Question

Jee main aspirants

answer given is A

solve..​

Answer 1

Answer :

$\large\bold\red{(d)\:(\frac{a}{3},\frac{a}{3})}$

Step-by-step Explanation:

Given,

3 rods of same mass 'm' and are placed to form a right angled triangle.

Therefore,

length of rods along X and Y axis will be a units and that along hypotenus will be √2a units.

Now,

we know that,

Centre of mass of a rod lies at the midpoint of its length.

Therefore,

the centre of mass for rod along X axis lies on

$\large\bold{(\frac{a}{2} ,0)}$

Similarly,

the centre of mass for rod along Y axis lies on

$\large\bold{(0,\frac{a}{2} )}$

Also,

Centre of mass for the rod along hypotenus lies on

$\large\bold{(\frac{a}{2} ,\frac{a}{2})}$

Now,

for the system,

$= > x_{com} = \frac{m( \frac{a}{2} ) + m(0) + m( \frac{a}{2} )}{m + m + m} \\ \\ = > x_{com} = \frac{am}{3m} \\ \\ = > x_{com} = \frac{a}{3}$

Similarly,

$= > y_{com} = \frac{m(0) + m( \frac{a}{2}) + m( \frac{a}{2} )}{m + m + m} \\ \\ = > y_{com} = \frac{am}{3m} \\ \\ = > y_{com} = \frac{a}{3}$

Hence,

The centre of mass for the system lies at,

$\large\bold{(d)\:(\frac{a}{3},\frac{a}{3})}$

Answer 2

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