A vessel contains oil (density=0.8 g/cm3) over mercury (density=13.6 g/cm3). A homogeneous sphere floats with half…

Q: A vessel contains oil (density=0.8 g/cm3) over mercury (density=13.6 g/cm3). A homogeneous sphere floats with half its volume immersed in mercury and the other half in oil. The density of the material of the sphere in g/cm3 is

(a) 3.3

(b) 6.4

(c)7.2

(d) 12.8

Ans: (c)

Sol: Let ρ = density of material of sphere

Applying Law of Floatation ,

Weight = Upthrust

$\large V \rho g = \frac{V}{2} \rho_o g + \frac{V}{2} \rho_m g $

$\large \rho = \frac{\rho_o}{2} + \frac{\rho_m}{2}$

$\large \rho = \frac{0.8}{2} + \frac{13.6}{2}$

= 7.2 g/cm3

A U-tube of uniform cross-section is partially filled with a liquid I. Another liquid II which does not mix…

Q: A U-tube of uniform cross-section is partially filled with a liquid I. Another liquid II which does not mix with liquid I is poured into one side. It is found that the liquid levels of the two sides of the tube are the same, while the level of liquid I has risen by 2cm. If the specific gravity of liquid I is 1.1, the specific gravity of liquid II must be

(a) 1.12

(b)1.1

(c) 1.05

(d) 1.0

Ans: (b)

Sol: By applying Pascal’s law

P1 = P2

P0 + ρI g h = P0 + ρII g h

ρI = ρII

A body floats in a liquid contained in a beaker. The whole system as shown in figure falls freely under gravity…

Q: A body floats in a liquid contained in a beaker. The whole system as shown in figure falls freely under gravity. The upthrust on the body is

Numerical

(a)Zero

(b) Equal to the weight of liquid displaced

(c) Equal to the weight of the body in air

(d) Equal to the weight of the immersed portion of the body

Ans:(a)

Sol: In a free falling system geff = g – g = 0

Let Vimm = immersed volume

Upthrust = Vimm ρL geff

Upthrust = 0

A metal ball immersed in alcohol weight W1 at 0°C and W2 at 50°C. the coefficient of cubical expansion…

Q: A metal ball immersed in alcohol weight W1 at 0°C and W2 at 50°C. the coefficient of cubical expansion of the metal is less than that of the alcohol. Assuming that the density of the metal is large compared to that of alcohol, it can be shown that

(a) W1 > W2

(b) W1 = W2

(c)W1 < W2

(d) All of these

Ans: (c)

Sol: Let Vs = volume of solid & ρL = density of liquid

$\large W_{apparent} = W_{actual} – Upthrust $

Upthrust F = Vs ρL g

At higher temperature ,

Upthrust F’ = Vs‘ ρL‘ g

$\large \frac{F’}{F} = \frac{V_s’}{V_s} .\frac{\rho_L’}{\rho_L} = \frac{1+\gamma_s \Delta \theta}{1+\gamma_L \Delta \theta}$

Since , $\large \gamma_s < \gamma_L $

Hence, $\large F’ < F $

Therefore , $\large W’_{app} > W_{app} $

$\large W_2 > W_1 $

A paper disc of radius R from which a hole of radius r is cut out is floating in a liquid of the surface tension S…

Q: A paper disc of radius R from which a hole of radius r is cut out is floating in a liquid of the surface tension S. The force on the disc due to the surface tension is

(a) S × 2π R

(b) S × 2πr

(c) S × 2π (R – r)

(d) S × 2π (R + r)

Ans: (d)

Sol: Surface tension S = F/l  ⇒ F = S l

F = S (2 π R + 2 π r )

F = S × 2 π(R + r)