Question 1 |

\lim_{x\rightarrow 0}\frac{\sin^{2}x}{x}
is equal to

0 | |

\infty | |

1 | |

-1 |

Question 1 Explanation:

\lim_{x \to 0}frac{\sin^2 x}{x}

=\lim_{x \to 0}\left (frac{\sin x}{x} \right )^2 x

=1 \times 0 =0

=\lim_{x \to 0}\left (frac{\sin x}{x} \right )^2 x

=1 \times 0 =0

Question 2 |

The accuracy of Simpson's rule quadrature for a step size h is

O(h^{2}) | |

O(h^{3}) | |

O(h^{4}) | |

O(h^{5}) |

Question 3 |

For the matrix \begin{bmatrix} 4 & 1\\ 1 & 4 \end{bmatrix}
the Eigen values are

3 and -3 | |

-3 and -5 | |

3 and 5 | |

5 and 0 |

Question 3 Explanation:

\small A=\left[\begin{array}{ll}4 & 1 \\ 1 & 4\end{array}\right]

\text{Now }\quad|A-\lambda I|=0

\text{where} \quad \lambda= \text{ eigen value}

\therefore \quad\left|\begin{array}{cc}4-\lambda & 1 \\ 1 & 4-\lambda \\ (4-\lambda)^{2}-1 & =0\end{array}\right|=0

\text{or,} \quad(4-\lambda)^{2}-(1)^{2}=0

\text{or,} (4-\lambda+1)(4-\lambda-1)=0

\text{or,} \quad(5-\lambda)(3-\lambda)=0

\therefore \quad \lambda=3,5

\text{Now }\quad|A-\lambda I|=0

\text{where} \quad \lambda= \text{ eigen value}

\therefore \quad\left|\begin{array}{cc}4-\lambda & 1 \\ 1 & 4-\lambda \\ (4-\lambda)^{2}-1 & =0\end{array}\right|=0

\text{or,} \quad(4-\lambda)^{2}-(1)^{2}=0

\text{or,} (4-\lambda+1)(4-\lambda-1)=0

\text{or,} \quad(5-\lambda)(3-\lambda)=0

\therefore \quad \lambda=3,5

Question 4 |

The second moment of a circular area about the diameter is given by (D is the diameter).

\frac{\pi D^{4}}{4} | |

\frac{\pi D^{4}}{16} | |

\frac{\pi D^{4}}{32} | |

\frac{\pi D^{4}}{64} |

Question 4 Explanation:

Polar moment of inertia perpendicular to the plane of paper

I_{p}=\frac{\pi D^{4}}{32}

By 'Perpendicular Axis" theorem

\begin{aligned} I_{x x}+I_{y y} &=I_{p} \quad\left[\because I_{x x}=I_{y y}\right] \\ 2 I_{x x} &=\frac{\pi D^{4}}{32} \\ \therefore \quad I_{x x} &=I_{y y}=\frac{\pi D^{4}}{64} \end{aligned}

I_{p}=\frac{\pi D^{4}}{32}

By 'Perpendicular Axis" theorem

\begin{aligned} I_{x x}+I_{y y} &=I_{p} \quad\left[\because I_{x x}=I_{y y}\right] \\ 2 I_{x x} &=\frac{\pi D^{4}}{32} \\ \therefore \quad I_{x x} &=I_{y y}=\frac{\pi D^{4}}{64} \end{aligned}

Question 5 |

A concentrated load of P acts on a simply supported beam of span L at a distance \frac{L}{3}
form the left support. The bending moment at the point of application of the load is given by

\frac{PL}{3} | |

\frac{2PL}{3} | |

\frac{PL}{9} | |

\frac{2PL}{9} |

Question 5 Explanation:

Taking moment about A, we have

\begin{aligned} R_{B} \times L &=P \times \frac{L}{3} \\ \therefore \quad R_{B} &=\frac{P}{3}\\ \text{and }\quad R_{A}=P-R_{B}&=P-\frac{P}{3}\\ R_{A}&=\frac{2 P}{3} \end{aligned}

Bending moment at the point of application of load,

M=\frac{2 P}{3} \times \frac{L}{3}=\frac{2 P L}{9}

Question 6 |

Two identical circular rods of same diameter and same length are subjected to same magnitude of axial tensile force. One of the rods is made out of mild steel having the modulus of elasticity of 206 GPa. The other rod is made out of cast iron having the modulus of elasticity of 100 GPa. Assume both the materials to be homogeneous and isotropic and the axial force causes the same amount of uniform stress in both the roads. The stresses developed are within the proportional limit of the respective materials. Which of the following observation is correct?

