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An accelration produces a narrow beam of protons, each having an initial speed of v_(0). The beam is directed towards an initially uncharges distant metal sphere of radius R and centered at point O. The initial path of the beam is parallel to the axis of the sphere at a distance of (R//2) from the axis, as indicated in the diagram. The protons in the beam that collide with the sphere will cause it to becomes charged. The subsequentpotential field at the accelerator due to the sphere can be neglected. The angular momentum of a particle is defined in a similar way to the moment of a force. It is defined as the moment of its linear momentum, linear replacing the force. We may assume the angular momentum of a proton about point O to be conserved. Assume the mass of the proton as m_(P) and the charge on it as e. Given that the potential of the sphere increases with time and eventually reaches a constant velue. If the initial kinetic energy of a proton is 2.56 ke V, then the final potential of the sphere is

Answer» <html><body><p>`2.56 kV`<br/>`1.92 kV`<br/>greater than `2.56 kV`<br/>needs more information</p>Solution :From <a href="https://interviewquestions.tuteehub.com/tag/previous-592857" style="font-weight:bold;" target="_blank" title="Click to know more about PREVIOUS">PREVIOUS</a> we can see that the final potential <a href="https://interviewquestions.tuteehub.com/tag/energy-15288" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGY">ENERGY</a> of the <a href="https://interviewquestions.tuteehub.com/tag/sphere-1222094" style="font-weight:bold;" target="_blank" title="Click to know more about SPHERE">SPHERE</a> is equal to the `3//4` of initial <a href="https://interviewquestions.tuteehub.com/tag/kinetic-533291" style="font-weight:bold;" target="_blank" title="Click to know more about KINETIC">KINETIC</a> energy <br/> `rArr 3/4xx2.56=1.92 kV`</body></html>


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