m, is denoted by Qhalf. The ratio Qfull/Qhalf is:
(A) 1
Answer:
(B)
2
(C) 2
(D) 4
(C)
2
Exp:
2 1
1
1
1 π
 D 3
Q = .AR 3 S 2 ⇒ Q full = . .D 2 .  s 2
n
n 4
4
2
Q half
1 πD 2  D  3 12
= .
.  s
n 8 4
Q full
=2
Q half
5.
Which of the following statements is NOT correct?
(A) Loose sand exhibits contractive behavior upon shearing
(B) Dense sand when sheared under undrained condition, may lead to generation of negative
pore pressure
(C) Black cotton soil exhibits expansive behavior
(D) Liquefaction is the phenomenon where cohesionless soil near the downstream side of
dams or sheet-piles loses its shear strength due to high upward hydraulic gradient
Answer: (D)
Exp: Liquefaction is due to cyclic loads, not due to high hydraulic gradient
6.
A fine-grained soil has 60% (by weight) silt content. The soil behaves as semi-solid when
water content is between 15% and 28%. The soil behaves fluid-like when the water content is
more than 40%. The ‘Activity’ of the soil is
(A) 3.33
Answer:
Exp:
(B) 0.42
(C) 0.30
(D) 0.20
(C)
IP = WL-WP = 40-28 = 12%
Activity =
Ip
C
=
12
= 0.3
100 − 60
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
5
CE-GATE-2015 PAPER|
7.
For steady incompressible flow through a closed-conduit of uniform cross-section, the
direction of flow will always be:
(A) from higher to lower elevation
(B) from higher to lower pressure
(C) from higher to lower velocity
(D) from higher to lower piezometric head
Answer:
8.
(B)
Two triangular wedges are glued together as shown in the following figure. The stress acting
normal to the interface, σn is __________ MPa.
Answer:
Exp:
www.gateforum.com
0
σn =
=
σx + σy
2
+
σx − σy
2
cos 2θ
100MPa
100 − 100 100 − (−100)
+
cos90 = 0
2
2
σn
100MPa
100MPa
45°
100MPa
9.
In a closed loop traverse of 1 km total length, the closing errors in departure and latitude are
0.3 m and 0.4 m, respectively. The relative precision of this traverse will be;
(A) 1:5000
Answer:
Exp:
(B) 1:4000
(D) 1:2000
(D)
e = 2 + d2 =
( 0.3) + ( 0.4 )
5
Relative precision =
10.
(C) 1:3000
2
= 0.5 m
0.5
= 1: 2000
1000
Solid waste generated from an industry contains only two components, X and Y as shown in
the table below
Component Composition
( % weight )
X
Y
c1
c2
Density
( kg
m3 )
ρ1
ρ2
Assuming (c1+c2) = 100, the composite density of the solid waste (ρ) is given by:
(A)
Answer:
100
 c1 c 2 
 + 
 ρ1 ρ2 
ρ ρ 
(B) 100  1 + 2 
 c1 c2 
(C) 100 ( c1ρ1 + c 2ρ2 ) (D) N77o 50′E
(A)
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
6
CE-GATE-2015 PAPER|
Exp:
www.gateforum.com
Let density of sludge is ρ
c1 + c2 c1 c 2
= +
ρ
ρ1 ρ2
⇒ρ=
11.
100
c1 c 2
+
ρ1 ρ2
The two columns below show some parameters and their possible values.
Parameter
Value
P-Gross Command Area
I-100 hectares/cumec
Q-Permanent Wilting Point
II-6℃
R-Duty of canal water
III-1000 hectares
S-Delta of wheat
IV-1000 cm
V-40 cm
VI-0.12
Which of the following options matches the parameters and the values correctly?
(A) P-I, Q-II, R-III, S-IV
(B) P-III, Q-VI, R-I, S-V
(C) P-I, Q-V, R-VI, S-II
(D) P-III, Q-II, R-V, S-IV
Answer:
(B)
Exp:
P-Gross Command Area=1000 ha
Q-Permanent Wilting Point=0.12
R-Duty of canal water=100 ha/cumec
S-Delta of wheat=40 cm
12.
In an unconsolidated undrained triaxial test, it is observed that an increase in cell pressure
from 150 kPa to 250 kPa leads to a pore pressure increase of 80 kPa. It is further observed
that, an increase of 50 kPa in deviatoric stress results in an increase of 25 kPa in the pore
pressure. The value of Skempton’s pore pressure parameter B is;
(A) 0.5
Answer:
Exp:
B=
(B) 0.625
(C) 0.8
(D) 1.0
(C)
∆U 80
=
= 0.8
∆σ3 100
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
7
CE-GATE-2015 PAPER|
13.
www.gateforum.com
Which of the following statements is TRUE for degree of disturbance of collected soil
sample?
(A) Thinner the sampler wall, lower the degree of disturbance of collected soil sample
(B) Thicker the sampler wall, lower the degree of disturbance of collected soil sample
(C) Thickness of the sampler wall and the degree of disturbance of collected soil sample are
unrelated
(D) The degree of disturbance of collected soil sample is proportional to the inner diameter of
sampling tube
Answer: (A)
Exp: As thickness of sampler increases, disturbance increases
14.
Which of the following statements is FALSE?
(A) Plumb line is along the direction of gravity
(B) Mean Sea Level (MSL) is used as a reference surface for establishing the horizontal
control
(C) Mean Sea Level (MSL) is a simplification of the Geoid
(D) Geoid is an equi-potential surface of gravity
Answer: (B)
Exp: Mean Sea Level (MSL) is used as a reference surface for establishing the vertical control and
not horizontal control
15.
For what value of p the following set of equations will have no solution?
2x+3y = 5
3x+py = 10
Answer: 4.5
Exp: Given 2x+3y=5
3x + py = 10
 2 3   x  5 
⇒
  =  
3 p   y  10 
AX = B
2 3 5 
Augmented matrix [A / B] = 

