Thus, the pressure exerted would be P = (981 X 1000) / 1 or P = 9.81 X 10 5 pa
Mass= density x volume = 8.5 x(3x3)=229.5gwt
Therefore second spring balance of range 0-250 gwt with least count 5gwt will be suitable.
Ans: The substance will float if its density is less than water and will sink if its greater.so, density of solid body is d = mass/volume
or d = 150/250 = 0.6 gm/cm3 which is less than the density of water (1 gm/3).So, the solid body will float on water.
so, F1 = 50 N F2 = 40 Nso, F1/F2 = 50/40 or relative masses m1/m2 = 5/4 anddensity = mass/volume and as volume remains constant, Relative density = d1/d2 = 5/4
V = Ö2gh = Ö 2x10x2 = 6.32 m/s
Buoyancy force by water try to stop the ball.
Buoyancy force = weight of displaced water = dx Vxg
where d = density of water V = volume of the ball , g = 10 m/s2deceleration of the body by buoyancy force, a = (dVg)/ m
where m= d'V d' = density of block
a = dVg/(d' V) = dg/d' =(d/d')*g =g/(0.8)= 10/0.8 (Given, d'/d = 0.8)= 12.5 m/s2Net deceleration of ball,a' = a-g = 2.5 m/s2Final speed of ball v' = 0
Use v' 2 = v2 + 2a's s= depth of ball in the water
=> 40 = 0 + 2x2.5xs => s = 8m
Q. Equal masses of water and a liquid of relative density 2 are mixed together. Then, the mixture has a relative density of (in g/cm3) a)2/3 b)4/3 c)3/2 d)3
Let the relative densities of water and liquid be ρ1 and ρ2 respectively.
The volume of the two be V1 and V2, of water and liquid respectively.
The volume of the mixture would be, V = V1 + V2 (1)
also, volume = mass/density
2m/ρ (V) = m/ρ1 (V 1 ) + m/ρ2 (V 2 )
here ρ1 = 1,ρ2 = 2 and ρ is the relative density of the mixture.
2/ρ = 1/ρ1 + 1/ρ2
by substituting the values, we
ρ/2 = 2/3
or, the relative density of the combined liquid will be, ρ=4/3