Learn

Saturday, 31 October

JEE Main 2021 · PhysicsMultiple choiceSingle correctMediumConceptual

JEE Main 17 March 2021, Shift 1, Physics Q15

Question 15 of 90 in this shift, Physics question 15 of 30, Section A.

Two ideal polyatomic gases at temperatures T1T_1 and T2T_2 are mixed so that there is no loss of energy. If F1F_1 and F2F_2, m1m_1 and m2m_2, n1n_1 and n2n_2 be the degrees of freedom, masses, number of molecules of the first and second gas respectively, the temperature of mixture of these two gases is :
  1. (1)n1F1T1+n2F2T2n1+n2\frac{n_1F_1T_1+n_2F_2T_2}{n_1+n_2}
  2. (2)n1F1T1+n2F2T2F1+F2\frac{n_1F_1T_1+n_2F_2T_2}{F_1+F_2}
  3. (3)n1F1T1+n2F2T2n1F1+n2F2\frac{n_1F_1T_1+n_2F_2T_2}{n_1F_1+n_2F_2}Official answer
  4. (4)n1T1+n2T2n1+n2\frac{n_1T_1+n_2T_2}{n_1+n_2}

Official answer

Option 3

NTA final key.