1
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 11th January Evening Slot

A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of 3, keeping the number of turns of the coil per unit length of the frame the same, then the self inductance of the coil:
A
decreases by a factor of $9\sqrt 3$
B
increases by a factor of 27
C
decreases by a factor of 9
D
increases by a factor of 3

Explanation

Total length L will remain constant

L = (3a) N        (N = total turns)

and length of winding = (d) N

(d = diameter of wire)

self inductance = $\mu$0n2A$\ell$

= $\mu$0n2$\left( {{{\sqrt 3 {a^2}} \over 4}} \right)$ dN

$\propto$ a2 N $\propto$ a

So self inductance will become 3 times
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 11th January Evening Slot

Seven capacitors, each of capacitance 2 $\mu$F, are to be connected in a configuration to obtain an effective capacitance of $\left( {{6 \over {13}}} \right)\mu F.$ Which of the combinations, shown in figures below, will achieve the desired value
A
B
C
D

Explanation

Ceq = ${6 \over {13}}$$\mu F Therefore three capacitors most be in parallel to get 6 in {1 \over {{C_{eq}}}} = {1 \over {3C}} + {1 \over C} + {1 \over C} + {1 \over C} + {1 \over C} Ceq = {{3C} \over {13}} = {6 \over {13}}$$\mu$F

3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 11th January Evening Slot

In the circuit shown, the potential difference between A and B is :

A
6 V
B
3 V
C
2 V
D
1 V

Explanation

Potential difference across AB will be equal to battery equivalent across CD

VAB $=$ VCD $=$ ${{{{{E_1}} \over {{r_1}}} + {{{E_2}} \over {{r_2}}} + {{{E_3}} \over {{r_3}}}} \over {{1 \over {{r_1}}} + {1 \over {{r_2}}} + {1 \over {{r_3}}}}} = {{{1 \over 1} + {2 \over 1} + {3 \over 1}} \over {{1 \over 1} + {1 \over 1} + {1 \over 1}}}$

$=$ ${6 \over 3}$ $=$ 2V
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 12th January Morning Slot

In the figure shown, after the switch 'S' is turned from position 'A' to position 'B', the energy dissipated in the circuit in terms of capacitance 'C' and total charge 'Q' is :

A
${1 \over 8}{{{Q^2}} \over C}$
B
${5 \over 8}{{{Q^2}} \over C}$
C
${3 \over 4}{{{Q^2}} \over C}$
D
${3 \over 8}{{{Q^2}} \over C}$

Explanation

Vi = ${1 \over 2}$CE2

Vf = ${{{{\left( {CE} \right)}^2}} \over {2 \times 4c}}$ = ${1 \over 2}{{C{E^2}} \over 4}$

$\Delta$E = ${1 \over 2}$CE2 $\times$ ${3 \over 4}$ = ${3 \over 8}$ CE2

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