Physics 05 – RAY OPTICS

1. If incident ray MP and reflected ray QN are parallel to each other, find the angle θ between the mirrors.  

(a) 60°       (b) 45°        (c) 90°          (d) 180°

Answer: (c)
The deviation produced by first mirror = 180 – 2i1 and the deviation produced by second mirror = 180 – 2i2. The total deviation is 180° as MP and QN are anti parallel

SO , 180° = 360 ° – 2( i1 + i2 )

∴i1 +i2 = θ =90°

2. A 16 cm long image of an object is formed by a convex lens on a screen placed normal to its principal axis. On moving the lens towards the screen, without changing the positions of the object and the screen, a 9 cm long image is formed again on the screen. The size of the object is:

(a) 9 cm      (b) 11 cm     (c) 12 cm    (d) 13 cm.

Answer: (c) 

y= √( y 1 × y2  

= √(16 × 9) 

=4 × 3 =12cm

3. If x and y be the distances of the object and image formed by a concave mirror from its focus and f be the focal length then

(a) xf = y2                                   (b) xy = f2
(c) x/y = f                                    (d) x/y = f2

Answer: (b)

According to Newton’s formula

xy= f2 .
Note that m = f /  f  – u   = f -v /f 

or ) f/x  = y/x  ⇒ XY = f2

4. What is the number of images of an object formed by two mirrors if the angle between the mirrors is in between 2π/4 and 2π/5?

(a) 10              (b) 8                  (c) 6                    (d) 4

Answer: (d)
As we know, if angle between two mirrors is between 2π / n and 2π / n+1  then the no. of images formed is ‘n’

So here in the question it is 2π /4 and 2π /4 +1

∴ no. of images will be 4.

5. A convergent lens of 6 diopters is combined with a diverging lens of -2 diopters. Find the power of combination?

(a) 4 diopter                                 (b) 6 diopter
(c) 8 diopter                                  (d) 10 diopter

Answer: (a)
HereP1 = 6 diopters, P2 = -2 diopters
Using the formula P = P1 + P2 = 6 – 2 = 4 diopters

6. A prism is made up of flint glass whose dispersive power is 0.053. Find the angle of dispersion if the mean refractive index of flint glass is 1.68 and the refracting angle of prism is 3°.


(a) 20.08°                                           (b) 10.08°
(c) 0.208°                                            (d) 0.108°

Answer: (d)
As the refracting angle is small deviation = (µ – 1) A
angle of dispersion =Dv – Df

= ( mv – 1 ) A – ( m– 1 ) A 

= ( mv – m) A

= (μv – μf/ μmean – 1 )A (μmean – 1 ) 

= wA ( mmean -1 ) 

= ( 0.053 ) × 3 × ( 1.68 -1) =0.108°


7. If f1 = 12 cm, f2 = – 18 cm find the maximum focal length of the combination?

(a) 36 cm (b) 34 cm (c) 32 cm (d) 30 cm

Answer: (a)
Put the numerical values in the expression in the earlier question and we get 36 cm.

8. In continuation of fourth questions determine the minimum focal length of the combination

(a) 21.6 cm    (b) 24 cm    (c) 30 cm   (d) 35 cm

Answer: (a)
Put the numerical values in the above expression which we found and we will get 21.6 cm

9. What value of d gives f = 30 cm?


(a) 12 cm (b) 10 cm (c) 1.2 cm (d) 14 cm

Answer: (c)
f = 30 cm then 30 = 216 / 6 + d 
6 + d = 7.2 hence d = 1.2 cm

10. Convex lens of 10 cm focal length is combined with a concave lens of 6 cm focal length. Find the focal length of the combination.

