JEE Exam > JEE Notes > DC Pandey Solutions for JEE Physics > DC Pandey Solutions (JEE Advance): Current Electricity- 1
DC Pandey Solutions (JEE Advance): Current Electricity- 1 | DC Pandey Solutions for JEE Physics PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
Current Electricity
For JEE Advanced
Objective Questions
Single Correct Option
Q 1. Two cells A and B of emf 1.3 V and 1.5 V respectively are arranged as shown in figure. The
voltmeter reads 1.45 V. Which cell has the higher internal resistance and how many times? The
voltmeter is assumed to be ideal
(a) r
1
= 2r
2
(b) r
1
= 3r
2
(c) r
2
= 2r
1
(d) r
2
= 3r
1
Q 2. A voltmeter connected in series with a resistance R
1
to a circuit indicates a voltage V
1
= 198 V
When a series resistor R
2
= 2R
1
is used, the voltmeter indicates a voltage V
2
= 180 V If the
resistance of the voltmeter is R
V
= 900 ? then, the applied voltage across A and 5 is
(a) 210 V (b) 200 V (c) 220 V (d) 240 V
Q 3. All bulbs in the circuit shown in figure are identical. Which bulb glows most brightly?
(a) B (b) A (c) D (d) C
Q 4. A student connects an ammeter A and a voltmeter V to measure a resistance R as shown in figure.
If the voltmeter reads 20 V and the ammeter reads 4 A, then R is
(a) equal to 5 ? (b) greater than 5 ? (c) less than 5 ?
(d) greater or less than 5 ? depending upon the direction of current
Q 5. The given figure represents an arrangement of potentiometer for the calculation of internal
resistance (r) of the unknown battery (E). The balance length is 70.0 cm with the key opened and
60.0 cm with the key closed. R is 132.40 ?. The internal resistance (r) of the unknown cell will be
Page 2
Current Electricity
For JEE Advanced
Objective Questions
Single Correct Option
Q 1. Two cells A and B of emf 1.3 V and 1.5 V respectively are arranged as shown in figure. The
voltmeter reads 1.45 V. Which cell has the higher internal resistance and how many times? The
voltmeter is assumed to be ideal
(a) r
1
= 2r
2
(b) r
1
= 3r
2
(c) r
2
= 2r
1
(d) r
2
= 3r
1
Q 2. A voltmeter connected in series with a resistance R
1
to a circuit indicates a voltage V
1
= 198 V
When a series resistor R
2
= 2R
1
is used, the voltmeter indicates a voltage V
2
= 180 V If the
resistance of the voltmeter is R
V
= 900 ? then, the applied voltage across A and 5 is
(a) 210 V (b) 200 V (c) 220 V (d) 240 V
Q 3. All bulbs in the circuit shown in figure are identical. Which bulb glows most brightly?
(a) B (b) A (c) D (d) C
Q 4. A student connects an ammeter A and a voltmeter V to measure a resistance R as shown in figure.
If the voltmeter reads 20 V and the ammeter reads 4 A, then R is
(a) equal to 5 ? (b) greater than 5 ? (c) less than 5 ?
(d) greater or less than 5 ? depending upon the direction of current
Q 5. The given figure represents an arrangement of potentiometer for the calculation of internal
resistance (r) of the unknown battery (E). The balance length is 70.0 cm with the key opened and
60.0 cm with the key closed. R is 132.40 ?. The internal resistance (r) of the unknown cell will be
(a) 22.1 ? (b) 113.5 ? (c) 154.5 ? (d) 10 ?
Q 6. Switch S is closed at time t = 0. Which one of the following statements is correct?