Both rods elongate by the same amount | |

Mild steel rod elongates more than the cast iron rod | |

Cast iron rod elongates more than the mild steel rod | |

As the stresses are equal strains are also equal in both the rods |

Question 6 Explanation:

\begin{aligned} \text{Given,}\quad E_{S}&=206 \mathrm{GPa} \\ \text{and}\quad E_{I}&=100 \mathrm{GPa} \end{aligned}

Elongation in mild steel,

\Delta L_{s}=\frac{P L}{A E_{s}} \qquad\cdots(i)

Elongation in cast Iron,

\Delta L_{1}=\frac{P L}{A E_{I}} \qquad\cdots(ii)

Dividing Eq. (ii) by Eq. (i), we get

\frac{\Delta L_{s}}{\Delta L_{l}}=\frac{E_{I}}{E_{S}}

since diameter and length of both shafts are equal

\therefore \quad \frac{\Delta L_{S}}{\Delta L_{I}}=\frac{100}{206} \lt 1

Elongation in mild steel,

\Delta L_{s}=\frac{P L}{A E_{s}} \qquad\cdots(i)

Elongation in cast Iron,

\Delta L_{1}=\frac{P L}{A E_{I}} \qquad\cdots(ii)

Dividing Eq. (ii) by Eq. (i), we get

\frac{\Delta L_{s}}{\Delta L_{l}}=\frac{E_{I}}{E_{S}}

since diameter and length of both shafts are equal

\therefore \quad \frac{\Delta L_{S}}{\Delta L_{I}}=\frac{100}{206} \lt 1

Question 7 |

Two beams, one having square cross section and another circular cross-section, are subjected to the same amount of bending moment. If the cross sectional area as well as the material of both the beams are the same then

maximum bending stress developed in both the beams is the same | |

the circular beam experiences more bending stress than the square one | |

the square beam experiences more bending stress than the circular one | |

as the material is same both the beams will experience same deformation |

Question 7 Explanation:

Let, d= diameter of circular cross-section

a= side of square cross-section

since cross-sectional area of square and circular cross-section are equal

\therefore \quad \frac{\pi}{4} d^{2}=a^{2}\quad \cdots(i)

For circular cross-section,

\sigma_{c}=\frac{M y}{I}=\frac{M \cdot \frac{d}{2}}{\frac{\pi d^{4}}{64}}=\frac{32 M}{\pi d^{3}}

Square cross-section,

\begin{aligned} \sigma_{s} &=\frac{M \cdot y}{I} \\ &=\frac{M \frac{a}{2}}{\frac{a^{4}}{12}}=\frac{6 M}{a^{3}} \\ \text{Now}\quad \frac{\sigma_{c}}{\sigma_{s}} &=\frac{\frac{32 M}{\pi d^{3}}}{\frac{6 M}{a^{3}}}=\frac{32 a^{3}}{6 \pi d^{3}}\\ &=\frac{32}{6 \pi}\left(\frac{a}{d}\right)^{2}\left(\frac{a}{d}\right) \\ &=\frac{32}{6 \pi} \times\left(\frac{\pi}{4}\right) \times\left(\sqrt{\frac{\pi}{4}}\right) \\ &=\frac{32}{24} \sqrt{\frac{\pi}{4}}=\frac{32 \sqrt{\pi}}{48}>1 \\ \sigma_{c}&> \sigma_{s} \end{aligned}

a= side of square cross-section

since cross-sectional area of square and circular cross-section are equal

\therefore \quad \frac{\pi}{4} d^{2}=a^{2}\quad \cdots(i)

