Question

pls solve this!!!!!!!​

Answer 1

Solution:-clearly,the rod of length L,will.move in a Circle of radius centred at Onand radius of.L

tension at the centre=mv^2/r

now here

v=rw=lW

where W is the angular velocity

putting this value we get

tension at o=m×l^2×W^2/l

=mlW^2

now let the length upto P be y

mass of length y=m/l ×y=my/l

tension at P=Tension at O due to length l-tension at O due to length y....i)

tension at O due to y length=my/l×y^2×W^2/y

=m(yW)^2/l

tension at P=mlW^2-m(yW)^2/l

also, tension at O=4×tension at P

mlW^2=4×mlW^2-4×m(yW)^2/l

(3mlW^2)/4=m×y^2×W^2/l

3l/4=y^2/l

y^2=3l^2/4

y=√3l/2

hence length will be √3l/2

{Note:-If we focus on equation i) ,tension at point p depends on the mass and speed of the length of rod,which is forward to it,,i.e from p to opening end

so,

tension created on P,by that part= tension created by total part-tension created by half part OP

after this as give in question,,I had eqaute tension at O by 4 times tension at P(which is given to make eqaul}

{hope it helps you}.

Answer 2

halloo

here's the explanation of your last question.

please tell which are the other questions I have to explain :-)

I have given the mechanism for friedel craft acylation/alkylation too.