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

• Physics (2018) - Solved questions 2018

One mole of a monoatomic ideal gas undergoes four thermodynamic processes as shown schematically in the pV - diagram below. Among these four processes, one is isobaric, one is isochoric , one is isothermal and one is adiabatic. Match the processes mentioned in List -I with the corresponding statements in List -II

Answer:

Explanation:

$(\frac{\text{d}p}{\text{d}V})_{adiabatic} =\gamma (\frac{\text{d}p}{\text{d}V})_{isothermal}$

 List -I

  (P) Process I     $\Rightarrow$   Adiabatic $\Rightarrow$ Q =0

  (Q) Process II  $\Rightarrow$ Isobaric

   W =pΔV= 3 p₀ (3 V₀ -V₀)= 6 p₀ V₀

    (R) Process III $\Rightarrow$ Isobaric $\Rightarrow$ W=0

      (S) Process (IV)  $\Rightarrow$ Isothermal $\Rightarrow$Temperature$\Rightarrow$ Constant

Consider a hydrogen -like ionised atom with atomic number Z , with  a single electron. In the emission spectrum  of atom, the photon emitted  in the n= 2 to n=1 transition has energy 74.8 eV higher than the photon emitted in the n=3 to n=2 transition.The ionisation energy of the hydrogen atom is 13.6 eV. The value of Z is..........

Answer:

Explanation:

$\triangle E_{2-1}=13.6\times Z^{2}[1-\frac{1}{4}]=13.6 \times Z^{2}[\frac{3}{4}]$

$\triangle E_{3-2}=13.6\times Z^{2}[\frac{1}{4}-\frac{1}{9}]=13.6 \times Z^{2}[\frac{5}{36}]$

 $\therefore \triangle E_{2}=\triangle E_{3-2}+74.8$

$=13.6\times Z^{2}[\frac{3}{4}]=13.6 \times Z^{2}[\frac{5}{36}]$ + 74.8

$=13.6\times Z^{2}[\frac{3}{4}-\frac{5}{36}]=74.8$

$ Z^{2}=9$

Z=3

In a photoelectric experiment , a parallel beam of monochromatic light with power of 200 W is incident on a perfectly absorbing cathode of work function 6.25 eV  . The frequency of light is just above the threshold frequency, so that the photoelectrons are emitted with negligible kinetic energy.  Assume that the photoelectron emission  efficiency is 100% . A potential difference of 500 V  is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force F= n × 10^-4 N due to the impact of the eletrons^.  The value of n is .................(Take mass of the electron, m_e =9× 10^-31 kg and eV = 1.6 × 10^-19 J)

Answer:

Explanation:

$\therefore$  Power= nhf

       (where , n=number of photons incident per second)

Since, KE= 0, hf = work-function W

    200=nW= n[6.25×1.6 × 10^-19 ]

$\Rightarrow$    $n=\frac{200}{1.6\times10^{-19}\times6.25}$

  As photon is just above threshold frequency KE_max is zero and they are accelerated by potential difference of 500V.

$\therefore$              KE_f =qΔV

                            $\frac{P^{2}}{2m}=q\triangle V$

$\Rightarrow$                  $P= \sqrt{2mq\triangle V}$

 Since,  efficiency is 100% , number of eletrons emitted per second= number of photons incident per second.

    As, photon  is completely absorbed, force excerted

$=n(mV)=nP=n \sqrt{2mq\triangle V}$

= $\frac{200}{6.25\times 1.6\times10^{-19}}\times\sqrt{2(9\times10^{-31})\times1.6\times10^{-19}\times500}$

= 24

One mole of monoatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is 100K and the universal gas constant  R= 8.0 j mol^-1 K^-1 , the decrease in its internal energy in joule, is ..............

Answer:

Explanation:

n=1,  $r=\frac{5}{3}$

T-V equation in adiabatic process is

TV^r-1 = constant

T₁ V₁^r-1   =T₂ V₂^r-1

$\Rightarrow$ T₂ = T₁  $(\frac{V_{1}}{V_{2}})^{r-1}$   = $100\times (\frac{1}{8})^{\frac{2}{3}}$

$\Rightarrow$     T₂ =25 K

$C_{v}=\frac{3}{2}R $ for monoactomic gas

$\therefore$   $\triangle U = n C_{v}\triangle T= n\times (\frac{3R}{2}) (T_{2}-T_{1})$

=  $1 \times \frac{3}{2}\times 8 \times (25-100)$

=  -900 J

Therefore, Decreases in internal energy = 900 J

A steel wire of diameter 0.5 mm and Young's modulus 2 × 10¹¹ N m^-2   carries a load of mass m.  The length of the wire with the load is 1.0 m. A  vernier scale with 10 divisions is attached to the end of this wire. Next to the steel wire is reference wire to which a main scale, of least count 1.0 mm , is attached. The 10 divisions of the vernier scale correspond to 9 divisions of the main scale. Initially , the zero of vernier scale coincides with the zero  of main scale . If the load on the steel wire is increased by 1.2 kg, the vernier scale division which concides with a main scale division is ...........   Take , g= 10 m s^-1 and π =3.2)

Answer:

Explanation:

Given , d= 0.5mm

   Y = 2 × 10¹¹   Nm^-2 

   l= 1 m

  $\triangle l= \frac{Fl}{AY}= \frac{mgl}{\frac{\pi d^{2}}{4}}Y$

    =   $\frac{1.2 \times 10 \times 1}{\frac{\pi}{4}\times (5\times 10^{-4})^{2}\times2\times 10^{11}}$

   = 0.3 mm

  LC of vernier = $(1-\frac{9}{10})mm = 0.1mm$

   So, 3rd division of vernier scale will coincide with main scale.

• Physics (2018) - Solved questions 2018