(a) Current in the resistance R increases if E
1
r
2
< E
2
(R + r
1
)
(b) Current in the resistance R increases if E
1
r
2
> E
2
(R + r
1
)
(c) Current in the resistance R decreases if E
1
r
2
> E
2
(R + r
1
)
(d) Current in the resistance R decreases if E
1
r
2
= E
2
(R + r
1
)
Q 7. A, B and C are voltmeters of resistances R,1.5R and 3R respectively. When some potential
difference is applied between x and y, the voltmeter readings are V
A
, V
B
and V
C
, then
(a)V
A
= V
B
= V
C
(b)V
A
? V
B
= V
C
(c) V
A
= V
B
? V
C
(d) V
A
+ V
B
= V
C
Q 8. In the circuit shown, the voltage drop across the 15 ? resistor is 30 V having the polarity as
indicated. The ratio of potential difference across 5 ? resistor and resistance R is
(a) 2/7 (b) 0.4 (c) 5/7 (d) 1
Q 9. In an experiment on the measurement of internal resistance of a cell by using a potentiometer,
when the key K is kept open then balancing length is obtained at y metre. When the K is closed
and some resistance R is inserted in the resistance box, then the balancing length is found to be x
metre. Then the
internal resistance is
Page 3
Current Electricity
For JEE Advanced
Objective Questions
Single Correct Option
Q 1. Two cells A and B of emf 1.3 V and 1.5 V respectively are arranged as shown in figure. The
voltmeter reads 1.45 V. Which cell has the higher internal resistance and how many times? The
voltmeter is assumed to be ideal
(a) r
1
= 2r
2
(b) r
1
= 3r
2
(c) r
2
= 2r
1
(d) r
2
= 3r
1
Q 2. A voltmeter connected in series with a resistance R
1
to a circuit indicates a voltage V
1
= 198 V
When a series resistor R
2
= 2R
1
is used, the voltmeter indicates a voltage V
2
= 180 V If the
resistance of the voltmeter is R
V
= 900 ? then, the applied voltage across A and 5 is
(a) 210 V (b) 200 V (c) 220 V (d) 240 V
Q 3. All bulbs in the circuit shown in figure are identical. Which bulb glows most brightly?
(a) B (b) A (c) D (d) C
Q 4. A student connects an ammeter A and a voltmeter V to measure a resistance R as shown in figure.
If the voltmeter reads 20 V and the ammeter reads 4 A, then R is
(a) equal to 5 ? (b) greater than 5 ? (c) less than 5 ?
(d) greater or less than 5 ? depending upon the direction of current
Q 5. The given figure represents an arrangement of potentiometer for the calculation of internal
resistance (r) of the unknown battery (E). The balance length is 70.0 cm with the key opened and
60.0 cm with the key closed. R is 132.40 ?. The internal resistance (r) of the unknown cell will be
(a) 22.1 ? (b) 113.5 ? (c) 154.5 ? (d) 10 ?
Q 6. Switch S is closed at time t = 0. Which one of the following statements is correct?
(a) Current in the resistance R increases if E
1
r
2
< E
2
(R + r
1
)
(b) Current in the resistance R increases if E
1
r
2
> E
2
(R + r
1
)
(c) Current in the resistance R decreases if E
1
r
2
> E
2
(R + r
1
)
(d) Current in the resistance R decreases if E
1
r
2
= E
2
(R + r
1
)
Q 7. A, B and C are voltmeters of resistances R,1.5R and 3R respectively. When some potential
difference is applied between x and y, the voltmeter readings are V
A
, V
B
and V
C
, then
(a)V
A
= V
B
= V
C
(b)V
A
? V
B
= V
C
(c) V
A
= V
B
? V
C
(d) V
A
+ V
B
= V
C
Q 8. In the circuit shown, the voltage drop across the 15 ? resistor is 30 V having the polarity as
indicated. The ratio of potential difference across 5 ? resistor and resistance R is
(a) 2/7 (b) 0.4 (c) 5/7 (d) 1
Q 9. In an experiment on the measurement of internal resistance of a cell by using a potentiometer,
when the key K is kept open then balancing length is obtained at y metre. When the K is closed
and some resistance R is inserted in the resistance box, then the balancing length is found to be x
metre. Then the
internal resistance is
(a)
xy
R
y
?
(b)
yx
R
x
?
(c)
yx
R
y
?
(d)
xy
R
x
?
