MHT CET 2016 :: Physics :: General questions

• Physics - General questions

A progressive wave is repesented by y= 12 sin (5t-4x)  cm. On this wave , how far away are the two points having phase difference of 90° ?

Answer:

Explanation:

According  to question , the progressive wave is represents by y=12 sin (5t-4x) cm 

Comparing this equatiuon with standard equation of progressive wave,

 $y= A \sin(\omega t-kx)$

 So, we have    A=12

$\omega$  = 5

$\Rightarrow$    k=4

 Here, ($\omega $ t- kx)  is phase difference = $\frac {\pi}{2}$

$\therefore$  5t-4x = $\frac{\pi}{2}$

When    t=0, 4x= $\frac{\pi}{2}$

   $\therefore$   x= $\frac{\pi}{8}$ cm

If the end correction of an open pipe  is 0.8 cm, then the inner radius of that pipe will be

Answer:

Explanation:

For an open organ pipe, the relation between end correction and inner radius of the organ pipe is given by

$\triangle l= 1.2 \times r$

[$\triangle l= $  end correction, r= inner radius]

 So, $r= \frac{\triangle l}{1.2}=\frac{0.8}{1.2} [\because \triangle l=0.8]$

 = $\frac{2}{3} $ cm

Two strings A and B of same material are stretched by same tension. The radius of the string A is double  the radius of string B . Transverse wave travels on string A with speed v_A and on string B with speed v_B . The ratio $\frac {v_{A}} {v_{B}}$ is 

Answer:

Explanation:

 According to question, the velocity of wave travelling on string is given by

$v=n\lambda=\frac{\lambda}{2L}\sqrt{\frac{T}{\mu}}$                    $\left(\because n=\frac{1}{2L}\sqrt{\frac{T}{\mu}}\right)$

 $\therefore$  $v=\sqrt {\frac{T}{m/l}}\Rightarrow v=\sqrt{\frac{Tl}{m}}$

 and Young's modulus , $Y= \frac{T\times l}{A \triangle l}$

$\therefore$   $T \times l$=$Y A \triangle L$

$\therefore$      $V\propto \sqrt{A}$       (A= Area)

 For string A, radius is 2r and for string B radius  is r

 So,     $\frac{V_{A}}{V_{B}}=\sqrt{\frac{4 r^{2}}{r^{2}}}=\sqrt{4}=2$

 [ $\because$    Y for both the strings is same]

A mass m₁ connected to a horizontal  spring performs SHM with amplitude A. While mass m₁ is passing through mean position, another mass m₂ is placed on it so that both the masses move together  with amplitude A₁.The ratio 

$\frac{A_{1}}{A}is (m_{2}<m_{1})$

Answer:

Explanation:

 The potential energy of oscillating mass is given by

 PE= $\frac{1}{2}kx^{2}$

When  only one block is oscillating at x= A

E_max= $\frac{1}{2} kA^{2}$ and at  mean position x=0  , So E=0

  $\therefore$   $A\propto\frac{1}{\sqrt{m}}$ ..........(i)

 when mass m₂ is placed on top of the mass m. Then , total mass is (m₁+m₂) and E =0 at this point , as x=0

When they reaches at x= A,

then  $A_{1}\propto\frac{1}{\sqrt{m_{1}+m_{2}}}$  ......(ii)

Dividing Eq.(ii)  by Eq.(i),

 $\frac{A_{1}}{A}=\left(\frac{m_{1}}{m_{1}+m_{2}}\right)^{1/2}$

Assuming the expression for the pressure exerted  by the gas on the walls of the container, it can be shown that pressure is 

Answer:

Explanation:

According to kinetic theory of gases, the pressure exerted by the gas on the walls of container is 

p= p₀+p₁+p₂

 i.e, $p= p_{0}+\frac{1}{3}8v^{2}+3gh$

 For a container,

 $p_{1}=\frac{1}{3}\rho v^{2}= \frac{1}{3}\frac{m}{v}.v^{2}\times\frac{2}{2}=\frac{2}{3}.\frac{1}{2}mv^{2}.\frac{1}{v}$

  = $\frac{2}{3v} KE   $                     $\left[\because KE= \frac{1}{2}mv^{2}\right]$

• Physics - General questions