TS EAMCET 2017 :: Physics :: General questions

• Physics - General questions

Which of the following  principles is being used in Sonar Technology ?

Answer:

Explanation:

SONAR technology  is based on the reflection of ultrasonic waves

An amplitude modulated signal consists of a message signal of frequency 1KHz and peak voltage of 5V, modulating a carrier frequency of 1 MHz and peak voltage of 15 V  . The correct description of this signal is 

Answer:

Explanation:

 Given ,

 Frequency  of message signal  $(f_{m})$ =1 KHz= $1 \times 10^{3} Hz$

 Peak voltage  of messages signal ($E_{m}$)=5V

 Carrier frequency $(f_{c})$ =1 MHz= $1 \times 10^{6} Hz$

 Peak voltage of carrier $(E_{c})$=15 V

 We know that, The equation of amplitude modulated wave is given by

     $e(t)= E_{c}\left[1+\frac{E_{m}}{E_{c}}\sin \omega_{m}t\right]\sin  \omega_{c}t$

     $= E_{c}\left[1+\frac{E_{m}}{E_{c}}\sin (2 \pi f_{m})t\right]\sin  2\pi f_{c}t$

$= 15\left[1+\frac{5}{15}\sin (2 \pi \times10^{3})t\right]\sin  2\pi \times 10^{6}t$

  $= 15\left[1+\frac{1}{3}\sin (2 \pi 10^{3}t)\right]\sin ( 2\pi  10^{6})t$

 A beam of light propagating at an angle $\alpha_{1}$  from a medium 1 through to another medium 2 at an angle $\alpha_{2}$. If the wavelength of light in medium 1 is $\lambda_{1}$ , then the wavelength of light in medium 2, ($\lambda_{2})$ , is 

Answer:

Explanation:

 We know that ,

  $\lambda= \frac{\lambda_{0}}{\mu}$

 According to the question,

    $\frac{\sin \alpha_{1}}{\sin \alpha_{2}}=\frac{\mu_{2}}{\mu_{1}}=\frac{\lambda_{1}}{\lambda_{2}}$

  $\lambda_{2}=\lambda_{1}\frac{\sin \alpha_{2}}{\sin \alpha_{1}}$

A piece of copper and a piece of germanium are cooled from room temperature to 80K, Then, which one of the following is correct?

Answer:

Explanation:

 Copper is a conductor and germanium is a semiconductor. when both are cooled from room temperature to 80 k. Then, the resistance of copper will decrease while that of germanium will increase.

A horizontal pipeline carrying gasoline has a cross-sectional diameter of 5 mm. If the viscosity and density of the gasoline are $6 \times 10^{-3}$  Poise and 720 kg/m³ respectively, the velocity after which the flow becomes turbulent is 

Answer:

Explanation:

 Given,

 Diameter of pipe (d)= 5mm= $5 \times 10^{-3}$ m

 Density of gasoline ($\rho$)=720 kg/m³

 Viscosity  of gasoline ($\eta$)= $6 \times 10^{-3}$ Poise

 We know that,

 $V_{c}=\frac{\eta}{\rho d}=\frac{6 \times 10^{-3}}{720 \times 5 \times 10^{-3}}$

$=\frac{6 \times 10^{-3}}{3600\times10^{-3}}=\frac{1}{600}=\frac{1}{6}\times10^{-2}$

   $=1.66 \times 10^{-3} m/s$

• Physics - General questions