so, Force, F = mg = 1000 X 1000 gm.cm/s2
and area, A = 1x1 cm2 = 1 cm2
Thus, the pressure exerted would be
P = (1000 X 1000) / 1 or P = 1 x 10 6 pa
Q. The mass of a solid iron cube of side 3cm is to be determined usig a spring balance. If the of iron is approximately 8.5 g/cm3, the best suited spring balance for determining weight of the solid would be of
2. range 0-250gwt ; least count 5gwt
3. range 0-1000gwt ; least count 5gwt
4. range 0-1000gwt ; least count 10gwt
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.
Density of turpentine oil = 840 kg/ m3 ( given).
Density of water at 4 0c = 1000 kg/ m3Relative density of turpentine oil = Density of turpentine oil /
Density of water at 4 0c = (840 / 1000 ) kg m-3/ kg m-3 = 0.84
Since, the relative density of the turpentine oil is less than 1, therefore it will float in water.
Q. A solid body of mass 150 g and volume 250cm3 is put in water . will following substance float or sink if the density of water is 1 gm-3?
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
Q. A ball of relative density 0.8 falls into water from a height of 2m. find the depth to which the ball will sink ?
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/s2
Net 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.
now, 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
More Questions are solved Here: http://jsuniltutorial.weebly.com/
Sure shoot MCQ for class 9 science term-2 ClicK Here
IX Thrust and Pressure, Archimedes’ Principle, Relative Density
CBSE Class 9 - Science - Chapter 10: Flotation: Notes and Quest
MCQ: Flotation: Thrust, Pressure, Buoyancy and Density
Thrust and Pressure, Archimedes’ Principle, Relative Density key point
Notes : Flotation: Thrust, Pressure, Buoyancy and Density
Physics Flotation Term-II Class IX Buoyant force Detail Study
9th Physics Solved Numerical Floating bodies