Q 10. A source of emf E= 10 V and having negligible internal resistance is connected to a variable
resistance. The resistance varies as shown in figure. The total charge that has passed through the
resistor R during the time interval from t
1
to t
2
is
(a) 401og
e
4 (b) 30log, 3 (c) 201og
e
2 (d) 101og
e
2
Q 11. In order to increase the resistance of a given wire of uniform cross section to four times its value, a
fraction of its length is stretched uniformly till the full length of the wire becomes
3
2
times the
original length. What is the value of this fraction ?
(a)
1
4
(b)
1
8
(c)
1
16
(d)
1
6
Q 12. The figure, shows a meter bridge circuit, with AB = 100 cm, X = 12 ? and R = 18 ? and the
jockey J in the position of balance. If R is now made 8 ?, through what distance will J have to be
moved to obtain balance?
(a) 10 cm (b) 20 cm (c) 30 cm (d) 40 cm
Q 13. A milliammeter of range 10 mA and resistance 9 ? is joined in a circuit as shown. The meter
gives full-scale deflection for current I when A and B are used as its terminals, i.e., current enters
at A and leaves at B(C is left isolated). The value of I is
Page 4
Current Electricity
For JEE Advanced
Objective Questions
Single Correct Option
Q 1. Two cells A and B of emf 1.3 V and 1.5 V respectively are arranged as shown in figure. The
voltmeter reads 1.45 V. Which cell has the higher internal resistance and how many times? The
voltmeter is assumed to be ideal
(a) r
1
= 2r
2
(b) r
1
= 3r
2
(c) r
2
= 2r
1
(d) r
2
= 3r
1
Q 2. A voltmeter connected in series with a resistance R
1
to a circuit indicates a voltage V
1
= 198 V
When a series resistor R
2
= 2R
1
is used, the voltmeter indicates a voltage V
2
= 180 V If the
resistance of the voltmeter is R
V
= 900 ? then, the applied voltage across A and 5 is
(a) 210 V (b) 200 V (c) 220 V (d) 240 V
Q 3. All bulbs in the circuit shown in figure are identical. Which bulb glows most brightly?
(a) B (b) A (c) D (d) C
Q 4. A student connects an ammeter A and a voltmeter V to measure a resistance R as shown in figure.
If the voltmeter reads 20 V and the ammeter reads 4 A, then R is
(a) equal to 5 ? (b) greater than 5 ? (c) less than 5 ?
(d) greater or less than 5 ? depending upon the direction of current
Q 5. The given figure represents an arrangement of potentiometer for the calculation of internal
resistance (r) of the unknown battery (E). The balance length is 70.0 cm with the key opened and
60.0 cm with the key closed. R is 132.40 ?. The internal resistance (r) of the unknown cell will be
(a) 22.1 ? (b) 113.5 ? (c) 154.5 ? (d) 10 ?
Q 6. Switch S is closed at time t = 0. Which one of the following statements is correct?
(a) Current in the resistance R increases if E
1
r
2
< E
2
(R + r
1
)
(b) Current in the resistance R increases if E
1
r
2
> E
2
(R + r
1
)
(c) Current in the resistance R decreases if E
1
r
2
> E
2
(R + r
1
)
(d) Current in the resistance R decreases if E
1
r
2
= E
2
(R + r
1
)
Q 7. A, B and C are voltmeters of resistances R,1.5R and 3R respectively. When some potential
difference is applied between x and y, the voltmeter readings are V
A
, V
B
and V
C
, then
(a)V
A
= V
B
= V
C
(b)V
A
? V
B
= V
C
(c) V
A
= V
B
? V
C
(d) V
A
+ V
B
= V
C
Q 8. In the circuit shown, the voltage drop across the 15 ? resistor is 30 V having the polarity as
indicated. The ratio of potential difference across 5 ? resistor and resistance R is
(a) 2/7 (b) 0.4 (c) 5/7 (d) 1
Q 9. In an experiment on the measurement of internal resistance of a cell by using a potentiometer,
when the key K is kept open then balancing length is obtained at y metre. When the K is closed
and some resistance R is inserted in the resistance box, then the balancing length is found to be x
metre. Then the
internal resistance is
(a)
xy
R
y
?
