Current Electricity — Page 4 | NEET Physics

Q31

A charged particle having drift velocity 7.5

\times

10 -4 ms -1 in an electric field of 3

\times

10 -10 Vm -1 has a mobility in m 2 V -1 s -1 of :

A 2.5

\times

10 6

B 2.5

\times

10 -6

C 2.25

\times

10 -15

D 2.25

\times

10 15

Correct Answer

Option A

Solution

The relation between the mobility and drift velocity is, Mobility,

\mu = {{{\upsilon _d}} \over E}

= {{7.5 \times {{10}^{ - 4}}} \over {3 \times {{10}^{ - 10}}}}

= 2.5 \times {10^6}{m^2}{V^{ - 1}}{s^{ - 1}}

Q32

The solids which have the negative temperature coefficient of resistance are :

A insulators only

B semiconductors only

C insulators and semiconductors

D metals

Correct Answer

Option C

Solution

The temperature coefficient of resistance is negative for semiconductor and insulators, and it is positive for metals.

Q33

In the circuits shown below, the readings of voltmeters and the ammeters will be :

A V 1 = V 2 and i 1 = i 2

B V 1 > V 2 and i 1 > i 2

C V 1 > V 2 and i 1 = i 2

D V 1 = V 2 and i 1 > i 2

Correct Answer

Option A

Solution

Resistance for ideal voltmeter = $\infty$ Resistance for ideal ammeter = 0 For 1 st circuit, V 1 = i 1 $\times$ 10 =

{{10} \over {10}} \times 10

= 10 volt For 2 nd circuit, 10

\Omega

is in series with ideal voltmeter. Therefore it will not affect the circuit. So, V 2 = i 2 $\times$ 10 =

{{10} \over {10}} \times 10

= 10 volt $\therefore$ V 1 = V 2 Hence, i 1 = i 2 =

{{10\,V} \over {10\,\Omega }}

= 1 A

Q34

Six similar bulbs are connected as shown in the figure with a DC source of emf E, and zero internal resistance. The ratio of power consumption by the bulbs when (i) all are glowing and (ii) in the situation when two from section A and one from section B are glowing, will be :

A 4 : 9

B 9 : 4

C 1 : 2

D 2 : 1

Correct Answer

Option B

Solution

(i) When all bulbs are glowing, from the given figure. Equivalent resistance, R eq =

{R \over 3} + {R \over 3} = {{2R} \over 3}

Initial Power, (P i ) =

{{3{E^2}} \over {2R}}

......(1) (ii) When two section A and one from section B are glowing. R eq =

{R \over 2} + R

=

{{3R} \over 2}

Final Power, (P f ) =

{{2{E^2}} \over {3R}}

$\therefore$

{{{P_i}} \over {{P_f}}} = {{3{E^2}} \over {2R}} \times {{3R} \over {2{E^2}}}

= 9 : 4

Q35

Which of the following acts as a circuit protection device?

A inductor

B conductor

C switch

D fuse

Correct Answer

Option D

Solution

Fuse is the protective device because of low melting point of the wire that can save other device on heating.

Q36

A carbon resistor of (47

\pm

4.7) k

\Omega

is to be marked with rings of different colours for its identification. The colour code sequence will be

A Violet – Yellow – Orange – Silver

B Yellow – Violet – Orange – Silver

C Yellow – Green – Violet – Gold

D Green – Orange – Violet – Gold

Correct Answer

Option B

Solution

Colour code for carbon resistor 0 - Black 1 - Brown 2 - Red 3 - Orange 4 - Yellow 5 - Green 6 - Blue 7 - Violet 8 - Grey 9 - White Tolerance : ± 5% Gold ± 10% Silver ± 20% No colour (47 $\pm$ 4.7) k

\Omega

= 47 $\times$ 10 3 $\pm$ 10%

\Omega

Which shows the colour code Yellow – Violet – Orange – Silver.

Q37

A set of n equal resistors, of value R each, are connected in series to a battery of emf E and internal resistance R. The current drawn is I. Now, the n resistors are connected in parallel to the same battery. Then the current drawn from battery becomes 10I. The value of n is

A 10

B 11

C 20

D 9

Correct Answer

Option A

Solution

n series grouping equivalent resistance R series = nR In parallel grouping equivalent resistance R parallel =

{R \over n}

Current drawn from a battery when n resistors are connected in series is I =

{E \over {nR + R}}

......(i) Current drawn from same battery when n resistors are connected in parallel is 10I =

{E \over {{R \over n} + R}}

......(ii) On dividing eqn. (ii) by (i), 10 =

{{\left( {n + 1} \right)R} \over {\left( {{1 \over n} + 1} \right)R}}

Solving we get, n = 10

Q38

A potentiometer is an accurate and versatile device to make electrical measurements of EMF because the method involves

A potential gradients

B a condition of no current flow through the galvanometer

C a combination of cells, galvanometer and resistances

D cells

Correct Answer

Option B

Solution

A potentiometer is an accurate and versatile device to make electrical measurements of E.M.F. because the method involves zero deflection without any current in galvanometer.

Q39

The resistance of a wire is 'R' ohm. If it is melted and stretched to 'n' times its original length. its new resistance will be

A

{R \over n}

B n 2 R

C

{R \over {{n^2}}}

D nR

Correct Answer

Option B

Solution

Resistance

R = \rho l/A

l' = nl

and

A' = {A \over n},R' = {{\rho l} \over {A'}}

R' = {{\rho nl} \over {A/n}} = {{\rho l} \over A}{n^2}

R' = R{n^2}

Q40

A filament bulb (500 W, 100 V) is to be used in a 230 V main supply. When a resistance R is connected in series, it works perfectly and the bulb consumes 500 W. The value of R is

A 230

\Omega

B 46

\Omega

C 26

\Omega

D 13

\Omega

Correct Answer

Option C

Solution

Resistance of bulb,

{R_B} = {{{V^2}} \over P} = {{{{\left( {100} \right)}^2}} \over {500}} = 20\Omega

Power of the bulb in the circuit, P = VI

I = {P \over {{V_B}}}

= {{500} \over {100}} = 5A

V R = IR $\Rightarrow$ (230 – 100) = 5 × R $\therefore$ R = 26

\Omega