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(B) The orbital radius of the satellite is $r = R + h = R + \frac{R}{3} = \frac{4R}{3}$.The orbital velocity $v_0$ is given by $v_0 = \sqrt{\frac{GM}{

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$3$

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(B) The orbital radius of the satellite is $r = R + h = R + \frac{R}{3} = \frac{4R}{3}$.The orbital velocity $v_0$ is given by $v_0 = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{4R/3}} = \sqrt{\frac{3GM}{4R}}$.The angular momentum $L$ of the satellite is given by $L = m v_0 r$,where $m = \frac{M}{2}$ is the mass of the satellite.Substituting the values,we get:$L = \left(\frac{M}{2}\right) \cdot \sqrt{\frac{3GM}{4R}} \cdot \left(\frac{4R}{3}\right)$$L = \frac{M}{2} \cdot \sqrt{\frac{3GM}{4R}} \cdot \sqrt{\frac{16R^2}{9}}$$L = \frac{M}{2} \cdot \sqrt{\frac{3GM}{4R} \cdot \frac{16R^2}{9}}$$L = \frac{M}{2} \cdot \sqrt{\frac{4GMR}{3}}$$L = M \cdot \sqrt{\frac{4GMR}{3 \cdot 4}} = M \sqrt{\frac{GMR}{3}}$.Comparing this with the given expression $M \sqrt{\frac{GMR}{x}}$,we find $x = 3$.

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