A solid weigh 10 .00 N in air, 6 N when fully immersed in water and 7 . 0 N when fully immersed in a liquid X. Calculate the relative density of the liquid, X.
5/3
4/3
3/4
7/10
Explanation
Video Explanation
No video available
Post your Contribution
Post your Contribution
Discussions (36)
The answer is C but I have the explanation to it here's Þhe formula (m1-m2/m1-m3) = mass one =10 mass two =7 mass three =6 so 10-7/10-6= 3/4 answer.
R.D=upthrust in liq/upthrust in H20
upthrust means loss in weight
upthrust in liq=10-7=4
upthrust in H20=10-6=3
R.D=3/4
r.d =upthrust in liquid/upthrust in water.
upthrust in li
quid=weight of object in air -weight of object in liquid.10-7=3
upthrust in water=10-6.
Wa= weight in air,10, Ww=weight in water,6, Wx=weight,7, in unknown liquid, formula is ; Wa-Wx/Wa-Ww ,so, 10-7/10-6 =3/4.... Done :)
D ansa is C...bcos R.D of a liquid = upthrust in liquid / upthrust in water.therefore,10.00-7.0/10.00-6 =3/4
R.D=Upthrust in liquid/upthrust in water.upthrust in liquid=10-7=3.upthrust in water=10-6=4.R.D=3/4.option D is correct
