Velocity (υ) and acceleration (a) in two system of units 1 and 2 are related as υ2=nm2υ1 and a2=a1mn respectively. Here m and n are constants. The relations for distance and time in two system respectively are:

Let, L1 & L2 one distance for system (1) & (2) respectively T1 & T2 are time for system (1) & (2) respectively Given : v2 = nm2v1 ⇒(L2T2)=nm2(4T1) [?v2=L2T2,v1=L1T1] ⇒L2L1=(nm2)[T2T1]−−(i) And a2 = a1mn⇒v2T2=v1T1mn ⇒T1T2=v1v2(1mn) [?v1v2=m2n] ⇒n2mT1=T2 from (1) L2L1=nm2[n2m][?T2T1=n2m] ⇒n3m3L1=L2

A ball is spun with angular acceleration a = 6t2 – 2t where t is in second and a is in rads-1. At t = 0, the ball has angular velocity of 10 rads-1 and angular position of 4 rad. The most appropriate expression for the angular position of the ball is:

a = 6t2 – 2t, of t 20, w = 10 rad/sec , q = 4 rad α=dωdt ⇒∫10ωdω=∫t−0tαdt ⇒ω−10=∫0t (6t2−2t)dt ω=dθdt dθ=ω dt ∫4θdθ=∫0t (2t3−t2+10)dt = t42−t33+10t+4

The circuit in figure shows two cells connected in opposition to each other. Cell E1is of emf 6 V and internal resistance 2 Ω; the cell E2 is of emf 4 V and internal resistance 8 Ω. Find the potential difference between the points A and B.

This is a Short Answer Type Questions as classified in NCERT Exemplar Explanation- total resistance = 2+8= 10 ohm I= 6-42+8 = 0.2A The direction of flow of current is always from high potential to low potential If VB > VA VB-4V-0.2 * 8 = VA VB-VA= 3.6V

Physics NCERT Exemplar Solutions Class 12th Chapter Three Current Electricity

Q: A block of mass 2kg moving on a horizontal surface with speed of 4 ms-1 enters a rough surface ranging from x = 0.5m to x = 1.5m. The retarding force in this range of rough surface is related to distance by F = -kx where k = 12Nm-1. The speed of the block as it just crosses the rough surface will be:

by, WET ∫Fdx=Δk.6 ∫− kx dx=12m [vt2−vi2] ⇒−k2 [xf2−xi2]0.51.5=12×2 [vf2−16] [? vi=4m/sec] −122 [1.52−0.52]= [vf2−16] vf2=16−12=4 vf = 2m/sec

Q: A 3 4 m long ladder weighing 10kg leans on a frictionless wall. Its feet rest on the floor 3m away from the wall as shown in the figure. If Ff and Fw are the reaction forces of the floor and the wall, then ratio of Fw / Ff will be: (Use g = 10 m/s2)

mg = 10 × 10 = 100 of NA = Fω reaction force offered by the wall N32+fB2=Ff reaction force offered by the floor. By NLM 1 Translational fB = NA NB = mg = 100N Ac2 + Bc2 = AB2 Ac2 = 34 – 9 Ac = 5m Rotational equilibrium τB=0 mg l2cosθ=NAlsinθ 100cosθ2=NAsinθ NA = 50 cot Now cot = 35 NA = 50×35=30N Therefore, NA = Fω=30N - (i) Ff=NB2+fB2=1002+302=10109N - (2) fωFf=3010109=3109

Q: Water falls from a 40m height dam at the rate of 9 × 104 kg per hour. Fifty percentage of gravitational potential energy can be converted into electrical energy. Using this hydro electric energy number of 100 W lamps, that can be lit, is: (Take g = 10 ms-2)

Given : water flow rate = (mo) = 9 × 104 kg/hr = 9×104kg/sec60×60 Potential energy available on water per unit time ⇒mogh=9×10460×60×10×40 mogh=104watt 50% of it converts into electrical energy = 50% of mogh=1042watt Let η be the number of bulbs n = 100 = 1042 n = 50

Updated on Jul 22, 2025 14:51 IST

By Pallavi Pathak, Assistant Manager Content

Shiksha's experts have developed NCERT Exemplar Solutions Class 12 Chapter Three Current Electricity for Class 12 students who are preparing for their CBSE Board examination. Students who will prepare from this exemplar will have a deeper understanding of the theoretical concepts of the Current Electricity NCERT and will also develop problem-solving skills. The solutions given in the exemplar cover all types of questions, including short answer, long answer, and multiple choice, with detailed explanations.
Class 12 Chapter Three Current Electricity exemplar is an excellent resource for CBSE Board exam preparation and also for entrance exam preparation, such as JEE and NEET. Students will also get the Current Electricity Class 12 NCERT PDF here, which they can download and study offline from anywhere and at any time. The PDF has well-structured questions that cover all the important topics of the Current Electricity Class 12 NCERT.

Chapter Three Current Electricity is one of the most fundamental chapters of Class 12. It includes the study of electric charges in motion and explores various concepts like resistance, Ohm's law, electric current, resistivity, and the combination of resistors. It also covers the most advanced topics such as electromotive force (emf), internal resistance of cells, and Kirchhoff's laws.

Also read:

NCERT Solutions for Class 12 Physics

Table of contents

Common Mistakes and Tips for NCERT Physics Exemplar Chapter Three
Important Formulas Related to Physics Chapter Three NCERT Exemplar
Current Electricity Questions and Answers
Chapters List
Current Electricity Short Answer Type Questions
Current Electricity Multiple Choice Questions
24th June 2022 (Second Shift)
JEE Mains Solutions 2022,28th june , Physics ,Second shift

Common Mistakes and Tips for NCERT Physics Exemplar Chapter Three

There are some common mistakes the students make while solving questions related to Current Electricity. The following are these common mistakes:

Many students incorrectly use Ohm's law. They apply this law without considering whether the material is ohmic or non-ohmic.
There is confusion between Terminal Voltage and electromotive force (EMF).
In Kirchhoff's laws, the students make sign convention errors. They frequently misuse the positive/negative for voltage or current direction.
They make incorrect identification of series and parallel arrangements of resistors, which results in wrong calculations of equivalent resistance.
Students also sometimes don't convert all quantities to SI units. It results in wrong final answers.
They use the wrong formula for electric power depending on what's given. The formula for electric power is

Following are some of the tips to solve the NCERT Exemplar Questions on Current Electricity and avoid these common mistakes:

Before attempting the advanced problems of Chapter Three, try to understand the basic principles such as voltage, resistance, current, and resistivity.
Practice drawing and simplifying circuits of the Current Electricity NCERT. Label all currents, resistances, and voltages to avoid any confusion.
Use Kirchhoff’s rules systematically by applying the KVL and KCL step-by-step. Even if you later find the assumed direction is negative, always assign the current direction consistently.
To find whether your answer has correct physical units, use dimensional analysis.
Revise key formulas like resistance combinations, Ohm's Law, and power equations frequently, and keep a formula sheet handy.
Practice from Chapter Three NCERT Exemplars to be ready for the JEE and NEET examinations. Many of the questions from this chapter are important from an entrance exam point of view. It provides exam-ready preparation.
It is also recommended to practice 2-3 numerical problems daily and focus on the power calculations and resistive networks.

Important Formulas Related to Physics Chapter Three NCERT Exemplar

The following are the important formulas related to Class 12 Physics Chapter 3:

Ohm’s Law

$V = I R$
Resistance of a Conductor

$R = ρ \frac{l}{A}$
Current

$I = \frac{Q}{t}$
Drift Velocity

$v_{d} = \frac{I}{n A e}$
Ohm’s Law in Terms of Resistivity

$E = ρ j$
Electric Power

$P = V I = I^{2} R = \frac{V^{2}}{R}$
Work Done by Current

$W = V I t$
Resistors in Series

$R_{eq} = R_{1} + R_{2} + R_{3} + \dots$
Resistors in Parallel

$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \dots$
Kirchhoff’s Current Law

$\sum I_{in} = \sum I_{out}$
Kirchhoff’s Voltage Law

$\sum V = 0$
Terminal Voltage of a Cell

$V = ε - I r$
Resistivity and Temperature Relation

$ρ_{t} = ρ_{0} (1 + α ∆ T)$

Current Electricity Questions and Answers

1. A room AC runs for 5 a day at a voltage of 220 V. 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?

Explanation – power consumption in a day i.e in 5 = 10 units

Power consumption per hour = 2 units

Power consumption = 2 units =2KW= 2000J/s

Also power =V $\times$ I

2000W= 220V $\times$ l or l= 9A approx.

R= $\frac{ρ l}{A}$

Power consumption in first current carrying wire

P= I²R

$\frac{ρ l}{A}$ l²= 1.7 $\times$ 10^{-8
$\times \frac{10}{π \times 10^{- 6}} \times 81$}j/s = 4J/s approx.

Loss due to joule heating in first wire = $\frac{4}{2000} \times$ 100=0.2%

Power loss in Al wire =1.6 $\times$ 4= 6.4J/s

Fractional loss due to joule heating in second wire = $\frac{6.4}{2000} \times$ 100= 0.32%

2. In an experiment with a potentiometer, VB = 10 V. R is adjusted to be 50 Ω (figure). A student wanting to measure voltage E₁of a battery (approx. 8 V) finds no null point possible. He then diminishes R to 10 Ω and is able to locate the null point on the-last (4th) segment of the potentiometer. Find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.

Explanation – let R’ be the resistance of potentiometer wire.

Effective resistance of potentiomter and variable resistor r=50ohm is 500+R’

Effective voltage across potentiometer = 10V

The current through main circuit I= $\frac{V}{50 + R}$ = $\frac{10}{50 + R}$

Potential difference across wire of potentiometer

IR’= $\frac{10 R'}{50 + R}$

Since with 50 ohm resistor , null point is not obtained it is possible when

$\frac{10 \times R'}{50 + R} < 8$

10R’ $< 400 + 8 R'$

2R’<400 or R’<200 ohm

Similarly with 10 ohm resistor , null point is obtained its is only possible when

$\frac{10 \times R'}{50 + R} < 8$

2R’>40

R’>40

$\frac{10 \times \frac{3}{4} R'}{10 + R'} < 8$

7.5R’<80+8R’

R’>160

160

Any R’ between 160 ohm and 200 ohm will achieve.

Since the null point on the last 4^th segment of the potentiometer , therefore potential drop across 400cm of wire >8V

Potential gradient K $\times 400 c m > 8 V$

K×4m>8V

K>2V/m

Similarly potential drop across 300 cm wire <8V

k×3m<8V, K<2×2/3V/m

2×2/3V/m>k>2V/m

3. (a) Consider circuit in Fig 3.10. 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 RI² per second to the thermal energy. What time scale would one associate with energy in problem (a)? n = no of electron/volume = 10²⁹/m3, length of circuit = 10 cm, cross-section = A = (1mm)²

Explanation- according to ohm’s law V= IR

I= 6/6 = 1A

I= AneV_d or V_d= i/neA

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

Explanation- applying kirchhoff’s junction rule I₁ = I+I₂

Applying kirchhoff’s rule in outer loop containing 10V cell

10= IR+10I₁……………..(1)

Applying kirchhoff’s rule in outer loop containing 2V cell

2= 5I₂-RI= 5(I₁-I)-RI

4= 10I₁-10I-RI…………(2)

From 1 and 2

6=3RI+10I

2=I(R+10/3)

V= I(R+R_eff)

After comparing V=2V, R_eff= 10/3 ohm

Since effective internal resistance R_eff of two cells 10/3 ohm , being the parallel combination 5 ohm and 10 ohm . the equivalent circuit is

Commonly asked questions

The circuit in figure shows two cells connected in opposition to each other. Cell E₁is of emf 6 V and internal resistance 2 Ω; the cell E₂ is of emf 4 V and internal resistance 8 Ω. Find the potential difference between the points A and B.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation-

total resistance = 2+8= 10 ohm

I= $\frac{6 - 4}{2 + 8}$ = 0.2A

The direction of flow of current is always from high potential to low potential

If VB $>$ VA

VB-4V-0.2 $*$ 8 = VA

V_B-V_A= 3.6V

In an experiment with a potentiometer, VB = 10 V. R is adjusted to be 50 Ω (figure). A student wanting to measure voltage E₁of a battery (approx. 8 V) finds no null point possible. He then diminishes R to 10 Ω and is able to locate the null point on the-last (4th) segment of the potentiometer. Find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation – let R' be the resistance of potentiometer wire.

Effective resistance of potentiomter and variable resistor r=50ohm is 500+R'

Effective voltage across potentiometer = 10V

The current through main circuit I= $\frac{V}{50 + R}$ = $\frac{10}{50 + R}$

Potential difference across wire of potentiometer

IR'= $\frac{10 R'}{50 + R}$

Since with 50 ohm resistor, null point is not obtained it is possible when

$\frac{10 * R'}{50 + R} < 8$

10R' $< 400 + 8 R'$

2R'<400 or R'<200 ohm

Similarly with 10 ohm resistor, null point is obtained its is only possible when

$\frac{10 * R'}{50 + R} < 8$

2R'>40

R'>40

$\frac{10 * \frac{3}{4} R'}{10 + R'} < 8$

7.5R'<80+8R'

R'>160

160

Any R' between 160 ohm and 200 ohm will achieve.

Since the null point on the last 4^th segment of the potentiometer, therefore potential drop across 400cm of wire >8V

Potential gradient K $* 400 c m > 8 V$

K*4m>8V

K>2V/m

Similarly potential drop across 300 cm wire <8V

k*3m<8V, K<2*2/3V/m

2*2/3V/m>k>2V/m

A room AC runs for 5 a day at a voltage of 220 V. 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?

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation – power consumption in a day i.e in 5 = 10 units

Power consumption per hour = 2 units

Power consumption = 2 units =2KW= 2000J/s

Also power =V $\times$ I

2000W= 220V $\times$ l or l= 9A approx.

R= $\frac{ρ l}{A}$

Power consumption in first current carrying wire

P= I²R

$\frac{ρ l}{A}$ l²= 1.7 $\times$ 10^{-8 $\times \frac{10}{π \times 10^{- 6}} \times 81$}j/s = 4J/s approx.

Loss due to joule heating in first wire = $\frac{4}{2000} \times$ 100=0.2%

Power loss in Al wire =1.6 $\times$ 4= 6.4J/s

Fractional loss due to joule heating in second wire = $\frac{6.4}{2000} \times$ 100= 0.32%

(a) Consider circuit in Fig 3.10. How much energy is absorbed by electrons from the initial state of no current (ignore thermal motion) to the state of drift velocity?

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- according to ohm's law V= IR

I= 6/6 = 1A

I= AneV_d or V_d= i/neA

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

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- applying kirchhoff's junction rule I₁ = I+I₂

Applying kirchhoff's rule in outer loop containing 10V cell

10= IR+10I₁……………. (1)

Applying kirchhoff's rule in outer loop containing 2V cell

2= 5I₂-RI= 5 (I₁-I)-RI

4= 10I₁-10I-RI………… (2)

From 1 and 2

6=3RI+10I

2=I (R+10/3)

V= I (R+R_eff)

After comparing V=2V, R_eff= 10/3 ohm

Since effective internal resistance R_eff of two cells 10/3 ohm, being the parallel combination 5 ohm and 10 ohm . the equivalent circuit is

Is the motion of a charge across junction momentum conserving? Why or why not?

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation – In the circuit when an electron approaches a junction, in addition to the uniform E that faces it normally (which keep the drift velocity fixed), as drift velocity (v_d) is directly proportional to Electric field (E). That's why there are accumulation of charges on the surface of wires at the junction.
These produce additional electric fields. These fields alter the direction of momentum. Thus, the motion of a charge across junction is not momentum conserving

The relaxation time T is nearly independent of applied field E 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 p with temperature. Elaborate why?

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- relaxation time =mean free path/rms velocity of electron

Also $ρ$ = $\frac{1}{σ}$ = $\frac{m}{n e 2 t}$ relaxation time is inversely proportional to velocities

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?

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- In a Wheatstone bridge the main advantage of null point method is that the resistance of galvanometer does not affect the balance point, there is no need to determine current in resistances and the internal resistance of a galvanometer. It is convenient and easy method for observer.
The R unknown can calculated applying Kirchhoff's rules to the circuit. We would need additional accurate measurement of all the currents in resistances and galvanometer and internal resistance of the galvanometer.

The necessary and sufficient condition for balanced Wheatstone bridge is P/Q = R/S where P and Q are ratio arms and R is known resistance and S is unknown resistance.

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

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- Metallic strips have negligible resistance and need not to be counted in the length l₁, of the null point of potentiometer. That's why the thick metallic strips are used in potentiometer. It is for the convenience of experimenter as he measures only their lengths along the straight segments each of lengths 1 m.

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

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- For the selection of metal for wiring in home the main criterion are: the availability, conductivity and the cost of the metal. The Cu wires or Al wires are used for wiring in the home. The main considerations are involved in this process are cost of metal and good conductivity of metal.

Why are alloys used for making standard resistance coils?

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- Alloys are used for making standard resistance coil because they have low temperature coefficient of resistance with less temperature sensitivity. The alloys also have high resistivity and hence high resistance, because for given length and cross-section area of conductor .

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.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation – power consumed through wire is P=V^2/R =I²R . for transmitting the power there are two types either current is large or voltage is less or vice versa. But when current in the circuit is higher then power loss is also higher . so we transmit the power through high voltage circuits.

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

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- If the value of R is increased, the current through the wire will decrease which in turn decreases the potential difference across AB, and hence potential gradient (k) across AB decreases.
Since, at neutral point, for given emf of cell, l increases as potential gradient (k) across AB has decreased because E' = kl
Thus, with the increase of l, which will result in increase in balance length. So, jockey J will shift towards B

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 R; (ii) the deflection increased, while the jockey was moved towards the end D.

(i) Which terminal positive or negative of the cell E₁ is connected at X in case (i) and how is E₁ related to E?
(ii) Which terminal of the cell E₁ is connected at X in case (ii)?

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- If the current in auxiliary circuit (lower circuit containing primary cell) decreases, and potential difference across A and jockey/increases. Then deflection in galvanometer is one sided and the deflection decreased, while moving from one end 'A ' of the wire to the end 'S'.
And clearly this is possible only when positive terminal of the cell E1 is connected at X and E₁>E.
(ii) If the current in auxiliary circuit increases, and potential difference across A and jockey J increases. Then also deflection in galvanometer shows one sided deflection.
And this is possible only when negative terminal of the cell E1 is connected at X

A cell of emf E and internal resistance r is connected across an external resistance R. Plot a graph showing the variation of potential difference across R, versus R.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- according to the relation V=e-Ir and I=e/r+R

The graphical relation between voltage and resistance.

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 / 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 V?

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation – in series combination of resistors current I =e/R+r

10I= $\frac{e}{R + \frac{r}{n}}$

$\frac{1 + n}{1 + \frac{1}{n}}$ $\Rightarrow$ 10 = ( $\frac{1 + n}{1 + n}$ )n $\Rightarrow$ n=10

Let there be n resistors R₁ . . . 𝑅 𝑛 with 𝑅 max = max ( R₁ . . . . 𝑅 ) and R min = min { R₁ . . . R n }. Show that when they are connected in parallel the resultant resistance R p = R min 𝑅 𝑝 = 𝑅 min and when they are connected in series, the resultant resistance R_s > R_max . Interpret the result physically.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- when all resistance are in parallel

$\frac{1}{R_{p}} = \frac{1}{R_{1}} + \dots \dots \dots \dots . . + \frac{1}{R_{n}}$ by multiplying this equation by R_min we have

$\frac{R_{m i n}}{R_{p}} = \frac{R_{m i n}}{R_{1}} + \dots \dots \dots \dots . . + \frac{R_{m i n}}{R_{n}}$

But there exist a term in RHS $\frac{R_{m i n}}{R_{m i n}} = 1$ and other terms are positive so we have

$\frac{R_{m i n}}{R_{p}} = \frac{R_{m i n}}{R_{1}} + \dots \dots \dots \dots . . + \frac{R_{m i n}}{R_{n}}$ >1

This shows that equivalent resistance is less than maximum resistance available.

But when all resistance are in series

R_s =R₁+R₂………R_n

here must be R_max value in RHS

R_s= R₁+……R_max+…..R_n

And R_s> R_max

So equivalent resistance is less than R_max

Two cells of same emf E but internal resistance r₁ and r₂ 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?

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- according to ohms law

I= $\frac{E + E}{R + r 1 + r 2}$

The potential difference across terminal is

V=e-Ir= E- $\frac{2 E}{R + r 1 + r 2}$ r1=0

E= $\frac{2 E r 1}{R + r 1 + r 2}$

1= $\frac{2 r 1}{R + r 1 + r 2}$

R+r1+r2 = 2r1

R= r1-r2

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

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- R= $ρ l$ /A

Resistance of first conductor, RA= $\frac{ρ l}{π {(10}^{- 3} \times 0.5)}$ 2

Resistance of second conductor, RB= $\frac{ρ l}{π {{(10}^{- 6}) - (0.5 \times 0.5 \times 10^{- 6})}}$

Now ratio $\frac{R_{A}}{R_{B}} = \frac{\frac{ρ l}{π {(10}^{- 3} \times 0.5)} 2}{\frac{ρ l}{π {{(10}^{- 6}) - (0.5 \times 0.5 \times 10^{- 6})}}}$ = 3:1

Suppose there is a circuit consisting of only resistances and batteries. Suppose one is to double (or increase it to n-times) all voltages and all. resistances. Show that currents are unaltered. Do this for circuit of Examples 3, 7 in the NCERT Text Book for Class XII.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- according to ohms law

I= $\frac{V_{e f f}}{R_{e f f} + R}$

If voltage and resistance increase

V'= nVeff, R'= nReff

I'= $\frac{n V e f f}{n R e f f + n R} =$ $\frac{V_{e f f}}{R_{e f f} + R}$ =

Consider a current carrying wire (current 7) 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.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b)

Explanation- as we know that J=E, and current density is directly proportional to electric field, so electric field produced by charges accumulated on the surface of wire.

Two batteries of emf e1 and e2 (e2>e1) and internal resistance r1 and r2 respectively are connected in parallel as shown in fig

(a) Two equivalent emf e_eq of the cells is between e1 and e2 i.e. e1eq

(b) The equivalent emf e_eq is samller than e₁

(c) The e_eq is given by e_eq=e₁+e₂ always

(d) E_eq is independent of internal resistance r₁ and r₂

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation- eeq= $\frac{e_{2} r_{1 +} e_{2} r_{1}}{r_{1} + r_{2}}$

A resistance R is to be measured using a meter bridge, a student chooses the standard resistance S to be 100 Ω He finds the null point at l₁= 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 1000 ? and repeat the experiment
(c) He should change S to 3 ? and repeat the experiment
(d) He should have given up hope of a more accurate measurement with a meter bridge.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (c)

Explanation – R/S= (l₁/100-l₁)= 100 (2.9/100-2.9)= 100/97.1=2.98ohm

So he should change S to almost 3 ohm and repeat the experiment.

Two cells of emfs approximately 5 V and 10 V are to be accurately compared using a potentiometer of length 400 cm.4.
(a) The battery that runs the potentiometer should have voltage of 8 V.
(b) The battery of potentiometer can have a voltage of 15 V and R adjusted so that the potential drop across the wire slightly exceeds 10 V.

(c) The first portion of 50 cm of wire itself should have a potential drop of 10 V.
(d) Potentiometer is usually used for comparing resistances and not voltages.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b)

Explanation-The potential drop along the wires of potentiometer should be greater than emfs of cells.
In a potentiometer experiment, the emf of a cell can be measured if the potential drop along the potentiometer wire is more than the emf of the cell to be determined. Here, values of emfs of two cells are given as 5 V and 10 V, therefore, the potential drop along the potentiometer wire must be more than 10 V.

A metal rod of length 10 cm and a rectangular cross-section of 1 cm x 1/2 cm is connected to a battery across opposite faces. The resistance will be
(a) Maximum when the battery is connected across 1 cm x 1/2 cm faces
(b) Maximum when the battery is connected across 10 cm x 1 cm faces
(c) Maximum when the battery is connected across 10 cm x 1/2 cm faces
(d) Same irrespective of the three faces

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation- R= $\frac{δ l}{A}$ for greater value of R, A should be less and its possible value when connected across 1cm $\times$ 1/2cm faces.

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

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation- I=AneV_d , current is directly proportional to drift velocity.

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.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (b, d)

Explanation-Algebraic sum of the currents flowing towards any point in an electric network is zero. It also tell us about the conservation of charge.

Consider a simple circuit shown in figure stands for a variable resistance R’. R’ can vary from R₀to infinity, r is internal resistance of the battery (r <(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 ≥ V/r + R always;

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (a, d)

Explanation- potential drop across AB is independent to R' also when we decrease the value of R the current will also very large. According to the relation I=e/R+r

Temperature dependence of resistivity ρ(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.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Explanation- according to the relation $ρ$ = $\frac{m}{n e 2 t}$ also T $\propto \frac{1}{t}$

The measurement of an unknown resistance R is to be carried out using Wheatstone bridge as given in the figure. Two students perform an experiment in two ways. The first student takes R₂= 10 Ω and R₁ = 5 Ω. The other student takes R₂ = 1000 Ω, and R₁ = 500 Ω. In the standard arm, both take R₃ = 5 Ω.
Both find R = R₂/R₁, R₃ = 100 Ω 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 R₂ and R₁ can be measured
(c) If the student uses large values of R₂ and R₁, 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 measurementAnswer – (b,c)

Answer – (b, c)

Explanation – According to the relation that p/q=r/s if galvanometer shows no deflection. In the above equation $\frac{R 1}{R 2} = \frac{R 3}{R 4}$ the value of R4 depends upon R1 and R2 . if the value of R1 and R2 will be feeble the value of R4 would be affected.

In a meter bridge, the point D is a neutral point (figure).
(a) The meter bridge can have no other neutral. A 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

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (a, c)

Explanation- In case of meter bridge, the resistance wire AC is 100 cm long. Varying the position of tapping point B, bridge is balanced. If in balanced position of bridge AB = l, BC = (100 – l) so that Q/P= (100-l)/l. Also P/Q=R/S=>S= (100-l)/l R
When there is no deflection in galvanometer there is no current across the galvanometer, then points B and D are at same potential. That point at which galvanometer shows no deflection is called null point, then potential at B and neutral point D are same. When the jockey contacts a point on the meter wire to the right of D, the potential drop across AD is more than potential drop across AB, which brings the potential of point D less than that of B, hence current flows from B to D in the galvanometer wire.

Chapters List

Chapters Name
Electric Charges And Fields	Electrostatic Potential And Capacitance	Current Electricity	Moving Charges And Magnetism	Magnetism And Matter
Electromagnetic Induction	Alternating Current	Electromagnetic Waves	Ray Optics And Optical Instruments	Wave Optics
Dual Nature Of Radiation And Matter	Atoms	Nuclei	Semiconductor Electronics	Communication Systems

Current Electricity Short Answer Type Questions

For reinforcing fundamental concepts like internal resistance, Ohm's law, and resistance in series/parallel, practicing such short-answer questions is ideal. They make students solve the numerals quickly and also make them clearer and precise while expressing concise explanations. Regular practice of the short-answer type questions helps students handle time-bound theoretical questions efficiently and gives them confidence in recalling key points.

For understanding the important topics of Class 11 NCERT Solutions Physics, and downloading the free PDFs, read here NCERT Solutions Physics Class 11th.

1. Is the motion of a charge across junction momentum conserving? Why or why not?

Explanation – In the circuit when an electron approaches a junction, in addition to the uniform E that faces it normally (which keep the drift velocity fixed), as drift velocity (v_d) is directly proportional to Electric field (E). That’s why there are accumulation of charges on the surface of wires at the junction.
These produce additional electric fields. These fields alter the direction of momentum. Thus, the motion of a charge across junction is not momentum conserving

2. The relaxation time T is nearly independent of applied field E 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 p with temperature. Elaborate why?

Explanation- relaxation time =mean free path/rms velocity of electron

Also $ρ$ = $\frac{1}{σ}$ = $\frac{m}{n e 2 t}$ relaxation time is inversely proportional to velocities

3. 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?

Explanation- In a Wheatstone bridge the main advantage of null point method is that the resistance of galvanometer does not affect the balance point, there is no need to determine current in resistances and the internal resistance of a galvanometer. It is convenient and easy method for observer.
The R unknown can calculated applying Kirchhoff’s rules to the circuit. We would need additional accurate measurement of all the currents in resistances and galvanometer and internal resistance of the galvanometer.

The necessary and sufficient condition for balanced Wheatstone bridge is P/Q = R/S where P and Q are ratio arms and R is known resistance and S is unknown resistance.

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

Explanation- Metallic strips have negligible resistance and need not to be counted in the length l₁, of the null point of potentiometer. That’s why the thick metallic strips are used in potentiometer. It is for the convenience of experimenter as he measures only their lengths along the straight segments each of lengths 1 m.

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

6. Why are alloys used for making standard resistance coils?

7. 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.

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

Explanation- If the value of R is increased, the current through the wire will decrease which in turn decreases the potential difference across AB, and hence potential gradient (k) across AB decreases.
Since, at neutral point, for given emf of cell, l increases as potential gradient (k) across AB has decreased because E’ = kl
Thus, with the increase of l, which will result in increase in balance length. So, jockey J will shift towards B

9. 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 R; (ii) the deflection increased, while the jockey was moved towards the end D.

Explanation- If the current in auxiliary circuit (lower circuit containing primary cell) decreases, and potential difference across A and jockey/increases. Then deflection in galvanometer is one sided and the deflection decreased, while moving from one end ‘A ’ of the wire to the end ‘S’.
And clearly this is possible only when positive terminal of the cell E1 is connected at X and E₁>E.
(ii) If the current in auxiliary circuit increases, and potential difference across A and jockey J increases. Then also deflection in galvanometer shows one sided deflection.
And this is possible only when negative terminal of the cell E1 is connected at X

10. A cell of emf E and internal resistance r is connected across an external resistance R. Plot a graph showing the variation of potential difference across R, versus R.

Explanation- according to the relation V=e-Ir and I=e/r+R

The graphical relation between voltage and resistance.

11. 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 / 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 V?

Explanation – in series combination of resistors current I =e/R+r

10I= $\frac{e}{R + \frac{r}{n}}$

$\frac{1 + n}{1 + \frac{1}{n}}$ $\Rightarrow$ 10 = ( $\frac{1 + n}{1 + n}$ )n $\Rightarrow$ n=10

12. The circuit in figure shows two cells connected in opposition to each other. Cell E₁is of emf 6 V and internal resistance 2 Ω; the cell E₂ is of emf 4 V and internal resistance 8 Ω. Find the potential difference between the points A and B.

Explanation-

total resistance = 2+8= 10 ohm

I= $\frac{6 - 4}{2 + 8}$ = 0.2A

The direction of flow of current is always from high potential to low potential

If VB $>$ VA

VB-4V-0.2 $\times$ 8 = VA

V_B-V_A= 3.6V

13. Let there be n resistors R₁ . . . 𝑅 𝑛 with 𝑅 max = max ( R₁ . . . . 𝑅 ) and R min = min { R₁ . . . R n }. Show that when they are connected in parallel the resultant resistance R p = R min 𝑅 𝑝 = 𝑅 min and when they are connected in series, the resultant resistance R_s > R_max . Interpret the result physically.

Explanation- when all resistance are in parallel

$\frac{1}{R_{p}} = \frac{1}{R_{1}} + \dots \dots \dots \dots . . + \frac{1}{R_{n}}$ by multiplying this equation by R_min we have

$\frac{R_{m i n}}{R_{p}} = \frac{R_{m i n}}{R_{1}} + \dots \dots \dots \dots . . + \frac{R_{m i n}}{R_{n}}$

But there exist a term in RHS $\frac{R_{m i n}}{R_{m i n}} = 1$ and other terms are positive so we have

$\frac{R_{m i n}}{R_{p}} = \frac{R_{m i n}}{R_{1}} + \dots \dots \dots \dots . . + \frac{R_{m i n}}{R_{n}}$ >1

This shows that equivalent resistance is less than maximum resistance available.

But when all resistance are in series

R_s =R₁+R₂………R_n

here must be R_max value in RHS

R_s= R₁+……R_max+…..R_n

And R_s> R_max

So equivalent resistance is less than R_max

14. Two cells of same emf E but internal resistance r₁ and r₂ 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?

Explanation- according to ohms law

I= $\frac{E + E}{R + r 1 + r 2}$

The potential difference across terminal is

V=e-Ir= E- $\frac{2 E}{R + r 1 + r 2}$ r1=0

E= $\frac{2 E r 1}{R + r 1 + r 2}$

1= $\frac{2 r 1}{R + r 1 + r 2}$

R+r1+r2 = 2r1

R= r1-r2

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

Explanation- R= $ρ l$ /A

Resistance of first conductor , RA= $\frac{ρ l}{π {(10}^{- 3} \times 0.5)}$ 2

Resistance of second conductor , RB= $\frac{ρ l}{π \{{(10}^{- 6}) - (0.5 \times 0.5 \times 10^{- 6})}}$

Now ratio $\frac{R_{A}}{R_{B}} = \frac{\frac{ρ l}{π {(10}^{- 3} \times 0.5)} 2}{\frac{ρ l}{π \{{(10}^{- 6}) - (0.5 \times 0.5 \times 10^{- 6})}}}$ = 3:1

16. Suppose there is a circuit consisting of only resistances and batteries. Suppose one is to double (or increase it to n-times) all voltages and all. resistances. Show that currents are unaltered. Do this for circuit of Examples 3, 7 in the NCERT Text Book for Class XII.

Explanation- according to ohms law

I= $\frac{V_{e f f}}{R_{e f f} + R}$

If voltage and resistance increase

V’= nVeff , R’= nReff

I’= $\frac{n V e f f}{n R e f f + n R} =$ $\frac{V_{e f f}}{R_{e f f} + R}$ = I

Current Electricity Multiple Choice Questions

1. Consider a current carrying wire (current 7) 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)

Explanation- as we know that J=E , and current density is directly proportional to electric field , so electric field produced by charges accumulated on the surface of wire.

2. Two batteries of emf e1 and e2 (e2>e1) and internal resistance r1 and r2 respectively are connected in parallel as shown in fig

(a) Two equivalent emf e_eq of the cells is between e1 and e2 i.e. e1 eq

(b) The equivalent emf e_eq is samller than e₁

(d) E_eq is independent of internal resistance r₁ and r₂

Answer- (a)

Explanation- eeq= $\frac{e_{2} r_{1 +} e_{2} r_{1}}{r_{1} + r_{2}}$

3. A resistance R is to be measured using a meter bridge, a student chooses the standard resistance S to be 100 Ω He finds the null point at l₁= 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 1000 Ω and repeat the experiment
(c) He should change S to 3 Ω and repeat the experiment
(d) He should have given up hope of a more accurate measurement with a meter bridge.

Answer –(c)

Explanation – R/S=(l₁/100-l₁)= 100(2.9/100-2.9)= 100/97.1=2.98ohm

So he should change S to almost 3 ohm and repeat the experiment.

4. Two cells of emfs approximately 5 V and 10 V are to be accurately compared using a potentiometer of length 400 cm.4.
(a) The battery that runs the potentiometer should have voltage of 8 V.
(b) The battery of potentiometer can have a voltage of 15 V and R adjusted so that the potential drop across the wire slightly exceeds 10 V.

(c) The first portion of 50 cm of wire itself should have a potential drop of 10 V.
(d) Potentiometer is usually used for comparing resistances and not voltages.

Answer- (b)

5. A metal rod of length 10 cm and a rectangular cross-section of 1 cm x 1/2 cm is connected to a battery across opposite faces. The resistance will be
(a) maximum when the battery is connected across 1 cm x 1/2 cm faces
(b) maximum when the battery is connected across 10 cm x 1 cm faces
(c) maximum when the battery is connected across 10 cm x 1/2 cm faces
(d) same irrespective of the three faces

Answer-(a)

Explanation- R= $\frac{δ l}{A}$ for greater value of R , A should be less and its possible value when connected across 1cm $\times$ 1/2cm faces.

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-(a)

Explanation- I=AneV_d , current is directly proportional to drift velocity.

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 – (b,d)

Explanation-Algebraic sum of the currents flowing towards any point in an electric network is zero. It also tell us about the conservation of charge.

8. Consider a simple circuit shown in figure stands for a variable resistance R’. R’ can vary from R₀to infinity, r is internal resistance of the battery (r < (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 ≥ V/r + R always;

Answer – (a,d)

Explanation- potential drop across AB is independent to R’ also when we decrease the value of R the current will also very large. According to the relation I=e/R+r

9. Temperature dependence of resistivity ρ(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.

Explanation- according to the relation $ρ$ = $\frac{m}{n e 2 t}$ also T $\propto \frac{1}{t}$

10. The measurement of an unknown resistance R is to be carried out using Wheatstone bridge as given in the figure. Two students perform an experiment in two ways. The first student takes R₂= 10 Ω and R₁ = 5 Ω. The other student takes R₂ = 1000 Ω, and R₁ = 500 Ω. In the standard arm, both take R₃ = 5 Ω.
Both find R = R₂/R₁, R₃ = 100 Ω 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 R₂ and R₁ can be measured
(c) If the student uses large values of R₂ and R₁, 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 measurementAnswer – (b,c)

Answer – (b,c)

Explanation – According to the relation that p/q=r/s if galvanometer shows no deflection. In the above equation $\frac{R 1}{R 2} = \frac{R 3}{R 4}$ the value of R4 depends upon R1 and R2 . if the value of R1 and R2 will be feeble the value of R4 would be affected.

11. In a meter bridge, the point D is a neutral point (figure).
(a) The meter bridge can have no other neutral. A 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 –(a,c)

Explanation- In case of meter bridge, the resistance wire AC is 100 cm long. Varying the position of tapping point B, bridge is balanced. If in balanced position of bridge AB = l, BC = (100 – l) so that Q/P= (100-l)/l. Also P/Q=R/S=>S=(100-l)/l R
When there is no deflection in galvanometer there is no current across the galvanometer, then points B and D are at same potential. That point at which galvanometer shows no deflection is called null point, then potential at B and neutral point D are same. When the jockey contacts a point on the meter wire to the right of D, the potential drop across AD is more than potential drop across AB, which brings the potential of point D less than that of B, hence current flows from B to D in the galvanometer wire.

24th June 2022 (Second Shift)

Commonly asked questions

Identify the pair of physical quantities that have same dimensions

Velocity gradient = $\frac{d v}{d x} = \frac{L T^{- 1}}{L} = T^{- 1}$

Decay constant $(λ) = \frac{0.693}{T_{\frac{1}{2}}} = T^{- 1}$

The distance between Sun and Earth is R. The duration of year if the distance between Sun and Earth becomes 3R will be

$T^{2} α R^{3}$

$\begin{array}{l} \Rightarrow \frac{{T_{1}}^{2}}{{T_{2}}^{2}} = \frac{{R_{1}}^{3}}{{R_{2}}^{3}} \\ \Rightarrow \frac{{T_{1}}^{2}}{{T_{2}}^{2}} = \frac{R^{3}}{{(3 R)}^{3}} \\ T_{2} = 3 \sqrt[]{3} y e a r s \end{array}$

A stone of mass m, tied to a string is being whirled in a vertical circle with a uniform speed. The tension in the string is

Is minimum at the highest position of the circular path.

Two identical charged particles each having a mass 10g and charge 2.0 × 10^-7 C are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g = 10ms^-2]

$μ m g = \frac{k q_{1} q_{2}}{L^{2}}$

$\Rightarrow 0.25 \times 0.01 \times 10 = \frac{9 \times 1 0^{9} \times 2 \times 1 0^{- 7} \times 2 \times 1 0^{- 7}}{L^{2}}$

$= \frac{9 \times 2^{2} \times 1 0^{- 5 + 4}}{25 \times 10}$

L = 12 cm

A carnot engine takes 5000 kcal of heat from a reservoir at $727 ° C$ and gives heat to a sink at $127 ° C .$ The work done by the engine is

T₁ = 727 + 273 = 1000k

T₂ = 127° + 273 = 400 k

Q₁ = 5 × 10³ k cal

$\Rightarrow \frac{600}{1000} = \frac{w}{5000}$

w = 12.6 × 10⁶ J

Two massless springs with spring constants 2k and 9k, carry 50g and 100g masses at their free ends. These two masses oscillate vertically such that their maximum velocities are equal. Then the ratio of their respective amplitudes will

$ω_{1} = \sqrt[]{\frac{k}{m}} = \sqrt[]{\frac{2 k}{0.05}}$

$\begin{array}{l} \Rightarrow A_{1} ω_{1} = A_{2} ω_{2} \\ \frac{A_{1}}{A_{2}} = \frac{ω_{2}}{ω_{1}} = \sqrt[]{\frac{9 k}{0.1}} \times \sqrt[]{\frac{0.05}{2 k}} \end{array}$

$\frac{A_{1}}{A_{2}} = \frac{3}{2}$

What will be the most suitable combination of three resistors A = $2 Ω$ , B = $4 Ω$ , C = $6 Ω$ so that $(\frac{22}{3}) Ω$ is equivalent resistance of combination

Given

$\begin{array}{l} A = 2 Ω \\ B = 4 Ω \\ C = 6 Ω \\ R_{e q} = \frac{2 \times 4}{2 + 4} + 6 \end{array}$

$R_{e q} = \frac{22}{3} Ω$

The soft-iron is a suitable material for making an electromagnet. This is because soft iron has

High permeability and low retentivity

A proton, a deuteron and an α - particle with same kinetic energy enter into a uniform magnetic field at right angle to magnetic field. The ratio of the radii of their respective circular paths is

$r = \frac{m v}{q B} = \frac{\sqrt[]{2 m k . E}}{q B}$

$r \propto \frac{\sqrt[]{m}}{q}$

m_p = m

q_p = q

m_d = 2m

q_d = q

m_a = 4m

q_a = 2q

$\frac{r_{p}}{r_{d}} = \frac{\sqrt[]{m_{p}} \times q_{d}}{q_{P} \times \sqrt[]{m_{d}}}$

$\frac{r_{d}}{r_{α}} = \sqrt[]{2}$

$r_{p} : r_{d} : r_{α} = 1 : \sqrt[]{2} : 1$

Statement – I : The reactance of an ac circuit is zero. It is possible that the circuit contains a capacitor and an inductor.

Statement – II : In ac circuit, the average power delivered by the source never becomes zero.

In the light of the above statements, choose the correct answer from the options given below

Z = $\sqrt[]{R^{2} + {(x_{C} - x_{L})}^{2}},$ if only L & C are present then R = 0 then p = 0

Potential energy as a function of r is given by $U = \frac{A}{r^{10}} - \frac{B}{r^{5}}$ , where r is the interatomic distance, A and B are positive constant. The equilibrium distance between the two atoms will be

$\frac{d U}{d r} = 0 \Rightarrow \frac{d}{d r} [\frac{A}{r^{10}}] - \frac{d}{d r} [\frac{B}{r^{5}}]$

$\Rightarrow \frac{d}{d r} [A r^{- 10}] - \frac{d}{d r} [B r^{- 5}] = 0$

$\Rightarrow - 10 A r^{- 11} + 5 B r^{- 6} = 0$

$r = {(\frac{2 A}{B})}^{\frac{1}{5}}$

An object of mass 5kg is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of 10 N throughout the motion. The ratio of time of ascent to the time of descent will be equal to : [Use g = 10ms^-2].

$a = \frac{m g + 10}{g} = \frac{5 \times 10 + 10}{10}$

$\Rightarrow a = 6 m / s e c^{2}$

v = u + at

0 = u – 6t

$t_{A} = \frac{u}{6}$ (time of Ascent)

$= u \times \frac{u}{6} - \frac{u}{2} \times 6 \times \frac{u}{6} \times \frac{u}{6}$

= $\frac{u^{2}}{6} - \frac{u^{2}}{12}$

$h = \frac{u^{2}}{12}$

for time of descent

$a = \frac{50 - 10}{10}$

a = 4 m /sec²

$\Rightarrow \frac{u^{2}}{12} = 0 \times t + \frac{1}{2} \times 4 t^{2}$

$t_{B} = \frac{u}{\sqrt[]{24}}$

$\frac{t_{A}}{t_{B}} = \frac{u \times 24}{6 \times u} =$ $\sqrt[]{2}$

A fly wheel is accelerated uniformly from rest and rotates through 5 rad in the first second. The angle rotated by the fly wheel in the next second, will be

$θ = ω_{0} t + \frac{1}{2} α t^{2}$

$5 = 0 \times 1 + \frac{1}{2} \times α \times 1^{2}$

= 10 rad/sec²

(ln 2 seconds) = $ω_{0} t + \frac{1}{2} α t^{2}$

$θ_{2} = 20 r a d$

Angle rotated by wheel in 2^nd second

= $θ_{2} - θ$

= 20 – 5

= 15 rad

A 100g of iron nail is hit by a 1.5 kg hammer striking at a velocity of 60ms^-1. What will be the rise in the temperature of the nail if one fourth of energy of the hammer goes into heating the nail?

Heating energy $= \frac{1}{4} t h$ energy of the hammer

$= \frac{1}{4} \times \frac{1}{2} m v^{2}$

$= \frac{1}{4} \times \frac{1}{2} \times 1.5 \times 60 \times 60$

$= \frac{15 \times 6 \times 60}{4 \times 2}$

= 675 J

$Δ T = \frac{675}{42} = 16.07 ° C$

If the charge on a capacitor is increased by 2C, the energy stored in it increases by 44%. The original charge on the capacitor is (in C)

$U_{i} = \frac{Q^{2}}{2 C}$

Uf = 1.44 Ui

Qf = Q + 2

$1.44 U_{i} = \frac{{(Q + 2)}^{2}}{2 C}$

$\Rightarrow 1.44 \frac{Q^{2}}{2 C} = \frac{{(Q + 2)}^{2}}{2 C}$

$\Rightarrow 1.2 Q = Q + 2$

0.2 Q = 2

Q = 10 C

A long cylindrical volume contains a uniformly distributed charge of density $ρ$ . The radius of cylindrical volume is R. A charge particle (q) revolves around the cylinder in a circular path. The kinetic energy of the particle is

$F_{C} = F_{q}$

$\Rightarrow \frac{m v^{2}}{R} = \frac{ρ R}{2 \in_{0}} q$

$m v^{2} = \frac{ρ R^{2} q}{2 ε_{0}} \Rightarrow \frac{1}{2} m v^{2} = k . E . = \frac{1}{4} \frac{ρ R^{2} q}{\in_{0}}$

An electric bulb is rated as 200W. What will be the peak magnetic field at 4m distance produced by the radiations coming from this bulb? Consider this bulb as a point source with 3.5% efficiency.

I_R = I_B (due to magnetic field)

(due to Radiation)

$\frac{P}{A} \times E f f i c i e n c y = \frac{B_{0}^{2}}{2 μ_{0}} c$

$\Rightarrow \frac{200}{4 π \times 4^{2}} \times \frac{3.5}{100} = \frac{B_{0}^{2}}{2 \times 4 π \times 1 0^{- 7}} \times 3 \times 1 0^{8}$

$\Rightarrow B_{0} = 1.71 \times 1 0^{- 8} T$

The light of two different frequencies whose photons have energies 3.8 eV and 1.4 eV respectively, illuminate a metallic surface whose work function is 0.6 eV successively. The ratio of maximum speed of emitted electrons for the two frequencies respectively will be

$h υ_{1} = h υ_{0} + K . E_{1}$

$\Rightarrow 3.8 e V = 0.6 e V + K . E_{1}$

K. E₁ = 3.2 eV (i)

$\Rightarrow h υ_{2} = h υ_{01} + K . E_{2}$

1.4 = 0.6 + K.E₂

KE₂ = 0.8 eV (ii)

$\frac{K . E_{1}}{K . E_{2}} = \frac{3.2}{0.8} = 4$

$\frac{v_{1}}{v_{2}} = 2 : 1$

Two light beams of intensities in the ratio of 9 : 4 are allowed to interfere. The ratio of the intensity of maxima and minima will be

$\frac{l_{1}}{I_{2}} = \frac{9}{4}$ $\frac{l_{m a x}}{l_{m a x}} = \frac{l_{1} + l_{2} + 2 \sqrt[]{l_{1} l_{2}}}{l_{1} + l_{2} - 2 \sqrt[]{l_{1} l_{2}}}$

$= \frac{\frac{9}{4} + 1 + 2 \sqrt[]{\frac{9}{4}}}{\frac{9}{4} + 1 - 2 \sqrt[]{\frac{9}{4}}}$

$= \frac{\frac{13}{4} + \frac{2 \times 3}{2}}{\frac{13}{4} - 2 \times \frac{3}{2}} = \frac{25}{1}$

In Bohr’s atomic model of hydrogen, let K, P and E are the kinetic energy, potential energy and total energy of the electron respectively. Choose the correct option when the electron undergoes transitions to a higher level

Based on theory

A body is projected from the ground at an angle of $45 °$ with horizontal. Its velocity after 2s is 20 ms^-1. The maximum height reached by the body during its motion is ______m. (use g = 10ms^-2)

after 2 sec, v = 20 m/sec

$\begin{array}{l} \vec{v} = u c o s 45 ° \hat{i} + (u s i n 45 ° - g t) \hat{j} \\ v = 20 = \sqrt[]{{(u c o s 45 °)}^{2} + {(u s i n 45 ° - 20)}^{2}} \end{array}$

$\Rightarrow 400 = u^{2} + 400 - 40 u s i n 45 °$

u = 40 sin 45°

$H_{m a x} = \frac{u^{2} s i n^{2} 45}{2 g} = \frac{4 0^{2} s i n^{2} 45 ° \times s i n^{2} 45}{2 g}$

$= \frac{40 \times 40}{2 \times 10} \times \frac{1}{2} \times \frac{1}{2} = 20 m$

An antenna is placed in a dielectric medium of dielectric constant 6.25. If the maximum size of that antenna is 5.0 mm, it can radiate a signal of minimum frequency of ________ GHz.

(Given µ_r = 1 for dielectric medium)

$v = \frac{C}{\sqrt[]{μ_{r} ε_{r}}} = \frac{3 \times 1 0^{8}}{\sqrt[]{6.25 \times 1}} = \frac{3 \times 1 0^{8}}{2.5} = 1.25 \times 1 0^{8} m / s e c$

$A s f λ = f (5 \times 1 0^{- 3} \times 4) = 1.25 \times 1 0^{8} \Rightarrow f = 6.25 G H z$

So, $f \approx 6 .$

A potentiometer wire of length 10m and resistance 20 $Ω$ is connected in series with a 25 V battery and an external resistance $30 Ω$ . A cell of emf E in secondary circuit is balanced by 250cm long potentiometer wire. The value of E (in volt) is $\frac{x}{10} .$ The value of x is ______.

AB = 10 m

$R_{A B} = 20 Ω$

$L_{A C} = 250 c m$

= 2.5 m

$l = \frac{25}{20 + 30}$

$E = \frac{x}{10} = \frac{25}{10}$

Two travelling waves of equal amplitudes and frequencies move in opposite directions along a string. They interfere to produce a stationary wave whose equation is given by y = $(10 c o s π x s i n \frac{2 π t}{T}) c m$

The amplitude of the particle at x = $\frac{4}{3}$ cm will be _______cm.

$y = (10 c o s π x s i n \frac{2 π t}{T}) c m$

$a t x = \frac{4}{3}$

$y = 10 c o s (\frac{4 π}{3}) s i n (\frac{2 π t}{T}) c m$

$= - 5 s i n (\frac{2 π t}{T}) c m$

Then Amplitude will be 5 cm.

In the given circuit, the value of current I_Lwill be _________mA. (When R_L = 1 k $Ω$ )

$I_{L} = \frac{V}{R_{L}} = \frac{5}{1 k Ω}$

$l_{L} = 5 \times 1 0^{- 3} A = 5 m A$

A sample contains 10^-2kg each of two substances A and B with half lives 4s and 8s respectively. The ration of their atomic weights is 1 : 2. The ratio of the amounts of A and B after16s is $\frac{x}{100}$ . The value of x is ______.

$f o r A, t_{\frac{1}{2}} = 4 s e c$

$= m A_{0} e^{-} \frac{0.693}{4} \times 16$

$m_{A} = m_{A 0} e^{- 4 \times 0.693}$ - (1)

for B,

for B, $t_{\frac{1}{2}} = 8 s e c$

$m_{B} = m_{B_{0}} e^{- 2 \times 0.693}$ - (2)

$\frac{m_{A}}{m_{B}} = \frac{m_{A O}}{m_{B O}} \frac{e^{- 4} \times 0.693}{e^{- 2} \times 0.693}$

$\frac{m_{A}}{m_{B}} = \frac{25}{100} = \frac{x}{100} x = 25$

A ray of light is incident at an angle of incidence $60 °$ on the glass slab of refractive index $\sqrt[]{3} .$ After refraction, the light ray emerges out from other parallel faces and lateral shift between incident ray and emergent ray is $4 \sqrt[]{3}$ cm. The thickness of the glass slab is _______cm.

$d = 4 \sqrt[]{3}$ cm (Lateral shift)

By Snell’s law

$μ_{a i r} s i n 60 ° = μ_{g} s i n θ$

$\Rightarrow 1 \times \frac{\sqrt[]{3}}{2} = \sqrt[]{3} s i n θ$

$s i n θ = \frac{1}{2}$

= 30°

$t a n 30 ° = \frac{d}{t} = \frac{4 \sqrt[]{3}}{t}$

$\Rightarrow \frac{1}{\sqrt[]{3}} = \frac{4 \sqrt[]{3}}{t}$

t = 12 cm

A circular coil of 1000 turns each with area 1m² is rotated about its vertical diameter at the rate of one revolution per second in a uniform horizontal magnetic field of 0.07 T. The maximum voltage generation will be ________V.

given N = 1000 turns

A = 1m²

$ω = 1 r e v / s e c$

$= 1 \times 2 π r a d / s e c$

$V = \frac{- d ?_{B}}{d t} = \frac{- d}{d t} (N B A c o s ω t)$

= $140 \times \frac{22}{7}$

= 20 × 22

= 440 volts

A monoatomic gas performs a work of $\frac{Q}{4}$ where Q is the heat supplied to it. The molar heat capacity of the gas will be ________ R during this transformation. Where R is the gas constant.

$Q = Δ υ + w \Rightarrow n C Δ T = n C_{v} Δ T = \frac{n C Δ T}{4} = \frac{3}{4} n C Δ T = n C_{v} Δ T$

$C = \frac{4}{3} C_{v} = \frac{4}{3} \times \frac{3}{2} R = 2 R$

In an experiment to verigy Newton’s law of cooling, a graph is plotted between, the temperature difference $(Δ T)$ of the water and surrounding and time as shown in figure. The initial temperature of water is taken as $80 ° C .$ The value of t₂ as mentioned in the graph will be _______.

$\frac{d T}{d t} = - k (T - T_{0})$

$\frac{d T}{(T - T_{0})} = - k d t$

$\Rightarrow k = - \frac{1}{6} l n (\frac{2}{3})$ - (1)

Now $\frac{d T}{d t} = - k (T - T_{0})$

t = 10.257

t₂ = 6 + 10.257 = 16.257 minutes

16.25 and or 16

JEE Mains Solutions 2022,28th june , Physics ,Second shift

Commonly asked questions

Velocity $(υ)$ and acceleration (a) in two system of units 1 and 2 are related as $υ_{2} = \frac{n}{m^{2}} υ_{1} a n d a_{2} = \frac{a_{1}}{m n}$ respectively. Here m and n are constants. The relations for distance and time in two system respectively are:

Let, L₁ & L₂ one distance for system (1) & (2) respectively

T₁ & T₂ are time for system (1) & (2) respectively

Given : v2 = $\frac{n}{m^{2}} v_{1}$

$\Rightarrow (\frac{L_{2}}{T_{2}}) = \frac{n}{m_{2}} (\frac{4}{T_{1}})$ $[? v_{2} = \frac{L_{2}}{T_{2}}, v_{1} = \frac{L_{1}}{T_{1}}]$

$\Rightarrow \frac{L_{2}}{L_{1}} = (\frac{n}{m_{2}}) [\frac{T_{2}}{T_{1}}] - - (i)$

And a2 = $\frac{a_{1}}{m n} \Rightarrow \frac{v_{2}}{T_{2}} = \frac{v_{1}}{T_{1} m n}$

$\Rightarrow \frac{T_{1}}{T_{2}} = \frac{v_{1}}{v_{2}} (\frac{1}{m n})$ $[? \frac{v_{1}}{v_{2}} = \frac{m^{2}}{n}]$

$\Rightarrow \frac{n^{2}}{m} T_{1} = T_{2}$

from (1)

$\frac{L^{2}}{L_{1}} = \frac{n}{m^{2}} [\frac{n^{2}}{m}] [? \frac{T_{2}}{T_{1}} = \frac{n^{2}}{m}]$

$\Rightarrow \frac{n^{3}}{m^{3}} L_{1} = L_{2}$

A ball is spun with angular acceleration a = 6t² – 2t where t is in second and a is in rads^-¹. At t = 0, the ball has angular velocity of 10 rads^-¹ and angular position of 4 rad. The most appropriate expression for the angular position of the ball is:

a = 6t² – 2t, of t 20, w = 10 rad/sec

, q = 4 rad

$α = \frac{d ω}{d t}$

$\Rightarrow \int_{10}^{ω} d ω = \int_{t - 0}^{t} α d t$

$\Rightarrow ω - 10 = \int_{0}^{t} (6 t^{2} - 2 t) d t$

$ω = \frac{d θ}{d t}$

$d θ = ω d t$

$\int_{4}^{θ} d θ = \int_{0}^{t} (2 t^{3} - t^{2} + 10) d t$

= $\frac{t^{4}}{2} - \frac{t^{3}}{3} + 10 t + 4$

A block of mass 2kg moving on a horizontal surface with speed of 4 ms^-¹ enters a rough surface ranging from x = 0.5m to x = 1.5m. The retarding force in this range of rough surface is related to distance by F = -kx where k = 12Nm^-¹. The speed of the block as it just crosses the rough surface will be:

by, WET $\int F d x = Δ k . 6$

$\int - k x d x = \frac{1}{2} m [v_{t}^{2} - v_{i}^{2}]$

$\Rightarrow \frac{- k}{2} {[x_{f}^{2} - x_{i}^{2}]}_{0.5}^{1.5} = \frac{1}{2} \times 2 [v_{f}^{2} - 16]$ $[? v_{i} = 4 m / s e c]$

$\frac{- 12}{2} [1 . 5^{2} - 0 . 5^{2}] = [v_{f}^{2} - 16]$

$v_{f}^{2} = 16 - 12 = 4$

vf = 2m/sec

A $\sqrt[]{34} m$ long ladder weighing 10kg leans on a frictionless wall. Its feet rest on the floor 3m away from the wall as shown in the figure. If F_f and F_w are the reaction forces of the floor and the wall, then ratio of F_w / F_f will be:

(Use g = 10 m/s²)

mg = 10 × 10

= 100 of

NA = $F_{ω}$ reaction force offered by the wall

$\sqrt[]{N_{3}^{2} + f_{B}^{2}} = F_{f}$ reaction force offered by the floor.

By NLM 1

Translational fB = NA

NB = mg = 100N

Ac² + Bc² = AB²

Ac² = 34 – 9

Ac = 5m

Rotational equilibrium $τ_{B} = 0$

mg $\frac{l}{2} c o s θ = N_{A} l s i n θ$

$100 \frac{c o s θ}{2} = N_{A} s i n θ$

NA = 50 cot

Now cot = $\frac{3}{5}$

N_A= $50 \times \frac{3}{5} = 30 N$

Therefore, NA = $F_{ω} = 30 N$ - (i)

$F_{f} = \sqrt[]{N_{B}^{2} + f_{B}^{2}} = \sqrt[]{10 0^{2} + 3 0^{2}} = 10 \sqrt[]{109} N$ - (2)

$\frac{f_{ω}}{F_{f}} = \frac{30}{10 \sqrt[]{109}} = \frac{3}{109}$

Water falls from a 40m height dam at the rate of 9 × 10⁴ kg per hour. Fifty percentage of gravitational potential energy can be converted into electrical energy. Using this hydro electric energy number of 100 W lamps, that can be lit, is:

(Take g = 10 ms^-2)

Given : water flow rate = $(\overset{o}{m})$ = 9 × 10⁴kg/hr

= $\frac{9 \times 1 0^{4} k g / s e c}{60 \times 60}$

Potential energy available on water per unit time

$\Rightarrow \overset{o}{m} g h = \frac{9 \times 1 0^{4}}{60 \times 60} \times 10 \times 40$

$\overset{o}{m} g h = 1 0^{4} w a t t$

50% of it converts into electrical energy = 50% of $\overset{o}{m} g h = \frac{1 0^{4}}{2} w a t t$

Let $η$ be the number of bulbs n = 100 = $\frac{1 0^{4}}{2}$

n = 50

Two objects of equal masses placed at certain from each other attracts each other with a force of F. If one-third mass of one object is transferred to the other, then the new force will be:

F = $\frac{G m_{2}}{d^{2}}$ - (i)

Now, $F' = \frac{G m_{1} m_{2}}{d^{2}} = \frac{G 2 m}{3} \times \frac{4 m}{3 d^{2}}$

$k' = \frac{8}{9} G \frac{m^{2}}{d^{2}}$

$∴ F' = \frac{8}{G} F$

A water drop of radius $1 μ m$ falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is 1.8 × 10^-⁵ Nsm^-² and its density is negligible as compared to that of water 10⁶ gm^-³. Terminal velocity of the water drop is:

(Take acceleration due to gravity = 10 ms^-²)

NLM 1

$ρ_{ω} \frac{4}{3} π r^{3} g = 6 π η r V_{T}$

$\frac{ρ_{ω} 4}{3} π r^{2} g =$ $6 π η v_{T}$

$V_{T} = \frac{4}{3} \frac{ρ_{ω} π r^{2} g}{6 π η}$

$= \frac{2}{9} \times 1 0^{3} \times \frac{{(1 0^{- 6})}^{2} \times 10}{1.8 \times 1 0^{- 5}}$

$= 0.1234 \times 1 0^{- 3}$

vT = 123.4 × 10^-6m/sec

A sample of an ideal gas is taken through the cyclic process ABCA as shown in figure. It absorbs, 40 J of heat during the part AB, no heat during BC and rejects 60 J of heat during CA. A work of 50 J is done on the during the part BC. The internal energy of the gas at A is 1560 J. The work done by the gas during the part CA is:

Given

Q_AB = +40J

Q_BC = 0J

Q_CA = -60J

W_CA =?

W_BC = -50J (on the gas)

U_A = 1560J

1^st Law on path A to B

W_AB + $Δ U = Q$

Þ 0 + $[U_{B} - U_{A}] = 40$

Þ U_B = U_A + 40

= 1560 + 40

U_B = 1600 J

1^st Low on path B to C

Q_BC = $Δ u + w_{B C}$

Þ 0 = U_c – U_D – 50

U_C = U_B + 50

= 1600 + 50

= 1650 J

U_A – U_C = $Δ U$ (Path C to A) = 1560-1650

= -90 J

1st Low for path C to A

Q_CA = $Δ U_{C A} + w_{C A} + w_{C A}$

Þ -60 = -90 + w_CA

W_CA = +30J

What will be the effect on the root mean square velocity of oxygen molecules if the temperagture is doubled and oxygen molecule dissociates into atomic oxygen?

V_rms = $\sqrt[]{\frac{3 R T}{M}}$

V_rms’ = $\sqrt[]{\frac{3 R T_{f}}{m_{f}}} = \sqrt[]{\frac{3 R \times 2 T \times 2}{M}}$ $[\begin{matrix} G i v e n, T_{f} = 2 T & M_{f} = \frac{M}{2} \end{matrix}]$

$= 2 \sqrt[]{\frac{3 R T}{M}}$

V_rms’ = 2V_rms

Two point charges A and B of magnitude + 8 × 10^-6 and -8 × 10^-6C respectively are placed at a distance d apart. The electric field at the middle point O between the charges is 6.4 × 10⁴ NC^-1. The distance ‘d’ between the point charges A and B is:

E₁ (done to q₁) = $\frac{k, q_{1}}{{(d / 2)}^{2}} = \frac{[9 \times 1 0^{9} \times 8 \times 1 0^{- 6}]}{d^{2}} \times 4$

E₂ (done to q₂) = $\frac{k q_{2}}{{(d / 2)}^{2}} = \frac{[9 \times 1 0^{9} \times 8 \times 1 0^{- 6}]}{d^{2}} \times 4$

Enet = E₁ + E₂

$= 2 [\frac{4 \times 9 \times 1 0^{9} \times 8 \times 1 0^{- 6}}{d^{2}}]$

Given Enet = 6.4 × 104

$\Rightarrow 2 \times [\frac{4 \times 9 \times 1 0^{9} \times 8 \times 1 0^{- 6}}{d^{2}}] = 6.4 \times 1 0^{4}$

d² = 9 × 10^-6 + 6

d= 3m

Resistance of the wire is measured as at 10°C and 30°C respectively. Temperature co-efficient of resistance of the material of the wire is:

R_T = R₀ $(1 + α Δ T)$

at t = 10°C, R_T = 2 $Ω$

2 = R₀ (1 + [10 – T0])- (1)

at T = 30°C, R_T= 3 $Ω$

3 = $R_{0} [1 + α (30 - T_{0})]-$ (2)

from (1)

$2 = R_{0} (1 + α 10)-$ (3)

from (2) 3 = R₀ (1 + 30) - (4)

$(4) \div (3)$

$\frac{3}{2} = \frac{1 + 30 α}{1 + 10 α}$

3 + 30 = 2T 60α = $\frac{1}{30} = 0.033$

The space inside a straight current carrying solenoid is filled with a magnetic material having magnetic susceptibility equal to 1.2 × 10^-5. What is fractional increase in the magnetic field inside solenoid with respect to air medium inside the solenoid?

$χ (S u s c e p t i b i l i t y) = 1.2 \times 1 0^{- 5}$

fractional change = $= \frac{Δ B}{B} = \frac{B i n s i d e - B o u t}{B o u t}$

$= \frac{μ_{m} H - μ_{0} H}{μ_{0} H} = \frac{(μ_{0} μ_{r} - μ_{0}) H}{μ_{0} H}$ $[? μ_{m} = b_{0} μ_{0}]$

$= μ_{r} - 1$

= $= (1 + χ) - 1 = χ = 1.2 \times 1 0^{- 5}$

Two parallel, long wires are kept 0.20 m apart in vacuum, each carrying current of x A in the same direction. If the force of attraction per meter of each wire is 2 × 10^-6N, then the value of x is approximately:

$\frac{F}{l} = \frac{μ_{0}}{2 π} \frac{i_{1} i_{2}}{d}$

$2 \times 1 0^{- 6} = \frac{4 π \times 1 0^{- 7}}{2 π} \times \frac{x \times x}{d}$

x² = 2x = $\sqrt[]{2}$ = 1.4x = 1.4

A coil is placed in a time varying magnetic field. If the number of turns in the coil were to be halved and the radius of wire doubled, the electrical power dissipated due to the current induced in the coil would be:

(Assume the coil to be short circuited.)

$P = \frac{E^{2}}{R} = \frac{{(N A \frac{d B}{d t})}^{2} \times 4 A C}{ρ l}$

$p' = {(\frac{N A}{2} \frac{d B}{d t})}^{2} \times 4 A C$

$p' = 2 P$

An EM wave propagating in x-direction has wavelength of 8 mm. The electric field vibrating y-direction has maximum magnitude of 60 Vm^-1. Choose the correct equations for electric and magnetic fields if the EM wave is propagating in vacuum:

Given, $λ = 8 m m, v = 8 m m, v = 3 \times 1 0^{8}$

$\vec{E} = ε_{0} s i n [k x - ω t] \hat{j}$

$k = \frac{2 π}{λ} = \frac{2 π}{8 \times 1 0^{- 3}} = \frac{π}{4} \times 1 0^{3}$

$\vec{E} = 60 s i n [k (x - \frac{ω}{k} t)] \hat{j}$

$\vec{E} = 60 s i n [\frac{π}{4} \times 1 0^{3} [x - v t]] \hat{j} [? \frac{ω}{k} = v]$

$\vec{E} = 60 s i n [\frac{π}{4} \times 1 0^{3} [x - 3 \times 1 0^{8} t]] \hat{j} v / m$

Now $\frac{E_{0}}{B_{0}} = C$

B0 = $\frac{60}{3 \times 1 0^{8}} = \frac{20}{1 0^{8}} = 2 \times 1 0^{- 7}$ $\vec{B} = 2 \times 1 0^{- 7} [\frac{π}{4} \times 1 0^{3} [x - 3 \times 1 0^{8} t]] \hat{k} T$

In young’s double slit experiment performed using a monochromatic light of wavelength $λ,$ when a glass plate $(μ = 1.5)$ of thickness $x λ$ is introduced in the path of the one of the interfering beams, the intensity at the position where the central maximum occurred previously remains unchanged. The value of x will be:

Þ AC = d + t $(μ - 1) \to$ path length

BC = d

path different = AC – BC = $n λ$

$t (μ - 1) = μ λ$

for contal maxima

$x λ (1.5 - 1) = n λ$

0.5x = x

n = 0, 1,2- -

n = 0 Not possible

Now, n = 1

0.5x = 1

x = 2

Let K₁and K₂ be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $λ_{1} a n d λ_{2},$ respectively are incident on a metallic surface. If $λ_{1} = 3 λ_{2}$ then:

E Energy with which photons are incident on a metallic surface.

$λ_{1} = 3 λ_{2} \Rightarrow λ_{1} > λ_{2}$

E $α \frac{1}{λ},$ E₁ < E₂

We know E₁ = $?_{0} + k_{1}$

$\frac{h c}{λ_{1}} = ? + k_{1}$

k₁ = $\frac{h c}{λ_{1}} - ?_{0} - - - - (i)$

& E₂ = $?_{0} + k_{2}$

$\frac{h c}{λ_{2}} = ?_{0} + k_{2}$

$\frac{3 h c}{λ_{1}} = ? + k_{2}$

k₂ = $\frac{3 h c}{λ_{1}} - ?_{0}$ (i)

from (1) & (2)

$k_{z} = 3 [k_{1} + ?_{0}] - ?_{0} = 3 k_{1} + 2 ?_{0}$

k₂ > $3 k_{1} \Rightarrow \frac{k_{2}}{3} > k_{1}$

Following statements related to radioactivity are given below:

(A) Radioactivity is a random and spontaneous process and is dependent on physical and chemical conditions.

(B) The number of un-decayed nuclei in the radioactive sample decays exponentially with time.

(C) Slope of the graph of log_e (no. of undecayed nuclei) Vs, time represents the reciprocal of mean life time $(τ) .$

(D) Product of decay constant $(λ)$ and half-life time $(T_{1 / 2})$ is not constant.

We know, N = $N_{0} e^{- λ t}$

Where, N Number of un-decayed Nuclei

No Initial No. Of Nuclei

N $e^{- λ t}$

$N = N_{0} {\bar{e}}^{λ t}$

Taking log both side log N = log N0 + loge $e^{- λ t}$

log N = log No $λ t$

Slope = $λ = \frac{1}{t_{a v}}$

In the given circuit the input voltage V_in is shown in figure. The cut-in voltage of p-n junction diode (D₁ or D₂) is 0.6 V. Which of the following output voltage (V₀) waveform across the diode is correct?

Amplitude modulated wave is represented by V_AM= $10 [1 + 0.4 c o s (2 π \times 1 0^{4} t)]$ $c o s (2 π \times 1 0^{7} t) .$ Total bandwidth of the amplitude modulated wave is:

$V_{A M} = 10 [1 + 0.4 c o s (2 π \times 1 0^{4} t)] c o s (2 π \times 1 0^{7} t)$

Main wave frequency (carrier frequency)

Bandwidth = 2 fm

= $2 \times 1 0^{4} H_{2} = 20 \times 1 0^{3} H_{2} = 20 K H_{2}$

A student in the laboratory measures thickness of a wire using screw gauge. The readings are 1.22mm, 1.23mm, 1.19mm and 1.20mm. The percentage error is $\frac{x}{121} % .$ The value of x is________.

Given x₁ = 1.22mm

x₂ = 1.23mm

x₃ = 1.19mm

& x₄ = 1.20mm

$x_{m e a n} = \frac{x_{1} + x_{2} + x_{3} + x_{4}}{4} = \frac{1.22 + 1.19 + 1.20}{4} = 1.21$

$| Δ x_{1} | = | x_{m e a n} - x_{1} | = | 1.21 - 1.22 | = 0.01$

$| Δ x_{2} | = | x_{m e a n} - x_{2} | = | 1.2 | - | 1.23 | = 0.02$

$| Δ x_{3} | = | x_{m e a n} - x_{3} | = | 1.21 - 1.19 | = 0.02$

$| Δ_{4} | = | x_{m e a n} - x_{4} | = | 1.21 - 1.20 | = 0.01$

${(Δ x)}_{m e a n} = \frac{0.01 + 0.02 + 0.02 + 0.01}{4}$

$= \frac{0.06}{4}$

$% e r r o r = \frac{Δ x_{m e a n}}{x_{m e a n}} = \frac{0.06}{4 \times 1.21} \times 100$

$= \frac{600}{4 \times 121}$

$= \frac{150}{121} %$

x = 150

A zener of breakdown voltage V_Z = 8V and maximum zener current, I_ZM = 10mA is subjected to an input voltage V_i = 10V with series resistance R = 100 $Ω$ In the given circuit R_L represents the variable load resistance. The ratio maximum and minimum value of R_L is__________.

Across zener diode & RL

8 = I_LR_L

8 = $(20 - 10) R_{L (m a x)}$

$R_{L m a x} = \frac{8}{10} k Ω$

At max current in loop (1)

10 = i_max R + 8

i_max= $\frac{2}{R} = \frac{2}{100} = 0.02 A$

= 20 mA

R_L (max) = $\frac{8}{10} k Ω$

At minimum curre3nt through zener

i_max R_L (minimum) = 8

R_L (minimum) = $\frac{8}{20} k Ω$

$\frac{R_{L m a x}}{R_{L m i n}} = 2$

In a Young’s double slit experiment, an angular width of the fringe is 0.53° on a screen place at 2m away for particular wavelength of 450 nm. The angular width of the fringe, when whole system is immersed in a medium of refractive index 7/5, is $\frac{1}{α} .$ The value of a is__________.

we know, = $\frac{λ}{d}$ $μ = \frac{C}{V} = \frac{C}{υ λ}$

$λ_{1} = \frac{C}{μ_{1} υ_{1}}$ [with change in medium frequency refnain constaint]

$λ_{2} = \frac{C}{μ_{2} υ_{2}}$

$\frac{λ_{2}}{λ_{1}} = \frac{μ_{1}}{μ_{2}} = \frac{5}{7}$ $λ_{2} = \frac{5}{7} λ_{1}$

$θ_{1} = \frac{λ_{1}}{d_{1}}$

2 = $\frac{λ_{2}}{d_{2}}$

$\frac{θ_{2}}{θ_{1}} = \frac{λ_{2}}{λ_{1}} [? d_{1} = d_{2}]$

2 $= \frac{5}{7} \times θ_{1}$

$θ_{2} = \frac{5}{7} \times 0.35$

$θ_{2} = \frac{1}{4} = \frac{1}{α}$

= 4

In the given circuit, the magnitude of V_L and V_C are twice that of V_R. Given that f = 50 Hz. the inductance of the coil is $\frac{1}{K π} m H .$ The value of K is_________.

Given

V_L = V_C= 2V_R, f = 50Hz

$L = \frac{1}{k π} m H$

since, V_L = V_C.

then

$V_{n e t} = \sqrt[]{{(V_{L} - V_{C})}^{2} + {(V_{R})}^{2}}$

V_R = V_Net= 220V

I = $\frac{V_{N e t}}{Z} = \frac{220}{\sqrt[]{{(X_{L} - X_{C})}^{2} + 1 2^{2}}}$

$I = 44 A$

$V_{L} = I x_{L} = 2 v_{R}$

= I _xL = 440

x_L = 10

WL = 10

2πfL = 10

$L = \frac{1}{\frac{1}{100} π} k = \frac{1}{100} = 0.01 \approx 0$

All resistances in figure are $1 Ω$ each. The value of current ‘I’ is $\frac{a}{5} A .$ The value of a is___________.

I = $\frac{3 \times 8}{15 R} = \frac{8}{5 R} = \frac{a}{5}$

$\Rightarrow \frac{8}{5 \times 1} = \frac{a}{5}$

a = 8

A capacitor C₁ of capacitance $5 μ F$ is charged to a potential of 30V using a battery. The battery is then removed and the charged capacitor is connected to an uncharged capacitor C₂ of capacitance $10 μ F$ as shown in figure. When the switch is closed charge flows between the capacitors. At equilibrium, the charge on the capacitor C₂ is__________ $μ C .$

Q = C₁v

= 5 × 30

Q = 150 $μ c$

Now it is connected to an unchanged capacitor C₂

Let change on C₁ ® q₁

& 4 C₂ ® q₂

Apply kvL

$\frac{q_{1}}{q} - \frac{q_{2}}{C_{2}} = 0$

Þ q₂ = 2q₁ – (1)

& conservation of change

q₁ + q₂ = 150 – (2)

Þ from (1) & (2)

3q₁ = 150

q₁ = 50 $μ c$

$q_{2} = 100 μ c$

$q_{2} = 100 μ c o n c_{2}$

A running fork of frequency 340 Hz resonates in the fundamental mode with a air column of length 125cm in a cylindrical tube closed at one end. When water is slowly poured in it, the minimum height of water required for observing resonance once again is __________cm.

(Velocity of sound in air is 340 ms^-1)

First resonating length = $λ = \frac{V}{f} = \frac{340}{340} = 1 m$ $\frac{}{}$

Second resonating length = $\frac{3 λ}{4} = 75 c m$ $\frac{}{}$

Third resonating length = $\frac{5 λ}{4} = 125 c m$ $\frac{}{}$

Height of water required = 125 – 75 = 50cm

View other answers

A uniform disc with mass M = 4 kg and radius R = 10 cm is mounted on a fixed horizontal axle as shown in figure. A block with mass m = 2kg hangs from a massless cord that is wrapped around the rim of the disc. During the fall of the block, the cord does not slip and there is no friction at the axle. The tension in the cord is_________ N.

NLM² for (1)

mg – T = ma

Þ 2 × 10 – T = 2a

Þ T + 2a = 20 - (1)

Rotation equestion for (2)

T × R = I_cma

Þ T R = $I_{C M} \frac{a}{R}$

$[α = \frac{a}{R}, N o s l i p p i n g c o n d i t i o n]$

T = $\frac{M a}{2}$

$T = \frac{4}{2} a$

T = 2a - (2)

from (1) & (2)

2T = 20

T = 10N

A car cover AB distance with first one-third at velocity $υ_{1} m s^{- 1},$ second one-third at $υ_{2} m s^{- 1}$ and last one-third at $υ_{3} m s^{- 1} .$ If then the average velocity of the car is__________ ms^-1.

$t_{1} = \frac{L}{3 V_{1}}$

$t_{2} = \frac{L}{3 V_{2}}$

$t_{3} = \frac{L}{3 V_{3}}$

$V_{a m g} = \frac{L}{t_{1} + t_{2} + t_{3}} = \frac{L}{\frac{L}{3 V_{1}} + \frac{L}{3 V_{2}} + \frac{L}{3 V_{3}}}$

$V_{a m g} = \frac{3}{\frac{1}{v_{1}} + \frac{1}{v_{2}} + \frac{1}{v_{3}}}$

$= \frac{3}{\frac{1}{11} + \frac{1}{2 \times 11} + \frac{1}{3 \times 11}}$

= $\frac{3 \times 11 \times 6}{6 + 3 + 2}$

= 3 × b

= 18 m/sec

A liquid of density 750 kgm^-3 flows smoothly through a horizontal pipe that tapers in cross-sectional area from A₁ = 1.2 × 10^-2m² to A₂ = $\frac{A_{1}}{2} .$ The pressure difference between the wide and narrow sections of the pipe is 4500 Pa. The rate of flow of liquid is_________ × 10^-3m³s^-1.

Given A₂ = $\frac{A_{1}}{2}$

Bernoulli’s b/w (1) & (2)

$\frac{P_{1}}{ρ g} + \frac{v_{1}^{2}}{2 g} + z_{1} = \frac{ρ_{2}}{ρ g} + \frac{v_{2}^{2}}{2 g} + z_{2}$

z₁ = z₂]

$\frac{P_{1} - P_{2}}{ρ g} = \frac{v_{2}^{2} - v_{1}^{2}}{2 g}$

[Q = A₁V₁ = A₂V₂]

$\frac{4500}{750} = \frac{{(\frac{Q}{A_{2}})}^{2}}{2} - {(\frac{Q}{A_{1}})}^{2}$

$12 = Q^{2} [\frac{1}{A_{2}^{2}} - \frac{1}{A_{1}^{2}}]$

$Q = 2 A_{1} = 2 \times 1.2 \times 1 0^{- 2}$

= 2.4 × 10^-2 m³ /sec

= 24 × 10^-3 m³ /sec

= 24

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