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121.

A wooden block is floating on water kept in a beaker, 40% of the block is above the water surface. Now the beaker is kept inside a lift that starts going upward with acceleration equal to $\frac{\mathrm{g}}{2}$. The block will then

sink

float with 10% above the water surface

float with 40% above the water surface

float with 70% above the water surface

B.

float with 10% above the water surface

According to Archimedes' principle

Weight of wooden block = weight of liquid displaced

$\Rightarrow \mathrm{mg}=\left(\frac{60}{100}\right)\mathrm{V}\times {\mathrm{\rho}}_{\mathrm{l}}\times \mathrm{g}\phantom{\rule{0ex}{0ex}}=\frac{3}{5}\mathrm{V}\times {\mathrm{\rho}}_{\mathrm{l}}\times \mathrm{g}.....\left(\mathrm{i}\right)$

(where, ρ_{l} = density of the liquid and V = total volume of the block)

Now, when the beaker is kept in the lift, net weight of the block

$=\mathrm{mg}+\frac{\mathrm{mg}}{2}=\frac{3\mathrm{mg}}{2}\left[\because \mathrm{R}\mathrm{mg}=\frac{\mathrm{mg}}{2}\right]$

Net weight of the block = Buoyant force

$\Rightarrow \frac{3\mathrm{mg}}{2}={\mathrm{V}}_{\mathrm{l}}\times {\mathrm{\rho}}_{\mathrm{l}}\times \mathrm{g}......\left(\mathrm{ii}\right)$

V_{l} = Volume of block inside liquid

Now from Eqs. (i) and (ii), we get

$\left(\frac{3}{2}\right)\times \frac{3}{5}\left(\mathrm{V}\times {\mathrm{\rho}}_{\mathrm{L}}\times \mathrm{g}\right)={\mathrm{V}}_{\mathrm{l}}\times {\mathrm{\rho}}_{\mathrm{l}}\times \mathrm{g}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{V}}_{\mathrm{l}}=\frac{9}{10}\mathrm{V}\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{the}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{block}\mathrm{above}\mathrm{the}\mathrm{liquid}=\frac{1}{10}\mathrm{V}=10\%\mathrm{of}\mathrm{total}\mathrm{volume}.$

122.

A steam of a liquid of density ρ flowing horizontally with speed v rushes out of a tube of radius r and hits a vertical wall nearly normally. Assuming that the liquid does not rebound from the wall, the force exerted on the wall by the impact of the liquid is given by

πrρv

πrρv

^{2}πr

^{2}ρvπr

^{2}ρv^{2}

D.

πr^{2}ρv^{2}

Cross-sectional area A = πr^{2}

Volume of liquid flowing per second = AV = πr^{2}v

Mass of the liquid flowing out per second = πr^{2}vρ

Initial momentum of liquid per second = mass of liquid flowing x speed of liquid

= πr^{2}vρ x v = πr^{2}v^{2}ρ

Since the liquid does not rebound after impact, the momentum after impact is zero.

Therefore, the rate of change of momentum = πr^{2}v^{2}ρ

According to Newton's second law, the force exerted on wall = rate of change of momentum

= πr^{2}ρv^{2}

**Assertion: ** A needle placed carefully on the surface of water may float, whereas the ball of the same material will always sink.

**Reason: ** The buoyancy of an object depends both on the materials and shape of the object.

If both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true but reason is not the correct explanation of assertion.

If assertion is true but reason is false.

If both assertion and reason are false.

C.

If assertion is true but reason is false.

When a needle is placed carefully on the surface of water, it floats on the surface of water due to the surface tension of water, which does not allow the needle to sink.

In case of a steel ball, the surface tension of the water is not sufficient to keep it floating, so it sinks down.

124.

A liquid is allowed into a tube of truncated cone shape. Identify the correct statement from the following.

The speed is high at the wider end and low at the narrow end

The speed is low at the wider end and high at the narrow end.

The speed is same at both ends in a streamline flow.

The liquid flows with the uniform velocity in the tube.

B.

The speed is low at the wider end and high at the narrow end.

For an incomressible liquid equation of continuity.

AV = = constant or A ∝(1/V)

Therefore, at the wider end speed below and at the narrow end speed will be high.

125.

