This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .

Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.

Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold

- bulleted - list

bulleted

list

1. numbered 2. list

numbered

list

Note: you must add a full line of space before and after lists for them to show up correctly

@A Former Brilliant Member
–
@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!!!

@A Former Brilliant Member
–
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.

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

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

Log in to reply

@Rohit Gupta Two questions

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

Log in to reply

@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!!!

Log in to reply

@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

Log in to reply

option 3 is correct..

Try some justification...

It is an interesting problem...

Log in to reply

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

Log in to reply

@Rohit Gupta It can be shown for box 2

SimilarlyLog in to reply

@Kalash Verma

Correct...!! Well DoneLog in to reply

Log in to reply

Log in to reply

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

Log in to reply

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?

Log in to reply

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

Log in to reply

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

Thankyou

Log in to reply

@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!!

Log in to reply

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.

Log in to reply

The Buoyant force will also change...!!

Log in to reply

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

Log in to reply

It's a wonderful question ....

Log in to reply

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

Log in to reply