3 p 10 
5
2 3
R 2 → 2R 2 − 3R 1 

0 2p − 9 5
system will have no solution if ρ(A / B) ≠ ρ(A)
⇒ 2p − 9 = 0
⇒p=
16.
9
= 4.5
2
In a two-dimensional steady flow field, in a certain region of the x-y plane, the velocity
1
component in the x-direction is given by v x = x 2 and the density varies as ρ = . Which of
x
the following is a valid expression for the velocity component in the y-direction, vy?
(A) v y = − x / y
(B) v y = x / y
(C) v y = − xy
(D) v y = xy
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
8
CE-GATE-2015 PAPER|
www.gateforum.com
Answer: (C)
Exp: Continuity equation
∂
∂
( ρY ) + ( ρV ) = 0
∂x
∂y
⇒
∂
∂ 1
( x ) +  .V  = 0
∂x
∂y  x 
∂ V
∂ V
  = 0 ⇒   = −1
∂y  x 
∂y  x 
⇒ V = − xy
⇒1+
17.
Workability of concrete can be measured using slump, compaction factor and Vebe time.
Consider the following statements for workability of concrete:
(i) As the slump increases, the Vebe time increases
(ii) As the slum increases, the compaction factor increases
Which of the following is TRUE?
(A) Both (i) and (ii) are True
(B) Both (i) and (ii) are False
(C) (i) is True and (ii) is False
(D) (i) is False and (ii) is True
Answer: (D)
Exp: As the slump increases, the Vebe time decreases
18.
Consider the following probability mass function (p.m.f) of a random variable X:
if X = 0
q

