A) 3V
B) 4V
C) 2V
D) 2.5V
Correct Answer: C
Solution :
If \[{{\omega }_{1}}\] and \[{{\omega }_{2}}\] be the angular speed of the cylinder and sphere after time t, then \[{{\omega }_{1}}={{\omega }_{0}}+{{\alpha }_{1}}t\] ... (iv) and \[{{\omega }_{2}}={{\omega }_{0}}+{{\alpha }_{2}}t\] \[={{\omega }_{0}}={{\omega }_{0}}+2.5{{\alpha }_{1}}t\] ... (v) where, \[{{\omega }_{0}}=\] initial angular speed \[\therefore \] From Eq. (iv) and (v), it is clear that \[{{\omega }_{2}}>{{\omega }_{1}}\] \[\therefore \] The sphere will acquire more angular speed as compared to that of the cylinder after a given time. In photoelectric effect, energy is conserved. In photoelectric effect, Einsteins equation is given by Photon energy = kinetic energy of electron + work function i.e. \[hv=e{{V}_{s}}+h{{v}_{0}}\] where, \[{{V}_{s}}\]is the stopping potential and \[{{v}_{0}}\] is the threshold frequency \[\therefore \] \[{{V}_{s}}=\frac{h}{e}(v-{{v}_{0}})\] ... (i) Given, \[{{v}_{0}}=3.3\times {{10}^{14}}Hz,\,v=8.2\times {{10}^{14}}Hz\] \[h=6.6\times {{10}^{-34}}J-s,e=1.6\times {{10}^{-19}}C\] Substituting these values in Eq. (i), we get \[{{V}_{s}}=\frac{6.6\times {{10}^{-34}}(8.2\,\,{{10}^{14}}-3.3\times {{10}^{14}})}{1.6\times {{10}^{-19}}}\] \[{{V}_{s}}=\frac{6.6\times 4.9}{1.6}\times {{10}^{-1}}=2\,\,V\] \[\Rightarrow \] \[{{V}_{s}}=2\,\,V\]
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