Archimedes Principle in Hindi
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- čas přidán 21. 08. 2024
- 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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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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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
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.
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..
No 2 , 3 galat hai
Q1.
(i)loss in weight= weight in air -weigh in water
200-170=30
(ii), upward force=loss weight
30=30
Q2 weight of object when it is floating on liquid or emerged in liquid is called apparent weight
Q3
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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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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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Hello sir I am subham nandi
Answers ~
Q1. i. 30 gf
ii. 30 gf
Q2. 170 gf
Q3. i. 1000 cm^3
ii. 4000 gf / 4 kgf
THANK YOU SIR...
Bro please explain the answer
I think 2 answer is wrong coz in that we just have to write what is the exact meaning of apparent weight
@@vvvcam5275 Yes I was wrong. Thanks for informing me.
(Ans)2. The apparent weight of a floating body is the weight of the body in air - weight of the liquid displaced by the volume of the body.
THANKS.....😃👍
Pls explain how to take out the apparent weight of the body in water that is Q3 (ii)
Q2. Apparent force = real force
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
Sorry 40 N OR Kg m/s2
Bro pls explain it answer
4 kg hoga
From Pakistan likhna jaruri tha bro😂
Apparent weight = real weight
Sir your way of teaching is excellent
Ans1-(i) 30gf.
(ii) buoyant force is same as the upward force=30gf.
Ans2- apparent weight=real weight -loss in weight.
Ans3-(i) 1/200cm^3.
(ii) 1000cm^3.
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The explanation was easy and simple as I understand the topic verry well
1.i)30gf
ii) 30gf
2. The apparent weight of a floating body is the weight of the body fully immersed in fluid.
1.a) 30 gf
b)30gf
2) zero for any floating body
3.a) 1000cm cube
b)4kgf
How 4kgf?
1. (i)30gf
(ii)30gf
2. Zero
3. (i)1000cm3
(ii)4kgf
Super! All your answers are correct :) Also try the Quiz for this video on my Website. Quiz button is below the video: www.manochaacademy.com/physics-hindi/archimedes-principle/l/archimedes-principle
@Suchitrakumari Can u plz explain how volume of body is 1000cm^3 in Q.3(i)
Q1. (i)- 200gf-170gf =30gf.
(ii)- Upthrust= loss in weight
Q2. 0
Q3. (i)-Volume in m will be 10^3m-3
Now, volume of body = volume of water displaced by body when fully immerged in water 1000 m3.
(ii)- Apparent weight=
Real weigth-Upthrust
Apparent weight=
5kgf- 1kgf = 4kgf
FB= Vdg =1kgf.
Hope this is helful for you.
Thank you 👍🏻👍🏻
How?
3. (ii) kaise kiya h
@@lagandevmahto5963 Dekho bro ,
Upthrust = Weight in air- Weight in water ,hota hai na
Now, Apparent weight KO hm log Weight in water bhi khete hai.
So, Weight in water = Weight in air - Upthrust , ho jay ga ok na
Ab value put kr doo
@@rounakverma8127 apne ques 3 ke 1st part m 1000mcube kyu likha h
Plz tell
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1. 30gf
2. Apparent wt. =real wt. - upthrust
Thank you so much for this wonderful video 😊
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Ans1(i) 30gf (ii) 30gf
Ans2 Apparent Wt= Real wt-upthrust(loss in wt).
Ans3(i) 1000cm3 (ii) 4kg
Hii
Bro can u tell me questions number 3
The reading of a digital balance is 500g when a beaker partially filled with water is placed on it. A piece of iron sinks completely when dropped in that beaker and no water spills out. Now the reading becomes 600g. The piece of iron is now taken out, tied to a string, and the hanging piece is completely immersed in water in the beaker without touching any wall or the bottom of the beaker, while the other end of the string is held by hand. The reading in the balance is expected to be
1.loss I weight =real weight -apparent weight =200gf-170gf=30gf
2.apparent weight =170gf
3.1.volume of body =v of liquid displaced as body is fully immersed
2.apparent weight =1000cm3 =1000g = 5-1=4kg
Excellent
Sir great!!!!
But please give the answers
Q.3
To find (1) volume of the body
(2) apparent weight of the body in water.
solution: according to Archimedes principle,
(1) volume of water displaced by the body = volume of the body.
here volume of water displaced by the body is 1000 cm³
so, volume of the body = 1000 cm³ =
1000 × (10-2 m)³
= 10-3 m³
(2) apparent weight = true weight - upthrust
first we have to find upthrust.
upthrust = volume of water displaced by the body density of water g X
= 10-3 m³ x 103 kg/m³ x 10 m/s²
= 10 N
now apparent weight = 5kg x 10 m/s² - 10
N
= 50 N-10 N
= 40 N
Therefore apparent weight of the body in water is 40 N
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Hindi me samajh me jadi aata hai
Thanks sir
Answer of 3rd question
i) 1000cm^3
ii) 1kg
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Thankyou sir your explanation is really awesome ..
Concept clear
Q1 : ¡) 30gf
ii) 30 gf
Q2 : apparent weight = real weight ( Because loss in weight = upward force = weight of the water displaced = 0)
Q3 : i) 1000 cm^3 or 1 letre ( just think, when 1000 cm^3 will go downward then 1000 cm^3 will go upward ☺)
ii) apparent weight =real weight - loss in weight
= real weight - upward force
= real weight - weight of the water displaced
= 5 kg × 9.8 m/s^2 - 1 kg × 9.8 m /s^2 ( A litre of liquid water has a mass almost exactly equal to one kilogram)
= 49 N - 9.8 N
= 39.2 N
Hope you all understand and helpful.
Wrong
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