fluids revision

Q2 ka ans kya hai??

sorry for delayed response i didnt see ur question before[2][2][2]

sorry yaar if m0 is given only and not h then u can rite AρH = M0 .....

so ull get height in terms of given variables for the second Q (of Q1)solution above!!!!!!!!!!!!!

ya i know..... plz try 2nd question..

Q2 minimum ya maximum???????

σgLÎ R² = mg + ρgLÎ R² ......

σLÎ R² - ρLÎ R²= m ............

LÎ R²(σ-ρ) = m ..........................

this is maximum value!!!!!!!!!!

u want minimum ya maximum?????

@eureka ur approach and ans. for accn. are rite but ur way dosnt justify that a is a constant with time which is req. for the second part ........... just my view [1]

MINIMUM VALUE DEPENDS...........IF σ/ρ<=1 ITS 0!!!!!!!!!!!!!!

given in question that σ>ρ

asish(1) balance d force of gravity n buoyant force (2)apply centre of gravity s y coordinate above centre of mass coordinate solving u ll get d ans i think
did u try it???/

koi to solve karo ise...

@eureka ve u got d ans???

(in the fig right side it is.. mL
2(m+m₁)

No for stable equilibrium P should be above A..

x/2≥ml/2(m+m₁).. (i)

m is mass of rod.. m₁ is mass of point mass..

also xAσg=(m+m₁)g

x=(m+m₁)/Aσ

from (i) for minimum m₁..

x=ml/(m+m₁)

(m+m₁) = ml/(m+m₁)
Aσ

(m+m₁)=√mlAσ

m=Alρ

m₁=lA(√ρσ-ρ)
A=Ï€R²

so m₁=LÏ€R²(√ρσ-ρ).. ans

[1][2] par pink kyun nahi hota mera.. :(

;)

wat abt (ii) part? and Q2.?

find answer..... (s) both of them

Q2. A wooden stick of length L, radius R and density ρ has a small piece of metal of mass m (of negligible volume) attached to its one end. Find the minimum value for the mass m (in terms of given parameters) that would make the stick float vertically in equilibrium in a liquid of density σ (>ρ)

yes

[1][1][1][1][1].........

sankara.. acceleration is given as g/50