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Three identical rigid circular cylinders A, B and C are arranged on smooth inclined surfaces as shown in figure. Find the least value of theta that prevent the arrangement from collapse. PLS EXPLAIN...!!!

Three identical rigid circular cylinders A, B and C are arranged on smooth inclined surfaces as shown in figure. Find the least value of theta that prevent the arrangement from collapse.


PLS EXPLAIN...!!!


1724_78139_cylinder.jpg

Grade:10

2 Answers

Baldeep Singh Chhabra
11 Points
6 years ago
If the arrangement is in equilibrium, the net force on each cylinder is zero. The arrangement is symmetrical so the forces on B are the same as those on A. There is no friction between the cylinders and the inclined plane, so the reaction from the plane on A is purely normal, directed through its centre. I assume also that there is no friction between the cylinders, so the forces between cylinders are also normal - ie directed through their centres. If the arrangement of cylinders is on the point of collapsing, then A and B are on the point of separating, so the force between A and B will be zero. Suppose the weight of each cylinder is W, the radius of each is r and the angle between the centres ACB is 2β - to allow for the case in which A and B are separated. The weight of C is supported by contact forces N from A and B. So 2Ncosβ = W. The normal reaction on A from the inclined plane is R and that between A and B (if in contact) is S. So the vertical and horizontal components of force on A are : Rcosθ - Ncosβ - W = 0 ... (i) Rsinθ - Nsinβ - S = 0 ... (ii). Since Ncosβ = W/2 then Nsinβ = (W/2)tanβ. Substituting in (i) Rcosθ = (3W/2) hence Rsinθ = (3W/2)tanθ. Substituting in (ii) (3W/2)tanθ = (W/2)tanβ+S. If A and B are on the point of separating then S = 0. Hence 3tanθ = tanβ. At this point the angle 2β = 60 degrees, β = 30 degrees, so tanβ = 1/√3. Therefore tanθ = 1/3√3 ANSWER: we require θ > 10.9 degrees to keep the arrangement from collapsing. If A and B have already separated so that 30
hmhm
16 Points
2 years ago
ii 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  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  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  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 mgThree 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 mgThree 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 mgThree 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 mgThree 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 mgThree 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 mgThree 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 mgThree 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 mgThree 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 mgThree 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 mgThree 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 mgThree 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 mgThree 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 mgThree 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  

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