p(x,q) = 1 − q if X = 1
0
otherwise

If q=0.4, the variance of X is ________ .
Answer: 0.24
Exp: p(x,q) = q
if X = 0
= 1 − q if X = 1
0
otherwise
given q = 0.4
⇒ p ( x,q ) = 0.4 if X = 0
= 0.6 if X = 1
=0
otherwise
⇒
X
0
1
p ( X = x ) 0.4 0.6
Required value = V ( X ) = E ( X 2 ) − {E ( X )}
2
E ( X ) = Σx i pi = 0 × 0.4 + 1 × 0.6 = 0.6
E ( X 2 ) = Σx i2 pi = 02 × 0.4 + 12 × 0.6 = 0.6
V ( X ) = 0.6 − ( 0.6 )
2
= 0.6 − 0.36
= 0.24
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
9
CE-GATE-2015 PAPER|
www.gateforum.com
19.
Which of the following statements CANNOT be used to describe free flow speed (uf) of a
traffic stream?
(A) uf is the speed when flow is negligible
(B) uf is the speed when density is negligible
(C) uf is affected by geometry and surface conditions of the road
(D) uf is the speed at which flow is maximum and density is optimum
Answer:
(D)
Exp:
Free flow speed → speed when flow is negligible
→ speed when density is negligible
→ affected by Geometry, deriver’s perception, roadway condition etc.
20.
Consider the singly reinforced beam shown in the figure below:
P
P
X
X
L
L/2
L
At cross-section XX, which of the following statement is TRUE at the limit state?
(A) The variation of stress is linear and that of strain is non-linear
(B) The variation of strain is linear and that of stress is no-linear
(C) The variation of both stress and strain is linear
(D) The variation of both stress and strain is non-linear
Answer:
Exp:
(B)
At
Strain variation Stress variation
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
10
CE-GATE-2015 PAPER|
21.
www.gateforum.com
For the beam shown below, the stiffness coefficient K22 can be written as
2
L
3
Note :1, 2 and 3
are the d.o.f
A, E, I
6EI
L2
(B)
(A)
Answer:
(B)
1
12EI
L3
3EI
L
(C)
(D)
EI
6L2
Exp:
2
∆
6EI∆
L2
3
1
12EI∆
L3
K 22 =
22.
12EI
L3
The penetration value of a bitumen sample tested at 25o C is 80. When this sample is heated
to 60o C and tested again, the needle of the penetration test apparatus penetrates the bitumen
sample by d mm. The value of d CANNOT be less than ____ mm.
Answer:
23.
The development length of a deformed reinforcement bar can be expressed as
(1 / k)(φσs / τbd ). From the IS:456-2000, the value of k can be calculated as ____ .
Answer:
Exp:
8
6.4
Ld =
φσs
But for deformed bars τbd is increased by 60%.
4τ bd
So,
Ld =
φσst
φσs
=
4 × 1.6 × τbd 6.4 τbd
So, k=6.4.
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
11
CE-GATE-2015 PAPER|
24.
www.gateforum.com
Total Kjeldahl Nitrogen (TKN) concentration (mg/L as N) in domestic sewage is the sum of
the concentrations of:
(A) organic and inorganic nitrogen in sewage
(B) organic nitrogen and nitrate is sewage
(C) organic nitrogen and ammonia is sewage
(D) ammonia and nitrate in sewage
Answer:
(C)
Exp:
Total Kjeldahl Nitrogen (TKN) = Ammonia (60%) + Organic Nitrogen (40%)
25.
For the beam shown below, the value of the support moment M is ____ kN-m.
20 kN
M
3m
1m
1m
EI
EI
EI
3m
Internal hinge
Answer:
5
10kN
Exp:
3m
1m
10×1
= 5kN−m
2
Q.No-26-55 Carry Two Marks Each
26.
The directional derivative of the field u(x,y,z) = x 2 − 3yz in the direction of the vector
ˆ at po int (2, −1, 4) is ____ .
(iˆ + ˆj − 2k)
Answer:
-5.72
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
12
CE-GATE-2015 PAPER|
Exp:
www.gateforum.com
Let u(x,y,z) = x2-3yz
a = i + j − 2k and P ( 2, −1, 4 )
∇u = i
∂u
∂u
∂u
+ j +k
∂x
∂y
∂z
= i2x + j ( −3z ) + k ( −3y )
∇u ( 2,−1,4) = 4i − 12 j + 3k
a = 1+1+ 4 = 6
directionalderivative = ∇u.a
= ( 4i − 12 j + 3k ) .
( i + j − 2k )
6
4 − 12 − 6
=
6
−14
=
= −5.72
6
27.
For formation of collapse mechanism in the following figure, the minimum value of Pu is
cMp/L. M p and 3M p denote the plastic moment capacities of beam sections as shown in this
figure. The value of c is ____ .
Pu
1m
1m
Mp
3 Mp
2m
Answer:
Exp:
13.33
Pu
Mechanism-I
L
3M P . θ + M P ( 2θ ) + MP.θ = PU × ×θ
4
3M P
θ
θ
L
⇒ 6M P .θ = PU . .θ
4
⇒ PU = 24
MP
L
MP
2θ
MP
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
13
CE-GATE-2015 PAPER|
www.gateforum.com
Mechanism-II
1.φ = 3.θ ⇒ φ = 3θ
Pu
L
3M P .θ + M P ( θ + φ ) + M P .φ = PU × φ
4
3M P
L
⇒ 3M P θ + M P ( θ + 3θ ) + M P .3θ = PU × 3θ ×
4
L
⇒ 10M P .θ = PU × 3θ ×
4
M
40 M P
⇒ PU =
.
=13.33 P
3
L
L
So, C =13.33
28.
θ
MP
φ
θ+φ
MP
A tapered circular rod of diameter varying from 20 mm to 10 mm is connected to another
uniform circular rod of diameter 10 mm as shown in the following figure. Both bars are made
of same material with the modulus of elasticity, E = 2 × 105 MPa. When subjected to a load
P = 30π kN, the deflection at point A is ____ mm.
d1 = 20 mm
2m
d 2 = 10 mm
1.5 m
A
P = 30 πkN
Answer:
Exp:
15
∆1 =
30π
4P.L
πd1d 2 × E
20mm
4 × 30π × 2 × 106
π × 20 × 10 × 2 × 105
= 6 mm
=
10mm
30π
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
14
CE-GATE-2015 PAPER|
∆2 =
P×L
AE
www.gateforum.com
30π
4 × 30π × 1.5 × 106
π × 10 × 10 × 2 × 105
= 9 mm
=
∆ = ∆1 + ∆ 2 = 15mm
30π
29.
A water tank is to be constructed on the soil deposit shown in the figure below. A circular
footing of diameter 3m and depth of embedment 1m has been designed to support the tank.
The total vertical load to be taken by the footing is 1500 kN. Assume the unit weight of water
as 10kN/ m 3 and the load dispersion pattern as 2V:1H. The expected settlement of the tank
due to primary consolidation of the clay layer is ____ mm.
2m
Silty Sand
Bulk unit weight = 15 kN / m3
Saturated unit weight = 18kN / m3 .
Sand
6m
10m
Normally
consolidated
clay
Saturated unit weight = 18 kN / m3
Compression index = 0.3
Initial void ratio = 0.7
Coefficient of consolidation = 0.004cm 2 / s
Dense Sand
Answer:
Exp:
53.236
Settlement =
 σ + ∆σ 
Cc
H 0 log  0

1 + e0
 σ0 
σ0 = 15 × 2 + (18 − 10 ) × 6 + (18 − 10 ) *5
= 118kN m 2
1500
∆σ0 =
= 8.488kN m 2
π
2
(3 + 6 + 6)
4
1
2
3
6
12
10
6
6
15
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
15
CE-GATE-2015 PAPER|
www.gateforum.com
0.3
 118 + 8.488 
× 10log10 

1 + 0.7
118


= 0.0532 m
∆H =
∆H = 53.236 mm
30.
Consider the following differential equation:
y
y
= y ( xdy − ydx ) sin
x
x
Which of the following is the solution of the above equation (c is an arbitrary constant)?
x(ydx + xdy) cos
(A)
x
y
cos = c
y
x
(B)
x
y
sin = c
y
x
(C) xy cos
y
=c
x
(D) xysin
y
=c
x
Answer: (C)
Exp: Given D.E
y
y
= y ( xdy − ydx ) sin
x
x
y 
y
⇒ x ( ydx + xdy ) cos +  − sin  y ( xdy − ydx ) = 0
x 
x
x ( ydx + xdy ) cos
y  y ( xdy − ydx )
 y 
⇒ ( ydx + xdy ) cos   +  − sin 
=0
x
x
x 
y  xdy − ydx 
 y

⇒ ( ydx + xdy ) cos   + ( xy )  − sin 
=0
x 
x2
x


y

By observing, the above equation is d  (xy) cos  = 0
x

y
Byintegrating, xy cos   = c
x
31.
The composition of an air-entrained concrete is given below:
:184 kg / m3
Water
Ordinary Portland Cement(OPC) : 368 kg / m3
Sand
: 606kg / m3
Coarse aggregate
:1155 kg / m 3
Assume the specific gravity of OPC, sand and coarse aggregate to be 3.14, 2.67 and 2.74,
respectively, The air content is_________ liters/ m 3 .
Answer:
Exp:
51
MC MS Ma
+
+
+ Vw + Va = 1
ρC
ρS
ρa
⇒
368
606
1155
184
+
+
+
+ VV = 1.0
3.14 × 1000 2.67 × 1000 2.74 × 1000 1000
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
16
CE-GATE-2015 PAPER|
www.gateforum.com
⇒ 0.117 + 0.227 + 0.421 + 0.184 + VV = 1.0
⇒ VV = 0.051
= 0.051 × 1000 = 51 50.32 l m 3
32.
An earth embankment is to be constructed with compacted cohesionless soil. The volume of
the embankment is 5000 m 3 and the target dry unit weight is 16.2 kN / m3 . Three nearby
sites (see figure below) have been identified from where the required soil can be transported
to the construction site. The void ratios (e) of different sites are shown in the figure. Assume
the specific gravity of soil to be 2.7 for all three sites. If the cost of transportation per km is
twice the cost of excavation per m 3 of borrow pits, which site would you choose as the most
economic solution? (Use unit Weight of water = 10kN / m 3 ) .
Site X
e = 0.6
140 km
Construction
site
Site Z
e = 0.64
(A) Site X
Answer:
Exp:
(B) Site Y
Site Y
e = 0.7
(C) Site Z
(D) Any of the sites
(B)
Vy
V
V
V
= x =
= z
1 + e 1 + e1 1 + e2 1 + e3
γd =
G
2.67
.γ w ⇒ 16 =
× 10
1+ e
1+ e
⇒ e = 0.67
Vy
5000 Vx
V
=
=
= z
1.67 1.6 1.64 1.7
⇒ Vx = 4790.42 m 3
Vy = 4910.18m 3
Vz = 5089.82 m 3
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
17
CE-GATE-2015 PAPER|
www.gateforum.com
C x = C × 479042 + 2 × C × 140
= 5070.42 C
C y = C × 4910.18 + 2 × C × 80 = 5070.18C
C z = C × 5089.82 + 2 × C × 100 = 5289.82C
33.
The concentration of Sulfur Dioxide (SO 2 ) is ambient atmosphere was measured as
30µg / m3 . Under the same conditions, the above SO 2 concentration expressed in ppm is
___.
Given : P / (RT) = 41.6 mol / m 3 ; where P=Pressure; T=Temperature ; R=universal gas
constant; Molecular weight of SO 2 = 64.
Answer: 0.0133
Exp:
1 m3 of air has 30 mg SO2
106 m 3 of air has 30g SO 2
30
mol SO 2
64
nRT
n
30 64 mol
V=
=
=
P
P RT 41.6 mol m 3
30
=
= 0.0113m3
64 × 41.6
Concentration of SO 2 in ppm = 0.0113ppm
=
34.
The 4-hr unit hydrograph for a catchment is given in the table below. What would be the
maximum ordinate of the S-curve (in m 3 / s) derived from this hydrograph?
Time(hr)
0
2
4
6
8
10
12
14
16
18
20
22
24
Unit
hydrograph
ordinate(m3/s)
0
0.6
3.1
10
13
9
5
2
0.7
0.3
0.2
0.1
0
Answer: 22
Exp:
Time
0
2
4
6
8
10
12
14
16
18
UHO
0
0.6
3.1
10
13
9
5
2
0.7
0.3
S-curve Addition
0
0.6
3.1
10.6
16.1
19.6
21.1
21.6
SA
0
0.6
3.1
10.6
16.1
19.6
21.1
21.6
21.8
21.9
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
18
CE-GATE-2015 PAPER|
20
0.2
21.8
22
0.1
21.9
24
0
22
Maximum S-curve ordinate is 22.
35.
(A)
FD
µ
and
VD
ρVD
(B)
FD
ρVD
and
2
ρVD
µ
(C)
FD
ρVD
and
2 2
ρV D
µ
(D)
FD
µ
and
3 3
ρV D
ρVD
(C)
ρvD
= Re → dim ensionles parameter
µ
FD ( kg − m s 2 )
2
 kg   m 
ρ  3  V2  2  × D2 ( m2 )
m   s 
36.
22
22
22
The drag force, FD, on a sphere kept in a uniform flow field depends on the diameter of the
sphere, D; flow velocity, V; fluid density, ρ; and dynamic viscosity, µ . Which of the
following options represents the non-dimensional parameters which could be used to analyze
this problem?
Answer:
Exp:
www.gateforum.com
→ dim ensionless parameter
Consider the singly reinforced beam section given below (left figure). The stress block
parameters for the cross-section from IS:456-2000 are also given below (right figure). The
moment of resistance for the given section by the limit state method is _________ kN-m.
0.42x u
M25
xu
300 mm
0.36 f ck x u
d
4-12 Φ
Fe415
x u,max = 0.4d
for Fe415
200 mm
Answer:
Exp:
42.82
π
A st = 4 × × (12) 2 = 453 mm 2
4
0.36 fck.b.xu = 0.87 fy Ast
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
19
CE-GATE-2015 PAPER|
⇒ xν =
0.87 f y A st
0.36f ck .b
=
www.gateforum.com
0.87 × 415 × 453
0.36 × 25 × 200
= 90.86 mm
x ν , max = 0.48d
= 0.4 × 300 = 120 mm
x ν < x ν , max so U.R.section
M ν = 0.87 × f y × A st × ( d − 0.42 x ν )
= 0.87 × 415 × 453 × ( 300 − 42 × 90.86 ) = 42.82 kNm
37.
Two reservoirs are connected through a 930 m long, 0.3 m diameter pipe, which has a gate
valve. The pipe entrances is sharp (loss coefficient = 0.5) and the valve is half-open (loss
coefficient = 5.5). The head difference between the two reservoirs is 20 m. Assume the
friction factor for the pipe as 0.03 and g = 10 m/s2. The discharge in the pipe accounting for
all minor and major losses is _________ m3/s.
Answer: 0.1413
Exp: Total loss = 20 m
⇒ 20 =
0.5 × v 2 f × L v 2 5.5v 2 v 2
+
×
+
+
29
d
2g
2g
2g
⇒ 20 × 2 × 10 = 0.5 v 2 +
0.03 × 930 × v 2
+ 5.5v 2 + v 2
0.3
400
=4
100
⇒ v =2m s
⇒ v2 =
π
π
2
θ = × d 2 × v = × ( 0.3) × 2 = 0.1413 m 3 s
4
4
38.
A sign is required to be put up asking drivers to slow down to 30 km/h before entering Zone
Y (see figure). On this road, vehicles require 174 m to slow down to 30 km/h (the distance of
174 m includes the distance travelled during the perception-reaction time of drivers). The sign
can be read by 6/6 vision drivers from a distance of 48 m. The sign is placed at distance of x
m from the start of Zone Y so that even a 6/9 vision driver can slow down to 30 km/h before
entering the zone. The minimum value of x is _________ m.
Direction of vehicle movement
Start of Zone Y
Sign
Road
x
Zone Y
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
20
CE-GATE-2015 PAPER|
Answer:
Exp:
www.gateforum.com
142
For a 6/6 person, driver can see from a distance of 48 m.
6
= 32m
9
The vehicle requires 174 m to slow down to 30 km/hr
For a 6/9 person, driver can see from distance = 48 ×
So, minimum distance, X=174-32=142 m.
39.
The quadric equation x 2 − 4x + 4 = 0 is to be solved numerically, starting with the initial
guess x 0 = 3. The Newton-Raphson method is applied once to get a new estimate and then
the Secant method is applied once using the initial guess and this new estimate. The estimated
value of the root after the application of the Secant method is ___________.
Answer: 2.333
Exp: f(x) = x2-4x+4
x0 = 3
f ' ( x ) = 2x − 4
By Newton Raphson method x1 = x 0 −
f ( x0 )
f '( x0 )
1
= 2.5
2
For secant method let x0 = 2.5 and x1 = 3
= 3−
By secant method x 2 = x1 −
x1 − x 0
f ( x1 )
f ( x1 ) − f ( x 0 )
= 3−
( 3 − 2.5) f 3
( )
f ( 3) − f ( 2.5 )
= 3−
0.5
×1
1 − ( 0.25 )
0.5
0.75
= 3 − 0.6667
= 3−
= 2.333
40.
Consider the following complex function:
f (z) =
9
( z − 1)( z + 2 )
2
Which of the following one of the residues of the above function?
(A) −1
Answer:
(B) 9/16
(C) 2
(D) 9
(A)
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
21
CE-GATE-2015 PAPER|
Exp:
f ( 3) =
www.gateforum.com
9
( z − 1)( z + 2 )
2
z = 1 is a simple pole
z = -2 is a pole of order 2
 Res f ( z )  z =1 = lim ( z − 1)
z →1
=
9
( z − 1)( z + 2 )
2
9
=1
9

1
d 
9
2
 Resf ( z )  z =−2 = lim ( z + 2 ) .

