8 - Fluid at rest Questions Answers
Dear Sir kindly help provide the solution along with right approach to solve such questions
On a horizontal floor, two identical cylindrical vessels A and B connected by a thin tube near their bottoms contain some water. When an ice block of volume 100cm^3 is gently put inside the vessel A, it gets half submerged in the water. How much mass of water will flow through the connecting tube during the process of melting of the ice cube? (Density of water=1000kg/m^3 and Density of ice=900kg/m^3)
WHEN TWO SUBSTANCES OF DENSITYρ1 & ρ2 MIXED IN EQUAL VOLUME THEN RELATIVE DENSITY OF MIXTURE IS 4 , WHEN THEY R MIXED IN EQUAL MASSES THEN RELATIVE DENSITY =3. THEN ρ1 X ρ 2 =?
ANS 12
in first case 4 = [ρ1V + ρ2V]/2V
and 3 = [m+m]/ [m/ρ1 + m/ρ2]
solve now
A BALL FLOATS ON D SURFACE OF H2O IN A CONTAINER EXPOSED TO ATMOSPHERE . IF D CONTAINER IS COVERED & AIR IS COMPRESSED , D BALL WILL RISE . WHY?
beacause due to compression density of air will increase and upthrust by air will make some contribution in the net upthrust so some extra part of the body will come outside water.
my previous ques from fluid topic is not solved yet so pls reply that ques & that of chem of some other one
very soon you will get the answers
A SUBMERGED BLOCK RISES BY 2cm WHEN MASS OF 10kg IS REMOVED ( BE4 THAT 10kg MASS WAS KEPT ON IT)
WHAT WILL BE THE VOLUME OF THE BLOCK
According to the given conditions
in first case block is completely inside water by putting 10 kg in it means density of block will be less than that of water
according to the law of floatation we know that
wt. of solid = upthrust
so 10g + AL(0.01)Dg = AL(0.01)dg
in second case
AL(0.01)Dg = A[L-2](0.01)dg
substracting 2nd eq. by 1st eq. we will get A = 1/2
now initial lenght in start is not provided by you, if it is given then calculate V = AL(0.01)
In thid question d and D are densities of liquid and solid, A and L are area of bottom of block and its length in cm. 0.01 is for converting cm. into m.
if more air is pushed in soap bubble the pressure in it decreases WHY?
since inside soap bubble pressure is given by the formula P0 + 4T/R and on pushing air R will increase so pressure will decrease.