MHT CET 2016 :: Physics :: General questions

• Physics - General questions

Two coherent sources P and Q produce interference at point A on the screen where there is a dark band that is formed between 4th bright band and 5th bright band. The wavelength of light used is 6000 Å . The path difference between  PA and QA is 

Answer:

Explanation:

 Path difference $\triangle $x is given by  $\frac{\lambda\delta}{2\pi}$

  [ where, $\lambda$= wave length of light , $\delta$ = phase difference ]

 for n= 4th dark fringe

                     $\delta$ =(2n-1)$\pi$

 $\Rightarrow$    $\delta = (8+1)\pi=9\pi$

 $\triangle x=\frac{\lambda 9\pi}{2\pi}=\frac{9}{2}\lambda=\frac{9}{2}\times6\times 10^{-7}$

$ 2.7 \times 10^{-6}$ m= $ 2.7 \times 10^{-4}$ cm

A galvanometer of resistance 30$\Omega$  is connected to a battery of emf 2 V with 1970 $\Omega$ resistance in series. A full scale deflection of 20 divisions is obtained in the galvanometer. To reduce the deflection to 10 divisions, the resistance in series required is 

Answer:

Explanation:

According to question

 Net current through  the galvanometer is given by

$l=\frac{V}{R_{eff}}=\frac{2}{1970+30}=\frac{2}{2000}= 10^{-3} A$

As we know, this current provides full scale deflection (i.e, 20 div) . In order to limit the deflection  to 10 divisions , the resistance needed to connect such that the current  reduces can be obtained as

 $\theta= \frac{nlAB}{k}(i.e, \theta \propto l)$

   [symbols have their usual meanings ]

$\Rightarrow $    $\frac{\theta_{1}}{\theta_{2}}=\frac{l_{1}}{l_{2}}=2$

$\Rightarrow$   $l_{2}=\frac{l_{1}}{2}=\frac{10^{-3}}{2}=5 \times 10^{-4} A$

Again      $l= \frac{V}{R_{eff}+R_{5}}$

 $\Rightarrow$     $R_{5}= \frac{V}{l}-R_{eff}$

     $=\frac{2}{5\times10^{-4}}-2000$   

    = $4 \times 10 ^{3}-2000$=2000 $\Omega$

 So, the resistance  of 1970 $\Omega$  is to be replaced by 1970+2000=3970 $\Omega$

The L-C  parallel resonant circuit 

Answer:

Explanation:

 In case of a parallel resonance L-C circuit the capacitive and inductive resistance equal to each other, the total impedance of the circuit increases to infinity meaning the circuit draws no current from the AC power source.

In Bohr 's theory of hydrogen atom,  the electron jumps from higher orbit n to lower orbit p. The wavelength will be minimum for the transition

Answer:

Explanation:

 According top question,

From Bohr's theory , the wavelength of radiation emitted is given by 

$\frac{1}{\lambda}=R\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$

   [Symbols have their usual meanings]

 as the energy difference $\triangle E=E_{5}-E_{4}$ is very small, so minimum wavelength will take place for transition of electron from n=5 to p=4

A disc of radius R and thickness  $\frac {R}{6}$  has moment of interia l about an axis passing through its centre and perpendicular to its plane. Disc  is melted and recast into a solid sphere. The moment  of interia of a sphere about its diameter is 

Answer:

Explanation:

According to question, Moment of inertia of disc is given by $l= \frac{MR^{2}}{2}$

 [symbols have their usual meanings]

 When the disc is remoulded into solid sphere, then volume remains same.

 i.e, Volume of disc= Volume of solid sphere

i.e,   $\pi R^{2}\times\frac{R}{6}=\frac{4}{3}\pi r^{3}$

 $\Rightarrow$    $r^{3}=\frac{R^{3}}{8}\Rightarrow r=\frac{R}{2}$

 Now,moment of inertia of solid sphere is given by $\frac{2}{5}m r^{2}$

 $=\frac{2}{5}\times m\times\frac{R^{2}}{4}=\frac{mR^{2}}{10}$

 =$\frac{l}{5}$  [l= moment of inertia of disc]

• Physics - General questions