MHT CET 2016 :: Physics :: General questions

• Physics - General questions

The magnetic field (B)  inside a long solenoid  having n turns per unit length and carrying current l when the iron core is kept  in it is ($\mu _{0}=$ permeability  of vacuum, $\chi$= magnetic susceptibility)

Answer:

Explanation:

According to question, charge in magnetic field due to insertion of iron core is given by

 B'=$\mu$ B

=  $\mu_{0}(1+\chi)B  [\because\mu=\mu_{0}(1+\chi)]$

where, B= magnetic field  in air 

 $\mu _{0}$= permeability of vacuum

 $\chi$= magnetic susceptibility

As we know magnetic field inside the solenoid is given by $ B= \mu_{0} nl$

So, B'= $\mu _{0}(1+\chi)$ nl 

Two particles  X and Y have equal charges after being accelerated through same potential difference enter a region of uniform magnetic field and describe circular paths of radii  r₁ and r₂ respectively. The ratio of the mass of X to that of Y is

Answer:

Explanation:

 According to question, force acting on the particle inside the magnetic field is given by

$F_{B}=qvB\sin \theta$

 [where  $\theta$= angle between  v and B]

 This force F_B provides necessary centripetal force for circular motion of the charged particle.

 So,  $\frac{mv^{2}}{r}=qvB\sin \theta$

 Now, for particles x and y and $\theta$ = 90°

 $\frac{m_{x}v_{x}^{2}}{r_{1}}=qv_{x}B$...................(i)

$\frac{m_{y}v_{y}^{2}}{r_{2}}=qv_{y}B$...........(ii)

 From Eqs.(i) and (ii) , we get

 $\frac{m_{x}v_{x}^{}}{m_{y}v_{y}^{}}=\frac{r_{1}}{r_{2}}$

 as, $\frac{v_{x}^{}}{v_{y}^{}}=1$

 So,   $\frac{m_{x}^{}}{m_{y}^{}}=\frac{r_{1}}{r_{2}}$

Interference fringe are produced on a screen by using two light sources of intensities l and 9l. The phase difference between the beams is $\frac{\pi}{2}$  at point P and $\pi$  at point Q on the screen. The difference between the resultant  intensities at point P and Q is 

Answer:

Explanation:

According to question  , resultant  intensity of interferring wave is given by

 $l_{p}=l_{1}+l_{2}+2\sqrt{l_{1}l_{2}\cos \phi}$

 For  $\phi= \frac{\pi}{2},l_{p}=l+9l=10l$

 again at point Q, resultant intensity is given by

$l_{Q}=l_{1}+l_{2}+2\sqrt{l_{1}l_{2}\cos \phi}$

 For ,  $\phi= \pi, l_{Q}=l+9l+(-2\sqrt{9(l)^{2}}$)

 =10l-6l=4l

 Now, difference between the resultant intensity is given by 

$\triangle l=l_{P}-l_{Q}=10l-4l=6l$

An alternating current of peak value ( $\frac {2}{\pi}$) ampere flows through the primary coil of the transformer. The coefficient of mutual  inductance  between primary and secondary  coil is 1H, the peak emf induced in the secondary coil is (Frequency of Ac= 50 Hz)

Answer:

Explanation:

according to question , peak value of current

$l_{0}=\sqrt{2}\times l_{rms}=\frac{2}{\pi}A$

Coefficient of mutual inductance =1H

as, we know  induced emf in secondary coil is given by

  $E_{s}=M.\frac{dl}{dt}  $       [ where , l= l₀ sinωt]

 $E_{s}=M.\omega l_{0}\cos (\omega t)$

  = $1\times 2\pi\times50\times\frac{2}{\pi}\cos(2\pi\times50\times t)$

                                               $(\therefore\omega=2\pi n)$

 For t=0, we have

 E_e = 4 x 50=200 V

A iron rod is placed parallel to the magnetic field of intensity 2000 Am^-1. The magnetic flux through the rod is $6 \times 10^{-4} Wb $ and its cross-sectional area is 3 cm² . The magnetic permeability  of the  rod in Wb A^-1m^-1 is 

Answer:

Explanation:

According to question  , magnetic

permeability of the rod is given by , $\mu_{r}=\frac{B}{B_{0}}$

where, B= magnetic field without iron rod

B₀ = magnetic field after insertion of iron  rod

so,  $\mu_{r}=\frac{B}{2000Am^{-1}}$

 $\left( \because B.A=\phi\Rightarrow B=\frac{\phi}{A}=\frac{6 \times 10 ^{-4}Wb}{3\times 10^{-4}m^{2}}=2 Wbm^{-2}\right)$

 So,   $\mu_{r}=\frac{2}{2000}=10^{-3}$

• Physics - General questions