MHT CET 2017 :: Physics :: General questions

• Physics - General questions

A disc of the moment of interia 'l₁' is rotating in a horizontal plane about an axis passing through a centre and perpendicular to its plane with constant angular speed '$\omega_{1}$'. Another disc of moment of interia 'l₂'  having zero angular speed is placed co-axially on a rotating disc . Now, both the discs are rotating with constant angular speed '$\omega_{2}$ . The energy lost by the initial rotating disc is 

Answer:

Explanation:

Net  external torque  on the system is zero. Therefore, angular momentum of  the system will remain  same.

$\Rightarrow $    $l_{1}\omega_{1}=(l_{1}+l_{2})\omega_{2}$

$\frac{\omega_{2}}{\omega_{1}}=\frac{l_{1}}{l_{1}+l_{2}}$ .....(i)

 The energy lost . $E_{1}-E_{2}$

$=\frac{1}{2}l_{1}\omega_{1}^{2}-\frac{1}{2}(l_{1}+l_{2})\omega_{2}^{2}$

$=\frac{1}{2}\omega_{1}^{2}\left[l_{1}-(l_{1}+l_{2})\frac{\omega_{2}^{2}}{\omega_{1}^{2}}\right]$

 $=\frac{1}{2}\omega_{1}^{2}\left[l_{1}-(l_{1}+l_{2})\frac{l_{1}^{2}}{(l_{1}+l_{2})^{2}}\right]$

                              [ $\because  $ Eq.(i)]

 $=\frac{1}{2}\omega_{1}^{2}\left[\frac{l_{1}^{2}+l_{1}l_{2}-l_{1}^{2}}{(l_{1}+l_{2})^{}}\right]$

  $=\frac{1}{2}\left[\frac{l_{1}l_{2}}{l_{1}+l_{2}^{}}\right]\omega_{1}^{2}$

The resistivity of potentiometer wire  is  40 x 10^-8 ohm  -metre and its area of cross-section is 8 x 10^-6 m² ,If 0.2-ampere current is flowing through the wire , then the potential gradient of the wire is 

Answer:

Explanation:

Given , $\rho =40 \times 10^{-8}$ $\Omega$m

 A= $8 \times 10^{-6} m^{2}$; l= 0.2 A

 as resistance is given as R= $\frac{ \rho l}{A}$

$\therefore$    $\frac{R}{l}=\frac{\rho}{A}=\frac{40\times 10^{-8}}{8 \times 10^{-6}}=5 \times 10^{-2}$

The potential gradient of the were  is 

$\frac{V}{l}=\frac{lR}{l}=0.2\times5\times 10^{-2}= 10^{-2}  V/m$

A parallel plate air capacity 'C' farad, potential 'V' volt and energy 'E' joule. When the gap between the plates is completely filled with dielectric.

Answer:

Explanation:

Note We have considered that the battery is kept disconnected from the capacitor.

When a dielectric being introduced inside the parallel plate capacitor, then potential difference gets reduced.

i.e,       V= $\frac {V_{0}}{k}$

 The energy decreases. i.e,   U= $\frac {U_{0}}{k}$

where, k is the dielectric constant

According to the de-Broglie hypothesis, the wavelength associated with moving electron of mass 'm' is $\lambda_{e}$ '. Using mass-energy relation and planck's quantum theory, the wavelength associated with a photon is '$\lambda_{p}$ '. If the energy (E)  of electron and the photon is same, then the relation between '$\lambda_{e}$' and '$\lambda_{p}$' is 

Answer:

Explanation:

The energy of photon  is given as,

  $E_{p}=\frac{hc}{\lambda_{p}}$

$\therefore$     $\lambda_{p}=\frac{hc}{E_{p}}$   ........(i)

 Energy of an moving electron is given as,

 $E_{e}=mc^2=pc\Rightarrow p=\frac{E_{p}}{c}$

$\therefore$    $\lambda_{e}=\frac{h}{p}=\frac{hc}{E_{e}}$   ........(ii)

 Given, Ep= E_e

 Therefore from Eqs.(i) and (ii) , we get   $\lambda_{p}\propto \lambda_{e}$

On a photosensitive material, when the frequency of incident radiation is increased by 30%, kinetic energy of emitted photoelectrons increases from  0.4 eV to 0.9 eV. The work function of the surface is 

Answer:

Explanation:

 According to the Einstein equation,

  $KE_{max}=hv_{0}-\phi _{0}$

 where,  $KE_{max}$   = maximu kinetic energy and $\phi_{0}$ 

                                               = work function

 Initially , hv= 0.4+ $\phi _{0}$   ........(i)

When the frequency of incident radiation  is increased  by 30%  , then 

     1.3 hv=0.9 +$\phi _{0}$          .........(ii)

 Solving Eqs.(i) and (ii) , we get

$0.3\phi _{0}=0.9-1.3(0.4)$

$\therefore$      $\phi _{0}=\frac{0.38}{0.3}=1.267 eV$

• Physics - General questions