TS EAMCET 2018 :: Physics :: General questions

• Physics - General questions

A particle moves in a circle  with speed v varying with time as v(t)=2t. The total acceleration of particle after it completes 2 rounds of circle 

Answer:

Explanation:

 Instantaneous velocity of particle in circular motion is 

 $v=r\omega=at$  , (let v=at)

Then , $\omega=\frac{d\theta}{dt}=\frac{at}{r}$

So,   $\int_{0}^{2 \pi n}\theta=\int_{0}^{t} \frac{at}{r} dt$

 n= number of rounds=2(given)

$\Rightarrow$   $2\pi n=\frac{at^{2}}{2r}\Rightarrow t^{2}=\frac{4\pi nr}{a}$

Radial acceleration , $a_{r}=\frac{v^{2}}{r} = \frac{a^{2}t^{2}}{r}$

 $\Rightarrow$   $a_{r}=\frac{a^{2}}{r} \times \frac{4 \pi nr}{a}$

 $\Rightarrow$    $a_{r}=4 \pi na$

Tangential acceleration , $a_{t}=\frac{dv}{dt}=a$

 $\therefore$   Total acceleration,

$a=\sqrt{a_{t}^{2}+a_{r}^{2}}=\sqrt{a^{2}+(4\pi na)^{2}}$

$=a\sqrt{1+(4\pi n)^{2}}$

$=2\sqrt{1+64 \pi^{2}}$    [given a=2,n=2]

A particle  of mass $m_{1}$ moving along X-axis collides with a stationary particle  of mass $m_{2}$ and deviates by an angle $30^{0}$ to the X-axis as shown inthe figure.If the percentage change in kinetic energy of the combined system of these two particles reduce by 50% , then the ratio  of the mass $\frac{m_{2}}{m_{1}}$ is 

Answer:

Explanation:

Momentum is conserved in both x and y directions separately

$\Rightarrow$    $m_{1}u=m_{2}v_{2}\cos 30^{0}$ ........(i)

and   $m_{1}v_{1}=m_{2}v_{2}\sin 30^{0}$  ........(ii)

 Also  , loss of K.E=50%

$\Rightarrow$     $50=\frac{\frac{1}{2}m_{1}u^{2}-\frac{1}{2}m_{1}v_{1}^{2}-\frac{1}{2}m_{2}v_{2}^{2}}{\frac{1}{2}m_{1}u^{2}}$ .......(iii)

 Solving eqs.(i), (ii) and (iii) , we get

 $\frac{m_{2}}{m_{1}}=8$

Consider a wheel rotating around a fixed axis. If the rotation angle $\theta$ varies with time as $\theta=at^{2}$ , then the  total acceleration of a point A on the rim of the wheel is (v being the tangential velocity)

Answer:

Explanation:

Given , $\theta=at^{2}$

 so  angular velocity ,$\omega=\frac{d\theta}{dt}=2at$

 linear tangential velocity,

 $v=\omega r=2atr $       $\frac{v}{t}=2ar$

So, tangential linear acceleration of particle is 

 $a_{1}=\frac{dv}{dt}=2ar$

and angular acceleration of partcle  is 

$\alpha=\frac{d\omega}{dt}=2a$

 Also, normal or radial acceleration of particle  is

$a_{n}=\frac{v^{2}}{r}= \frac{4a^{2}t^{2}r^{2}}{r}=4a^{2}t^{2}r$

Total acceleration of particle is

 $a_{total}=\sqrt{a_{t}^{2}+a_{n}^{2}}=\sqrt{4a^{2}r^{2}+16a^{4}t^{4}r^{2}}$

$=2ar\sqrt{1+4a^{2}t^{4}}$

$=\frac{v}{t}\sqrt{1+4a^{2}t^{4}}$

A small object  is thrown at an angle $45^{0}$ to the horizontal with an initial velocity $v_{0}$ .The velocity is averaged for first $\sqrt{2}$  s  and the magnitude of average velocity comes out to be the same as that of initial velocity,i.e,$ |v_{0}|$. The magnitude $|v_{0}|$ will be  (take ,g=10 m/s²)

Answer:

Explanation:

Let object is at B(x,y) after $t=\sqrt{2}s$

Then, $x=u_{x} \times t=v_{0}\cos 45^{0} \times \sqrt{2}=v_{0}$

and $y= u_{y}t-\frac{1}{2}a_{y}t^{2}=v_{0}(\sin 45^{0}) \sqrt{2}-\frac{1}{2}(10)\times  (\sqrt{2})^{2}$

   =$(v_{0}-10)m$

 Displacement OB of particle is

$OB=\sqrt{OA^{2}+AB^{2}}$= $\sqrt{v_{0}^{2}+(v_{0}-10)^{2}}$

So, $v_{avg}= \frac{OB}{t}=v_{0}$  $OB=v_{0}t$

 $\Rightarrow$   $\sqrt{v_{0}^{2}+(v_{0}-10)^2}=v_{0}\sqrt{2}\Rightarrow v_{0}^{2}+(v_{0}-10)^{2}=2v_{0}^{2}$

$\Rightarrow $     $v_{0}-10=\pm v_{0}$         $2v_{0}=10\Rightarrow v_{0}=5 ms^{-1}$

 If $V_{0}$ is the volume of a standard unit cell of germanium for crystal containing $N_{0}$ atoms, then the expression for the mass m of volume V in terms of $V_{0}$, $N_{0}$, $M_{mol}$ and $N_{A}$  is [here, M is the molar mass of germanium and $N_{A}$ is the Avogadro's constant ]

Answer:

Explanation:

Number of unit cells in volume V

= Total volume/ Volume of a unit cell = $\frac{V}{V_{0}}$

 Nu,ber of atoms in volume V

 = Number of unit cells x number of atoms in 1 unit cell =$\frac{V}{v_{0}} \times N_{0}$

Number of moles in volume V

     = Number  of atoms/ Avagradro number= $\frac{V}{V_{0}} \times \frac{N_{0}}{N_{A}}$

 Mass n of  given sample volume= Number of moles x Molar mass

 $\Rightarrow$    m=$\frac{V}{V_{0}} \times \frac{N_{0}}{N_{A}} \times M$

• Physics - General questions