VITEEE 2014 :: Physics :: General questions

• Physics - General questions

An eye can detect $5 \times 10^{4}$ photons per square meter per sec of green light ($\lambda \alpha=5000Å)$ while the ear can detect $10^{-13}$W/m² . The factor by which the eye is more sensitive as a power detector the ear is close to 

Answer:

Explanation:

Energy = $\frac{12375}{5000}=2.475Ev=4 \times 10^{-19}J$

 Minimum intensity to which the eye can respond 

$I_{eye}=$(photon flux) x energu of photon

$I_{eye}=(5 \times 10^{4}) \times 4 \times 10^{-19}$

=$2 \times 10^{-14} W/m^{2}$

Now, lesser the intensity required by a detector for detection more sensitive it will  be

  $\frac{I_{ear}}{I_{eye}}=\frac{10^{-13}}{2 \times 10^{-14}}=5$

An L-C-R  circuit contains $R= 50 \Omega$ , L=1 mH and C=$0.1 \mu F$ . The impedance of the circuit will be minimum for a frequency of 

Answer:

Explanation:

Impedence  of L-C-R  circuit  will be minimum for a resonant frequency so,

$v_{0}=\frac{1}{2\pi\sqrt{LC}}$

  = $v_{0}=\frac{1}{2\pi\sqrt{1\times10^{-3}\times0.1\times10^{-6}}}=\frac{10^{5}}{2 \pi}Hz$

 Two coils have a mutual inductance of 0.55H. The current changes in the first coil according to equation  $I=I_{0}\sin \omega t$.

where ,$I_{0}=10A$ and $\omega=100\pi$ rad/s

The maximum value of emf in the second coil is 

Answer:

Explanation:

E.M.F e= $M \frac{di}{dt}=0.005 \times \frac{d}{dt} (i_{0}\sin \omega t)$

=$0.005 \times i_{0} \cos \omega t$

$e _{max}=0.005 \times 10 \times 100 \pi=5 \pi$

The incorrect statement regarding the lines of force of the magnetic field B is 

Answer:

Explanation:

 Inside a magnet, magnetic lines of force move from south pole to north pole

A circular current-carrying coil has a radius R.The distance from the centre of the coil on the axis, where the magnetic induction will be $\frac{1}{8}$ th to its value at the centre of the coil is 

Answer:

Explanation:

 Here the ratio, $\frac{B_{centre}}{B_{axis}}=\left(1+\frac{x^{2}}{R^{2}}\right)^{3/2}$

Also, $B_{axis}=\frac{1}{8}B_{centre}$

$\frac{8}{1}=\left(1+\frac{x^{2}}{R^{2}}\right)^{3/2} \Rightarrow   4=1+\frac{x^{2}}{R^{2}}$

$3=\frac{x^{2}}{R^{2}} \Rightarrow X^{2}=3R^{2}$

or ,$x= \sqrt{3}R$

• Physics - General questions