Q. a cube of mass 1kg with each side of 1cm is lying on the table. find the pressure exerted by the block on the table. take g=10 m/s

^{2}
Ans: Pressure is given as force/area so, Force, F = mg = 1000 X 981 gm.m/s

^{2}and area, A = 1x1 cm^{2}= 1 cm^{2 }Thus, the pressure exerted would be P = (981 X 1000) / 1 or P = 9.81 X 10^{5 }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/cm

^{3}, the best suited spring balance for determining weight of the solid would be of 1. range 0-250gwt ; least count 1gwt 2. range 0-250gwt ; least count 5gwt3. range 0-1000gwt ; least count 5gwt 4. range 0-1000gwt ; least count 10gwt
Ans: Edge=3 cm , Density=8.5 g/cm

Therefore second spring balance of range 0-250 gwt with least count 5gwt will be suitable.

^{3}Mass= density x volume = 8.5 x(3x3)=229.5gwtTherefore second spring balance of range 0-250 gwt with least count 5gwt will be suitable.

Q. The density of turpentine oil is 840 Kg/ m3. What will be its relative density. (Density of water at 4 degree C is 10 cube kg minus cube)

Ans:Relative Density = Density of Substance/ Density of water at 4 0cDensity 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 = 0.84Since, 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.

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

Q. A body weighs 50 N in air and when immersed in water it weighs only 40 N. Find its relative density.

Ans: the relative density would be ratio of the density of the body with respect to air and the density of the body with respect to 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 ?

Ans: Speed of the ball

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/s

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/s

Use v'

=> 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/cm

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/s

^{2}deceleration of the body by buoyancy force, a = (dVg)/ mwhere 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/s

^{2}Net deceleration of ball,a' = a-g = 2.5 m/s^{2}Final speed of ball v' = 0Use v'

^{2 }= v^{2}+ 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/cm

^{3}) a)2/3 b)4/3 c)3/2 d)3Let 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

thus,

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

## No comments:

## Post a Comment