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An ideal gas undergoes a quasi-static reversible process in which its molar heat capacity C remains constant. If during this process the relation of  pressure p and volume  V is given by pVⁿ = constant, then n is given by (here C_p and C_v are molar specific heat at constant pressure and constant volume respectively)

Answer:

Explanation:

$\triangle Q= \triangle U+\triangle W$

 In the process pVⁿ  = constant, molar heat capacity is given by

$C= \frac{R}{\gamma -1}+\frac{R}{1-n}=C_{v}+\frac{R}{1-n}$

$C-C_{v}= \frac{R}{1-n}$

 $\Rightarrow  1-n=\frac{C_{p}-C_{v}}{C-C_{v}}$

  $\therefore  n= 1-(\frac{C_{p}-C_{v}}{C-C_{v}})$

$\frac{(C-C_{v})-(C_{p}-C_{v})}{C-C_{v}}$

$n=\frac{C-C_{p}}{C-C_{v}}$

A pendulum clock loses  12 s a day if the temperature is 40° C and gains 4 s a day if the temperature is 20° C. The temperature at which the clock will show the correct time and the coefficient of linear expansion α of the metal of the pendulum shaft are, respectively

Answer:

Explanation:

$T_{0}=2\pi\sqrt{\frac{L}{g}}$

$T^{'}=T_{0}+\triangle T=2\pi\sqrt{\frac{L+\triangle L}{g}}$

$T^{'}=T_{0}+\triangle T=2\pi\sqrt{\frac{L(1+\propto\triangle\theta) }{g}}$

$= \left\{2\pi \sqrt{\frac{L}{g}}\right\} (1+\alpha \triangle \theta)^{\frac{1}{2}}=T_{0}(1+\frac{\alpha\triangle\theta}{2})$

$\therefore$       $\triangle T=T'-T_{0}=\frac{\alpha\triangle \theta T_{0}}{2}$

$        \frac{\triangle T_{1}}{\triangle T_{2}}=\frac{\alpha \triangle \theta_{1}T_{0}}{\alpha \triangle \theta_{2}T_{0}}$

$\Rightarrow$   $\frac{12}{4}=\frac{40^{0}-\theta}{\theta-20^{0}}$

$\Rightarrow$   $3(\theta -20^{0})=40^{0}-\theta$

$\Rightarrow$     $4\theta=100^{0}$

$\Rightarrow$     $\theta=25^{0}C$

 Time gained or lost given by

    $\triangle T= (\frac{\triangle T}{T_{0}+\triangle T})t=\frac{\triangle t}{T_{0}}t$

From Eq.(i0.

   $\frac{\triangle T}{T_{0}}=\frac{\alpha \triangle \theta}{2}$

$\triangle t= \frac{\alpha (\triangle \theta)t}{2}$

$12= \frac{\alpha(40^{0}-25^{0})(24\times 3600)}{2}$

$\alpha =1.85\times 10^{-5}/^{0}C$

A satellite is revolving in a circular orbit at a height h from the Earth's surface (radius of earth R, h<<R). The minimum I lease in its orbital velocity required so that the satellite could escape from the earth's gravitational field, is close to (neglect the effect of the atmosphere.)

Answer:

Explanation:

$v_{orbital}= \sqrt{\frac{GM}{R}}=\sqrt{gR}$

$v_{escape}= \sqrt{2gR}$

Extra velocity required

           $=v_{escape}-v_{orbital}=\sqrt{gR}(\sqrt{2}-1)$

A roller is made by joining together two corners at their vertices O. It is kept on two rails AB and CD which are placed asymmetrically (see the figure), with its axis perpendicular to CD and its centre O at the centre of the line joining AB and CD (see the figure). It is given a light push so that it starts rolling with its centre O moving parallel to CD  in the direction shown. As it moves, the roller will tend to 

Answer:

Explanation:

At distance x, from O,v= ω R

distance less than x_0,v > ω R

 Initially, there is pure rolling at both the contacts  As the cone moves forward slipping at AB will start in the forward direction, as the radius  at left contact decreases.

               Thus, the cone will start turning towards the left. As it moves further slipping at CD will start in the backward direction which will also turn the cone towards left.

A person trying to lose weight by burning fat lifts a mass of 10kg up to a height of 1 m  1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies 3.8 × 10⁷ J of energy per kg which is converted to mechanical energy with a 20% efficiency rate (Take, g=9.8 ms^-2)

Answer:

Explanation:

 Work done in lifting the mass

                         =(10 × 9.8 × 1)× 1000

if m is mass of far burnt, then energy

                   = $m\times 3.8\times 10^{7} \times\frac{20}{100}$

Equating the two , we get

      $\therefore$   $m= \frac{49}{3.8}=12.89 \times 10^{-3}$ kg 

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