@Kalash Verma
–
@Kalash Verma , The diagram itself suggest the necessary information needed to be known about the boxes...
Secondly, A constant velocity of the container doesnot affect the tension at all..so it wont matter whether the container is at rest or has some velocity...you need to focus on acceleration and that is given!!!

@Kalash Verma
–
Obviously the boxes cant be identical...the upper one is denser than liquid and the one below is less dense than liquid..
option 3 is correct..
Try some justification...
It is an interesting problem...

@Rohit Gupta
–
Justification is easy.
Let d1 = density of block 1 and its Volume be v1 and d2 =density of liquid and g= downward acceleration due to gravity.

$T1=Vg(d1-d2$ now as acceleration increases downward acceleration of the block increases hence T1 increases.

If system starts accelerating upwards,then psuedo force will act on both the boxes in downward direction.
Hence,T1 increases and T2 decreases @Rohit Gupta,please tell me where i am wrong?

@Akhil Bansal
–
@Akhil Bansal Buoyancy depends upon effective gravity.. when the container starts accelerating in the upward direction then the water molecules at the bottom pushes the water molecules at the top and hence increases the pressure every where. It may be seen in another way that when the container is accelerated upwards then the effective weight of the liquid displaced is increased..!! Thus buoyancy also increases!!

T1 increases and T2 decreases. Since the system is accelerating upwards, the downward acceleration on the boxes would increase, thus increasing their weight. the buoyant force due to the liquid remains constant.Thus, resulting in the (1) situation.

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## Comments

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TopNewestI'm not much of a "Fluids" guy but guess no. 1 is correct?

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I am Afraid you are wrong..But i will still love to hear you justification..

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@Rohit Gupta Two questions

1) Are the two box identical? 2) The system given above is at rest ???

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@Kalash Verma , The diagram itself suggest the necessary information needed to be known about the boxes... Secondly, A constant velocity of the container doesnot affect the tension at all..so it wont matter whether the container is at rest or has some velocity...you need to focus on acceleration and that is given!!!

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@Rohit Gupta Is the option 3 correct ? In which both T1 and T2 increases.

I have assumed that the boxes are not identical and T1 and T2 have some magnitude.If the boxes are identical then T1 = T2 = 0

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option 3 is correct..

Try some justification...

It is an interesting problem...

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$T1=Vg(d1-d2$ now as acceleration increases downward acceleration of the block increases hence T1 increases.

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@Rohit Gupta It can be shown for box 2

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@Kalash Verma

Correct...!! Well DoneLog in to reply

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Both will simply increase hence we are doing nothing but increasing the effective gravity hence everything will increase

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If system starts accelerating upwards,then psuedo force will act on both the boxes in downward direction.

Hence,

T1 increasesandT2 decreases@Rohit Gupta,please tell me where i am wrong?

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Hi @Akhil Bansal, You are missing the effect of acceleration on buoyancy!!!

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Could you please explain how pseudo force effect buoyant force in liquid.

Thankyou

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@Akhil Bansal Buoyancy depends upon effective gravity.. when the container starts accelerating in the upward direction then the water molecules at the bottom pushes the water molecules at the top and hence increases the pressure every where. It may be seen in another way that when the container is accelerated upwards then the effective weight of the liquid displaced is increased..!! Thus buoyancy also increases!!

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T1 increases and T2 decreases. Since the system is accelerating upwards, the downward acceleration on the boxes would increase, thus increasing their weight. the buoyant force due to the liquid remains constant.Thus, resulting in the (1) situation.

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The Buoyant force will also change...!!

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@Ansh Bhatt I am sry to tell you that your answer is wrong...

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It's a wonderful question ....

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Thanks @Tushar Panda I am glad you like it..!!

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