TS EAMCET 2019 :: Physics :: General questions

• Physics - General questions

 The magnitude of the force vector acting on a unit length of a thin wire carrying a current I= 8A at a point O, if the wire  is bent as shown in the figure with a radius , $R=10 \pi$ cm is 

Answer:

Explanation:

Given, current flowing  through wire, I= 8 A and radius of circular wire,

 R=$10 \pi $ cm= $10 \pi \times 10^{-2}$ m

 According to the question,

$\therefore$  Magnetic field produced by semi-circular current carrying thin wire at centre O.

 $B=\frac{\mu_{0}I}{4R}$

 Putting the given values, we get

 $B=\frac{4 \pi \times 10 ^{-7} \times 8}{4 \times 10 \pi \times 10^{-2}}$

$\Rightarrow$      B= $ 8 \times 10 ^{-6} $ T

 $\therefore$  Magnitude  of the force on thin wire per unit length.

   F= IB

 Putting the given values,

 = $8 \times 8 \times 10^{-6}$

= $64 \times 10^{-6}$ N/m=64$\mu$ N/m

For  the circuits A and B  as shown in the figure, identify the correct option

Answer:

Explanation:

In the given circuits, A represents an ammeter, which is used to measured the current passing through the resistor it is not a low resistive device, so that it does not affect the circuit. It is connected in series with resistor.

Similarly, V represents a voltmeter , which is used to measure the potential drop across the resistor. It is usually a high resistive device, so that the negligible current passes through it . Thus , does not affect any current passing through the resistor.It is connected in parallel with  to the resistor, This implies  that, these devices are used measure. Current  and voltage drop across a resistor , without actually affecting any of these quantities because of their addition in the circuit.

 So, the given circuits A and B as shown below,

 We can conclude that the circuit A has been correctly connected. Therefore , it can be used to measure R accurately , which can be calculated with the help of ammeter and voltmeter reading as ,

 R= $\frac{ voltage (V)}{ current (I)}$      [ from ohm's law]

 However, when the arrangement of ammeter  and voltmeter  is as per given in circuit B. The circuit then is usually used for measuring higher resistance as compare to circuit A.

 The space between the two large parallel plates is filled with a  material of uniform charge density $\rho$ . Assume that one of the plate is kept at x=0 . The potential at any point x between these plates  is given by (A and B are constants) 

Answer:

Explanation:

According to the question, the space between the two large parallel plates is filled with a material of uniform charge density $\rho$  i.e, it is a homogeneous medium , therefore the potential  at any point between these plates is calculated by Poisson's equation as given below,

$\triangledown ^{2} V= -\frac{\rho}{\epsilon _{0}}$    ..........(i)

 Since, we have to calculate the potential at any point along x- direction , hence the derivative of the potential along y and z directions is zero.

 Hence, from Eq.(i)  , we get

$\frac{d^{2} V}{dx^{2}}= -\frac{\rho}{\epsilon _{0}}$

Integreate  on both the sides, we get

$\Rightarrow$      $\int \frac{d^{2} V}{dx^{2}}dx= -\int \frac{\rho}{\epsilon _{0}}dx\Rightarrow \frac{d V}{dx}=\frac{-\rho x}{\epsilon _{0}}-A$

 Integrate  both sides again, we get

 $\Rightarrow$    $\int \frac{d^{} V}{dx^{}}dx= \int \frac{-\rho x}{\epsilon _{0}}dx-\int A dx$

 $V= \frac{-\rho x^{2}}{2\epsilon _{0}}- A x-B$

   $V=- \left(\frac{+\rho x^{2}}{2\epsilon _{0}}+ A x+B\right]$

In Young's double slit experiment, mth order and n th order of bright fringes are  formed at point P on a distant  screen if  monochromatic  source of wavelength 400 nm  and 600 nm are used respectively . The  minimum  vale of m  and n are respectively ,

Answer:

Explanation:

Let $m ^{th} $ bright fringe of wavelength  $\lambda_{m}$ and $n ^{th}$  bright fringe  of wavelength $\lambda_{n}$ formed at a  distance $x_{n}$  from central maxima at same point P, 

i.e, they coincide  , therefore ,$x_{m}=x_{n}$

 $\frac{m \lambda_{m} D}{d}=\frac{n \lambda_{n} D}{d}$

$\Rightarrow$     $\frac{m}{n}=\frac{ \lambda_{n} }{\lambda_{m}}$

 Putting  the given values, we get

 = $\frac { 600 \times 10^{-9}}{400 \times 10 ^{-9}}$

 [ Given , $\lambda_{m}$= $400 \times 10 ^{-9} $ m, $\lambda_{n}$ = $600 \times 10 ^{-9} $ m]

 $\therefore $  $\frac{m}{n} = \frac{3}{2}$

 So, least integral values of m and n satisfying  above requirement are m=3 and n=2.

One mole of nitrogen gas being initially  at a temperature of T₀ =300K  is adiabatically  compressed to increase its pressure 10 times .The final gas temperature after compression  is (Assume , nitrogen gas molecules as rigid diatomic and $100^{1/7}$= 1.9)

Answer:

Explanation:

given, initial  temperature of 1 mole of $N_{2}$ gas

   $T_{0}$ =300K

 initial pressure of gas , $p_{1}$=p

 and final pressure of gas , $p_{2}=10p$

 By adiabatic process  relation between p and T.

 $p_{1}^{1-\gamma}T_{1}^{\gamma}=p_{2}^{1-\gamma}T_{2}^{\gamma}$

$\left(\frac{p_{1}}{p_{2}}\right)^{1-\gamma}=\left(\frac{T_{2}}{T_{1}}\right)^{\gamma}\Rightarrow\left(\frac{p}{10p}\right)^{1-\gamma}=\left(\frac{T_{2}}{300}\right)^{\gamma}$

$10^{\gamma-1}=\frac{T_{2}^{\gamma}}{300^{\gamma}}\Rightarrow T_{2}^{\gamma}= 10^{\gamma-1}\times 300^{\gamma}$

 $\Rightarrow$     $T_{2}=10^{\frac{\gamma-1}{\gamma}}\times 300= 10^{1-\frac{1}{\gamma}}\times300$

 $= 10^{1-\frac{1}{7/5}}\times300\left[\because For diatomic,\gamma=\frac{7}{5}\right]$

 $= 10^{2/7}\times300= (100)^{1/7} \times 300$

 =$1.9 \times 300$              [$\because  100^{3/7}=1.9$]

 =570 K

Therefore, the final gas temperature after compression is 570 K

• Physics - General questions