Current Electricity — Page 5 | NEET Physics

Q41

The charge flowing through a resistance R varies with time t is Q = at

-

bt 2 , where

a

and

b

are positive constants. The total heat produced in R is

A

{{{a^3}R} \over {2b}}

B

{{{a^3}R} \over {b}}

C

{{{a^3}R} \over {6b}}

D

{{{a^3}R} \over {3b}}

Correct Answer

Option C

Solution

Given: Charge Q = at – bt 2 $\therefore$ Current

i = {{\partial Q} \over {\partial t}} = a - 2bt

\left\{ {for\,i = 0 \Rightarrow t = {a \over {2b}}} \right\}

From joule's law of heating, heat produced dH = i 2 Rdt

H = \int\limits_0^{a/2b} {{{\left( {a - 2bt} \right)}^2}Rdt}

H = \left. {{{{{\left( {a - 2bt} \right)}^2}R} \over { - 3 \times 2b}}} \right|_0^{{a \over {2b}}} = {{{a^3}R} \over {6b}}

Q42

A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf's is

A 3 : 4

B 3 : 2

C 5 : 1

D 5 : 4

Correct Answer

Option B

Solution

When two cells are connected in series i.e., E 1 + E 2 the balance point is at 50 cm.

And when two cells are connected in opposite direction i.e., E 1 – E 2 the balance point is at 10 cm.

According to principle of potential

{{{E_1} + {E_2}} \over {{E_1} - {E_2}}} = {{50} \over {10}}

\Rightarrow {{2{E_1}} \over {2{E_2}}} = {{50 + 10} \over {50 - 10}}

\Rightarrow {{{E_1}} \over {{E_2}}} = {3 \over 2}

Q43

Two metal wires of identical dimensions are connected in series. If

\sigma

1 and

\sigma

2 are the conductivity of the combination is

A

{{{\sigma _1} + {\sigma _2}} \over {{\sigma _1}{\sigma _2}}}

B

{{{\sigma _1}{\sigma _2}} \over {{\sigma _1} + {\sigma _2}}}

C

{{2{\sigma _1}{\sigma _2}} \over {{\sigma _1} + {\sigma _2}}}

D

{{{\sigma _1} + {\sigma _2}} \over {2{\sigma _1}{\sigma _2}}}

Correct Answer

Option C

Solution

In figure, two metal wires of identical dimension are connected in series

{R_{eq}} = {1 \over {{\sigma _1}A}} + {1 \over {{\sigma _2}A}} = {{{l_{eq}}} \over {{\sigma _{eq}}{A_{eq}}}}

{{2l} \over {{\sigma _{eq}}A}} = {1 \over A}\left( {{{{\sigma _1} + {\sigma _2}} \over {{\sigma _1}{\sigma _2}}}} \right)

\therefore {\sigma _{eq}} = {{2{\sigma _1}{\sigma _2}} \over {{\sigma _1} + {\sigma _2}}}

Q44

A circuit contains an ammeter, a battery of 30 V and a resistance 40.8 ohm all connected in series. If the ammeter has a coil of resistance 480 ohm and a shunt of 20 ohm, the reading in the ammeter will be

A 2A

B 1A

C 0.5 A

D 0.25 A

Correct Answer

Option C

Solution

From circuit diagram Resistance of ammeter =

{{480 \times 20} \over {480 + 20}} = 19.2\Omega

Total resistance R = 40.8 + 19.2 = 60

\Omega

Reading in the ammeter

i = {V \over R}

= {{30} \over {40.8 + 19.2}} = 0.5A

Q45

A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. E 0 and a resistance r 1 . An unknown e.m.f. E is balanced at a length

l

of the potentiometer wire. The e.m.f. E will be given by

A

{{{E_0}l} \over L}

B

{{L{E_0}r} \over {\left( {r + {r_1}} \right)l}}

C

{{L{E_0}r} \over {l{r_1}}}

D

{{{E_0}r} \over {\left( {r + {r_1}} \right)}}.{l \over L}

Correct Answer

Option D

Solution

The current through the potentiometer wire is

I = {{{E_0}} \over {\left( {r + {r_1}} \right)}}

and the potential difference across the wire is

V = Ir = {{{E_0}r} \over {\left( {r + {r_1}} \right)}}

The potential gradient along the potentiometer wire is

k = {V \over L} = {{{E_0}r} \over {\left( {r + {r_1}} \right)L}}

As the unknown e.m.f. E is balanced against length l of the potentiometer wire, $\therefore$

E = kl = {{{E_0}r} \over {\left( {r + {r_1}} \right)}}{l \over L}

Q46

Across a metallic conductor of non-uniform cross section a constant potential difference is applied. The quantity which remains constant along the conductor is

A drift velocity

B electric field

C current density

D current

Correct Answer

Option D

Solution

The area of cross section of conductor is non uniform so current density will be different but the flow of electrons will be uniform so current will be constant.

Q47

A potentiometer wire has length 4 m and resistance 8

\Omega

. The resistance that must be connected in series with the wire and an accumulator of e.m.f. 2 V, so as to get a potential gradient 1 mV per cm on the wire is

A 44

\Omega

B 48

\Omega

C 32

\Omega

D 40

\Omega

Correct Answer

Option C

Solution

Total potential difference across potentiometer wire = 10 –3 × 400 volt = 0.4 volt potential gradient =

{{1mv} \over {cm}}

= {10^{ - 3}}v/cm = {10^{ - 1}}{v \over m}

Let resistance of R

\Omega

connected in series. $\therefore$

{2 \over {R + 8}} = {{{{10}^{ - 1}} \times 4} \over 8} = {1 \over {20}}

$\Rightarrow$ R + 8 = 40 or, R = 32

\Omega

Q48

A, B and C are voltmeters of resistance R, 1.5 R and 3R respectively as shown in the figure. When some potential difference is applied between X and Y, the voltmeter readings are V A , V B and V C respectively, Then

A V A = V B

\ne

V C

B V A

\ne

V B

\ne

V C

C V A = V B = V C

D V A

\ne

V B = V C

Correct Answer

Option C

Solution

Effective resistance of B & C = (1.5R)(3R)/(1.5R + 3R) = R In series sequence voltage across A = voltage across B & C Now B & C are parallel, so V B = V C , $\therefore$ V A = V B = V C

Q49

The resistances in the two arms of the meter bridge are 5

\Omega

and R

\Omega

respectively. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6

l

1 . The resistance R is

A 10

\Omega

B 15

\Omega

C 20

\Omega

D 25

\Omega

Correct Answer

Option B

Solution

This is a balanced wheatstone bridge condition,

{5 \over R} = {{{\ell _1}} \over {100 - {\ell _1}}}

and

{5 \over {R/2}} = {{1.6{\ell _1}} \over {100 - 1.6{\ell _1}}}

\Rightarrow R = 15\Omega

Q50

Two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is 8 volt and the average resistance per km is 0.5

\Omega

. The power loss in the wire is

A 19.2 W

B 19.2 kW

C 19.2 J

D 12.2 kW

Correct Answer

Option B

Solution

Total voltage drop across the wire : V = 150 × 8= 1200 volt Total resistance of wire, (R) = 150 × 0.5 = 75

\Omega

Hence, current through wire I = V/R $\Rightarrow$ I = 1200/75 =16 A Finally the power loss will be, P = I 2 R = (16) 2 × 75 = 19200 W or 19.2 kW