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 5gwt

3. range 0-1000gwt ; least count 5gwt 4. range 0-1000gwt ; least count 10gwt

Ans: Edge=3 cm , Density=8.5 g/cm^{3}

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.

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 0c

Density of turpentine oil = 840 kg/ m3 ( given).

Density of water at 4 0c = 1000 kg/ m3

Relative 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.

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 N

so, F1/F2 = 50/40 or relative masses m1/m2 = 5/4 and

density = 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^{2}

deceleration 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/s^{2}

Net deceleration of ball,a' = a-g = 2.5 m/s^{2}

Final speed of ball v' = 0 Use 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)3

Ans: he masses of two liquids are equal, let it be m.

Let the relative densities of water and liquid be ρ_{1} and ρ_{2} respectively.

The volume of the two be V_{1} and V_{2}, of water and liquid respectively.

The volume of the mixture would be, V = V_{1} + V_{2 }(1)

also, volume = mass/density

thus,

2m/ρ *(V) *= m/ρ_{1} *(V* _{1}_{ }*)* + m/ρ_{2} *(V* _{2}_{ }*)*

here ρ_{1} = 1,ρ_{2} = 2 and ρ is the relative density of the mixture.

now,

2/ρ = 1/ρ_{1} + 1/ρ_{2}

by substituting the values, we

ρ/2 = 2/3

**or, the relative density of the combined liquid will be, ρ=4/3**

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