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Match list I (electromagnetic wave type)  with List II (Its association/application  ) and select  the correct option from the choices given below the lists

Answer:

Explanation:

(a) Infrared rays are used to treat muscular strain

(b) Radiowaves are used for broadcasting purpose

(c) X-rays are used to detect fracture of bones.

 (d) Ultraviolet  rays are absorbed by ozone.

A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?

Answer:

Explanation:

If student measures 3.50 cm. it means that there is an uncertanily of order 0.01cm. For vernier scale with 1MSD=1mm

 and                          9MSD=10VSD

LC of vernier caliper =1 MSD-1VSD

                  = $\frac{1}{10}\left(1-\frac{9}{10}\right)$

                =  $\frac{1}{100}$ cm

Hydrogen (₁H¹), deuterium (₁H²), singly ionised helium   $(_{2}He^{4})^{+}$   and doubly ionised lithium$(_{3}Li^{8})^{++}$ all have one electron around the nucleus. Consider an electron transition from n=2 to n=1 . If the wavelength s of emitted radiation are   $\lambda_{1},\lambda_{2},\lambda_{3}$ and   $\lambda_{4}$, respectively for four elements, then approximately which one of the following is correct?

Answer:

Explanation:

For hydrogen atom, we get

 $\frac{1}{\lambda}=Rz^{2}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)\Rightarrow\frac{1}{\lambda_{1}}=R(1)^{2}\left(\frac{3}{4}\right)$

 $\frac{1}{\lambda_{2}}=R(1)^{2}\left(\frac{3}{4}\right)$

$\Rightarrow$          $\frac{1}{\lambda_{3}}=R(2)^{2}\left(\frac{3}{4}\right)$

                    $\frac{1}{\lambda_{4}}=R(3)^{2}\left(\frac{3}{4}\right)$

                       $\Rightarrow\frac{1}{\lambda_{1}}=\frac{1}{4\lambda_{3}}=\frac{1}{9\lambda_{4}}=\frac{1}{\lambda_{2}}$

The radiation corresponding to 3→ 2 transition of hydrogen atom falls on a metal surface to produce photo electrons . These electrons are made toi enter a magnetic field of  $3\times 10^{-4}T$ . If the radius of the largest circular path followed by these electrons is 10.0mm. the work function of the metal is close to 

Answer:

Explanation:

The problem is based on frequency dependence of photoelectric emission. When incident light with certain frequency (greater than on the threshold frequency is focused on a metal surface) then some electrons are emitted from the metal with substantial initial speed.

 When an electron moves in a circular path, then

               $r=\frac{mv}{eB}$

$\Rightarrow $       $\frac{r^{2}e^{2}B^{2}}{2}=\frac{m^{2}v^{2}}{2}$

$KE_{max}=\frac{(mv)^{2}}{2m}$

             $\Rightarrow$         $\frac{r^{2}e^{2}B^{2}}{2m}=KE_{max}$

 Work function  of the metal(W)

i.e,  $W=hv=KE_{max}$

  $1.89-\phi= \frac{r^{2}e^{2}B^{2}}{2m}\frac{1}{2}eV$

                             $= \frac{r^{2}e^{}B^{2}}{2m}eV$

[hv→1.89 eV, for the transition on from third to second orbit of H-atom]

=  $\frac{100\times10^{-6}\times1.6\times 10^{-19}\times9\times 10^{-8}}{2\times9.1\times10^{-31}}$

$\phi=1.89-\frac{1.6\times9}{2\times9.1}$

   =1.89-0.79=1.1 eV

Two beams, A and B , of plane  polarised light with mutually perpendicular planes of polarisation are seen through a polarised. From the position  when the beam  A has  maximum intensity (and beam B has  zero intensity)  , a rotation of polarised through 30° makes the two beams appear equally bright. If the initial intensities  of the  two beams are I_A and I_B  respectively, then I_A/I_B equals

Answer:

Explanation:

By law of Malus  i.e,   $I=I_{0}\cos^{2}\theta$

Now,   $I_{A'}=I_{A}\cos^{2}30^{0}$

            $I_{B'}=I_{B}\cos^{2}60^{0}$

As,           $I_{A'}=I_{B'}$

           $ I_{A}\cos^{2}30^{0}$=  $ I_{B}\cos^{2}60^{0}$

     $\Rightarrow$          $ I_{A} \frac{3}{4}=I_{B}\frac{1}{4}$

$\Rightarrow$       $\frac{I_{A}}{I_{B}}=\frac{1}{3}$

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