^{2}

^{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

^{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

^{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.

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

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

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

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

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