Advertisement

Electromagnetic Induction Class 12th Physics Part I CBSE Solution

Class 12th Physics Part I CBSE Solution
Exercises
  1. Predict the direction of induced current in the situations described by the following…
  2. Use Lenz’s law to determine the direction of induced current in the situations described…
  3. A long solenoid with 15 turns per cm has a small loop of area 2.0 cm^2 placed inside the…
  4. A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region…
  5. A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s-1 about an…
  6. A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with…
  7. A horizontal straight wire 10 m long extending from east to west is falling with a speed…
  8. Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V…
  9. A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil…
  10. A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage…
Additional Exercises
  1. Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet…
  2. A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a…
  3. It is desired to measure the magnitude of field between the poles of a powerful loud…
  4. Figure 6.20 shows a metal rod PQ resting on the smooth rails AB and positioned between the…
  5. An air - cored solenoid with length 30 cm, area of cross - section 25 cm^2 and number of…
  6. Obtain an expression for the mutual inductance between a long straight wire and a square…
  7. Now assume that the straight wire carries a current of 50 A and the loop is moved to the…
  8. A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass M and…

Exercises
Question 1.

Predict the direction of induced current in the situations described by the following Figs. 6.18(a) to (f).



Answer:

The direction of induced current will be given by Lenz’s law.

According to the Lenz’s law: The direction of induced current by change in magnetic field (Faraday’s induction) will be such that it opposes the change that caused it.
As seen from the given figures, the direction of the induced current is shown when the north pole of the magnet is moved towards the closed loop or away from the closed loop.




Question 2.

Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.19:

(a) A wire of irregular shape turning into a circular shape;

(b) A circular loop being deformed into a narrow straight wire.



Answer:

According to the Lenz’s law: The direction of induced current by change in magnetic field (Faraday’s induction) will be such that it opposes the change that caused it.


(a) In the figure given below, the direction of magnetic field is perpendicular into the paper.



Now, when we change the shape of irregular loop into a circular loop, then the area of the loop will increase. Because, area of circle is considered to be greatest.


Since, area of loop increases, ∴ the magnetic flux will also increase.


Now, the current will be induced in the loop such that the magnetic flux decreases. And, magnetic field will decrease when the area will decrease i.e. the wires should be pulled in the inwards directions, to decrease the net magnetic field.


∴ the current will be along “adcba”.


(b) In the figure given below, the direction of magnetic field is perpendicular coming out of the paper.



Now, when circular loop is transformed into a narrow straight wire, then the area of the loop will decrease. Since, area of loop decreases, ∴ the magnetic flux will also decrease.


Now, the current will be induced in the loop such that the net magnetic flux increases in the wire. ∴ the current flow will be along “adcba” which produces the induced magnetic field outwards the paper.



Question 3.

A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?


Answer:

Given:

No. of turns in the solenoid coil, N = 15 turns/cm = 15000 turns/m


(∵ 1m = 100 cm)


∴ no. of turns in the solenoid coil per unit length, n = 15000 turns


Area of the solenoid coil = 2.0 cm2


In m2, Area will be 2 × 10 - 4 m2


Since current carried by the solenoid coil changes from 2.0 to 4.0 A in 0.1 seconds


Therefore, change in the current in the coil of solenoid = final current – initial current


di = 4.0 – 2.0 = 2.0 A


The change in time “dt” is given as ,dt = 0.1 seconds


Applying Faraday’s law, induced e.m.f can be calculated as follows:


 …(1)


Where Ф is flux induced in the loop


And, flux is given as, Ф = BA


Where “B” is the magnetic field and “A” is the area of the loop.


For a solenoid, B= μ0nI


Therefore, the equation (1) becomes:



Or 


Or E = Aμ0 n (dI/dt)…(2)


∵ A, μo and n are constants


Substituting the values in equation (2), we get,



⇒ E = 7.54 × 10 - 6 V


Or E = 7.54 μV


(∵ 10-6 = μ)


Hence, the induced voltage in the loop is 7.54 μV



Question 4.

A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s–1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?


Answer:

Given:

Length of rectangular wire = 8 cm


Breadth of rectangular wire = 2 cm



The area of the rectangular wire can be calculated as follows:


A = l × b


A = 8 cm × 2 cm = 16 cm2


In m2, area will be A = 16 × 10 - 4 m2


Magnitude of magnetic field B = 0.3T


Velocity of the loop, v =1 cms - 1 = 0.01ms - 1


(a) The emf developed across the cut can be calculated as follows:


e = Blv …(1)


where B is the magnetic field


l is the length of the loop and,


v is the velocity of the rectangular loop


substituting the values in equation (1), we get,


e = (0.3 T × 0.08m × 0.01ms - 1)


e = 2.4 × 10 - 4 V


Time taken for travelling along the width of the loop can be calculated as follows:


T =distance travelled/velocity with which distance is travelled


T = b(width of the loop)/v (velocity)


T = 0.02 m/0.01 ms - 1


T= 2 seconds


This means that the induced voltage will last for very short direction of 2 seconds.


(b) The emf developed across the cut if the velocity of the loop is 1 cms–1 in a direction normal to the shortest side of the loop can be calculated as follows:


e = Bbv…(2)


where B is the magnetic field


b is the width of the loop and,


v is the velocity of the rectangular loop


substituting the values in equation (2), we get,


e = (0.3 T × 0.02 m × 0.01 ms - 1)


e = 0.6 × 10 - 4 V


Time taken for travelling along the width of the loop can be calculated as follows:


T =distance travelled/velocity with which distance is travelled


T = l(length of the loop)/v


T = 0.08 m/0.01ms - 1


T = 8 seconds


This means that the induced voltage will last for 8 seconds.



Question 5.

A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s–1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.


Answer:

Given:

Length of the metallic rod = 1.0 m


Angular frequency = 400 rads - 1


Strength of magnetic field B= 0.5 T


Since it is clear that one end of the metallic rod has zero linear velocity whereas the other end of the rod has linear velocity of lω


∴ The average linear velocity of the metallic rod can be calculated as follows:


v = (lω + 0)/2


⇒ v = lω /2


The e.m.f. developed between the centre and the ring can be calculated as follows:


e = Blv ...(1)


where B is the magnetic field


l is the length of the loop and,


v is the velocity of the rectangular loop


⇒ Substituting the value of v in above equation, we get


e = B × l × (lω/2)


e = (Bl2ω)/2


e = (0.5 T × 1m2 × 400 rads - 1)/2


On calculating, we get


e = 100 V


Hence, the emf developed across the ring is 100 V.



Question 6.

A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s–1 in a uniform horizontal magnetic field of magnitude 3.0 × 10–2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 W, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating.

Where does this power come from?


Answer:

Given:

Radius of the circular coil r = 8.0 cm = 0.08 m


No. of turns in the coil, N =20


Angular frequency ω = 50 rad s - 1


Magnitude of magnetic field B =3.0 × 10–2 T


Resistance of the closed loop = 10 Ω


The area of the coil can be calculated using the formula


A =Ï€r2


A = 3.14 × (0.08 m)2


The maximum induced e.m.f is calculated as follows:


e = N ω A B


e = 20 × 50 rad s - 1 × 3.14 × (0.08 m)2 × 3.0 × 10–2 T


On calculating we get


e = 0.603 V


Since for a full cycle, the average induced emf will be zero.


The maximum current for the circular coil can be calculated as follows:


I = e/R


Substituting the values


I = 0.603 V/10 Ω


I = 0.0603A


The average power loss due to Joule’s heating effect in the circular coil will be given by:


P = (eI)/2


P = (0.603 V × 0.0603 A)/2


On calculating we get


P = 0.018 W


A torque was produced due to current induced in the coil. This torque opposes the rotation of the circular coil. Since a rotor is the external agent in the coil and it should apply the torque so that the coil keeps rotating in the uniform manner. Therefore, the dissipated power comes from the external rotor.



Question 7.

A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s–1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 × 10–4 Wb m–2.

(a) What is the instantaneous value of the emf induced in the wire?

(b) What is the direction of the emf?

(c) Which end of the wire is at the higher electrical potential?


Answer:

Given:

Length of the wire, l = 10 m


Speed of wire, v = 5.0 ms - 1


Magnetic field B = 0.30 × 10–4 Wb m–2


The diagram is:



(a) The e.m.f induced in the wire can be calculated as follows:


e = Blv ...(1)


where B is the magnetic field


l is the length of the loop and,


v is the velocity of the rectangular loop


substituting the values in above equation, we get,


e = 0.30 × 10–4 Wb-m - 2 × 10 m × 5ms-1


e = 1.5 × 10–3 V


(b) The direction of the induced emf will be from west to east in accordance with the Fleming’s right - hand rule.


According to Fleming’s right - hand rule: When the right hand is held with the thumb, first finger and second finger mutually perpendicular to each other, as shown below.


Then, the thumb points in the direction in which the conductor moves; the first finger gives the direction of the magnetic field (N ⇒ S); and the second finger gives the direction of the induced current.



(c) The east end of the straight wire will be having high electrical potential.



Question 8.

Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self - inductance of the circuit.


Answer:

Given: Initial current I1 = 5.0 A

Final current I2 = 0.0 A


Time = 0.1 seconds


Average e.m.f = 200 V


Change in the current can be calculated as:


dI = I1 - I2 = 5.0A – 0.0A


dI = 5 A


The self-conductance of the circuit is calculated using the formula:



or 


substituting the values, we get



∴ L = 4 Henry



Question 9.

A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?


Answer:

Given: Mutual conductance, μ = 1.5 H

Initial current I1= 0


Final current I2 = 20 A


Time taken t = 0.5 seconds


Change in the current can be calculated as:


dI = I1 - I2 = 20 – 0.0


dI = 20 A


The mutual conductance of the circuit is calculated using the formula:



or 


As we know that 


Substituting it in equation, we get






Question 10.

A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 × 10–4 T and the dip angle is 30°.


Answer:

Given: Speed of the jet plane towards west = 1800 km/h

In m/s, speed of the jet plane towards west = 500 m/s


Length of the span of the jet plane = 25 m


Magnetic field = 5 × 10–4 T


Dip angle =30°


The vertical component of Earth’s magnetic field can be calculated as follows:


Bv= B sinδ


On substituting the values, we get,


Bv= 5 × 10–4 T × sin30°


⇒ Bv= 5 × 10–4 T × 1/2


∴ Bv = 2.5 × 10–4 Tesla


The voltage difference developed between the ends of the jet can be calculated using the formula:


e = Bv × l × v


On substituting the values, we get


e =2.5×10–4 T× 25m × 500m/sec


on calculating, we get


e = 3.125 V


∴ the voltage difference developed between the ends of the wings is approximately 3V.




Additional Exercises
Question 1.

Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T s–1. If the cut is joined and the loop has a resistance of 1.6Ω , how much power is dissipated by the loop as heat? What is the source of this power?


Answer:

Given: Length of rectangular wire = 8 cm

Breadth of rectangular wire = 2 cm


The area of the rectangular wire can be calculated as follows:


A = l × b


A = 8cm × 2cm = 16 cm2= 16 × 10-4 m2


Initial Magnitude of magnetic field B = 0.3T


Velocity of the loop, v =1 cms-1


The rate of decrease of the magnetic field i.e. 


The e.m.f developed in the rectangular loop is given as:



dФ is the change in the flux across the loop = AB


⇒ 


Or 


Therefore, substituting the values in above equation, we get:


e = 16×10-4 m2× 0.02 Ts-1


⇒ e = 0.32 × 10-4V


The current induced in the loop can be calculated as follows:


i = e/r


i = (0.32 × 10 - 4)/1.6


⇒ i = 2 × 10 - 5A


The Power loss or dissipation can be calculated as follows:


P = i2R


Where, R is resistance = 1.6Ω


⇒ P = (2 × 10-5A)2 × 1.6Ω


⇒ P = 6.4×10-10 W


The source of the power is an external agent. This is due to change in the magnetic field with time.



Question 2.

A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s–1 in the positive x - direction in an environment containing a magnetic field in the positive z - direction. The field is neither uniform in space nor constant in time. It has a gradient of 10–3 T cm–1 along the negative x - direction (that is it increases by 10–3 T cm–1 as one moves in the negative x - direction), and it is decreasing in time at the rate of 10–3 T s–1. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mW.


Answer:

Given: side of square loop = 12 cm

In metre, the side of square loop = 0.12m


Area of the square loop = 0.12 × 0.12 = 144 × 10 - 4m2


Velocity of the loop = 8 cms-1


In metre/s, the velocity of the loop, v= 0.08m/s


Since gradient of electric field is along negative x - direction



In Tm - 1


∴ The rate of decrease of magnetic field in time is given as:



Resistance of the square loop = 4.50 mΩ =4.5 × 10 - 3Ω


Due to change in the motion of the square loop in presence of non - uniform magnetic field, the decrease in the magnetic flux is given as:





Due to explicit time variation in magnetic field, the rate of change of flux is given by:



Substituting values in above, we get




The total emf induced in the square loop is :


e =1.44 × 10–5 V+ 11.52× 10-5 V


e = 12.96 × 10-5 V


The induced current can be written as:


i =e/R


⇒ 


⇒ i = 2.88 × 10 - 2A


The current direction will be in the way that there will be increase in the flux along positive z direction.



Question 3.

It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area 2 cm2 with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region.

Equivalently, one can give it a quick 90° turn to bring its plane parallel to the field direction). The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50. Estimate the field strength of magnet.


Answer:

Given: Area of small flat search coil = 2 cm2

Total number of turns in the coil n = 25


Total charge measured by the ballistic galvanometer = 7.5 mC=7.5 ×10-3C


combined resistance of the coil and the galvanometer R= 0.50


The current induced in the coil can be calculated as follows:


i = e/R


also,


substituting in the equation we get



⇒ …………..(i)


Initial flux through the coil, Ф1 = AB


Final flux through the coil, Ф2= 0


Integrating equation (i) on both sides, we get



Total charge 


Therefore, 



⇒ Q = (NBA)/R


Therefore, B = (QR)/NA


So, substituting all the values in above equation, We get



On solving, we get


B = 0.75 T



Question 4.

Figure 6.20 shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 9.0 mW. Assume the field to be uniform.



(a) Suppose K is open and the rod is moved with a speed of 12 cm s–1 in the direction shown. Give the polarity and magnitude of the induced emf.

(b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed?

(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain.

(d) What is the retarding force on the rod when K is closed?

(e) How much power is required (by an external agent) to keep the rod moving at the same speed (=12cm s–1) when K is closed? How much power is required when K is open?

(f) How much power is dissipated as heat in the closed circuit? What is the source of this power?

(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?


Answer:

Given: Length of the rod = 15 cm

Length of rod in metres l = 0.15 m


Magnetic field strength B = 0.50 T


Resistance of the loop R = 9 mΩ = 9 × 10 - 3 â„¦


Emf induced in the loop = 9 mV


Speed of the rod, v= 12 cm/s


Speed of the rod in m/s, v= 0.12m/s


It is shown as below:



(a) Induced emf can be calculated as follows:


e = Bvl


e = 0.50T× 0.12ms-1× 0.15m


e = 9× 10-3V = 9mV


The polarity of the induced emf is such that the end P is positive and end Q is negative.


(b) Yes, there is an excess charge built up at the ends of the rods when K is open. When K is closed, the excess charge present will be maintained by the flow of current continuously.


(c) The net force on the rod PQ is zero still they experience magnetic force due to motion of the rod magnetic force is cancelled by the electric force set up. This happens because of the excess charge on both the opposite ends of the rod.


(d) The retarding force acting on the rod when K is closed can be calculated as follows:


F = iBl……………(i)


Where i= induced emf(e)/Resistance (RΩ)


I = 9mV/9×10-3Ω


I= 1A


Substituting the values in equation (i), we get


F = 1A× 0.50T× 0.15m


F = 75×10-3 N


(e) Power required when K is closed:


Power = Fv


P = 75×10-3 N× 0.12ms-1


P = 9×10 - 3 W


Power = 9 mW


When key K is open, there is no power required.


(f) Power dissipated H = I2R


H = 1A 2× 9×10-3Ω = 9 mW


9mW power is dissipated as heat in the closed circuit and the source of this power is external agent.


(g) The induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular is zero because the motion of the rod doesn’t cross across the lines and therefore induced emf will be zero.



Question 5.

An air - cored solenoid with length 30 cm, area of cross - section 25 cm2 and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10–3 s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.


Answer:

Given: Length of the solenoid = 30 cm=0.30 m

Area of cross - section = 25 cm2 = 25m× 10-4m-2


No. of turns in the solenoid = 500


Current, i = 2.5A


Time for which the current flows t = 10–3 s


The average back emf induced across the ends of the open switch in the circuit is as follows:


 ………………(i)


where dФ is the change in the flux and is equal to NAB


B is the magnetic field = (μ0Ni)/l


Substituting in equation (i), we get



Substituting the values in above equation , we get



On solving, we get


e = 6.5 V


∴ the emf induced in the solenoid is 6.5V



Question 6.

Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig. 6.21.



Answer:

Let a small element dy in the loop at a distance of y as shown in the given figure:



Magnetic field associated with the the element taken i.e. dy can be calculated as follows:


dФ = BdA ………………..(i)


where dA is the area of the element dy i.e. ady


B is the magnetic field intensity of the element dy at a distance of y and can be written as:



Putting in the equation (i)




Putting limits of y from x to a + x




Mutual conductance is given by the relation:


Ф = MI


Where MI = 


Or M = 



Question 7.

Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m/s.

Calculate the induced emf in the loop at the instant when x = 0.2 m.

Take a = 0.1 m and assume that the loop has a large resistance.


Answer:

The emf induced in the coil is given as:


e = B’ av


e = 


substituting the given values in above relation, we get


e = 


On calculating, we get


e = 5 × 10-5V



Question 8.

A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non - conducting spokes and is free to rotate without friction about its axis (Fig. 6.22). A uniform magnetic field extends over a circular region within the rim. It is given by,

B = – B0 k (r ≤ a; a < R)

= 0 (otherwise)

What is the angular velocity of the wheel after the field is suddenly switched off?



Answer:

Line charge per unit length is given by the following relation:

Total charge/ length = Q/2Ï€ r


Where r is the distance of the point within the wheel


Let m and R are mass and radius of the wheel


Magnetic field is given by the following relation:



The magnetic force at balanced by the centripetal force at a distance of r


i.e. BQv = mv2/r


v is the linear velocity of the wheel and is equal to v= 2π rλ


⇒ B × 2Ï€ rλ = mv/r


Or 


As we know that angular velocity is given as: ω = v/r


⇒ w = 


When R> a and r ≤ a


ω = 


PDF FILE TO YOUR EMAIL IMMEDIATELY PURCHASE NOTES & PAPER SOLUTION. @ Rs. 50/- each (GST extra)

SUBJECTS

HINDI ENTIRE PAPER SOLUTION

MARATHI PAPER SOLUTION

SSC MATHS I PAPER SOLUTION

SSC MATHS II PAPER SOLUTION

SSC SCIENCE I PAPER SOLUTION

SSC SCIENCE II PAPER SOLUTION

SSC ENGLISH PAPER SOLUTION

SSC & HSC ENGLISH WRITING SKILL

HSC ACCOUNTS NOTES

HSC OCM NOTES

HSC ECONOMICS NOTES

HSC SECRETARIAL PRACTICE NOTES

2019 Board Paper Solution

HSC ENGLISH SET A 2019 21st February, 2019

HSC ENGLISH SET B 2019 21st February, 2019

HSC ENGLISH SET C 2019 21st February, 2019

HSC ENGLISH SET D 2019 21st February, 2019

SECRETARIAL PRACTICE (S.P) 2019 25th February, 2019

HSC XII PHYSICS 2019 25th February, 2019

CHEMISTRY XII HSC SOLUTION 27th, February, 2019

OCM PAPER SOLUTION 2019 27th, February, 2019

HSC MATHS PAPER SOLUTION COMMERCE, 2nd March, 2019

HSC MATHS PAPER SOLUTION SCIENCE 2nd, March, 2019

SSC ENGLISH STD 10 5TH MARCH, 2019.

HSC XII ACCOUNTS 2019 6th March, 2019

HSC XII BIOLOGY 2019 6TH March, 2019

HSC XII ECONOMICS 9Th March 2019

SSC Maths I March 2019 Solution 10th Standard11th, March, 2019

SSC MATHS II MARCH 2019 SOLUTION 10TH STD.13th March, 2019

SSC SCIENCE I MARCH 2019 SOLUTION 10TH STD. 15th March, 2019.

SSC SCIENCE II MARCH 2019 SOLUTION 10TH STD. 18th March, 2019.

SSC SOCIAL SCIENCE I MARCH 2019 SOLUTION20th March, 2019

SSC SOCIAL SCIENCE II MARCH 2019 SOLUTION, 22nd March, 2019

XII CBSE - BOARD - MARCH - 2019 ENGLISH - QP + SOLUTIONS, 2nd March, 2019

HSC Maharashtra Board Papers 2020

(Std 12th English Medium)

HSC ECONOMICS MARCH 2020

HSC OCM MARCH 2020

HSC ACCOUNTS MARCH 2020

HSC S.P. MARCH 2020

HSC ENGLISH MARCH 2020

HSC HINDI MARCH 2020

HSC MARATHI MARCH 2020

HSC MATHS MARCH 2020


SSC Maharashtra Board Papers 2020

(Std 10th English Medium)

English MARCH 2020

HindI MARCH 2020

Hindi (Composite) MARCH 2020

Marathi MARCH 2020

Mathematics (Paper 1) MARCH 2020

Mathematics (Paper 2) MARCH 2020

Sanskrit MARCH 2020

Sanskrit (Composite) MARCH 2020

Science (Paper 1) MARCH 2020

Science (Paper 2)

Geography Model Set 1 2020-2021


MUST REMEMBER THINGS on the day of Exam

Are you prepared? for English Grammar in Board Exam.

Paper Presentation In Board Exam

How to Score Good Marks in SSC Board Exams

Tips To Score More Than 90% Marks In 12th Board Exam

How to write English exams?

How to prepare for board exam when less time is left

How to memorise what you learn for board exam

No. 1 Simple Hack, you can try out, in preparing for Board Exam

How to Study for CBSE Class 10 Board Exams Subject Wise Tips?

JEE Main 2020 Registration Process – Exam Pattern & Important Dates


NEET UG 2020 Registration Process Exam Pattern & Important Dates

How can One Prepare for two Competitive Exams at the same time?

8 Proven Tips to Handle Anxiety before Exams!

BUY FROM PLAY STORE

DOWNLOAD OUR APP

HOW TO PURCHASE OUR NOTES?

S.P. Important Questions For Board Exam 2022

O.C.M. Important Questions for Board Exam. 2022

Economics Important Questions for Board Exam 2022

Chemistry Important Question Bank for board exam 2022

Physics – Section I- Important Question Bank for Maharashtra Board HSC Examination

Physics – Section II – Science- Important Question Bank for Maharashtra Board HSC 2022 Examination

Important-formula



THANKS