(a) 15 cm                       (b) – 15 cm
(c) 30 cm                      (d) – 30 cm

Answer: (b)
Here f1 = 10 cm, f2 = -6 cm, F = ?
Use the formula 1/F  =  1/ f1 + 1/f2 =1/10   –   1/6  = -1/15
∴ =− F 15 cm

11. Focal length of a convex lens of refractive index 1.5 is 2 cm. Focal length of lens when immersed in a liquid of
refractive index 1.25 will be


(a) 5 cm                                      (b) 7.5 cm
(c) 10 cm                                    (d) 2.5 cm

Answer: (a)

fp / fa  = ( μ-1) / ( μ0 / μ) – 1  ⇒ f= (  5/2  ) × fa

( 5/2 ) × 2 cm = 5cm 

12. A double convex lens of focal length 20 cm is made of glass of refractive index 3/2. When placed completely in water (aμϖ   =  4/3 ) , its focal length will be

(a) 80 cm                                  (b) 15 cm
(c) 17.7 cm                                (d) 22.5 cm

Answer: (a)
fw /fa = ( μ– 1) /( ( μg  / μ) -1 ) ⇒ fw /fa = 4
fw = 4 × fa = 4 × 20 = 80 cms.

13. Inside Cutson glass


(a) Orange light travels faster than yellow light
(b) Yellow light travels faster than orange light
(c) Orange and yellow light ravels equally fast
(d) We cannot determine which colour of light travels faster

Answer: (a)
The velocity of light is inversely proportional to the index of refraction v = c / n , So, the lower the n, the higher the v. According to table, Cutson glass has a slightly lower n for orange light than it has for yellow light. Therefore, inside Cutson glass, orange light travels slightly faster.

14. The time required for light to pass through a glass slab of 2 mm thickness is glass  µglass =( 1.5)

(a) 10–8 s                                  (b) 10–6 s
(c) 10–11 s                                 (d) None of these

Answer: (c)

t= d/v  = (d / c/μ )  = dμ /c  = (2×10 -3  × 1.5 / 3×10 8 ) sec = 10-11 sec 

15. A spherical surface of curvature R separates air (µ=1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O and PO = OQ. The distance PO is equal to

(a) 2R                                       (b) 5R
(c) 3R                                       (d) 1.5R

Answer: (b)

As (μ2 /v ) – (μ2 / μ )= μ – 1 / R

Or ( 1.5 /x ) –  (1/x) =( 1.5 – 1/ R ) ⇒ x = 2.5R /0.5   = 5R 

16. If the tube length of astronomical telescope is 105 cm and magnifying power is 20 for normal setting, calculate the focal length of the objective

(a) 100 cm                                           (b) 10 cm
(c) 20 cm                                             (d) 25 cm

Answer: (a)
L = f0 + fe = 105 cm and M = 0 = f0 / f=20

⇒ f0 = 100 cm and fe = 5 cm.

17. An achromatic convergent doublet has power of +2D. If power of convex lens is + 5D, then ratio of the dispersive powers of a convergent and divergent lenses will be


(a) 3 : 5                                           (b) 2 : 5
(c) 2 : 3                                           (d) None of these

Answer: (a)
P = +2D   ∴ p1 = +5D    ∴ p2 = -3D

As p2 /p1    = f1 /f2 = – w1 /w = -3/5 ⇒ w1/w2  =3/5

18. A candle flame 3cm is placed at distance of 3m from a wall. How far from wall must a concave mirror be placed in order that it may form an image of flame 9cm high on the wall

(a) 425cm                                  (b) 300cm
(c) 450cm                                  (d) 650cm

Answer: (c)
Let the mirror be placed at a distance x from wall By using 

l/ o  =  -v / u  ⇒ -9 / 3  = -(-x) / -(x-3) (image will be inverted)

⇒ x= -4.5m =-450 cm

19. A thin rod of length f/3 lies along the axis of a concave mirror of focal length f. One end of its magnified image touches an end of the rod. The length of the image is

(a) f                                            (b) 1/2 f

(c) 2f                                          (d) 1/4 f

Answer: (b)

One end of its image touches an end
of the rod. So, one end of the rod is present at the centre of curvature of mirror.
Image is magnified. So, the other end
A is between C and focus.
Lets assume that end A of rod has its image at A’

