A) \[14\,\,m\]
B) \[28\,\,m\]
C) \[35\,\,m\]
D) \[7\,\,m\]
Correct Answer: C
Solution :
Key Idea: The sphere will fly off when the normal reaction at the surface of sphere is zero. Let \[P\] be the instantaneous position of the particle during its sliding from the top\[A\]. \[N\]be the normal reaction at the surface of sphere, thus supplies the necessary centripetal force thus. \[mg\,\,\cos \,\theta -N=\frac{m{{v}^{2}}}{R}\] When\[N=0\], particle will fly off \[mh\,\,\cos \,\,\theta \frac{m{{v}^{2}}}{R}\] \[\frac{1}{2}mg\,R\,\cos \,\,\theta =\frac{1}{2}m{{v}^{2}}\] ?(1) Fall in height in sliding from \[A\] to \[P\] is \[AC=AO-CO=R-R\,\,\cos \,\,\theta \] Loss in potential energy is \[mg\,\,R\left( 1-\cos \,\,\theta \right)=\frac{1}{2}m{{v}^{2}}\] ?(2) From Eqs. (1) and (2), we get \[\frac{1}{2}mg\,\,R\,\,\cos \,\,\theta =mg\,\,R\left( 1-\cos \,\,\theta \right)\] \[\Rightarrow \] \[\cos \,\,\theta =\frac{2}{3}\] Vertical height of point \[P\] from bottom \[BC=BO+OC=R+R\,\,\cos \,\,\theta \] =\[=R+R\left( \frac{2}{3} \right)=\frac{5}{3}R\] Given, \[R=21\,\,m\] \[BC=\frac{5}{3}\times 21=35\,\,m\]You need to login to perform this action.
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