### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2002

If ${\theta _1},$ is the inversion temperature, ${\theta _n}$ is the neutral temperature, ${\theta _c}$ is the temperature of the cold junction, then
A
${\theta _i} + {\theta _c} = {\theta _n}$
B
${\theta _i} - {\theta _c} = 2{\theta _n}$
C
${{{\theta _i} + {\theta _C}} \over 2} = {\theta _n}$
D
${\theta _c} - {\theta _i} = 2{\theta _n}$

## Explanation

${\theta _n} = {{{\theta _i} + {\theta _c}} \over 2}.$
2

### AIEEE 2002

A wire when connected to $220$ $V$ mains supply has power dissipation ${P_1}.$ Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is ${P_2}.$ Then ${P_2}:{P_1}$ is
A
$1$
B
$4$
C
$2$
D
$3$

## Explanation

Case 1 : ${P_1} = {{{V^2}} \over R}$

Case 2 : The wire is cut into two equal pieces. Therefore the resistance of the individual wire is ${R \over 2}.$ These are connected in parallel

$\therefore$ ${{\mathop{\rm R}\nolimits} _{eq}} = {{R/2} \over 2} = {R \over 4}$

$\therefore$ ${P_2} = {{{V^2}} \over {R/4}} = 4\left( {{{{V^2}} \over R}} \right) = 4{P_1}$
3

### AIEEE 2002

If an ammeter is to be used in place of a voltmeter, then we must connect with the ammeter a
A
low resistance in parallel
B
high resistance in parallel
C
high resistance in series
D
low resistance in series

## Explanation

KEY CONCEPT : To convert a galvanometer into a voltmeter we connect a high resistance in series with the galvanometer.

The same procedure needs to be done if ammeter is to be used as a voltmeter.