A) \[\frac{R}{3}\]
B) \[\frac{R}{2}\]
C) \[\frac{R}{4}\]
D) \[\frac{R}{9}\]
Correct Answer: B
Solution :
Since mass of the object remains same \[\therefore \]Weight of object will be proportional to 'g' (acceleration due to gravity) Given \[\frac{{{W}_{earth}}}{{{W}_{planet}}}=\frac{9}{4}=\frac{{{g}_{earth}}}{{{g}_{planet}}}\] Also, \[g{{ & }_{surface}}=\frac{GM}{{{R}^{2}}}\](M is mass planet, G is universal gravitational constant, R is radius of planet) \[\therefore \]\[\frac{9}{4}=\frac{G{{M}_{earth}}R_{planet}^{2}}{G{{M}_{planet}}R_{earth}^{2}}=\frac{{{M}_{earth}}}{{{M}_{planet}}}\times \frac{R_{planet}^{2}}{R_{earth}^{2}}=9\frac{R_{planet}^{2}}{R_{earth}^{2}}\] \[\therefore \]\[R{{ & }_{planet}}=\frac{{{R}_{earth}}}{2}=\frac{R}{2}\]
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