1

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

### 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
3

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

A potentiometer wire AB having length L and resistance 12 r is joined to a cell D of emf $\varepsilon$ and internal resistance r. A cell C having emf $\varepsilon$/2 and internal resistance 3r is connected. The length AJ at which the galvanometer as shown in figure shows no deflection is –

A
${{11} \over {12}}L$
B
${{13} \over {24}}L$
C
${{5} \over {12}}L$
D
${{11} \over {24}}L$

## Explanation

$i = {\varepsilon \over {13r}}$

$i\left( {{x \over L}12r} \right) = {\varepsilon \over 2}$

${\varepsilon \over {13r}}\left[ {{x \over L}.12r} \right] = {\varepsilon \over 2}$ $\Rightarrow \,\,\,\,x = {{13L} \over {24}}$
4

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

Four equal point charges Q each are placed in the xy plane at (0, 2), (4, 2), (4, –2) and (0, –2). The work required to put a fifth charge Q at the origin of the coordinate system will be -
A
${{{Q_2}} \over {4\pi {\varepsilon _0}}}$
B
${{{Q^2}} \over {2\sqrt 2 \pi {\varepsilon _0}}}$
C
${{{Q_2}} \over {4\pi {\varepsilon _0}}}\left( {1 + {1 \over {\sqrt 3 }}} \right)$
D
${{{Q_2}} \over {4\pi {\varepsilon _0}}}\left( {1 + {1 \over {\sqrt 5 }}} \right)$

## Explanation

Potential at origin = ${{KQ} \over 2} + {{KQ} \over 2} + {{KQ} \over {\sqrt {20} }} + {{KQ} \over {\sqrt {20} }}$

(Potential at $\infty$ = 0)

= KQ$\left( {1 + {1 \over {\sqrt 5 }}} \right)$

$\therefore$  Work required to put a fifth charge Q

at origin is equal to ${{{Q^2}} \over {4\pi {\varepsilon _0}}}\left( {1 + {1 \over {\sqrt 5 }}} \right)$