8 - Fluid at rest Questions Answers

Dear Sir kindly help provide the solution along with right approach to solve such questions 

Dear Sir kindly help provide the solution along with right approach to solve such questions 

Asked By: RK GUPTA
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Joshi sir comment
Dear Sir kindly help provide the solution along with right approach to solve such questions 
Asked By: RK GUPTA
is this question helpfull: 0 0 submit your answer
Joshi sir comment

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)

Asked By: PARTHSARTHY
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Joshi sir comment

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

Asked By: SARIKA
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Joshi sir comment

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?

Asked By: SARIKA
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Joshi sir comment

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 

Asked By: SARIKA SHARMA
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Joshi sir comment

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

Asked By: SARIKA SHARMA
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Joshi sir comment

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?

Asked By: SARIKA
is this question helpfull: 2 1 read solutions ( 1 ) | submit your answer
Joshi sir comment

since inside soap bubble pressure is given by the formula P0 + 4T/R and on pushing air  R will increase so pressure will decrease.

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