Tag: properties of matter

Questions Related to properties of matter

Blood vessel is $0.10\ m$ in length and has a radius of $1.5\times{10}^{-3}m$. Blood flows at rate of ${10}^{-7}{m}^{-3}/s$ through this vessel. The pressure difference that must be maintained in this flow, between the two ends of the vessel is $20\ Pa$. What is the viscosity sufficient of blood?

  1. $2\times{10}^{-3}\ Pa-s$

  2. $1\times{10}^{-3}\ Pa-s$

  3. $4\times{10}^{-3}\ Pa-s$

  4. $5\times{10}^{-4}\ Pa-s$


Correct Option: C

A U-tube having identical limbs is partially filled with water. An immiscible oil having a density of 0.8 g/cc is poured into one side until the water rises by 25 cm on the other side. the level of oil will stand higher than the water level? 

  1. 6.25 cm

  2. 75 cm

  3. 22.5 cm

  4. 12.5 cm


Correct Option: A

A small sphere of mass M and density $D _1$ is dropped in a vessel filled with glycerine. If the density of glycerine is $D _2$ then the viscous force acting on the ball will be in Newton.

  1. $M D _1 D _2$

  2. $Mg \displaystyle \left [ 1- \frac {D _2}{D _1} \right ]$

  3. $\displaystyle \frac {M D _1 g}{D _2}$

  4. $\displaystyle \frac {M}{g} \left ( D _1 + D _2 \right)$


Correct Option: B
Explanation:

When sphere is in glycerine, three forces acts on it. which balances each other.

  • weight $(W)$
  • buoyant force $(F _B)$
  • viscous force $(F _V)$
and $W=F _B+F _V$
$\Rightarrow Mg=V _g D _2g +F _V$
$\Rightarrow Mg=V _s D _2g +F _V           \because V _g=V _s$
$\Rightarrow Mg=\frac{M}{D _1} D _2g +F _V             \because V _s=\frac{M}{D _1}$
$\Rightarrow F _V=Mg \left [ 1- \frac {D _2}{D _1} \right ]$

The viscous drag on a spherical body moving with a speed V is proportional to:

  1. $\sqrt V$

  2. $V$

  3. $\displaystyle \frac{1}{\sqrt V}$

  4. $V^{2}$


Correct Option: B
Explanation:

The viscous drag on a spherical body is given as $F=6\pi \eta RV$. Here $\eta$ is the coefficient of viscosity, R is the radius of the sphere and V is its velocity. 

An air bubble of radius $1 \,cm$ is found to rise in a cylindrical vessel of large radius at a steady rate of $0.2 \,cm$ per second. If the density of the liquid is $1470 \,kg \,m^{-3}$, then coefficient of viscosity of liquid is approximately equal to

  1. $163$ poise

  2. $163$ centi-poise

  3. $140$ poise

  4. $140$ centi-poise


Correct Option: B

A capillary tube of area of cross-section A is dipped in water vertically. The amount of heat evolved as the water rises in the capillary tube up to height h is: (The density of water is $\rho$)

  1. $\dfrac{A\rho gh^2}{2}$

  2. $Agh^2\rho$

  3. $2Agh^2\rho$

  4. None of these


Correct Option: A

Viscous force is somewhat like friction as it opposes, the motion and is non-conservative but not exactly so, because

  1. It is velocity dependent while friction is not

  2. It is velocity independent while friction is not

  3. It is temperature dependent while friction is not

  4. It is independent of area is like surface tension while friction is dependent


Correct Option: A,C
Explanation:

Using the relation $\tau=\mu\displaystyle\frac{du}{dy}$ we see that viscosity is velocity dependent. Also viscosity decreases with increasing temperature. Thus A and C are correct.

A liquid flows between two parallel plates along the x-axis. The difference between the velocity of two  layers separated by the distance $dy$ is $dv$. If $A$ is the area of each plate, then Newton's law of viscosity may be written as:

  1. $F=-\eta A\dfrac{dv}{dx}$

  2. $F=+\eta A\dfrac{dv}{dx}$

  3. $F=-\eta A\dfrac{dv}{dy}$

  4. $F=+\eta A\dfrac{dv}{dy}$


Correct Option: C
Explanation:

The Newton's viscous force acting between two liquid surfaces with relative velocity $dv$ and distance $dy$ between the layers is given as $-\eta A\dfrac{dv}{dy}$

If the shearing stress between the horizontal layers of water in a river is $1.5 mN/ m^{2}$ and $\eta  _{water}= 1\times10^{-3}Pa-s$ , The velocity gradient is:

  1. $1.5$

  2. $3$

  3. $0.7$

  4. $1$


Correct Option: A
Explanation:

Shearing stress  $=\eta \dfrac{dv}{dy}=1.5 \times 10^{-3} N /m^{2}$


$\Rightarrow 1 \times 10^{-3} \dfrac{dv}{dy}=1.5 \times 10^{-3}$

$\Rightarrow \dfrac{dv}{dy}=1.5 \ s^{-1}$

An air bubble of radius $1 mm$ moves up with uniform velocity of $0.109ms^{-1}$ in a liquid column of density $14.7 \times 10^{3} kg/m^{3}$, then coefficient of viscosity will be ($g = 10ms^{-2}$)

  1. $1.3 Pa- s$

  2. $300 Pa -s$

  3. $15 Pa- s$

  4. $150 Pa- s$


Correct Option: A
Explanation:

If the bubble is moving up with uniform velocity, the frictional force acting downwards is equal to the buoyant force acting upwards.

$\implies V\rho g=6\pi\eta rv$
$\implies \eta=\dfrac{V\rho g}{6\pi rv}$
$=0.3Pa.s$