Question 1 |

An undamped spring-mass system with mass m and spring stiffness k is shown
in the figure. The natural frequency and natural period of this system are \omega rad/s and T \;s, respectively. If the stiffness of the spring is doubled and the mass is halved, then the natural frequency and the natural period of the modified system, respectively, are

2\omega rad/s and T/2 s | |

\omega /2 rad/s and 2T s | |

4\omega rad/s and T/4 s | |

\omega rad/s and T s |

Question 1 Explanation:

This is question of undamped free vibration of
sinle degree of freedom system.
For the above figure it is given that natural
frequency and natural period are w rad/s and T s,
respectively.
The equation for soof system is

\omega =\sqrt{\frac{k}{m}}, T=\frac{2\pi}{\omega }

For the modified system, stiffness of the spring is doubled and the mass is halved.

Let us assume k'=2k and m'=m/2, here k' and m' represent the stiffness and mass of the modified system respectively.

\omega '=\sqrt{\frac{k'}{m}}=\sqrt{\frac{2k}{\left ( \frac{m}{2} \right )}}=\sqrt{\frac{4k}{m}}=2\sqrt{\frac{k}{m}}=2\omega \;\;\; \left ( \because \omega = \sqrt{\frac{k}{m}}\right )

T'=\frac{2\pi}{\omega '}=\frac{2 \pi}{2\omega }=\frac{1}{2}\left ( \frac{2 \pi}{\omega } \right )=\frac{T}{2} \;\;\left ( T=\frac{2 \pi}{\omega } \right )

\omega ' and T' are natural frequency and natural period of the modified system.

hence, the natural frequency of the modified system is doubled and natural period is halved compared to the original given system. Hence option (A) is correct.

\omega =\sqrt{\frac{k}{m}}, T=\frac{2\pi}{\omega }

For the modified system, stiffness of the spring is doubled and the mass is halved.

Let us assume k'=2k and m'=m/2, here k' and m' represent the stiffness and mass of the modified system respectively.

\omega '=\sqrt{\frac{k'}{m}}=\sqrt{\frac{2k}{\left ( \frac{m}{2} \right )}}=\sqrt{\frac{4k}{m}}=2\sqrt{\frac{k}{m}}=2\omega \;\;\; \left ( \because \omega = \sqrt{\frac{k}{m}}\right )

T'=\frac{2\pi}{\omega '}=\frac{2 \pi}{2\omega }=\frac{1}{2}\left ( \frac{2 \pi}{\omega } \right )=\frac{T}{2} \;\;\left ( T=\frac{2 \pi}{\omega } \right )

\omega ' and T' are natural frequency and natural period of the modified system.

hence, the natural frequency of the modified system is doubled and natural period is halved compared to the original given system. Hence option (A) is correct.

Question 2 |

A wedge M and a block N are subjected to forces P and Q as shown in the figure. If force P is sufficiently large, then the block N can be raised. The weights of the wedge and the block are negligible compared to the forces P and Q. The coefficient of friction (\mu) along the inclined surface between the wedge and the block is 0.2. All other surfaces are frictionless. The wedge angle is 30^{\circ}.

The limiting force P, in terms of Q, required for impending motion of block N to just move it in the upward direction is given as P=\alpha Q. The value of the coefficient '?' (rounded off to one decimal place) is

The limiting force P, in terms of Q, required for impending motion of block N to just move it in the upward direction is given as P=\alpha Q. The value of the coefficient '?' (rounded off to one decimal place) is

0.6 | |

0.5 | |

2.0 | |

0.9 |

Question 2 Explanation:

\begin{aligned} \Sigma Y&=0 \Rightarrow \\ N_{2} \sin 60^{\circ}-0.2 N_{2} \sin &30^{\circ}-Q=0 \\ Q&=0.766 \mathrm{~N}_{2} \end{aligned}

\begin{aligned} \Sigma X &=0 \\ \Rightarrow \qquad 0.2 N_{2} \cos 30^{\circ}+N_{2} \cos 60^{\circ}-P &=0\\ P &=0.67 N_{2} \\ \Rightarrow \qquad \qquad \qquad \qquad \qquad \qquad \qquad P &=0.67 \times \frac{Q}{0.766}\\ \Rightarrow \qquad \qquad \qquad \qquad \qquad \qquad \qquad P&=0.875 \mathrm{Q} \simeq 0.9 Q \\ P&=\alpha Q \\ \Rightarrow \qquad \qquad \qquad \qquad \qquad \qquad \qquad \alpha&=0.9 \end{aligned}

Question 3 |

A simple mass-spring oscillatory system consists of a mass m, suspended from a spring of stiffness k. Considering z as the displacementof the system at any time t, the equation of motion for the free vibration of the system is m\ddot{z}+kz=0. The natural frequency of the system is

\frac{k}{m} | |

\sqrt{\frac{k}{m}} | |

\frac{m}{k} | |

\sqrt{\frac{m}{k}} |

Question 3 Explanation:

\begin{aligned} m\ddot{z}+kz&=0 \\ \ddot{z}+\frac{k}{m}z&=0 \\ \text{Comparing with} \\ \ddot{z}+\omega _n^2 z&=0 \\ \text{We get}\;\; \omega _n&=\sqrt{\frac{k}{m}} \end{aligned}

Question 4 |

A cable PQ of length 25 m is supported at two ends at the same level as shown in the figure. The horizontal distance between the supports is 20 m. A point load of 150 kN is applied at point R which divides it into two equal parts.

Neglecting the self-weight of the cable, the tension (in kN, in integer value) in the cable due to the applied load will be _____

Neglecting the self-weight of the cable, the tension (in kN, in integer value) in the cable due to the applied load will be _____

100 | |

125 | |

200 | |

250 |

Question 4 Explanation:

\begin{aligned} 12.5^{2} &=x^{2}+10^{2} \\ \Rightarrow \; \; x &=7.5\; \text{m} \end{aligned}

Bending moment at S=0 {consider the left part}

\begin{aligned} H_{P}\times 7.5 &=75\times 10 \\ H_{P} &=100\; \text{kN} \end{aligned}

Tension in cable

\begin{aligned} &=\sqrt{H_{P}^{2}+V_{P}^{2}}=\sqrt{75^{2}+100^{2}} \\ &=125\; \text{kN} \end{aligned}

Question 5 |

Two rigid bodies of mass 5 kg and 4 kg are at rest on a frictionless surface until acted upon by a force of 36 N as shown in the figure. The contact force generated between the two bodies is

4.0 N | |

7.2 N | |

9.0 N | |

16.0 N |

Question 5 Explanation:

As after action of 36 N, both blocks will move with same acceleration so considering 5 kg and 4 kg togather in a system and applying Newton's 2nd law.

a=\frac{36}{9}=4\; m/s^{2}

\begin{aligned} N &=ma\; \; \; \; \; \text{(Newton's 2nd law)} \\ N&=4\times 4=16\; \text{N} \end{aligned}

Question 6 |

An aircraft approaches the threshold of a runway strip at a speed of 200 km/h. The pilot decelerates the aircraft at a rate of 1.697 m/s^{2}and takes 18 s to exit the runway strip. If the deceleration after exiting the runway is 1 m/s^{2}, then the distance (in m, up to one decimal place) of the gate position from the location of exit on the runway is ______

312.8 | |

360.4 | |

210.4 | |

111.8 |

Question 6 Explanation:

Speed of aircraft,

\begin{aligned} u_{i} &=200\; \text{km/hr} \\ &=\frac{200\times 1000}{3600}=55.56\; \text{m/s} \end{aligned}

Deceleration of aircraft on the runway =1.697 m/s^{2}

Aircraft takes 18 sec to exit the runway strip.

Speed of the aircraft at the exit of the runway

\begin{aligned} u_{f} &=u_{i}+at \\ &=55.56-1.697\times 18 \\ &=25.014\; \text{m/s} \end{aligned}

After runway aircraft decelerate with 1 m/s^{2}

Total distance travelled by the aircraft from the location of exit on the runway

\begin{aligned} U^{2} &=u^{2}+2aS \\ 0 &=25.014^{2}-2\times 1\times S \\ S &=312.8\; \text{m} \end{aligned}

\begin{aligned} u_{i} &=200\; \text{km/hr} \\ &=\frac{200\times 1000}{3600}=55.56\; \text{m/s} \end{aligned}

Deceleration of aircraft on the runway =1.697 m/s^{2}

Aircraft takes 18 sec to exit the runway strip.

Speed of the aircraft at the exit of the runway

\begin{aligned} u_{f} &=u_{i}+at \\ &=55.56-1.697\times 18 \\ &=25.014\; \text{m/s} \end{aligned}

After runway aircraft decelerate with 1 m/s^{2}

Total distance travelled by the aircraft from the location of exit on the runway

\begin{aligned} U^{2} &=u^{2}+2aS \\ 0 &=25.014^{2}-2\times 1\times S \\ S &=312.8\; \text{m} \end{aligned}

Question 7 |

A cylinder of radius 250 mm and weight, W = 10 kN is rolled up an obstacle of height 50 mm by applying a horizontal force P at its centre as shown in the figure.

