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

A body weighs W1 gf in air and when immersed in a liquid it weighs W2 gf, while it weighs W3 gf on immersing it in water. Find: (i) volume of the body (ii) upthrust due to liquid (iii)

relative density of the solid and (iv)relative density of the liquid

Given:

W1 = weight of the body in air

W2 = weight of the body when immersed in liquid

W3 = weight of the body when immersed in water

(i) Volume of the body = W1-W3 cm3

(ii) Upthrust due to liquid = W1-W2 gf

\text { (iii) } \quad \text { relative density of the solid }=\frac{\text { weight of the solid in air }}{\text { weight in air-weight in water }}=\frac{W_{1}}{W_{1}-W_{3}}

\text { (iv) relative density of the liquid }=\frac{w_{1}-w_{2}}{w_{1}-w_{3}}

Video transcript
"hello everyone welcome to lido this is priya and here we have a question to solve let us look into the question now a body weighs w1 gf in 8 and when it is immersed in a liquid it weighs w2gf while it weighs w3gf on immersing it in water find volume of the body afters due to liquid and three relative density of the solid and four relative density of the liquid so let us give the first one volume of the body volume of the body can be given as can be given as w 1 minus w3 centimeter cube so what is w1 it's a weight of the object in air and w3 is the weight immersed in the water so these two when we subtract these two will get the volume of object and the second one we need to find out address due to liquid for this a thrust due to liquid equal to w1 minus w to gf so what is w1 weight of the object in air and w2 is the weight of the object in liquid so the uprest will be w1 minus w2 and let us see the third one we need to find out relative density of solid for that relative density of solid equal to weight of solid in air by weight of solid in air minus weight of solid in water weight of solid in a minus weight of solid in water so what is the weight of a solid in air it is w1 and weight of solid in air is again w1 minus w3 will gives the relative density of solid and the fourth one to find out the relative density of liquid relative density of liquid equal to w1 minus w2 by w 1 minus w 3 so if we subtract w 1 minus w 2 by w 1 minus w 3 will get the relative density of liquid hope you understand this video if you like my video please subscribe to a little channel and if you have any doubts you can keep a message in comment box thank you"
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