KVPY SA Exam 2014 Question Paper

Question : 66 of 80

Marks: +1, -0

PHYSICS

A uniform square wooden sheet of side a has its center of mass located at point O as shown in the figure on the left. A square portion of side b of this sheet is cut out to produce an L-shaped sheet as shown in the figure on the right.

The center of mass of the L-shaped sheet lies at the point P (in the diagram) when

$a/b = (√5 -1)/2$
$a/b = (√5 + 1) /2$
$a/b = ( √3 - 1 ) / 2$
$a/b = (√ 3 + 1)/2$

Validate

Solution:

X_\cm ={m_1 x _1-m_2 x_2}/{m_1-m_2}

_1

is the mass of square wooden sheet of side a & m

_2

is the mass of removed square portion of side b.
x-coordinate of C.O.M. of remaining L-shaped sheet.

λ

is areal mass density ⇒

m_1 = λa^ 2 , m_2 = λb^2

X_\cm={λ(a^2)(a/2)-λ(b^2)(b/2)}/{λa^2-λb^2}

X_\cm=1/2({a^3-b^3}/{a^2-b^2})

similarly

Y_\cm =1/2[{a^3-b^3}/{a^2-b^2}]

centre of mass lies on point P (b, b)
⇒

X_\cm = b\ &\ Y_\cm = b

1/2[{a^2+b^2+ab}/{a+b}]=b

a^ 2 + b^2 + ab = 2ab + 2b^2

a^ 2 = ab + b^2

(a/b)^2=(a/b)+1

Let

x =a/b

x^ 2 -x-1 = 0

x = {1±√{1+4}}/2

x={√5+1}/2

a/b={√5+1}/2

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