# When two soap of different diameter are connected by a tube the small bubbles gradual collapse and large bubbles grow bigger

hmhm
16 Points
2 years ago

Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg
hmhm
16 Points
2 years ago
Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg