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A square hole is punched out of a circul...

A square hole is punched out of a circular lamina, the diagonal of the square being the radius of the circle. If `'a'` be the diameter of the sircle, find the distance of centre of mass of the remainder from the centre of the circle.

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To find the distance of the center of mass of the remaining part of the circular lamina after a square hole is punched out, we can follow these steps: ### Step 1: Define the Problem We have a circular lamina with a diameter \( a \). The radius \( r \) of the circle is given by: \[ r = \frac{a}{2} \] A square hole is punched out from this circular lamina, and the diagonal of the square is equal to the radius of the circle. ...
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Knowledge Check

  • A square hole is punched out of a circular lamina of diameter a. The diagonal of the square is equal to the radius of the circle and one of the corners is at the centre of the circle . The distance of the C.G. of the remainder from the centre of circle is

    A
    `(a)/(8pi+4)`
    B
    `(a)/(8pi-4)`
    C
    `(a)/(5pi-4)`
    D
    `(a)/(5pi+4)`
  • A square hole is punched out of a circular lamina of diameter 4 cms., the diagonal of the square being a radius of the circle. Centroid of the remainder from the centre of the circle is at a distance

    A
    `(1)/(2pi+1)`
    B
    `(1)/(2pi-1)`
    C
    `(1)/(pi+1)`
    D
    `(1)/(pi-1)`
  • From a circle of radius a, an isosceles right angles triangle with the hypotenuse as the diameter of the circle is removed. The distance of the centre of mass of the remaining portion from the centre of the circle is

    A
    `3(pi-1)a`
    B
    `((pi-1)a)/(6)`
    C
    `(a)/(3(pi-1))`
    D
    `(a)/(3(pi+1))`
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