TS EAMCET 2017 :: Physics :: General questions

• Physics - General questions

A ball is thrown at a speed of 20 m/s at an angle of $30^{0}$ with the horizontal. The maximum height reached by the ball is

[Use g=10 m/s²]

Answer:

Explanation:

 Given,

 speed of ball (u) =20 m/s

 Angle ($\theta$)= $30^{0}$

 We know that,

   $H= \frac{u^{2}\sin^{2}\theta}{2g}$

$H= \frac{(20)^{2}(\sin30)^{2}}{2\times10}$

 $H= \frac{20\times20\times\left(\frac{1}{2}\right)^{2}}{20}$

 $H= 20 \times \frac{1}{4}$

 H=5 m

An office room contains about 2000 moles of air. The change in the internal energy of this much air when it is cooled from $34^{0}C$ to $24^{0}C$ at a constant pressure of 1.0 atm is 

[ Use $\gamma_{air}$=1.4 and universal gas constant =8.314 J/mol-K]

Answer:

Explanation:

 Given, 

 Number  of moles of air in room (n)=2000= $2 \times 10^{3}$

 Temperature difference (dT)=24-34=$-10^{0} C$

 We know that,

 $dQ=nC_{v}dT= 2 \times 10^{3} \times \frac{R}{0.4}[-10]$

 = $\frac{-2 \times 10^{3} \times 8.314 \times 10}{0.4}$

  $=\frac{2 \times 8.314 \times 10^{5}}{4}= \frac{16.628}{4} \times  10^{5}$

   $=-4.2 \times 10^{5}$ J

Consider a light source placed at a distance of 1.5 m along the axis facing the convex side of a spherical mirror of radius of curvature  1 m. The  position  (S'), nature and magnification  (m) of the image  are 

Answer:

Explanation:

 Given,

 Distance of light source (u) =-1.5 m=$-\frac{-3}{2}m$

 Radius of curvature  of mirror (R)= 1m

 we know that

   $\frac{1}{f}= \frac{1}{v}+\frac{1}{u}$          [ $\because$  $f=\frac{R}{2}$]

$\therefore$     $\frac{2}{R}=\frac{1}{v}+\frac{1}{u} \Rightarrow \frac{2}{1}= \frac{1}{v}+ \frac{1}{\frac{-3}{2}}$

   $\frac{2}{1}= \frac{1}{v}-\frac{2}{3} \Rightarrow  \frac{1}{v}=2+\frac{2}{3}$

 $\frac{1}{v}= \frac{6+2}{3} \Rightarrow \frac{1}{v}= \frac{8}{3} \Rightarrow v= \frac{3}{8}=0.375 m$

 The image is virtual

 Magnification (m)= $\frac{-v}{u}= \frac{\frac{-3}{8}}{\frac{-3}{2}}=\frac{1}{4}=0.25$

Consider a reversible engine of efficiency  $\frac{1}{6}$. When the temperature of the sink is reduced by $62^{0}C$ , its efficiency gets doubled. The temperature of the source and sink  respectively are

Answer:

Explanation:

 Given

           $\eta_{1}=\frac{1}{6}$

 According to the question,

$\eta_{1}=\frac{T_{1}-T_{2}}{T_{1}}$

 $\frac{T_{1}-T_{2}}{T_{1}}=\frac{1}{6}$......(i)

$\eta_{2}=\frac{T_{1}-(T_{2}-62)}{T_{1}}\Rightarrow 2 \times \eta_{1}=\frac{T_{1}-T_{2}+62}{T_{1}}$

$2\times \frac{1}{6}=\frac{T_{1}-T_{2}-62}{T_{1}}\Rightarrow \frac{1}{3}=\frac{T_{1}-T_{2}+62}{T_{1}}$ .....(ii)

 from  Eq.(i) we ,get,

 $T_{1}-T_{2}=\frac{T_{1}}{6}$

 Substituting this value in Eq(11), we get

   $ \frac{1}{3}=\frac{\frac{T_{1}}{6}+62}{T_{1}}\Rightarrow \frac{1}{3}= \frac{T_{1}+372}{6T_{1}}$

  $6T_{1}=3T_{1}+1116\Rightarrow 3T_{1}=1116 \Rightarrow T_{1}=372 K$

 From Eq.(i) , we get

      $\frac{372-T_{2}}{372}= \frac{1}{6}$

 $6(372- T_{2})=372$

 $2232-6T_{2}=372$

   $6T_{2}=2232-372$

  $6T_{2}=1860$

  $T_{2}=\frac{1860}{6}$

    $T_{2}=310 K$

Consider the motion  of a particle described by $x= a\cos t$ , $y= a \sin t$  and z=t . The trajectory traced by the particle as a  function of time is 

Answer:

Explanation:

 Given,

   $x= a \cos t$

  $y=a \sin t$

and z=t

 Path is circular in xy-plane,

  $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}}= \cos^{2} \theta+\sin^{2} \theta$

   $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

 In one revolution, the particle moves a distance of unit along x-axis 

 $\frac{dz}{dt}=1$

 Hence, the path is a helix

• Physics - General questions