Question

A uniform square plate and a disc have same mass per unit area are kept in contact as shown in figure. Diameter of disc is equal to side length of square. The position of centre of mass is a) At distance na from centre of square on line 4 joining centre of objects WE b) At distance from centre of square on line joining centre of objects c) At mid-point from centre of square with the (4- t)a d) At distance from centre of square on 2 line joining centre of objects​

Answer 1

$\large{\frak{\red{Answer}}}$

Let mass of square plate be M & mass of disc be m

${\bf{\frac{M}{a {}^{2}}{\implies{\orange{\frac{m}{Ta {}^{2} /4}}}}}}{\implies}{\purple{K}}$

$\bf{M}{\implies}{\orange{Ka {}^{2} }}$

$\bf{m}{\implies}{\orange{ \frac{πKa{}^{2}}{4}}}$

w.r.t centre of square

${\large{X}}cm{\implies}{\bf{\orange{ \frac{M(0) + m(a)}{M + m}}}}$

$\large{\implies{\orange{ \frac{( \frac{πKa {}^{2}}{4})a}{a {}^{2}(K + \frac{πK}{4})}}}}$

$\bf{\implies{\orange{ \frac{πKa}{4K + πK}}}}$

$\bf{\implies{\orange{ \frac{πa}{4 + π}}}}$

${\large{\implies{\orange{Y}}}}{\orange{cm}}\implies{0}$

$\pmb{\mathbb{\fbox{\red{Hope it's help you}}}}$