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51.

A river `10 m` deep is flowing at `5ms^(-1)`. The shearing stress between horizontal layers of the rivers is (`eta=10^-(3) SI` units)A. `10^(-3)N//m^(2)`B. `0.8xx10^(-3)N//m^(2)`C. `0.5xx10^(-3)N//m^(2)`D. `1 N//m^(2)`

Answer» Correct Answer - C
Shearing stress`=(|"viscous force"|)/("area")`
`=(etaA(dv)/(dx))/A=eta(dv)/(dx)`
`10^(-3)xx5/10N/m^(2)=05.5xx10^(-3)N//m^(2)`
Discussion
52.

Three rods of uniform area of cross section `A =10^(-7)m^(2)` are arranged as shown in Fig Find out the shift in point `B, C` and `D`.

Answer» `/_L_(B)=/_L_(AB)=(FL)/(AL)=(MgL)/(AY)`
`=(10x10x0.1)/(10^(-7)xx2.5xx10^(10))=4xx10^(-3)m=4mm`
`/_L_(C)=/_L_(B)+/_L_(BC)`
`=4xx10^(-3)+(10xx0.2)/(10^(-7)xx4xx10^(10))`
`/_L_(C)=/_L_(B)+/_L_(BC)`
`=4xx10^(-3)+(100xx0.2).(10^(-7)x10^(-3)+5xx10^(-3))=9mm`
`=4xx10^(-3)+5xx10^(-3)=24mm`
`/_L_(D)=/_L_(C)+/_L_(CD)`
`=9xx10^(-3)+(100xx0.15)/(10^(-7)xx1xx10^(10))`
`=9xx10^(-3)+15xx10^(-3)=24mm`
Discussion
53.

Find out longitudinal stress and tangential stress on a fixed block.

Answer» Longitudinal or normal stress
`sigma_(1)=(100sin30^@)/(5xx2)=5N//m^(2)`
Tangential stress
`sigma_(1)=(100cos30^@)/(5xx2)=5sqrt(3)N//m^(2)`
Discussion
54.

A film of water is formed between two straight paralel wires each `10 cm` long and at seperation `0.5 cm`. Calculate the work requied to increase `1mm` distance between the wires. Surface tension of water `=72xx10^(-3)N//m`.

Answer» Initial surface area `=2xx"length"xx"separation"`
`=2xx10cmxx0.5cm=10cm^(2)=10xx10^(-4)m^(2)`
Final surface area `=2xx10cmxx(0.5+0.1)cm`
`=2xx10xx0.6cm^(2)=12xx10^(-4)m^(2)`
The required work
`WT/_A=72xx10^(-3)xx(12xx10^(-4)-10xx10^(-4))J`
`=72xx10^(-3)xx2xx10^(-4)=144xx10^(-7)J`
Discussion
55.

A film of water is formed between two straight parallel wires each `10 cm` long and at a seperation of `0.5 cm`. Calculate the work required to increase `1mm` distance between the wires. Surface tension of water `=72xx10^(-3)N//m`.A. `1.44xx10^(-5)xx10^(-5)J`B. `1.72xx10^(-5)J`C. `1.44xx10^(-4)J`D. `1.72xx10^(-4)J`

Answer» Correct Answer - A
As we increase the separation between the wire by `1mm`, the increase in surface area of the film is
`/_A=2xx0.1xx10^(-3)=2xx10^(-4)m^(2)`
(as there are two surfaces of film)
So, increase in suface energy `/_U=S/_A=1.44xx10^(-5)J`
Work done`=/_J`
Discussion
56.

A solid sphere of radius R made of a material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless pistion of area A floats on the surface of the liquid. When a mass M is placed on the piston to compress the liquid the fractional change in the radius of the sphere, `deltaR//R`, is .............A. `(Mg)/(AK)`B. `(Mg)/(3AK)`C. `(3Mg)/(AK)`D. `(Mg)/(2AK)`

Answer» Correct Answer - B
Change in pressure due to placing of mass of prism is `/_p=(Mg)/A`
From bulk modulus definition `K=(-dp)/(dV//V)`
`|(dV)/V|=(/_P)/K=(Mg)/(AK)`
From `V=4/3pir^(3)`
`(dV)/V=(3dR)/Rimplies(dR)/R=1/3(dV)/V=(Mg)/(3AK)`
Discussion
57.

A straw `6 cm` long floats on water. The water film on one side has surface tension of `50 dyn//cm`. On the other slide, camphor reduces the surface tension to `40 dyn//cm`. The resultant force acting on the straw isA. `(50xx6-40xx6)dyn`B. `10 dyn`C. `(50/6-40/6)dyn`D. `90dyn`

Answer» Correct Answer - A
`F=(sigma_(1)-sigma_(2))l`
Discussion
58.

A water drop is divided into eight equal droplets. The pressure difference between inner and outer sides of big dropA. will be the same as for smaller dropletB. will be half of that for smaller dropletC. will be one-fourth of that for smaller dropletD. will be twice of that for smaller droplet

Answer» Correct Answer - B
Suppose `R=`radius of water drop
`:. 4/3piR^(3)=8x4/3pir^(3)`
(since volume remains constant)
`:. r=R/2`
since excess pressure inside drop `=(2T)/R`
(T- surface tension, `R` -radius)
Therefore, pressure difference between inner and outer surface of bid drop will be half of that for smaller droplet.
Discussion