Question 1 |

The wheels and axle system lying on a rough surface is shown in the figure.

Each wheel has diameter 0.8 m and mass 1 kg. Assume that the mass of the wheel is concentrated at rim and neglect the mass of the spokes. The diameter of axle is 0.2 m and its mass is 1.5 kg. Neglect the moment of inertia of the axle and assume g=9.8 m/s^2. An effort of 10 N is applied on the axle in the horizontal direction shown at mid span of the axle. Assume that the wheels move on a horizontal surface without slip. The acceleration of the wheel axle system in horizontal direction is ________m/s^2(round off to one decimal place).

Each wheel has diameter 0.8 m and mass 1 kg. Assume that the mass of the wheel is concentrated at rim and neglect the mass of the spokes. The diameter of axle is 0.2 m and its mass is 1.5 kg. Neglect the moment of inertia of the axle and assume g=9.8 m/s^2. An effort of 10 N is applied on the axle in the horizontal direction shown at mid span of the axle. Assume that the wheels move on a horizontal surface without slip. The acceleration of the wheel axle system in horizontal direction is ________m/s^2(round off to one decimal place).

2.8 | |

3.6 | |

5 | |

6.4 |

Question 1 Explanation:

\begin{aligned} I_G&=2 \times (M \times r^2) \\&=2 \times 1 \times 0.4^2=0.32\; kg/m^2 \\ \Sigma F&=m\cdot a\\ &\Rightarrow 10-f=3.5 \times a\;...(i)\\ \Sigma T&=I\cdot \alpha \\ &\Rightarrow 10 \times 0.1 -f\times 0.4=0.32 \times \frac{a}{0.4} \;...(ii)\\ \end{aligned}

[As there is no slip \therefore \;\;a=r\alpha ]

Solving (i) and (ii),

\therefore \;\;a=5\; m/s^2

Question 2 |

Consider the mechanism shown in the figure. There is rolling contact without slip between the disc and ground.

Select the correct statement about instantaneous centers in the mechanism.

Select the correct statement about instantaneous centers in the mechanism.

Only points P, Q, and S are instantaneous centers of mechanism | |

Only points P, Q, S and T are instantaneous centers of mechanism | |

Only points P, Q, R, S, and U are instantaneous centers of mechanism | |

All points P, Q, R, S, T and U are instantaneous centers of mechanism |

Question 2 Explanation:

Points P, Q, R, S, T and U are instantaneous centers of mechanism.

Question 3 |

The number of qualitatively distinct kinematic inversions possible for a Grashof chain with four revolute pairs is

1 | |

2 | |

3 | |

4 |

Question 3 Explanation:

They are:

1. Double crank mechanism

2. Crank-rocker mechanism

3. Double rocker mechanism

1. Double crank mechanism

2. Crank-rocker mechanism

3. Double rocker mechanism

Question 4 |

The 2 kg block shown in figure (top view) rests on a smooth horizontal surface and is
attached to a massless elastic cord that has a stiffness 5 N/m.

The cord hinged at O is initially unstretched and always remains elastic. The block is given a velocity v of 1.5 m/s perpendicular to the cord. The magnitude of velocity in m/s of the block at the instant the cord is stretched by 0.4 m is

The cord hinged at O is initially unstretched and always remains elastic. The block is given a velocity v of 1.5 m/s perpendicular to the cord. The magnitude of velocity in m/s of the block at the instant the cord is stretched by 0.4 m is

0.83 | |

1.07 | |

1.36 | |

1.5 |

Question 4 Explanation:

Energy conservation

\begin{aligned} \frac{1}{2} m V_{1}^{2} &=\frac{1}{2} m V_{0}^{2}+\frac{1}{2} k x^{2} \\ \Rightarrow \quad 2 \times 1.5^{2} &=2 \times V_{0}^{2}+5 \times 0.4^{2} \\ V_{0} &=1.360 \mathrm{m} / \mathrm{s} \end{aligned}

\begin{aligned} \frac{1}{2} m V_{1}^{2} &=\frac{1}{2} m V_{0}^{2}+\frac{1}{2} k x^{2} \\ \Rightarrow \quad 2 \times 1.5^{2} &=2 \times V_{0}^{2}+5 \times 0.4^{2} \\ V_{0} &=1.360 \mathrm{m} / \mathrm{s} \end{aligned}

Question 5 |

A four bar mechanism is shown below

For the mechanism to be a crank-rocker mechanism, the length of the link PQ can be

For the mechanism to be a crank-rocker mechanism, the length of the link PQ can be

80 mm | |

200 mm | |

300 mm | |

350 mm |

Question 5 Explanation:

For Crank-Rocker mechanism, shortest link must be crank and adjacent to fixed as well as Grashoff's law must be satisfied.

If l = 80 mm then shortest will be = 80 mm

as well as (80 + 600) \lt(400 + 300)

680 \lt 700

Therefore law is satisfied.

\Rightarrow l = 80 mm

There are 5 questions to complete.

Sir,7 the question som pyq present plz change otherwise great sir this practice sheet

Plz sir, do check ans of Q19 again, it should be shaping machine, ans QRMM used in it.