Current Electricity — Page 7 | JEE Physics

Q61

In a building there are 15 bulbs of 45 W, 15 bulbs of 100 W, 15 small fans of 10 W and 2 heaters of 1 kW. The voltage of electric main is 220 V. The minimum fuse capacity (rated value) of the building will be :

A 15 A

B 20 A

C 25 A

D 10 A

Correct Answer

Option B

Solution

Total power is = (15 × 45) + (15 × 100) + (15 × 10) + (2 × 1000) = 4325 W $\therefore$ Current =

{{4325} \over {220}}

= 19.66 A

\simeq

20 A

Q62

A galvanometer having a coil resistance 100

\Omega

gives a full scale deflection when a current of 1 mA is passed through it. What is the value of the resistance which can convert this galvanometer into a voltmeter giving full scale deflection for a potential difference of 10 V?

A 8.9 k

\Omega

B 10 k

\Omega

C 9.9 k

\Omega

D 7.9 k

\Omega

Correct Answer

Option C

Solution

ig = 1 mA , Rg = 100

\Omega

V = ig(R + Rg) $\Rightarrow$ 10 = 1 × 10–3 (R + 100) $\Rightarrow$ R = 9.9 k

\Omega

Q63

A current through a wire depends on time as i =

\alpha

0t +

\beta

t2 where

\alpha

0 = 20 A/s and

\beta

= 8 As

-

2. Find the charge crossed through a section of the wire in 15 s.

A 2250 C

B 2100 C

C 260 C

D 11250 C

Correct Answer

Option D

Solution

Given,

i = {\alpha _0}t + \beta {t^2}

where, $\alpha$ 0 = 20 A/s, $\beta$ = 8 A/s2 We know that,

i = {{dq} \over {dt}}

\Rightarrow {{dq} \over {dt}} = i = {\alpha _0}t + \beta {t^2} = 20t + 8{t^2}

\Rightarrow dq = (20t + 8{t^2})dt

On integrating both sides, we get

\int\limits_0^q {dq = \int\limits_0^{15} {(2t + 8{t^2})dt} }

q = \left[ {{{20{t^2}} \over 2} + {{8{t^3}} \over 3}} \right]_0^{15} = 10 \times {(15)^2} + {8 \over 3} \times {(15)^3}

$\therefore$ q = 11250 C

Q64

A wire of 1

\Omega

has a length of 1 m. It is stretched till its length increases by 25%. The percentage change in resistance to the nearest integer is :

A 76%

B 12.5%

C 25%

D 56%

Correct Answer

Option D

Solution

R0 = 1

\Omega

R1 = ? l0 = 1m l1 = 1.25 m A0 = A As volume of wire remains constant so A0l0 = A1l1 $\Rightarrow$ A1 =

{{{l_0}{A_0}} \over {{l_1}}}

Now Resistance (R) =

{{pl} \over A}

{{{R_0}} \over {{R_1}}} = {{{l_0}} \over {{A_0}}}\left( {{{{l_0}{A_0}} \over {{l_1} \times {l_1}}}} \right)

$\Rightarrow$

{R_1} = {{l_1^2} \over {l_0^2}} = 1.5625 \,\Omega

So % change in resistance

= {{{R_1} - {R_0}} \over {{R_0}}} \times 100\%

= {{1.5625 - 1} \over 1} \times 100\%

= 56.25\%

Q65

A conducting wire of length 'l', area of cross-section A and electric resistivity

\rho

is connected between the terminals of a battery. A potential difference V is developed between its ends, causing an electric current. If the length of the wire of the same material is doubled and the area of cross-section is halved, the resultant current would be :

A

4{{VA} \over {\rho l}}

B

{3 \over 4}{{VA} \over {\rho l}}

C

{1 \over 4}{{VA} \over {\rho l}}

D

{1 \over 4}{{\rho l} \over {VA}}

Correct Answer

Option C

Solution

We know that

R = \rho {l \over A}

Now, new length :

l' = 2l

new area of cross section :

A' = A/2

$\therefore$ New resistance :

R' = \rho .{{2l} \over {A/2}}

\Rightarrow R' = 4{{\rho l} \over A}

\Rightarrow R' = 4R

$\therefore$ Resultant current :

I = {V \over {4R}}

$\Rightarrow$

I = {1 \over 4}{{VA} \over {\rho l}}

Q66

In the experiment of Ohm's law, a potential difference of 5.0 V is applied across the end of a conductor of length 10.0 cm and diameter of 5.00 mm. The measured current in the conductor is 2.00 A. The maximum permissible percentage error in the resistivity of the conductor is :

A 3.9

B 8.4

C 7.5

D 3.0

Correct Answer

Option A

Solution

V = I \times \rho {l \over A}

\Rightarrow \rho = {{VA} \over {Il}} = {\pi \over 4}{{V{d^2}} \over {Il}}

{{\Delta \rho } \over \rho } = {{2\Delta d} \over d} + {{\Delta V} \over V} + {{\Delta I} \over I} + {{\Delta l} \over l}

= 2\left( {{{0.01} \over 5}} \right) + {{0.1} \over 5} + {{0.01} \over 2} + {{0.1} \over {10}}

$\Rightarrow$

{{\Delta \rho } \over \rho } = 0.039 = 3.9\%

Q67

The combination of two identical cells, whether connected in series or parallel combination provides the same current through an external resistance of 2

\Omega

. The value of internal resistance of each cell is

A 2

\Omega

B 4

\Omega

C 6

\Omega

D 8

\Omega

Correct Answer

Option A

Solution

From diagram

{i_p} = {E \over {2 + {r \over 2}}}

and

{i_s} = {{2E} \over {2 + 2r}}

given

{i_p} = {i_s}

{1 \over {2 + {r \over 2}}} = {1 \over {1 + r}}

1 + r = 2 + {r \over 2}

r = 2\,\Omega

Q68

A current of 5 A is passing through a non-linear magnesium wire of cross-section 0.04 m2. At every point the direction of current density is at an angle of 60

^\circ

with the unit vector of area of cross-section. The magnitude of electric field at every point of the conductor is : (Resistivity of magnesium

\rho

= 44

\times

10

-

8

\Omega

m)

A 11

\times

10

-

5 V/m

B 11

\times

10

-

3 V/m

C 11

\times

10

-

7 V/m

D 11

\times

10

-

2 V/m

Correct Answer

Option A

Solution

Given, current, I = 5A Area of cross-section of wire, A = 0.04 m2 We know that,

J = {I \over A}

\Rightarrow I = JA

or

I = J\,.\,A

or

I = JA\cos \theta

where, J = current density.

\Rightarrow 5 = J\left( {{4 \over {100}}} \right) \times \cos (60^\circ )

[ $\because$ Given, $\theta$ = 60

^\circ

]

J = 500 \times {1 \over 2}

[ $\because$ cos60

^\circ

=

{1 \over 2}

] $\Rightarrow$ J = 250 Am $-$ 2 The relation between electric field, current density and resistivity can be given as, E = $\rho$ .

J = 44 $\times$ 10 $-$ 8 $\times$ 250 [ $\because$ Resistivity, $\rho$ = 44 $\times$ 10 $-$ 8

\Omega

-m] = 11 $\times$ 10 $-$ 5 V/m

Q69

The resistance of a conductor at 15

^\circ

C is 16

\Omega

and at 100

^\circ

C is 20

\Omega

. What will be the temperature coefficient of resistance of the conductor?

A 0.010

^\circ

C

-

1

B 0.033

^\circ

C

-

1

C 0.003

^\circ

C

-

1

D 0.042

^\circ

C

-

1

Correct Answer

Option C

Solution

16 = R0 [1 + $\alpha$ (15 $-$ T0)] 20 = R0 [1 + $\alpha$ (100 $-$ T0)] Assuming T0 = 0

^\circ

C, as a general convention. $\Rightarrow$

{{16} \over {20}} = {{1 + \alpha \times 15} \over {1 + \alpha \times 100}}

$\Rightarrow$ $\alpha$ = 0.003

^\circ

C $-$ 1

Q70

What equal length of an iron wire and a copper-nickel alloy wire, each of 2 mm diameter connected parallel to give an equivalent resistance of 3

\Omega

? (Given resistivities of iron and copper-nickel alloy wire are 12

\mu

\Omega

and 51

\mu

\Omega

cm respectively)

A 82 m

B 97 m

C 110 m

D 90 m

Correct Answer

Option B

Solution

{{{R_1}{R_2}} \over {{R_1} + {R_2}}} = 3

{{{{(12 \times {{10}^{ - 6}} \times {{10}^{ - 2}})} \over {\pi {{(2)}^2} \times {{10}^{ - 6}}}} \times {{(51 \times {{10}^{ - 6}} \times {{10}^{ - 2}})l \times 4} \over {\pi {{(2)}^2} \times {{10}^{ - 6}}}}} \over {{{63 \times {{10}^{ - 6}} \times {{10}^{ - 2}} \times l \times 4} \over {\pi {{(2)}^2} \times {{10}^{ - 6}}}}}}

= 3 $\Rightarrow$ l = 97 m