Question 1 |
The area moment of inertia about the y-axis of a linearly tapered section shown in
the figure is ______ m^4.
(Answer in integer)
1254 | |
3625 | |
3024 | |
2542 |
Question 1 Explanation:
Above figure is symmetrical about {x} axis so, moment of inertia of whole section will be two times of the one section above x-axis.
Moment of inertia of rectangular section OABC about y-axis i.e. OC.
=\frac{1}{3} \times \mathrm{OC} \times(\mathrm{AB})^{3}
=\frac{1}{3} \times d \times b^{3}
\mathrm{I}_{\mathrm{OC}}=\frac{1}{3} \times 3 \times(12)^{3}=1728
Moment of inertia of triangular section BCD about y-axis i.e. about DC
\mathrm{I}_{D C}=\frac{1}{12} \times(D C) \times(B C)^{3} =\frac{1}{12} \times 1.5 \times(12)^{3}=216
So, moment of inertia of tapered section OABD about y-axis:
\mathrm{I}_{\mathrm{y}^{\prime} \mathrm{y}^{\prime}}=\mathrm{I}_{\mathrm{OC}}-\mathrm{I}_{\mathrm{DC}}
=1728-216=1512
So, moment of inertia of whole tapered section
I_{y y}=2 l_{y^{\prime} y^{\prime}}=2 \times 1512=3024 \mathrm{~m}^{4}
Moment of inertia of rectangular section OABC about y-axis i.e. OC.
=\frac{1}{3} \times \mathrm{OC} \times(\mathrm{AB})^{3}
=\frac{1}{3} \times d \times b^{3}
\mathrm{I}_{\mathrm{OC}}=\frac{1}{3} \times 3 \times(12)^{3}=1728
Moment of inertia of triangular section BCD about y-axis i.e. about DC
\mathrm{I}_{D C}=\frac{1}{12} \times(D C) \times(B C)^{3} =\frac{1}{12} \times 1.5 \times(12)^{3}=216
So, moment of inertia of tapered section OABD about y-axis:
\mathrm{I}_{\mathrm{y}^{\prime} \mathrm{y}^{\prime}}=\mathrm{I}_{\mathrm{OC}}-\mathrm{I}_{\mathrm{DC}}
=1728-216=1512
So, moment of inertia of whole tapered section
I_{y y}=2 l_{y^{\prime} y^{\prime}}=2 \times 1512=3024 \mathrm{~m}^{4}
Question 2 |
The rod PQ of length L=\sqrt{2} m,and uniformly distributed mass of M=10 kg, is released from rest at the position shown in the figure. The ends slide along the frictionless faces OP and OQ. Assume acceleration due to gravity, g=10 \:m/s^{2}. The mass moment of inertia of the rod about its centre of mass and an axis perpendicular to the plane of the figure is (ML^{2}/12). At this instant, the magnitude of angular acceleration (in radian/s^{2} ) of the rod is _____
15 | |
10.5 | |
7.5 | |
4 |
Question 2 Explanation:
\begin{aligned} L&=\sqrt{2} m \\ M&=10 \mathrm{kg}\\ g&=10 \mathrm{m} / \mathrm{s}^{2} \\ \end{aligned}
Both the surfaces are smooth
\begin{aligned} \ddot{\theta} &=\alpha=\frac{3 g}{2 L} \sin \theta \\ \alpha &=\frac{3 \times 10}{2 \times \sqrt{2}} \sin 45^{\circ} \quad\left[\because \theta=45^{\circ}\right] \\ a &=\frac{3 \times 10}{2 \times \sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{30}{4} \\ &=7.5 \mathrm{rad} / \mathrm{s}^{2} \end{aligned}
Both the surfaces are smooth
\begin{aligned} \ddot{\theta} &=\alpha=\frac{3 g}{2 L} \sin \theta \\ \alpha &=\frac{3 \times 10}{2 \times \sqrt{2}} \sin 45^{\circ} \quad\left[\because \theta=45^{\circ}\right] \\ a &=\frac{3 \times 10}{2 \times \sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{30}{4} \\ &=7.5 \mathrm{rad} / \mathrm{s}^{2} \end{aligned}
Question 3 |
An inextensible massless string goes over a frictionless pulley. Two weights of 100 N and 200 N are attached to the two ends of the string. The weights are released from rest, and start moving due to gravity. The tension in the string (in N) is __________
182.56 | |
133.33 | |
4.256 | |
2.56 |
Question 3 Explanation:
Tension in string
\begin{aligned} &=\left[\frac{2 m_{1} m_{2}}{m_{1}+m_{2}}\right] g \\ &=\left[\frac{2 \times \frac{100}{g} \times \frac{200}{g}}{\frac{100}{g}+\frac{200}{g}}\right] g \\ &=\left[\frac{\frac{2 \times 100 \times 200}{g^{2}}}{\frac{(100+200)}{g}}\right] \times g \\ &=\frac{2 \times 100 \times 200}{300} \\ &=\frac{400}{3}=133.33 \mathrm{N} \end{aligned}
Question 4 |
Figure shows a wheel rotating about O_{2}. Two points A and B located along the radius of wheel have speeds of 80 m/s and 140 m/s respectively. The distance between the points A and B is 300 mm. The diameter of the wheel (in mm) is ________
1400mm | |
1200mm | |
1600mm | |
1800mm |
Question 4 Explanation:
\begin{aligned} \omega_{A} &=\omega_{B} \\ \frac{V_{A}}{R-0.3} &=\frac{V_{B}}{R} \\ \frac{80}{R-0.3} &=\frac{140}{R} \\ 80 R &=140 R-42 \\ 42 &=60 R \\ R &=0.7 \mathrm{m} \\ \therefore \qquad D &=2 R=1.4 \mathrm{m}=1400 \mathrm{mm} \end{aligned}
Question 5 |
For the same material and the mass, which of the following configurations of flywheel will have maximum mass moment of inertia about the axis of rotation OO' passing through the center of gravity.
A | |
B | |
C | |
D |
Question 5 Explanation:
I=m r^{2}
Here for the rimmed wheel, the mass m will be most far away distribution from the axis 0-0.
\therefore \;(B) is the correct answer.
Here for the rimmed wheel, the mass m will be most far away distribution from the axis 0-0.
\therefore \;(B) is the correct answer.
There are 5 questions to complete.