All interfaces are assumed frictionless. The minimum value of P is

All interfaces are assumed frictionless. The minimum value of P is

4.5 kN | |

5.0 kN | |

6.0 kN | |

7.5 kN |

Question 7 Explanation:

Given: r=250 mm, W=10 kN

Note:

1. When the cylinder will be about to move out of the cylinder, it will loose its contact at point A, only contact will be at point B.

2. Considering equation of cylinder of that instant under P, W and R_{B} (contact force at B).

\begin{aligned} \sum \vec{M}_{B} &=0 \\ P\times OC-W\times BC &=0 \end{aligned}

In \Delta OCB

\begin{aligned} OC^{2}+CB^{2} &=OB^{2} \\ CB &=\sqrt{250^{2}-200^{2}}=150 \\ P &=\frac{10\times 10^{3}\times 150}{200} \\ &=2.5\times 3\: \text{kN}=7.5\: \text{kN} \end{aligned}

Question 8 |

An assembly made of a rigid arm A-B-C hinged at end A and supported by an elastic rope C-D at end C is shown in the figure. The members may be assumed to be weightless and the lengths of the respective members are as shown in the figure.

Under the action of a concentrated load P at C as shown, the magnitude of tension developed in the rope is

Under the action of a concentrated load P at C as shown, the magnitude of tension developed in the rope is

\frac{3P}{\sqrt{2}} | |

\frac{P}{\sqrt{2}} | |

\frac{3P}{8} | |

\sqrt{2}P |

Question 8 Explanation:

\begin{aligned} \sum M_{A}&=0\\ \Rightarrow \; R_{D}\times 2 L-P\times L&=0\\ \Rightarrow R_{D}&=\frac{P}{2} \end{aligned}

At joint D:

\begin{aligned} \sum F_{y}&=0\\ \Rightarrow \; T \cos 45^{\circ}&=\frac{P}{2}\\ \therefore \; T&=\frac{P}{\sqrt{2}} \end{aligned}

Question 9 |

The magnitudes of vectors P, Q and R are 100 kN, 250 kN and 150 kN, respectively as shown in the figure

The respective values of Magnitude(in KN) and the Direction (with the respect to the x-axis) of the resultant vector are

The respective values of Magnitude(in KN) and the Direction (with the respect to the x-axis) of the resultant vector are

290.9 and 96.0^{\circ} | |

368.1 and 94.7^{\circ} | |

330.4 and 118.9^{\circ} | |

400.1 and 113.5^{\circ} |

Question 9 Explanation:

Resultant force in x-direction,

\begin{aligned} R_{x}&=P\cos 60^{\circ}-Q\sin 15^{\circ}-R\sin 15^{\circ} \\ &=100\cos 60^{\circ}-250\sin 15^{\circ}-150\sin 15^{\circ} \\ &=-159.59\, kN\left ( \rightarrow \right ) \\ &=-159.59\, kN\left ( \rightarrow \right ) \\ &=159.59\, kN\left ( \leftarrow \right ) \end{aligned}

Resultant force in y-direction,

\begin{aligned} R_{y}&=P\sin 60^{\circ}+Q\cos 15^{\circ}-R\sin 15^{\circ} \\ &=100\sin 60^{\circ}+250\cos 15^{\circ}-150\sin 15^{\circ} \\ &=289.26\, kN \left ( \uparrow \right ) \end{aligned}

Final resultant force,

\begin{aligned} R&=\sqrt{R^{2}_{x}+R^{2}_{y}}=330.4\, kN \\ \theta &=\tan ^{-1}\left ( \frac{159.59}{289.56} \right ) \\ \theta &=28.9^{\circ} \end{aligned}

Direction with respect to x-axis

\begin{aligned} &=\theta+90^{\circ} \\ &=28.9^{\circ}+90^{\circ} \\ &=118.9^{\circ} \end{aligned}

Question 10 |

Polar moment of inertia (I_{p}), in cm^{4}, of a rectangular section having width, b=2 and depth, d=6 cm is _____

22 | |

44 | |

20 | |

40 |

Question 10 Explanation:

Polar moment of inertia,

\begin{aligned} I_p&= I_x+I_y\\ &=\frac{bd^3}{12}+\frac{db^3}{12}\\ &=\frac{bd}{12}(b^2+d^2) \\ &= \frac{2 \times 6}{12}(2^2+6^2)\\ &=40\; cm^4 \end{aligned}

\begin{aligned} I_p&= I_x+I_y\\ &=\frac{bd^3}{12}+\frac{db^3}{12}\\ &=\frac{bd}{12}(b^2+d^2) \\ &= \frac{2 \times 6}{12}(2^2+6^2)\\ &=40\; cm^4 \end{aligned}

There are 10 questions to complete.