In the figure shown ABC is a uniform wire. If centre of mass of wire lies vertically below point A, then `(BC)/(AB)` is close to :

Options

A. `1.85`
B. `1.37`
C. `1.5`
D. 3

Answer

B

Explanation

Taking origin at point B and X axis along BC and Y axis perpendicular to it. Let AB = y and BC = x
Thus, the coordinates of points A, B and Care `(ycos60^(@),ysin60^(@)),(0,0)` and `(xcos60^(@),0)` respectively.
Or, `A,((y)/(2),(ysqrt3)/(2)),B:(0,0),C:(x,0)`
As ABC is a uniform wire, thus the individual centres of mass for the parts AB and BC lie at their respective geometric centres `O_(1)` and `O_(2)`, respectively as shown in the figure.
Their coordinates are:
`O_(1):((y)/(4),(ysqrt3)/(4)),O_(2):((x)/(2),0)`
As it is given that the centre of mass of the wire ABC lies just below point A, that means the X-coordinate of the centre of mass must be same as that of the point A.
Let the mass per unit length for the wire ABC bek. Thus, the masses of the parts AB and BC are ky and kx, respectively.
The X-coordinate of the centre of mass of the wire ABC is given by:
Using the formula:
`X_(cm)=((m_(1)x_(1))+(m_(2)x_(2)))/(m_(1)+m_(2))`
Here, `X_(cm)=(y)/(2),x_(1)=(y)/(4)andx_(2)=(x)/(2),m_(1)=kyandm_(2)=kx`
`(y)/(2)=((ky)((y)/(4))+(kx)((x)/(2)))/(ky+kx)`
`(y)/(2)=((y)((y)/(4))+(x)((x)/(2)))/(y+x)`
`(y)/(2)=(((y^(2))/(4))+((x^(2))/(2)))/(y+x)`
`(y)/(2)=(y^(2)+2x^(2))/(4(y+x))`
`2y^(2)+2xy=y^(2)+2x^(2)`
`y^(2)+2xy-2x^(2)=0" "......(i)`
Dividing both sides of equation (i) by `y^(2)`
`1+2(x)/(y)-2(x^(2))/(y^(2))=0`
`2(x^(2))/(y^(2))-2(x)/(y)-1=0`
Taking `(x)/(y)=z`
`2z^(2)-2z-1=0`
Applying quadratic formula
`z=(2pmsqrt((-2)^(2)-4(2)(-1)))/(2(2))`
`z=(2pmsqrt(4+8))/(4)`
Taking only positive sing
`z=(2+sqrt12)/(4)=(2+2sqrt3)/(4)=1.365`
`z=(x)/(y)=(BC)/(AB)~~1.37`