Q196 / 200 · WBJEE 2014 · Physics

$A$ heating element of resistance $r$ is fitted inside an adiabatic cylinder which carries a frictionless piston of mass $m$ and cross section A as shown in diagram. The cylinder contains one mole diatomic current the element temperatures time $t$ as (a and $\beta$ are constants), while pressure remains constant. The atmospheric pressure above the piston is $P_{0}$. Then

  • Correct answer
    A
    the rate of increase in internal energy is $\frac{5}{2} R(\alpha+\beta t)$
  • B
    the current flowing in the element is $\sqrt{\frac{5}{2 r} R(\alpha+\beta t)}$
  • C
    the piston moves upwards with constant acceleration
  • D
    the piston moves upwards with constant speed

Solution

We know that Internal energy, $U=\frac{n f R T}{2}=\frac{5 R}{2}\left(\alpha t+\frac{1}{2} \beta t^{2}\right)$ Differentiate with respect to $t$ $$ \frac{d U}{d t}=\frac{5 R}{2}[\alpha+\beta t] $$ But, $d Q=n C_{p} d T$ $$ \begin{array}{l} \therefore \quad \frac{d Q}{d t}=n C_{p} \times \frac{d T}{d t} \Rightarrow i^{2} r=\frac{7}{2} R \times[\alpha+\beta t] \\ \quad i=\sqrt{\frac{7 R}{2 r}(\alpha+\beta t)} \\ p V=n R T \end{array} $$ or $\quad V=\frac{n R T}{p}=\frac{n R}{p}\left(\alpha t+\frac{1}{2} \beta t^{2}\right)$ $X=\frac{n R}{p A}\left(\alpha t+\frac{1}{2} \beta t^{2}\right)$ $$ v=\frac{n R}{p A}(\alpha+\beta t) $$ and acceleration $=\frac{n R}{p A} \times \beta$ So, the rate of increase in internal energy is $\frac{5}{2} R(\alpha+\beta t)$ and the piston moves upwards with constant acceleration.

Hence, the correct answer is option: 'A'.

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