TS EAMCET 2019 :: Physics :: General questions

• Physics - General questions

In Young's double-slit experiment , a thin sheet of refractive index  1.6 is used to cover one slit while a thin the sheet of refractive index  1.3 is used to cover the second slit. The thickness of both the sheets are same and the wavelength of light used in 600nm. If the central point on the screen is now occupied by what had been the 10th bright fringe (m=10)  , then the thickness of covering sheets is 

Answer:

Explanation:

Given, refractive index of first sheet $\mu_{1}$=1.6  refractive index of second sheet $\mu_{2}$=1.3 and  wavelength of light $\lambda$=600nm=$600 \times 10^{-9}$ m

 A young's double-slit experiment is shown below, in which two thin sheets have covered the slits . So, the path difference introduced in slit 1.

 $\triangle x_{1}=(\mu_{1}-1)t= (1.6-1)t=0.6t$

 Similarly , path difference introduced in slit 2,

 $\triangle x_{2}= (\mu _{2}-1)t=(1.3-1)t=0.3t$

 So, the net path difference introduced  in central maxima,

$\triangle x_{central maxima}=\triangle x_{1}-\triangle x_{2}$=0.6t-0.3t=0.3t

 For central maxima, which occupied the $10^{th}$ bright fringe,

$\triangle x_{cen}=10 \lambda$

$\Rightarrow$                 t= $\frac{10 \times 600 \times 10^{-9}}{0.3}-20\mu m$

Hence, the correct option is (c)

An ideal gas is placed in a tank at $27^{0} C$. The pressure is initially 600 kPa. One-fourth of the gas is then released from the tank and thermal equilibrium is established. What will be the pressure if the temperature is $327^{0}$ C?

Answer:

Explanation:

 As, we know that , the ideal gas equation is given as, 

          pV=nRT

It can written as,

 $\frac{p_{1}V_{1}}{n_{1}T_{1}}=\frac{p_{2}V_{2}}{n_{2}T_{2}}$  ........(i)    ($\because$  R= constant)

 Given, $T_{1}=273+27^{0}=300K $ and $p_{1}=600 KPa$

If one fourth of gas is released from tank,

then , $n_{2}= \frac{3}{4}n_{1}$ and   $V_{2}=V_{1}$    ($\therefore  $  No change)

Given, $T_{2}$=327+273=600K

So, from Eq.(i)  , we get

$\frac{600V_{1}}{n_{1}300}=\frac{p_{2}V_{2}}{\frac{3}{4}n_{1}600}\Rightarrow p_{2}=900KPa$

 So, the correct option is (a)

A body cools from $ 70^{0}$ C to $40^{0}$ C in 5 min. Calculate the time it takes to cool from $60^{0}$ C to $30^{0}$C . The temperature of the surrounding is $20^{0}$ C,

Answer:

Explanation:

 Body cools from 70° C to 40°C in 5 minute, hence , $T_{1}=70^{0}C, T_{2}=40^{0}C$ and t=5 minute 

Temperature of surrounding , $T_{0}=20^{0}C$

 By Newton's law of cooling,

$\frac{T_{1}-T_{2}}{t}=k\left[\frac{T_{1}+T_{2}}{2}-T_{0}\right]$

or  $\frac{70-40}{5}=k\left[\frac{70+40}{2}-20\right]\Rightarrow k= \frac{6}{35}$  ...........(i)

 Again, let body cools from $60^{0}$C  to $30^{0} C$ in time t minutes.

 i.e, $T_{1}'=60^{0} C$ and $T_{2}'=30^{0}C$,

 $\therefore$   By Newton's law of cooling

$\frac{T_{1}'-T_{2}'}{t}=k\left[\frac{T_{1}'+T_{2}'}{2}-T_{0}\right]$

$\frac{60-30}{t}=\frac{6}{35}\left[\frac{60+30}{2}-20\right]$        [ $\because $ From eq.(i)]

 $\frac{30}{t}=\frac{6}{35}[25] \Rightarrow $ t=7 minutes

 Hence, option (b) is correct

Statement A Convection involves the flow of matter within a fluid due to unequal temperature of its parts

Statement B A hot bar placed under a running tap water loses heat due to the effect of convection with in water

Statement C  Heat transfer always involves temperature different between two systems. Identify  the correct option

Answer:

Explanation:

(A)   Convection involved flow of matter within a fluid due to heat transferred from one point to another by the actual motion of matter from a region  of high-temperature to a region of lower temperature

(B)    A hot bar placed under a running top water loses heat with in water.The warmer water rises up , these leads to circulation of heat from bar to water by the process of convection.

However,  heat will be transferred inside the molecules  of the bar also by conduction. Thus,  the heat transfer when a hot bar placed under a running top water is due to both conduction and convection.

(C)  Heat can only be transferred between two system due to their temperature difference in the form of radiation, molecular vibrations, molecular displacement etc.

A copper wire of cross-sectional area 0.01 cm² is under a tension of 22 N. Find the percentage change in the cross-sectional area, (Young's  modulus of copper =$1.1 \times 10^{11}  N/m^{2}$ and Poisson's ratio =0.32)

Answer:

Explanation:

 Given,  cross section area of copper wire A=0.01 cm² =10^-6 m²  and tension force.F=22N

Posson's  ratio , $\sigma$= 0.32, and Young's modulus  Y= $1.1 \times 10^{11}  N/m^{2}$

Since,Posson's ratio, $\sigma$= lateral strain/ Longitudinal strain

$\sigma=\frac{\frac{\triangle D}{D}}{\frac{\triangle L}{L}}$  ........(i)

$\because$   Area ,A= $\pi D^{2}$

$\Rightarrow$    $  \frac{\triangle A}{A}=2\frac{\triangle D}{D}$  .......(Iii)

 From Eqs,(i) and (ii) , we get

 $\Rightarrow$     $  \frac{\triangle A}{A}=2\sigma\frac{\triangle L}{L}$    ........(iiii)

 Young's modulus ,   $Y= \frac{FL}{A\triangle L}$   .......(iv)

From eqs.(iii) and (iv)  , we get

$\Rightarrow$          $\frac{\triangle A}{A}$=$\frac{2 \sigma F}{YA}$

Now , putting the given values

$\frac{\triangle A}{A}=\frac{2 \times 0.32 \times 22}{1.1 \times 10^{11} \times 10^{-6}}=12.8 \times  10^{-3}$

$\Rightarrow\frac{\triangle A}{A}$% =$12.8 \times 10^{-3}$=$12.6 \times 10^{-3}$

 So, option (a) is correct

• Physics - General questions