TS EAMCET 2017 :: Physics :: General questions

• Physics - General questions

 The Young's  modulus of a material $2 \times 10^{11}$ N/m² and its elastic limit is $1 \times 10^{8} N/m^{2}$ . For a wire of 1 m length of this material, the maximum elongation achievable is 

Answer:

Explanation:

Given,

 Young's modulus of material= $2 \times 10^{11}  N/m^{2}$

 Elastic limit or stress = $1 \times 10^{8} N/m^{2}$

 Length of the wire =1 m

 We know that ,

    $Y= \frac{stress}{Strain}\Rightarrow Y= \frac{stress}{\frac{\triangle L}{l}}$

Here, l= original length of the wire,

ΔL= charge in the length of the wire

ΔL = $\frac{Stress \times  l}{Y}$=$\frac{1\times 10^{8}1}{2\times 10^{11}}$

 = $0.5 \times 10^{-3}$=0.5 mm

In the given circuit, a charge of +80 $\mu$ C is given to upper plate of a $4 \mu $F capacitor. At a steady state, the charge  on the upper plate  of the $3 \mu F$  capacitor is

Answer:

Explanation:

 The net charge on upper plates both the capacitor   $(2 \mu F$ and $3\mu F)$ will be $+80 \mu C$  . So charge on upper plate of

$3 \mu C$  capacitor will be

          $q=\left(\frac{3}{3+2}\right)(80\mu C)$

 $ q= \frac{3}{5} \times 80$= $48 \mu C$

The energy that should be added to an electron  to reduce its de- Broglie wavelength from 1 nm to 0.5 nm is 

Answer:

Explanation:

 We know that,

 We have,   $\lambda=\frac{h}{\sqrt{2mk}}$

 $\therefore$    Kinetic energy (KE)   $\propto \frac{1}{\lambda^{2}}$

 Hence,   $\frac{KE_{2}}{KE_{1}}= \left(\frac{\lambda_{1}}{\lambda_{2}}\right)^{2}$

 Here, $\lambda_{1}$=1 nm

   $\lambda_{2}$=0.5 nm

$\frac{KE_{2}}{KE_{1}}= \left(\frac{1}{0.5}\right)^{2}$

KE₂ = 4KE₁

 Hence, the kinetic energy is increasing three times

 $\triangle KE=3KE_{1}$

Each of the six ideal batteries of emf 20 V is connected to an external resistance of $4 \Omega$ as shown in the figure. The circuit through the resistance is 

Answer:

Explanation:

 Resistance =$4\Omega$

 The equivalent emf of the combination

  $E_{eq}=nE=3 \times 20=60$

 The cell have no internal resistance

 $\therefore$   Main current flowing through the load

       $i= \frac{nE}{R}=\frac{60}{4}=15 A$

Consider a metal ball of radius  r moving at a constant velocity v in a uniform  magnetic field of induction  $\overline{B}$, Assuming that the direction of velocity forms an angle $\alpha$  with the direction of $\overline{B}$, the maximum potential  difference between  points  on the ball is 

Answer:

Explanation:

 Given , 

The radius of metal ball=r

 Velocity =v

Angle between direction of velocity and magnetic field (B)= $\alpha$

 We know that,

                           $e= Bl v \sin \alpha$

 Here, l=2r,          $e=2r|B||v| \sin \alpha$

• Physics - General questions