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Ray And Wave Optics Mcq
Quiz 3
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Q.1
A rectangular glass slab ABCD of refractive index n1 is immersed in water of refractive index n2 ( n1 > n2). A ray of light is incident at the surface AB of the slab as shown. The maximum value of the angle of incident αmax such that the ray comes out only from the other surface CD is given by [ IIT 2000]
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a)
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b)
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c)
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d)
Explanation
See figure. the ray will come out from CD if it suffers total internal reflection at surface AD. i.e. it strikes the surface AD at critical angle C Applying Snell's law at P n1sinC=n2 sinC=n2 / n1 C=sin-1(n2 / n1)Applying Snell's law at Q n2α=n1cosCThus Answer: (a)
Q.2
In a double slit experiment instead of taking slits of equal width, one slit is made twice as wide as the other. Then, in the interference pattern [ IIT 2000]
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a)the intensities of both the maxima and the minima increase
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b) the intensity of the maxima increases and the minima has zero intensity
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c)the intensity of the maxima decreases and that of the minima increases
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d)the intensity of the maxima decreases and the minima has zero intensity
Explanation
When slits are equal width Imax ∝ (a+a)2=4a2 Imin=(a-a)2=0When one slit's width is twice that of other Answer: (a)
Q.3
In a compound microscope, the intermediate image is [ IIT 2005]
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a) virtual, erect and magnified
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b) real, erect and magnified
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c)real, inverted and magnified
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d)virtual, erect and reduced
Explanation
Answer: (c)
Q.4
Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. the phase difference between the beams is π/2 at point A and π at point B. Then the difference between the resultant intensities at A and B is [ IIT 2001]
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a) 2I
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b) 4I
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c)5I
0%
d)7I
Explanation
We know that I ∝ I1 + I2 + 2√I1× √I2 cosφ --eq(1)Applying equation (1) when phase difference is π/2 Iπ/2 ∝ I+4I=5I Applying equation (1) when phase difference is π Iπ ∝ I+4I+2√I×√4Icosπ=I ∴ Iπ/2 - Iπ ∝ 4I Answer:(b)
Q.5
In a Young's double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen when light of wave length 600nm is used. If the ave length of light is changed to 400 nm, number of fringes observed in the same segment of the screen is given by [ IIT 2001]
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a) 12
0%
b) 18
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c) 24
0%
d) 30
Explanation
In Young's double slit experiment Fringe Width=λD / d Given that 12 fringe occupy the segment of screen y=12λ1D / d Here λ1=600 nm Also k fringes of wavelength λ2 occupy the same segment of screen thus y=kλ2D / d where λ2=400 nm From above Answer: (b)
Q.6
A ray of light passes through four transparent media with refractive indices n1, n2, n3 and n4 as shown in the figure. the surface of all media are parallel. if the emergent ray CD is parallel to the incident ray AB, we must have [ IIT 2001]
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c)n3=n4
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a)n1=n2
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b) n2=n3
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d)n4=n1
Explanation
Emergent ray is parallel to incident ray hence refractive index n1=n4Answer: (d)
Q.7
A given ray of light suffers minimum deviation in an equilateral prism P. Additional prism Q and R of identical shape and of the same material as P are now added as shown in the figure. The will now suffer [ IIT 2001]
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a) greater deviation
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b) no deviation
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c)same deviation as before
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d)total internal reflection
Explanation
prism P and Q both together form a glass slab with opposite faces parallel. Thus ray emerging from Q will not suffer any deviation. Thus prisR will cause deviation . Hence ay will now have the same deviation.Answer: (c)
Q.8
An observer can see through a pin-hole the top end of a thin rod of height h, placed as shown in figure. The beaker height is 3h and its radius h. When the beaker is filled with a liquid up to a height 2h, he can see the lower end of the rod. Then the refractive index of the liquid is [ IIT 2002]
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a)5/2
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b)√(5/2)
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c)√(3/2)
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d)3/2
Explanation
For the image pint P to be seen by the observer, it should be formed at point Q ΔAQM,QA=AM=h∴ ∠AQM=45°∴ ∠ r=45° sin r=1/√2ΔAQMPB=h and BM=2h∴ sin i=1/√5from snell's law at M n2sin i=n1sin rn2 (1 / √5)=1× ( 1/√2)n2=√ (5/2) Answer:(b)
Q.9
Which of the following spherical lenses does not exhibit dispersion? The radii of curvature of the surface of the lenses are as given in the diagram [ IIT 2002]
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a)
0%
b)
0%
c)
0%
d)
Explanation
option "c" is correct as both surfaces have same radius of curvature on the same side, no dispersion will occur Answer: (c)
Q.10
In the ideal double slit experiment, when a glass-plate ( refractive index 1.5) of thickness 't' is introduced in the path of one of the interfering beams ( wave-length λ), the intensity at the position where the central maximum occurred previously remains unchanged. the minimum thickness of the glass-plate is [ IIT 2002]
0%
a)2λ
0%
b) 2λ / 3
0%
c)λ / 3
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d)λ
Explanation
Path difference=( n -1)t=nλFor minimum t, n=1∴ t=2λAnswer: (a)
Q.11
Two plane mirrors A and B are aligned parallel to each other, as shown in figure. A light ray is incident at an angle 30° at a point just inside one end of A. the plane of incidence coincides with the plane of the figure. the maximum number of times the ray undergoes reflections ( including the first one) before it emerges out is [ IIT 2002]
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a) 28
0%
b) 30
0%
c)32
0%
d)34
Explanation
Maximum reflection=l / xwhere x=0.2tan30=0.2 × (1 / √3) Number of reflection=2√3 / ( 0.2/√3)=30Answer: (b)
Q.12
In the adjacent diagram, CP represents a wavefront and AO and BP, the corresponding two rays. Find the condition on θ for constructive interference at P between the ray BP and reflected ray OP [ IIT 2003]
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a)cosθ=3λ / 2d
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b) cosθ=λ / 4d
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c)secθ - cosθ=λ / d
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d)secθ - cosθ=4λ / d
Explanation
From figure OP=d / cosθ from ΔCOP CO=dcos2θ / cosθFrom the condition for constructive interference Path difference=nλ for n=1Path difference=CO + OP + λ/2=λλ/2 is added as on reflection phase changes by π which is equal to λ/2 Answer:(b)
Q.13
A ray of light traveling in water is incident on its surface open to air. The angle of incident is i, which is less than the critical angle. Then there will be .. [ IIT 2007]
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a)only a reflected ray and no refracted ray
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b) only a refracted ray and no reflected ray
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c) a reflected ray and refracted ray and the angle between them would be less than 180° - 2i
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d)a reflected ray and a refracted ray and the angle between them would be greater than 180° -i
Explanation
The ray is partly reflected and partly refracted ∠MOB=180 -2i Answer: (c)
Q.14
The size of the image of an object, which is at infinity, as formed by a convex lens of focal length 30cm is 2cm. If a concave lens of focal length 20cm is placed between the convex lens and the image at a distance of 26 cm from the convex lens, calculate the new size of the image [ IIT 2003]
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a) 1.25 cm
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b) 2.5 cm
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c) 1.05 cm
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d) 2.0 cm
Explanation
Convex lens froms the image at second focus at I1. size of image is 2cm. I1 acts as a virtual object for concave lens. Concave lens forms image of I1 at I2 From lens formula for concave lens 5 cm is the distance of image I2 from concave lens Now magnification=v/u=5/4 magnification=size of image / size of object 5/4=size of image / 2 size of image=2.5 cm Answer: (b)
Q.15
A ray of light is incident at the glass-water interface at an angle i, it emerges finally parallel to the surface of water, then the value of ng would be [ IIT 2003]
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a)(4/3)sin i
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b) 1 / sini
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c)4/3
0%
d)1
Explanation
