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NEET 2013 Free Practice and Sample Questions

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Medical Entrance Exams 2015 Free Practice and Sample Questions

It is said that Practice makes a man perfect. This is more so true in case of medical entrance exams like those of AIIMS and AIPMT.The mantra to crack medical entrance exam is - the more you practice, more you achieve. We have a complete set of practice tests for medical entrance exams. These practice tests have more questions than you would ever need to clear the AIIMS, AIPMT and other medical entrance exams 2015.

Following are a few questions picked up from our Sample Papers for AIIMS/AIPMT 2015. Click Here To Attempt Complete NEET 2015 Sample Test

Sample/Practice  Questions for AIIMS/AIPMT 2015 Physics

AIIMS Sample Question : 1 Which of the following graphs represents variation of gravitational potential with distance due to uniform solid sphere

(a)                                     (b)

(c)                                (d)

Answer Option : (C)

Answer Explanation : As we know that Gravitational Potential at any internal
point of the solid sphere will be

V = -GM/2R3 (3R2 - r2)--------------------(i)

But outside the it will be
V = -GM/r---------------------------------------(ii)

Hence above equation satisfied the third graph
therefore answer is 3rd option.

AIIMS Sample Question : 2 The distance of the object from the focus of convex lens is 4 cm whose real image is formed at  a distance of 9 cm from the focus then focal length of the lens will be

(a) 6 cm                                (b) 2 cm                    (c) 3 cm                    (d) 5 cm

Answer Option :

Answer Explanation : According to Newtons formula
   xy = f2
Here x = Distance of object from focus
        y = Distance of Image from focus

AIIMS Sample Question : 3 Two point charges  of masses m kg and 2m kg respectively are connected with each other by means of massless non conducting rod to form an electric dipole of length 2l . The dipole is placed in a uniform electric field of field strength  . Initially the axis of the dipole is parallel to the field . if the dipole axis is slightly tilted and released the system performs angular oscillations about its centre of mass , the frequency of angular oscillation will be

(a)                         (b)                 (c)                 (d)

Answer Option :

Answer Explanation :

AIIMS Sample Question : 4 The energy stored in the the capacitor in the given network in the steady state

(a) 90 mJ                        (b) 30 mJ                        (c) 20 mJ                        (d) 10 mJ

Answer Option : (A)

Answer Exaplanation :
i = V/R = 40/20
i = 2Amp

P.D across C = iR1 = 2X15

                    = 30 V
E = 1/2 CV2 = 1/2 X 2 X10-6 X 900
                   = 9 X 10-4
                   = 0.9X10-3J
                   = 90 mJ

AIIMS Sample Question : 5 A particle slides on the surface of a fixed smooth sphere starting from the highest point , then the angle rotated by the radius vector through the particle when it leaves contact with the air will be

                    (b)                     (c)                     (d)

Answer Option : (A)

Answer Explanation :

Let the require angle be θ then
to loose contact N = 0
mgcosθ = mv2/r,   v2 = grcosθ-------------------(1)
According to laws of conservation of energy
mg(r-rcosθ) = 1/2mv2
2gr(1-cosθ) = v2
From eq (1) & (2)
2gr(1-cosθ) = grcosθ
2-2cosθ = cos θ
2 = 3cosθ
cosθ = 2/3
θ= cos-1(2/3)

AIIMS Sample Question : 6 The speed of sound in gas in which two waves of wave lengths 1 m and 1.02 m produces 10 beats per second will be

(a) 510 m/s
                    (b) 200 m/s                    (c) 100 m/s                    (d) 50 m/s

Answer Option : (A)

Answer Explanation :

AIIMS Sample Question : 7 The charge +q is uniformly distributed over a thin half ring of radius R then the electric field intensity at the centre of the ring will be

                    (b)                     (c)                     (d)

Answer Option :

Answer Explanation :

Sample/Practice  Questions for NEET 2015 Chemistry

AIIMS Sample Question :8 pH of 10-8M Ca(OH)2 is

(a) 6.3
                            (b)7.7                                (c) 7.08                            (d) 8

Answer Option : (C)

Answer Explanation : [(CaCOH)2] = 10-8 M
[OH-] = 2X10-8 + 10-7 M
So pOH=-log10(2x10-8 + 10-7)
             = -log10(12x10-8)
            =8-log10 12
So  pH= 14-pOH
          =14- 6.92

AIIMS Sample Question :9 Extraction of zinc from zinc blende is achieved by

(a) Electrolytic reduction                                                                            (b) Roasting followed by reduction with carbon
(c) Roasting followed by reduction with another metal                           (d) Roasting followed by self reduction

Answer Option : (D)

Answer Explanation : σn o-2, no. of valence electrons = 6+6+1 =13
So Its moleculare orbital electronic is
        σIs2 σ*Is2 σ2s2 σ*2s2σ2p22π2p2xπ2py2π*2p2
  O-2 has one unpaired electron
Therefore , KO2 is paramagnetic.

