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1.

A body weights 75 gf in air, when completely immersed in a liquid and 67 gf when completely immersed in water. Find the density of the liquid

Answer» Given, weight of the body in air, `w_(1)=75" gf"`
Weight of the body in water, `w_(2)=67` gf
Weight of the body in the given liquid, `w_(3)=51` gf
`:.` Apparent loss of weight of the body in the given liquid `=w_(1)-w_(3)=75-51=24` gf
Apparent loss of weight of the body in water `w_(1)-w_(2)-75-67=8` gf.
Relative density of the liquid `=(w_(1)-w_(3))/(w_(1)-w_(2))=(24)/(8)=3`
`:.` Density of the liquid = 3 gf `cm^(-3)`
Discussion
2.

When equal quantities of an oil, water mercury are poured into a beaker, the order in which the liquids arrange themselves from bottom to top isA. mercury, water, oilB. water, mercury, oilC. water, oil, mercury.D. mercury, oil water.

Answer» Correct Answer - A
`p_("oil")ltp_("water")ltp_(Hg)`.
Thus
oil `to` top layer
water `to` middle layer
mercury `to` bottom layer
Discussion
3.

An object floats in three immiscible liquids A, B and C of densities `3"g cm"^(-3),2"g cm"^(-3)`, respectively as shown in the figure. When the object is placed in the liquids, the levels of liquid A, B and C rise by 3 cm, 5 cm and 8 cm, respectively. The areas of cross-sections of the container and the object are 10 `cm^(2)and5cm^(2)`, respectively. Calculate the density of the object.

Answer» Weight of the floating body = Weight of the liquids A,B and C displaced.
Volume of the body immersed in a liquid = (Rise in level) `xx` (area of cross-section of the container)
`V_(A)=(3cm)xx(10cm^(2))`
`V_(A)+V_(B)=(5cm)xx(10cm^(2))`
`V_(A)+V_(B)+V_(C)=(8cm)xx(10cm)`= Total volume of the body
Weight of A displaced `=V_(A)xxd_(A)`
Similarly, determine the weights of liquids B and C displaced.
Determine the density from the definition,
`"Density"=("mass")/("volume")`
Discussion
4.

A container is filled with two immiscible liquids A and B of densities `"2 g cm"^(-3) and "3 g cm"^(-3)` , respectively. A wooden cube of side 1 cm floats on the surface of liquid A such that one-fourth of its total length is immersed in this liquid (A). Now, the wooden cube is completely immersed in liquid A by suspending a sinker of volume `"10 cm"^(3)` which is completely submerged in liquid B. Determine the weight of the sinker.

Answer» Apparent weight of the sinker = extra upthrust.
Discussion
5.

In a mercury barometer, if the tube containing mercury is tilted, thenA. cury is height of mercury column remains same.B. the length of mercury column in the tube increses.C. the vertical height of the mercury column decreases.D. Both (a) and (b)

Answer» Correct Answer - D
When a mercury barometer is tilted, then the vertical height remains constant but length of mercury column increases.
Discussion
6.

A small thin cylindrical glass beaker having an area of cross section `"6 cm"^(2)` weighs 10 gwt. If floats vertically in water upto a certain depth with 10 lead shots in it, is 8 cm, how many such lead shots can be added further into it before it sinks.

Answer» Mass of a beaker, `m_(1)=10g`
The no. of lead shots = 10
The mass of each lead shot = 2 g
`:.` the mass of total lead shots = 20 g
`:.` the total mass of the beaker and the lead shots = 30 g
`:.` the volume of water displaced by the beaker with 10 lead shots in it = `30cm^(3)`
area of cross section of the beaker, `a=6cm^(2)`
total length of the beaker, l = 8 cm
`:.` total volume of the beaker `6xx8=48 cm^(3)`
`:.` volume of the beaker above the water surface `=48-30=18cm^(3)`
if the beaker has to sink in water, the total beaker should be immersed in water.
`rArr18cm^(3)` of water has to be displaced further
`rArr18` gwt force should be further added
`:.` number of lead shots that has to be added further
`(18gwt)/(2gwt)=9`
Discussion