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Selina Solutions Class 9 Physics Solutions for Exercise 5B in Chapter 5 - Chapter 5 Upthrust In Fluids Archimedes Principle And Floatation

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Question 26 Exercise 5B
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A body of volume 100cm3 weighs 1 kgf in air. Find: (i) its weight in water and (ii) its relative density.

Answer:

Given:

Volume of the body = 100cm3

Weight of the body in the air, W1= 1kgf or 1000gf

Let Weight of the body in water be W2

We know that R.D of water is 1 and that of a solid is 10

(i) To find the weight of the body in water

\text { Relative density of the body }=\frac{w_{1}}{W_{1}-W_{2}} x \text { relative density of water }

\begin{array}{l} 10=\frac{1000}{1000-W_{2}} \times 1 \\ 10\left(1000-W_{2}\right)=1000 \end{array}

W2 = 1000 – 100 = 900 gf

(ii) To find the relative density of the body

Density of the body = mass/volume

= 1000/100 = 10gf

Relative density is equivalent to density in C.G.S. with no unit, hence R.D. of the body is 10.

Video transcript
"hello everyone welcome to lido this is priya and here we have a question about the salt let us look into the question a body of volume 100 cubic centimeters weighs 1 kgf in air find its weight in water and its relative density so first we try to solve the first one its weight in water so what we know from this given question is the volume of object volume of the object which is 100 cubic centimeters and we know the weight of the object in the air weight of the object in air which is one k g f and let's take the weight of the object in water as w2 and this is w1 and let let us assume let us assume weight of the body in weight of the body in water as w2 and as we know the relative density of the object in what um as we know that the relative density of water is 1 and that of a solid is 10 so to find out the weight of the object in water the relative density of the body relative density of body equal to w 1 by w 1 minus w 2 so by substituting this 10 equal to 1000 by 1000 minus w2 into 1 which comes 10 equal to 1000 by thousand minus w two into one that implies ten into thousand minus w2 equal to 1000 then w2 equal to thousand minus 100 which is 900 k g f which is a 1000 gf so weight in water so the object weight in water is 900 gf now let us try to find out its relative density as we know the relative density equal to density without units then first we need to find out the density so we have a formula density as mass per volume so what's the mass we have mass is 1000 and volume is 10 so the density is 10 g f so the relative density has no units so the relative density of the object will be temp hope you understand this video if you like my video please subscribe to our leader channel and if you have any doubts you can keep a message in comment box thank you"
Related Questions
Exercises
Exercise 5A
Exercise 5B
Exercise 5C
Chapters
Chapter 1 – Measurements and Experimentation
Chapter 2 Motion In One Dimension
Chapter 3 Laws Of Motion
Chapter 4 Pressure In Fluids And Atmospheric Pressure
Chapter 5 Upthrust In Fluids Archimedes Principle And Floatation
Chapter 6 Heat And Energy
Chapter 7 Reflection Of Light
Chapter 8 Propogation Of Sound Waves
Chapter 9 Current Electricity
Chapter 10 Magnetism
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