Home
Class 11
PHYSICS
A wooden rod weighing 25N is mounted on ...

A wooden rod weighing `25N` is mounted on a hinge below the free surface of water as shown. The rod is `3m` long and uniform in cross section and the support is 1.6m below the free surface. At what angle `alpha` rod is in equilibrium? The cross-section of the rod is `9.5 xx10^(-4)m^(2)` in area. Density of water is `1000kg//m^(3)`. Assume buoyancy to act at centre of immersion. `g=9.8m//s^(2)`. Also find reaction on the hinge in this position.
.

Text Solution

Verified by Experts

The correct Answer is:
A, C, D
Let `G` be the mid-point of `AB` and `E` the mid point of `AC`(i.e. the centre of buoyancy)

`AC=1.6 cosec alpha`
Volume of `AC`
`(1.6xx9.5xx10^(-4)) cosec (alpha)`
Weight of water displaced by `AC`
`=(1.6xx9.5xx10^(-4))cosec alpha`
`=14.896 cosec alpha`
Hence, the buoyant force is `14.896 cosed alpha` vertically upwards at` E`. While the weight of the rod is `25N` acting vertically downwards at `G`. Taking moments about A.
`(14.896 cosec alpha)(AE cos alpha)=(25)(AG cos alpha)`
or, `(14.896 cosec alpha)((1.6 cosec alpha)/(2))=25xx(3)/(2)`
or, `sin^(2)alpha =0.32`
`:. sinalpha=0.56`
or. `alpha=34.3^(@)`
Further. Let` F` be the reaction at hinge in vertically downward direction. Then, considering the tranlatory equilibrium of in vertical direction we have,
`F+`weight of the rod=upthrust
`F=`upthrust -weight of the rod
`=14.896 cosec(34.3^(@))-25`
`=26.6-25`
` :. F=1.6N`.
Promotional Banner

Topper's Solved these Questions

  • FLUID MECHANICS

    DC PANDEY|Exercise Exercise16.1|6 Videos

  • FLUID MECHANICS

    DC PANDEY|Exercise Introductory Exercise|1 Videos

  • FLUID MECHANICS

    DC PANDEY|Exercise Solved Examples|21 Videos

  • EXPERIMENTS

    DC PANDEY|Exercise Subjective|15 Videos

  • GENERAL PHYSICS

    DC PANDEY|Exercise INTEGER_TYPE|2 Videos

Similar Questions

Explore conceptually related problems

If the young's modulus of the material of the rod is 2 xx 10^(11) N//m^(2) and its density is 8000 kg//m^(3) then the time taken by a sound wave to traverse 1m of the rod will be

Find the change in density of water in occean at depth of 700m below the surface. The denity of water at the surface is 1000kg//m^(3) and the bulk modulus of water is 4.9xx10^(9) Nm^(-2)

A uniform rod of mass M and length L, area of cross section A is placed on a smooth horizontal surface. Forces acting on the rod are shown in the digram Total elastic potential energy stored in the rod is :

For a steel rod, the Young's modulus of elasticity is 2.9xx10^(11)N//m^(2) , and the density of steel is 8xx10^(3)kg//m^(3) .What is the velocity of longitudinal waves in steel?

At the mount of the tap area of cross-section is 2.0cm^(2) and the speed of water is 3m//s . The area of cross-section of the water column 80 cm below the tap is (use g=10m//s^(2))

The thin rod shown below has mases M and length L. A force F acts at one end as shown and the rod is free to rotate about the other end in horizontal plane. Initial angular acceleration of the rod is :

A uniform rod of mass M and length L, area of cross section A is placed on a smooth horizontal surface. Forces acting on the rod are shown in the digram Ratio of elongation in section PQ of rod and section QR of rod is

A uniform rod of mass M and length L, area of cross section A is placed on a smooth horizontal surface. Forces acting on the rod are shown in the digram Ratio of elastic potential energy stored in section PQ and section QR of the rod is

A wooden rod of a uniform cross section and of length 120 cm is hinged at the bottom of the tank which is filled with water to a height of 40 cm . In the equilibrium position, the rod makes an angle of 60^@ with the vertical. The centre of buoyancy is located on the rod at a distance (from the hinge) of