Punjab Medical Punjab - MET Solved Paper-2003

  • question_answer
    The ratio of electrostatic and gravitational forces acting between electron and proton separated by a distance\[5\times {{10}^{-11}}m\], will be: (Charge on electron\[=1.6\times {{10}^{-19}}C\], mass of electron\[=9.1\times {{10}^{-31}}kg\], mass of proton\[=1.6\times {{10}^{-27}}\,\,kg\],\[G=6.7\times {{10}^{-11}}N{{m}^{2}}/k{{g}^{2}}\])

    A) \[2.36\times {{10}^{39}}\]                           

    B) \[2.36\times {{10}^{40}}\]

    C) \[2.34\times {{10}^{41}}\]                           

    D) \[2.34\times {{10}^{42}}\]

    Correct Answer: A

    Solution :

    Gravitational force\[F=\frac{G{{M}_{1}}{{M}_{2}}}{{{r}^{2}}}\] \[{{F}_{G}}=\frac{6.7\times {{10}^{-11}}\times 9.1\times {{10}^{-31}}\times 1.6\times {{10}^{-27}}}{{{(5\times {{10}^{-11}})}^{2}}}\]                 \[=3.9\times {{10}^{-47}}N\] Electrostatic force\[{{F}_{E}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] \[{{F}_{e}}=\frac{9\times {{10}^{9}}\times 1.6\times {{10}^{-19}}\times 1.6\times {{10}^{-19}}}{{{(5\times {{10}^{-11}})}^{2}}}\] So,\[\frac{\text{Electrostatic}\,\,\text{force}}{\text{Gravitational}\,\,\text{force}}\text{=}\frac{{{\text{F}}_{\text{E}}}}{{{\text{F}}_{\text{G}}}}\]                                            \[=\frac{9.22\times {{10}^{-8}}}{3.9\times {{10}^{-47}}}\]                                            \[=2.36\times {{10}^{39}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner