Tag: option b: engineering physics

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Questions Related to option b: engineering physics

Multiple choice viscosity option b: engineering physics properties of matter physics

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The potential energy gained by the water column is m * g * h_cm = (A * h * rho) * g * (h/2) = (A * rho * g * h^2) / 2. This energy is provided by the surface tension work, and the heat evolved is the difference between the work done by surface tension and the potential energy gained.

Multiple choice viscosity option b: engineering physics properties of matter physics

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
Reveal answer Fill a bubble to check yourself
A,C Correct answer
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.

Multiple choice viscosity option b: engineering physics properties of matter physics

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. $F=+\eta A\dfrac{dv}{dy}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

Multiple choice viscosity option b: engineering physics properties of matter physics

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. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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


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

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

Multiple choice viscosity option b: engineering physics properties of matter physics

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. $150 Pa- s$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

Multiple choice viscosity option b: engineering physics properties of matter physics

Match List I with List II and select the correct answer using the codes given below the lists :

List I List II
p. Boltzmann constant 1. $[ML^2T^{-1}]$
q. Coefficient of viscosity 2. $[ML^{-1}T^{-1}]$
r. Planck constant 3. $[MLT^{-3}K^{-1}]$
s. Thermal conductivity 4. $[ML^2T^{-2}K^{-1}]$

  1. P - 3, Q - 1, R - 2, S - 4

  2. P - 3, Q - 2, R - 1, S - 4

  3. P - 4, Q - 2, R - 1, S - 3

  4. P - 4, Q - 1, R - 2, S - 3
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$c. p \rightarrow (4); q \rightarrow (2); r \rightarrow (1); s \rightarrow (3)$

Multiple choice viscosity option b: engineering physics properties of matter physics

The space between two large horizontal metal plates 6 cm apart, is filled with 
liquid of viscosity 0.8 $N/m^2.$ A thin plate of surface area 0.01 $m^2$ is moved  parallel to the length of the plate such that the plate is at a distance of 2 m  from one of the plates and 4 cm from the other. If the plate moves with a  constant speed of 1 m $s^{-1}$, then

  1. the layer of the fluid, which is having the maximum velocity, is lying mid-way between the plates

  2. the layers of the fluid, which is in contact with the moving plate, is having the maximum velocity

  3. the layer of the fluid, which is in Contact with the moving plate and is on the side of farther plate, is moving with the maximum velocity

  4. the layer of the fluid, which is in contact with the moving plant and is on the Side of nearer plate, is moving with the maximum velocity
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The profile of the velocity of fluid as shown in the figure The velocity of the layer of fluid ,which in contact with metal plates (fixed), is zero. As we move towards the centre from either plate the velocity of the layer of fluid increases and it becomes maximum at the location of moving plate. This maximum value is same as that of the velocity of plate

Multiple choice viscosity option b: engineering physics properties of matter physics

A solid ball of density half that of water falls freely under gravity from a height of 19.6 m and then enters the water. Up to what depth will the ball go? How much time will it take to come again to the water surface. Neglect air resistance and viscosity effects in water. ($
g=9.8 \mathrm{ms}^{-2}
$)

  1. 4 s

  2. 8 s

  3. 6 s

  4. 2 s
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The ball falls 19.6 m, reaching velocity v = sqrt(2gh) = sqrt(2 * 9.8 * 19.6) = 19.6 m/s. In water, the buoyant force exceeds gravity (density is half of water), so it decelerates. The time to stop and return is calculated using the effective acceleration in water.

Multiple choice viscosity option b: engineering physics properties of matter physics

A spherical ball of radius $3\times 10^{-4}\ m$ and density $10^{4}\ kg\ m^{-3}$ falls freely under gravity through a distance $h$ before entering a tank of water. If after entering the water, the velocity of the ball does not change, then the value of $h$ is (Given, $viscosity >of> water=9.8\times 10^{-6}\ Nsm^{-2}$ and $\rho _{water}=10^{3}\ kgm^{-3}$)

  1. $1650\ m$
  2. $165\ m$
  3. $1050\ m$
  4. $105\ m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If the velocity does not change upon entering water, the drag force must balance the net weight in water. F_drag = (m_ball - m_displaced) * g. Using Stokes' Law for F_drag and equating, we solve for the velocity v, then use v^2 = 2gh to find h.

Multiple choice physics option b: engineering physics introduction to sound free, forced and damped oscillations resonance

Assertion (A): In damped vibrations, amplitude of oscillation decreases
Reason (R): Damped vibrations indicate loss of energy due to air resistance

  1. Both A and R are true and R is the correct explanation of A

  2. Both A and R are true and R is not the correct explanation of A

  3. A is true and R is false

  4. A is false and R is true
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Damped vibrations in which an oscillating system has the effect of reducing, restricting or preventing its oscillations.