1
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Morning Slot

A parallel plate capacitor is made of two square plates of side 'a', separated by a distance d (d < < a). The lower triangular portion is filled with a dielectric of dielectric constant K, as shown in the figure. Capacitance of this capacitor is :

A
$${{K{ \in _0}{a^2}} \over {2d(K + 1)}}$$
B
$${{K{ \in _0}{a^2}} \over {d(K - 1)}}\ln K$$
C
$${{K{ \in _0}{a^2}} \over d}\ln K$$
D
$${1 \over 2}{{K{ \in _0}{a^2}} \over d}$$

Explanation





Let the capacitance of the upper part of the strip of length dx is dC1 and lower part of the strip is dC2

$$ \therefore $$   dC1 = $${{{\varepsilon _0}\,adx} \over {d - y}}$$

and dC2 = $${{K{\varepsilon _0}\,adx} \over y}$$

Here dC1 and dC2 are in series.

So, equivalent capacitance,

$${1 \over {dC}}$$ = $${1 \over {d{C_1}}} + {1 \over {d{C_2}}}$$

$$ \Rightarrow $$    $${1 \over {dC}}$$ = $${{d - y} \over {{\varepsilon _0}a\,dx}} + {y \over {{\varepsilon _0}K\,adx}}$$

$$ \Rightarrow $$   $${1 \over {dC}}$$ = $${1 \over {{\varepsilon _0}\,adx}}$$ ( d $$-$$ y + $${y \over K}$$)

$$ \Rightarrow $$   dC = $${{{\varepsilon _0}\,adx} \over {\left( {d - y} \right) + {y \over K}}}$$

We can divide entire parallel plate capacitor into similar part like the strip of length dx and all the strips will have common end A and B. So they are in parallel.

So equivalent capacitance is the sum of all the strips capacitance.

$$ \therefore $$   $$\int {dC} $$ = $$\int\limits_0^a {{{{\varepsilon _0}\,adx} \over {\left( {d - y} \right) + {y \over K}}}} $$

$$ \Rightarrow $$   C = $$\int\limits_0^a {{{{\varepsilon _0}\,a\,dx} \over {\left( {d - y} \right) + {y \over K}}}} $$

From above diagram you can find this $$ \to $$



tan$$\theta $$ = $${y \over x}$$

Also tan$$\theta $$ = $${d \over a}$$

$$ \therefore $$   $${y \over x}$$ = $${d \over a}$$

$$ \Rightarrow $$   y = $${d \over a}$$ x

By putting this value of y in the integration we get,

C = $$\int\limits_0^a {{{{\varepsilon _0}\,a\,dx} \over {d - {d \over a}x + {d \over {Ka}}x}}} $$

= $$\int\limits_0^a {{{{\varepsilon _0}\,a\,dx} \over {d + \left( {{1 \over K} - 1} \right){d \over a}x}}} $$

= $$\int\limits_0^a {{{{\varepsilon _0}\,{a^2}\,dx} \over {da + \left( {{{1 - K} \over K}} \right)xd}}} $$

= $$\int\limits_0^a {{{K{\varepsilon _0}\,{a^2}\,dx} \over {Kda + \left( {1 - K} \right)xd}}} $$

= $${{K{\varepsilon _0}{a^2}} \over {d\left( {1 - k} \right)}}\left[ {\ln \left( {\left( {1 - K} \right)x + Ka} \right)} \right]_0^a$$

= $${{K{\varepsilon _0}{a^2}} \over {d\left( {1 - k} \right)}}\left[ {\ln \left( a \right) - \ln \left( {Ka} \right)} \right]$$

= $${{K{\varepsilon _0}\,{a^2}} \over {d\left( {1 - k} \right)}}\ln \left( {{a \over {Ka}}} \right)$$

= $${{K{\varepsilon _0}\,{a^2}} \over {d\left( {1 - k} \right)}}\ln \left( {{1 \over K}} \right)$$

= $${{K{\varepsilon _0}\,{a^2}} \over {d\left( {1 - k} \right)}}\ln \left( K \right)$$
2
MCQ (Single Correct Answer)

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
3
MCQ (Single Correct Answer)

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%
4
MCQ (Single Correct Answer)

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

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

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

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