Current Electricity — Page 14 | JEE Physics

Q131

Current passing through a wire as function of time is given as

I(t)=0.02 t+0.01 \mathrm{~A}

. The charge that will flow through the wire from

t=1 \mathrm{~s}

to

t=2 \mathrm{~s}

is

A 0.02 C

B 0.07 C

C 0.06 C

D 0.04 C

Correct Answer

Option D

Solution

To find the total charge $q$ that flows through the wire from $t = 1 \, \text{s}$ to $t = 2 \, \text{s}$ , we can integrate the current function $I(t) = 0.02t + 0.01 \, \text{A}$ over this interval.

The formula for the charge $q$ flowing through the wire is given by the integral of the current with respect to time: $q = \int I(t) \, dt$ We integrate the function from $t = 1$ to $t = 2$ : $q = \int_1^2 (0.02t + 0.01) \, dt$ To solve this, first find the antiderivative: $q = \left[ 0.02 \dfrac{t^2}{2} + 0.01t \right]_1^2$ Calculate the expression at the upper and lower bounds: $= \left[ 0.01t^2 + 0.01t \right]_1^2$ Plug in the values: For $t = 2$ : $= 0.01(2)^2 + 0.01(2) = 0.04 + 0.02 = 0.06$ For $t = 1$ : $= 0.01(1)^2 + 0.01(1) = 0.01 + 0.01 = 0.02$ Subtract the result at $t=1$ from the result at $t=2$ : $q = 0.06 - 0.02 = 0.04 \, \text{C}$ Therefore, the charge that flows through the wire from $t = 1 \, \text{s}$ to $t = 2 \, \text{s}$ is $0.04 \, \text{C}$ .

Q132

A 50

\Omega

resistance is connected to a battery of 5 V. A galvanometer of resistance 100

\Omega

is to be used as an ammeter to measure current through the resistance, for this a resistance rs is connected to the galvanometer. Which of the following connections should be employed if the measured current is within 1% of thecurrent without the ammeter in the circuit ?

A rs = 0.5

\Omega

in parallel with the galvanometer

B rs = 0.5

\Omega

in series with the galvanometer

C rs = 1

\Omega

in series with galvanometer

D rs =1

\Omega

in parallel with galvanometer

Correct Answer

Option A

Solution

Current in the circuit without ammeter I $=$

{5 \over {50}} = 0.1

A $\therefore$ With ammeter current $=$ 0.1 $\times$

{{99} \over {100}}

= 0.099 A With ammeter equivalent resistance, Req = 50 +

{{100\,{r_s}} \over {100 + {r_s}}}

$\therefore$ 0.099 =

{5 \over {50 + {{100\,{r_s}} \over {100 + {r_s}}}}}

$\Rightarrow$ 50 +

{{100\,{r_s}} \over {100 + {r_s}}}

=

{5 \over {0.099}}

$\Rightarrow$

{{100\,{r_s}} \over {100 + {r_s}}} = 0.5

$\Rightarrow$ 100 rs = 50 + 0.5rs $\Rightarrow$ 99.5 rs = 50 $\Rightarrow$ rs $=$

{{50} \over {99.5}} = 0.5\,\Omega

Q133

If

{\theta _1},

is the inversion temperature,

{\theta _n}

is the neutral temperature,

{\theta _c}

is the temperature of the cold junction, then

A

{\theta _i} + {\theta _c} = {\theta _n}

B

{\theta _i} - {\theta _c} = 2{\theta _n}

C

{{{\theta _i} + {\theta _C}} \over 2} = {\theta _n}

D

{\theta _c} - {\theta _i} = 2{\theta _n}

Correct Answer

Option C

Solution

{\theta _n} = {{{\theta _i} + {\theta _c}} \over 2}.

Q134

Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross section 5 mm2, is

\upsilon

. If the electron density in copper is 9

\times

1028/m3 the value of

\upsilon

. in mm/s is close to (Take charge of electron to be = 1.6

\times

10

-

19C)

A 0.02

B 3

C 2

D 0.2

Correct Answer

Option A

Solution

We know, I = neAVd $\therefore$ Vd =

{{\rm I} \over {neA}}

=

{{1.5} \over {9 \times {{10}^{28}} \times 1.6{ \times ^{ - 19}} \times 5 \times {{10}^{ - 6}}}}

= 0.02 m/s

Q135

The Kirchhoff's first law

\left( {\sum i = 0} \right)

and second law

\left( {\sum iR = \sum E} \right),

where the symbols have their usual meanings, are respectively based on

A conservation of charge, conservation of momentum

B conservation of energy, conservation of charge

C conservation of momentum, conservation of charge

D conservation of charge, conservation of energy

Correct Answer

Option D

Solution

NOTE : Kirchhoff's first law is based on conservation of charge and Kirchhoffs second law is based on conservation of energy.

Q136

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

Correct Answer

Option C

Solution

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

Q137

In a circuit for finding the resistance of a galvanometer by half deflection method, a 6 V battery and a high resistance of 11 k

\Omega

are used. The figure of merit of the galvanometer is 60

\mu A/

division. In the absence of shunt resistance, the galvanometer produces a deflection of

\theta

= 9 divisions when current flows in the circuit. The value of the shunt resistance that can cause the deflection of

\theta /2,

is closest to :

A 500

\Omega

B 220

\Omega

C 55

\Omega

D 110

\Omega

Correct Answer

Option D

Solution

Current required by unit deflection is 60 $\mu$ A.

For, $\theta$ = 9 current is I = 9 $\times$ 60 $\mu$ A $\Rightarrow$ I = 540 $\mu$ A = 540 $\times$ 10 $-$ 6 A Let G is resistance of galvanometer.

Then,

540 \times {10^{ - 6}} = {6 \over {(11000 + G)}}

[11000 + G] 90 $\times$ 10 $-$ 6 = 1 99000 + 9G = 105 9G = 100000 $-$ 99000 9G = 1000

G = {{1000} \over 9}\Omega

Also, in half deflection method,

G = {{RS} \over {R - S}} \Rightarrow {{1000} \over 9} = {{11000\,S} \over {11000 - S}}

{1 \over 9} = {{11S} \over {11000 - S}} \Rightarrow 11000 - S = 99\,S

100 S = 11000 $\Rightarrow$ S = 110

\Omega

Q138

An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the lengths and radii are in the ratio of

{4 \over 3}

and

{2 \over 3}

, then the ratio of the current passing through the wires will be

A

8/9

B

1/3

C

3

D

2

Correct Answer

Option B

Solution

{i_1}{R_1} = {i_2}{R_2}\,\,\,\,\,\,\,\,\,\,

(same potential difference) V = I1R1 = I1 $\times$

{{\rho {l_1}} \over {\pi r_1^2}}

Also V = I2R2 = I2 $\times$

{{\rho {l_2}} \over {\pi r_2^2}}

$\therefore$ I1 $\times$

{{\rho {l_1}} \over {\pi r_1^2}}

= I2 $\times$

{{\rho {l_2}} \over {\pi r_2^2}}

$\Rightarrow$

{{{I_1}} \over {{I_2}}} = {{{\ell _1}} \over {{\ell _2}}} \times {{r_1^2} \over {r_2^2}}

= {3 \over 4} \times {4 \over 9} = {1 \over 3}\,\,

Q139

Two electric bulbs marked

25W

-

220

V

and

100W

-

220V

are connected in series to a

440

V

supply. Which of the bulbs will fuse?

A Both

B

100

W

C

25

W

D Neither

Correct Answer

Option C

Solution

The current upto which bulb of marked

25W

-

220V,

will not fuse

{I_1} = {{{W_1}} \over {{V_1}}} = {{25} \over {220}}Amp

Similarly,

{I_2} = {{{W_2}} \over {{V_2}}} = {{100} \over {220}}\,Amp

The current flowing through the circuit

I = {{440} \over {{{\mathop{\rm R}\nolimits} _{eff}}}},\,\,{{\mathop{\rm R}\nolimits} _{eff}} = {R_1} + {R_2}

{R_1} = {{V_1^2} \over {{P_1}}} = {{{{\left( {220} \right)}^2}} \over {25}};\,\,\,

{R_2} = {{V_2^2} \over P} = {{{{\left( {220} \right)}^2}} \over {100}}

I = {{440} \over {{{{{\left( {220} \right)}^2}} \over {25}} + {{{{\left( {220} \right)}^2}} \over {100}}}}

= {{440} \over {{{\left( {220} \right)}^2}\left[ {{1 \over {25}} + {1 \over {100}}} \right]}}

I = {{40} \over {220}}\,\,Amp

as

{I_1}\left( { = {{25} \over {220}}A} \right) < I\left( { = {{40} \over {220}}A} \right) < {I_2}\left( { = {{100} \over {200}}A} \right)

Thus the bulb marked

25W

-

220

will fuse.

Q140

A uniform wire of length 1 and radius r has a resistance of 100

\Omega

. It is recast into a wire of radius

{r \over 2}.

The resistance of new wire will be :

A 1600

\Omega

B 400

\Omega

C 200

\Omega

D 100

\Omega

Correct Answer

Option A

Solution

Resistance of a wire of length l and radius r is given by R =

{{\rho l} \over A}

=

{{\rho l} \over A} \times {A \over A} = {{\rho V} \over {{A^2}}} = {{\rho V} \over {{\pi ^2}{r^4}}}

$\Rightarrow$ R $\propto$

{1 \over {{r^4}}}

$\therefore$

{{{R_1}} \over {{R_2}}} = {\left( {{{{r_2}} \over {{r_1}}}} \right)^4}

Given, R1 = 100

\Omega

, r1 = r, r2 =

{r \over 2}

, R2 = ? $\therefore$ R2 = R1

{\left( {{{{r_1}} \over {{r_2}}}} \right)^4}

= 16R1 = 1600

\Omega