Question 1 |

Match the items in columns I and II.

P-1 Q-4 R-3 S-2 | |

P-1 Q-4 R-2 S-3 | |

P-1 Q-3 R-2 S-4 | |

P-4 Q-1 R-2 S-3 |

Question 1 Explanation:

(P) Gauss- Seidal method \rightarrow Linear algebraic equation

(Q) Forward Newton \rightarrow Gauss method Interpolation

(R) Runge-Kutta method \rightarrow Non-linear differential equations

(S) Trapezoidal Aule \rightarrow Numerical integration

(Q) Forward Newton \rightarrow Gauss method Interpolation

(R) Runge-Kutta method \rightarrow Non-linear differential equations

(S) Trapezoidal Aule \rightarrow Numerical integration

Question 2 |

The solution of the differential equation \frac{\mathrm{d} y}{\mathrm{d} x}+ 2xy = e^{-x^{2}}
with y(0)=1
is:

(1+x) e^{+x^{2}} | |

(1+x) e^{-x^{2}} | |

(1-x) e^{+x^{2}} | |

(1-x) e^{-x^{2}} |

Question 2 Explanation:

Given equation

\frac{d y}{d x}+2 x y=\sigma^{x^{2}}

Integrating factor

\begin{aligned} \text { I.F. } &=\mathrm{e}^{\int 2 x \;dx}=e^{x^{2}} \\ \text { Solution is } y x^{e^{x^{2}}}&=\int e^{-x^{2}} \cdot e^{x^{2}} d x \\ \therefore \quad y x^{ e^{x^{2}}}&=x+c\\ d t x &=0 \\ y &=1\\ \therefore \; \; c&=1 \\ y&=(1+x) e^{-x^{2}} \end{aligned}

\frac{d y}{d x}+2 x y=\sigma^{x^{2}}

Integrating factor

\begin{aligned} \text { I.F. } &=\mathrm{e}^{\int 2 x \;dx}=e^{x^{2}} \\ \text { Solution is } y x^{e^{x^{2}}}&=\int e^{-x^{2}} \cdot e^{x^{2}} d x \\ \therefore \quad y x^{ e^{x^{2}}}&=x+c\\ d t x &=0 \\ y &=1\\ \therefore \; \; c&=1 \\ y&=(1+x) e^{-x^{2}} \end{aligned}

Question 3 |

Let x denote a real number. Find out the INCORRECT statement.

S = \left \{ x : x \gt 3 \right \}
represents the set of all real numbers greater than 3 | |

S=\left \{ x : x^{2} \lt 0 \right \}
represents the empty set | |

S=\left \{ x : x \in A \; and \; x \in B \right \}
represents the union of set A and set B | |

S = \left \{ x : a \lt x \lt b \right \}
represents the set of all real numbers between a and b, where a and b are real numbers. |

Question 3 Explanation:

The incorrect statement is, S=\left \{ x : x \in A \; and \; x \in B \right \} represents the union of set A and set B

As and represents the intersection not union.

As and represents the intersection not union.

Question 4 |

A box contains 20 defective items and 80 non-defective items. If two items are selected at random without replacement, what will be the probability that both items are defective.?

\frac{1}{5}
| |

\frac{1}{25} | |

\frac{20}{99} | |

\frac{19}{495} |

Question 4 Explanation:

Probability of first item being defective is

P_{1}=\frac{20}{100}

Probability that second item being defective is

P_{2}=\frac{19}{99}

Probability that both are defective

P=P_{1} P_{2}=\frac{20}{100} \times \frac{19}{99}=\frac{19}{495}

P_{1}=\frac{20}{100}

Probability that second item being defective is

P_{2}=\frac{19}{99}

Probability that both are defective

P=P_{1} P_{2}=\frac{20}{100} \times \frac{19}{99}=\frac{19}{495}

Question 5 |

For a circular shaft of diameter d subjected to torque T, the maximum value of the shear stress is:

\frac{64T}{\pi d^{3}} | |

\frac{32T}{\pi d^{3}} | |

\frac{16T}{\pi d^{3}} | |

\frac{8T}{\pi d^{3}} |

Question 5 Explanation:

\begin{aligned} \frac{T}{J} &=\frac{\tau}{d / 2} \\ \Rightarrow \qquad \frac{T}{\frac{\pi}{32} d^{4}} &=\frac{\tau}{d / 2} \\ \therefore \qquad & \tau=\frac{16 T}{\pi d^{3}} \end{aligned}

Question 6 |

For a four-bar linkage in toggle position, the value of mechanical advantage is:

0 | |

0.5 | |

1 | |

\infty |

Question 6 Explanation:

\omega_{4} of the output link DC becomes zero at the extreme positions. The extreme positions of the linkage are known as "Toggle position".

\therefore Mechanical advantage =\frac{\omega_{\text {input }}}{\omega_{\text {output }}}

\because \quad \omega_{\text {output }}=0

\therefore Mechanical advantage =\infty

Question 7 |

The differential equation governing the vibrating system is:

m\ddot{x} + c\dot{x} + k(x-y) = 0 | |

m(\ddot{x}-\ddot{y}) + c(\dot{x}-\dot{y}) + kx = 0 | |

m \ddot{x} + c(\dot{x}-\dot{y}) + kx = 0 | |

m(\ddot{x}-\ddot{y}) + c(\dot{x}-\dot{y}) + k(x-y) = 0 |

Question 7 Explanation:

Differential equation governing the above vibration

system is given by

\begin{array}{l} \Rightarrow \frac{m d^{2} x}{d t^{2}}+C\left(\frac{d x}{d t}-\frac{d y}{d t}\right)+k x=0 \\ \Rightarrow m \ddot{x}+c(\dot{x}-\dot{y})+k x=0 \end{array}

Question 8 |

A pin-ended column of length L, modulus of elasticity E and second moment of the cross-sectional area I is loaded centrically by a compressive load P. the critical buckling load (P_{cr})
is given by

P_{cr}= \frac{EI}{\pi ^{2}L^{2}} | |

P_{cr}= \frac{\pi ^{2} EI}{3L^{2}} | |

P_{cr}= \frac{\pi EI}{L^{2}} | |

P_{cr}= \frac{\pi^{2} EI}{L^{2}} |

Question 8 Explanation:

P_{E}=\frac{n \pi^{2} E I}{L^{2}}

For pin-ended column

n=1

Here n= End fixity coefficient

E = Modulus of elasticity

I = Second moment of area

L = Actual length of column

For pin-ended column

n=1

Here n= End fixity coefficient

E = Modulus of elasticity

I = Second moment of area

L = Actual length of column

Question 9 |

The number of inversions for a slider crank mechanism is:

6 | |

5 | |

4 | |

3 |

Question 9 Explanation:

No. of links of a slider crank mechanism =4

So there are four inversion of slider crank mechanism.

So there are four inversion of slider crank mechanism.

Question 10 |

For a Newtonian fluid

shear stress is proportional to shear strain | |

rate of shear stress is proportional to shear strain | |

shear stress is proportional to rate of shear strain | |

rate of shear stress is proportional to rate of shear strain |

Question 10 Explanation:

For a Newtonian fluid

\tau \propto \frac{d u}{d y}

\tau=\mu \frac{d u}{d y}

where \quad \tau= shear stress

\frac{d u}{d y}= rate of shear strain

\tau \propto \frac{d u}{d y}

\tau=\mu \frac{d u}{d y}

where \quad \tau= shear stress

\frac{d u}{d y}= rate of shear strain

There are 10 questions to complete.