8 - Fluid at rest Questions Answers

1.)Why only mercury is used in barometers?

2.)Is pressure scalar or vector? How?

Asked By: PRATYUSH SINHA 3 year ago
A uniform rod AB, 4m long and weighing 12 kg, is supported at end A, with a 6kg lead weight at B. The rod floats as shown in the figure with one-half of its length submerged. The buoyant force on the lead mass is negligible as it is of negligible volume. Find the tension in the cord and total volume of the rod.
Asked By: P. NAVIN 3 year ago
A uniform rod AB, 4m long and weighing 12 kg, is supported at end A, with a 6kg lead weight at B. The rod floats as shown in the figure with one-half of its length submerged. The buoyant force on the lead mass is negligible as it is of negligible volume. Find the tension in the cord and total volume of the rod.
Asked By: P. NAVIN 3 year ago

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 7 year ago

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 7 year ago

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 7 year ago

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 7 year ago

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 7 year ago