JEE 2011 :: General Questions :: Physics

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5.6 L of helium gas at STP is adiabatically compressed to 0.7 L. Taking the initial temperature to be $T_{1}$, the work done in the process is

Answer:

Explanation:

 At STP  ,22.4 L of any gas is 1 mole

 $\therefore$ $5.6 L= \frac{5.6}{22.4}=\frac{1}{4} moles =n$

 In adiabatic process

  $T V ^{\gamma-1}= constant$

 $\therefore$    $T_{2} V_{2}^{\gamma-1}= T_{1} V_{1}^{\gamma-1}$

 or     $T_{2}=T_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma-1}$

 $\gamma= \frac{C_{p}}{C_{V}}=\frac{5}{3}$ for monoatomic He gas

 $\therefore$ $T_{2}= T_{1}\left(\frac{5.6}{0.7}\right)^{5/3-1}=4T_{1}$

 $\because$ Further in adiabatic process

 $Q=0$

 $\therefore$   $W+ \triangle U=0$

 or $W= -\triangle U$

 = $-n C_{v} \triangle T$

 =$-n\left(\frac{R}{\gamma-1}\right)(T_{2}-T_{1})$

  =$-\frac{1}{4}\left(\frac{R}{\frac{5}{3}-1}\right)(4T_{1}-T_{1})$

 = $-\frac{9}{8} RT_{1}$

 $\therefore$ correct option is (a) 

Analysis of Question
(i) From a calculation point of view question is moderately tough.
(ii) Volume of gas is decreasing. Therefore, work done by the gas should be negative.

A $2 \mu F$  capacitor i scharge as shown in the figure . The percentage of its stored energy dissipated after the switch  S is turned to position 2 , is 

Answer:

Explanation:

 $q_{i}=C_{i}V=2V=q(say)$

This charge will remain constant after switch is shifted from position 1 to position 2.

 $U_{i}= \frac{1}{2} \frac{q^{2}}{C_{i}}=\frac{q^{2}}{2 \times 2}=\frac{q^{2}}{4}$

$U_{f}= \frac{1}{2} \frac{q^{2}}{C_{f}}=\frac{q^{2}}{2 \times 10}=\frac{q^{2}}{20}$

 $\therefore$ Energy Dissipated = $U_{i}-U_{f}= \frac{q^{2}}{5}$

 This energy dissipated $\left(=\frac{q^{2}}{5}\right)$ is 80%

 of the initial stored energy  $\left(=\frac{q^{2}}{4}\right)$

Analysis of Question
(i) This question is moderately tough'
(ii) In a capacitor circuit, redistribution of charge takes Place under the following three conditions'
(a) A switch is closed.
(b) A closed switch is opened.
(c) A switch is shifted from one Position to another Position.
In the redistribution of charge, energy is dissipated

A ball of mass (m)0.5 kg is attached to the end of a string having a length (L)0.5 m. The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball (in rad/s) is

Answer:

Explanation:

T $\cos \theta$ component will cancel mg

 $T \sin \theta$  component will provide necessary centripetal force to the ball towards  centre C

 $\therefore$   $T \sin \theta =mr \omega^{2} = m(l \sin \theta) \omega^{2}$

 $\therefore$   $\omega= \sqrt{\frac{T}{ml}}$

 or    $\omega_{max}=\sqrt{\frac{T_{max}}{ml}}=\sqrt{\frac{324}{0.5 \times 0.5}}$

 =36 rad/s

 $\therefore$   correct option is (d) 

Analysis of Question
(i) Question is simple

(ii) This is called the conical pendulum

(iii)The interesting fact in this problem is that ω or T is independent of  $\theta$

 $\omega \propto \sqrt{T}$

 If $\omega$  is increased, T will also increase

 Consider an electric field $E= E_{0} \widehat{x}$ where  $\overline{ E_{0}}$  is a constant. The flux through the shaded area (as shown in the figure) due to the field is 

Answer:

Explanation:

Electric  flux , E.S , or $\phi= ES \cos \theta$

 here, $\theta$ is the angle between E and S 

 In this question $\theta= 45^{0}$ , because S is perpendicular to the surface 

 $E= E_{0}$

 $\Rightarrow$   $S= (\sqrt{2a} ) (a) = \sqrt{2} a^{2}$

 $\therefore$   $\phi = ( E_{0}) (\sqrt{2}a^{2}) \cos 45^{0}=E_{0}a^{2}$

 $\therefore$ correct option is (c)

(i) Question is moderatelY tough,
(ii) The given shaded area is a rectangle not a square. One side of this rectangle is a and other side is  $\sqrt{2} a$
(iii) Electric field is uniform, whose magnitude is E₀ and direction is positive x. In uniform electric field we can use , $\phi= ES \cos \theta$

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