using Snells law at glass water surface ng sin i=nw sin r --eq(1)using Snells law at water air surface nw sin r=nasin 90 --eq(2)from equation 1 and 2ng sin i=nasin 90ng=1/ sin iAnswer: (b)
Q.16
A beam of white light is incident on glass air interface from glass to air such that green light just suffers total internal reflection. The colors of the light which will come out to air are [ IIT 2004]
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a) Violet, Indigo, Blue
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b) All colors except green
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c)Yellow, Orange, Red
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d)White light
Explanation
sin C=1/n and Refractive index n ∝ (1/λ) since sin C ∝ λ For higher value of λ, the angle C also increases Thus colors having more wave length than green will come out option (c)Answer: (c)
Q.17
An equilateral prism is placed on a horizontal surface. A ray PQ is incident onto it. For minimum deviation [ IIT 2004]
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a) PQ is horizontal
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b) QR is horizontal
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c)RS is horizontal
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d)Any one will be horizontal
Explanation
For minimum deviation, incident angle is equal to emerging angle possible when QR is horizontal Answer:(b)
Q.18
Monochromatic light of wavelength 400nm and 560nm are incident simultaneously and normally on double slits apparatus whose slits separation is 0.1 mm and screen distance is 1m. Distance between areas of total darkness will be [ IIT 2004]
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a) 4 mm
0%
b) 5.6 mm
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c) 14 mm
0%
d) 28 mm
Explanation
At the area of total darkness minima will occur for both the wave lengths path difference for minima=(2n+1)λ / 2 by inspection for m=2 and n=3 and for m=7 , n=10, the distance between them will be th distance between such points put n2=10, n1=3On solving we get Δx=28 mm Answer: (d)
Q.19
A point object is placed at the centre of glass sphere of radius 6cm and refractive index 1.The distance of virtual image from the surface is [ IIT 2004]
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a)6 cm
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b) 4cm
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c)12cm
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d)9 cm
Explanation
Here u=- 6cmR=-6cmn1=1 (air)n2=1.5Using formula of spherical refracting surface Answer: (a)
Q.20
In Yong's double slit experiment intensity at a point is (1/4) of the maximum intensity. Angular position of this point is [ IIT 2005]
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a) sin-1 (λ/d)
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b) sin-1 (λ/2d)
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c)sin-1 (λ/3d)
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d)sin-1 (λ/4d)
Explanation
Let P be the point on the central maxima whose intensity is one fourth of the maximum intensity For interference we know that Now I1=I2=I and Imax=4I Thus I=I+I+2IcosΦ ⇒ -(1/2)=cosΦ ∴ Φ=2π/3 For a phase difference of 2π/ 3, the path difference is But the path difference ( in terms of P and Q ) is dsin θ as shown in figure ∴ d sinθ=λ/3∴ θ=sin-1( λ / 3d)Answer: (c)
Q.21
A convex lens is in contact with concave lens. The magnitude of ratio of their power is 2/Their equivalent focal length is 30 cm. What are their individual focal lengths? [ IIT 2005]
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a) -15, 10
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b) -10, 15
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c)75, 50
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d)-75, 50
Explanation
Focal length of their combination Answer:(a)
Q.22
A container is filled with water ( n=1.33) upto a height of 33.25 cm. A concave mirror is placed 15cm above the water level and the image of an object placed at the bottom is formed 25 cm below the water level. Focal length of the mirror is [ IIT 2005]
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a) 15 cm
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b) 20 cm
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c) -18.31 cm
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d) 10 cm
Explanation
i) the image I' for first refraction ( when rays comes out of liquid) is at the depth of Apparent depth=Real Depth / n as observer is in air Apparent depth=33.25 / 1.33=25 cm ii) Now, reflection will occur at concave mirror. For this I' behaves as an object u=-(15+25)=-40 cm Using mirror formula Given image is formed at 25 cm below liquid thus real image=25/1.33=18.79 ( observer is in liquid ) v=15 + 18.79=- 33.79 Answer: (c)
Q.23
Focal length of the plano-convex lens is 15 cm. A small object is placed at A as shown in figure. The plane surface is silvered. The image will be formed at [ IIT 2006]
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a)60cm to left of the lens
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b) 12 cm to the left of the lens
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c)60 cm to right of lens
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d)30 cm to the left of lens
Explanation
The focal length of equivalent mirror is since f has positive value, the combination behaves as a converging mirrorNOTE : The sign convention for the above formula i) Focal length of converging lens or mirror is positiveii) Focal length of diverging lens or mirror is negative Use mirror formula for converging mirror to get answerAnswer: (b)
Q.24
the graph shows relationship between object distance and image distance for equiconvex lens. Then focal length of the lens is [ IIT 2006]
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a) 0.50±0.05 cm
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b) 0.50±0.10 cm
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c)5.00±0.05 cm
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d)5.00±0.10 cm
Explanation
We know that in case of convex lens when object is placed at C, the image is obtained at C. This situation is represented in the graph by the point corresponding to u=-10 cm and v=-10 cm.Therefore R=10 cm and f=R/2=10/2=5 cmfrom lens formulaTherefore, the focal length=( 5.00±0.05 cm)Answer: (c)
Q.25
Rays of light from Sun falls on a biconvex lens of focal length f and the circular image of Sun of radius r is formed on the focal plane of the lens. Then [ IIT 2006]
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a) Area of image is πr2 and area is directly proportional to f
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b) Area of image is πr2 and area is directly proportional to f2
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c)Intensity of image increases if f is increased
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d)If lower half of the lens is covered with black paper are will become half
Explanation
From figure in ΔABC , tanβ=AB / AC AB=AC tanβ 2r=f tanβ ⇒ Area of image=πr2 ∝ f2 Answer:(b)
Q.26
Two beams of red and violet colors are made to pass separately through a prism ( angle of the prism is 60°). In the position of minimum deviation, the angle of refraction will be [ IIT 2008]
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a) 30° for both the colors
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b) greater for violet colors
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c)greater for red colors
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d)equal but not 30° for both the colors.
Explanation
For a minimum deviation the ray in the prism is parallel to the base of the prism. This condition does not depend on the colors ( or wave length ) of incident radiation. So in both the cases, by geometry r=30°. So option (a) is correctAnswer: (a)
Q.27
A light beam is traveling from region I to IV( figure). The refractive index in regions I, II, II and IV are no , no/2, no/6, no/8 respectively. The angle of incident θ for which the beam just misses entering region IV is .. [ IIT 2008]
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a) sin-1(3/4)
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b) sin-1(1/8)
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c)sin-1(1/4)
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d)sin-1(1/3)
Explanation
Light ray do not enter region IV. It implies that angle of refraction must be 90° at the surface separating region III and IV Thus no sinθ=(no/8) sin 90sinθ=(1/8) θ=sin-1(1/8) Answer:(b)
Q.28
A ball is dropped from height of 20 m above the surface of water in a lake. the refractive index of water is 4.A fish inside the tank, in the line of fall of the ball, is looking at the ball. At an instant, when the ball is 12.8 m above the water surface, the fish sees the speed of ball as [ Take g=10 m/s2] [ IIT 2009]
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a) 9 m/s
0%
b) 12 m/s
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c) 16 m/s
0%
d) 21.33 m/s
Explanation
Displacement in air=20.0 - 12.8=7.2 m Velocity of ball in air V2 - u2=2gs v2=2×10×7.2 v=12 m/s Apparent velocity in water by definition of refractive index of air with respect to water velocity in water / velocity in air=4/3 velocity in water=velocity in air ×(4/3) velocity in water=12 × (4/3)=16 m/s Answer: (c)
Q.29
An astronomical telescope has a large aperture to [ IIT 2002]
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a)reduce spherical aberration
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b) high resolution
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c)increase span of observation
