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Physics

Mechanics

Easy

Question

A disc (of radius r cm) of uniform thickness and uniform density $sigma$ has a square hole with sides of length $ell equals fraction numerator r over denominator square root of 2 end fraction c m$ . One corner of the hole is located at the centre of the disc and centre of the hole lies on y-axis as shown. Then the y -coordinate of position of centre of mass of disc with hole (in cm) is

$- \frac{r}{2 (π - 1 / 4)}$
$- \frac{r}{4 (π - 1 / 4)}$
$- \frac{r}{4 (π - 1 / 2)}$
$- \frac{3 r}{4 (π - 1 / 4)}$

The correct answer is: $negative fraction numerator r over denominator 4 left parenthesis pi minus 1 divided by 2 right parenthesis end fraction$

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