All Exams >
Mechanical Engineering >
Strength of Materials (SOM) >
All Questions
All questions of Principal Stresses & Strains (Mohr`s Circle) for Mechanical Engineering Exam
A bar of square cross - section has b ee n subjected to an axial tensile load. A plane normal to the axis of loading will have- a)maximum shear stress and minimum normal stress
- b)maximum shear stress as well as maximum normal stress
- c)maximum normal stress and zero shear stress
- d)minimum normal stress and zero shear stress
Correct answer is option 'C'. Can you explain this answer?
A bar of square cross - section has b ee n subjected to an axial tensile load. A plane normal to the axis of loading will have
a)
maximum shear stress and minimum normal stress
b)
maximum shear stress as well as maximum normal stress
c)
maximum normal stress and zero shear stress
d)
minimum normal stress and zero shear stress
|
Sanvi Kapoor answered |
For the plane normal to the axis of loading,
A rectangular strain rosette, shown in figure gives following reading in a strain measurement task ϵ1 = 1000 × 10−6, ϵ2 = 800 × 10−6 and ϵ3 = 600 × 10−6. The direction of the major principal strain with respect to gauge 1 is
- a) 0°
- b) 15°
- c) 30°
- d) 45°
Correct answer is option 'A'. Can you explain this answer?
A rectangular strain rosette, shown in figure gives following reading in a strain measurement task ϵ1 = 1000 × 10−6, ϵ2 = 800 × 10−6 and ϵ3 = 600 × 10−6. The direction of the major principal strain with respect to gauge 1 is
a)
0°
b)
15°
c)
30°
d)
45°
|
Gate Funda answered |
tan 2θp = 
= 
= 0
The principal stress induced at a point in a machine element made of steel (Syt = 460 MPa) are as follows σ1 = 210 MPa and σ2 = 145 MPa, the factor of safety according to Hencky’s theory of failure is- a) 1.64
- b) 4.26
- c) 2.47
- d) 3.23
Correct answer is option 'C'. Can you explain this answer?
The principal stress induced at a point in a machine element made of steel (Syt = 460 MPa) are as follows σ1 = 210 MPa and σ2 = 145 MPa, the factor of safety according to Hencky’s theory of failure is
a)
1.64
b)
4.26
c)
2.47
d)
3.23
|
Zoya Sharma answered |
Σ1 = 210 MPa σ2 = 145 MPa According to Hencky’s Theory (Von-Mises)
(210)2 + (145)2 − 210 × 145 ≤ 
N = 2.47
Which point on the stress strain curve occurs after yield plateau?- a)Ultimate point
- b)lower yield point
- c)Upper yield point
- d)Breaking point
Correct answer is option 'A'. Can you explain this answer?
Which point on the stress strain curve occurs after yield plateau?
a)
Ultimate point
b)
lower yield point
c)
Upper yield point
d)
Breaking point
|
Jaya Yadav answered |
Ultimate Point on the Stress-Strain Curve:
The ultimate point on the stress-strain curve is the point that occurs after the yield plateau. This point represents the maximum stress that a material can withstand before it fails or fractures. It is also known as the ultimate tensile strength (UTS) or simply the breaking point.
Explanation:
The stress-strain curve is a graphical representation of the relationship between the applied stress and the resulting strain in a material. It provides valuable information about the mechanical properties and behavior of the material under different loading conditions.
Yield Plateau:
The yield plateau is the region on the stress-strain curve where the material undergoes plastic deformation without experiencing a significant increase in stress. It indicates the onset of yield, which is the point where the material starts to undergo permanent deformation.
Lower Yield Point:
The lower yield point is the first drop in stress that occurs after the elastic region on the stress-strain curve. It represents the point where the material transitions from elastic to plastic deformation. However, this point is not relevant to this question as it occurs before the yield plateau.
Upper Yield Point:
The upper yield point is the highest stress value reached on the stress-strain curve before the yield plateau. It is typically observed in certain materials, such as low-carbon steel, and represents a temporary softening of the material due to localized necking. However, this point is also not relevant to this question as it occurs before the yield plateau.
Breaking Point:
The breaking point or fracture point is the point on the stress-strain curve where the material fails and fractures. It occurs after the ultimate point and represents the maximum stress that the material can withstand before complete failure.
Conclusion:
In summary, the ultimate point on the stress-strain curve occurs after the yield plateau. It represents the maximum stress that a material can withstand before it fails or fractures. The lower yield point, upper yield point, and breaking point are all distinct points on the stress-strain curve, but they occur either before or after the yield plateau. Therefore, the correct answer to the question is option 'A' - Ultimate point.
The ultimate point on the stress-strain curve is the point that occurs after the yield plateau. This point represents the maximum stress that a material can withstand before it fails or fractures. It is also known as the ultimate tensile strength (UTS) or simply the breaking point.
Explanation:
The stress-strain curve is a graphical representation of the relationship between the applied stress and the resulting strain in a material. It provides valuable information about the mechanical properties and behavior of the material under different loading conditions.
Yield Plateau:
The yield plateau is the region on the stress-strain curve where the material undergoes plastic deformation without experiencing a significant increase in stress. It indicates the onset of yield, which is the point where the material starts to undergo permanent deformation.
Lower Yield Point:
The lower yield point is the first drop in stress that occurs after the elastic region on the stress-strain curve. It represents the point where the material transitions from elastic to plastic deformation. However, this point is not relevant to this question as it occurs before the yield plateau.
Upper Yield Point:
The upper yield point is the highest stress value reached on the stress-strain curve before the yield plateau. It is typically observed in certain materials, such as low-carbon steel, and represents a temporary softening of the material due to localized necking. However, this point is also not relevant to this question as it occurs before the yield plateau.
Breaking Point:
The breaking point or fracture point is the point on the stress-strain curve where the material fails and fractures. It occurs after the ultimate point and represents the maximum stress that the material can withstand before complete failure.
Conclusion:
In summary, the ultimate point on the stress-strain curve occurs after the yield plateau. It represents the maximum stress that a material can withstand before it fails or fractures. The lower yield point, upper yield point, and breaking point are all distinct points on the stress-strain curve, but they occur either before or after the yield plateau. Therefore, the correct answer to the question is option 'A' - Ultimate point.
