MHT CET 2019 :: Physics :: General questions

• Physics - General questions

In the biprism experiment, the distance between source and eyepiece is 1.2 m, the distance between two virtual sources is 0.84 mm. Then the wavelength of light  used if the eyepiece is to be moved transversely through a distance of 2.799 cm to shift 30 fringes is 

Answer:

Explanation:

 In a biprism  experiment, wavelength of the light is given as 

                $\lambda=\frac{d\beta}{D}$  .......(i)

 where, $\beta$ is the fringe width, 

d is the distance between the two sources and D is the distance between the source and eyepeice.

 Given  , D=1.2 m

    d= 0.84 mm=0.84 x 10^-3 m and 

   $\beta=\frac{2.799\times 10^{-2}}{30}=9.33 \times 10^{-4}$

  Substituting these values in Eq.(i) , we get

$\lambda=\frac{0.84 \times 10^{-3}\times9.33 \times 10^{-4}}{1.2}=6.531 \times 10^{-7}m =$

     =6531  Å

A clock pendulum having a coefficient of linear expansion.  $\alpha= 9\times 10^{-7}/^{0}C^{-1}$ has a period of 0.5 s at $20^{0}$ C. If the clock is used in a climate, where the temperature is  $30^{0}$ C, how much time does the clock lose in each oscillation?

Answer:

Explanation:

 Given,  the coefficient of linear expansion,

  $\alpha=9\times10^{-7}{^{o}C^{-1}}$

 Initial time period  , T₀  =0.5 s, initial  temperature T_i= 20° C and final temperature , T_f= 30° C

 Expansion  in length , $\triangle l=l\propto(T_{f}-T_{i})$

$\Rightarrow$    $\triangle l=l\times 9\times 10^{-7}(30-20)$

 Now, the time period of pendulum,

$T=2\pi\sqrt{\frac{l}{g}}$

 error in time period,

$\Rightarrow $    $ \frac{\triangle t}{T}=\frac{1}{2}\frac{\triangle l}{l}+\frac{1}{2}\frac{\triangle g}{g}$

 since, $\triangle g$ =0

$\Rightarrow $    $ \frac{\triangle T}{T}=\frac{1}{2}\frac{\triangle l}{l}$

  Now, substituting  values in the above equation, we get

$\Rightarrow$      $  \frac{\triangle T}{0.5}=\frac{1}{2}\left[\frac{l\times 9 \times 10^{-7}}{l}(30-20)\right]$

$\Rightarrow$    $ \triangle T=\frac{0.5\times 9 \times 10^{-7}\times10}{2}$

$\Rightarrow$    $ \triangle T=2.25 \times 10^{-6}s$

 A galvanometer has a resistance of 100 $\Omega$ and a current of 10 mA produces full-scale deflection in it. The resistance to being connected in series, to get a voltmeter of range 50 volt is 

Answer:

Explanation:

 A galvanometer is converted  into a voltmeter by connecting  a resistance in series with the galvanometer as shown in the circuit diagram.

 where, r is resistance of the resistor connected in series.

 Given galvanometer resistance ,$ R_{G}$  = 100 $\Omega$

 Voltmeter range V_max = 50 V and full deflection current l_fl  = 10 mA

 So, by applying the KVL in above  circuit diagram,

 $V_{AB}=100l_{fl}+Rl_{fl}$

 $\Rightarrow V_{AB}=(100+R)l_{fl}$

 $\therefore$    For a 50 V voltmeter  range there must be

    $V_{AB}$= 50 V  and  $l_{fl}$= 10 mA

 Now,  substituting the values of V_AB  and l_fl  in Eq.(i)

 we get,

 $50=(100+R)10\times10^{-3}$

$\therefore$    R=4900 $\Omega$

  Hence, the resistance to be connected in series to get a voltmeter of range 50 V is 4900 $\Omega$ 

 In air, a charged soap bubble of radius 'R'  breaks into 27 small soap bubbles of equal radius 'r'. Then the ratio of mechanical force acting per unit area of a big soap bubble to that of a small soap bubble is 

Answer:

Explanation:

 Key Idea The force per unit area is pressure. The pressure inside a soap bubble of radius R, is given by

$P=\frac{4T}{R}$

 where T= surface tension

 and R= radius of the drop

 the pressure inside a soap bubble, $P=\frac{4T}{R}$

 If a bubble is broken into 27 small soap bubbles then the volume of single bubble of radius  R and the combined volume of 27 bubbles of radius r would be constant.

 27 x volume of small bubbles = volume of larger bubble

 $\Rightarrow$     $ 27\left(\frac{4}{3}\pi r^{3}\right)=\frac{4}{3}\pi R^{3}$

 $\Rightarrow $    $27 r^{3}= R^{3}$

 $\Rightarrow $    $r=\frac{R}{3}$   .............(i)

Now, the pressure inside smaller soap bubble,

 $P_{small}=\frac{4T}{r}=\frac{12 T}{R}$   (using the relation)

 and similarly    $P_{large}=\frac{4t}{R}$

 $\therefore$    Ratio  of pressure  of the smaller and larger  soap bubble is given as, 

$\frac{P_{larger}}{P_{small}}=\frac{4T}{R}\times\frac{R}{12T}=\frac{1}{3}$

  $P_{larger}:{P_{small}}=1:3$

 Hence, the ratio of mechanical  force acting per unit area of big soap bubble to that of a small bubble is 1:3

 The equation of simple harmonic progressive wave is given by y=a sin 2 $\pi$  (bt-cx). The maximum particle velocity will be twice the  wave velocity  if 

Answer:

Explanation:

 Given wave equation  , y= a sin 2 $\pi$  (bt-cx) 

 Compairing the above equation with the general equation of the progressive  wave which is given as 

$y=A_{0} \sin 2 \pi  (ft-\frac{x}{\lambda})$

 We get,  frequency , f=b, wavelength, $\lambda=\frac{1}{c}$ and amplitude of the wave , A₀=a

 As, we know , that the maximum velocity of the particle,

   $V_{max}=A_{0}\omega= a\times 2\pi b$  ......(i)

 Wave velocity,  $V_{wave}=f\lambda$

 $\Rightarrow$    $ V_{wave}=\frac{b}{c}$ .......(ii)

 it is given that, 

$ V_{max}=2V_{wave}$

 So, by substituting the values from 

 Eqs. (i) and (ii)  in the above realation , we get

$a2\pi b=2\frac{b}{c}$

$\therefore$    $c=\frac{1}{a\pi}$

• Physics - General questions