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
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..
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.
Brilliant! So clear, I'm an English only speaker, and I understood the concept clearly! Thanks so much for explaining floatation using Archimedes insight!
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
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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
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
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
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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@@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.....😃👍
Before this video i usually think this is that topic which i could never understand but after this video i can say this is that topic wich i can never forget in my life❤
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
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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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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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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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
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1.a) 30 gf
b)30gf
2) zero for any floating body
3.a) 1000cm cube
b)4kgf
How 4kgf?
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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)
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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)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
1. 30gf
2. Apparent wt. =real wt. - upthrust
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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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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
Concept clear
It is easy to understand
Thank you sir for clearing my doubts
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Ans1(i) 30gf (ii) 30gf
Ans2 Apparent Wt= Real wt-upthrust(loss in wt).
Ans3(i) 1000cm3 (ii) 4kg
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Before this video i usually think this is that topic which i could never understand but after this video i can say this is that topic wich i can never forget in my life❤
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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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Answer of 3rd question
i) 1000cm^3
ii) 1kg
Sir your video is so nice but give me the answer of 2, 3 questions 😄
1.i)30gf
ii) 30gf
2. The apparent weight of a floating body is the weight of the body fully immersed in fluid.
Thank you sir . you are best