The osmotic pressure of 2.22%(w/v) solution of CaCl2 at 300 K is

2.22/100×8.314 ×300=55.37124 (approximate) Where 22.22/100 is concentration and R,is gas constant 300 is given temp

Dear Student,
Please find below the solution to your problem.

Let the total angle twisted by A and B is (not mentioned in question) and twist angle made by A and B will beθ1andθ2respectively. henceθ=θ1+θ2.......1>(totalangletwistediscontributionofindividualangulartwist) now the force applied produce the torque on B and a restoring torque is produce in A to counter torque of B this continuous till both the rods does not archives the condition of equilibrium. using the formula of torque in both cylinder A and B
Torque clockwose by B = torque anticlockwise by A
n* pi* (2r)4 theta2/ 2L = n * pi* (r)4 theta1/ 2L
Theta1 = 16 * theta2
Hence
Theta2 + 16 theta2 = theta
Theta2 = theta/17
And
Theta1 = 16 theta/17

Thanks and Regards