1
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Morning Slot

A resistance is shown in the figure. Its value and tolerance are given respectively by :

A
270 $$\Omega $$, 10 %
B
27 k$$\Omega $$, 10 %
C
27 k$$\Omega $$, 20 %
D
270 $$\Omega $$, 5 %

Explanation

From color code table :

For Red value is 2

For Violet value is 7

For Orange multiplier is 103

For Silver tolarence is 10%

$$ \therefore $$ Resistance and tolerance is

= 27 $$ \times $$ 103 $$ \pm $$ 10%

= 27 k$$\Omega $$ $$ \pm $$ 10%
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Evening Slot

Two point charges q1$$\left( {\sqrt {10} \mu C} \right)$$ and q2($$-$$ 25 $$\mu $$C) are placed on the x-axis at x = 1 m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is,
[take $${1 \over {4\pi { \in _0}}}$$ = 9 $$ \times $$ 109 Nm2C$$-$$2]
A
$$\left( {63\widehat i - 27\widehat j} \right) \times {10^2}$$
B
$$\left( { - 63\widehat i + 27\widehat j} \right) \times {10^2}$$
C
$$\left( {81\widehat i - 81\widehat j} \right) \times {10^2}$$
D
$$\left( { - 81\widehat i + 81\widehat j} \right) \times {10^2}$$

Explanation


Electric field due to $$\sqrt {10} \,\mu C$$ charge :

$$\overrightarrow {{E_1}} = - $$ E1 sin$$\theta $$1 $$\widehat i$$ + E1 cos$$\theta $$1 $$\widehat j$$

Where,

E1 $$ = {1 \over {4\pi {\varepsilon _0}}} \times {{\left| {{q_1}} \right|} \over {{r_1}^2}}$$

$$ = 9 \times {10^9} \times {{\sqrt {10} \times {{10}^{ - 6}}} \over {{{\left( {\sqrt {{1^2}} + {3^2}} \right)}^2}}}$$

$$ = {{9 \times {{10}^3}} \over {\sqrt {10} }}\,v/m$$

sin $$\theta $$1 $$=$$ $${1 \over {\sqrt {10} }}$$

and cos$$\theta $$1 = $${3 \over {\sqrt {10} }}$$

$$ \therefore $$   $$\overrightarrow {{E_1}} = {{9 \times {{10}^3}} \over {\sqrt {10} }}\,\left( { - {1 \over {10}}\widehat i + {3 \over {\sqrt {10} }}\widehat j} \right)$$

$$ = 9 \times {10^2}\left( { - \widehat i + 3\widehat j} \right)$$

Electric field due to $$-$$ 25 $$\mu $$C charge,



$$\overrightarrow {{E_2}} = $$ E2 sin$$\theta $$2$$\widehat i$$ $$-$$ E2 cos$$\theta $$2 $$\widehat j$$

where

E2 $$ = {1 \over {4\pi {\varepsilon _0}}} \times {{\left| {{9_2}} \right|} \over {r_2^2}}$$

$$ = 9 \times {10^9} \times {{25 \times {{10}^{ - 6}}} \over {{{\left( {\sqrt {{4^2} + {3^2}} } \right)}^2}}}$$

$$ = 9 \times {10^3}$$ V/m

sin$$\theta $$2 = $${4 \over 5}$$

and cos$$\theta $$2 = $${3 \over 5}$$

$$ \therefore $$   $$\overrightarrow {{E_2}} = 9 \times {10^3}\,\,\left( {{4 \over 5}\widehat i - {3 \over 5}\widehat j} \right)$$

$$ = 18 \times {10^2}\left( {4\widehat i - 3\widehat j} \right)$$

$$ \therefore $$   Net electric field,

$$\overrightarrow E $$ = $${\overrightarrow E _1}$$ + $${\overrightarrow E _2}$$

$$ = \left( {63\widehat i - 27\widehat j} \right) \times {10^2}\,\,V/m$$
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Evening Slot

Charge is distributed within a sphere of radius R with a volume charge density $$\rho \left( r \right) = {A \over {{r^2}}}{e^{ - {{2r} \over s}}},$$ where A and a are constants. If Q is the total charge of this charge distribution, the radius R is :
A
a log $$\left( {1 - {Q \over {2\pi aA}}} \right)$$
B
$${a \over 2}$$ log $$\left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)$$
C
a log $$\left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)$$
D
$${a \over 2}$$ log $$\left( {1 - {Q \over {2\pi aA}}} \right)$$

Explanation



Volume of this spherical layer,

dv = (4$$\pi $$r2)dr

charge present in this layer,

dq = $$\rho $$ (4$$\pi $$r2 dr)

= $${A \over {{r^2}}}{e^{ - {{2r} \over a}}}\,\,\left( {4\pi {r^2}dr} \right)$$

= $$A\,{e^{ - {{2r} \over a}}}\left( {4\pi dr} \right)$$

$$ \therefore $$  Total charge in the sphere,

Q= $$\int\limits_0^R {4\pi A{e^{ - {{2r} \over a}}}} \,dr$$

= 4$$\pi $$A$$\int\limits_0^R {{e^{ - {{2r} \over a}}}} \,dr$$

= 4$$\pi $$A$$\left[ {{{{e^{ - {{2r} \over a}}}} \over { - {2 \over a}}}} \right]_0^R$$

= 4$$\pi $$A $$\left( { - {a \over 2}} \right)\left( {{e^{ - {{2R} \over a}}} - 1} \right)$$

$$ \therefore $$  Q = 2$$\pi $$aA $$\left( {1 - {e^{ - {{2R} \over a}}}} \right)$$

$$ \Rightarrow $$  $${1 - {e^{ - {{2R} \over a}}}}$$ = $${Q \over {2\pi aA}}$$

$$ \Rightarrow $$  $${{e^{ - {{2R} \over a}}}}$$ = 1 $$-$$ $${Q \over {2\pi aA}}$$

$$ \Rightarrow $$  $${e^{{{2R} \over a}}}$$ = $${1 \over {1 - {Q \over {2\pi aA}}}}$$

$$ \Rightarrow $$  $${{2R} \over a} = \log \left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)$$

$$ \Rightarrow $$  R = $${a \over 2}$$ log $$\left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)$$
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Evening Slot

In the given circuit the the internal resistance of the 18 V cell is negligible. If R1 = 400 $$\Omega $$, R3 = 100 $$\Omega $$ and R4 = 500 $$\Omega $$ and the reading of an ideal voltmeter across R4 is 5V, then the value of R2 will be :

A
300 $$\Omega $$
B
450 $$\Omega $$
C
550 $$\Omega $$
D
230 $$\Omega $$

Explanation

Voltage accross resistance R4 = 5 V

$$ \therefore $$   IR4 = 5 V

$$ \Rightarrow $$   500 $$ \times $$ I = 5

$$ \Rightarrow $$   I = $${1 \over {100}}$$ A

$$ \therefore $$   Voltage across resistor R3 = $${1 \over {100}}\left( {100} \right)$$ = 1 A

$$ \therefore $$   Total drop in resistance R3 and R4 = 5 + 1 = 6V

So, voltage accross R2 resistance is also 6V as R3, R4 and R2 are in parallel

$$ \therefore $$   Voltage accross R1 resistor R1 resistor = 18 $$-$$ 6 = 12 V

$$ \therefore $$   Current through R1 resistor = $${{12} \over {400}}$$ = $${3 \over {100}}$$ A

$$ \therefore $$   Current through R2 resistor

= $${3 \over {100}} - {1 \over {100}}$$

= $${2 \over {100}}$$ A

$$ \therefore $$   $$\left( {{2 \over {100}}} \right)$$ R2 = 6

$$ \Rightarrow $$   R2 = 300 $$\Omega $$

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