list out some questions based on vectors , dot and cross product

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  

 

If you want to do questions for dot and cross product then there is special chapter in amit agrawal book .(Ch2).

Its a good book you can try from it.