TS EAMCET 2018 :: Physics :: General questions

• Physics - General questions

Let  $S_{1}$  be the amount of Rayleigh scattered light of wavelength  $\lambda_{1}$ and $S_{2}$ that of light of wavelength $\lambda_{2}$  from a particle  of size a. Which of the following statement is true?

Answer:

Explanation:

 Condition for scattering is 

   $\lambda$  >> a

 and according to Rayleigh , intensity (or amount ) of  scattered wavelength  is inversely proportional to fourth power of the wavelength  of incident wave, 

 i.e,

    $s \propto\frac{1}{\lambda^{4}}\Rightarrow \frac{S_{1}}{S_{2}}=\left(\frac{\lambda_{2}}{\lambda_{1}}\right)^{4}$ and $\lambda>>a$

An ideal gas in a cylinder is compressed adiabatically to one -third of its original volume. A work of 45 J is done on the gas by the process. The change in internal energy of the gas and the heat flowed into the gas, respectively are

Answer:

Explanation:

For adiabatic process, $\triangle Q$=0

Now, from first law of thermodynamics , we have

 $\triangle Q= \triangle U+ \triangle W$

 $\triangle W=$ -45 J  or work done on the gas

 $\Rightarrow$    $ 0= \triangle U-45$

 $\Rightarrow$    $\triangle U= 45 J$

In a tensile test on a metal bar of diameter 0.015 m and length  0.2 m , the relation between the load and elongation within the proportional limit is found to be F=97.2 x 10⁶ ($\triangle $  L) , where  F is  the load (in N) and $\triangle L$  is the elongation  (in m) . The youngs' modulus  of the material in GPa is 

Answer:

Explanation:

Given that, diameter  of metal bar (d)= 0.15 m

length (L)=0.2 m

 $F=97.2 \times  10^{6}  \times \triangle L$

 We know that, $Y=\frac{FL}{A.\triangle L}$

 $\Rightarrow$     $Y= \frac{97.2\times10^{6}\times0.2\times4}{3.14\times (0.015)^{2}}$

  = $110063.69 \times 10^{6}$≈ 110 GPa

A car of mass 1200kg (together with the driver) is moving with a constant acceleration of 2 m/s². How much power does the engine generate at the instance, when the speed reaches 20 m/s ? (Assume that the coefficient  of friction between the car  and the road is 0.5)

Answer:

Explanation:

Power  of engine overcomes friction and provides necessary acceleration to the car

 Now friction, $f=\mu mg=0.5 \times 1200 \times 10$=6000 N

 and accelerating  force, F=ma=$1200 \times 2$=2400N

 So, total force produced by engine

  =$F_{r}$=8400 N

We know that, the formula pf power is 

 $P= F_{r}v$

 Then, $ 8400 \times 20 =168000 W $

 A small block starts sliding down an inclined plane forming an angle $45^{0}$ horizontal. The coefficient of friction $\mu$ varies with distance s as  $\mu=cs^{2}$ , where c is a constant of appropriate dimensions, then distance covered by the block before it stops is 

Answer:

Explanation:

Given that

 $\mu= Cs^{2}$      [ $\because$ s= distance and C = constant ]

 net force on the block is 

 $Mg\sin \theta- f=Ma$

$ Mg sin \theta - \mu  Mg \cos  \theta=Ma$

  $g \sin \theta- \mu g \cos \theta$=a

 $g[\sin \theta -Cs^{2} \cos \theta]=a$    ..........(i)

  $\because$        $a= \frac{dv}{dt}=v \frac{dv}{ds}$

 $ads=vdv$         ...........(ii)

 From eqs.(i) and (ii)  , we get

 $\Rightarrow$     $g[\sin \theta-Cs^{2} \cos \theta ]ds=vdv$

 Integreating  both sides , we get

 $\Rightarrow$     $(g\sin\theta)s-g\frac{Cs^{3}}{3}\cos\theta=\frac{v^{2}}{2}+K$  .....(iii)

 For  $\theta=45^{0}$, we have

$\left(\frac{g}{\sqrt{2}}\right)s-\left(\frac{Cs^{3}}{3}\right)\frac{g}{\sqrt{2}}=\frac{v^{2}}{2}+K$

Initially t=0, s=0 ,u=0 substitutingthese , we get k=0

 So,   $\frac{g}{\sqrt{2}}\left(s-\frac{Cs^{3}}{3}\right)=\frac{v^{2}}{2}$

$\Rightarrow  s-\frac{Cs^{3}}{3}=\frac{v^{2}}{2}\times \frac{\sqrt{2}}{g}$

 The body stops, when v=0  $\Rightarrow s-\frac{Cs^{3}}{3}=0$

 $\Rightarrow$     $s= \frac{Cs^{3}}{3}\Rightarrow s=\sqrt{\frac{3}{C}}$

• Physics - General questions