Current Electricity — Page 10 | NEET Physics

Q91

Resistance n, each of r ohm, when connected in parallel give an equivalent resistance of R ohm. If these resistances were connected in series, the combination would have a resistance in ohms, equal to

A n 2 R

B R/n 2

C R/n

D nR

Correct Answer

Option A

Solution

R = {r \over n} \Rightarrow r = nR

When connected in series,

{R_{eq}} = nr

= n (nR) = n 2 R

Q92

when three identical bulbs of 60 watt, 200 volt rating are connected in series to a 200 volt supply, the power drawn by them will be

A 60 watt

B 180 watt

C 10 watt

D 20 watt

Correct Answer

Option D

Solution

The resistance of each bulb

= {{{V^2}} \over P} = {{{{\left( {200} \right)}^2}} \over {60}}\Omega

When three bulbs are connected in series their resultant resistance

= {{3 \times {{\left( {200} \right)}^2}} \over {60}}

Thus power drawn by bulb when connected across 200 V supply

P = {{{V^2}} \over {{R_{re}}}} = {{{{\left( {200} \right)}^2}} \over {3 \times {{\left( {200} \right)}^2}/60}} = 20W

Q93

Fuse wire is a wire of

A high resistance and high melting point

B high resistance and low melting point

C low resistance and low melting point

D low resistance and high melting point

Correct Answer

Option B

Solution

Fuse wire should have high resistance and low melting point.

Q94

An electric kettle has two heating coils. When one of the coils is connected to an a.c. source, the water in the kettle boils in 10 minutes. When the other coil is used the water boils in 40 minutes. If both the coils are connected in parallel, the time taken by the same quantity of water to boil will be

A 8 minutes

B 4 minutes

C 25 minutes

D 15 minutes

Correct Answer

Option A

Solution

Let R 1 and R 2 be the resistance of the two coils and V be the voltage supplied.

Effective resistance of two coils in parallel =

{{{R_1}{R_2}} \over {{R_1} + {R_2}}}

Let H be the heat required to begin boiling in kettle. Then H = Power $\times$ time =

{{{V^2}{t_1}} \over {{R_1}}} = {{{V^2}{t_2}} \over {{R_2}}}

For parallel combination, H =

{{{V^2}\left( {{R_1} + {R_2}} \right){t_p}} \over {{R_1}{R_2}}}

\Rightarrow {1 \over {{t_p}}} = \left( {{{{t_2} + {t_1}} \over {{t_2}{t_1}}}} \right)

\therefore {t_p} = {{{t_2}{t_1}} \over {{t_2} + {t_1}}} = {{10 \times 40} \over {10 + 40}} = 8

minute

Q95

Two 220 volt, 100 watt bulbs are connected first in series and then in parallel. Each time the combination is connected to a 220 volt a.c. supply line. The power drawn by the combination in each case respectively will be

A 50 watt, 100 watt

B 100 watt, 50 watt

C 200 watt, 150 watt

D 50 watt, 200 watt

Correct Answer

Option D

Solution

R = {{{V^2}} \over P} = {{220 \times 220} \over {100}} = 484\Omega

In series, R eq = 484 + 484 = 968

\Omega

\therefore {P_{eq}} = {{{V^2}} \over {968}} = {{220 \times 220} \over {968}} = 50\,watt

In parallel,

{R_{eq}} = 242\Omega

\therefore {P_{eq}} = {{{V^2}} \over {242}} = {{220 \times 220} \over {242}} = 200\,watt

Q96

In a Wheatstone's bridge all the four arms have equal resistance R. If the resistance of the galvanometer arm is also R, the equivalent resistance of the combination as seen by the battery is

A R/4

B R/2

C R

D 2R

Correct Answer

Option C

Solution

In balance Wheatstone bridge, the galvanometer arm can be neglected so equivalent resistance = R.

Q97

Specific resistance of a conductor increases with

A increase in temperature

B increase in cross-section area

C increase in cross-section and decrease in length

D decrease in cross secton area.

Correct Answer

Option A

Solution

Resistance of a conductor is given by R =

\rho {l \over A}

, where $\rho$ is the specific resistance,

l

is the length and A is the cross-sectional area of the conductor. Now, when

l

= 1 and A = 1, R = $\rho$ .

So specific resistance or resistivity of a material may be defined as the resistance of a specimen of the material having unit length and unit cross-section.

Hence, specific resistance is a property of a material and it will increase with the increase of temperature, but will not vary with the dimensions (length, crosssection) of the conductor.

Q98

For a cell terminal potential difference is 2.2V when circuit is open and reduce to 1.8 V when cell is connected to a resistance of R = 5

\Omega

. Determine internal resistance of cell (r)

A

{{10} \over 9}\Omega

B

{{9} \over 10}\Omega

C

{{11} \over 9}\Omega

D

{{5} \over 9}\Omega

Correct Answer

Option A

Solution

Terminal potential difference is 2.2 V when circuit is open. $\therefore$ e.m.f. of the cell = E = 2.2 volt Now, when the cell is connected to the external resistance, circuit current I is given by

I = {E \over {R + r}} = {{2.2} \over {5 + r}}

ampere, where r is the internal resistance of the cell. Potential difference across the cell = IR

= {{2.2} \over {5 + r}} \times 5 = 1.8

$\therefore$ 5 + r = 11/1.8.

\therefore r = {{11} \over {1.8}} - 5 = {{110 - 90} \over {18}} = {{10} \over 9}\Omega

Q99

Copper and silicon is cooled from 300 K to 60 K , the specific resistance

A decrease in copper but increase in silicon

B increase in copper but decrease in silicon

C increase in both

D decrease in both

Correct Answer

Option A

Solution

For metals specific resistance decrease with decrease in temperature whereas for semiconductors specific resistance increases with decrease in temperature.

Q100

If specific resistance of a potentiometer wire is 10 -7

\Omega

m and current flow through it is 0.1 amp., cross-sectional area of wire is 10

-

6 m 2 then potential gradient will be

A 10

-

2 volt/m

B 10

-

4 volt/m

C 10

-

6 volt/m

D 10

-

8 volt/m.

Correct Answer

Option A

Solution

Potential gradient = Potential fall per unit length. In this case resistance of unit length.

R = {{\rho l} \over A} = {{{{10}^{ - 7}} \times 1} \over {{{10}^{ - 6}}}} = {10^{ - 1}}\Omega

Potential fall across R is V = I.R = 0.1 $\times$ 10 -1 = 0.01volt/m.

= {10^{ - 2}}

volt/m