### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2007

Photon of frequency $v$ has a momentum associated with it. If $c$ is the velocity of light, the momentum is
A
$hv/c$
B
$v/c$
C
$h$ $v$ $c$
D
$hv/{c^2}$

## Explanation

Energy of a photon of frequency $v$ is given by $E = hv.$

Also, $E = m{c^2},\,\,m{c^2} = hv$

$\Rightarrow mc = {{hv} \over C} \Rightarrow p = {{hv} \over c}$
2

### AIEEE 2007

If ${g_E}$ and ${g_M}$ are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan's oil drop experiment could be performed on the two surfaces, one will find the ratio
${{electro\,\,ch\arg e\,\,on\,\,the\,\,moon} \over {electronic\,\,ch\arg e\,\,on\,\,the\,\,earth}}\,\,to\,be$
A
${g_M}/{g_E}$
B
$1$
C
$0$
D
${g_E}/{g_M}$

## Explanation

electronic charge does does not depend on acceleration due to gravity as it is a universal constant.

So, electronic charge on earth

$=$ electronic charge on moon

$\therefore$ Required ratio $=1.$
3

### AIEEE 2006

The anode voltage of a photocell is kept fixed. The wavelength $\lambda$ of the light falling on the cathode is gradually changed. The plate current ${\rm I}$ of the photocell varies as follows
A
B
C
D

## Explanation

As $\lambda$ decreases, $v$ increases and hence the speed of photo electron increases. The chances of photo electron to meet the anode increases and hence photo electric current increases.
4

### AIEEE 2006

The time taken by a photoelectron to come out after the photon strikes is approximately
A
${10^{ - 4}}\,s$
B
${10^{ - 10}}\,s$
C
${10^{ - 16}}\,s$
D
${10^{ - 1}}\,s$

## Explanation

Emission of photo-electron starts from the surface after incidence of photons in about ${10^{ - 10}}s.$