Apiece of thin uniform wire of mass m and length 3b is bent into an equilaeral triangle.

find the moment of inertia of the wire triangle about an axis perpendicular to the plane of the triangle and passing through one of its vertices.

a) 7/12 mb^2

b)1/3 mb^2

c) 1/2 mb^2

d) 2/3 mb^2

e) 7/4 mb^2

Option (C) is the correct answer.

Explanation:

Formula for moment of inertia is as follows.

                 M.I =

Hence, moment of inertia of two rods is as follows.

      M.I of two rods =

                                =    

As third rod have no connection with vertices. So, moment of inertia of a rod along an axis passing through its center is as follows.

           M.I =

                 =

                 =

Using parallel axis theorem moment of inertia through vertices is as follows.

                  =

                h =

Now, we will calculate the moment of inertia of third rod about vertices is as follows.

            =

            = 5

Therefore, total moment of inertia is calculated as follows.

            Total M.I =

                            =

Thus, we can conclude that the moment of inertia of the wire triangle about an axis perpendicular to the plane of the triangle and passing through one of its vertices is .

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .