NCERT Exemplar Class 12 Physics Solutions Chapter 3 Current Electricity 2021

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NCERT Exemplar Class 12 Physics Solutions Chapter 3 Current Electricity

Edited By Safeer PP | Updated on Sep 14, 2022 12:28 PM IST | #CBSE Class 12th

NCERT Exemplar Class 12 Physics solutions chapter 3 is from the point of view of exams. It also gives students an insight on how current flows in working of different types of devices in real life such as conductors. The previous chapters dealt with charges that were considered as at rest, whereas this chapter introduces various concepts relating to steady currents. NCERT Exemplar Class 12 Physics chapter 3 solutions includes the idea of electric energy and power, electric resistivity, and conductivity, along with a detailed explanation of carbon resistors and their colour codes. The concepts discussed in the NCERT Exemplar Class 12 Physics solutions chapter 3 are important from the point of view of board exams as the JEE Main exam. The students can also make use of NCERT Exemplar Class 12 Physics solutions chapter 3 PDF download, for further learning.

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This Story also Contains

NCERT Exemplar Class 12 Physics Solutions Chapter 3 MCQI
NCERT Exemplar Class 12 Physics Solutions Chapter 3 MCQII
NCERT Exemplar Class 12 Physics Solutions Chapter 3 Very Short Answer
NCERT Exemplar Class 12 Physics Solutions Chapter 3 Short Answer
NCERT Exemplar Class 12 Physics Solutions Chapter 3 Long Answer
NCERT Exemplar Class 12 Physics Solutions Chapter 3 Current Electricity Main Subtopics Covered
NCERT Exemplar Class 12 Physics Chapter Wise Links
Important Topics To Cover For Exams From NCERT Exemplar Class 12 Physics Solutions Chapter 3 Current Electricity

Also see - NCERT Solutions for Class 12 Physics

NCERT Exemplar Class 12 Physics Solutions Chapter 3 MCQI

Question:1

Consider a current carrying wire (current I) in the shape of a circle. Note that as the current progresses along the wire, the direction of j (current density) changes in an exact manner, while the current I remain unaffected. The agent that is essentially responsible for is
A. source of emf.
B. electric field produced by charges accumulated on the surface of wire.
C. the charges just behind a given segment of wire which push them just the right way by repulsion.
D. the charges ahead.

Answer:

b)
Current density (J) is the current per unit area and is given by $\overrightarrow{J}=\sigma\overrightarrow{ E}$ , where conductivity $\left (\sigma =\frac{l}{AR} \right )$ and electric field $\left ( \overrightarrow{E} \right )$ .

Question:2

Two batteries of emf $\varepsilon _{1}$ and $\varepsilon _{2}\left ( \varepsilon_ {2}>\varepsilon _{1} \right )$ and internal resistances r₁ and r₂respectively are connected in parallel as shown in Figure.

capture-32
A. The equivalent emf $\varepsilon _{eq}$ of the two cells is between $\varepsilon _{1}$ and $\varepsilon _{2}$ , i.e. $\varepsilon _{1}<\varepsilon _{eq}<\varepsilon _{2}$
B. The equivalent emf $\varepsilon _{eq}$ is smaller than 1.
C. The $\varepsilon _{eq}$ is given by $\varepsilon _{eq}=1+2$ always.
D. $\varepsilon _{eq}$ is independent of internal resistances r₁ and r₂.

Answer:

The correct answer is the option (a)
Explanation: Equivalent EMF in parallel is given as:
$\varepsilon _{eq}=\frac{\varepsilon _{1}r_{2}+\varepsilon _{2}r_{1}}{r_{1}+r_{2}}$
This value will always lie between $\varepsilon _{1}$ and $\varepsilon _{2}$

Question:3

A resistance R is to be measured using a meter bridge. Student chooses the standard resistance S to be 100. He finds the null point at $l_1 = 2.9 cm$ . He is told to attempt to improve the accuracy. Which of the following is a useful way?

A. He should measure l1 more accurately.
B. He should change S to 1000and repeat the experiment.
C. He should change S to 3 and repeat the experiment.
D. He should give up hope of a more accurate measurement with a meter bridge.

Answer:

The correct answer is the option (c)
Explanation:
$R=S\left (\frac{l_1}{100-l_1} \right )=100\left (\frac{2.9}{97.1} \right )=2.98\Omega$
He should aim to get the balance point near the centre (l1=50 cm)
$2.98\Omega =S\left (\frac{50}{50} \right )$
$S\cong 3\Omega$

Question:4

Two cells of emf’s approximately 5V and 10V are to be accurately compared using a potentiometer of length 400cm.
A. The battery that runs the potentiometer should have a voltage of 8V.
B. The battery of potentiometer can have a voltage of 15V and R adjusted so that the potential drop across the wire slightly exceeds 10V.
C. The first portion of 50 cm of the wire itself should have a potential drop of 10V.
D. Potentiometer is usually used for comparing resistances and not voltages.

Answer:

The correct answer is the option (b)
Explanation: The potential drop along the wires of the potentiometer should be greater than the emfs of cells. The battery of potentiometer must be in excess of 10V.

Question:5

A metal rod of length 10 cm and a rectangular cross-section of $1cm \times \frac{1}{2} cm$ is connected to a battery across opposite faces. The resistance will be
A. maximum when the battery is connected across $1cm \times \frac{1}{2} cm$ faces.
B. maximum when the battery is connected across $10cm \times 1 cm$ faces.
C. maximum when the battery is connected across $10cm \times \frac{1}{2} cm$ faces.
D. same irrespective of the three faces.

Answer:

The correct answer is the option (a)
Explanation: $R=\rho\frac{l}{A}$
Resistance is directly proportional to length and inversely proportional to the cross-sectional area. Resistance will be maximum when l is maximum and A minimum. Both these conditions are met simultaneously when the battery is connected across the $1cm \times \frac{1}{2} cm$ faces.

Question:6

Which of the following characteristics of electrons determines the current in a conductor?
A. Drift velocity alone.
B. Thermal velocity alone.
C. Both drift velocity and thermal velocity.
D. Neither drift nor thermal velocity

Answer:

The correct answer is the option (a)
$I=Anev_d$

NCERT Exemplar Class 12 Physics Solutions Chapter 3 MCQII

Question:7

Kirchhoff ’s junction rule is a reflection of
A. conservation of current density vector.
B. conservation of charge.
C. the fact that the momentum with which a charged particle approaches a junction is unchanged (as a vector) as the charged particle leaves the junction.
D. the fact that there is no accumulation of charges at a junction.

Answer:

The correct answers are the options (b,d)
Explanation: According to the law of conservation of charges, the charge of a system will stay constant. From this, it can be derived that net charge flowing into a junction is equal to the net charge going out of the system.

Question:8

Consider a simple circuit shown in Figure capture-33 variable resistance R’. R’ can vary from $R_0$ to infinity. r is internal resistance of the battery $(r<<R<<R_0).$
capture-34
A. Potential drop across AB is nearly constant as R’ is varied.
B. Current through R is nearly a constant as R’ is varied.
C. Current I depends sensitively on R’.
D. $I\geqslant \frac{V}{r+R}$ always.

Answer:

The correct answers are the options (a,d)
Explanation:
$I=\frac{V}{r+R_{eq}}$
As $r+R<r+R_{eq}$
$I=\frac{V}{r+R_{eq}}\leq \frac{V}{r+R}$

Question:9

Temperature dependence of resistivity $\rho (T)$ of semiconductors, insulators and metals is significantly based on the following factors:
A. number of charge carriers can change with temperature T.
B. time interval between two successive collisions can depend on T.
C. length of material can be a function of T.
D. mass of carriers is a function of T.

Answer:

The correct answers are the options (a,b)
Explanation: Resistivity is a function of relaxation time $(\tau )$ and mass of charge carrier (m). Mass of charge carrier is independent of temperature, whereas length also does not fluctuate significantly over a temperature range.

Question:10

The measurement of an unknown resistance R is to be carried out using Wheat stones bridge (see Figure of NCERT Book). Two students perform an experiment in two ways. The first students take $R_2 = 10\Omega$ and $R_1 = 5\Omega$ . The other student takes $R_2 = 1000\Omega$ and $R_1 = 500\Omega$ . In the standard arm, both take $R_3= 5\Omega$ . Both find $R=\frac{R_{2}}{R_{1}}R_{3}=10\Omega$ within errors.

