262i think since the sand particles will fall with some momentum ,the force will increase momentarily to a eqm. value and then decrease in the end.
Explain the Process and what should be the correct answer and why?
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49Look this is what happens.
When the dripping process starts till it ends, the same amount of mass is in free fall state.
Since they are freely falling, no force other than the weight is acting on the sand. In other words, the system or the balance is providing no force on the sand. Thus, by newton's third law of motion, the sand is also not giving any reaction. Thus, although the sand is present in the system, it is not felt by the balance.
When the dripping process ends, the balance has to counter the entire weight of the sand + hourglass.
So, the force experienced decreases when the process starts and increases when it ends. And the increase and decrease is continuous process (not quantised). Try to think of the reason of this.
71Someone? Try this then only the next one could be posted.
26271I had already said- answer doesn't matter. Explanation does. Although your answer is correct.
158The sand particles fall with some acceleration(gravity) as well as momentum. So when the process begins,the electronic balance experiences an increase in force. But as time progresses,the balance is experiencing no considerable change in the magnitude of force(the sand particles are microscopic particles which are unable to cause any change in force when they are dripped in single quantities),hence,the force on the balance remains constant.But when the entire heap of sand is taken into account,it is responsible for generating a large amount of force on the balance.Thus,when the process ends,the acceleration and momentum with which the sand particles are dripping,nullifies to zero,and hence, the balance experiences a decrease in the force.From there on,no further force is experienced by the balance as no further actions is taken...........is the explanation correct?
71OOps sorry, My mistake. The CORRECT ANSWER IS (e)
So, you might need to rework your answer.
158is my explanation sumwat correct??..or rather....is my approach to the question correct??
1i think,
first through the clot dust particles will get down slowly,a small mass of dust particles falling down will
have no effect on meter scale since they are taken to be very fine..but at the same time mass from upper part of the hourglass is being deducted..so force will decrease first and slowly as more mass of dust particles will accumulate it will increase and at some time t will touch equilibrium value .and when there is no motion at last then it will show original mass of the system so it would be (e)
1i think force decreases and attain eqm. after process.
when dripping take place,F<weight.as aacc.is downward.
after process is over,f=weight.
so(b)must be correct.