A) WON
B) SOON
C) 200 N
D) 100 N
Correct Answer: A
Solution :
From the law of gravitation, the force of attraction acting on the body due to the earth is given by \[F=G\frac{{{M}_{e}}m}{R_{e}^{2}}\] where, \[{{M}_{e}}\]is mass of earth, \[{{R}_{e}}\]the radius, and m the mass of object. Also, acceleration due to gravity g in the body arises due to the force F. From Newton's second law, we have \[F=mg\] \[\Rightarrow \] \[F=mg=\frac{G{{M}_{e}}m}{R_{e}^{2}}\] \[\Rightarrow \] \[g=\frac{G{{M}_{e}}}{R_{e}^{2}}\] For planet, \[{{g}_{P}}=\frac{G{{M}_{p}}}{R_{p}^{2}}\] Given, \[{{R}_{p}}=\frac{{{R}_{e}}}{2},{{M}_{p}}=\frac{{{M}_{e}}}{7}\] \[\therefore \] \[{{g}_{p}}=\frac{g({{M}_{e}}/7)}{{{({{R}_{e}}/2)}^{2}}}\] ?(i) For earth, \[{{g}_{e}}=\frac{G{{M}_{e}}}{R_{e}^{2}}\] ?(ii) Dividing Eqs. (ii) by (i), wse have \[\frac{{{g}_{e}}}{{{g}_{p}}}=\frac{{{M}_{e}}}{\frac{1}{7}{{M}_{e}}}\times \frac{{{\left( \frac{1}{2}{{R}_{e}} \right)}^{2}}}{R_{e}^{2}}=\frac{7}{4}\] \[\Rightarrow \] \[{{g}_{p}}=\frac{4}{7}{{g}_{e}}\] Hence, weight of body \[=\frac{4}{7}\times 700=400\,\text{N}\]
You need to login to perform this action.
You will be redirected in
3 sec
