1

### JEE Main 2019 (Online) 10th January Morning Slot

A uniform metallic wire has a resistance of 18 $\Omega$ and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is -
A
12 $\Omega$
B
2 $\Omega$
C
4 $\Omega$
D
8 $\Omega$

$\Omega$

## Explanation Req berween any two vertex will be

${1 \over {{{\mathop{\rm R}\nolimits} _{eq}}}} = {1 \over {12}} + {1 \over 6} \Rightarrow {{\mathop{\rm R}\nolimits} _{eq.}} = 4\Omega$
2

### JEE Main 2019 (Online) 10th January Morning Slot

A 2 W carbon resistor is color coded with green, black, red and brown respectively. The maximum current which can be passed through this resistor is -
A
0.4 mA
B
20 mA
C
63 mA
D
100 mA

## Explanation

P = i2R.

$\therefore$   for imax, R must be minimum

from color coding R = 50 $\times$ 102$\Omega$

$\therefore$   imax = 20mA
3

### JEE Main 2019 (Online) 10th January Morning Slot

A charge Q is distributed over three concentric spherical shells of radii a, b, c (a < b < c) such that their surface charge densities are equal to one another. The total potential at a point at distance r from their common centre, where r < a, would be -
A
${{Q\left( {{a^2} + {b^2} + {c^2}} \right)} \over {4\pi {\varepsilon _0}\left( {{a^3} + {b^3} + {c^3}} \right)}}$
B
${Q \over {4\pi {\varepsilon _0}\left( {a + b + c} \right)}}$
C
${Q \over {12\pi {\varepsilon _0}}}{{ab + bc + ca} \over {abc}}$
D
${{Q\left( {a + b + c} \right)} \over {4\pi {\varepsilon _0}\left( {{a^2} + {b^2} + {c^2}} \right)}}$

## Explanation Potential at point P, V = ${{k{Q_a}} \over a} + {{k{Q_b}} \over b} + {{k{Q_c}} \over c}$

$\because$  Qa : Qb : Qc : : a2 : b2 : c2

{since $\sigma$a = $\sigma$b = $\sigma$c}

$\therefore$  Qa = $\left[ {{{{a^2}} \over {{a^2} + {b^2} + {c^2}}}} \right]$Q

Qb = $\left[ {{{{b^2}} \over {{a^2} + {b^2} + {c^2}}}} \right]$ Q

Qc = $\left[ {{{{c^2}} \over {{a^2} + {b^2} + {c^2}}}} \right]$ Q

V = ${Q \over {4\pi { \in _0}}}\left[ {{{\left( {a + b + c} \right)} \over {{a^2} + {b^2} + {c^2}}}} \right]$
4

### JEE Main 2019 (Online) 10th January Morning Slot

In the given circuit the cells have zero internal resistance. The currents (in Amperes) passing through resistance R1 and R2 respectively, are - A
0.5, 0
B
0, 1
C
1, 2
D
2, 2

## Explanation

i1 = ${{10} \over {20}}$ = 0.5A

i2 = 0