| Two particles A and B are connected by a rigid rod AB. The rod slides along perpendicular rails as shown here. The velocity of A to the right is \[10\text{ }m/s\]. What is the velocity of B when angle\[\alpha ={{60}^{o}}\]? [AIPMT 1998] |
 |
A) 9.8 m/s
B) 10 m/s
C) 5.8 m/s
D) 17.3 m/s
Correct Answer: D
Solution :
| Let the velocity along x and y axes be \[{{v}_{x}}\] and \[{{v}_{y}}\]respectively. |
| \[\therefore \] \[{{v}_{x}}=\frac{dx}{dt}\] and \[{{v}_{y}}=\frac{dy}{dy}\] |
| From figure, |
| \[\tan \alpha =\frac{y}{x}\] |
| \[\Rightarrow \] \[y=x\tan \alpha \] |
| Differentiating Eq. (i) w.r.t. \[t\], we get |
| \[\frac{dy}{dt}=\frac{dx}{dt}\tan \alpha \] |
| \[\Rightarrow \] \[{{v}_{y}}={{v}_{x}}\tan \alpha \] |
| Here, \[{{v}_{x}}=10\,m/s\,,\,\alpha ={{60}^{o}}\] |
| \[\therefore \] \[{{v}_{y}}=10\,\tan {{60}^{o}}=10\sqrt{3}=17.3\,m/s\] |
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