For it,    u =- ( 2f    –     f/3 )   = -5f /3

So, using 1/ f   =  1/v    +   1/u        ⇒1/-f  =  1/v    +     (1/-5f/3 )  ⇒ v = (-5/2) f

∴ Length of image = ( 5/2 )f  –  2f = f/2

20. A concave mirror is placed on a horizontal table with its axis directed vertically upwards. Let O be the pole of the mirror and C its centre of curvature. A point object is placed at C. It has a real image, also located at C. If the mirror is now filled with water, the image will be


(a) Real, and will remain at C
(b) Real, and located at a point between C and ∞
(c) Virtual and located at a point between C and O
(d) Real, and located at a point between C and O

21. An object is placed in front of a convex mirror at a distance of 50cm. A plane mirror is introduced covering the lower half of the convex mirror. If the distance between the object and plane mirror is 30cm, it is found that the images formed by two mirrors are along same line. Radius of curvature of  mirror will be


(a) 12.5cm      (b) 25cm        (c) 50/3 cm       (d) 18cm

Answer: (b)
According to property of plane mirror it will form an image at a distance of 30 cm behind it. Hence, for convex mirror

u  =  –50cm, v =10  cm

By using 1/ f    =   1/v   +  1/u    ⇒  1/f  =  1/10       +     1/-50      = 4/50 

⇒  f =25 /2 cm        ⇒  R =2f =25cm 

22. Consider the situation shown in figure. Water ( μ= 4/3 )  filled in a breaker upto a height of 10cm. A plane mirror is fixed at a height of 5cm from the surface of water. Distance of the image of an object O at the bottom of the beaker formed by the mirror after reflection from it is

(a) 15cm    (b) 12.5cm    (c)7.5cm    (d) 10cm

Answer: (b) 

D apparent  = D actual / μrel    =(  10 /4/3  ) = 7.5 cm

So, the object appears to be raised by 10cm –  7.5cm  = 2.5cm 
Hence, distance between mirror and O’  = 5 +  7.5=  12.5cm
So, final image will be formed at 12.5 cm behind the plane mirror.

23. A glass prism of refractive index 1.5 is
immersed in water (μ=  4 / 3) A light beam incident normally on the face AB is totally reflected to reach the face BC if

(a) sin θ > 8 / 9  
(b) 2 / 3 < sinθ  <  8 / 9 
(c) sin θ  ≤ 2 / 3 
(d) cos θ  ≥ 8 / 9

Answer: (a)

From the figure, it is clear that TIR takes place at AC, only if θ > c 

⇒ sin θ >sin c                      ⇒ sin θ > 1/ ωμg                    ⇒ sin θ > 1/9/8            ⇒ sin θ >8/9          

24. A beam of light consisting of red, green and blue colour is incident on a right angled prism. The refractive indices of the material of the prism for the above red, green and blue
wavelength are 1.39, 1.44 and 1.47 respectively. The prism will

(a) Separate part of red colour from  the green and the blue colour
(b) Separate part of the blue colour from the red and green colour
(c) Separate all the colours from one another
(d) Not separate even partially any colour from the other two  colours

Answer: (a)
At face AB, i = 0. So, r = 0, i.e., no
bending will take place. So, light beam will be incident on face AC at an angle 45o. The face AC will not transmit the light for which i >θc , i.e.,  sin i  >  sin θc

0r sin 45° > (1/μ)  i.e.,                            μ  > √2  (=1.41)

Now as µR < µ  while µG and  µB >µ, so red will be transmitted through the face AC while green and blue will be reflected. So, the prism will separate red colour from green and blue.

