VITEEE 2014 :: Physics :: General questions

• Physics - General questions

A battery is charged at a potential of 15 V in 8 h  when the current flowing is 10A. The battery on discharge supplies a current of 5 A for 15 h. The mean terminal voltage  during discharge is 14 V . The watt-hour efficiency of the battery is 

Answer:

Explanation:

Power of battery, when  charged  is given by   $P_{1}=V_{1}I_{1}$

Electrical  energy dissipated id given by $E_{1}=P_{1}t_{1}$

$E_{1}=V_{1}I_{1}t_{1}=15  \times 10  \times 8=1200Wh$

Similarly , the electrical energy dissipated during the discharge of battery is given by $E_{2}=V_{2}I_{2}t_{2}$

=$14 \times 5 \times 15=1050 Wh$

 Hence, watt-hour efficiency of the battery 

$\Rightarrow$  $\eta= \frac{E_{2}}{E_{1}} \times 1000.875 \times 100=87.5$

When a resistor  of $11 \Omega$ is connected in series with a electric cell. The current following in it is 0.5A .In stead when a resistor of $5 \Omega$ is connected to the same electric  cell in series , the current  increases by 0.4A. The internal resistance of the cell  is        

Answer:

Explanation:

Here, $i= \frac{E}{R+r}$

$0.5 =\frac{E}{11+r} \Rightarrow E=5.5+0.5 r$

$0.9 =\frac{E}{5+r}$ or $E=4.5+0.9r$

 On solving  we get r=2.5$\Omega$

Following process is known as $hv \rightarrow e^{+}+e^{-}$

Answer:

Explanation:

The creation of an elementary particle and its antiparticle usually from a photon (or another neutral boson) is called Pair production. This is allowed, provided there is enough energy available to create the pair 

If the 3 distance of 100W lamp is increased from a photocell, the saturation current in the photocell  varies with the distance d as 

Answer:

Explanation:

 As we know, photoelectric current depends on the intensity of incident radiation i.e, $i \propto I$

But, intensity  of radiation  $I \propto \frac{1}{d^{2}}$ so, $i \propto  \frac{1}{d^{2}}$

Silver has a work function of 4.7 eV. When ultraviolet light of wavelength 100 nm is incident on it a potential of 7.7  V is required to stop the photoelectrons from reaching the collector plate. How much potential will be required to stop photoelectrons, when the light of wavelength 200 nm is incident on it?

Answer:

Explanation:

From Einsyein's photoelectric  equation,

$ev_{0}+\phi=\frac{hc}{\lambda}$

and $ev_{0}+\phi_{0}=\frac{hc}{\lambda'}$

$\frac{ev_{0}'+\phi}{eV_{0}+\phi}=\frac{\lambda}{\lambda'}=\frac{100}{200}=\frac{1}{2}$

$2ev_{0}'+2\phi=ev_{0}+\phi$

$ev_{0}'=\frac{ev_{0}-\phi}{2}=\frac{7.7-4.7}{2}=1.5V$

• Physics - General questions