1

### JEE Main 2019 (Online) 9th January Evening Slot

A carbon resistance has a following colour code. What is the value of the resistance ?

A
530 k$\Omega$ $\pm$ 5%
B
5.3 M$\Omega$ $\pm$ 5%
C
6.4 M$\Omega$ $\pm$ 5%
D
64 k$\Omega$ $\pm$ 10%

## Explanation

From colour coding table,
Green line represents number = 5
Orange line represents number = 3
Yellow line represent multiplier = 104
Golden line represents tollerence = $\pm$ 5%

$\therefore$   Resistance = 53 $\times$ 104 $\pm$ 5%

= 530 k$\Omega$ $\pm$ 5%
2

### JEE Main 2019 (Online) 9th January Evening Slot

A parallel plate capacitor with square plates is filled with four dielecytrics of dielectrics constants K1, K2, K3, K4 arranged as shown in the figure. The effective dielectric constant K will be :

A
$K = {{{K_1}{K_2}} \over {{K_1} + {K_2}}} + {{{K_3}{K_4}} \over {{K_3} + {K_4}}}$
B
$K = {{\left( {{K_1} + {K_2}} \right)\left( {{K_3} + {K_4}} \right)} \over {2\left( {{K_1} + {K_2} + {K_3} + {K_4}} \right)}}$
C
$K = {{\left( {{K_1} + {K_2}} \right)\left( {{K_3} + {K_4}} \right)} \over {{K_1} + {K_2} + {K_3} + {K_4}}}$
D
$K = {{\left( {{K_1} + {K_4}} \right)\left( {{K_2} + {K_3}} \right)} \over {2\left( {{K_1} + {K_2} + {K_3} + {K_4}} \right)}}$

## Explanation

Here, ${C_1} = {{{K_1}{\varepsilon _0}.L.{L \over 2}} \over {{d \over 2}}} = {{{K_1}{\varepsilon _0}{L^2}} \over d}$

${C_2}{{{K_2}{\varepsilon _0}.L.{L \over 2}} \over {{d \over 2}}} = {{{K_2}{\varepsilon _0}{L^2}} \over d}$

${C_3} = {{{K_3}{\varepsilon _0}{L^2}} \over d}$

${C_4} = {{{K_4}{\varepsilon _0}{L^2}} \over d}$

Here  C1 C2 are in series and C3, C4 are in series.

Equivalent capacitance point A and B is,

Ceq $=$ ${{{C_1}{C_2}} \over {{C_1} + {C_2}}} + {{{C_3}{C_4}} \over {{C_3} + {C_4}}}$

$= {C_{12}} + {C_{34}}$

${C_{12}}\,\, = \,\,{{{{{K_1}{\varepsilon _0}{L^2}} \over d} \times {{{K_2}{\varepsilon _0}{L^2}} \over d}} \over {{{{K_1}{\varepsilon _0}{L^2}} \over d} + {{{K_2}{\varepsilon _0}{L^2}} \over d}}}$

$= \,\,{{{K_1}{K_2}} \over {{K_1} + {K_2}}} \times {{{\varepsilon _0}{L^2}} \over d}$

Similarly,

${C_{34}} = \,\,{{{K_3}{K_4}} \over {{K_3} + {K_4}}} \times {{{\varepsilon _0}{L^2}} \over d}$

$\therefore$  Ceq $=$ $\left[ {{{{K_1}{K_2}} \over {{K_1} + {K_2}}} + {{{K_3}{K_4}} \over {{K_3} + {K_4}}}} \right]{{{\varepsilon _0}{L^2}} \over d}$

$\Rightarrow$  ${{K{\varepsilon _0}{L^2}} \over d}$  $= \left[ {{{{K_1}{K_2}} \over {{K_1} + {K_2}}} + {{{K_3}{K_4}} \over {{K_3} + {K_4}}}} \right]{{{\varepsilon _0}{L^2}} \over d}$

$\Rightarrow$  $K = {{{K_1}{K_2}} \over {{K_1} + {K_2}}} + {{{K_3}{K_4}} \over {{K_3} + {K_4}}}$
3

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

A parallel plate capacitor is of area 6 cm2 and a separation 3 mm. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants K1 = 10, K2 = 12 and K3 = 14. The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be -

A
12
B
36
C
14
D
4

## Explanation

Let dielectric constant of material used be K.

$\therefore$  ${{10{ \in _0}\,A/3} \over d} + {{12{ \in _0}\,A/3} \over d} + {{14{ \in _0}\,A/3} \over d} = {{K{ \in _0}A} \over d}$

$\Rightarrow$    K $=$ 12
4

### 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

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