A) \[\text{2400Vc}{{\text{m}}^{\text{-1}}}\]
B) \[\text{600Vc}{{\text{m}}^{\text{-1}}}\]
C) \[\text{1200Vc}{{\text{m}}^{\text{-1}}}\]
D) \[\text{12000Vc}{{\text{m}}^{\text{-1}}}\]
Correct Answer: C
Solution :
Cathode rays are composed of electrons, when they move in electric field a force F = eE ..(i) Acts on them, this provides the necessary centripetal force to the particles. \[\therefore \]\[F=\frac{m{{v}^{2}}}{r}\] ?(ii) Form Eqs. (i) and (ii), we get \[eE=\frac{m{{v}^{2}}}{r}\] \[\Rightarrow \]\[r=\frac{m{{v}^{2}}}{eE}=\frac{m{{({{10}^{6}})}^{2}}}{e(300)}\] ?(iii) When velocity is doubled same circular path is followed, hence radius is same\[r=\frac{m{{(2\times {{10}^{6}})}^{2}}}{eE}\] Equating Eqs. (iii) and (iv), we get \[r=\frac{m\times {{({{10}^{6}})}^{2}}}{300e}=\frac{m\times {{(2\times {{10}^{6}})}^{2}}}{eE}\] \[\Rightarrow \]\[E=300\times 4=1200V\,c{{m}^{-1}}\]
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