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Saturday, 31 October

JEE Main 2022 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 25 June 2022, Shift 2, Physics Q54

Question 54 of 90 in this shift, Physics question 24 of 30, Section B.

Moment of Inertia (M.I.) of four bodies having same mass ‘MM’ and radius ‘2R2R’ are as follows : I1\mathrm{I}_1 = M.I. of solid sphere about its diameter I2\mathrm{I}_2 = M.I. of solid cylinder about its axis I3\mathrm{I}_3 = M.I. of solid circular disc about its diameter I4\mathrm{I}_4= M.I. of thin circular ring about its diameter If 2(I2+I3)+I4=x⋅I12(\mathrm{I}_2 + \mathrm{I}_3) + \mathrm{I}_4 = x \cdot \mathrm{I}_1 then the value of xx will be ____________.

Official answer

5

NTA final key.