A) \[\therefore \]
B) \[{{v}^{2}}=\frac{2eV}{m}\]
C) \[\Rightarrow \]
D) \[v=\sqrt{\frac{2eV}{m}}\]
Correct Answer: D
Solution :
When an electron of charge e and mass m is placed in a uniform electric field E, then force F external on the electron is \[(\Delta A)\] If the electron is left free it will perform accelerated motion Under the force F Acceleration \[\therefore \] Let velocity become v after travelling a distance x in the field Then, \[T=\frac{W}{\Delta A}\] For \[{{A}_{1}}=10\times 6\times {{10}^{-4}}=60\times {{10}^{-4}}{{m}^{2}}\], \[{{A}_{2}}=10\times 11\times {{10}^{-4}}=110\times {{10}^{-4}}{{m}^{2}}\] \[\Delta A={{A}_{2}}-{{A}_{1}}=50\times {{10}^{-4}}{{m}^{2}}\] For potential v, we have \[=2\times 50\times {{10}^{-4}}{{m}^{2}}\] \[=100\times {{10}^{-4}}{{m}^{2}}\] \[\therefore \] \[T=\frac{W}{\Delta A}=\frac{2\times {{10}^{-4}}}{100\times {{10}^{-4}}}=2\times {{10}^{-2}}N{{m}^{-1}}\] \[y=a\sin (lx-\omega t)\] Alternative: Kinetic energy of electron \[\omega \] \[y=a\sin (0.01x-2t)\] \[k=0.01\] \[\omega =2\] \[\frac{\omega }{k}=\frac{2}{0.01}=200cm/s\]
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