⚙️Mechanical Engineering GATE Pattern Questions (1-55)
Section A: Engineering Mathematics (Questions 1-5)
Question 1 (1 Mark - MCQ)
If A= ( 24 13), then the value of the determinant of its inverse, ¿ A −1
∨¿, is:
(A) 1/2 (B) 2 (C) 5 (D) 1/5
Solution 1
Concept: Determinant of the inverse of a matrix.
Formula: ¿ A− 1∨¿ 1/¿ A∨¿ .
Calculation: ¿ A∨¿(2)(3)−(1)(4 )=6 − 4=2 ¿ A− 1∨¿ 1/¿ A∨¿ 1 /2
Answer: (A) 1/2
Question 2 (2 Marks - NAT)
π /2
The value of the definite integral ∫ sin 2(x )d x is π / K . Find the value of K .
0
Solution 2
Concept: Definite integration.
1− cos(2 x )
Formula: Use the identity sin ( x )= .
2
2
[ ]
π /2 π /2 π/2
1 −cos (2 x) sin(2 x)
Calculation: ∫ sin (x )d x= ∫ d x ¿ 1 x−
2
0 0 2 2 2 0
¿
1
2 [( π sin (π)
2
−
2 )(
− 0−
sin (0)
2 )] [ ¿
1 π
2 2 ]
−0 − 0 =
π
4
π /2
Comparison: Given the result is π / K . Since ∫ sin 2(x )d x=π / 4, we have K=4.
0
Answer: 4
,Question 3 (2 Marks - MCQ)
2 2
∂ u ∂ u
The partial differential equation 2
+ 2 =0 is classified as:
∂x ∂y
(A) Parabolic (B) Hyperbolic (C) Elliptic (D) Non-linear
Solution 3
Concept: Classification of second-order linear PDEs.
2 2
∂ u ∂ u
Analysis: For 2
+ 2 =0 (Laplace's equation), we have A=1, B=0 , C=1.
∂x ∂y
2 2
B − 4 A C=0 − 4 (1)(1)=− 4
Conclusion: Since the discriminant is less than zero, the equation is Elliptic.
Answer: (C) Elliptic
Question 4 (1 Mark - NAT)
A bag contains 5 red and 5 blue balls. If two balls are drawn at random without
replacement, what is the probability that they are of different colors? (Round to
two decimal places).
Solution 4
Concept: Probability of drawing two different colored balls without
replacement.
Calculation: P(Different Colors)=P(R then B)+ P(B then R )
¿( 5 5
× +
10 9 )( 5 5
× =2× =
10 9 ) 25 5
90 9
Decimal Value: 5/9 ≈ 0.5555 . . .
Answer (to 2 decimal places): 0.56
Question 5 (2 Marks - MCQ)
The Taylor series expansion of ln (x) around x=1 is given by:
∞ ∞ ∞ ∞
(A) ∑ ¿¿ ¿ (B) ∑ ¿ ¿¿ (C) ∑ ¿¿ ¿ (D) ∑ ¿ ¿¿
n =1 n=0 n =1 n=0
Solution 5
Concept: Taylor series expansion for f (x) around a .
, ∞
Result (for f (x)=ln(x ) and a=1): ln (x)=(x − 1)−¿ ¿ ¿ ∑ ¿ ¿ ¿
n=1
∞
Answer: (A) ∑ ¿¿ ¿
n =1
Section B: Applied Mechanics & Design (Questions 6-15)
Question 6 (1 Mark - MCQ)
The deflection of a spring is directly proportional to the applied load and is
governed by Hooke's Law. The stiffness of a spring is defined as:
(A) The ratio of the deflection to the applied load. (B) The ratio of the applied load
to the deflection. (C) The product of the applied load and the deflection. (D) The
square of the applied load.
Solution 6
Concept: Definition of spring stiffness (k ).
Definition: Stiffness k =F /δ . It is the load required per unit deflection.
Answer: (B) The ratio of the applied load to the deflection.
Question 7 (2 Marks - NAT)
A circular shaft of diameter D=50 mm is subjected to a torque T =2 kNm . The
maximum shear stress (in MPa) developed in the shaft is: (Round to two decimal
places).
Solution 7
T R 16 T
Concept: Torsion equation for a circular shaft: τ m a x = J = 3.
πD
Parameters: T =2000 Nm , D=0.05 m .
16 ×2000 Nm
Calculation: τ m a x = τ m a x ≈ 81.49 MPa
π ¿¿
Answer: 81.49
Question 8 (1 Mark - MCQ)
For a bolted joint subjected to fluctuating external load, the bolts are prestressed
(pre-tensioned) primarily to:
Section A: Engineering Mathematics (Questions 1-5)
Question 1 (1 Mark - MCQ)
If A= ( 24 13), then the value of the determinant of its inverse, ¿ A −1
∨¿, is:
(A) 1/2 (B) 2 (C) 5 (D) 1/5
Solution 1
Concept: Determinant of the inverse of a matrix.
Formula: ¿ A− 1∨¿ 1/¿ A∨¿ .
Calculation: ¿ A∨¿(2)(3)−(1)(4 )=6 − 4=2 ¿ A− 1∨¿ 1/¿ A∨¿ 1 /2
Answer: (A) 1/2
Question 2 (2 Marks - NAT)
π /2
The value of the definite integral ∫ sin 2(x )d x is π / K . Find the value of K .
0
Solution 2
Concept: Definite integration.
1− cos(2 x )
Formula: Use the identity sin ( x )= .
2
2
[ ]
π /2 π /2 π/2
1 −cos (2 x) sin(2 x)
Calculation: ∫ sin (x )d x= ∫ d x ¿ 1 x−
2
0 0 2 2 2 0
¿
1
2 [( π sin (π)
2
−
2 )(
− 0−
sin (0)
2 )] [ ¿
1 π
2 2 ]
−0 − 0 =
π
4
π /2
Comparison: Given the result is π / K . Since ∫ sin 2(x )d x=π / 4, we have K=4.
0
Answer: 4
,Question 3 (2 Marks - MCQ)
2 2
∂ u ∂ u
The partial differential equation 2
+ 2 =0 is classified as:
∂x ∂y
(A) Parabolic (B) Hyperbolic (C) Elliptic (D) Non-linear
Solution 3
Concept: Classification of second-order linear PDEs.
2 2
∂ u ∂ u
Analysis: For 2
+ 2 =0 (Laplace's equation), we have A=1, B=0 , C=1.
∂x ∂y
2 2
B − 4 A C=0 − 4 (1)(1)=− 4
Conclusion: Since the discriminant is less than zero, the equation is Elliptic.
Answer: (C) Elliptic
Question 4 (1 Mark - NAT)
A bag contains 5 red and 5 blue balls. If two balls are drawn at random without
replacement, what is the probability that they are of different colors? (Round to
two decimal places).
Solution 4
Concept: Probability of drawing two different colored balls without
replacement.
Calculation: P(Different Colors)=P(R then B)+ P(B then R )
¿( 5 5
× +
10 9 )( 5 5
× =2× =
10 9 ) 25 5
90 9
Decimal Value: 5/9 ≈ 0.5555 . . .
Answer (to 2 decimal places): 0.56
Question 5 (2 Marks - MCQ)
The Taylor series expansion of ln (x) around x=1 is given by:
∞ ∞ ∞ ∞
(A) ∑ ¿¿ ¿ (B) ∑ ¿ ¿¿ (C) ∑ ¿¿ ¿ (D) ∑ ¿ ¿¿
n =1 n=0 n =1 n=0
Solution 5
Concept: Taylor series expansion for f (x) around a .
, ∞
Result (for f (x)=ln(x ) and a=1): ln (x)=(x − 1)−¿ ¿ ¿ ∑ ¿ ¿ ¿
n=1
∞
Answer: (A) ∑ ¿¿ ¿
n =1
Section B: Applied Mechanics & Design (Questions 6-15)
Question 6 (1 Mark - MCQ)
The deflection of a spring is directly proportional to the applied load and is
governed by Hooke's Law. The stiffness of a spring is defined as:
(A) The ratio of the deflection to the applied load. (B) The ratio of the applied load
to the deflection. (C) The product of the applied load and the deflection. (D) The
square of the applied load.
Solution 6
Concept: Definition of spring stiffness (k ).
Definition: Stiffness k =F /δ . It is the load required per unit deflection.
Answer: (B) The ratio of the applied load to the deflection.
Question 7 (2 Marks - NAT)
A circular shaft of diameter D=50 mm is subjected to a torque T =2 kNm . The
maximum shear stress (in MPa) developed in the shaft is: (Round to two decimal
places).
Solution 7
T R 16 T
Concept: Torsion equation for a circular shaft: τ m a x = J = 3.
πD
Parameters: T =2000 Nm , D=0.05 m .
16 ×2000 Nm
Calculation: τ m a x = τ m a x ≈ 81.49 MPa
π ¿¿
Answer: 81.49
Question 8 (1 Mark - MCQ)
For a bolted joint subjected to fluctuating external load, the bolts are prestressed
(pre-tensioned) primarily to: