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1

JEE Main 2019 (Online) 12th January Morning Slot

MCQ (Single Correct Answer)
In a meter bridge, the wire of length 1 m has a non-uniform cross-section such that, the variation $${{dR} \over {d\ell }}$$ of its resistance R with length $$\ell $$ is $${{dR} \over {d\ell }}$$ $$ \propto $$ $${1 \over {\sqrt \ell }}$$. Two equal resistances are connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point P. What is the length AP ?

A
0.3 m
B
0.25 m
C
0.35 m
D
0.2 m

Explanation

For the given wire :

dR = C $${{d\ell } \over {\sqrt \ell }}$$,

where C = constant.

Let resistance of part

AP is R1 and PB is R2

$$ \therefore $$  $${{R'} \over {R'}} = {{{R_1}} \over {{R_2}}}$$   or    R1 = R2     

By balanced WSB concept.

Now   $$\int {dR = c\int {{{d\ell } \over {\sqrt \ell }}} } $$

$$ \therefore $$  R1 = C$$\int\limits_0^\ell {{\ell ^{ - 1/2}}d\ell } $$ = C.2.$${\sqrt \ell }$$

R2 = C$$\int\limits_\ell ^1 {{\ell ^{ - 1/2}}d\ell } $$ = C.(2 $$-$$ 2$${\sqrt \ell }$$)

Putting R1 = R2

C2$${\sqrt \ell }$$ = C(2 $$-$$ 2$${\sqrt \ell }$$)

$$ \therefore $$  2$${\sqrt \ell }$$ = 1

$${\sqrt \ell }$$ = $${1 \over 2}$$

i.e. $$\ell $$ = $${1 \over 4}$$ m     $$ \Rightarrow $$  0.25 m
2

JEE Main 2019 (Online) 12th January Morning Slot

MCQ (Single Correct Answer)
An ideal battery of 4 V and resistance R are connected in series in the primary circuit of a potentiometer of length 1 m and esistance 5 $$\Omega $$. The value of R, to give a potential difference of 5 mV across 10 cm of potentiometer wire, is :
A
480 $$\Omega $$
B
495 $$\Omega $$
C
490 $$\Omega $$
D
395 $$\Omega $$

Explanation


Let current flowing in the wire is i.

$$ \therefore $$  i = $$\left( {{4 \over {R + 5}}} \right)A$$

If resistance of 10 m length of wire is x

then x = 0.5 $$\Omega $$ = 5 $$ \times $$ $${{0.1} \over 1}$$ $$\Omega $$

$$ \therefore $$   $$\Delta $$V = P.d. on wire = i. x

5 $$ \times $$ 10$$-$$3 = $$\left( {{4 \over {R + 5}}} \right)$$.(0.5)

$$ \therefore $$   $${{4 \over {R + 5}}}$$ = 10$$-$$2

or   R + 5 = 400 $$\Omega $$

$$ \therefore $$   R = 395 $$\Omega $$
3

JEE Main 2019 (Online) 12th January Morning Slot

MCQ (Single Correct Answer)
In the figure shown, a circuit contains two identical resistors with resistance R = 5$$\Omega $$ and an inductance with L = 2mH. An ideal battery of 15 V is connected in the circuit. What will be the current through the battery long after the switch is closed ?

A
6 A
B
7.5 A
C
3 A
D
5.5 A

Explanation

Ideal inductor will behave like zero resistance long time after switch is closed



I = $${{2\varepsilon } \over R}$$ = $${{2 \times 15} \over 5} = 6A$$
4

JEE Main 2019 (Online) 12th January Morning Slot

MCQ (Single Correct Answer)
The galvanometer deflection, when key K1 is closed but K2 is open, equals $$\theta $$0 (see figure). On closing K2 also and adjusting R2 to 5$$\Omega $$, the deflection in galvanometer becomes $${{{\theta _0}} \over 5}.$$ . The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery] :

A
5 $$\Omega $$
B
25 $$\Omega $$
C
12 $$\Omega $$
D
22 $$\Omega $$

Explanation

case I :

ig = $${E \over {220 + {R_g}}}$$ = C$$\theta $$0     . . .(i)

Case II :

ig = $$\left( {{E \over {220 + {{5{R_g}} \over {5 + {R_g}}}}}} \right)$$ $$ \times $$ $${5 \over {\left( {{R_g} + 5} \right)}}$$ = $${{C{\theta _0}} \over 5}$$       . . .(ii)

$$ \Rightarrow $$  $${{5E} \over {225{R_g} + 1100}}$$ = $${{C{\theta _0}} \over 5}$$       . . .(ii)

$${E \over {220 + {R_g}}} = C\theta $$       . . .(i)

$$ \Rightarrow $$  $${{225{R_g} + 1100} \over {1100 + 5{R_g}}} = 5$$

$$ \Rightarrow $$  5500 + 25Rg = 225Rg + 1100

200Rg = 4400

Rg = 22$$\Omega $$

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