TS EAMCET 2019 :: Physics :: General questions

• Physics - General questions

A carnot  engine  with efficiency $\eta$ operates between two heat  reservoirs with temperatures $T_{1}$ and $T_{2}$  , where $T_{1} >T_{2}$ . If only $T_{1}$ is changed by 0.4% , the  change in efficiency is $\triangle\eta_{1}$ , whereas if only  $T_{2}$  is changed by 0.2% , the efficiency is changed by $\triangle\eta_{2}$ . The ratio  $\frac{\triangle\eta_{1}}{\triangle\eta_{2}}$ is approximately

Answer:

Explanation:

Key idea  Efficiency of carnot engine

   $\eta=1-\frac{T_{2}}{T_{1}}$

where, $T_{1}$= source temperature and $T_{2}$= sink temperature

If $T_{1}$ is changed by 0.4%, then

 $\frac{\triangle \eta_{1}}{\eta}= \frac{ \triangle T_{1}}{T_{1}}+\frac{\triangle T_{2}}{T_{2}}$

   ( from combination of the error)

$\Rightarrow$      $\frac{\triangle \eta_{1}}{\eta}= \frac{0.4}{100}+0$  .........(i)

Similarly , $T_{2}$ is changed  by 0.2% then

  $\frac{\triangle \eta_{2}}{\eta}= 0+\frac{0.2}{100}$  .........(ii)

 So, from Eq.(i) and (ii) , we get

 $\frac{\triangle \eta_{1}}{\triangle \eta_{2}}=\frac{0.4}{0.2}=2$

Hence, option (a) is correct

A water tank kept on the ground has an orifice of 2 mm diameter on the vertical side. What is the minimum height of the water above the orifice for which the output flow of water is found to be turbulent? [ Assume  g=10 m/s², $\rho_{water}$= 10³ kg/m³, viscosity =1 centi-poise]

Answer:

Explanation:

 Here, D=2mm, $\eta=1 cent-poise= 10^{-3} Pa-s$

 and density of the water , $\rho$ = $10^{-3} kg/m^{3}$

 for flow to be just turbulent , $R_{e}=3000$

 $\therefore$      $v= \frac{R_{e} \eta}{ \rho D}= \frac{ 3000 \times 10^{-3}}{10^{3} \times 2 \times 10^{-3}}=1.5$

 We know that the velocity head, $h= \frac{v^{2}}{2g}$

 $\Rightarrow$            $h = \frac{(1.5)^{2}}{2 \times 10}=0.1125=11 cm$

 SO, no option is matched

A body of mass 0.3 kg hangs by a spring with a force constant of 50N/m . The amplitude of oscillations is damped and reaches $\frac{1}{e}$ of its original value in about 100 oscillations. If $\omega$ and $\omega'$ are the angular frequencies of undamped and damped oscillations respectively , then percentage of  $\left(\frac{\omega-\omega'}{\omega}\right)$ is 

Answer:

Explanation:

Hint  

    The dispalcement  of a spring mass oscillator , $X= Ae^{-bt/2m} \cos (\omega't+\phi)$

 where, $\omega'$ = angular freequency of damped

oscillation  and    $\omega'= \omega_{0}\sqrt{1-\frac{b^{2}}{4m^{2}\omega_{0}^{2}}}$

A rocket motor consumes 100 kg of fuel per second exhausting it with a speed of 5 km/s. The speed of the rocket when its mass is  reduced to  $\frac{1}{20}^{th}$  of its initial mass, is [Assume initial speed  to be zero and ignored gravitational and viscous forces]

Answer:

Explanation:

 Key idea    Velocity ofa rocket at any time t.

    $v= u\left(\frac{m_{0}}{m}\right) -gt$

 where, u= speed of exhausted gases

  $m_{0}$= initial  mass of the rocket

 and m= mass of the rocket at time t

 Given, fuel burned rate $(\frac{dm}{dt})$=100 kg/s, u=5 km/s

 Then, $v= 5ln( \frac{m_{0}}{m})-gt$

 As, it is given that the gravitational force is ignored and mass of rocket is reduced to $\frac{1}{20}$ th of 

it's initial mass  i.e, m= $\frac{1}{20}  m_{0}$

 or    $\frac{m_{0}}{m}=20$

so, v=5 ln(20) km/s

 Hence, the correct option  is (c)

A particle A moves along the line, y=30m with a constant velocity, v parallel to the x-axis. At the momemt particle A passes the y-axis, particle  B starts from the origin  witrh zero  initial speed and a constant acceleration  .$a=0.40 m/sec^{2}$ . The angle between a and y-axis is $60^{0}$. If the particles A and B collide after sometimes , then the value of |v|  will be

Answer:

Explanation:

 As shown in the figure, two particles are  moving with velocity v and acceleration a.

 Let the time  of collision  of two particles be t, then the equation of motion

 $s=ut+\frac{1}{2} at^{2}$

 For particle A,

     $0m =(0) t +\frac{a \cos 60^{0} t^{2}}{2}$

 $\Rightarrow$         $30= \frac{0.4}{2} \times \frac{1}{2} \times t^{2}$

   $t= \sqrt {300}$

 Gence, in time t = $\sqrt{300}$  , both travelled equal distance in horizontal direction , so that the collision takes place.

 $\Rightarrow$     $ut=\frac{1}{2}  a \cos 30^{0} t^{2}( \because  30^{0}=90^{0}-60^{0}$)

  $\Rightarrow$     $u= \frac{1}{2}\times 0.4 \times\frac{\sqrt{3}}{2}\times\sqrt{300}$

 $\Rightarrow$      u=3 m/s

 Hence , the correct option is (b)

• Physics - General questions