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
Question 6 |
A four bar mechanism is shown in the figure. The link numbers are mentioned near the links. Input link 2 is rotating anticlockwise with a constant angular speed \omega _2. Length of different links are:
O_2O_4=O_2A=L,
AB=O_4B=\sqrt{2}L.
The magnitude of the angular speed of the output link 4 is \omega _4 at the instant when link 2 makes an angle of 90^{\circ} with O_2O_4 as shown. The ratio \frac{\omega _4}{\omega _2} is ______(round off to two decimal places).
O_2O_4=O_2A=L,
AB=O_4B=\sqrt{2}L.
The magnitude of the angular speed of the output link 4 is \omega _4 at the instant when link 2 makes an angle of 90^{\circ} with O_2O_4 as shown. The ratio \frac{\omega _4}{\omega _2} is ______(round off to two decimal places).
0.78 | |
0.42 | |
0.28 | |
1.25 |
Question 6 Explanation:
\begin{aligned} &\mathrm{O}_{2} \mathrm{A}=\mathrm{O}_{2} \mathrm{O}_{4}=\mathrm{L}\\ &\mathrm{AB}=\mathrm{O}_{4} \mathrm{B}=\mathrm{L} \sqrt{2}\\ &\omega_{2} \mathrm{I}_{12} \mathrm{I}_{24}=\omega_{4} \mathrm{I}_{14} \mathrm{I}_{24}\\ &\frac{\omega_{4}}{\omega_{2}}=\frac{\mathrm{I}_{12} \mathrm{I}_{24}}{\mathrm{I}_{14} \mathrm{I}_{24}}\\ &\text { from triangle } \mathrm{O}_{2} \mathrm{AI}_{24}\\ &\frac{I_{12} I_{24}}{\sin 75}=\frac{L}{\sin 15}\\ &\mathrm{I}_{12} \mathrm{I}_{24}=\mathrm{L} \frac{\sin 75}{\sin 15}=3.73 \mathrm{L}\\ &\mathrm{I}_{14} \mathrm{I}_{24}=3.73 \mathrm{L}+\mathrm{L}=4.73 \mathrm{L}\\ &\frac{\omega_{4}}{\omega_{2}}=\frac{\mathrm{I}_{12} \mathrm{I}_{24}}{\mathrm{I}_{14} \mathrm{I}_{24}}=\frac{3.73 \mathrm{L}}{4.73 \mathrm{L}}=0.789 \end{aligned}
Question 7 |
A thin-walled cylindrical can with rigid end caps has a mean radius R=100 mm and a wall thickness of t=5 mm . The can is pressurized and an additional tensile stress of 50 MPa is imposed along the axial direction as shown in the figure. Assume that the state of stress in the wall is uniform along its length. If the magnitudes of axial and circumferential components of stress in the can are equal, the pressure (in MPa) inside the can is ___________ (correct to two decimal places).
2 | |
3 | |
4 | |
5 |
Question 7 Explanation:
Circumferential stress, \sigma_{n}=\frac{P R}{t}
\text { Axial stress, } \sigma_{L}=\frac{P R}{2 t}+50 \mathrm{MPa}
\begin{aligned} {Now,}\quad \sigma_{n} &=\sigma_{L} \\ \frac{P R}{t} &=\frac{P R}{2 t}+50 \mathrm{MPa} \\ \therefore\quad \frac{P R}{2 t} &=50 \mathrm{MPa}\\ P&=\frac{50 \times 2 \times 5}{100}=5 \mathrm{MPa} \end{aligned}
\text { Axial stress, } \sigma_{L}=\frac{P R}{2 t}+50 \mathrm{MPa}
\begin{aligned} {Now,}\quad \sigma_{n} &=\sigma_{L} \\ \frac{P R}{t} &=\frac{P R}{2 t}+50 \mathrm{MPa} \\ \therefore\quad \frac{P R}{2 t} &=50 \mathrm{MPa}\\ P&=\frac{50 \times 2 \times 5}{100}=5 \mathrm{MPa} \end{aligned}
Question 8 |
A four bar mechanism is made up of links of length 100, 200, 300 and 350 mm. If the 350 mm link is fixed, the number of links that can rotate fully is __________.
0 | |
1 | |
2 | |
4 |
Question 8 Explanation:
\begin{array}{c} s=100, p=200, l=350, q=300 \\ (s+l)=350+100=450 \lt (p+q) \\ 450 \lt 200+300 \\ 450 \lt 500 \end{array}
Grashot law is satisfied.
350 mm link is fixed.
Then, shortest link = 100mm s adjacent to fixed, will give crank only.
Question 9 |
Block 2 slides outward on link 1 at a uniform velocity of 6 m/s as shown in the figure. Link 1 is rotating at a constant angular velocity of 20 radians/s counterclockwise. The magnitude of the total acceleration (in\, m/s^{2}) of point P of the block with respect to fixed point O is ___.
243.31 | |
247.33 | |
258.37 | |
260.44 |
Question 9 Explanation:
\begin{array}{l} V=6 \mathrm{m} / \mathrm{s} \\ \omega=20 \mathrm{rad} / \mathrm{s} \end{array}
Here the centripital and corialis acceleration will
be there
Total Acceleration:
\begin{aligned} a_{p} &=\sqrt{(40)^{2}+(240)^{2}} \\ &=243.310 \mathrm{m} / \mathrm{s}^{2} \end{aligned}
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.
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.