Complimentary shear stresses are- a)equal both in magnitude and sign
- b)equal in magnitude but opposite in sign
- c)unequal in magnitude but of same sign
- d)equal in magnitude, and the direction may be same or opposite
Correct answer is option 'B'. Can you explain this answer?
Complimentary shear stresses are
a)
equal both in magnitude and sign
b)
equal in magnitude but opposite in sign
c)
unequal in magnitude but of same sign
d)
equal in magnitude, and the direction may be same or opposite
|
Kajal Tiwari answered |
**Complimentary Shear Stresses:**
Complimentary shear stresses refer to shear stresses acting on two mutually perpendicular planes in a solid material. These shear stresses are equal in magnitude but opposite in sign.
**Explanation:**
When a solid material is subjected to shear stress, it deforms and experiences internal forces. These internal forces can be resolved into normal stresses and shear stresses acting on different planes within the material.
When considering shear stresses on two mutually perpendicular planes, they are called complimentary shear stresses. The magnitudes of these shear stresses are equal, but their signs are opposite.
**Example:**
Let's consider a simple example to understand this concept. Suppose a rectangular block is subjected to shear stress along the x-axis. The shear stress acting on the xz-plane will be positive, indicating a shear force in the positive x-direction. Simultaneously, the shear stress acting on the yz-plane will be negative, indicating a shear force in the negative x-direction.
This means that the shear stress acting on the xz-plane and the shear stress acting on the yz-plane are equal in magnitude but opposite in sign. They are complimentary shear stresses.
**Importance:**
Complimentary shear stresses are essential in understanding the distribution of internal forces within a material. They help in analyzing the deformation and failure of structural components when subjected to shear stress. By considering the complimentary shear stresses, engineers can design structures that can withstand these forces without failure.
In conclusion, complimentary shear stresses are equal in magnitude but opposite in sign. They play a significant role in understanding the behavior of materials under shear stress and are essential in engineering design and analysis.
Complimentary shear stresses refer to shear stresses acting on two mutually perpendicular planes in a solid material. These shear stresses are equal in magnitude but opposite in sign.
**Explanation:**
When a solid material is subjected to shear stress, it deforms and experiences internal forces. These internal forces can be resolved into normal stresses and shear stresses acting on different planes within the material.
When considering shear stresses on two mutually perpendicular planes, they are called complimentary shear stresses. The magnitudes of these shear stresses are equal, but their signs are opposite.
**Example:**
Let's consider a simple example to understand this concept. Suppose a rectangular block is subjected to shear stress along the x-axis. The shear stress acting on the xz-plane will be positive, indicating a shear force in the positive x-direction. Simultaneously, the shear stress acting on the yz-plane will be negative, indicating a shear force in the negative x-direction.
This means that the shear stress acting on the xz-plane and the shear stress acting on the yz-plane are equal in magnitude but opposite in sign. They are complimentary shear stresses.
**Importance:**
Complimentary shear stresses are essential in understanding the distribution of internal forces within a material. They help in analyzing the deformation and failure of structural components when subjected to shear stress. By considering the complimentary shear stresses, engineers can design structures that can withstand these forces without failure.
In conclusion, complimentary shear stresses are equal in magnitude but opposite in sign. They play a significant role in understanding the behavior of materials under shear stress and are essential in engineering design and analysis.
Two wooden joists 50mm and 100mm are glued together along the joist AB as shown. Determine the normal and tangential stress in the glue if P = 200 kN
- a) 165 MPa and 197 MPa
- b) 16.5 MPa and 19.7 MPa
- c) 1.65 MPa and 1.97 MPa
- d) 160 MPa and 190 MPa
Correct answer is option 'B'. Can you explain this answer?
Two wooden joists 50mm and 100mm are glued together along the joist AB as shown. Determine the normal and tangential stress in the glue if P = 200 kN
a)
165 MPa and 197 MPa
b)
16.5 MPa and 19.7 MPa
c)
1.65 MPa and 1.97 MPa
d)
160 MPa and 190 MPa
|
Manish Aggarwal answered |
σx = 
= 40N/mm2
= 40N/mm2Normal stress, σθ
σθ = 
= 
σθ = 16.5N/mm2
Tangential stress, τθ
τθ = 
= 19.7MPa
If a small concrete cube is submerged deep in still water in such a way that the pressure exerted on all faces of the cube is P, then the maximum shear stress developed inside the cube is- a) 0
- b) P/2
- c) P
- d) 2P
Correct answer is option 'A'. Can you explain this answer?
If a small concrete cube is submerged deep in still water in such a way that the pressure exerted on all faces of the cube is P, then the maximum shear stress developed inside the cube is
a)
0
b)
P/2
c)
P
d)
2P
|
Avinash Sharma answered |
Maximum shear stress
τmax = 
σ1, σ2 = 
= 
σ1, σ2 = P
τmax =
= 0
= 0A Mohr’s circle reduces to a point when the body is subjected to- a)pure shear
- b)uniaxial stress only
- c)equal and opposite axial stresses on two mutually perpendicular planes, the planes being free of shear
- d)equal axial stresses on two mutually perpendicular planes, the planes being free of shear
Correct answer is option 'D'. Can you explain this answer?
A Mohr’s circle reduces to a point when the body is subjected to
a)
pure shear
b)
uniaxial stress only
c)
equal and opposite axial stresses on two mutually perpendicular planes, the planes being free of shear
d)
equal axial stresses on two mutually perpendicular planes, the planes being free of shear
|
Neha Joshi answered |
When the normal stresses on the two mutually perpendicular planes are equal and alike then radius of Mohr’s circle will be zero.
i.e.
Hence Mohr’s circle reduces to a point.
i.e.
Hence Mohr’s circle reduces to a point.
The state of stresses on an element is as shown in following figure.
The necessary and sufficient condition for its equilibrium is- a)τ1 - τ3 and τ2 = τ4
- b)τ1= τ4 and τ2 = τ3
- c)τ1 = 2τ4 and τ2 = 2 τ3
- d)τ1 = τ2 = τ3 = τ4
Correct answer is option 'D'. Can you explain this answer?
The state of stresses on an element is as shown in following figure.
