Current Electricity
528 Questions
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 1st September Evening Shift
Due to cold weather a 1 m water pipe of cross-sectional area 1 cm2 is filled with ice at $-$10$^\circ$C. Resistive heating is used to melt the ice. Current of 0.5A is passed through 4 k$\Omega$ resistance. Assuming that all the heat produced is used for melting, what is the minimum time required? (Given latent heat of fusion for water/ice = 3.33 $\times$ 105 J kg$-$1, specific heat of ice = 2 $\times$ 103 J kg$-$1 and density of ice = 103 kg/m3
A.
0.353 s
B.
35.3 s
C.
3.53 s
D.
70.6 s
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 1st September Evening Shift
Two resistors R1 = (4 $\pm$ 0.8) $\Omega$ and R2 = (4 $\pm$ 0.4) $\Omega$ are connected in parallel. The equivalent resistance of their parallel combination will be :
A.
(4 $\pm$ 0.4) $\Omega$
B.
(2 $\pm$ 0.4) $\Omega$
C.
(2 $\pm$ 0.3) $\Omega$
D.
(4 $\pm$ 0.3) $\Omega$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 31st August Evening Shift
The equivalent resistance of the given circuit between the terminals A and B is :
A.
0$\Omega$
B.
3$\Omega$
C.
${9 \over 2}$$\Omega$
D.
1$\Omega$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 31st August Morning Shift
Consider a galvanometer shunted with 5$\Omega$ resistance and 2% of current passes through it. What is the resistance of the given galvanometer ?
A.
300 $\Omega$
B.
344 $\Omega$
C.
245 $\Omega$
D.
226 $\Omega$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th August Evening Shift
The Colour coding on a carbon resistor is shown in the given figure. The resistance value of the given resistor is :
A.
(5700 $\pm$ 285) $\Omega$
B.
(7500 $\pm$ 750) $\Omega$
C.
(5700 $\pm$ 375) $\Omega$
D.
(7500 $\pm$ 375) $\Omega$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th August Evening Shift
For full scale deflection of total 50 divisions, 50 mV voltage is required in galvanometer. The resistance of galvanometer if its current sensitivity is 2 div/mA will be :
A.
1$\Omega$
B.
5$\Omega$
C.
4$\Omega$
D.
2$\Omega$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th August Morning Shift
Five identical cells each of internal resistance 1$\Omega$ and emf 5V are connected in series and in parallel with an external resistance 'R'. For what value of 'R', current in series and parallel combination will remain the same?
A.
1 $\Omega$
B.
25 $\Omega$
C.
5 $\Omega$
D.
10 $\Omega$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th August Evening Shift
If you are provided a set of resistances 2$\Omega$, 4$\Omega$, 6$\Omega$ and 8$\Omega$. Connect these resistances so as to obtain an equivalent resistance of ${{46} \over 3}$$\Omega$.
A.
4$\Omega$ and 6$\Omega$ are in parallel with 2$\Omega$ and 8$\Omega$ in series
B.
6$\Omega$ and 8$\Omega$ are in parallel with 2$\Omega$ and 4$\Omega$ in series
C.
2$\Omega$ and 6$\Omega$ are in parallel with 4$\Omega$ and 8$\Omega$ in series
D.
2$\Omega$ and 4$\Omega$ are in parallel with 6$\Omega$ and 8$\Omega$ in series
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th August Evening Shift
An electric bulb of 500 watt at 100 volt is used in a circuit having a 200 V supply. Calculate the resistance R to be connected in series with the bulb so that the power delivered by the bulb is 500 W.
A.
20 $\Omega$
B.
30 $\Omega$
C.
5 $\Omega$
D.
10 $\Omega$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th August Morning Shift
In the given figure, the emf of the cell is 2.2 V and if internal resistance is 0.6$\Omega$. Calculate the power dissipated in the whole circuit :
A.
1.32 W
B.
0.65 W
C.
2.2 W
D.
4.4 W
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th August Morning Shift
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)
(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
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th July Evening Shift
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
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th July Morning Shift
In the given figure, a battery of emf E is connected across a conductor PQ of length 'l' and different area of cross-sections having radii r1 and r2 (r2 < r1).
Choose the correct option as one moves from P to Q :
Choose the correct option as one moves from P to Q :
A.
Drift velocity of electron increases.
B.
Electric field decreases.
C.
Electron current decreases.
D.
All of these
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th July Evening Shift
In the given potentiometer circuit arrangement, the balancing length AC is measured to be 250 cm. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes 400 cm. The ratio of the emf of two cells, ${{{\varepsilon _1}} \over {{\varepsilon _2}}}$ is :
A.
${5 \over 3}$
B.
${8 \over 5}$
C.
${4 \over 3}$
D.
${3 \over 2}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th July Evening Shift
The given potentiometer has its wire of resistance 10$\Omega$. When the sliding contact is in the middle of the potentiometer wire, the potential drop across 2$\Omega$ resistor is :
A.
10 V
B.
5 V
C.
${{40} \over 9}$ V
D.
${{40} \over 11}$ V
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th July Morning Shift
In the given figure, there is a circuit of potentiometer of length AB = 10 m. The resistance per unit length is 0.1 $\Omega$ per cm. Across AB, a battery of emf E and internal resistance 'r' is connected. The maximum value of emf measured by this potentiometer is :
A.
5 V
B.
2.25 V
C.
6 V
D.
2.75 V
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 22th July Evening Shift
A Copper (Cu) rod of length 25 cm and cross-sectional area 3 mm2 is joined with a similar Aluminium (Al) rod as shown in figure. Find the resistance of the combination between the ends A and B.
(Take Resistivity of Copper = 1.7 $\times$ 10$-$8 $\Omega$m and Resistivity of Aluminium = 2.6 $\times$ 10$-$8 $\Omega$m)
(Take Resistivity of Copper = 1.7 $\times$ 10$-$8 $\Omega$m and Resistivity of Aluminium = 2.6 $\times$ 10$-$8 $\Omega$m)
A.
0.0858 m$\Omega$
B.
1.420 m$\Omega$
C.
0.858 m$\Omega$
D.
2.170 m$\Omega$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 20th July Morning Shift
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)
(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
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 20th July Morning Shift
The value of current in the 6 $\Omega$ resistance is :
A.
4A
B.
8A
C.
10A
D.
6A
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Morning Shift
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
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Evening Shift
Two cells of emf 2E and E with internal resistance r1 and r2 respectively are connected in series to an external resistor R (see figure). The value of R, at which the potential difference across the terminals of the first cell becomes zero is
A.
r1 $-$ r2
B.
${{{r_1}} \over 2} - {r_2}$
C.
${{{r_1}} \over 2} + {r_2}$
D.
r1 + r2
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Evening Shift
The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of 15$\Omega$ resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10V is maintained across AC.
A.
4.87 $\mu$A
B.
4.87 mA
C.
2.44 mA
D.
2.44 $\mu$A
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Morning Shift
A current of 10A exists in a wire of cross-sectional area of 5 mm2 with a drift velocity of 2 $\times$ 10$-$3 ms$-$1. The number of free electrons in each cubic meter of the wire is ___________.
A.
625 $\times$ 1025
B.
1 $\times$ 1023
C.
2 $\times$ 1025
D.
2 $\times$ 106
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 16th March Evening Shift
A resistor develops 500 J of thermal energy in 20 s when a current of 1.5A is passed through it. If the current is increased from 1.5A to 3A, what will be the energy developed in 20 s.
A.
1000 J
B.
2000 J
C.
1500 J
D.
500 J
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 16th March Morning Shift
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 :
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.
${1 \over 4}{{\rho l} \over {VA}}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Evening Shift
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%
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Morning Shift
Five equal resistances are connected in a network as shown in figure. The net resistance between the points A and B is :
A.
${{3R} \over 2}$
B.
${{R} \over 2}$
C.
2R
D.
R
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Morning Shift
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.
i = $\alpha$0t + $\beta$t2
where $\alpha$0 = 20 A/s and $\beta$ = 8 As$-$2. Find the charge crossed through a section of the wire in 15 s.
A.
2250 C
B.
2100 C
C.
260 C
D.
11250 C
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Morning Shift
A cell E1 of emf 6V and internal resistance 2$\Omega$ is connected with another cell E2 of emf 4V and internal resistance 8$\Omega$ (as shown in the figure). The potential difference across points X and Y is :
A.
10.0 V
B.
2.0 V
C.
5.6 V
D.
3.6 V
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 1st September Evening Shift
A uniform heating wire of resistance 36$\Omega$ is connected across a potential difference of 240 V. The wire is then cut into half and potential difference of 240V is applied across each half separately. The ratio of power dissipation in first case to the total power dissipation in the second case would be 1 : x, where x is ____________
Correct Answer: 4
Explanation:
For Case I,
The potential difference of the uniform wire, V = 240 V
The resistance of the uniform wire, R1 = 36 $\Omega$
The power dissipation in the first case,
${P_1} = {{{V^2}} \over {{R_1}}} = {{{{(240)}^2}} \over {36}}$
For Case II,
The resistance of each half, ${R_2} = {{{R_1}} \over 2} = {{36} \over 2} = 18\Omega $
${P_2} = {{{V^2}} \over {{R_2}}} + {{{V^2}} \over {{R_2}}} = {{{{(240)}^2}} \over {18}} + {{{{(240)}^2}} \over {18}} = {{{{(240)}^2}} \over 9}$
Thus, the ratio of the total power dissipation in the first case to the second case
${{{P_1}} \over {{P_2}}} = {{{{(240)}^2}/36} \over {{{(240)}^2}/9}} \Rightarrow {{{P_1}} \over {{P_2}}} = {1 \over 4}$
Comparing with, ${{{P_1}} \over {{P_2}}} = {1 \over x}$
The value of the x = 4.
The potential difference of the uniform wire, V = 240 V
The resistance of the uniform wire, R1 = 36 $\Omega$
The power dissipation in the first case,
${P_1} = {{{V^2}} \over {{R_1}}} = {{{{(240)}^2}} \over {36}}$
For Case II,
The resistance of each half, ${R_2} = {{{R_1}} \over 2} = {{36} \over 2} = 18\Omega $
${P_2} = {{{V^2}} \over {{R_2}}} + {{{V^2}} \over {{R_2}}} = {{{{(240)}^2}} \over {18}} + {{{{(240)}^2}} \over {18}} = {{{{(240)}^2}} \over 9}$
Thus, the ratio of the total power dissipation in the first case to the second case
${{{P_1}} \over {{P_2}}} = {{{{(240)}^2}/36} \over {{{(240)}^2}/9}} \Rightarrow {{{P_1}} \over {{P_2}}} = {1 \over 4}$
Comparing with, ${{{P_1}} \over {{P_2}}} = {1 \over x}$
The value of the x = 4.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Evening Shift
A resistor dissipates 192 J of energy in 1s when a current of 4A is passed through it. Now, when the current is doubled, the amount of thermal energy dissipated in 5s in _________ J.
Correct Answer: 3840
Explanation:
E = i2Rt
192 = 16 (R) (1)
R = 12$\Omega$
E1 = (8)2 (12) (5)
= 3840 J
192 = 16 (R) (1)
R = 12$\Omega$
E1 = (8)2 (12) (5)
= 3840 J
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Morning Shift
A square shaped wire with resistance of each side 3$\Omega$ is bent to form a complete circle. The resistance between two diametrically opposite points of the circle in unit of $\Omega$ will be ___________.
Correct Answer: 3
Explanation:
Req = 3$\Omega$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Morning Shift
The voltage drop across 15$\Omega$ resistance in the given figure will be ______________ V.
Correct Answer: 6
Explanation:
$\Rightarrow$ effective circuit diagram will be
Point drop across 6$\Omega$ = 1 $\times$ 6 = 6 = VAB
$\Rightarrow$ Hence point drop across 15$\Omega$ = 6 volt = VAB
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Evening Shift
The ratio of the equivalent resistance of the network (shown in figure) between the points a and b when switch is open and switch is closed is x : 8. The value of x is ___________.
Correct Answer: 9
Explanation:
${R_{eq\,open}} = {{3R} \over 2}$
${R_{eq\,closed}} = 2 \times {{R \times 2R} \over {3R}} = {{4R} \over 3}$
${{{R_{eq\,open}}} \over {{R_{eq\,closed}}}} = {{3R} \over 2} \times {3 \over {4R}} = {9 \over 8}$
$\therefore$ $x = 9$
${R_{eq\,closed}} = 2 \times {{R \times 2R} \over {3R}} = {{4R} \over 3}$
${{{R_{eq\,open}}} \over {{R_{eq\,closed}}}} = {{3R} \over 2} \times {3 \over {4R}} = {9 \over 8}$
$\therefore$ $x = 9$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Morning Shift
First, a set of n equal resistors of 10 $\Omega$ each are connected in series to a battery of emf 20V and internal resistance 10$\Omega$. A current I is observed to flow. Then, the n resistors are connected in parallel to the same battery. It is observed that the current is increased 20 times, then the value of n is ............... .
Correct Answer: 20
Explanation:
In series
${R_{eq}} = nR = 10n$
${i_s} = {{20} \over {10 + 10n}} = {2 \over {1 + n}}$
In parallel
${R_{eq}} = {{10} \over n}$
${i_p} = {{20} \over {{{10} \over n} + 10}} = {{2n} \over {1 + n}}$
${{{i_p}} \over {{i_s}}} = 20$
${{\left( {{{2n} \over {1 + n}}} \right)} \over {\left( {{2 \over {1 + n}}} \right)}} = 20$
$n = 20$
${R_{eq}} = nR = 10n$
${i_s} = {{20} \over {10 + 10n}} = {2 \over {1 + n}}$
In parallel
${R_{eq}} = {{10} \over n}$
${i_p} = {{20} \over {{{10} \over n} + 10}} = {{2n} \over {1 + n}}$
${{{i_p}} \over {{i_s}}} = 20$
${{\left( {{{2n} \over {1 + n}}} \right)} \over {\left( {{2 \over {1 + n}}} \right)}} = 20$
$n = 20$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Evening Shift
For the circuit shown, the value of current at time t = 3.2 s will be _________ A.
[Voltage distribution V(t) is shown by Fig. (1) and the circuit is shown in Fig. (2)]
[Voltage distribution V(t) is shown by Fig. (1) and the circuit is shown in Fig. (2)]
Correct Answer: 1
Explanation:
From graph voltage at t = 3.2 sec is 6 volt.
i = ${{6 - 5} \over 1}$
$ \Rightarrow $ i = 1 A
i = ${{6 - 5} \over 1}$
$ \Rightarrow $ i = 1 A
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Evening Shift
A 16 $\Omega$ wire is bend to form a square loop. A 9V supply having internal resistance of 1$\Omega$ is connected across one of its sides. The potential drop across the diagonals of the square loop is _______________ $\times$ 10$-$1 V
Correct Answer: 45
Explanation:
Here assume current as
By KVL in outer loop
9 $-$ 12i $-$ 4i = 0
16i = 9
8i = ${9 \over 2}$ = 4.5
= 45 $\times$ 10-1
By KVL in outer loop
9 $-$ 12i $-$ 4i = 0
16i = 9
8i = ${9 \over 2}$ = 4.5
= 45 $\times$ 10-1
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Morning Shift
An electric bulb rated as 200 W at 100 V is used in a circuit having 200 V supply. The resistance 'R' that must be put in series with the bulb so that the bulb delivers the same power is _____________ $\Omega$.
Correct Answer: 50
Explanation:
Power, $P = {{{V^2}} \over {{R_B}}}$
${R_B} = {{{V^2}} \over P} = {{100 \times 100} \over {200}}$
${R_B} = 50\Omega $
To produce same power, same voltage (i.e. 100 V) should be across the bulb.
Hence, R = RB
R = 50 $\Omega$
${R_B} = {{{V^2}} \over P} = {{100 \times 100} \over {200}}$
${R_B} = 50\Omega $
To produce same power, same voltage (i.e. 100 V) should be across the bulb.
Hence, R = RB
R = 50 $\Omega$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 22th July Evening Shift
In an electric circuit, a cell of certain emf provides a potential difference of 1.25 V across a load resistance of 5$\Omega$. However, it provides a potential difference of 1 V across a load resistance of 2$\Omega$. The emf of the cell is given by ${x \over {10}}V$. Then the value of x is ______________.
Correct Answer: 15
Explanation:
In case (a) $\varepsilon = {{1.25} \over 5}(5 + r)$
$ \Rightarrow 4\varepsilon = 5 + r$ ..... (1)
In case (b), $\varepsilon = {1 \over 2}(2 + r)$
$ \Rightarrow 2\varepsilon = 2 + r$ ..... (2)
From equation (1) & (2)
$2\varepsilon = 3 \Rightarrow \varepsilon = 1.5$
or x = 15
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Evening Shift
In the given figure switches S1 and S2 are in open condition. The resistance across ab when the switches S1 and S2 are closed is _____________ $\Omega$.
Correct Answer: 10
Explanation:
When switch S1 and S2 are closed
Rab = ${{12 \times 6} \over {12 + 6}} + 2 + {{6 \times 12} \over {6 + 12}}$
= ${{72} \over {18}} + 2 + {{72} \over {18}} = 4 + 2 + 4 = 10\Omega $
Rab = ${{12 \times 6} \over {12 + 6}} + 2 + {{6 \times 12} \over {6 + 12}}$
= ${{72} \over {18}} + 2 + {{72} \over {18}} = 4 + 2 + 4 = 10\Omega $
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Evening Shift
Two wires of same length and thickness having specific resistances 6$\Omega$ cm and 3$\Omega$ cm respectively are connected in parallel. The effective resistivity is $\rho$$\Omega$ cm. The value of $\rho$, to the nearest integer, is ____________.
Correct Answer: 4
Explanation:
Let length of each wire is l and area A. When they are connected in parallel then their effective area 2A.
From formula we know,
${R_{eq}} = {{{R_1}{R_2}} \over {{R_1} + {R_2}}}$
$ \Rightarrow {{\rho l} \over {2A}} = {{{\rho _1}{l \over A} \times {\rho _2}{l \over A}} \over {{\rho _1}{l \over A} + {\rho _2}{l \over A}}}$
$ \Rightarrow {\rho \over 2} = {{{\rho _1} \times {\rho _2}} \over {{\rho _1} + {\rho _2}}}$
$ \Rightarrow {\rho \over 2} = {{6 \times 3} \over {6 + 3}} = 2$
$ \Rightarrow \rho = 4$
From formula we know,
${R_{eq}} = {{{R_1}{R_2}} \over {{R_1} + {R_2}}}$
$ \Rightarrow {{\rho l} \over {2A}} = {{{\rho _1}{l \over A} \times {\rho _2}{l \over A}} \over {{\rho _1}{l \over A} + {\rho _2}{l \over A}}}$
$ \Rightarrow {\rho \over 2} = {{{\rho _1} \times {\rho _2}} \over {{\rho _1} + {\rho _2}}}$
$ \Rightarrow {\rho \over 2} = {{6 \times 3} \over {6 + 3}} = 2$
$ \Rightarrow \rho = 4$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Evening Shift
Consider a 72 cm long wire AB as shown in the figure. The galvanometer jockey is placed at P on AB at a distance x cm from A. The galvanometer shows zero deflection.
The value of x, to the nearest integer, is ___________.
The value of x, to the nearest integer, is ___________.
Correct Answer: 48
Explanation:
As galvanometer shows zero deflection so it act's as balanced wheatstone bridge.
$ \therefore $ ${{12} \over 6} = {x \over {72 - x}}$
$ \Rightarrow $ 14 $-$ 2x = x
$ \Rightarrow $ 3x = 144
$ \Rightarrow $ x = 48 cm
$ \therefore $ ${{12} \over 6} = {x \over {72 - x}}$
$ \Rightarrow $ 14 $-$ 2x = x
$ \Rightarrow $ 3x = 144
$ \Rightarrow $ x = 48 cm
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Morning Shift
The voltage across the 10$\Omega$ resistor in the given circuit is x volt.
The value of 'x' to the nearest integer is _________.
The value of 'x' to the nearest integer is _________.
Correct Answer: 70
Explanation:
${R_{eq}} = {{50 \times 20} \over {50 + 20}}$
${R_{eq}} = {{1000} \over {70}} = {{100} \over 7}\Omega $
${R_{ef}} = 10 + {{100} \over 7} = {{170} \over 7}\Omega $
$I = {V \over R} = {{170/7} \over {170}} = {{170} \over {170/7}}$
I = 7A
Potential Across 10$\Omega$ resister
V10 = IR = 7 $\times$ 10 = 70V
$ \Rightarrow $ V10 = 70V
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Morning Shift
The equivalent resistance of series combination of two resistors is 's'. When they are connected in parallel, the equivalent resistance is 'p'. If s = np, then the minimum value for n is ____________. (Round off to the Nearest Integer)
Correct Answer: 4
Explanation:
$s = np$
${R_1} + {R_2} = n\left[ {{{{R_1}{R_2}} \over {{R_1} + {R_2}}}} \right]$
$ \Rightarrow $ $R_1^2 + R_2^2 + 2{R_1}{R_2} = n{R_1}{R_2}$
$ \Rightarrow $ $R_1^2 + (2 - n){R_1}{R_2} + R_2^2 = 0$
For real roots, b2 - 4ac $ \ge $ 0
${[(2 - n){R_2}]^2} - 4 \times 1 \times R_2^2$ $ \ge $ 0
$ \Rightarrow $ ${(2 - 4)^2}R_2^2 \ge 4R_2^2$
$ \Rightarrow $ 2 $-$ n $\ge $$\pm$2
$ \Rightarrow $ 2 $-$ n $\ge$ $-$2
$ \Rightarrow $ n $\ge$ 4
So, minimum value for n = 4
${R_1} + {R_2} = n\left[ {{{{R_1}{R_2}} \over {{R_1} + {R_2}}}} \right]$
$ \Rightarrow $ $R_1^2 + R_2^2 + 2{R_1}{R_2} = n{R_1}{R_2}$
$ \Rightarrow $ $R_1^2 + (2 - n){R_1}{R_2} + R_2^2 = 0$
For real roots, b2 - 4ac $ \ge $ 0
${[(2 - n){R_2}]^2} - 4 \times 1 \times R_2^2$ $ \ge $ 0
$ \Rightarrow $ ${(2 - 4)^2}R_2^2 \ge 4R_2^2$
$ \Rightarrow $ 2 $-$ n $\ge $$\pm$2
$ \Rightarrow $ 2 $-$ n $\ge$ $-$2
$ \Rightarrow $ n $\ge$ 4
So, minimum value for n = 4
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Evening Shift
The energy dissipated by a resistor is 10 mJ in 1 s when an electric current of 2 mA flows through it. The resistance is ___________$\Omega$. (Round off to the Nearest Integer)
Correct Answer: 2500
Explanation:
Given, energy dissipated by a resistor, H = 10 mJ = 10 $\times$ 10$-$3 J
Time, t = 1 s
Electric current, I = 2 mA = 2 $\times$ 10$-$3 A
Resistance, R = ?
According to Joule's law of heating,
H = I2Rt
$ \Rightarrow R = {H \over {{I^2}T}}$ ....... (i)
Substituting the given values in Eq. (i), we get
$R = {{10 \times {{10}^{ - 3}}} \over {{{(2 \times {{10}^{ - 3}})}^2} \times 1}}$
$ \Rightarrow R = {{{{10}^{ - 2}}} \over {4 \times {{10}^{ - 6}}}} \Rightarrow R = 0.25 \times {10^4}$
$ \Rightarrow R = 2500\,\Omega $
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Morning Shift
In the figure given, the electric current flowing through the 5 k$\Omega$ resistor is 'x' mA.
The value of x to the nearest integer is ____________.
The value of x to the nearest integer is ____________.
Correct Answer: 3
Explanation:
$I = {{21} \over {5 + 1 + 1}}$
$ \therefore $ I = 3 mA
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Morning Shift
In an electrical circuit, a battery is connected to pass 20C of charge through it in a certain given time. The potential difference between two plates of the battery is maintained at 15V. The workdone by the battery is __________J.
Correct Answer: 300
Explanation:
Given, charge passing through circuit, q = 20 C
Potential difference between two plates,
V = 15 V
Let W be the amount of work done by battery.
$\therefore$ W = qV = 20 $\times$ 15 = 300 J
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Evening Shift
A current of 6A enters one corner P of an equilateral triangle PQR having 3 wires of resistance 2$\Omega$ each and leaves by the corner R. The currents i1 in ampere is _________.
Correct Answer: 2
Explanation:
The current ${i_1} = \left( {{{{R_2}} \over {{R_1} + {R_2}}}} \right)i$
$ = \left( {{2 \over {4 + 2}}} \right) \times 6$
${i_1} = 2A$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Morning Shift
In the given circuit of potentiometer, the potential difference E across AB (10 m length) is larger than E1 and E2 as well. For key K1 (closed), the jockey is adjusted to touch the wire at point J1 so that there is no deflection in the galvanometer. Now the first battery (E1) is replaced by second battery (E2) for working by making K1 open and K2 closed. The galvanometer gives then null deflection at J2. The value of ${{{E_1}} \over {{E_2}}}$ is ${a \over b}$, where a = _________.
Correct Answer: 1
Explanation:
Length of AB = 10 m
For battery E1, balancing length is l1
l1 = 380 cm [from end A]
For battery E2, balancing length is l2
l2 = 760 cm [from end A]
Now, we know that ${{{E_1}} \over {{E_2}}} = {{{l_1}} \over {{l_2}}}$
$ \Rightarrow {{{E_1}} \over {{E_2}}} = {{380} \over {760}} = {1 \over 2} = {a \over b}$
$ \therefore $ a = 1 & b = 2
$ \therefore $ a = 1
For battery E1, balancing length is l1
l1 = 380 cm [from end A]
For battery E2, balancing length is l2
l2 = 760 cm [from end A]
Now, we know that ${{{E_1}} \over {{E_2}}} = {{{l_1}} \over {{l_2}}}$
$ \Rightarrow {{{E_1}} \over {{E_2}}} = {{380} \over {760}} = {1 \over 2} = {a \over b}$
$ \therefore $ a = 1 & b = 2
$ \therefore $ a = 1
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Evening Shift
A cylindrical wire of radius 0.5 mm and conductivity 5 $\times$ 107 S/m is subjected to an electric field of 10 mV/m. The expected value of current in the wire will be x3$\pi$ mA. The value of x is _________.
Correct Answer: 5
Explanation:
We know that current density,
J = $\sigma$E
$ \Rightarrow $ J = 5 $\times$ 107 $\times$ 10 $\times$ 10$-$3
$ \Rightarrow $ J = 50 $\times$ 104 A/m2
Current flowing;
I = J $\times$ $\pi$R2
I = 50 $\times$ 104 $\times$ $\pi$(0.5 $\times$ 10$-$3)2
I = 5 $\times$ 104 $\times$ $\pi$ $\times$ 0.25 $\times$ 10$-$6
I = 125 $\times$ 10$-$3$\pi$
$ \therefore $ x = 5
J = $\sigma$E
$ \Rightarrow $ J = 5 $\times$ 107 $\times$ 10 $\times$ 10$-$3
$ \Rightarrow $ J = 50 $\times$ 104 A/m2
Current flowing;
I = J $\times$ $\pi$R2
I = 50 $\times$ 104 $\times$ $\pi$(0.5 $\times$ 10$-$3)2
I = 5 $\times$ 104 $\times$ $\pi$ $\times$ 0.25 $\times$ 10$-$6
I = 125 $\times$ 10$-$3$\pi$
$ \therefore $ x = 5