Question 1 |

A real square matrix A is called skew-symmetric if

A^{T}=A | |

A^{T}=A^{-1} | |

A^{T}=-A | |

A^{T}=A+A^{-1} |

Question 1 Explanation:

A is skew-symmetric

\therefore \quad A^{T}=-A

\therefore \quad A^{T}=-A

Question 2 |

\lim_{x\rightarrow 0}\frac{log_{e}(1+4x)}{e^{3x}-1} is equal to

0 | |

1/12 | |

4/3 | |

1 |

Question 2 Explanation:

\lim_{x \rightarrow 0} \frac{\ln (1+4 x)}{e^{3 x}-1}

\lim_{x \rightarrow 0} \frac{\frac{1}{1+4 x} \cdot 4}{3 e^{3 x}}=\frac{4}{3}

\lim_{x \rightarrow 0} \frac{\frac{1}{1+4 x} \cdot 4}{3 e^{3 x}}=\frac{4}{3}

Question 3 |

Solutions of Laplace's equation having continuous second-order partial derivatives are called

biharmonic functions | |

harmonic functions | |

conjugate harmonic functions | |

error functions |

Question 3 Explanation:

Solution of laplace equation having continuous

Second order partial derivating

\therefore \nabla^{2} \phi =0

\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{\partial^{2} \phi}{\partial y^{2}} =0

\therefore \; \phi is harmonic function.

Second order partial derivating

\therefore \nabla^{2} \phi =0

\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{\partial^{2} \phi}{\partial y^{2}} =0

\therefore \; \phi is harmonic function.

Question 4 |

The area (in percentage) under standard normal distribution curve of random variable Z within limits from -3 to +3 is __________

55.2 | |

88.6 | |

99.8 | |

44.6 |

Question 4 Explanation:

Question 5 |

The root of the function f(x)=x^{3}+x-1 obtained after first iteration on application of NewtonRaphson scheme using an initial guess of x_{0}=1 is

0.682 | |

0.686 | |

0.75 | |

1 |

Question 5 Explanation:

\begin{aligned} f(x)&=x^3+x-1\\ f(1)&=1+1-1=1 \\ f'(x)&=3x^2+1\\ f'(1)&=3+1=4\\ x_1&=x_0-\frac{f(x_0)}{f'(x_0)}\\ &=1-\frac{1}{4}=1-0.25=0.75 \end{aligned}

Question 6 |

A force F is acting on a bent bar which is clamped at one end as shown in the figure.

The CORRECT free body diagram is

The CORRECT free body diagram is

A | |

B | |

C | |

D |

Question 6 Explanation:

When we draw a free body diagram we remove all the supports and force applied due to that support are drawn and force or moment will apply in that manner so that it resists forces in any direction as well as any tendency of rotation.

Option (B) is wrong because the ground should not be shown in FBD as the forces due to ground are already depicted.

Option (C) is wrong because x-component is not shown.

Option (D) is wrong because the rotation due to the force acting eccentrically causes moment which is not shown.

Option (B) is wrong because the ground should not be shown in FBD as the forces due to ground are already depicted.

Option (C) is wrong because x-component is not shown.

Option (D) is wrong because the rotation due to the force acting eccentrically causes moment which is not shown.

Question 7 |

The cross-sections of two solid bars made of the same material are shown in the figure. The square cross-section has flexural (bending) rigidity I1, while the circular cross-section has flexural rigidity I2. Both sections have the same cross-sectional area. The ratio I1/I2 is

1/\pi | |

2/\pi | |

\pi/3 | |

\pi/6 |

Question 7 Explanation:

\begin{aligned} \therefore \quad b^{2}&=\frac{\pi}{4} d^{2} \\ \Rightarrow \quad \frac{b^{2}}{d^{2}}&=\frac{\pi}{4} \\ I_{1} &=\frac{b^{4}}{12} \\ I_{2} &=\frac{\pi d^{4}}{64} \\ \therefore \qquad \frac{I_{1}}{I_{2}} &=\frac{b^{4}}{12} \times \frac{64}{\pi d^{4}}=\frac{16}{3 \pi} \frac{b^{4}}{d^{4}} \\ &=\frac{16}{3 \pi} \cdot \frac{\pi^{2}}{16}=\frac{\pi}{3} \end{aligned}

Question 8 |

The state of stress at a point on an element is shown in figure (a). The same state of stress is shown in another coordinate system in figure (b).

The components (\tau _{xx},\tau _{yy},\tau _{xy}) are given by

The components (\tau _{xx},\tau _{yy},\tau _{xy}) are given by

(p/\sqrt{2},-p/\sqrt{2},0 ) | |

(0,0,p ) | |

(p,-p,p/\sqrt{2} ) | |

(0,0,p/\sqrt{2} ) |

Question 8 Explanation:

The given plane is principal plane

\left(\sigma_{1}=P, \sigma_{2}=-P\right) . At 45^{\circ} from principal plane, plane of max shear occurs. On the plane of max shear.

\begin{aligned} \operatorname{both}, \quad \tau_{x y} &=\tau_{y y}=\frac{\sigma_{1}+\sigma_{2}}{2}=0 \\ \tau_{x y} &=\frac{\sigma_{1}+\sigma_{2}}{2}=P \end{aligned}

\left(\sigma_{1}=P, \sigma_{2}=-P\right) . At 45^{\circ} from principal plane, plane of max shear occurs. On the plane of max shear.

\begin{aligned} \operatorname{both}, \quad \tau_{x y} &=\tau_{y y}=\frac{\sigma_{1}+\sigma_{2}}{2}=0 \\ \tau_{x y} &=\frac{\sigma_{1}+\sigma_{2}}{2}=P \end{aligned}

Question 9 |

A rigid link PQ is undergoing plane motion as shown in the figure (V_{P} and V_{Q} are non-zero). V_{PQ} is the relative velocity of point Q with respect to point P.

Which one of the following is TRUE?

Which one of the following is TRUE?

V_{PQ} has components along and perpendicular to PQ | |

V_{PQ} has only one component directed from P to Q | |

V_{PQ} has only one component directed from Q to P | |

V_{PQ} has only one component perpendicular to PQ |

Question 9 Explanation:

To find relative velocity direction of V_{p} is reversed

and placed at fail of \vec{V}_{Q}

Ans. (D) Resultant V_{PQ} is perpendicular to link P Q.

Question 10 |

The number of degrees of freedom in a planar mechanism having n links and j simple hinge joints is

3(n-3)-2j | |

3(n-1)-2j | |

3n-2j | |

2j-3n+4 |

Question 10 Explanation:

Degrees of freedom is given by 3(n-1) - 2j

There are 10 questions to complete.