MHT CET 2017 :: Physics :: General questions

• Physics - General questions

The polarising angle for transparent medium is '$\theta$' and 'v'  is the speed of light in that medium. Then the relation between '$\theta$  and 'v' is 

(v=velocity of light in air)

Answer:

Explanation:

 From the relation , $\tan \theta= \mu=\frac{c}{v}$

$\Rightarrow$     $ \cot \theta=\frac{v}{c}$

$\therefore$     $\theta=\cot^{-1}\left(\frac{v}{c}\right)$

Two parallel plate air capacitors of same capacity C are connected in series to a battery of emf E. Then one of the capacitors is completely filled with dielectric material of constant K. The change in the effective capacity of the series combination is 

Answer:

Explanation:

initial effective capacity of the series combination is 

$\frac{1}{C_{1}}=\frac{1}{C_{}}+\frac{1}{C_{}}=\frac{2}{C_{}}$

$\Rightarrow$    $C_{1}=\frac{C}{2}$

 Effective capacity of the series combination  with dielectric material is 

 $\frac{1}{C_{1}}=\frac{1}{C}+\frac{1}{KC}$

 $\frac{1}{C_{2}}=\frac{1}{C_{}}\left[1+\frac{1}{K}\right]$

$\therefore$     $C_{2}=\frac{C}{\left(1+\frac{1}{K}\right)}=\frac{CK}{(K+1)}$

The change in the effective capacitance  is $\triangle C= C_{2}-C_{1}$

$=\frac{CK}{\left(1+K\right)}-\frac{C}{2}=C\left[\frac{K}{K+1}-\frac{1}{2}\right]$

$=C\left[\frac{2K-K-1}{( 2(K+1)}\right]=\frac{C}{2}\left[\frac{K-1}{K+1}\right]$

In Fraunhofer diffraction pattern, slit width is 0.2 mm and the screen is at 2m away from the lens,If wavelength of light used is 5000Å , then the distance between  the first minimum on either  side of the central maximum is 

($\theta$  is small and measured in radian)

Answer:

Explanation:

Given , a =0.2 x 10^-3 m, D=2 m

 $\lambda$  = 5 x 10^-7 m

 As,   $y=\frac{\lambda D}{a}=\frac{5\times 10^{-7}\times2}{0.2\times 10^{-3}}$

 $\therefore$    $y=\frac{5\times 10^{-7}}{ 10^{-4}}$

$\therefore$    $y=5\times 10^{-3}$ m

Distance between 1 st minima on either side

 $=5\times 10^{-3}+5\times 10^{-3}=10\times 10^{-3}=10^{-2}m$

The sensitivity of the moving coil galvanometer is 'S'.If  a shunt of ($\frac{1}{8}$) the of the resistance of galvanometer is connected to moving coil galvanometer, its sensitivity becomes

Answer:

Explanation:

Given, S= $\frac{G}{8}$ ............(i)

As, S=$ \frac{G}{n-1}$     ............(ii)

Comparing Eqs.(i) and (ii), we get

 $\frac{G}{8}=\frac{G}{n-1}$

 $\therefore$   8=n-1

$\therefore$    n=8+1=9

 Since the range of the galvanometer is increased by nine times, therefore its sensitivity reduces to $\frac{1}{9}$

$\therefore$   S'= $\frac{S}{9}$

The frequencies for series limit of Balmer and P aschen series respectively are 'v₁' and 'v₃' .If frequency of first line of Balmer series is 'v2' then the relation between 'v₁'. 'v₂' and 'v₃' is 

Answer:

Explanation:

 As we know, v= n$\lambda$

$\Rightarrow $   $\frac{1}{\lambda}=\frac{n}{v}\Rightarrow \frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$

$\Rightarrow $      $v=Rc\left(\frac{1}{n^{2}_{1}}-\frac{1}{n^{2}_{2}}\right)$

$\therefore$   

$v_{2}=Rc\left(\frac{1}{2^{2}}-\frac{1}{3^2}\right)=Rc\left(\frac{1}{4}-\frac{1}{9}\right)$ .....(i)

$v_{1}=Rc\left(\frac{1}{2^{2}}\right)=\frac{Rc}{4}$

 $v_{3}=Rc\left(\frac{1}{3^{2}}\right)=\frac{Rc}{9}$

 $\Rightarrow$        $v_{1}-v_{3}=Rc\left(\frac{1}{4}-\frac{1}{9}\right)$  ......(ii)

 From Eqs.(i) and (ii), we get

    v₁ -v₃  =v₂

 $\Rightarrow$      $v_{1}-v_{2}=v_{3}$

• Physics - General questions