25. An air bubble in a glass slab (µ = 1.5) is 6cm deep when viewed from one face and 4cm deep when viewed from the opposite face. The thickness of the glass plate is

(a) 10cm                                    (b) 6.67cm
(c) 15cm                                    (d) None of these

Answer: (c)

Let the thickness of slab be t and distance of air bubble from one side is x When viewed from side (1):

6 = x/ 1.5 ⇒ x =9cm 

When viewed from side (2): 4 = (t-x) /1.5 ⇒4 = ( t-9) /1.5 ⇒ t= 15cm

Formulaused : D apparent  = D actual / μ rel 

26. Figure given below shows a beam of light converging at point P. When a concave lens of focal length 16cm is introduced in the path of the beam at a place O shown by dotted line such that OP becomes the axis of the lens, the beam converges at a distance x from the lens. The value x will be equal to

(a)12cm                            (b) 24/3cm
(c)36/5cm                        (d) 48/7cm

Answer: (d)

From the figure shown it is clear that For lens: u = 12cm and v = x = ?

By using 1/ f  = 1/v     –     1/u  we have 

1/16 = 1/x   –   1/12    ⇒ x =48 /7 cm

27. An air bubble in a glass sphere having 4cm diameter appears 1cm from surface nearest to eye when looked along
diameter. If µg =  1.5, the distance of bubble from refracting surface is

(a) 1.2cm     (b) 3.2cm     (c) 2.8cm     (d) 1.6cm

Answer: (a)

By using μ2 /v   –  μ1 /u   =  μ2 – μ1 /R

where u = ? , v = − 1cm,

μ=1.5 ,  μ2 = 1,   R =-2cm

(1/-1)   –  1.5 / u  = (  1 – 1.5 )/-2    ⇒ u = -6/5 = -1.2cm

28. Two point light sources are 24cm apart. At what distance from one source, a convex lens of focal length 9cm be kept in between them, so that the images of both the sources are formed at the same place.


(a) 6cm    (b) 9cm    (c) 12cm     (d) 15cm

Answer: (a)

The given condition will be satisfied only if one source (S1) placed on one side such that u < f (i.e. it lies under the focus). The other source (S2 ) is placed on the other side of the lens such that u > f (i.e. it lies beyond the focus).

If S1 is the object for lens then

1/f = (1/-y)   –   (1/-x )  ⇒      1/y  =  (1/x ) –  (1/f ) …………( i )

If S2 is the object for lens then

1/f =( 1/y) –  ( 1/ -(24 – x)  )  ⇒   1/y = (1/f  )   –     1/(24 -x)……. ( ii)

From equation (i) and (ii)

1/x – 1/f = 1/f     –     (  1/ (24-x) )

⇒ 1/x  +  1/(24 -x)   =  2/f  = 2/9 

⇒ x2 -24x + 108 =0

On solving we have x = 18 cm,6cm

29. Aluminous object is placed at a distance of 30cm from the convex lens of focal length 20cm. On the other side of the lens, at what distance from the lens, a convex mirror of radius of curvature 10cm be placed in order to have an upright image of the object coinciding with it

(a) 12cm    (b) 30cm    (c) 50cm   (d) 60cm

Answer: (c)

For lens u =  30 cm, f = 20cm,  hence by using 1/f  =1/v   –   1/u we have 

1/20 = /v –  ( 1/ -30)    ⇒v-60cm

The final image will coincide with the object, if light ray falls normally on convex mirror as shown. From the figure it is clear that gap between lens and mirror is 60 – 10 = 50cm.

30. Angle of a prism is 30° and its refractive index is √2 and one of the surfaces is silvered. At what angle of incidence, a ray should be incident on one surface so that after reflection from the silvered surface, it retraces its  path

(a)  30°        (b)  60°     (c)      45°   (d)  sin-1  √ 1.5

Answer: (c)

From the figure, r1 = 30°  
Using snell’s law ,   i × sin i = μ sin r1

⇒ sin i = (√2) ×  sin 30º 

⇒i = 45º  

31. PQR is a right angled prism with other angles as o 60 and o 30 . Refractive index of prism is 1.5. PQ has a thin layer of liquid. Light falls normally on the face PR. For total internal reflection, maximum refractive index of liquid is


Answer: (c)
For TIR at PQ: θ > C
From the geometry of figure:  θ = 60° 

so, 60°   > c  ⇒ sin 60 °   > sin c 

⇒ √3/ 2  >  μ liquid / μ prism  ⇒ μ liquid  < √3 /2 × μ prism  ⇒ μ liquid < √3/ 2 × 1.5 ⇒ μ liquid <  1.3

32. A crow is sitting on a branch of a tree and on ground surface wheat grains are spread. The crow flies towards the ground and picks up a grain and sit on the front wall. The possible path are shown in diagram. Choose the correct path so that path length becomes minimum.

Given that BD = DE,

∠ AFB = EFC.

(a) ABC       (b) ADC     (c) AFC     (d) AEC

Answer: (c)

Light takes shortest path hence AFC’ = AFC.

33. For a prism kept in air it is found that for an angle of incidence 60° ,the angle of prism ‘A’, angle of deviation ‘ δ ‘ and angle of emergence ‘e’ become equal. Then the refractive index of the prism is

(a) 1.73     (b) 1.15     (c) 1.5       (d) 1.33

Answer: (a)
Given i  =  60° and A= δ = e

δ = i +e -A  ⇒ δ = i ( ⋅ e=A )

μ = sin  (  ( A+ δ ) /2   )  sin A/2

Here angle of deviation(δ )  is minimum (i =e)

μ = sin  (  ( 60°+ 60° ) /2   )  /  sin (60° /2 ) =1.732

34. A screen is placed a distance 40cm away from an illuminated object. A converging lens is placed between the source and the screen and it is attempted to form the image of the source on the screen. If no position could be found, the focal length of the lens

(a) must be less than 10cm
(b) must be greater than 10cm
(c) must not be greater than 20cm
(d) must not be less than 10cm.

Answer: (b)
Let the distance between source and lens be d. If image is formed on screen then v = 40 – d  Using lens equation 1/v  –   1/u = 1/f we have  (1/40 -d)  –   (1/-d) = 1/f      ⇒ f= (d- d2 /40 ) 

For ‘f’ to be maximum: df/dd = 0 

⇒ 1 – d/20 = 0      ⇒ d=20cm

∴ fmax = 20 – (20 )2 /40 = 10cm

This is maximum value of focal length for which image is formed on screen. But, in question image is not formed.  ∴ f > 10cm

35. The distance between an object and the screen is 100cm. A lens placed between them produces an image on the screen when the lens is placed at two positions 40cm apart. The power of the lens is nearly:

(a) 3 diopters             (b) 5 diopters
(c) 2 diopters             (d) 9 diopters

Answer: (b)

At first position of lens, let the distance of lens from object and screen be u and v respectively.

∴ x+y =100    ………..(1)

∴ At second position of lens the distance of lens from object and screen shall be v and u respectively.

∴ y-x =40   ………….(2) 

Solving equation (1) and (2) we get

y= 70 cm =70/100 = m and x=30cm =30/100 =m

∴ The power of lens is, p= 1/f = 1/y + 1/x =( 100/70 ) = 100 /21 = 5 diopters

36. When a light wave refracts from air into glass:

(a) the frequency increases and the wavelength decreases
(b) the frequency decreases and the wavelength increases
(c) the frequency does not change, but the wavelength decreases
(d) the frequency does not change, but the wavelength increases.

Answer: (c)
In denser medium, wavelength decreases. Frequency does not change on refraction.

37. In the figure shown a convex mirror of radius of curvature 20 cm is shown. An object O is placed in front of this mirror. Its ray diagram is shown. How many mistakes are there in the ray diagram (AB is its principal axis):

(a) 3                                      (b) 2
(c) 1                                       (d) 0

Answer: (b)

Error – I

Error – II

38. A ray of light is incident on a face of equilateral triangle at an incident angle 50. At this angle minimum deviation occurs. This deviation is

(a) 30   (b) 40    (c) 50      (d) None of these

Answer: (b)
A = 60°
At minimum deviation: i =e =50°

Also r1 = r2 = A/2 =30 °

δmin  = i +e -A =100° -60° =40 °

39. A hemispherical paper weight contains a small artificial flower of transverse size 2mm on its axis of symmetry at a distance of 4cm from its flat surface. What is the size of the flower as it appears to an observer when he looks at it along the axis of symmetry from the top? (Radius of the hemisphere is 10cm. Index of refraction of glass = 1.5).

