MHT CET 2016 :: Physics :: General questions

• Physics - General questions

A wire having tension 225 N  produces six beats  per second when it is tuned with a fork .When tension changes to 256 n , it is tuned with the same fork, the number of beats remains unchanged. The frequency of the fork will be 

Answer:

Explanation:

The fundamental frequency of wire vibrating  under tension T is given by

$f_{1}=\frac{1}{2}\sqrt{\frac{T}{\mu}}$

Here, $\mu$= mass of string per unit length

Let the frequency of tunning fork be x which beat 6 beats per second

Let (x-f)=$\pm$ 6

$\therefore$     $f_{1}=\frac{1}{2L}\sqrt{\frac{225}{\mu}}$ and $f_{2}=\frac{1}{2L}\sqrt{\frac{256}{\mu}}$

$\therefore$   $\frac{f_{1}}{f_{2}}=\frac{15}{16}$

$\therefore$    $f_{2}=\frac{16}{15}\times f_{1}$

$\Rightarrow$    $f_{2}=\frac{16}{15} (6+x)$

Taking two cases of f₁ and equating both , we have

$(x+6)=\frac{16}{15} (x-6)$

 $\therefore$      15x+90=16x-96

 $\therefore$     x=186 Hz

A ray of light travelling through rarer medium is incident at very small angle i on a glass slab and after refraction, its velocity is reduced by 20%. The angle of deviation  is

Answer:

Explanation:

According to question , at glass-air interface velocity is reduced by 20% of the velocity of light

So, deviation $\delta$ =20% of i

 =  $\frac{20\times i}{100}=\frac{i}{5}$

Light of wavelength  $\lambda$  which is less than threshold wavelength is incident on a photosensitive material. if the incident wavelength  is decreased so that emitted photoelectrons are moving with the same velocity, then stopping potential will

Answer:

Explanation:

According to photo electric equation,

$\frac{hc}{\lambda}$=$\phi$+E

 symbols have their usual  meaning

 E= kinetic energy

if E is constant, then $\frac {1}{\lambda}$ $\propto \phi$

If we decrease the wavelength $\lambda$  , then stopping potential $\phi$  will increase such that 

$\frac{hc}{\lambda}-\phi= constant$

 Three parallel  plate air capacitors are connected in parallel. Each capacitor  has plate area $\frac {A}{3}$  and the separation between the plates is d, 2d and 3d respectively. The equivalent  capacity  of combination  is  $(\epsilon _{0}=$ absolute permitivity of free space)

Answer:

Explanation:

 According to question, capacitance of parallel plate capacitor  is given by $C=\frac{\epsilon_{0 A}}{d}$

 For first capacitor  , $C_{1}=\frac{\epsilon_{0 A}}{3d}$

For second capacitor = $C_{2}=\frac{\epsilon_{0 A}}{6d}$

For third capacitor , $C_{3}=\frac{\epsilon_{0 A}}{9d}$

 Again they are arranged in parallel 

combination , so equivalent capacitor is given by

    $ C_{eq}= C_{1}+C_{2}+C_{3}$

 $=\frac{\epsilon_{0 A}}{d}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{9}\right)=\frac{\epsilon_{0 A}}{d}\times\frac{11}{18}=\frac{11\epsilon_{0 A}}{18d}$

In a balanced meter bridge, the resistance of the bridge wire is 0.1 $\Omega$ cm. Unknown resistance X is connected in left gap and 6 $\Omega$  in the right gap.  null point divides the wire in the ratio  2:3. find the current drawn  from the battery of 5 v having negligible resistance

Answer:

Explanation:

According to question

$\frac{l_{1}}{l_{2}}=\frac{2}{3}$            (given)

 For potentimeter,   $\frac{R_{1}}{R_{2}}=\frac{l_{1}}{l_{2}}=\frac{2}{3}$

$\frac{x}{6}=\frac{2}{3}$

$\Rightarrow$       x= 4 $\Omega$m^-1

 Total resistance of known and unknown resistance =6+4=10 $\Omega$

The resistance of wire is 0.1 x 100= 10 $\Omega$

 So, effective resistance R_eff is given by

 $\frac{1}{R_{eff}}=\frac{1}{10}+\frac{1}{10}$

 $\Rightarrow$   R_eff= 5 $\Omega$

 Hence  , current drawn from the battery is given by Ohm'sa law  , V=IR

 $\Rightarrow$      I= $\frac{V}{R}=\frac {5}{5}$ = 1 A

• Physics - General questions