33/2rogH + 1/4(3/4)(1/5)rogL - 3/4(1/4)(1/5)(rogL) [Q4]
10 Answers
106Leave the first two questions..
Q3. What is the pressure at a point just below the bottom surface ot the block?
Q4. What is the pressure at the bottom most point of the fluid system ?
31/2rogH + 1/4(3/4)(1/5)rogL - 3/4(1/4)(1/5)(rogL) + (1/4)2roLg [Q3]
106Ans 4. was Patm + ÏgH/2+2ÏgH/2 + M(body)g/A(body)
Why this extra term??
Ive seen this answer in atleast two books.. (arihant and GC Agarwal)
Ans 3. Seems fine
106Another similar question.
Consider a system in which liquid of density d is there upto a height h and base area of container A. A small object of mass M and base area a is floating in the container. What is the pressure at the bottom of the fluid system?
106Really its got me confused...
for Q3. its explanation was that acc, to law of floatation, P=P0 + dgh/2 + 2dgh/2 because this balances the weight of the body.
for Q4. It found total weight on bottom surface divided by area.
Which method is correct??
Finding total weight and dividing by base area
or finding pressure
that was my first question..
For calculating pressure can we replace the block by respective fluids (according to the volumes immersed in either) and dfind out the pressure at the bottom?
1357For question 3
answer is P=Patm+Ïg(H/2)+2Ïg(L/4)
For question 4
answer is P=Patm+Ïg(H/2)+2Ïg(H/2)
The question for the given answer(given by asish) is
Suppose you have liquids of density of Ï and 2Ï and height H/2 and H/2. Now drop a block of lenght L and density 5Ï/4. Then the answer given by asish is correct.
106Thank you sir,
The confusion arose because it was prev iit ques in which it had asked to find out density of block (the lengths were given) which i calc to be 5d/4. For the pressure at bottom most point Arihant (31yrs) had given ans what i have written so had GC aggarwal and Aakash.
Teen teen sources se ek hi answer se main confuse ho gayaa..
Anyways thanks to u and eureka