(b)
yx
R
x
?
(c)
yx
R
y
?
(d)
xy
R
x
?
Q 10. A source of emf E= 10 V and having negligible internal resistance is connected to a variable
resistance. The resistance varies as shown in figure. The total charge that has passed through the
resistor R during the time interval from t
1
to t
2
is
(a) 401og
e
4 (b) 30log, 3 (c) 201og
e
2 (d) 101og
e
2
Q 11. In order to increase the resistance of a given wire of uniform cross section to four times its value, a
fraction of its length is stretched uniformly till the full length of the wire becomes
3
2
times the
original length. What is the value of this fraction ?
(a)
1
4
(b)
1
8
(c)
1
16
(d)
1
6
Q 12. The figure, shows a meter bridge circuit, with AB = 100 cm, X = 12 ? and R = 18 ? and the
jockey J in the position of balance. If R is now made 8 ?, through what distance will J have to be
moved to obtain balance?
(a) 10 cm (b) 20 cm (c) 30 cm (d) 40 cm
Q 13. A milliammeter of range 10 mA and resistance 9 ? is joined in a circuit as shown. The meter
gives full-scale deflection for current I when A and B are used as its terminals, i.e., current enters
at A and leaves at B(C is left isolated). The value of I is
(a) 100 mA (b) 900 mA (c) 1A (d) 1.1 A
Q 14. A battery of emf E
0
= 12 Vis connected across a 4 m long uniform wire having resistance 4 ?/m
The cell of small emfs ?
1
= 2 V and ?
2
= 4 V having internal resistance 2 ? and 6 ? respectively
are connected as shown in the figure. If galvanometer shows no deflection at the point N, the
distance of point N from the point A is equal to
(a)
5
m
3
(b)
4
m
3
(c)
3
m
2
(d) None of these
Q 15. In the circuit shown, when keys K
1
and K
2
both are closed, the ammeter reads I
o
. But when K
1
is
open and K
2
is closed the ammeter reads I
o
/2 Assuming that ammeter resistance is much less than
R
2
, the values of rand R
1
in ? are
(a) 25, 50 (b) 25, 100 (c) 0, 100 (d) 0,50
Q 16. In the circuit shown in figure, V must be
(a) 50 V (b) 80 V (c) 100 V (d) 1290 V
Q 17. In the circuit shown in figure ammeter and voltmeter are ideal. If E = 4V, R = 9 ? and r = 1 ?,
then readings of ammeter and voltmeter are
Page 5
Current Electricity
For JEE Advanced
Objective Questions
Single Correct Option
Q 1. Two cells A and B of emf 1.3 V and 1.5 V respectively are arranged as shown in figure. The
voltmeter reads 1.45 V. Which cell has the higher internal resistance and how many times? The
voltmeter is assumed to be ideal
(a) r
1
= 2r
2
(b) r
1
= 3r
2
(c) r
2
= 2r
1
(d) r
2
= 3r
1
Q 2. A voltmeter connected in series with a resistance R
1
to a circuit indicates a voltage V
1
= 198 V
When a series resistor R
2
= 2R
1
is used, the voltmeter indicates a voltage V
2
= 180 V If the
resistance of the voltmeter is R
V
= 900 ? then, the applied voltage across A and 5 is
(a) 210 V (b) 200 V (c) 220 V (d) 240 V
Q 3. All bulbs in the circuit shown in figure are identical. Which bulb glows most brightly?
(a) B (b) A (c) D (d) C
Q 4. A student connects an ammeter A and a voltmeter V to measure a resistance R as shown in figure.
If the voltmeter reads 20 V and the ammeter reads 4 A, then R is
(a) equal to 5 ? (b) greater than 5 ? (c) less than 5 ?