(a) 1.5mm   (b) 2mm    (c) 2.5mm    (d) none of these

Answer: (c)

Here m =hi/ ho = μ1v/ μ2u

From the question,

μ1  = 1.5 , μ= 1

u = -6 cm ,v=?

∴ Using μ2/ v –  μ1/ u   =  μ– μ1 /R We have 

1/v  – (1.5 /-6) = (1- 1.5 ) / -10 


∴   hi/h0 = hi/2mm = 1.5 × -5 / 1 × -6  =1.25

⇒ h= 2.5mm

40. The mirror of length L moves horizontally as shown in the figure with a velocity v. The mirror is illuminated by a point source of light ‘P’ placed on the ground. The rate at which the length of  the  light spot on the ground increases is:

(a) v                                           (b) zero
(c) 2v                                        (d) 3v

Answer: (b)

At any general time  Length of light spot

=2(d +L) – 2d = 2L = const

41. Two thin slabs of refractive indices µ1 and µ2 are placed parallel to each other in the x-z plane. If the direction of propagation of a ray in  the two media are along the vectors

→                                        →

r1 = aiˆ + bjˆ  and r2     = ciˆ + djˆthen we have:


(a) µ1 a= µ2b                                         b) µ1 a / √a2 +b  =  µ2 c / √c2 +d2

c) µ1  (a2 +b)  =  µ2 (c2 +d2)

d)none of these

Answer: (b)

For snell’s law,

μ1 sin θ1 = μ2 sin θ2

μ1 a / √a2+b =  μ2 c / √c2+d2

42. A converging lens of focal length f is placed just above a water surface parallel to the surface without touching it. A point source of light S is placed inside the water vertically before the lens at a depth f from it. This arrangement will produce:

(a) a parallel beam of light converging from the lens
(b) a real image of S in air
(c) a virtual image of S in water
(d) a virtual image of S in air

Answer: (c)
Due to refraction at water-air interface, the object is at a distance less than the focal length of the lens, Thus, a virtual image is created inside water

43. A diverging lens of focal length –10 cm is moving towards right with a velocity 5 m/s.An object, placed on Principal axis is moving towards left with a velocity 3 m/s. Find the velocity of image at the instant when the lateral magnification produced is 1/2. All velocities are with respect
to ground.


(a) 3 m/s towards right                             (b) 3 m/s towards left
(c) 7 m/s towards right                             (d) 7 m/s towards left

Answer: (c)

dv / dt = v2/u2 . du /dt 

(where dv/dt denotes image speed w.r.t. lens and du/dt denotes object speed w.r.t. lens)

= m2.( du/dt )=( 1/4 ).8 =2

Therefore image speed w.r.t. lens is 2 m/s

Therefore image speed w.r.t. ground is 7 m/s towards right.

44. A converging lens is used to produce an image on a screen of an object. What change is needed for the image to be formed nearer to the lens?

(a) Increase the focal length of the lens (lens and position of object is fixed).
(b) Insert a diverging lens between the lens and the screen
(converging lens and position of object is fixed).
(c) Increase the distance of the object from the lens.
(d) Move the object closer to the lens.

Answer: (c)

if 1/v = 1/f    – 1/u 

if u → increases

1/u → decreases and

1/v→ increases

So v decreases.


