1
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 11th January Morning Slot

In the circuit shown,


the switch S1 is closed at time t = 0 and the switch S2 is kept open. At some later time(t0), the switch S1 is opened and S2 is closed. The behavior of the current I as a function of time 't' is given by :
A
B
C
D

Explanation

From time t = 0 to t = t0, growth of current takes place and after that decay of current takes place.


most appropriate is (A)
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 11th January Morning Slot

In a Wheatstone bridge(see fig.), Resistances P and Q are approximately equal. When R = 400 $$\Omega $$, the bridge is balanced. On interchanging P and Q, the value of R, for balance, is 405 $$\Omega $$. The value of X is close to :

A
402.5 ohm
B
401.5 ohm
C
403.5 ohm
D
404.5 ohm

Explanation

For a balanced bridge,

$${P \over Q} = {{400} \over X}$$ ......(i)

After interchanging P and Q,

$${Q \over P} = {{405} \over X}$$ ......(ii)

Multiplying equation (i) and (ii), we get

X2 = 400$$ \times $$405

$$ \Rightarrow $$ X = $$\sqrt {400 \times 405} $$ = 402.5 $$\Omega $$
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 11th January Evening Slot

An electric field of 1000 V/m is applied to an electric dipole at angle of 45o. The value of electric dipole moment is 10–29 C.m. What is the potential energy of the electric dipole?
A
- 7 $$ \times $$ 10–27 J
B
$$-$$ 9 $$ \times $$ 10–20 J
C
$$-$$ 10 $$ \times $$ 10–29 J
D
$$-$$ 20 $$ \times $$ 10–18 J

Explanation

U = $$-$$ $$\overrightarrow P .\overrightarrow E $$

= $$-$$ PE cos $$\theta $$

= $$-$$ (10$$-$$29) (103) cos 45o

= $$-$$ 0.707 $$ \times $$ 10$$-$$26 J

= $$-$$ 7 $$ \times $$ 10$$-$$27 J.
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 11th January Evening Slot

A galvanometer having a resistance of 20 $$\Omega $$ and 30 divisions on both sides has figure of merit 0.005 ampere/division. The resistance that should be connected in series such that it can be used as a voltmeter upto 15 volt, is:
A
120 $$\Omega $$
B
125 $$\Omega $$
C
80 $$\Omega $$
D
100 $$\Omega $$

Explanation

Rg = 20$$\Omega $$

NL = NR = N = 30

FOM = $${1 \over \phi }$$ = 0.005 A/Div.

Current sentivity = CS = $$\left( {{1 \over {0.005}}} \right)$$ = $${\phi \over {\rm I}}$$

$${\rm I}$$gmax = 0.005 $$ \times $$ 30

= 15 $$ \times $$ 10$$-$$2 = 0.15

15 = 0.15 [20 + R]

100 = 20 + R

R = 80

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