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1.

One end of a wire 2m long and 0.2 cm^2 in cross section is fixed in a ceilign and a load of 4.8 kg is attached to the free end. Find the extension of the wire. Young modulus of steel `=2.0xx10^11Nm^-2`. Take `g=10ms^-2`.

Answer» We have
`y=stress/strain=(T/A)/(l/L)`
with symbols having their usual meanings. The extension is
`l=(TL)/(AY)`
As the load is in equilibrium after the extension, the tension in the wire is equal to the weight of the load
`=4.8kgxx10ms^-2=48N`
Thus, `l=((48N)(2m))/((0.2xx10^-4m^2)xx(2.0xx10^11Nm^-2))`
`=2.4xx10^-5m`
Discussion
2.

Two blocks of masses 1 kg and 2kg are connect by a metal wire going over a smooth pulley as shown in figure. The breaking stress of the metal is `2xx10^9m^-2`. What should be the minmum radius of the wire used if it is not to break? Take `g=10ms^-2`

Answer» The stress in the wire `=Tension/(Area of cross section)`
To avoid breaking, this stress should not exceed the breaking stress.
Let the tension in the wire be T. The equations of motion of the two blocks are
`T-10N=(1kg)a`
`and 20N-T=(2kg)a`
ElimiN/Ating a from these equations
`T=(40/3)N`
`The stress=((40/3)N)/(pir^2)`
If the minimum radius needed to avoid breaking is r, `2xx10^9 N/m^2=((40/3)N)/(pir^2)`
`solving this
`r=4.6xx10^-5m`.
Discussion
3.

A vertical metal cylinder of radius 2 cm and length 2 m is fixed art the lower end and a load of 10 kg is put on it. find a. the stress b. the strain and c. the compression of the cylinder. Young modulus of the metl `=2xx10^11Nm^-2`

Answer» Correct Answer - A::B::C::D
`r=2cm`
`Apir^2=4picm^2`
`=4pixx10^-4m^2`
`L=2m`
`m=100kg`
`Y=2xx10^11N/m^2`
a. `rho=stress=(mg)/A`
`=((100xx10))/((4pixx10^-4))` ltbr `=7.96xx10^5N/m^2`
b. c=strain
`=rho/Y=((97.96xx10^5))/((2xx10^11))`
`=4xx10^-6`
c. Compression `Delta L=eL`
`=4xx10^-6xx2=8x10^-6m`
Discussion
4.

A load of 4.0 kg is suspended from a ceiling through a steel wire of radius 2.0 mm. find the tensile stress developed in the wire when equilibrium is achieved. Take `g=3.1pi m/s^-2`

Answer» tension in the wire is
`F=4.0x3.1piN`
`The area of cross section is
`A=pir^2=pix(2.0xx10^-2m)^2`
`=4.0pixx10^-6m^2`
Thus, the tensile stress developed.
`=F/A=(4.0xx3.1pi)/(4.0pixx10^-6)Nm^-2`
`=3.1xx10^6Nm^-2`
Discussion
5.

A steel wire of length 2.0 m/s is stretched through 2.0 mm. The cross sectional area of the wire is 4.0 mm^2. Calculate the elastic potential energy stored in the wire in the stretched condition. Young modulus of steel `=2.0x10^11Nm^-2`

Answer» The strain in the wire `(/_l)/l=(2.0mm)/(2.0m)=10^-3`
The stress in the wire `=YxxStrain`
`=2.0xx10^1Nm^-2xx10^-3=2.0xx10^8Nm^-2`
The volume of the wire `=(4xx10^-6m^2)xx(2.0m)`
`=8.0xx106-6m^3`
the elastic potential energy stored
`=1/2xxstressxxstrainxxvolume`
`=1/2xx2.0xx10^8Nm^-2xx8.0xx10^-6m^3`
`=0.8J`
Discussion
6.

Find the excess pressure inside a mercury drop of radius 2.0 mm. The surface tension oif mercury `0.464Nm^-1`.

Answer» The excess pressure inside the drop is
P_2-P_1=2S/R`
`=(2xx0.464Nm^-1)/(2.0xx10^-3m)=464Nm^-2`
Discussion
7.

A large wooden plate of area `10m^2` floating on the surface of river is made to move horizontally wilth a speed of `2ms^-1` by applying a tangential force. If the river is 1m deep and the water contact with the bed is stationary, find the tangential force needed to keep the plate moving. Coefficient of viscosity of water at the temperature of the river `=10^-2 poise.`

Answer» The velocity decreases from `2ms^-1` to zero in 1 m of perpendicular length. Hence, velocity gradient
`=dv/dx2s^-1`
Now, `eta=|(F/A)/(dv/dx)|`
or `10^-3(N-s)/m^2=F/((10m^2)(2s^-1))`
or `F=0.02N`.
Discussion
8.

The lower end of a capillary tube of radius 1 mm is dipped vertically into mercury. (a) Find the depression of mercury column in the capillary. (b) If the length dipped inside is half the answer of part (a), find the angle made by the mercury surface at the end of the capillary with the vertical. Surface tension of mercury `=0.465Nm^-1` and the contact angle of mercury with glass `=135^@`.

Answer» Correct Answer - A::B::C::D
`r=1 mm=10^-3m, theta=1.35^@`
`h=(2Tcostheta)/(r rhog)`
`=(2xx0.465xxcos135^@)/(10^-3xx13600xx(9.8))`
`=0.0053m=5.3mm`
Discussion
9.

A copper wire of cross sectional asrea 0.01 cm is under a tension of 20 N. Find the decrease in the cross sectional area. Young modulus of copper `=1.1xx10^11Nm^-2` and Poisson ratio 0.32.

Answer» Correct Answer - A::B::C::D
`A=0.01 cm^2`
`=10^-6m^2`
`=T=20N`
`Y=1.1xx10^11N/m^2`
we know that `Y=(FL)/(A/_\\L)`
`=(/_\\L)/L=F/(AY)`
`=20/(10^-6xx1.1x10^11)x18.18xx10^-5`
`d=((/_\\D)/D)/((/_\\L)/L)=0.32`
`So, (/_\\D)/D=(0.32)xx(KL)/L`
`=32xx(18.18)xx10^5=5.81xx10^-5`
Again `(/_\A)/A=(2/_\\r)/r`
`rarr /_\\A=(2/_\\r)/r`
`/_\\A=2xx(5.8xx10^-5)xx(0.01)`
`=1.16xx10^9N/m^2`
Discussion
10.

Two large glass plates are placed vertically and parallel to each other inside a tank of water with separation between the plates equal to 1 mm. Find the rise of water in the space between the plates. Surface tension of water `=0.075Nm^-1`.

Answer» Correct Answer - A::C
Given `T=0.075N/m, `
`d=1mm=10^-3m`
`rho=10^3kg/m^3`
. `Now , T(2L)=[1xx(10^-3)xxh]rhog`
`rarr h=(2xx(0.075))/(10^-3xx10^3xx10)`
`h=0.015m=1.5m
Discussion