TS EAMCET 2018 :: Physics :: General questions

• Physics - General questions

The output  F of the logic  circuit given below

Answer:

Explanation:

 Let.s  name intermediate  states as a and b as shown in figure.

 Now,       $a= \overline{Y}$

 $b=Za=z\overline{y}$

 $F= b+ X= Z\overline{Y}+X$

An active nucleus decays to $(\frac{1}{3})$ rd in 20 h . tHe fraction of original activity remaining after 80 h is 

Answer:

Explanation:

 Decay equation , $N=N_{0}exp(-\lambda t)$

 Given,  $N_{1}= N_{0}/3, N_{2}=  ?$

  $t_{1}=20 h$  , $t_{2}= 20 h$

 So,   $N_{1}=N_{0}exp(-\lambda t_{1})$

$\Rightarrow$        $\frac{1}{3}=e(-20 \lambda)\Rightarrow e^{20 \lambda}=3$

   Now,  $N_{2}=N_{0}exp(-\lambda t_{2})$  $\Rightarrow$  $\frac{N_{0}}{N_{2}}= e^{80 \lambda}$

 $\Rightarrow$     $\frac{N_{0}}{N_{2}}=(e^{20 \lambda})^{4}=(3)^{4}$

 $\Rightarrow$    $N_{2}= \frac{1}{(3)^{4}}N_{0}=   \frac{N_{0}}{81}$

A particle of mass m is attached to four springs with spring constant k , k,2k and 2k as shown in the figure. Four springs are attached to the four corners of a square and a particle is placed  at the centre . If the particle is pushed slightly  towards  any sides of the square and released, the period of oscillation will be 

Answer:

Explanation:

 along Y-axis , net force on the particle is 

  $F=-(kx \sin 45^{0}+kx \sin 45^{0}+2kx \sin 135^{0}+2kx \sin 135^{0})$

  =    $-\frac{6}{\sqrt{2}}$kx

 $\Rightarrow$     $m\frac{d^{2}x}{dt^{2}}=-3\sqrt{2}kx\Rightarrow\frac{d^{2}x}{dt^{2}}+\frac{3\sqrt{2}kx}{m}=0$

  This is equation of SHM

 Angular frequency,   $\omega^{2}=3\sqrt{2}\frac{k}{m}$

$\Rightarrow$     $f=\left(\frac{1}{2}\pi\right)\sqrt{\frac{3\sqrt{2}k}{m}}\Rightarrow T=2\pi\sqrt{\frac{m}{3\sqrt{2}k}}$

A solid sphere of radius R makes a perfect rolling down on a plane which is inclined to the horizontal axis at an angle $\theta$. If the radius  of gyration is k , then its  acceleration is 

Answer:

Explanation:

 By force balancing  perpendicular to the inclined plane

 N=$mg \cos \theta$

 By forces  balancing  parallel to inclined  plane,

$mg \sin \theta-f_{r}=m\frac{dv}{dt}$

  $\Rightarrow  \frac{dv}{dt}=\frac{mg \sin \theta-f_{r}}{m}\Rightarrow v(t)=\int g \sin \theta dt-\frac{1}{m}\int f_{r}.dt $

 Now, torque,    $\tau= I \frac{d\omega}{dt}= r \times F_{r} $   and    $ I=mk^{2}$

$\therefore$      $mk^{2}\frac{d \omega}{dt}=rF_{r}$    or   $ \frac{d\omega}{dt}= \frac{rF_{r}}{mk^{2}}$

$\Rightarrow$      $\omega(t)=\frac{R}{mk^{2}}\int f_{r}dt$

or    $\frac{1}{m}\int f_{r}  dt= \frac{k^{2}}{R}\omega(t)=\frac{k^{2}}{R^{2}}v(t)$   (using $\omega$=v/R)

  $\Rightarrow$     $v(t)=\int g \sin \theta dt- \frac{k^{2}}{R^{2}}v(t)$

 or    $v(t)=\frac{1}{\left(1+\frac{k^{2}}{R^{2}}\right)}\int g\sin \theta dt$

 So, acceleration, 

$a(t)=\frac{dv(t)}{dt}=\frac{g \sin \theta}{\left(1+\frac{k^{2}}{R^{2}}\right)}$

 Two objects are located at a height 10 m above the ground. At some point of time, the objects are thrown with initial velocity $2\sqrt{2}$ m/s at an angle $45^{0}$ and $135^{0}$ with the positive X-axis, respectively. assuming  g=10m/s² , the velocity vectors will be perpendicular to each other at a time is equal to 

Answer:

Explanation:

 In this projectile motion, both objects are projected at the angle difference of $90^{0}$  . They will again have same angle  difference when they are again at same height (of 10 m) from ground. Taking , plane $O_{2}OO_{1}$  as reference , time taken to reach  at point $O_{1}$  by first object is

 Time of flight  . $T_{1}= \frac{2u \sin \theta}{g}$

   $=\frac{2\times2\sqrt{2} \sin 45^{0}}{10}=0.4 s$

Time taken  by second object to reach at point $O_{2}$ is 

  $T_{2}=\frac{2\times2\sqrt{2} \sin 135^{0}}{10}=0.4 s$

 $\because$    $T_{1}=T_{2}$

 So, velocities of both objects will be perpendicular  to each other after 0.4 s

• Physics - General questions