MHT CET 2018 :: Physics :: General questions

• Physics - General questions

If $A= 3\hat{i}-2\hat{j}+\hat{k}, B= \hat{i}-3\hat{J}+5\hat{K} $and $C= 2\hat{i}+\hat{j}-4\hat{k}$ forn a right angled triangle, then out of the following which one is satisfied ?

Answer:

Explanation:

Given,

 $A= 3\hat{i}-2\hat{j}+\hat{k}, $

$B= \hat{i}-3\hat{J}+5\hat{K} $

$C= 2\hat{i}+\hat{j}-4\hat{k}$

 Here,  $|A|=\sqrt{(3)^{2}+(-2)^{2}+(1)^{2}}$

or  $|A|=\sqrt{9+4+1}=\sqrt{14}$  .....(i)

$|B|=\sqrt{(1)^{2}+(-3)^{2}+(5)^{2}}$

$|B|=\sqrt{1+9+25}=\sqrt{35}$   ......(ii)

and   

  $|C|=\sqrt{(2)^{2}+(1)^{2}+(-4)^{2}}$

$|C|=\sqrt{4+1+16}=\sqrt{21}$  .........(iii)

 From Eqs.(i) , (ii) and (iii) , we get

 B²= A²+C²

$(\sqrt{35})^{2}=(\sqrt{14})^{2}+(21)^{2}$

Now, A.C= ($3\hat{i}-2\hat{j}+\hat{k}).(2\hat{i}+\hat{j}-4\hat{k}$)

A.C= 6-2-4=0

 Hence, A and C  are perpendicular to each other 

$\therefore$ Resultant of A and C  is B

  B= A+C

 [according to triangle law]

A ray of light is incident normally on a glass slab of thickness 5 cm and refractive index 1.6. The time taken to travel by a ray from source to surface of the slab is the same as to travel through a glass slab. The distance of the source  from the surface is 

Answer:

Explanation:

 Time taken by light to cross the glass slab,

 t= Distance/ speed of light in glass

 =  $\frac{5\times 10^{-2}}{\frac{3}{1.6}\times 10^{5}}s=2.67 \times 10^{-10}s$

In same time distance travelled by light in air

   =c x t= 3 x10⁸  x 2.67 x 10^-10

 =8.02 x 10^-2 m= 8 cm

An alternating electric field of frequency v is applied across the dees (radius R) of a cyclotron to accelerate protons  (mass m). The operating magnetic field B used and KE of the proton beam produced by it are respectively(where, e= charge on proton).

Answer:

Explanation:

 In cyclotron frequncy, $v=\frac{eB}{2\pi m}$

 or   $B=\frac{2\pi m.v}{e}$

 Also, relation for radius (R)is

$R=\frac{ m.v}{e B}$ or $v=\frac{eBR}{m}$

$\therefore$    $K E =\frac{1}{2}mv^{2}=\frac{1}{2}m\times\frac{ e^{2}B^{2}R^{2}}{m^{2}}=\frac{ e^{2}B^{2}R^{2}}{2m}$

 = $\frac{ e^{2}R^{2}}{2m}\times\frac{4\pi^{2}m^{2}v^{2}}{e^{2}}=2\pi^{2}m R^{2}v^{2}$

With forwarding biased mode, the p-n junction diode.

Answer:

Explanation:

 In the forward biased mode, the p-n junction diode acts as a closed switch

The molar specific heat of an ideal gas at constant pressure and constant volume is C_P and CV respectively. If R is the universal gas constant and the ratio of C_P TO C_V is  $\gamma$, then C_V

Answer:

Explanation:

According to Mayer formula ,

   C_p  -C_v  = R  .....(i)

 Where, C_p = specific heat at constant pressure,

 C_V = specific heat at constant  volume

 and R= gas constant

  Now, $\gamma= \frac{C_{p}}{C_{v}}$

 $\Rightarrow C_{p}=\gamma C_{v}$            ............(ii)

 From  eqs,(i) and (ii) , we get

 $\Rightarrow \gamma C_{v}- C_{v}=R\Rightarrow  C_{v}(\gamma-1)=R$

 $\Rightarrow $   $ C_{v}= \frac{R}{(\gamma-1)}$

• Physics - General questions