Fluid mechanics conceptual question

  • #1
a beaker containing water is placed on the platform of a digital weighing machine. it reads 1100 grams. a metal body of density 8g/cc and mass 200 g is suspended in water in the beaker(without touching the walls of the beaker). it is attached by a suitable string fixed to some support. Now what will be the reading on the weighing machine. from the following options which one is correct? give reasons
a. 1100 g
b. 1125 g
c. 1275 g
d. 1300 g
 

Answers and Replies

  • #2
tiny-tim
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hi alphabeta1720! :wink:

show us what you've tried, and what you think the reason is, and then we'll know how to help! :smile:
 
  • #3
hi alphabeta1720! :wink:
show us what you've tried, and what you think the reason is, and then we'll know how to help! :smile:
my only confusion is whether it will be a total reading of 1300 or upthrust will be subtracted.
 
  • #4
tiny-tim
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hint: consider the beaker the water and the mass as a single body (mass M + m, say) …

what are the forces on it?

is there any tension in the string? :wink:
 
  • #5
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hint: consider the beaker the water and the mass as a single body (mass M + m, say) …

what are the forces on it?

is there any tension in the string? :wink:
that is the confusion i have about the tension of the spring. nothing else is giving in the problem and this is famous problem of KVPY. it is a competitive examination which is very prestigious at high school level. so any one tell me the answer and also explain me. I am alplebeta.
 
  • #6
tiny-tim
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hi alphabeta! :smile:
my only confusion is whether it will be a total reading of 1300 or upthrust will be subtracted.
that is the confusion i have about the tension of the spring. nothing else is giving in the problem and this is famous problem of KVPY.
what i was getting at is that the only external forces on the beaker-water-mass system are the weight, the tension in the string, and the reaction from the platform …

since the system is in equilibrium, they must add to zero

since common-sense tells you that there must be some tension in the string (you are ok with that, aren't you?), that shows that the reaction from the platform (which is the same as the reading on the weighing scale) must be less than the weight, doesn't it?

that should help you to work out the solution to the problem (as you know, on this forum we're not allowed to give you the answer, we only help you to find it yourself!) :smile:
 

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