 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2002

If a current is passed through a spring then the spring will
A
expand
B
compress
C
remains same
D
none of these

Explanation

When current is passed through a spring then current flows parallel in the adjacent turns.

NOTE : When two wires are placed parallel to each other and current flows in the same direction, the wires attract each other.

Similarly here the various turns attract each other and the spring will compress.
2

AIEEE 2002

The mass of product liberated on anode in an electrochemical cell depends on (where $t$ is the time period for which the current is passed).
A
${\left( {It} \right)^{1/2}}$
B
$It$
C
$I/t$
D
${I^2}t$

Explanation

According to Faraday's first law of electrolysis

$m = ZIt \Rightarrow m \propto It$
3

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

${\theta _n} = {{{\theta _i} + {\theta _c}} \over 2}.$
4

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

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