Recent content by Adi6677

Ah thanks finally got the answer it's quite embarassing to make soo many algebraic mistakes thanks a lot for ur time 😊

Ah sorry for that my bad but I reattempted that but still didn't got that ##17## Can u please tell what I did wrong?

Hi so here is a photo of the attempt to the question hope u get what I wrote its a little messy

Area of the top came to be equal to ##4A##

So I tried the question and reattempted the question 2-3 times but the answer I got still wasn't correct. after following ur advice the final answer I got was $$ V' = √(6gh)/√(1-16a^2/A^2)$$

So I tried the question and reattempted the question 2-3 times but the answer I got still wasn't correct. after following ur advice the final answer I got was $$ V' = √(6gh)/√(1-16a^2/A^2)$$

It's just a hypothetical case to help one get used to applying Bernoulli I guess it is a sample paper of an exam conducted after high school so I don't think much dept is given in it

Okay I tried but I don't know what's the relationship between them is it something to do with similarity ??

K thanks but the area of the opening of cone isn't given can it be found out in terms of A ??

My approach was that I consider the pressure of cross section A first Pa= P + dgh Now by writting Bernoulli's Equation between the cross-section A and the opening : $$ Pa + 2dgh + 2dV^2/2 = Pb + 2dV'^2/2$$ Where Pb is the pressure of the opening which is equal to the atmospheric pressure...