A) 1 : 2 : 8
B) 1 : 2 : 4
C) 1 : 2 : 2
D) 1 : 1 : 2
Correct Answer: A
Solution :
Force acting on charge particle in normal direction of motion. \[F=Eq\] Acceleration \[a=\frac{F}{m}=\frac{Eq}{m}\] If particle travels a distance \[d\] in time\[t\], then \[t=\frac{d}{v}\] Motion in normal direction \[y=\frac{1}{2}a{{t}^{2}}\] \[y=\frac{1}{2}\times \frac{Eq}{m}\times \frac{{{d}^{2}}}{{{v}^{2}}}\] But \[p=mv\] \[\therefore \] \[y=\frac{Eq{{d}^{2}}\cdot m}{2{{p}^{2}}}\] \[\therefore \] \[y\propto \frac{qm}{{{p}^{2}}}\] \[\therefore \] \[{{y}_{p}}:{{y}_{d}}:{{y}_{\alpha }}=\frac{{{q}_{p}}{{m}_{p}}}{p_{p}^{2}}=\frac{{{q}_{d}}{{m}_{d}}}{p_{d}^{2}}=\frac{{{q}_{\alpha }}{{m}_{\alpha }}}{p_{\alpha }^{2}}\] Given, \[{{p}_{p}}={{p}_{d}}={{p}_{\alpha }}\] \[{{q}_{p}}=e,\] \[{{q}_{d}}=e,\] \[{{q}_{\alpha }}=2e\] \[{{m}_{p}}=m,\] \[{{m}_{d}}=2m,\] \[{{m}_{\alpha }}=4m\] \[\therefore \] \[{{y}_{p}}:{{y}_{d}}:{{y}_{\alpha }}=e\times m:e\times 2m:2e\times 4m\] \[=1:2:8\]
You need to login to perform this action.
You will be redirected in
3 sec
