Q121 / 160 · AP EAMCET 2016 · Physics
Match the List-I with List-II.
The correct match in the following is
The correct match in the following is
- A$\begin{array}{llll} A & B & C & D \end{array}$ $\begin{array}{llll}\text { III } & \text { I } & \text { II } & \text { IV }\end{array}$
- B$\begin{array}{llll} A & B & C & D \end{array}$ $\begin{array}{llll}\text { III } & \text { II } & \text { I } & \text { IV }\end{array}$
- Correct answerC$\begin{array}{llll} A & B & C & D \end{array}$ $\begin{array}{llll}\text { IV } & \text { II } & \text { I } & \text { III}\end{array}$
- D$\begin{array}{llll} A & B & C & D \end{array}$ $\begin{array}{llll}\text { IV } & \text { I } & \text { II } & \text { III}\end{array}$
Solution
(A) The value of Boltzmann constant $=1.38 \times 10^{-23}$ $\mathrm{kg} \mathrm{m}^2 \mathrm{~s}^{-2} \mathrm{k}^{-1}$
So dimensions of Boltzmann constant $=\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right]$
(B) From stokes' law, $\mathrm{F}=6 \pi \eta v r$
Coefficient of viscosity $\eta=\frac{\mathrm{F}}{6 \pi v r}$
Coefficient of viscosity $=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]$
(C) Water equivalent $(\mathrm{W})$ is the mass of water which would absorb or evolve the same amount of heat as in done by the body in rising or falling through the same range of temperature.
Water equivalent $=\left[\mathrm{ML}^0 \mathrm{~T}^0\right]$
(D) Since $\frac{Q}{t}=\frac{K A\left(\theta_1-\theta_2\right)}{l}$
$\therefore$ Coefficient of thermal conductivity
$K=\frac{Q . l}{A\left(\theta_1-\theta_2\right) t} \text { so }$
Coefficient of thermal conductivity $=\left[\mathrm{MLT}^{-3} \mathrm{~K}^{-1}\right]$
Hence, the correct answer is option: 'C'.