JEE 2014 :: Advanced 2014 :: Physics 2014

• Advanced 2014 - Physics 2014
• Advanced 2014 - Chemistry 2014
• Advanced 2014 - Mathematics 2014

A glass capillary tube is of the shape of a truncated cone with an apex angle  $\alpha$  so that its two ends have cross-sections of different radii. When dipped in water vertically, water rises in its height h, where the radius of its cross-section in b. If the4 surface tension of water is S.  its density is ρ, and its contact angle with glass is θ, the value of h will be (g is the acceleration due to gravity)

Answer:

Explanation:

Using geometry 

$\frac{b}{R}=\cos(\theta+\frac{\alpha}{2})$

$\Rightarrow $   $ R=\frac{b}{\cos(\theta+\frac{\alpha}{2})}$

  Using pressure  equation along the path MNTK

$p_{0}-\frac{2S}{R}+h\rho g=p_{0}$

 Substituting the value of R , we get 

           $h=\frac{2S}{R\rho g}$

         $=\frac{2S}{b\rho g}\cos(\theta+\frac{\alpha}{2})$

 A planet of radius R=   $\frac{1}{10}\times$  (radius of earth)  has the same mass density as earth. Scientists dig a well of depth $\frac{R}{5}$  on it and lower a wire of the same length and of linear mass density 10^-3 kgm^-1 into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding  it in place is (take the radius of earth =  $6\times10^{6}$  m and the acceleration due to gravity of the earth is 10 ms^-2)

Answer:

Explanation:

 Given,   $R_{planet}=\frac{R_{earth}}{10}$

and density ,   $\rho= \frac{M_{earth}}{\frac{4}{3}\pi R_{earth}^{3}}$

       $\Rightarrow$            $ M_{planet}=\frac{M_{planet}}{10^{3}}$

g_{surface of planet}   $=\frac{GM_{planet}}{R^{2}_{planet}}$

$=\frac{GM_{e}.10^{2}}{10^{3}.R^{2}_{e}}$

     $=\frac{GM_{e}.}{10^{}.R^{2}_{e}}$

   =g_{surface of earth/10}

g _{depth of planet}= g _{surafce of planet     $\left(\frac{X}{R}\right)$}

 where x= distance from centre of planet.

  $\therefore$  Total force on wire

                      $F=\int_{4R/5}^{R} \lambda dx g\left(\frac{X}{R}\right)$

  = $\frac{\lambda g}{R}\left[\frac{X^{2}}{2}\right]_{4R/5}^{R}$

  Here  g= g _{surface of planet}

                    R= R_planet

 Substituting  the given values , we get   F=108N

If $\lambda_{Cu}$  is the wavelength of  $k_{\alpha}$ , X-ray line of copper (atomic number 29)  and $\lambda_{Mo}$  is the wavelength of the   $k_{\alpha}$ , X-ray line of molybdnenum (atomic number 42), then  the ratio  $\lambda_{Cu}$/ $\lambda_{Mo}$  is close to 

Answer:

Explanation:

$k_{\alpha}$   transition takes place from n₁=2 to n₂=1

$\therefore$     $  \frac{1}{\lambda}=R(Z-b)^{2}\left[\frac{1}{(1)^{2}}-\frac{1}{(2)^{2}}\right]$

 For K-series, b=1

$\therefore$         $  \frac{1}{\lambda}\propto (Z-1)^{2}$

 $\Rightarrow$        $ \frac{\lambda_{Cu}}{\lambda_{Mo}}=\frac{(z_{M0}-1)^{2}}{(z_{Cu}-1)^{2}}$

                     =$\frac{(42-1)^{2}}{(29-1)^{2}}=\frac{41\times41}{28\times 28}$

                       $=\frac{1681}{784}=2.144$

A metal surface is illuminated by light of two different wavelengths 248nm and 310 nm. The maximum speed of the photoelectrons corresponding to these wavelengths are u1  and u₂, respectively. If the ratio u1:u₂=2:1 and hc=1240 eV n m, the work  function of the meal  is nearby

Answer:

Explanation:

Energy corresponding to 248 nm wavelength

               $= \frac{1240}{248}eV=5eV$

 Energy corresponding to 310 nm wavelength

$= \frac{1240}{310}eV=4eV$

                $= \frac{KE_{1}}{KE_{2}}=\frac{\mu_1^2}{\mu_2^2}=\frac{4}{1}$

                               =  $\frac{5eV-W}{4eV-W}$

  $\Rightarrow$          $  16-4W=5-W$

$\Rightarrow$        11=3W

 $\Rightarrow$       $W=\frac{11}{3}=3.67 eV \cong 3.7eV$

Parallel rays of light of intensity I=912 Wm^-2 are incident on a spherical black body kept in surroundings of temperature 300K. Take Stefan constant   $\sigma=5.7\times 10^{-8}W m^{-2}K^{-4}$  and assume that the energy exchange with surrounds is only through radiation. The  final steady-state temperature of the black body is close to 

Answer:

Explanation:

In steady-state

Energy incident per second = Energy radiated per second

 $\therefore $     $ I\pi R^{2}=\sigma(T^{4}-T_{0}^{4})4$

 $\Rightarrow$          $ I=\sigma(T^{4}-T_{0}^{4})4$

 $\Rightarrow$             $T^{4}-T_{0}^{4}=40\times 10^{8}$

$\Rightarrow$        $T^{4}-81\times 10^{8}=40\times 10^{8}$

$\Rightarrow$          $T^{4}=121\times 10^{8}$

$\Rightarrow$              T=330K

• Advanced 2014 - Physics 2014
• Advanced 2014 - Chemistry 2014
• Advanced 2014 - Mathematics 2014