A wide hose pipe is held horizontally by a fireman. It delivers water through a nozzle at one litre per second. On increasing the pressure. this increases to two litres per second. The firman has now to

push forward twice as hard

push forward four times as hard

push backward four times as hard

push backward twice as hard

B.

push forward four times as hard

The rate of change of mass, dm/dt = avρ

Therefore, F =vdm/dt

= (avρ)v = av2ρ

A volume of liquid flowing per second = av

This is proportional to velocity. By doubling this volume of liquid flow per second the force becomes four times. when liquid flows forward, the hosepipe tens to come backwards. So to keep it intact, it should be pushed forward. Thus, the hosepipe should be pushed forward four times.

126.

Mercury boils at 367° C. However, mercury thermometers are made such that they can measure temperature upto 500°C. This is done by

maintaining vacuum above mercury column in the stem of the thermometer.

filling nitrogen gas at high pressure above the mercury column.

filling oxygen gas at a high pressure above mercury column.

filling nitrogen gas at a low pressure above the mercury column.

B.

filling nitrogen gas at high pressure above the mercury column.

If we fill nitrogen gas at a high pressure above mercury level, the boiling point of mercury is increased which can extend to the range upto 500°C

127.

A solid sphere of volume V and density p floats at the interface of two immiscible liquids of densities and p, respectively. If ρ_{1} < ρ < ρ_{2}, then the ratio of volume of the parts of the sphere in upper and lower liquid is

$\frac{\mathrm{\rho}-{\mathrm{\rho}}_{1}}{{\mathrm{\rho}}_{2}-\mathrm{\rho}}$

$\frac{{\mathrm{\rho}}_{2}-\mathrm{\rho}}{\mathrm{\rho}-{\mathrm{\rho}}_{1}}$

$\frac{\mathrm{\rho}+{\mathrm{\rho}}_{1}}{\mathrm{\rho}+{\mathrm{\rho}}_{2}}$

$\frac{\mathrm{\rho}+{\mathrm{\rho}}_{2}}{\mathrm{\rho}+{\mathrm{\rho}}_{1}}$

B.

$\frac{{\mathrm{\rho}}_{2}-\mathrm{\rho}}{\mathrm{\rho}-{\mathrm{\rho}}_{1}}$

V = Volume of solid sphere

Let V_{1} = Volume of the part of the sphere immersed in a liquid of density p_{1} and

V_{2} = Volume of the part of the sphere immersed in liquid of density p_{2}

The law of flotation says that for a floating object the weight of the object equals the weight of the fluid.

According to law of flotation,

Vρ g = V_{1} p_{1 }g + V_{2} p_{2 }g ....(i)

V_{1} ( ρ $-$ ρ_{1} ) g = V_{2} ( p_{2 }$-$ ρ )

⇒ $\frac{{\mathrm{V}}_{1}}{{\mathrm{V}}_{2}}=\frac{{\mathrm{\rho}}_{2}-\mathrm{\rho}}{\mathrm{\rho}-{\mathrm{\rho}}_{1}}$

128.

Two identical glass spheres filled with air are connected by a horizontal glass tube. The glass tube contains a pellet of mercury at its mid-point. Air in one sphere is at 0°C and the other is at 20°C. If both the vessels are heated through 10°C, then neglecting the expansions of the bulbs and the tube.

the mercury pellet gets displaced towards the sphere at a lower temperature.

The mercury pellet gets displaced towards the sphere at a higher temperature

The mercury pellet does not get displaced at all.

the temperature rise causes the pellet to expand with any displacement.

C.

The mercury pellet does not get displaced at all.

Let n1 and n2 be the number of moles in the bulbs at 0°C and 20°C, respectively, then,

pV = 273n_{1}R = 293n_{2}R

$\frac{{\mathrm{n}}_{1}}{{\mathrm{n}}_{2}}=\frac{293}{273}...\left(\mathrm{i}\right)$

when the vessels are heated, let the volume of low temperature be V_{1} and that of the other be V_{2}. Since pressure are same, hence

V_{1}p = 283 n_{1}R and V_{2}p = 303 n_{2}R

$\frac{{\mathrm{V}}_{1}}{{\mathrm{V}}_{2}}=\frac{283{\mathrm{n}}_{1}}{303{\mathrm{n}}_{2}}...\left(\mathrm{ii}\right)$

From Eqs. (i) and (ii), we get

$\frac{{\mathrm{V}}_{1}}{{\mathrm{V}}_{2}}=\frac{283}{303}\mathrm{x}\frac{293}{273}=1$

Thus, the mercury pellet remains at the same position.