45. A man starting from point P crosses a 4 km wide lagoon and reaches point Q in the shortest possible time by the path shown. If the person swims at a speed of 3 km/hr and walks at a speed of 4 km/hr, then his time of journey is(μsalt water = 4/3

(a) 4hr, 10 min                    (b) 4 hr and 30 min

(c) 3 hr and 50 min            (d) 5 hr and 10 min

Answer: (a)

As we know that light travels in a path such as to reach from one point to another in shortest possible time. Therefore, the man must travel along that path on which light would have travelled in moving from P to Q.
By Snell’s law;

sin i / sin r   =  μ2 / μ1

sin r   = (3/5 ) .  (4/3) =  4/5  ⇒ r= 53°

∴  AQ =10 Km

From P to A:  t1 =10/4  = 5/2

From A to Q: t2  =10/4  = 5/2

T= t1 + t2  = 5/3   +   5/2   =   25/6 hr

= (24/6   + 1/6 ) hr = ( 4hr + 1/6hr) = 4hr +10minutes


NERAM classroom is Tamil nadu’s one of the top NATA and JEE coaching center which made great change in students especially from small towns. Started in 2017,and successfully we have alumni who have been placed under top institutes such as NIT ,SPA ,AU and some other good institutes under Anna universities. NERAM classroom offers highly personalised and effective learning programs for aspirants of competitive exams like JEE, NATA etc . NERAM classroom has become one of the most preferred education platforms across India.



Our niche is creating personalised learning experiences for every type of learner. NERAM breaks away from a typical classroom teaching structure and uses technology and customisation to enable students to interact with their tutors.In India, students often feel too shy to clarify their doubts in a crowded coaching class. Many students travel long hours to get to class. NERAM breaks these shackles by liberating learning and teaching from the limits of location.Students are given full freedom to interact with teachers,experts and alumni as per their convenient time.



Traditionally, students rely on brick-and-mortar coaching centers, but companies like NERAM classroom are springing up and more students are supplementing offline courses with online tutoring.
NERAM classroom is an incredibly personalized tutoring platform for you, while you are staying at your home. We have grown leaps and bounds to be the best Online Tuition Website in India with immensely talented Master Teachers, from the most reputed institutions. We provide you year-long structured coaching classes for JEE and NATA entrance exam preparation at affordable tuition fees, with an exclusive session for clearing doubts, ensuring that neither you nor the topics remain unattended. Revision notes and formula sheets are shared with you, for grasping the toughest concepts. Assignments, Regular Homeworks, Subjective & Objective Tests promote your regular practice of the topics.



Traditionally, students rely on brick-and-mortar coaching centers, but companies like NERAM classroom are springing up and more students are supplementing offline courses with online tutoring.
Personalised Learning is the foundation of our unique programs for students. Using knowledge graphs, the program adapts and creates personalised learning journeys for students.Based on the student's progress, personal learning profiles are created which help analyse strengths and areas of improvement. Each concept gets tagged at different levels of difficulty to create a smooth learning curve which keeps students motivated through any new learning challenge. Whenever errors are made, motivating remedial methods are generated to strengthen and improve the student's learning experience.



At NERAM, we leverage technology to merge best practices like use of well verified study materials, videos, engaging content and quizzes with the best teachers so that every student across India has access to the best learning and guidance for their future .




NERAM classroom Learning Programs provide students a holistic learning experience. Students across regions can access the best teachers.Very few teams like BYJU’S, VEDANTA in India have a great sense for the education sector as well as how technology can be used for effective scaling, NERAM is one such team.
Hours and Hours of Study with no fun, is a bad idea for you, foreseeing the long run. To ensure that motivation is stirred in the best proportion for your clear understanding, a good number of quizzes and Objective tests are organized to impart knowledge. FREE Sample Papers and Important questions are extracted, solved and discussed, ensuring that you are 100% prepared before any exam.
Hours and Hours of Study with no fun, is a bad idea for you, foreseeing the long run. To ensure that motivation is stirred in the best proportion for your clear understanding, a good number of quizzes and Objective tests are organized to impart knowledge. FREE Sample Papers and Important questions are extracted, solved and discussed, ensuring that you are 100% prepared before any exam.

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