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d)have low dispersion
Explanation
Resolving power of telescope=D / 1.22λAnswer: (b)
Q.30
If two mirrors are kept at 60° to each other, then the number of images formed by them is [ AIEEE 2002]
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a) 5
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b) 6
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c)7
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d)8
Explanation
When two plane mirrors are inclined at each other at an angle θ then the number of images of a point object placed between the plane mirror is ( 360/θ) - 1, if 360/θ is even.∴ Number of images formed=(360/60 - 1)=5Answer: (a)
Q.31
Electromagnetic waves are transverse in nature is evident by [ AIEEE 2002]
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a) polarization
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b) interference
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c)reflection
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d)diffraction
Explanation
Answer:(a)
Q.32
Wavelength of light used in an optical instrument are λ1=4000 Å and λ2=5000 Å, then ratio of their respective resolving powers ( corresponding toλ1 and λ2) is [ AIEEE 2002]
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a) 16:25
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b) 9:1
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c) 4:5
0%
d) 5:4
Explanation
We know that Resolving power & 1 / λ Thus (R.P)1 / (R.P)2=λ2 / λ1 (R.P)1 / (R.P)2=5/4 Answer: (d)
Q.33
A lens is placed between a source of light and a wall. It forms image of area A1 and A2 on the wall, for its two different positions. The area of the source of light is [ CBSE 1995]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
The length of the object is geometric mean of the length of the two imagesAnswer: (c)
Q.34
Diameter of plano-convex lens is 6cm and thickness is 3 mm. If speed of light in the material of the lens is 2×108 metres per second, the focal length of the lens is [ CPMT 1989]
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a) 15 cm
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b) 20 cm
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c)30 cm
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d)10 cm
Explanation
Refractive index of the lens n=c/v=3×108 / 2×108=1.5From the geometry of figure, using Pythagoras we get OB2=ON2 + BN2Given BN=3 cm and OB=AO=radius=R AN=3mm=0.3cm R2=(R-0.3)2 + 32R2=R2 - 0.6R + .09 + 9 neglecting 0.09 we get R=15 cm Now from lens makers formula Answer: (c)
Q.35
A ray of light falls on the surface of spherical glass paper weight making an angle α with the normal and is refracted in the medium at an angle of β. The angle of deviation of emergent ray from the direction of incident ray from the direction of incident ray is
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a) (α -β)
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b)2(α -β)
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c)(α -β) / 2
0%
d)(β - α)
Explanation
In ΔABC , AC=BC=radius ∠CAB=∠ CBA=β Naturally ∠IBN=α δ1=α - β δ2=α - β ∴ δ=δ1 + δ2δ=2(α - β) Answer:(b)
Q.36
Which of the following is used in optical fibers [ AIEEE 2002]
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a) total internal reflection
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b) scattering
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c) diffraction
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d) refraction
Explanation
Answer: (a)
Q.37
Consider telecommunication through optical fibers. Which of the following statement is not true? [ AIEEE 2003]
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a)Optical fibers can be of graded refractive index
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b) Optical fibers are subject to electromagnetic interference from outside
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c)optical fibers have extremely low transmission loss
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d)Optical fibers may have homogeneous core with a suitable cladding
Explanation
Answer: (b)
Q.38
To demonstrate the phenomenon of interference, we require two sources which emit radiation [ AIEEE 2003]
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a) of nearly the same frequency
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b) of the same frequency
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c)of different wave lengths
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d)of the same frequency and having a definite phase relationship
Explanation
Answer: (d)
Q.39
The image formed by an objective of a compound microscope is [ AIEEE 2003]
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a) virtual and diminished
0%
b) real and diminished
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c)real and enlarged
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d)virtual and enlarged
Explanation
A real, inverted and enlarged image of the object is formed by the objective lens of compound microscope Answer:(c)
Q.40
To get three images of single object, one should have two mirror at an angle of [ AIEEE 2003]
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a) 60°
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b) 90°
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c) 120°
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d) 30°
Explanation
Since number of images is odd number (360 /θ) -1=number of images (360 /θ) -1=3θ=360/4=90 Answer: (b)
Q.41
A light ray is incident perpendicularly to one face of a 90° prism and is totally internally reflected at the glass-air interface. If the angle of reflection is 45°, we conclude that the refractive index n [ IIT 2004]
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a)n > 1 /√2
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b) n > √2
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c)n < 1 /√2
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d)n < √2
Explanation
The incident angle is 45° Incident angle > critical angle ∴ sin i > sin C sin C=i / n ∴ sin 45 > 1/n (1/√2) > (1/n) n > √ 2Answer: (b)
Q.42
A plano convex lens of refractive index 1.5 and radius of curvature 30cm. Is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens an object be placed in order to have a real image of size of object [ AIEEE 2004]
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a) 60cm
0%
b) 30 cm
0%
c)20 cm
0%
d)80 cm
Explanation
focal length of lens fl focal length of combination is by substituting values of fl The combination acts as a converging mirror, For image to be of the same size of object u=2f=20 cmAnswer: (c)
Q.43
The angle of incidence at which reflected light is totally polarized for reflection from air to glass ( refractive index n) is [ AIEEE 2004]
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a) tan-1 (1/n)
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b) sin-1(1/n)
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c) sin-1(n)
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d) tan-1 (n)
Explanation
The angle of incidence for total polarization is given by tanθ=n θ=tan-1n Answer:(d)
Q.44
The maximum number of possible interference maxima for slit-separation equal to twice the wave length in Young's double-slit experiment is [ AIEEE 2004]
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a) three
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b) five
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c) infinite
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d) zero
Explanation
For constructive interference dsinθ=nλ given d=2λ Thus sinθ=n /2 n=0, 1, -1, 2,-2 hence five maxima are possible Answer: (b)
Q.45
An electromagnetic wave of frequency f=3.0 MHz passes from vacuum into a dielectric medium with permittivity ε=4.Then [AIEEE 2004]
0%
a)Wave length is halved and frequency remains unchanged
0%
b) Wave length is doubled and frequency become half
0%
c)Wave length is doubled and the frequency remains unchanged
0%
d)Wave length and frequency both remains unchanged
Explanation
Frequency remains constant during refraction velocity in medium∴ wavelength is halved and frequency remains unchangedAnswer: (a)
Q.46
A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is 4/3 and fish is 12 cm below the surface, the radius of this circle in cm is [ AIEEE 2005]
0%
a) 36 / √7
0%
b) 36√7
0%
c)4√5
0%
d)36√5
Explanation
Fish observes circular horizon because of total internal reflection Answer: (a)
Q.47
The angle of prism is 30°. The rays incident at 60° on one refracting face suffer a deviation of 30°. Then the angle of emergence is [ EAMCT 1990]
0%
a) 0°
0%
b) 30°
0%
c)60°
0%
d)90°
Explanation
we know that r1=i -δr1=60 - 30 r1 + r2=A 30 + r2=30 r2=0 since angle of incidence on second face is zero , thus angle of emergence=0 Answer: (a)
Q.48
The correct relation between time interval δ and phase difference φ is
0%
a)δ=(T/2π)φ
0%
b) δ=(2π/T)φ
0%
c)δ=2πφ
0%
d)δ=φ/2π
Explanation
Answer: (a)
Q.49
Two point white dots are 1 mm apart on blank paper. They are viewed by eye pupil diameter 3 mm. Approximately what is the maximum distance at which these dots can be resolved by the eye? [ Take wavelength of light=500 nm] [AIEEE 2005]
0%
a)1 m
0%
b)5 m
0%
c)3 m
0%
d)6 m
Explanation
Distance between two points y=1 mm=10-3 m Diameter of pupil d=3 mm=3×10-3 mwavelength of light λ=500 nm=5×10-7 From formula Answer:(b)
Q.50
A thin glass ( refractive index 1.5) lens has optical power of -5D in air. Its optical power in a liquid medium with refractive index 1.6 will be [ AIEEE 2005]
0%
a) -1D
0%
b) 1D
0%
c) -25D
0%
d) 25D
Explanation
From lens makes' formula Answer: (b)
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