Answer:
Initially, \[g=\frac{GM}{{{R}^{2}}}\] and \[T=2\pi \sqrt{\frac{l}{g}}=2s\] (For a second's pendulum) When M and R are doubled, \[g'=\frac{G(2M)}{{{(2R)}^{2}}}=\frac{g}{2}\] and \[T'=2\pi \sqrt{\frac{l}{g/2}}=\sqrt{2}T=2\sqrt{2}s.\]
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