A circular disc of radius R is removed from a bigger circular disc of radius 2R, such that the circumferences of the discs coincide. The centre of mass of the new disc is α/R from the centre of the bigger disc. The value of α is

Question

A circular disc of radius R is removed from a bigger circular disc of radius 2R, such that the circumferences of the discs coincide. The centre of mass of the new disc is α/R from the centre of the bigger disc. The value of α is (A) (1/4) (B) (1/3) (C) (1/2) (D) (1/6)

Tardigrade Admin · Accepted Answer

In figure,, O is the centre of circular disc of radius

2 R

and mass

M . C_{1}

is centre of disc of radius R, which is removed. If

σ

is mass per unit area of disc, then

M = π (2 R)^{2} σ

Mass of disc removed,

M_{1} = π R^{2} σ = \frac{1}{4} M

Mass of remaining disc,

M_{2} = M - M_{1}

= M - \frac{1}{4} M = \frac{3}{4} M

Let centre of mass of remining disc be at

C_{2}

where

O C_{2} = x

As

M_{1} \times O C_{1} = M_{2} \times O C_{2}

∴ \frac{M}{4} R = \frac{3 M}{4} x

x = \frac{R}{3} = α R ∴ α = \frac{1}{3}

Q. A circular disc of radius $R$ is removed from a bigger circular disc of radius $2 R$ , such that the circumferences of the discs coincide. The centre of mass of the new disc is $α / R$ from the centre of the bigger disc. The value of $α$ is

$\frac{1}{4}$

$\frac{1}{3}$

$\frac{1}{2}$

$\frac{1}{6}$

Q. A circular disc of radius R is removed from a bigger circular disc of radius 2R, such that the circumferences of the discs coincide. The centre of mass of the new disc is α/R from the centre of the bigger disc. The value of α is

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31​

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Q. A circular disc of radius $R$ is removed from a bigger circular disc of radius $2 R$ , such that the circumferences of the discs coincide. The centre of mass of the new disc is $α / R$ from the centre of the bigger disc. The value of $α$ is

$\frac{1}{4}$

$\frac{1}{3}$

$\frac{1}{2}$

$\frac{1}{6}$