(d) greater or less than 5 ? depending upon the direction of current
Q 5. The given figure represents an arrangement of potentiometer for the calculation of internal
resistance (r) of the unknown battery (E). The balance length is 70.0 cm with the key opened and
60.0 cm with the key closed. R is 132.40 ?. The internal resistance (r) of the unknown cell will be
(a) 22.1 ? (b) 113.5 ? (c) 154.5 ? (d) 10 ?
Q 6. Switch S is closed at time t = 0. Which one of the following statements is correct?
(a) Current in the resistance R increases if E
1
r
2
< E
2
(R + r
1
)
(b) Current in the resistance R increases if E
1
r
2
> E
2
(R + r
1
)
(c) Current in the resistance R decreases if E
1
r
2
> E
2
(R + r
1
)
(d) Current in the resistance R decreases if E
1
r
2
= E
2
(R + r
1
)
Q 7. A, B and C are voltmeters of resistances R,1.5R and 3R respectively. When some potential
difference is applied between x and y, the voltmeter readings are V
A
, V
B
and V
C
, then
(a)V
A
= V
B
= V
C
(b)V
A
? V
B
= V
C
(c) V
A
= V
B
? V
C
(d) V
A
+ V
B
= V
C
Q 8. In the circuit shown, the voltage drop across the 15 ? resistor is 30 V having the polarity as
indicated. The ratio of potential difference across 5 ? resistor and resistance R is
(a) 2/7 (b) 0.4 (c) 5/7 (d) 1
Q 9. In an experiment on the measurement of internal resistance of a cell by using a potentiometer,
when the key K is kept open then balancing length is obtained at y metre. When the K is closed
and some resistance R is inserted in the resistance box, then the balancing length is found to be x
metre. Then the
internal resistance is
(a)
xy
R
y
?
(b)
yx
R
x
?
(c)
yx
R
y
?
(d)
xy
R
x
?
Q 10. A source of emf E= 10 V and having negligible internal resistance is connected to a variable
resistance. The resistance varies as shown in figure. The total charge that has passed through the
resistor R during the time interval from t
1
to t
2
is
(a) 401og
e
4 (b) 30log, 3 (c) 201og
e
2 (d) 101og
e
2
Q 11. In order to increase the resistance of a given wire of uniform cross section to four times its value, a
fraction of its length is stretched uniformly till the full length of the wire becomes
3
2
times the
original length. What is the value of this fraction ?
(a)
1
4
(b)
1
8
(c)
1
16
(d)
1
6
Q 12. The figure, shows a meter bridge circuit, with AB = 100 cm, X = 12 ? and R = 18 ? and the
jockey J in the position of balance. If R is now made 8 ?, through what distance will J have to be
moved to obtain balance?
(a) 10 cm (b) 20 cm (c) 30 cm (d) 40 cm
Q 13. A milliammeter of range 10 mA and resistance 9 ? is joined in a circuit as shown. The meter
gives full-scale deflection for current I when A and B are used as its terminals, i.e., current enters
at A and leaves at B(C is left isolated). The value of I is
(a) 100 mA (b) 900 mA (c) 1A (d) 1.1 A
Q 14. A battery of emf E
0
= 12 Vis connected across a 4 m long uniform wire having resistance 4 ?/m
The cell of small emfs ?
1
= 2 V and ?
2
= 4 V having internal resistance 2 ? and 6 ? respectively
are connected as shown in the figure. If galvanometer shows no deflection at the point N, the
distance of point N from the point A is equal to
(a)
5
m
3
(b)
4
m
3
(c)
3
m
2
(d) None of these
Q 15. In the circuit shown, when keys K
1
and K
2
both are closed, the ammeter reads I
o
. But when K
1
is
open and K
2
is closed the ammeter reads I
o
/2 Assuming that ammeter resistance is much less than
R
2
, the values of rand R
1
in ? are
(a) 25, 50 (b) 25, 100 (c) 0, 100 (d) 0,50
Q 16. In the circuit shown in figure, V must be
(a) 50 V (b) 80 V (c) 100 V (d) 1290 V
Q 17. In the circuit shown in figure ammeter and voltmeter are ideal. If E = 4V, R = 9 ? and r = 1 ?,
then readings of ammeter and voltmeter are
(a) 1A, 3V (b) 2 A, 3 V (c) 3A,4V (d) 4 A, 4 V
Q 18. A moving coil galvanometer is converted into an ammeter reading up to 0.03 A by connecting a
shunt of resistance
r
4
. What is the maximum current which can be sent through this galvanometer,
if no shunt is used. (Here r = resistance of galvanometer)
(a) 0.004 A (b) 0.005 A (c) 0.006 A (d) 0.008 A
Q 19. The potential difference between points A and B is
(a)
20
V
7
(b)
40
V
7
(c)
10
V
7
(d) zero
Q 20. Two wires A and B made of same material and having their lengths in the ratio 6 : 1 are connected
in series. The potential difference across the wires are 3 V and 2 V respectively. If r
A
and r
B
are
the radii of A and B respectively, then
B
A
r
r
is
(a)
1
4
(b)
1
2
(c) 1 (d) 2
Q 21. A galvanometer of resistance 50 ? is connected to a battery of 3 V along with resistance of 2950
? in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to
reduce this deflection to 20 divisions the above series resistance should be
(a) 4450 ? (b) 5050 ? (c) 5550 ? (d) 6050 ?
Q 22. Figure shows a potentiometer arrangement with R
AB
= 10 ? and rheostat of variable resistance x.
For x = 0 null deflection point is found at 20 cm from A. For unknown value of x null deflection
point was at 30 cm from A, then the value of x is
(a) 10 ? (b) 5 ? (c) 2 ? (d) 1 ?
Read More
FAQs on DC Pandey Solutions (JEE Advance): Current Electricity- 1 - DC Pandey Solutions for JEE Physics
| 1. What is the JEE Advance exam? |  |
Ans. The JEE Advance exam is a highly competitive entrance examination conducted in India for admission into prestigious engineering colleges, including the Indian Institutes of Technology (IITs). It is the second stage of the Joint Entrance Examination (JEE) and is considered one of the toughest engineering entrance exams in the country.
| 2. Who is DC Pandey and how does his book help with JEE Advance preparation? | |
Ans. DC Pandey is a renowned author of physics books for competitive exams like JEE. His book on Current Electricity provides detailed explanations and solutions for JEE Advance level questions. It covers the essential concepts, theory, and practice problems related to current electricity, which is an important topic in the JEE Advance syllabus. The book is a valuable resource for students preparing for the exam.
| 3. What are some important topics covered in the article "Current Electricity: JEE Advance(Part - 1) - Physics, Solution by DC Pandey JEE"? | |
Ans. Some important topics covered in the article include the JEE Advance exam, the book "Current Electricity" by DC Pandey, and its relevance to JEE Advance preparation. The article also mentions that it is Part 1 of a series, indicating that there may be subsequent parts covering other topics related to current electricity in the JEE Advance syllabus.
| 4. How can I effectively prepare for the JEE Advance exam? | |
Ans. Effective preparation for the JEE Advance exam requires a structured study plan, thorough understanding of concepts, regular practice of problem-solving, and proper time management. It is important to refer to reliable study materials, such as books by renowned authors like DC Pandey, and solve previous year question papers and mock tests to familiarize yourself with the exam pattern and improve your problem-solving skills.
| 5. Is current electricity a challenging topic in the JEE Advance exam? | |
Ans. Yes, current electricity is considered a challenging topic in the JEE Advance exam. It requires a clear understanding of concepts like Ohm's law, Kirchhoff's laws, and various circuit elements. Additionally, the exam often includes complex circuits and questions that test the application of these concepts. Thorough preparation and practice are essential to excel in this topic and score well in the JEE Advance exam.
Related Exams
About this Document
|
4.70/5 Rating |
|
Nov 23, 2024 Last updated |
Document Description: DC Pandey Solutions (JEE Advance): Current Electricity- 1 for JEE 2024 is part of DC Pandey Solutions for JEE Physics preparation.
The notes and questions for DC Pandey Solutions (JEE Advance): Current Electricity- 1 have been prepared according to the JEE exam syllabus. Information about DC Pandey Solutions (JEE Advance): Current Electricity- 1 covers topics
like and DC Pandey Solutions (JEE Advance): Current Electricity- 1 Example, for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for DC Pandey Solutions (JEE Advance): Current Electricity- 1.
Introduction of DC Pandey Solutions (JEE Advance): Current Electricity- 1 in English is available as part of our DC Pandey Solutions for JEE Physics
for JEE & DC Pandey Solutions (JEE Advance): Current Electricity- 1 in Hindi for DC Pandey Solutions for JEE Physics course.
Download more important topics related with notes, lectures and mock test series for JEE
Exam by signing up for free. JEE: DC Pandey Solutions (JEE Advance): Current Electricity- 1 | DC Pandey Solutions for JEE Physics
Description
Full syllabus notes, lecture & questions for DC Pandey Solutions (JEE Advance): Current Electricity- 1 | DC Pandey Solutions for JEE Physics - JEE | Plus excerises question with solution to help you revise complete syllabus for DC Pandey Solutions for JEE Physics | Best notes, free PDF download
Information about DC Pandey Solutions (JEE Advance): Current Electricity- 1
In this doc you can find the meaning of DC Pandey Solutions (JEE Advance): Current Electricity- 1 defined & explained in the simplest way possible. Besides explaining types of
DC Pandey Solutions (JEE Advance): Current Electricity- 1 theory, EduRev gives you an ample number of questions to practice DC Pandey Solutions (JEE Advance): Current Electricity- 1 tests, examples and also practice JEE
tests
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Extra Questions
,DC Pandey Solutions (JEE Advance): Current Electricity- 1 | DC Pandey Solutions for JEE Physics
,past year papers
,Viva Questions
,Objective type Questions
,Sample Paper
,study material
,practice quizzes
,Previous Year Questions with Solutions
,Important questions
,ppt
,video lectures
,Free
,MCQs
,DC Pandey Solutions (JEE Advance): Current Electricity- 1 | DC Pandey Solutions for JEE Physics
,mock tests for examination
,Semester Notes
,Summary
,DC Pandey Solutions (JEE Advance): Current Electricity- 1 | DC Pandey Solutions for JEE Physics
,Exam
,shortcuts and tricks
,
Additional Information about DC Pandey Solutions (JEE Advance): Current Electricity- 1 for JEE Preparation
DC Pandey Solutions (JEE Advance): Current Electricity- 1 Free PDF Download
The DC Pandey Solutions (JEE Advance): Current Electricity- 1 is an invaluable resource that delves deep into the core of the JEE exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the DC Pandey Solutions (JEE Advance): Current Electricity- 1 now and kickstart your journey towards success in the JEE exam.
Importance of DC Pandey Solutions (JEE Advance): Current Electricity- 1
The importance of DC Pandey Solutions (JEE Advance): Current Electricity- 1 cannot be overstated, especially for JEE aspirants.
This document holds the key to success in the JEE exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
DC Pandey Solutions (JEE Advance): Current Electricity- 1 Notes
DC Pandey Solutions (JEE Advance): Current Electricity- 1 Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to DC Pandey Solutions (JEE Advance): Current Electricity- 1.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, DC Pandey Solutions (JEE Advance): Current Electricity- 1 Notes on EduRev are your ultimate resource for success.
DC Pandey Solutions (JEE Advance): Current Electricity- 1 JEE Questions
The "DC Pandey Solutions (JEE Advance): Current Electricity- 1 JEE Questions" guide is a valuable resource for all aspiring students preparing for the
JEE exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study DC Pandey Solutions (JEE Advance): Current Electricity- 1 on the App
Students of JEE can study DC Pandey Solutions (JEE Advance): Current Electricity- 1 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the DC Pandey Solutions (JEE Advance): Current Electricity- 1,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of DC Pandey Solutions (JEE Advance): Current Electricity- 1 is prepared as per the latest JEE syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup