JEE 2018 :: Physics (2018) :: Solved questions 2018

• Physics (2018) - Solved questions 2018

Two moles of an ideal monoatomic gas occupies a volume V at 27° C. The gas expands adiabatically to a Volume 2 V. Calculate(i) the final temperature of the gas and (ii) change in its internal energy.

Answer:

Explanation:

For adiabatic process relation of temperature and volume is,

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

$T_{2}(2V_{})^{\frac{2}{3}}=300_{}(V_{})^{\frac{2}{3}}$

           [ $\gamma=\frac{5}{3} $ for monoatomic gases ]

$T_{2}=\frac{300}{2^{\frac{2}{3}}}\approx 189K$

 Also, in adiabatic process,

  $\triangle Q=0, \triangle U=-\triangle W$

 OR       $\triangle U=\frac{-nR(\triangle T)}{\gamma -1}$

  $= -2\times \frac{3}{2}\times \frac{25}{3}(300-189)$

    $\approx -2.7KJ$

     $T_{2}\approx 189K,\triangle U\approx2.7KJ$

Seven identical circular planar discs, each of mass M and radius R are welded symmetrically as show in figure. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is

Answer:

Explanation:

First we found moment of inertia (MI) of system using parallel axis theorem about centre of mass, then we use it to find moment of inertia about given axis.

Moment of inertia of an outer disc about the axis through centre is

= $\frac{MR^{2}}{2}+M(2K)^{2}=MR^{2}(4+\frac{1}{2})=\frac{9}{2}MR^{2}$

For 6 such discs,

  Moment of inertia = $6\times \frac{9}{2}MR^{2}=27MR^{2}$

 So, moment of inertia of system

           = $\frac{MR^{2}}{2}-27MR^{2}=\frac{55}{2}MR^{2}$

Hence,

   $I_{p}=\frac{55}{2}MR^{2}+(7M\times 9R^{2})$

    $I_{p}=\frac{181}{2}MR^{2}$

    $I_{system}=\frac{181}{2}MR^{2}$

A parallel plate capacitor of capacitance 90pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant K=$\frac{5}{3}$  is inserted between  the plates ,the magnitude of the induced charge will be

Answer:

Explanation:

Magnitude of induced charge is given by Q' = (K-1)CV₀

= $(\frac{5}{3}-1)90\times 10^{12}\times 20$

= $1.2\times 10^{-9}C$

$\Rightarrow$     Q'=1.2nC

A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 10¹² per second. What is the force constant of the bonds connecting one atom with other? ( Take, molecular weight of silver=108 and Avogadro  number=6.02×10²³ g mol^-1)

Answer:

Explanation:

 For a harmonic oscillator.

$T=2\pi \sqrt{\frac{m}{k}}$ , where k= force constant and

$T=\frac{1}{f}$

$\therefore$      $k =4\pi^{2}f^{2}m$

$=4\times (\frac{22}{7})^{2}\times (10^{12})^{2}\times \frac{108 \times 10^{-3}}{6.02\times 10^{23}}$

$\Rightarrow$        k=7.1 N/m

The mass of a hydrogen molecule is 3.32 x10^-27 kg. If 10²³hydrogen molecules strike per second, a fixed wall of area 2 cm² at an angle of 45° to the normal and rebound elastically with a speed of 10³ m/s, then the pressure on the wall is nearly

Answer:

Explanation:

Momentum imparted due to first collision

$\Rightarrow 2mv\sin45$ =$\sqrt{2mv}$      \[[\therefore sin45^{0}=\frac{1}{\sqrt{2}}]\]

$\therefore$    Pressure on surface

 = $\frac{n\sqrt{2}mv}{Area}=\frac{10^{23}\times \sqrt{2}\times 3.32\times 10^{-27} \times10^{3}}{(2\times  10^{-2})^{2}}$

$p =2.35\times 10^{3} N/m^{2}$

• Physics (2018) - Solved questions 2018