In the arrangement shown in the figure if the blocks of masses m and 2m are released from the state of rest tension in the string is (`mu` = coefficient of friction string is massless and inextensible pulley is frictionless )

Options

A. mg
B. `sqrt(2)` mg
C. `(2sqrt(2)mg)/(3)`
D. `(sqrt(2)mg)/(3)`

Answer

C

Explanation

Let `N_(1),N_(2)` are normal reaction force and `f_(1),f_(2)` are the friction force on two blocks. Acceleration is a and tension is T .
The respective FBD

Friction force `f_(1)=muN_(1), f_(2) = muN_(2)`
AS `N_(1)= "mg cos" 45^(@), f_(1)= (2)/(3) . "mg cos"45^(@)= (sqrt(2))/(3)` mg
and `N_(2)= 2mg cos 45^(@), f_(2)= (2)/(3).2mgcos 45^(@)= (2sqrt(2))/(3)` mg
`implies ` Now by second law of motion for 2m mass 2mg cos` 45^(@)- T -f_(2) =` 2ma
`sqrt(2),mg -T -(2sqrt(2))/(3) mg = 2ma " " cdots(i)`
For `m_(1), T-f_(1)-"mg cos"45^(@)= ma `
`T - (sqrt(2))/(3) mg - (mg)/(sqrt(2)) = ma " " cdots(ii)`
Multiplying by 2 in Eq . (ii) we get
`2T-(2sqrt(2))/(3) mg - sqrt(2)mg = 2ma " " cdots (iii)`
Subtracting Eq. (iii) from Eq. (i) we get
`3T -2sqrt(2) mg =0 implies T = (2sqrt(2mg))/(2)`