MHT CET 2020 :: Physics :: General questions

• Physics - General questions

 Two cells having unknown emf E₁ and E₂ (E₁  > E₂)  are connected in a potentiometer circuit,to assist each other. The null point obtained is at 490 cm from the higher potential end. When cell E₂ is connected, so as to oppose cell E₁, the null point is obtained at 90 cm from the same end, the ratio of the EMFs of two cells $\left(\frac{E_{1}}{E_{2}}\right)$ is 

Answer:

Explanation:

  The emf of a cell in a potentiometer is 

                   $E=\frac{V}{L}l$

 where, l= length of wire at  null point,

 V= voltage  of source

 and L= total length  of  potentiometer wire

 $\therefore$                $E\propto l$

 When two cell  of emf  E₁ and E₂   (E₁ > E₂) are connected to assist each other , then

 E₁+E₂=490   ........(i)

 when the cell of emf E₂ is connected , so as to  oppose cell E₁ then

 E₁-E₂=90    ..............(ii)

 from ERqs. (i) and (ii) , we get

 E₁=290 V  and E₂= 200 V

$\therefore$    $\frac{E_{1}}{E_{2}}=\frac{290}{200}=\frac{29}{20}=1.45$

A child starts running from rest along a circular track of radius r with constant tangential acceleration a. After a time the feels that slipping of shoes on the ground has started. The coefficient of friction between shoes and ground is 

[ g= acceleration due to gravity]

Answer:

Explanation:

 when a child moves in a circular track, he is acted upon by two forces as shown below.

 Here f_c =f sin $\theta$ and f_t= f cos $\theta$,

 As f_c is the centripetal force and f_t= is the tangential force. so

 $f_{c}=\frac{mv^{2}}{r}=f\sin\theta$

 and   $f_{t}= ma= f \cos\theta$

 $\therefore$    Resultant force,   $f_{R}= \sqrt{(f \sin\theta)^{2}+(f\cos\theta)^{2}}$

  $=\sqrt{\left(\frac{mv^{2}}{r}\right)^{2}+(ma)^{2}}$

Also, when the shoes starts slipping  , the friction becomes equal to resultant force.

 $\therefore$      f= f_R

        $\mu mg=\sqrt{\left(\frac{mv^{2}}{r}\right)^{2}+(ma)^{2}}$

 $\mu^{2} m^{2}g^{2}=\frac{m^{2}v^{4}}{r^{2}}+m^{2}a^{2}$         

$\mu^{2}g^{2}=\frac{(at)^{4}+a^{2}r^{2}}{r^{2}}$                   $\left( \because a=\frac{v}{t}\right)$

 or     $\mu^{}=\frac{(a^{4}t^{4}+a^{2}r^{2})^{1/2}}{gr}$

Water rises up to a height h in a capillary tube on the surface of the earth. The value of h will increases if the experimental setup is kept in [ g=acceleration due to gravity]

Answer:

Explanation:

 The height  of liquid column in capillary tube is 

 $h=\frac{2T\cos\theta}{\rho rg}\Rightarrow h\propto\frac{1}{g}$

 $\therefore$   when the setup is kept ina lift going down with acceleration a < g, then the height of water in the tube increases as net downward acceleration becomes (g-a)

If the maximum kinetic energy of emitted electrons in photoelectric effect is 3.2 x 10^-19 J and the work-function for metal is

6.63 x10^-19 J, then stopping potential and threshold wavelength respectively are

[Planck's constant,h=6.63 x10³⁴ J-s]

[ Velocity of light , c=3 x10⁸ m/s]

[charge on electron =1.6 x10^-19 C]

Answer:

Explanation:

 The maximum kinetic energy of emitted photo electron is 

K_max = eV_s

where V_s= stopping potential.

$\Rightarrow   V_{s}=\frac{K_{max}}{e}=\frac{3.2\times 10^{-19}}{1.6\times 10^{-19}}=2V$

 Also, the work function of a metal is

$\phi=\frac{1242}{\lambda_{0}(nm)}eV$

 $\lambda_{0}=\frac{1242\times1.6\times 10^{-19}}{\phi}nm$

$\Rightarrow$    $\frac{1242\times 1.6\times 10^{-19}}{6.63 \times 10^{-19}}nm$

 = 3000  Å

The graph of stopping potential V_s  against  frequency v of incident radiation is plotted for two different  metals P and Q as shown in the graph, $\phi_{P}$  and $\phi_{Q}$ are  work-functions of P and Q respectively, then

Answer:

Explanation:

 The work function of surface

 $\phi = h v_{0}$

 where h= Planck's comnstant

    and v₀= threshold frequency

 From graph it si clear that

 (v₀)_P   < (v₀) _Q

 $\therefore$   $\phi_{P} <  \phi_{Q}$

• Physics - General questions