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← JEE Advanced 2021 · Paper 2
PhysicsInteger+4Question 18 of 57

JEE Advanced 2021 · Paper 2, Physics question 18

The distance between two stars of masses 3MS3M_S and 6MS6M_S is 9R9R. Here RR is the mean distance between the centers of the Earth and the Sun, and MSM_S is the mass of the Sun. The two stars orbit around their common center of mass in circular orbits with period nTnT, where TT is the period of Earth's revolution around the Sun.

The value of nn is ____.

Answer and worked solution

Official answer: 9

Kepler's third law for a binary. The period depends on the total mass and the separation:
Tb2=4π2d3G(m1+m2).T_b^2=\frac{4\pi^2d^3}{G(m_1+m_2)}.

Compare with the Earth. For the Earth, T2=4π2R3GMST^2=\dfrac{4\pi^2R^3}{GM_S}. With d=9Rd=9R and total mass 9MS9M_S:
Tb2T2=939=81  ⟹  Tb=9T.\frac{T_b^2}{T^2}=\frac{9^3}9=81\implies T_b=9T.
So n=9n=9.