A. The errors of measurement of the two students are the same.
B. Errors of measurement do depend on the accuracy with which $R_{2}$ and $R_{1}$ can be measured.
C. If the student uses large values of $R_{2}$ and $R_{1}$ , the currents through the arms will be feeble. This will make determination of null point accurately more difficult.
D. Wheatstone bridge is a very accurate instrument and has no errors of measurement.

Answer:

The correct answers are the options (b,c)
Explanation: Using large values of $R_{1}$ and $R_{2}$ , will reduce the current flowing through the system, making it difficult to get accurate readings from the galvanometer.

Question:11

In a meter bridge the point D is a neutral point (Figure).

capture-35
A. The meter bridge can have no other neutral point for this set of resistances.
B. When the jockey contacts a point on meter wire left of D, current flows to B from the wire.
C. When the jockey contacts a point on the meter wire to the right of D, current flows from B to the wire through galvanometer.
D. When R is increased, the neutral point shifts to left.

Answer:

The correct answers are the options (a,c)
Explanation: No current will flow through the Galvanometer when the jockey is placed at D. Potential Difference between B and D is zero. If the jockey is placed to the right of D, Potential at B becomes greater than the potential at D, the current flows from B to D.

NCERT Exemplar Class 12 Physics Solutions Chapter 3 Very Short Answer

Question:12

Is the momentum conserved when charge crosses a junction in an electric circuit? Why or why not?

Answer:

As drift velocity is proportional to the electric field, when charge crosses a junction the momentum is not conserved.

Question:13

The relaxation time $\tau$ is nearly independent of applied E field whereas it changes significantly with temperature T. First fact is (in part) responsible for Ohm’s law whereas the second fact leads to variation of ρ with temperature. Elaborate why?

Answer:

Relaxation time decreases with increase in drift velocity. This increases the resistivity
$\rho=\frac{m}{ne^2\tau }$

Question:14

What are the advantages of the null-point method in a Wheatstone bridge? What additional measurements would be required to calculate $R_{unknown}$ by any other method?

Answer:

As no electricity will pass through the galvanometer of the Wheatstone bridge at the null point. It can be used to calculate the resistance of an unknown resistor.

Question:15

What is the advantage of using thick metallic strips to join wires in a potentiometer?

Answer:

Thick metallic strips have large cross section and thus offer less resistance

Question:16

For wiring in the home, one uses Cu wires or Al wires. What considerations are involved in this?

Answer:

The following considerations are made to decide on the wiring:

Conductivity of metal
Cost
Availability

Question:17

Why are alloys used for making standard resistance coils?

Answer:

Because of their low temperature sensitivity and low temperature coefficient of resistance, alloys are used to make standard resistance coils. Also, they have a lower conductivity or higher resistivity, this reduces the length required to make coils.

Question:18

Power P is to be delivered to a device via transmission cables having resistance $R_C$ . If V is the voltage across R and I the current through it, find the power wasted and how can it be reduced.

Answer:

P (Power transmitted)=VI
H (Heat Loss)= $I^{2}R$
High current flowing through the devices will lead to higher Energy Loss (as heat), so power should be transmitted at a high voltage and low current.

Question:19

AB is a potentiometer wire (Figure). If the value of R is increased, in which direction will the balance point J shift?
capture-36

Answer:

On increasing the applied resistance, the potential difference across AB will decrease. To balance the potential across the other battery, the length AJ' should increase. Balance point shift towards B.

Question:20

While doing an experiment with potentiometer (Figure ) it was found that the deflection is one sided and (i) the deflection decreased while moving from one end A of the wire to the end B; (ii) the deflection increased. While the jockey was moved towards the end B.

capture-37
(i) Which terminal +or –ve of the cell $E_{1}$ , is connected at X in case (i) and how is $E_{1}$ related to $E$ ?
(ii) Which terminal of the cell $E_{1}$ is connected at X in case (ii)?

