Archimedes Principle in Hindi

Archimedes’ Principle in Hindi! Watch till the end for a 'surprise' that will help you remember Archimedes Principle FOREVER!!
This video covers Archimedes’ Principle and buoyant force. Explained and visualized with experiments.
Try the Top 3 Questions at the end of video and write your answers and doubts in the comments below!
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[@kiranangadi3721]

Q1:
I) 200-170= 30
ii) buoyand force = upward force = loss in weight of the object = original wt - weight when immersed = 200-170 = 30
Q2= apparent wt = body weight in water
=
Body is floating means = body is equilibrium state , but has a weight
So it can answered in two way
1. based on Q1 , it is 170 , which is wt of body in water
2. I need wt in water and loss in weight ( upward /buoyand force) to find it .
Q3:
I) I have wt in air that body of mass 5kg, but I need density of body to find volume ( density = mass/ volume)
ii) Apparent weight: body weight in water, to find this
I have wt of body in air (5kg) and then I need loss in weight ( upward force/ buoyand force)
I have volume 1000cm3 of displaced water, but I need density of water to find wt of water, we can refer google to know density of water = 1 g/ cm3
Loss in wt = density x volume
= 1 g/cm3 x 1000 cm3
= 1000g = 1 kg
Apperant weight = real wt in air - loss in wt
= 5kg - 1 kg= 4kg

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Brilliant! So clear, I'm an English only speaker, and I understood the concept clearly! Thanks so much for explaining floatation using Archimedes insight!

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[@pankajgoyal3641]

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[@PirateCaptain666]

Sir please pin up the correct answers. Tried to solve the problems.
1. Weight in air 200 gf
Weight in liquid 170 gf
Loss in weight 30 gf
Buoyant force on object = Loss in weight = 30 gf
2. For a freely floating body,
Upward force = Total weight of body in air
Thus, apparent weight = 0
3. Mass of body 5 kg, fully immersed
Displaces 1000 cm 3
Find:
Vol of body
Apparent wt of body
Volume of a fully immersed body = Volume of water displaced = 1000 cm3
Mass of water displaced = Density x Volume = 1 x 1000 = 1000 g = 1 kg
Upward force experienced by ball = Loss in weight of the ball = Wt of water displaced
Real Weight of ball in air 5 kg
Upward force 1 kg
Apparent Weight of ball 4 kg

[@mitalisingh5544]

You have calculated mass of displaced water ie 1kg, not wt.
Then how could you say that wt.of displaced water(=upword force) is 1kg.

[@b_12_himanshubeck58]

Q1 i) 30gf
ii) 30gf
Q2 0
Q3) I) 5000 cm3
ii) 0.005 kg
Sir is this the right solution??
Sir plzz reply. And thanku sir for this video it's really very helpful..

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Q1 (i) 30 gf. Rest all question I am not able to answer. Please make a video on this questions. It is a bold request to you and I would appreciate you for making me understand this topic as my exam is tomorrow. Please make the video as soon as possible before 12 noon of tomorrow.

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[@sadaqatalikhan1217]

Well, Explain. My answers are ...
A (I) 30 gf
(II) 30 gf
B) Apparent Weight = Wt air - Upthrust
C) (I) Vol. of Body = 1000 cm3
(II) Apparent of Body in H2O = 40 kg
From Pakistan

[@sadaqatalikhan1217]

Sorry 40 N OR Kg m/s2

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4 kg hoga

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Apparent weight = real weight

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[@subharamasubbu5133]

1.a) 30 gf
b)30gf
2) zero for any floating body
3.a) 1000cm cube
b)4kgf

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How 4kgf?

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