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Physics

Mechanics

Easy

Question

$A space s q u a r e space l a m i n a space o f space s i d e space a space a n d space a space c i r c u l a r space l a m i n a space o f space d i a m e t e r space a space a r e space p l a c e d space t o u c h i n g space e a c h space o t h e r space a s space s h o w n space i n space f i g u r e. space F i n d space d i s tan c e space o f space t h e i r space c e n t r e space o f space m a s space f r o m space p o i n t space O comma space t h e space c e n t r e space o f space s q u a r e.$

$(\frac{π}{π + 4}) a$
$(\frac{2 π}{π + 4}) a$
$(\frac{π}{2 π + 2}) a$
$(\frac{π}{2 π + 4}) a$

The correct answer is: $open parentheses fraction numerator pi over denominator pi plus 4 end fraction close parentheses a$

$open parentheses fraction numerator pi over denominator pi plus 4 end fraction close parentheses a$

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Bodies of masses 4 Kg and 2 Kg are attached at the ends of a massless rod. The positions of 4Kg and 2 Kg are $left parenthesis 2 i with not stretchy hat on top plus 5 j with not stretchy hat on top right parenthesis m$ and $left parenthesis 4 i with not stretchy hat on top plus 2 j with not stretchy hat on top right parenthesis m$ . Then the position of centre of mass of the system is

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