2
1! z →−2 dz 
( z − 1)( z + 2 ) 
= lim
z →−2
=
41.
( z − 1)
2
−9
= −1
9
A short reach of a 2 m wide rectangular open channel has its bed level rising in the direction
of flow at a slope of 1 m in 10000. It carries a discharge of 4 m3/s and its Manning’s
roughness coefficient is 0.01. The flow in this reach is gradually varying. At a certain section
in this reach, the depth of flow was measured as 0.5m. The rate of change of the water depth
with distance, dy/dx, at this section is __________ (use g = 10 m/s2).
Answer:
Exp:
−9
0.0032
Adverse slope = −
1
10000
θ = 4 m 3 / s, n = 0.01, y = 0.5 m
dy S0 − Sf
=
dx 1 − Fr 2
Fr =
V
gy
=
Q
By gy
1
AR 2/3Sf 1/2
n
Q×n
Sf 1/2 =
=
A × R 2/3
=
4
2 × 0.5 × 10 × 5
= 1.79
Q=
4 × 0.01
 2 × 0.5 
2 × 0.5 × 

 2 +1 
2/3
Sf = 6.92 × 10−3
1
− 6.92 × 10−3
dy 10000
=
= 3.2 × 10−3 = 0.0032
2
dx
1 − (1.79)
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
22
CE-GATE-2015 PAPER|
42.
www.gateforum.com
In a survey work, three independent angles X, Y and Z were observed with weights WX, WY,
WZ, respectively. The weight of the sum of angles X, Y and Z is given by:
 1
 1
1
1 
1
1 
+
+
(B) 
+
+
(A) 1 


 WX WY WZ 
 WX WY WZ 
(C) WX + WY + WZ
Answer: (A)
(D) WX2 + WY2 + WZ2
43.
A hydraulic jump is formed in a 2m wide rectangular channel which is horizontal and
frictionless. The post-jump depth and velocity are 0.8 m and 1 m/s, respectively. The prejump velocity is ___________ m/s. (use = 10 m/s2).
Answer: 4.94
Exp: B=2m, y 2 = 0.8m, U 2 = 1m / s
F2 =
U2
g.y 2
=
1
= 0.35
10 × 0.8
y1
1 1
= − + . 1 + 8F22
y2
2 2
y1 −1 1
=
+ . 1 + 8 × (3.5) 2 = 0.203
0.8 2 2
⇒ y1 = 0.203 × 0.8 = 0.162 m
⇒
Q = B.y 2 .V2 = B.y1V1
⇒ 0.8 × 1 = 0.162 × V1
⇒ V1 = 4.94 m / s.
44.
A 20 m thick clay layer is sandwiched between a silty sand layer and a gravelly sand layer.
The layer experiences 30 mm settlement in 2 years.
Given
2

π U 
for U ≤ 60%



TV = 
4  100 
1.781 − 0.933 log (100 − U ) for U > 60%
10

Where Tv is the time factor and U is the degree of consolidation in %.
If the coefficient of consolidation of the layer is 0.003 cm2/s, the deposit will experience a
total of 50 mm settlement in the next ________ years.
Answer: 4.43
silty sand
Exp:
20m
clay
g ravel
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
23
CE-GATE-2015 PAPER|
TU =
003 × 2 × 86400 × 365
 20

 × 100 
2


2
www.gateforum.com
= 0.189
π 2
U = 0.189 ⇒ U = 0.49 ≤ 60%
4
30
= 61.2.2 mm
Consolidation =
0.49
Degree of consolidation for 50mm settlement
50
= 0.817 = 81.7%
61.22
⇒ Tv = 1.784 − 0.933 log10 (100 − U )
U=
= 0.608 =
CV × t
d2
0.608 × (10 )
0.608 × H 2
=
s
−4
0.003 × 10
0.003 × 10−4
= 202666667 s
= 6.43 yr
2
⇒t=
Additional number of years = 6.43 − 2 − 4.43 years
45.
A bracket plate connected to a column flange transmits a load of 100 kN as shown in the
following figure. The maximum force for which the bolts should be designed is __________
kN.
100 kN
75 75
600
75 75
Answer:
Exp:
All dimensions are in mm
156.20
FD =
P 100
=
= 20 kN
n
5
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
24
CE-GATE-2015 PAPER|
www.gateforum.com
75
75
θ
Ft θ F
b
Ft =
(P.d)r 100 × 600 × 75 2
=
= 141.42 kN
4 × (75 2) 2
∑ r2
FR = FD2 + Ft2 + 2 × FD × Ft cos θ
= (20) 2 + (141.42) 2 + 2 × 20 × 141.42 ×
1
2
= 156.20 kN
1
2
o
⇒ θ = 45 .
cos θ =
46.
Consider a primary sedimentation tank (PST) in a water treatment plant with surface
Overflow Rate (SOR) of 40 m3/m2/d. The diameter of the spherical particle which will have
90 percent theoretical removal efficiency in this tank is ___________ µm. Assume that
settling velocity of the particles in water is described by Stokes’s Law.
Given Density of water = 1000 kg/m3; Density of particle = 2650 kg/m3; g = 9.81 m/s2;
Kinematic viscosity of water ( v ) = 1.10 × 10−6 m 2 s
Answer:
Exp:
22.58
% removal =
Vs '
× 100
Vs
Vs ' = 0.9 Vs
=
0.9 × 40
ms
86400
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
25
CE-GATE-2015 PAPER|
⇒
www.gateforum.com
1
g
0.9 × 40
× d 2 × ( ρS − ρ w ) =
µ
18
86400
⇒d =
0.9 × 40 × 18 × V. ρ w
86480(G S − 1) × ρ w × g
⇒ d = 22.58 µm
47.
A non-homogenous oil deposit consists of a silt layer sandwiched between a fine-sand layer at
top and a clay layer below. Permeability of the silt layer is 10 times the permeability of the
clay layer and one-tenth of the permeability of the sand layer. Thickness of the silt layer is 2
time the thickness of the sand layer and two-third of the thickness of the clay layer. The ratio
of equivalent horizontal and equivalent vertical permeability of the deposit is ___________.
Answer:
10.967
Exp:
(1)
( 2)
( 3)
H1
H2
H3
K1 finesand
K2
silt
K 3 clay
1
k1
10
⇒ k1 = 10k 2
k 2 = 10k 3 =
= 10 × 10k 3
k1 = 100k 3
k1 = 10k 2
H2 = 2H1
2
3
3
H 2 = H 3 ⇒ H 3 = H 2 = × 2H1 = 3H1
3
2
2
H 3 = 3H1
K H + K 2 H 2 + K 3 H3
Kx = 1 1
=
H1 + H 2 + H 3
K1H1 +
1
1
K1 × 2H1 +
K1 × 3H1
10
100
H1 + 2H1 + 3H1
2
3 

1 + +
 K1H1
123
10 100 

Kx =
=
K1
6H1
100 × 6
Ky =
H1 + H 2 + H 3
6H1
=
H1 H 2 H 3 H1 2H1 × 10 3H1 × 100
+
+
+
+
K1 K 2 K 3 K1
K1
K1
Kx
123
321
=
×
= 10.967
K y 100 × 6 6
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
26
CE-GATE-2015 PAPER|
48.
www.gateforum.com
In a region with magnetic declination of 2o E, the magnetic Fore bearing (FB) of a line AB
was measured as N79o 50′E. There was a local attraction at A. To determine the correct
magnetic bearing of the line, a point O was selected at which there was no local attraction.
The magnetic FB of line AO and OA were observed to be S52o 40′E and N50o 20′W,
respectively. What is the true FB of line AB?
(A) N81o 50′E
(B) N82o10′E
(C) N84o10′E
(D) N77o 50′E
Answer: (C)
Exp:
δ = 2°E
Magnetic F.B. of AB = N79O50’E = 79O50.
Correct FB of OA = N50O20’W = 309O40’
∴ Correct B.B of OA = 129O 40'
∵ observed F.B. of AO = observed BB of OA
= 552O 40'E =127O 20'
Error = M.V- T.V = -2O20’
Correction + 2O20’
T.B. of FB of AB = N 79°50'E + 2 + 2°20'
= N84°10'E
49.
For the 2D truss with the applied loads shown below, the strain energy in the member XY is
___________ kN-m. For member XY, assume AE = 30 kN, where A is cross-section area and
E is the modulus of elasticity.
5kN
10 kN
3m
3m
X
Y
3m
3m
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
27
CE-GATE-2015 PAPER|
Answer:
Exp:
www.gateforum.com
5
5kN
D
10kN
E
3m
F
3m
G
3m
C
B
A
10kN
RA
3m
RG
RA×3 + 10×9 = 0
⇒ RA = -30 kN
RG = 35 kN
Taking joint A
30kN
10kN
10kN
30kN
Joint G
10 2
10kN
35kN
Joint B
Fx = 10 kN
U=
F × L 10 × 10 × 3
=
= 5kN − m
2A E
2 × 30
10
2
10 2
30
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
28
CE-GATE-2015 PAPER|
50.
www.gateforum.com
Two beam are connected by linear spring as shown in the following figure. For a load P as
shown in the figure, the percentage of the applied load P carried by the spring is
___________.
P
L
EI
EI
Answer:
( )
Ks = 3EI 2L3
33.33
Exp:
P
R
R
R
R
R
K
R
=
× 2L3
3EI
2R.L3
∆=
3EI
∆=
PL3 RL3 2RL3
−
=
3EI 3EI
3EI
3
3
PL 3RL
P
⇒
=
⇒ R = = 33.33%.
3EI 3EI
3
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
29
CE-GATE-2015 PAPER|
51.
www.gateforum.com
The smallest and largest Eigen values of the following matrix are:
 3 −2 2 
 4 −4 6 


