JEE 2013 :: Advanced 2013 :: Physics 2013

• Advanced 2013 - Physics 2013
• Advanced 2013 - Chemistry 2013
• Advanced 2013 - Mathematics 2013

A freshly prepared sample of a radioisotope of half-life 1386 s has activity 10³  disintegrations per second. Given that ln 2=0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first 80s after preparation of the sample is 

Answer:

Explanation:

Number of nuclei decayed in time t,

 $N_{d}=N_{0}(1-e^{-\lambda t)}$

  $\therefore$    % decayed=   $=\left(\frac{N_{d}}{N_{0}}\right)\times 100$

    =    $(1-e^{-\lambda t}d)$ x100     ......(i)

  Here,   $\lambda=\frac{0.693}{1386}=5 \times10^{-4}s^{-1}$

  $\therefore$   % decayed=   $(\lambda t)\times100$

  =  $(5 \times 10^{-4})(80)(100)=4$

A particle of mass 0.2 kg moving in one dimension under a force that delivers a  constant power of 0.5W to the particle. If the initial speed (in ms^-1)  of the particle is zero, the  speed (in ms^-1) after 5s  is 

Answer:

Explanation:

$W= \frac{1}{2}mv^{2}$

  $\therefore$     $Pt= \frac{1}{2}mv^{2}$     $\therefore $    $v=\sqrt{\frac{2Pt}{m}}$

   $=\sqrt{\frac{2\times0.5\times5}{0.2}}=5m/s$

A bob of mass m,  suspended by a string of length l₁,  is given a minimum velocity required to complete a full circle in the vertical plane, At the highest point, it collides elastically with another bob of mass m  suspended by a string of length l₂, which is initially at rest, Both the strings are massless and inextensible. If the second bob. After collision acquires the minimum  speed  required  to  complete  a full circle in the vertical  plane, the ratio  $\frac{l_{1}}{l_{2}}$  is 

Answer:

Explanation:

 Velocity of first bob at highesr point 

     $v_{1}=\sqrt{gR}=\sqrt{gl_{1}}$

  (to just completeb the vertical circle )

 = ve;ocity of second bob just after elastic collision,

  = velocity of second bob at the bottommost point     

=$\sqrt{5gl_{2}}$    $\therefore$    $\frac{l_{1}}{l_{2}}=5$

The work functions   of silver and sodium are 4.6 and 2.3 eV, respectively The  ratio of the slope of the stopping potential versus frequency plot for silver to that of sodium is 

Answer:

Explanation:

$eV_{0}=hf-W$

$\therefore$               $V_{0}=\left(\frac{h}{e}\right)f-\frac{W}{e}$

V₀   versus f graph  is a  straight line with slope=   $\frac{h}{e}$= a universal constant.

Therefore the ratio of two slopes sholud be 1

The uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad/s about its own axis, is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m, are genty placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis.  The new angular velocity (in rad s^-1) of the system is

Answer:

Explanation:

$I_{1}\omega_{1}=I_{2}\omega_{2}$

 $\therefore$                   $\omega_{2}=\left(\frac{I_{1}}{I_{2}}\right)\omega_{1}$

  $\left[\frac{\frac{1}{2}MR^{2}}{\frac{1}{2}MR^{2}+2(mr)^{2}}\right]\omega_{1}$

   = $\left[\frac{50(0.4)^{2}}{50(0.4)^{2}+8\times(6.25)\times(0.2)^{2}}\right](10)$

   = 8 rad/ s

• Advanced 2013 - Physics 2013
• Advanced 2013 - Chemistry 2013
• Advanced 2013 - Mathematics 2013