For circular cross-section,

\sigma_{c}=\frac{M y}{I}=\frac{M \cdot \frac{d}{2}}{\frac{\pi d^{4}}{64}}=\frac{32 M}{\pi d^{3}}

Square cross-section,

\begin{aligned} \sigma_{s} &=\frac{M \cdot y}{I} \\ &=\frac{M \frac{a}{2}}{\frac{a^{4}}{12}}=\frac{6 M}{a^{3}} \\ \text{Now}\quad \frac{\sigma_{c}}{\sigma_{s}} &=\frac{\frac{32 M}{\pi d^{3}}}{\frac{6 M}{a^{3}}}=\frac{32 a^{3}}{6 \pi d^{3}}\\ &=\frac{32}{6 \pi}\left(\frac{a}{d}\right)^{2}\left(\frac{a}{d}\right) \\ &=\frac{32}{6 \pi} \times\left(\frac{\pi}{4}\right) \times\left(\sqrt{\frac{\pi}{4}}\right) \\ &=\frac{32}{24} \sqrt{\frac{\pi}{4}}=\frac{32 \sqrt{\pi}}{48}>1 \\ \sigma_{c}&> \sigma_{s} \end{aligned}

Question 8 |

The mechanism used in a shaping machine is

a closed 4-bar chain having 4 revolute pairs | |

a closed 6-bar chain having 6 revolute pairs | |

a closed 4-bar chain having 2 revolute and 2 sliding pairs | |

an inversion of the single slider-crank chain |

Question 8 Explanation:

Crank and slotted lever mechanism is mostly used
in shaping machines, which is the inversion of
single slider-crank chain.

Question 9 |

The lengths of the links of a 4-bar linkage with revolute pairs only are p,q,r and s units. Given that p \lt q \lt r \lt s. Which of these links should be the fixed one, for obtaining a "double crank" mechanism?

link of length p | |

link of length q | |

link of length r | |

link of length s |

Question 9 Explanation:

Double crank mechanism is obtained when the
shortest link is fixed. Here, the shortest link is p.
Therefore link of length p should be fixed.

Question 10 |

Consider the arrangement shown in the figure below where J is the combined polar mass moment of inertia of the disc and the shafts. K_1,K_2,K_3 are the torsional stiffness of the respective shafts. The natural frequency of torsional oscillation of the disc is given by

\sqrt{\frac{{K_{1}+K_{2}+K_{3}}}{J}} | |

\sqrt{\frac{{K_{1}K_{2}+K_{2}K_{3}+K_{3}K_{1}}}{J(K_{1}+K_{2})}} | |

\sqrt{\frac{{K_{1}K_{2}K_{3}}}{J(K_{1}K_{2}+K_{2}K_{3}+K_{3}K_{1})}} | |

\sqrt{\frac{{K_{1}K_{2}+K_{2}K_{3}+K_{3}K_{1}}}{J(K_{2}+K_{3})}} |

Question 10 Explanation:

Equivalent stiffness

k_{\mathrm{eq}}=\frac{k_{1} k_{2}}{k_{1}+k_{2}}+k_{3}=\frac{k_{1} k_{2}+k_{1} k_{3}+k_{2} k_{3}}{k_{1}+k_{2}}

Now natural frequency

=\sqrt{\frac{k_{e q}}{J}}=\sqrt{\frac{k_{1} k_{2}+k_{1} k_{3}+k_{2} k_{3}}{J\left(k_{1}+k_{2}\right)}}

k_{\mathrm{eq}}=\frac{k_{1} k_{2}}{k_{1}+k_{2}}+k_{3}=\frac{k_{1} k_{2}+k_{1} k_{3}+k_{2} k_{3}}{k_{1}+k_{2}}

Now natural frequency

=\sqrt{\frac{k_{e q}}{J}}=\sqrt{\frac{k_{1} k_{2}+k_{1} k_{3}+k_{2} k_{3}}{J\left(k_{1}+k_{2}\right)}}

There are 10 questions to complete.