 2 −3 5 
(A) 1.5 and 2.5
Answer:
Exp:
(B) 0.5 and 2.5
(C) 1.0 and 3.0
(D) 1.0 and 2.0
(D)
 3 −2 2 
Let A =  4 −4 6 
 2 −3 5 
Characteristic equation is
|A-λI| = 0
3−λ
−2
2
⇒ 4
−4 − λ 6
2
−3
=0
5−λ
⇒ λ 3 − 4λ 2 + 5λ − 2 = 0
⇒ ( λ − 1) ( λ 2 − 3λ + 2 ) = 0
( λ − 1)( λ − 1)( λ − 2 ) = 0
λ = 1, 2
52.
A square footing (2mx 2m) is subjected to an inclined point load, P as shown in the figure
below. The water table is located well below the base of the footing. Considering one-way
eccentricity, the net safe load carrying capacity of the footing for a factor of safety of 3.0 is
__________ kN.
The following factors may be used.
Bearing capacity factors: Nq = 33.3, N γ = 37.16; Shape factors: Fqs = Fγs = 1.314; Depth
factors: Fqd = Fγd = 1.113; Inclination factors: Fqi = 0.444, Fγi = 0.02
P
GL
30o
1m
0.85m
Unit weight = 18kN m3
2m
Cohesion = 0
Friction angle = 35o
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
30
CE-GATE-2015 PAPER|
Answer:
Exp:
www.gateforum.com
450
q nu
3
q nu = cN c + qN q + 0.5γBN γ − 8∆
q safc =
C=0
q nu = q ( N q − 1) + 0.5γ BN γ
(
)
1
q ( N q − 1) Fqs × Fqd + Fqp + 0.5γBN γ × Fγs × Fγo × Fγp
3
1


1 18 × 1( 33.3 − 1) × 1.314 × 1.113 × 0.444 + × 2 × 18 × 
q ns =
2

3 
37.16 × 1.314 × 1.113 × 0.02 

397.03
=
= 132.364 kN m 2
3
q ns =
For one way shear (eccentricity) area to be reduced
2
0.85 0.85
2
Reduced area of footing = 2×1.7 = 3.4m2
Load carrying capacity = 132.364×3.4 = 450.kN
53.
In a catchment, there are four rain-gauge stations, P, Q, R, and S. Normal annual precipitation
values at these stations are 780 mm, 850 mm, 920 mm, and 980 mm, respectively. In the year
2013, stations Q, R, and S, were operative but P was not. Using the normal ratio method, the
precipitation at station P for the year 2013 has been estimated as 860 mm. If the observed
precipitation at stations Q and R for the year 2013 were 930 mm and 1010 mm, respectively;
what was the observed precipitation (in mm) at station S for that year?
Answer:
Exp:
1076.2
P
Ps 1  Pp
P 
= 
+ Q + R 
N s 3  N p N Q N R 
P
1  860 930 1010 
⇒ s = 
+
+
980 3  780 850 920 
⇒ Ps = 1076.20 mm
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
31
CE-GATE-2015 PAPER|
54.
www.gateforum.com
The acceleration-time relationship for a vehicle subjected to non-uniform acceleration is,
dv
= ( α − β v0 ) e −βt
dt
Where, v is the speed in m/s, t is the time in s, α and β are parameters, and v0 is the initial
speed in m/s. If the accelerating behavior of a vehicle, whose drive intends to overtake a slow
moving vehicle ahead, is described as,
dv
= ( α − βv )
dt
dv
= 1.3 m s 2 at t = 3 s, the distance (in m)
dt
travelled by the vehicle in 35 s is ____________.
Considering α = 2 m s 2 , β = 0.05s −1 and
Answer:
Exp:
900.83
dV
= ( α − β VO ) e −βT
dt
−βt
∫ dv = ∫ ( α − β VO ) × e .dt
=
( α − βVO ) e−βt
−β
t = 0, V = V0
⇒ V0 =
( α − βV0 ) + C
−β
α − βV0
α
C = V0 +
⇒C =
β
β
⇒V=
x=
α − ( αβ V0 ) × e−βt
β
αt 0 α − β V0 −βt 0
+
e −1
β
β2
(
)
dv
= ( α − β V0 ) e−3β = 1.3
dt t =3
1.3
e −3β
αt
1.3
x = 0 + 2 −3β e −βt 0 − 1
β β (e )
⇒ α − β V0 =
(
)
−35×0.05
− 1)
2 × 35 1.3 ( e
=
+
2
−3x 0.05
0.05 ( 0.05 ) ( e
)
=1400 − 499.17 = 900.83m
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
32
CE-GATE-2015 PAPER|
55.
www.gateforum.com
On a circular curve, the rate of super elevation is e. While negotiating the curve a vehicle
comes to a stop. It is seen that the stopped vehicle does not slide inwards (in the radial
direction). The coefficient of side friction is f. Which of the following is true?
(A) e ≤ f
(B) f < e < 2f
(C) e ≥ 2f
(D) None of the above
Answer:
(A)
Exp:
N
fN
mg sin θ
θ
mg cos θ
mg
f N ≥ mg sin θ
⇒ f ( mg cos θ ) = mg sin θ
⇒ f ≥ tan θ
⇒f ≥ e
⇒ e ≤f
India’s No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India
33