1

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

A series AC circuit containing an inductor (20 mH), a capacitor (120 $\mu$F) and a resistor (60 $\Omega$) is driven by an AC source of 24V/50 Hz. The energy dissipated in the circuit in 60 s is :
A
5.65 $\times$ 102J
B
2.26 $\times$ 103J
C
5.17 $\times$ 102 J
D
3.39 $\times$ 103 J

## Explanation

Energy dissipated in 60 Sec

= (Pavg) $\times$ 60

= Vrms $\times$ Irms $\times$ cos$\phi$ $\times$ 60

= Vrms $\times$ ${{{V_{rms}}} \over Z} \times$ cos$\phi$ $\times$ 60

XL = $\omega$L = 2$\pi$FL

= 2$\pi$(50)$\times$ 20 $\times$ 10$-$3

= 2$\pi$ $\Omega$

XC = ${1 \over {\omega C}}$

= ${1 \over {2\pi fC}}$

= ${1 \over {2\pi \left( {50} \right) \times 120 \times {{10}^{ - 6}}}}$

= 26.52 $\Omega$

$\therefore$   XC $-$ XL = 20.24 $\simeq$ 20

$\therefore$   Z = $\sqrt {{{\left( {{X_C} - {X_L}} \right)}^2} + {R^2}}$

= $\sqrt {{{\left( {20} \right)}^2} + {{60}^2}}$

= 20 $\sqrt {10}$ $\Omega$

Also cos$\phi$ = ${R \over Z}$ = ${{60} \over {20\sqrt {10} }} = {3 \over {\sqrt {10} }}$

$\therefore$   Energy dissipated in 60 sec

= ${{{{\left( {24} \right)}^2}} \over {20\sqrt {10} }} \times {3 \over {\sqrt {10} }} \times 60$

= 5.17 $\times$ 102 J
2

### 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%
3

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

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

$\Rightarrow$  $K = {{{K_1}{K_2}} \over {{K_1} + {K_2}}} + {{{K_3}{K_4}} \over {{K_3} + {K_4}}}$
4

### 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