Sample Question : 10 The IUPAC name of Na4[Fe(CN6)]

(a) Sodium Hexacyano Iron?                                                                     (b) Sodium Iron? hexacyano                   
(c) Tetra Sodium Iron? hexacyano                                                             (d) Sodium  hexacyano ferrate?

Answer Option : (D)

Answer Explanation : Na4[Fe(CN)6]
Oxidation no. of fe=6-4=2
So   Its IUPAC name is sodium hexacyanoferrate(II)

Sample Question :11
In which case (Ecell-Eocell) is zero

(a) CulCu2+ (0.01M)? Ag+(0.1M) I  Ag                                                                       (b) Pt(H2) |ph =1? Zn2+(0.01M)|Zn
(c) Pt(H2)ph =1? Zn2+(1M)|Zn                                                                                   (d) Pt(H2) |H+(0.01M)? Zn2+ (0.01  M)  |Zn

Answer Option : (A) ANd (B)

Answer Explanation : From nersl,s equation
            Ecell = Eocell -.059/n logQ
       =>Ecell-Eocell = - .059/n logQ
For Cu |Cu2+ (.01 M) || Ag+ (.1M) |Ag
     Ecell – Eocell = -.059/2 log  .01/(.1)2
For  Pt(H2) |pH = 1 | | Zn2+(.01 M) | Zn
        Ecell – Eocell = -.059/2 log (10-1)2/.01
                               = - .59/2 log .01/.01
                               = 0
For Pt (H2) |Ph = 1 || Zn2+ (1M) |Zn   
       Ecell- Eocell = -.059/n log Q
                             = -.059/2 log (10-1).2/1
                             =-.059/2 log 10-2
                             = - . 059
For  Pt(H2) |H+(.01M) | | Zn2+ (.01M) | Zn
        Ecell – Eocell = - . 059/2 log (.01)2/.01
                                =- . 59
Hence  (a) and (b) are correct.

Sample Question :12 The reactivity of an element becomes  th of its original value in 60 sec. Then half life period is

(a) 15 sec                        (b) 20 sec                        (c) 5 sec                        (d) 40 sec

Answer Option : (A)

Answer Explanation :

4XT1/2 =60
T1/2 =15sec

Sample Question : 13 Among the following processes for removal of hardness of water, which of the following uses ion-exchange process?

(a) Permutit process                (b) Clark’s process                (c) Calgon process                (d) None of the above

Answer Option : (A)

Answer Explanation : Clark process – Uses slaked lime
Calgon process – Uses sodium hexametaphosphate salt

Sample Question : 14 Match the following:

(a) Dynamite                                                    (p) cyclomethylenetrinitramine

(b) TNT                                                              (q) nitrocellulose

(c) RDX                                                              (r) nitroglycerin

(d) Guncotton                                                     (s) trinitrotoluene

Answer Option : a:r|b:s|c:p|d:q

Answer Explanation :

Dynamite – nitroglycerin
TNT - trinitrotoluene
RDX - cyclomethylenetrinitramine
Guncotton – nitrocellulose

Sample Question : 15 Which of the below mentioned alcohols will have the highest boiling point

(a) 1,2- dihydroxy benzene               
                    (b) 1,3- dihydroxy benzene               
(c) 1,4- dihydroxy benzene              
  (d) hydroxyl benzene

Answer Option : (C)

Answer Explanation : 1,4- dihydroxy benezene will have the highest boiling point because it forms intermolewlar hydrogen bonding.

Sample/Practice  Questions for NEET 2015 Biology

Sample Question : 16 Which of the following is not true for a species ?

(a) Members of a species can interbreed                                                

(b) Gene flow does not occur between  the populations of a species
(c) Each species is reproductively isolated from every other species
(d) Variations occur among  members of  a species

Answer Option : (C)

Answer Explanation : Gene flow does occur between  the populations of a species

Sample Question : 17 One endangered species of Indian medicinal plants is that of

(a) Podophyllum                        (b) Ocimum                        (c) Garlic                        (d) Nepenthes

Answer Option : (A)

Sample Question : 18 Which one of the following mineral elements plays an important role in biological nitrogen fixation?

(a) Molybdenum                        (b) Copper                         (c) Manganese                (d) Zinc

Answer Option : (A)

Sample Question : 19 What type of placentation is seen in sweet pea?

(a) Basal                                    (b) Axile                             (c) Free Central                (d) Marginal

Answer Option : (D)

Sample Question : 20 The animal with bilateral symmetry in young stage and radial pemamerous symmetry in the adult stage belong to the phylum

(a) Annelida                            (b) Mollusca                         (c) Cnidaria                        (d) Echinodermata

Answer Option : (D)
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