MHT CET 2020 :: Physics :: General questions

• Physics - General questions

A body of mass 64 g is made to oscillate turn by turn on two different springs A and B. spring A and B has a force constant 4$\frac{N}{m}$ and 16 $\frac{N}{m}$ respectively. If T₁ and T₂ are period of oscillations of spring A and B respectively , Then $\frac{T_{1}+T_{2}}{T_{1}-T_{2}}$ will be

Answer:

Explanation:

 Given  , m=64g g= 64 x 10^-3 kg

 k_A  =4 N/m and k_B⁼16 N/m

 The time period of oscillation of a spring

 $T=2\pi\sqrt{\frac{m}{k}}$

 $\Rightarrow T_{A}=T_{1}=2\pi\sqrt{\frac{m}{k_{A}}}=2\pi\sqrt{\frac{64\times 10^{-3}}{4}}$

                     =$2\pi\sqrt{16\times 10^{-3}}$  ...........(i)

 and  

and $T_{B}=T_{2}=2\pi\sqrt{\frac{m}{k_{B}}}=2\pi\sqrt{\frac{64\times 10^{-3}}{16}}$
$=2\pi\sqrt{4\times 10^{-3}}$..........(ii)

  Dividing Eq.(i) by Eq.(ii) , we get

$\frac{T_{1}}{T_{2}}=\sqrt{\frac{16\times 10^{-3} }{4\times 10^{-3}}}=2$

$\Rightarrow $     $T_{1}=2T_{2}$

 $\therefore$   $\frac{T_{1}+T_{2}}{T_{1}-T_{2}}=\frac{2T_{2}+T_{2}}{2T_{2}-T_{2}}=\frac{3T_{2}}{T_{2}}=\frac{3}{1}or 3:1$

 The Brewster's angle for the glass-air interface is (54.74) °. If a ray of  light passing from air to glass strikes at an angle of incidence 45°, then the angle of refraction  is

  $[\tan (54.74)^{0}=\sqrt{2},\sin 45=\frac{1}{\sqrt{2}}]$

Answer:

Explanation:

According  to Brewster'law , refarctive index of glass

$\mu_{2}=\tan(i_{p})$

 = $\tan (54.74)^{0}$= $\sqrt{2}$

 From Snell's law,

$\mu_{1}\sin i=\mu_{2}\sin r$

 $1\times \sin(45^{0})=\sqrt{2}\sin r $

$\frac{1}{\sqrt{2}\times\sqrt{2}}=\sin r$

 sin r=0.5

    r = sin^-1 (0.5)

 A uniform wire has length L mass M and density $\rho$  . it is under tension T and v is the speed of transverse wave along  the wire. The area of cross-section  of the wire is 

Answer:

Explanation:

 The speed of transverse wave along a wire is given by

               $v=\sqrt{\frac{T}{\mu}}$

 where $\mu$=  mass per unit length

                     = volume of unit length  x density

                      = area x density

   $\therefore$               $v=\sqrt{\frac{T}{A\rho}}\Rightarrow A=\frac{T}{v^{2}\rho}$

Two bodies  A and B of equal mass are suspended from two separate massless springs of force constant k₁ and k₂, respectively. The bodies oscillate vertically such that their maximum velocities are equal. The ratio of the amplitudes  of body  A to that of body B is 

Answer:

Explanation:

 The maximum velocity of body in SHM

                                   v= A $\omega$

 where , A is amplitude  of body and $\omega$  is the angular frequency.

 It is given that , the maximum velocities of bodies are equal

                       $V_{A}=V_{B}$

 $A_{A}\omega_{A}=A_{B}\omega_{B}$

 $A_{A}\sqrt{\frac{k_{1}}{m}}=A_{B}\sqrt{\frac{k_{2}}{m}}$           

                                                                                    $\left(\because \omega =\sqrt{\frac{k}{m}}\right)$

 $\frac{A_{A}}{A_{B}}=\sqrt{\frac{k_{2}}{k_{1}}}$

 The angle subtended  by the vector   $A=4\hat{i}+3\hat{j}+12\hat{k}$ with the X-axis is 

Answer:

Explanation:

 Given ,  $A=4\hat{i}+3\hat{j}+12\hat{k}$

 Also, A= A_x+A_y+A_z

 $\Rightarrow$    A_x=4

 The angle subtended by vector A with X- axis is 

$\cos \phi=\frac{A_{x}}{|A|}=\frac{4}{\sqrt{4^{2}+3^{2}+12^{2}}}$

 $\cos \phi=\frac{4}{13}$

$\phi=\cos^{-1}\left(\frac{4}{13}\right)$

• Physics - General questions