A) 0.25 cm
B) 2 cm
C) 4 cm
D) 2.5 cm
Correct Answer: D
Solution :
The various forces acting on the particle, are its weight mg acting vertically downwards, normal reaction N. Equating the vertical forces, we have \[N\,\sin \theta =mg\] ... (i) Also, centripetal force, \[\frac{m{{v}^{2}}}{R}=N\,\cos \theta \] ...(ii) From Eqs. (i) and (ii), we get \[\tan \theta =\frac{Rg}{{{v}^{2}}}\] ...(iii) Also, from triangle OAB, \[\tan \theta =\frac{R}{h}\] ...(iv) Equating Eqs. (iii) and (iv), we get \[h=\frac{{{v}^{2}}}{g}\] Given, \[v=0.5\text{ }m/s,\text{ }g=10\text{ }m/{{s}^{2}}\] \[\therefore \] \[h=\frac{{{(0.5)}^{2}}}{10}=0.025\,m\] Since, \[100\text{ }cm=1\text{ }m\] \[\therefore \] \[h=2.5\,cm\]You need to login to perform this action.
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