The necessary and sufficient condition for its equilibrium is
a)
τ1 - τ3 and τ2 = τ4
b)
τ1= τ4 and τ2 = τ3
c)
τ1 = 2τ4 and τ2 = 2 τ3
d)
τ1 = τ2 = τ3 = τ4
|
|
Sanvi Kapoor answered |
Moment about upper right corner = 0
Moment about lower left corner = 0
A spherical pressure shell 3.8 meters diameter and 3 mm thick is subjected to an internal pressure ‘P’. The elastic limit of the shell material in simple tension is 260 MPa. The factor of safety is to 2.8. The internal pressure, if the failure of shell is to be prevented, according to maximum shear stress theory is- a) 1.78 bar
- b) 2.932 bar
- c) 5.86 bar
- d) 2.33 bar
Correct answer is option 'B'. Can you explain this answer?
A spherical pressure shell 3.8 meters diameter and 3 mm thick is subjected to an internal pressure ‘P’. The elastic limit of the shell material in simple tension is 260 MPa. The factor of safety is to 2.8. The internal pressure, if the failure of shell is to be prevented, according to maximum shear stress theory is
a)
1.78 bar
b)
2.932 bar
c)
5.86 bar
d)
2.33 bar
|
Lavanya Menon answered |
σnoop = σ1 = σ2 =
Syt = 260 MPa
N = 2.8
According to MSST
Abs τmax ≤ 
Abs τmax = 
P = 2.932 bar
A square plate of side 4 units undergoes deformation. The angle between the diagonals becomes 60°. Calculate τ45° (4 decimal places)
(A) 0.5235(B) 0.5235
Correct answer is option ''. Can you explain this answer?
A square plate of side 4 units undergoes deformation. The angle between the diagonals becomes 60°. Calculate τ45° (4 decimal places)
(A) 0.5235
(B) 0.5235
|
Cstoppers Instructors answered |
Before deformation, the angle between diagonals is 90°
τ45° = 90° − 60° = 30°
=
=0.5235
Question_Type: 5
The shear stress along the principal plane subjected to maximum principal stress is- a) minimum
- b) maximum
- c) zero
- d) any value depending on loading
Correct answer is option 'C'. Can you explain this answer?
The shear stress along the principal plane subjected to maximum principal stress is
a)
minimum
b)
maximum
c)
zero
d)
any value depending on loading
|
Mehul Choudhury answered |
The shear stress along the principal plane subjected to the maximum principal stress is zero.
Explanation:
- When a material is subjected to external forces, it experiences stresses in different directions. These stresses can be decomposed into normal and shear stresses.
- The principal stresses are the maximum and minimum normal stresses experienced by a material at a particular point.
- The principal planes are the planes on which the principal stresses act. These planes are perpendicular to each other.
- The maximum principal stress is the highest normal stress experienced by the material, and it occurs on the principal plane with the highest normal stress value.
- The minimum principal stress is the lowest normal stress experienced by the material, and it occurs on the principal plane with the lowest normal stress value.
- Shear stress occurs when forces are applied parallel to a plane, causing one part of the material to slide over the adjacent part. It is perpendicular to the normal stress.
- Along the principal planes, the shear stress is zero. This is because the principal planes are oriented in such a way that the maximum and minimum normal stresses act purely in the normal direction and do not cause any shearing motion.
- The shear stress can be calculated using the formula: shear stress = (maximum principal stress - minimum principal stress) / 2. Since the minimum principal stress is zero on the principal plane subjected to the maximum principal stress, the shear stress is also zero.
In conclusion, the shear stress along the principal plane subjected to the maximum principal stress is zero.
Explanation:
- When a material is subjected to external forces, it experiences stresses in different directions. These stresses can be decomposed into normal and shear stresses.
- The principal stresses are the maximum and minimum normal stresses experienced by a material at a particular point.
- The principal planes are the planes on which the principal stresses act. These planes are perpendicular to each other.
- The maximum principal stress is the highest normal stress experienced by the material, and it occurs on the principal plane with the highest normal stress value.
- The minimum principal stress is the lowest normal stress experienced by the material, and it occurs on the principal plane with the lowest normal stress value.
- Shear stress occurs when forces are applied parallel to a plane, causing one part of the material to slide over the adjacent part. It is perpendicular to the normal stress.
- Along the principal planes, the shear stress is zero. This is because the principal planes are oriented in such a way that the maximum and minimum normal stresses act purely in the normal direction and do not cause any shearing motion.
- The shear stress can be calculated using the formula: shear stress = (maximum principal stress - minimum principal stress) / 2. Since the minimum principal stress is zero on the principal plane subjected to the maximum principal stress, the shear stress is also zero.
In conclusion, the shear stress along the principal plane subjected to the maximum principal stress is zero.
The maximum shear due to an axial compression of 100 MPa is- a) 50 MPa
- b) 100 MPa
- c) 150 MPa
- d) 75 MPa
Correct answer is option 'A'. Can you explain this answer?
The maximum shear due to an axial compression of 100 MPa is
a)
50 MPa
b)
100 MPa
c)
150 MPa
d)
75 MPa
|
|
Avinash Sharma answered |
Maximum shear stress,
τ = 
= 
τ = 50 MPa
For which one of the following two-dimensional states of stress will be Mohr’s stress circle degenerate into a point?- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
For which one of the following two-dimensional states of stress will be Mohr’s stress circle degenerate into a point?
a)
b)
c)
d)
|
Pioneer Academy answered |
Mohr’s circle will be a point.
Radius of the Mohr's circle
= 
∴ τxy = 0 and σx = σy = σ
On an aluminium sample, strain readings were recorded on 3-element delta rosette as follows. ε0° = −100 μs, ε120° = −500 μs and ε60° = +600 μs Then the maximum tensile strain is __________ μs.(A) 589(B) 600
Correct answer is option ''. Can you explain this answer?
On an aluminium sample, strain readings were recorded on 3-element delta rosette as follows. ε0° = −100 μs, ε120° = −500 μs and ε60° = +600 μs Then the maximum tensile strain is __________ μs.