Answer:

i) As jockey moves closer to B, the galvanometer potential difference decreases. This requires the positive terminal of $E_{1}$ to be at X.
ii) Exactly opposite to (i), at X $E_{1}^{'}$ s negative terminal is connected.

Question:21

A cell of emf E and internal resistance r is connected across an external resistance R. Plot a graph showing the variation of P.D. across R, verses R.

Answer:

$I=\frac{E}{R+r} and V=IR$
$V=\frac{ER}{R+r}=\frac{E}{1+\frac{r}{}R}$
An increase in R leads to an increase in V, till the point it approaches E
Following is the graphical representation:
capture-38

NCERT Exemplar Class 12 Physics Solutions Chapter 3 Short Answer

Question:22

First a set of n equal resistors of R each are connected in series to a battery of emf E and internal resistance R. 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 10 times. What is ‘n’?

Answer:

$R_{Series}=nR$
$I_{Series}=\frac{E}{R+nR}=\frac{E}{(n+1)R}$
$R_{Parallel}=\frac{R}{}n$
$I_{Parallel}=\frac{E}{R+\frac{R}{}n}=\frac{E}{}R\times \frac{n}{n+1}$
$I_{Parallel}=10\times I_{Series}$
$\frac{E}{}R\times \frac{n}{n+1}=10\times \frac{E}{}R\times \frac{1}{n+1}$
$n=10$

Question:23

Let there be n resistors $R_1 ............R_n$ with $R_{max} = max (R_1......... R_n)$ and $R_{min} = min (R_1 ..... R_n)$ . Show that when they are connected in parallel, the resultant resistance R_P< Rmin and when they are connected in series, the resultant resistance $R_S > R_{max}$ . Interpret the result physically.

Answer:

$\frac{1}{R_{parallel}}=\frac{1}{R_1}+\frac{1}{R_{2}}+.....+\frac{1}{R_n}$
$\frac{R_{min}}{R_{parallel}}=\frac{R_{min}}{R_1}+\frac{R_{min}}{R_2}....+\frac{R_{min}}{R_n}$
$\frac{R_{min}}{R_{parallel}}> 1$
$R_{series}= R_1+R_2...+R_n$
$R_{series}>R_{max}$
This can be represented physically as shown below:

Question:24

The circuit in Figure shows two cells connected in opposition to each other. Cell $E_{1}$ is of emf 6V and internal resistance 2 $\Omega$ ; the cell $E_{2}$ is of emf 4V and internal resistance 8 $\Omega$ . Find the potential difference between the points A and B.

capture-42

Answer:

Equivalent EMF of the Cells $=6V-4V=2V$
$R_{eq}=2\Omega +8\Omega=10\Omega$
$I=\frac{2V}{10 \Omega}=0.2A$
$V_B-V_A=4+8\times 0.2=5.6V$

Question:25

Two cells of same emf E but internal resistance $r_{1}$ and $r_{2}$ are connected in series to an external resistor R (Figure). What should be the value of R so that the potential difference across the terminals of the first cell becomes zero.

capture-44

Answer:

$EMF_{ef}=E+E=2E$
$R_{ef}=r_1+r_2$
$I=\frac{2E}{R+r_1+r_2}$
As net potential across the cell is 0,
$E-Ir_1=0$
$E-\frac{2Er_1}{R+r_1+r_2}=0$
$R=r_1-r_2$

Question:26

Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1mm. Conductor B is a hollow tube of outer diameter 2mm and inner diameter 1mm. Find the ratio of resistance to

Answer:

$Resistance(R)=\rho\frac{L}{A}$
$R_A=\rho _A\left (\frac{L_A}{A_A} \right ) and R_B=\rho _B\left (\frac{L_B}{A_B} \right )$
$\rho _A=\rho _B \ and \ L_A=L_B$ $\frac{R_A}{R_B}=\frac{A_B}{A_A}=\frac{\pi r_A^2}{\pi (r_{out}^2-r_{in}^2)}=\frac{1^2-\left (\frac{1}{2} \right )^{2}}{\left ( \frac{1}{2} \right )^{2}}=\frac{\frac{3}{4}}{\frac{1}{4}}=3$

Question:27

Suppose there is a circuit consisting of only resistances and batteries and we have to double (or increase it to n-times) all voltages and all resistances. Show that currents are unaltered. Do this for circuit of Example 3.7 in the NCERT Text Book for Class XII.

Answer:

Original:
$EMF = V_1$
$Resistance = R + R_{eff}$
$Current = \frac{V_1}{R+R_{eff}}$
New:
$EMF = nV_1$
$Resistance =n( R + R_{eff})$
$Current =\frac{n(V_1)}{n(R+R_{eff})}= \frac{V_1}{R+R_{eff}}$
Increasing the EMF and Resistance in the same ratio (say 2), will not change the current flowing through the system.

NCERT Exemplar Class 12 Physics Solutions Chapter 3 Long Answer

Question:28

Two cells of voltage 10V and 2V and internal resistances 10 $\Omega$ and 5 $\Omega$ respectively, are connected in parallel with the positive end of 10V battery connected to negative pole of 2V battery (Figure). Find the effective voltage and effective resistance of the combination.

capture-46

Answer:

capture-47
Kirchoff's Junction Law at $C: I_1=I_2+I_3$
Kirchoff's Loop Law for efbae: $10-I_1\times 10-I -3\times R=0$
$10-10I_2-(R+10)I_3=0$
Kirchoff's Loop Law for cbadc:- $2+5I_2-I_3R=0$
$(10-10I_2-(R+10)I_3)+(-2+5I_2-I_3R)\times 2=0$
$6-I_3R+10-2I_3R=0$
$6=3I_3R+10I_3$
$I_3=\frac{6-3I_3R}{10}$
Veff & Reff will fulfil the following:
$V_{eff}-I_3(R+R_{eff})=0$
$R_{eff}=\frac{1}{\frac{1}{10}+\frac{1}{15}}=\frac{10}{}3$
$V_{eff}-I_3\left (R+\frac{10}{}3 \right )=0$
$V_{eff}=I_3R+I_3\times \frac{10}{}3$
$V_{eff}=I_3R+\frac{6-3I_3R}{10}\times \frac{10}{}3=I_3R+6-\frac{3I_3R}{}3=2V$

Question:29

A room has AC run for 5 hours a day at a voltage of 220V. The wiring of the room consists of Cu of 1 mm radius and a length of 10 m. Power consumption per day is 10 commercial units. What fraction of it goes in the joule heating in wires? What would happen if the wiring is made of aluminium of the same dimensions?

$[\rho_ {Cu} = 1.7 \times -810\Omega m, \rho _{Al} = 2.7 \times 10^{-8} \Omega m]$

Answer:

Energy consumed day=10 kWh
Energy consumed hour=2 kWh
Power=2kW
P=VI
V=220V
$I=\frac{P}{}V=\frac{2000}{220}=\frac{100}{11}$
Heat $=I^2\times R_{Cu}$
$R_{Cu}=\rho\left ( \frac{L}{}A \right )=1.7\times 10^{-8}\times \left (\frac{10}{\pi \times \left (5\times 10^{-4} \right )^{2}} \right )$
$Heat=\left (\frac{100}{110} \right )^{2}\times 1.7\times 10^{-8}\times \left (\frac{10}{\pi \times \left (5\times 10^{-4} \right )^{2}} \right )=4.4 W$
fractional loss in heating
$=\frac{4.4}{2000}\times 100 =0.22\%$
$\frac{Heat_{Al}}{Heat_{Cu}}=\frac{I^2RAl}{I^2R_{Cu}}=\frac{\rho _{Al}}{\rho _{Cu }}$
Heat_Al=7W
fractional loss in heating
$=\frac{7}{2000}\times1000=0.35\%$

Question:30

In an experiment with a potentiometer, $V_{B}=10V$ R is adjusted to be $50 \Omega$ (Figure). A student wanting to measure voltage E₁ of a battery (approx. 8V) finds no null point possible. He then diminishes R to $10 \Omega$ and is able to locate the null point on the last (4^th) segment of the potentiometer. Find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.

Answer:

R_eq=50+R
V_eq=10V
$I=\frac{10}{50+R'}$
The potential difference across the potentiometer
$V'=IR'=\frac{10R'}{50+R'}$
V'<8
$10R'<8(50+R' )$
R'<200
$\frac{10R'}{10+R'}>8$
R'>40
At balance point:
$\frac{10\times \frac{3}{}4R'}{10+R'}<8$
R'>-160
As R’ can only be positive, R'>160
$k\times 400cm>8V$
k>2 V/m
As balance, point isn’t at 3m
$K\times 3<8$
$k<2\frac{2}{}3\frac{V}{}m$

Question:31

(a) Consider the circuit in Figure. How much energy is absorbed by electrons from the initial state of no current (ignore thermal motion) to the state of drift velocity?

(b) Electrons give up energy at the rate of $RI_2$ per second to the thermal energy. What time scale would one associate with energy in problem (a)? n = no of electron/volume = $10^{29}/m^3$ , length of circuit = 10 cm, cross-section = A $= (1mm)^2$
capture-49

Answer:

a)
$A=10^{-6}m^2$
$R=6\Omega$
$n=10^{29} m^{-3}$
$e=1.6\times 10^{-19}C$
$I=\frac{V}{}R=1A$
$m_e=9.1\times10^{-31}Kg$
$L=10^{-1}m$
a) $I=Anev_d$
$v_d=\frac{I}{Ane}=6.25\times 10^{-5}m/s$
Kinetic energy=
$\frac{1}{}2\times m_e\times v_d^2=2\times 10^{-17}J$
b) Power loss=
$I^2R=6W$
$P=\frac{E}{}t$
$t=\frac{E}{}P=10^{-17}s$

NCERT Exemplar Class 12 Physics Solutions Chapter 3 Current Electricity Main Subtopics Covered

Introduction
Electric Current
Electric Currents in Conductors
OHM’s Law
Drift of Electrons and the Origin of Resistivity
Limitations of OHM’s Law
Resistivity of Various Materials
Temperature Dependence of Resistivity
Electrical Energy, Power
Combination of Resistors- Series and Parallel
Cells, EMF, Internal Resistance
Cells in Series and in Parallel
Kirchhoff’s Rules
Wheatstone Bridge
Meter Bridge
Potentiometer

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NCERT Exemplar Class 12 Physics Chapter Wise Links

Chapter 1 Electric Charges and Fields

Chapter 2 Electrostatic Potential and Capacitance

Chapter 4 Moving Charges and Magnetism

Chapter 5 Magnetism and Matter

Chapter 6 Electromagnetic Induction

Chapter 7 Alternating Current

Chapter 8 Electromagnetic Waves

Chapter 9 Ray Optics and Optical Instruments

Chapter 10 Wave Optics

Chapter 11 Dual Nature of Radiation and Matter

Chapter 12 Atoms

Chapter 13 Nuclei

Chapter 14 Semiconductor Electronics

Chapter 15 Communication Systems

Important Topics To Cover For Exams From NCERT Exemplar Class 12 Physics Solutions Chapter 3 Current Electricity

· NCERT Exemplar solutions for Class 12 Physics chapter 3 introduces new principles/laws as OHM’s law, Kirchhoff’s rule, etc., which display and govern the flow of current in various devices, and how it can be determined with the help of other factors.

· Students will study different circuits in which a number of resistors and cells are interconnected for which different calculations need to be done to obtain different values of current, resistance, etc., with the help of Class 12 Physics NCERT Exemplar solutions chapter 3

· NCERT Exemplar Class 12 Physics solutions chapter 3 also introduces different devices such as potentiometer, meter bridge and exhibits the methods, and formulas for understanding current, resistance, etc., through these devices.

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Frequently Asked Questions (FAQs)

1. Is this chapter important from exam POV?

Yes, current electricity is one of the most crucial chapters that one needs to focus on for getting good marks in boards and entrance exams.

2. How many questions are solved in this chapter?

Our teachers have solved all questions from NCERT exemplar Class 12 Physics chapter 3 current electricity with complete steps and details.

3. How to use these solutions to the best?

One can make use of these solutions to understand the questions to the best and also understand how to solve questions in exams.

Also Read

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

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Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

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Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024

Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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kumardurgesh1802 10 July,2024

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Central Board of Secondary Education Class 12th Examination

JEE Test Series

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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