(A) 589
(B) 600
|
|
Lavanya Menon answered |
Ε0 = εxx = −100 μs ⋯ ①
= −83.33 μs
ε120° = −500 μs = 
−500 = −25 + 0.75 εyy − 0.433 γxy
or − 475 = 0.75 εxy − 0.433 γxy ⋯ ②
ε60° = 600 = 
= 25 + 0.75 εyy + 0.433 γxy
575 = 0.75 εyy + 0.433 γxy ⋯③
−475 = 0.75 εyy − 0.433 γxy ⋯④
100 = 1.5 εyy
or εyy = +66.67 μs
Putting the values in equation ③
575 = 0.75 × 66.67 + 0.433 γxy = 50 + 0.433 γxy
γxy = 1212.5 μs
Now
= −16.665 μs
= −16.665 μs = −83.33 μs= 
= √(−83.33)2 + (606.25)2
√6943.9 + 367539 = 611.95 μs
Principal strains εp1 = −16.665 + 611.95 = +595.285 μs
εp2 = −16.665 − 611.95 = −628.615 μs
Question_Type: 5
n a strained material, normal stresses on two mutually perpendicular planes are σx and σy (both alike) accompanied by a shear stress τxy, one of the principal stresses will be zero, only if,- a) τxy =
- b) τxy = √σx × σy
- c) τxy = σx × σy
- d) τxy =√σx2 × σy2
Correct answer is option 'B'. Can you explain this answer?
n a strained material, normal stresses on two mutually perpendicular planes are σx and σy (both alike) accompanied by a shear stress τxy, one of the principal stresses will be zero, only if,
a)
τxy =
b)
τxy = √σx × σy
c)
τxy = σx × σy
d)
τxy =√σx2 × σy2
|
Telecom Tuners answered |
σ1.2 = 
let σ2 = 0
Square both sides
τxy2 =
τxy2 = 
τxy2 =
τxy2 =
τxy = √σxσy
τxy = √σx × σy
The value of σ at failure according to strain energy theory- a) 100 MPa
- b) 75 MPa
- c) 90 MPa
- d) 150 MPa
Correct answer is option 'C'. Can you explain this answer?
The value of σ at failure according to strain energy theory
a)
100 MPa
b)
75 MPa
c)
90 MPa
d)
150 MPa
|
|
Zoya Sharma answered |
According to strain energy theory
σ12 + σ22 + σ32 − 2μ(σ1σ2 + σ2σ3 + σ3σ1) ≤ 
4.95 σ2 = (200)2
σ = 89.89 MPa ≈ 90 MPa
At the principal planes- a)the normal stress is maximum or minimum and the shear stress is zero
- b)the tensile and compressive stresses are zero
- c)the tensile stress is zero and the shear stress is maximum
- d)no stress acts
Correct answer is option 'A'. Can you explain this answer?
At the principal planes
a)
the normal stress is maximum or minimum and the shear stress is zero
b)
the tensile and compressive stresses are zero
c)
the tensile stress is zero and the shear stress is maximum
d)
no stress acts
|
Janhavi Datta answered |
Principal stresses are maximum or minimum normal stresses which may occur on a stressed body. In a 3-D body there may three principal planes which are mutually perpendicular to each other. The plane of principle stresses is called principal plane which always carries zero shear stress.
A spherical ball of volume 106 mm3 is subjected to a hydrostatic pressure of 90 MPa. If the bulk modulus for the material is 180 GPa, the change in the volume of the ball is- a) 50 mm3
- b) 250 mm3
- c) 100 mm3
- d) 500 mm3
Correct answer is option 'D'. Can you explain this answer?
A spherical ball of volume 106 mm3 is subjected to a hydrostatic pressure of 90 MPa. If the bulk modulus for the material is 180 GPa, the change in the volume of the ball is
a)
50 mm3
b)
250 mm3
c)
100 mm3
d)
500 mm3
|
|
Avinash Sharma answered |
Volume = 106 mm3 Hydrostatic pressure, P = 90 MPa
Bulk modulus, k = 180 kN/m3 + 2
k =
180 × 103 = 
δv = 500 mm3
If σ1 = 600N/mm2 (tensile) and σ2 = 400 N/mm2 (Compressive), the maximum shear stress is- a) 100 N/mm2
- b) 200 N/mm2
- c) 300 N/mm2
- d) 500 N/mm2
Correct answer is option 'D'. Can you explain this answer?
If σ1 = 600N/mm2 (tensile) and σ2 = 400 N/mm2 (Compressive), the maximum shear stress is
a)
100 N/mm2
b)
200 N/mm2
c)
300 N/mm2
d)
500 N/mm2
|
|
Zoya Sharma answered |
Σ1 = 600 N/mm2 σ2 = −400 N/mm2
τmax = 
= 500 N/mm2
Biaxial stress system is correctly shown in- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Biaxial stress system is correctly shown in
a)
b)
c)
d)
|
|
Sarita Yadav answered |
Complementary shear stresses are balanced Normal stress is balanced in x and y-direction.
Consider the below shown loading conditions. Which one will fail first according to distortion energy theory? [take, σ/τ = 2]I.
II. 
- a) I will fail first
- b) II will fail first
- c) Both I and II will fail at a time
- d) Insufficient data to decide.
Correct answer is option 'B'. Can you explain this answer?
Consider the below shown loading conditions. Which one will fail first according to distortion energy theory? [take, σ/τ = 2]
I.
II.
a)
I will fail first
b)
II will fail first
c)
Both I and II will fail at a time
d)
Insufficient data to decide.
|
|
Avinash Sharma answered |
According to distortion Energy Theory
Syt = 
Syt)I:
Syt =
=
=
= √3 × [4τ2 + 2τ2]
= √18 τ2
Syt)II:
=
= 
= 
= √7 × 4 × τ2 + 3τ2
= √31 τ2
As, Syt )II > Syt)I
II will fail first.
= 60 MPa A Plane stressed element is subjected to the state of stress given by σx = τxy = 100 kgf/cm2 and σy = 0. Maximum shear stress in the element is equal to- a) 50√3 kgf/cm2
- b) 100 kgf/cm2
- c) 50√5 kgf/cm2
- d) 150 kgf/cm2
Correct answer is option 'C'. Can you explain this answer?
A Plane stressed element is subjected to the state of stress given by σx = τxy = 100 kgf/cm2 and σy = 0. Maximum shear stress in the element is equal to
a)
50√3 kgf/cm2
b)
100 kgf/cm2
c)
50√5 kgf/cm2
d)
150 kgf/cm2
|
|
Neha Joshi answered |
Maximum shear stress,
τmax = 
= 
= √12, 500
= 50√5 kgf/cm2
Normal stresses of equal magnitude a but of opposite signs act at a point of a strained material in perpendicular direction. What would be the resultant normal stress on a plane inclined at 45° to the applied stresses?- a)2 σ
- b)0.5 σ
- c)0.25σ
- d)zero
Correct answer is option 'D'. Can you explain this answer?
Normal stresses of equal magnitude a but of opposite signs act at a point of a strained material in perpendicular direction. What would be the resultant normal stress on a plane inclined at 45° to the applied stresses?
a)
2 σ
b)
0.5 σ
c)
0.25σ
d)
zero
|
Anshu Patel answered |
If Normal stresses of same nature px and py and shear stress q is acting on two perpendicular planes and q = (pxpy)1/2, then the major and minor principal stresses respectively are- a) px + py and px − py
- b) px and px − py
- c) 0.5 (px + py)and 0.5 (px − py )
- d) px + py and zero
Correct answer is option 'D'. Can you explain this answer?
If Normal stresses of same nature px and py and shear stress q is acting on two perpendicular planes and q = (pxpy)1/2, then the major and minor principal stresses respectively are
a)
px + py and px − py
b)
px and px − py
c)
0.5 (px + py)and 0.5 (px − py )
d)
px + py and zero
|
Meera Bose answered |
Major and Minor Principal Stresses
The given question is asking for the major and minor principal stresses when normal stresses of the same nature (px and py) and a shear stress (q) are acting on two perpendicular planes. The relationship between these stresses is given as q = sqrt(px * py).
To determine the major and minor principal stresses, we need to understand the concept of principal stresses and how they relate to the given normal and shear stresses.
Principal stresses are the maximum and minimum values of normal stresses that act on a particular point in a material. They are represented by sigma1 (major principal stress) and sigma2 (minor principal stress).
To find the major and minor principal stresses, we can use the following equations:
sigma1 = (px + py)/2 + sqrt(((px-py)/2)^2 + q^2)
sigma2 = (px + py)/2 - sqrt(((px-py)/2)^2 + q^2)
Now, let's solve the given problem step by step:
Step 1: Given normal stresses:
px and py are normal stresses acting on two perpendicular planes.
Step 2: Shear stress:
q = sqrt(px * py)
Step 3: Major Principal Stress (sigma1):
sigma1 = (px + py)/2 + sqrt(((px-py)/2)^2 + q^2)
= (px + py)/2 + sqrt((px^2 - 2pxpy + py^2)/4 + pxpy)
= (px + py)/2 + sqrt((px^2 + 2pxpy + py^2)/4)
= (px + py)/2 + sqrt((px + py)^2)/2
= (px + py)/2 + (px + py)/2
= px + py
Therefore, the major principal stress (sigma1) is equal to px + py.
Step 4: Minor Principal Stress (sigma2):
sigma2 = (px + py)/2 - sqrt(((px-py)/2)^2 + q^2)
= (px + py)/2 - sqrt((px^2 - 2pxpy + py^2)/4 + pxpy)
= (px + py)/2 - sqrt((px^2 + 2pxpy + py^2)/4)
= (px + py)/2 - sqrt((px + py)^2)/2
= (px + py)/2 - (px + py)/2
= 0
Therefore, the minor principal stress (sigma2) is equal to zero.
Hence, the major and minor principal stresses are px + py and zero respectively, which corresponds to option D.
The given question is asking for the major and minor principal stresses when normal stresses of the same nature (px and py) and a shear stress (q) are acting on two perpendicular planes. The relationship between these stresses is given as q = sqrt(px * py).
To determine the major and minor principal stresses, we need to understand the concept of principal stresses and how they relate to the given normal and shear stresses.
Principal stresses are the maximum and minimum values of normal stresses that act on a particular point in a material. They are represented by sigma1 (major principal stress) and sigma2 (minor principal stress).
To find the major and minor principal stresses, we can use the following equations:
sigma1 = (px + py)/2 + sqrt(((px-py)/2)^2 + q^2)
sigma2 = (px + py)/2 - sqrt(((px-py)/2)^2 + q^2)
Now, let's solve the given problem step by step:
Step 1: Given normal stresses:
px and py are normal stresses acting on two perpendicular planes.
Step 2: Shear stress:
q = sqrt(px * py)
Step 3: Major Principal Stress (sigma1):
sigma1 = (px + py)/2 + sqrt(((px-py)/2)^2 + q^2)
= (px + py)/2 + sqrt((px^2 - 2pxpy + py^2)/4 + pxpy)
= (px + py)/2 + sqrt((px^2 + 2pxpy + py^2)/4)
= (px + py)/2 + sqrt((px + py)^2)/2
= (px + py)/2 + (px + py)/2
= px + py
Therefore, the major principal stress (sigma1) is equal to px + py.
Step 4: Minor Principal Stress (sigma2):
sigma2 = (px + py)/2 - sqrt(((px-py)/2)^2 + q^2)
= (px + py)/2 - sqrt((px^2 - 2pxpy + py^2)/4 + pxpy)
= (px + py)/2 - sqrt((px^2 + 2pxpy + py^2)/4)
= (px + py)/2 - sqrt((px + py)^2)/2
= (px + py)/2 - (px + py)/2
= 0
Therefore, the minor principal stress (sigma2) is equal to zero.
Hence, the major and minor principal stresses are px + py and zero respectively, which corresponds to option D.
At a point in a stressed material, the stress system is σx = + 800N/mm2 , σy = 0 and q = 0.The radius of the Mohr’s stress circle for this case is__________ units.- a) 800
- b) 600
- c) 400
- d) 200
Correct answer is option 'C'. Can you explain this answer?
At a point in a stressed material, the stress system is σx = + 800N/mm2 , σy = 0 and q = 0.
The radius of the Mohr’s stress circle for this case is__________ units.
a)
800
b)
600
c)
400
d)
200
|
|
Sanvi Kapoor answered |
Σx = 800 N/mm2, σy = 0, q = τxy = 0
Radius of Mohr’s circle = τmax
τmax = 
=400 N/mm2
=400 N/mm2The principal stresses at a point in an elastic material are 60 N/mm2 tensile, 20 N/mm2 tensile and 50 N/mm2 compressive. If the material properties are: μ = 0.35 and E = 105 N/mm2 then the volumetric strain of the material is- a)9 x 10-5
- b)3 x 10-4
- c)10.5 x 10-5
- d)21 x 10-5
Correct answer is option 'A'. Can you explain this answer?
The principal stresses at a point in an elastic material are 60 N/mm2 tensile, 20 N/mm2 tensile and 50 N/mm2 compressive. If the material properties are: μ = 0.35 and E = 105 N/mm2 then the volumetric strain of the material is
a)
9 x 10-5
b)
3 x 10-4
c)
10.5 x 10-5
d)
21 x 10-5
|
Debolina Chavan answered |
Volumetric strain,
A circle of diameter 500 mm is inscribed on a square plate of copper, before the application of stresses as shown, σ1 = 160 MPa, σ2 = 80 MPa, τ = 30 MPa. After the application of stresses, the circle is deformed into an ellipse. Then the major and minor axes of the ellipse will be Given E = 100 GPa, ν = 0.35.
- a) Major = 500.73 mm, Minor = 500.05 mm
- b) Major = 480.61 mm, Minor = 420.21 mm
- c) Major = 250.13 mm, Minor = 230.31 mm
- d) Major = 460.51 mm, Minor = 448.31 mm
Correct answer is option 'A'. Can you explain this answer?
A circle of diameter 500 mm is inscribed on a square plate of copper, before the application of stresses as shown, σ1 = 160 MPa, σ2 = 80 MPa, τ = 30 MPa. After the application of stresses, the circle is deformed into an ellipse. Then the major and minor axes of the ellipse will be Given E = 100 GPa, ν = 0.35.
a)
Major = 500.73 mm, Minor = 500.05 mm
b)
Major = 480.61 mm, Minor = 420.21 mm
c)
Major = 250.13 mm, Minor = 230.31 mm
d)
Major = 460.51 mm, Minor = 448.31 mm
|
|
Cstoppers Instructors answered |
= 120 MPa
= 40 MPa
where τ = 30 MPa
= √402 + 302 = 50 MPa
Principal stresses, p1 = 120 + 50 = 170 MPa p2 = 120 − 50 = 70 MPa
Principal strains,
e1 = 
= 
= 1455 × 10−6 = 1455 μ strain
e2 = 
= 105 × 10−6 = 105 μ strain
Major axis = 500 + 1455 × 10−6 × 500
= 500 + 0.7275 = 500.7275 mm
Minor axis = 500 + 105 × 10−6 × 500
= 500 + 0.0525 = 500.0525 mm
Principal angles, θ1 = 
= 
θ1 = −18.44 ° as shear stress on reference plane is positive
If a body carries two unlike principal stresses, what is the maximum shear stress?- a)Half the difference of magnitude of the principal stresses
- b)Half the sum of the magnitude of principal stresses
- c)Difference of the magnitude of principal stresses
- d)Sum of the magnitude of principal stresses
Correct answer is option 'B'. Can you explain this answer?
If a body carries two unlike principal stresses, what is the maximum shear stress?
a)
Half the difference of magnitude of the principal stresses
b)
Half the sum of the magnitude of principal stresses
c)
Difference of the magnitude of principal stresses
d)
Sum of the magnitude of principal stresses
|
|
Poulomi Khanna answered |
The principal stress given are unlike i.e if one is tensile and other is compressive. Let σ1 be tensile and σ2 be compressive, then
In a strained material the normal stress on one plane is 10t/cm2 tensile and the shear stress is 5t/cm2. On a plane perpendicular to this plane there is only shear stress. The center for Mohr’s circle- a) Coincides with the origin
- b) Is at a distance representing 10 t/mm2 on the positive side of the axis representing tensile stress
- c) Is at a distance representing 5 t/cm2 on the positive side of the axis representing tensile stress
- d) At a distance representing 5√2 t/cm2 on the positive side of the axis representing tensile stress
Correct answer is option 'C'. Can you explain this answer?
In a strained material the normal stress on one plane is 10t/cm2 tensile and the shear stress is 5t/cm2. On a plane perpendicular to this plane there is only shear stress. The center for Mohr’s circle
a)
Coincides with the origin
b)
Is at a distance representing 10 t/mm2 on the positive side of the axis representing tensile stress
c)
Is at a distance representing 5 t/cm2 on the positive side of the axis representing tensile stress
d)
At a distance representing 5√2 t/cm2 on the positive side of the axis representing tensile stress
|
|
Avinash Sharma answered |
Centre of Mohr’s circle
σ1 = 
= 
= 12.07 t/cm2
σ2 = 5 − √52 + 52 = −2.07 t/cm2
Centre of Mohr′s circle =
= 
= 5 t/cm2
The radius of Mohr’s circle for two unlike principal stresses of magnitude σ is- a) σ/2
- b) σ
- c) σ/4
- d) Zero
Correct answer is option 'B'. Can you explain this answer?
The radius of Mohr’s circle for two unlike principal stresses of magnitude σ is
a)
σ/2
b)
σ
c)
σ/4
d)
Zero
|
|
Neha Joshi answered |
Radius of Mohr’s circle, τmax = 
τmax = σ
The data obtained from a rectangular strain gage rosette attached to a stressed steel member are ε0 = −220 × 10−6, ε45° = 120 × 10−6, ε90° = +220 × 10−6.Given E = 2 × 105 N⁄mm2 and Poisson’s ratio = 0.3. The principal stress acting at the point is ___________ MPa(A) 37.5(B) 39.5
Correct answer is option ''. Can you explain this answer?
The data obtained from a rectangular strain gage rosette attached to a stressed steel member are ε0 = −220 × 10−6, ε45° = 120 × 10−6, ε90° = +220 × 10−6.Given E = 2 × 105 N⁄mm2 and Poisson’s ratio = 0.3. The principal stress acting at the point is ___________ MPa
(A) 37.5
(B) 39.5
|
|
Neha Joshi answered |
Εxx = ε0 = −220 μs
εyy = ε90o = +220 μs
Principal strain,
ε1 , ε2 = 
ε1, ε2 = 0 ± √(−220)2 + (120)2
= ±250.6 μs
p1 = 
= 
= 38.55 N/mm2
p2 = −38.55 N/mm2
Question_Type: 5
For an inclined plane in a rectangular block subjected to two mutually perpendicular normal stresses 1000 MPa and 400 MPa and shear stress 400 MPa, the maximum normal stress will be- a) 1200 MPa
- b) 700 MPa
- c) 600 MPa
- d) 200 MPa
Correct answer is option 'A'. Can you explain this answer?
For an inclined plane in a rectangular block subjected to two mutually perpendicular normal stresses 1000 MPa and 400 MPa and shear stress 400 MPa, the maximum normal stress will be
a)
1200 MPa
b)
700 MPa
c)
600 MPa
d)
200 MPa
|
|
Lavanya Menon answered |
σ1 = 
= 700 + √9 × 104 + 16 × 104
= 700 + 500 = 1200MPa
Consider the below shown loading conditions, which one will fail first according to principal strain energy theory [μ = 0.3]I.
II.
- a) I will fail first
- b) II will fail first
- c) Both I and II will fail at a time
- d) Insufficient data to decide
Correct answer is option 'B'. Can you explain this answer?
Consider the below shown loading conditions, which one will fail first according to principal strain energy theory [μ = 0.3]
I.
II.
a)
I will fail first
b)
II will fail first
c)
Both I and II will fail at a time
d)
Insufficient data to decide
|
|
Lavanya Menon answered |
Σx = σ = σ1 similarly, σ2 = 2σ
σ3 = 3σ
Syt = 
Syt )I =
= 
= 
= √7.4 σ2
∵ [σ1 = 2σ, σ2 = σ, σ3 = 4σ]
Syt )II =
= 
= 
=√12.6σ2
Syt )II > Syt )I
For which one of the following stress conditions the diameter of Mohr’s stress circle is zero?Where σ1 =major principal stress,σ2 = Minor principal stress andq = Shear stress- a) σ1 = +400 N/mm2 ; σ2 = +400 N/mm2 ; q = 400 N/mm2
- b) σ1 = +400 N/mm2 ; σ2 = −400 N/mm2 ; q = 400 N/mm2
- c) σ1 = +400 N/mm2 ; σ2 = +400 N/mm2 ; q = zero
- d) σ1 = −400 N/mm2 ; σ2 = + 400N/mm2 ; q = zero
Correct answer is option 'C'. Can you explain this answer?
For which one of the following stress conditions the diameter of Mohr’s stress circle is zero?
Where σ1 =major principal stress,
σ2 = Minor principal stress and
q = Shear stress
a)
σ1 = +400 N/mm2 ; σ2 = +400 N/mm2 ; q = 400 N/mm2
b)
σ1 = +400 N/mm2 ; σ2 = −400 N/mm2 ; q = 400 N/mm2
c)
σ1 = +400 N/mm2 ; σ2 = +400 N/mm2 ; q = zero
d)
σ1 = −400 N/mm2 ; σ2 = + 400N/mm2 ; q = zero
|
|
Pioneer Academy answered |
Diameter of Mohr’s circle
2(τmax) = 
Option (A):
2(τmax) = 
= 800 N/mm2
Option (B):
2(τmax) = 
= 1,131.37 N/mm2
Option (C):
2(τmax) = 
Option (D):
2(τmax) = 
= 800 N/mm2
A body is subjected to a tensile stress of 1200 MPa on one plane and another tensile stress of 600 MPa on a plane at right angles to the former. If is also subjected to shear stress of 400 MPa on the same planes. The maximum shear stress will be- a) 1400 MPa
- b) 400 MPa
- c) 900 MPa
- d) 500 MPa
Correct answer is option 'D'. Can you explain this answer?
A body is subjected to a tensile stress of 1200 MPa on one plane and another tensile stress of 600 MPa on a plane at right angles to the former. If is also subjected to shear stress of 400 MPa on the same planes. The maximum shear stress will be
a)
1400 MPa
b)
400 MPa
c)
900 MPa
d)
500 MPa
|
|
Lavanya Menon answered |
maximum Shear Stress , τmax
τmax = 
=
= 500 MPa
A square plate of side 4 units experiences the following strain components.ϵx = 0.001ϵy = 0.002τxy = 0.003Find the elongation of the diagonal (4 decimal places)(A) 0.0169(B) 0.0169
Correct answer is option ''. Can you explain this answer?
A square plate of side 4 units experiences the following strain components.
ϵx = 0.001
ϵy = 0.002
τxy = 0.003
Find the elongation of the diagonal (4 decimal places)
(A) 0.0169
(B) 0.0169
|
|
Lavanya Menon answered |
ϵ45° =?
As, ϵθ = 
ΔL = (0.003) × 4√2
ΔL = 0.0169 units
Question_Type: 5
For drawing Mohr’s circle, the required stresses are- a) two mutually perpendicular stress
- b) two shear stresses
- c) one normal stress one shear stress
- d) two stresses in two inclined planes
Correct answer is option 'A'. Can you explain this answer?
For drawing Mohr’s circle, the required stresses are
a)
two mutually perpendicular stress
b)
two shear stresses
c)
one normal stress one shear stress
d)
two stresses in two inclined planes
|
Paranthaman Sagadevan answered |
Actually to draw a mohr's circle we need to principal stresses . two mutually perpendicular stress means to principal stresses hence option A is correct
Identify the WRONG statement:- a)Principal plane is a plane on which shear stress is zero.
- b)The normal stress acting normal to the principal plane is called the principal stress.
- c)On the planes, other than principal planes, the normal stress is accompanied by the tangential stresses.
- d)Two principal planes carrying the maximum and minimum direct stresses are always collinear.
Correct answer is option 'D'. Can you explain this answer?
Identify the WRONG statement:
a)
Principal plane is a plane on which shear stress is zero.
b)
The normal stress acting normal to the principal plane is called the principal stress.
c)
On the planes, other than principal planes, the normal stress is accompanied by the tangential stresses.
d)
Two principal planes carrying the maximum and minimum direct stresses are always collinear.
|
Sagnik Choudhary answered |
Two principal planes carrying the maximum and direct stresses are always perpendicular to each other.
The value of σ at failure according to maximum principal stress theory- a) 100 MPa
- b) 250 MPa
- c) 150 MPa
- d) 50 MPa
Correct answer is option 'A'. Can you explain this answer?
The value of σ at failure according to maximum principal stress theory
a)
100 MPa
b)
250 MPa
c)
150 MPa
d)
50 MPa
|
Athul Kumar answered |
Understanding Principal Stresses
The problem involves calculating shearing and normal stresses based on given principal stresses. The principal stresses are 250 MPa (tensile) and 150 MPa (compressive).
Stress Components on a Plane
1. Normal Stress (σ₁):
- Given as 200 MPa (tensile).
2. Finding Normal Stress on the Orthogonal Plane:
- The orthogonal plane will experience a different normal stress, denoted as σ₂.
- To find σ₂, we can utilize the relationship between normal and shear stresses.
Calculating Shear Stress
1. Using the Mohr's Circle Concept:
- The maximum shear stress (τ_max) is calculated using the formula:
τ_max = (σ₁ - σ₂) / 2.
2. Applying the Principal Stresses:
- With σ₁ = 250 MPa (tensile) and σ₂ = -150 MPa (compressive), we find:
- The stress on the plane at 200 MPa tensile (σ₁ = 200) implies we need to calculate the orthogonal stress:
- σ₂ = 50 MPa (compressive).
Final Results
1. Shearing Stress Calculation:
- τ = (200 - 50) / 2 = 75 MPa (shear).
2. Normal Stress on the Orthogonal Plane:
- The normal stress is σ₂ = 100 MPa (compressive).
3. Conclusion:
- The shearing stress is 50√7 MPa, and the normal stress on the orthogonal plane is 100 MPa (compressive).
Thus, the correct answer is option 'C': 50√7 MPa and 100 MPa (compressive).
The problem involves calculating shearing and normal stresses based on given principal stresses. The principal stresses are 250 MPa (tensile) and 150 MPa (compressive).
Stress Components on a Plane
1. Normal Stress (σ₁):
- Given as 200 MPa (tensile).
2. Finding Normal Stress on the Orthogonal Plane:
- The orthogonal plane will experience a different normal stress, denoted as σ₂.
- To find σ₂, we can utilize the relationship between normal and shear stresses.
Calculating Shear Stress
1. Using the Mohr's Circle Concept:
- The maximum shear stress (τ_max) is calculated using the formula:
τ_max = (σ₁ - σ₂) / 2.
2. Applying the Principal Stresses:
- With σ₁ = 250 MPa (tensile) and σ₂ = -150 MPa (compressive), we find:
- The stress on the plane at 200 MPa tensile (σ₁ = 200) implies we need to calculate the orthogonal stress:
- σ₂ = 50 MPa (compressive).
Final Results
1. Shearing Stress Calculation:
- τ = (200 - 50) / 2 = 75 MPa (shear).
2. Normal Stress on the Orthogonal Plane:
- The normal stress is σ₂ = 100 MPa (compressive).
3. Conclusion:
- The shearing stress is 50√7 MPa, and the normal stress on the orthogonal plane is 100 MPa (compressive).
Thus, the correct answer is option 'C': 50√7 MPa and 100 MPa (compressive).
A solid cube is subjected to identical normal forces on all the faces. The ratio of volumetric strain to linear strain in any of the three axes is- a) 1
- b) 2
- c) 3
- d) √3
Correct answer is option 'C'. Can you explain this answer?
A solid cube is subjected to identical normal forces on all the faces. The ratio of volumetric strain to linear strain in any of the three axes is
a)
1
b)
2
c)
3
d)
√3
|
|
Neha Joshi answered |
For a solid cube and identical normal stresses εx = εy = εz and εv = εx + εy + εz = 3εx
The extremities of any diameter on Mohr’s circle represent- a) principal stresses
- b) normal stresses on planes at 45°
- c) shear stresses on planes at 45°
- d) None of the above
Correct answer is option 'D'. Can you explain this answer?
The extremities of any diameter on Mohr’s circle represent
a)
principal stresses
b)
normal stresses on planes at 45°
c)
shear stresses on planes at 45°
d)
None of the above
|
Mira Sharma answered |
Any diameter on Mohr’s circle represents states of stress on any pair of mutually perpendicular planes.
Following stress Tensor represents tri-axial state of stress at a point. Determine the distortion energy per unit volume in terms of kJ/m3 . Take μ = 0.3, E = 200 GPa[σ] = 
(A) 1.682(B) 1.682
Correct answer is option ''. Can you explain this answer?
Following stress Tensor represents tri-axial state of stress at a point. Determine the distortion energy per unit volume in terms of kJ/m3 . Take μ = 0.3, E = 200 GPa
[σ] =
(A) 1.682
(B) 1.682
|
|
Cstoppers Instructors answered |
Given,
σx = 10 MPa
σy = 20 MPa
σz = −10 MPa
τxy = 5 MPa
τxz = τyz = 0
σ1,2 =
= 
= 15 ± √52 + 52
σ1 = 22.07 MPa
σ2 = 3.82 MPa
σ3 = σ2 = −10 MPa [∵ τyz = τxz = 0]
Distortion energy per unit volume
= 
= 
+ +(3.82 + 10)2 + (−10 − 22.07)2]
+ +(3.82 + 10)2 + (−10 − 22.07)2]= 1.682 × 10−3 mJ/m3
= 1.682 kJ/m3
Question_Type: 5
For which of the following cases distortion energy theory is more liberal when compared to shear stress theory.I. σ1 , σ2 are equal and are tensile in nature.II. σ1 , σ2 are equal and are compressive in nature.III. σ1 is compressive and σ2 is tensile and the magnitudes are the same.IV. σ1 is tensile and σ2 is compressive and the magnitudes are same- a) I and II
- b) III and IV
- c) II and III
- d) I and IV
Correct answer is option 'B'. Can you explain this answer?
For which of the following cases distortion energy theory is more liberal when compared to shear stress theory.
I. σ1 , σ2 are equal and are tensile in nature.
II. σ1 , σ2 are equal and are compressive in nature.
III. σ1 is compressive and σ2 is tensile and the magnitudes are the same.
IV. σ1 is tensile and σ2 is compressive and the magnitudes are same
a)
I and II
b)
III and IV
c)
II and III
d)
I and IV
|
|
Zoya Sharma answered |
For I, II both the theories will give the same result. For III, IV distortion theory is more liberal than shear stress theory.
Chapter doubts & questions for Principal Stresses & Strains (Mohr`s Circle) - Strength of Materials (SOM) 2025 is part of Mechanical Engineering exam preparation. The chapters have been prepared according to the Mechanical Engineering exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Mechanical Engineering 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
Chapter doubts & questions of Principal Stresses & Strains (Mohr`s Circle) - Strength of Materials (SOM) in English & Hindi are available as part of Mechanical Engineering exam.
Download more important topics, notes, lectures and mock test series for Mechanical Engineering Exam by signing up for free.
Strength of Materials (SOM)
37 videos|41 docs|47 tests
|
Signup to see your scores go up within 7 days!
Study with 1000+ FREE Docs, Videos & Tests
10M+ students study on EduRev
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup