datavorous/entrance-exam-dataset huggingface dataset:
A short bar magnet placed with its axis at 30o with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 x 10-2 J. What is the magnitude of magnetic moment of the magnet ?

Options:
A 0.36 J/T
B 0.25 J/T
C 1.65 J/T
D 1.05 J/T

The correct answer is:
0.36 J/T

Solution:
Given, uniform magnetic field
B = 0.25 T

The magnitude of torque τ = 4 . 5 × 1 0 - 2 J
Angle between magnetic moment and magnetic field θ = 3 0 o
Torque experienced on a magnet placed in external magnetic field
τ = M × B
τ = MB sin θ âµ A × B = AB sin θ
4.5 x 10-2 = M x 0.25 x sin 30o
M = 4 . 5 × 1 0 - 2 0 . 2 5 × sin 3 0 o
= 4 . 5 × 1 0 - 2 × 2 0 . 2 5 × 1 âµ sin 3 0 o = 1 2
= 0.36 J/T.

Topic: Physics, Magnetic Effects of Current, NEET
A short electric dipole has a dipole moment of 16×10–9 C m. The electric potential due to the dipole at a point at a distance of 0.6 m from the centre of the dipole, situated on a line making an angle of 60° with the dipole axis is:
14πε0=9×109 N m2 C-2

Options:
A 200 V
B 400 V
C zero
D 50 V

The correct answer is:
200 V

Solution:
We know that,

Electric dipole potential, V=14πε∘pcosθr2, for large distance of r.

V=kPcosθr2=9×109×16×10−9(0.6)2×12

V=200 V

Topic: Physics, Electrostatics, NEET, NEET 2020 (Phase 1)
A short linear object of length b lies on the axis of a concave mirror of focal length f at a distance u from the pole. The length of the image will be.

Options:
A fu-fb
B u-ffb
C fu-f2b
D u-ff2b

The correct answer is:
fu-f2b

Solution:
The concave mirror formula is
1v+1u=1f .... (i)
Since, for a given concave mirror, focal length f is fixed, we partially differentiate Eq. (i) to get
-Î´vv2=-Î´uu2=0 or Î´v=-vu2 Î´u ...... (ii)
Multiplying Equation. (i) by u , we get
uv+1=uf or vu=fu-f ..... (iii)
Using Equation. (iii) in Equation. (ii), we get
Î´v=-fu-f2Î´u
Given Î´u=b. Therefore
Î´v=-fu-f2b
The negative sign shown that image is longitudinally inverted. The magnitude of the size of the image is
Î´v=bfu-f2

Topic: Physics, Ray Optics, NEET
A signature written with carbon pencil weighs 1 mg. what is the number of carbon atoms present in the signature?

Options:
A 6.02×1020
B 0.502×1020
C 5.02×1023
D 5.02×1020

The correct answer is:
0.502×1020

Solution:
12 gm carbon contain = 6.023× 10 23
10 −23  gm 1mg = 6.023× 10 23 12 × 10 −3
=0.502× 10 20

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
A silicon diode has a threshold voltage of 0.7 V. If an input voltage given by 2sin(Ït) is supplied to a half-wave rectifier circuit using this diode, the rectified output has a peak value of

Options:
A 2 V
B 1.4 V
C 1.3 V
D 0.7 V

The correct answer is:
1.3 V

Solution:
The peak value of rectified output voltage = peak value of input voltage – barrier voltage
= 2 - 0.7 = 1.3 V

Topic: Physics, Semiconductors, NEET
A silicon specimen is made into a p-type semiconductor by doping, on an average, one indium atom per 5×107 silicon atoms. If the number density of atoms in the silicon specimen is 5×1028 atoms m-3, then the number of acceptor atoms in silicon per cubic centimetre will be

Options:
A 2.5×1030 atom cm-3
B 2.5×1035 atom cm-3
C 1×1013 atom cm-3
D 1×1015 atom cm-3

The correct answer is:
1×1015 atom cm-3

Solution:
Number density of atoms in silicon specimen
= 5Ã1028 atoms-m-3=5Ã1022 atoms cm-3.
Since, 1 atom of indium is doped in 5Ã107 silicon atoms, so total number of indium atoms doped per cm3 of silicon will be
n=5Ã10225Ã107=1015 atom cm-3.

Topic: Physics, Semiconductors, NEET
A silicon specimen is made into a p-type semiconductor by doping, on an average, one indium atom per 5×107 silicon atoms. If the number density of atoms in the silicon specimen is 5×1028 atoms m-3, then the number of acceptor atoms in silicon per cubic centimeter will be

Options:
A 2.5×1030 atom cm-3
B 2.5×1035 atom cm-3
C 1×1013 atom cm-3
D 1×1015 atom cm-3

The correct answer is:
1×1015 atom cm-3

Solution:
Number density of atoms in silicon specimen
= 5Ã1028 atoms m-3=5Ã1022 atoms cm-3.
Since, 1 atom of indium is doped in 5Ã107 silicon atoms, so total number of indium atoms doped per cm3 of silicon will be
n=5Ã10225Ã107=1015 atom cm-3.

Topic: Physics, Semiconductors, NEET
A silver chromate solution at 298 K has its Ksp value =1.1Ã 10 â12 . Determine the solubility of the solution in mol L â1

Options:
A 6.5Ã10-5
B 2.4×10-2
C 3.6×10-3
D 8.9×10-4

The correct answer is:
6.5×10-5

Solution:
Ag 2 CrO 4 s â 2Ag + 2s + CrO 4 2â s ;â K sp =1.1Ã 10 â12 â
K sp = Ag + 2 Â· CrO 4 2 -
K sp = [2s] 2 . s = 4 s 3
s 3 = K sp 4
= 1.1Ã 10 â12 4
âs = 275Ã 10 â15 3 â
=6.5Ã 10 â5 mol L â1
âs=6.53Ã 10 â5

Topic: Chemistry, Ionic Equilibrium, NEET
A silver cup is plated with silver by passing 965 C of electricity. The amount of Ag deposited is

Options:
A 107.89 g
B 9.89 g
C 1.0002 g
D 1.08 g

The correct answer is:
1.08 g

Solution:
Ag++ e- ⟶ Ag
∵ 96500 C are required to deposite Ag = 108 g
∴ 965 C are required to deposite Ag
= 10896500 × 965 = 1.08 g

Topic: Chemistry, Electrochemistry, NEET
A silver wire has a temperature coefficient of resistivity 4Ã10-3 Â°C-1 and its resistance at 20 Â°C is 10 Î© . Neglecting any change in dimensions due to the change in temperature, its resistance at 40 Â°C is

Options:
A 0.8 Î©
B 1.8 Î©
C 10.8 Î©
D 11.6 Î©

The correct answer is:
10.8 Î©

Solution:
Given, α=4×10-3 °C-1
T1=20 °C
T2=40 °C
R( t 2 )=R( t 1 )( 1+aΔt )
∆T=40-20=20 °C
R(t1)=10 Ω
R(40 °C)=10(1+4×10−3×20) Ω
=101+80×10-3
=101.08=10.8 Ω

Topic: Physics, Current Electricity, NEET
A simple harmonic motion is represented by xt=sin2⁡ωt-2cos2⁡ωt. The angular frequency of oscillation is given by

Options:
A ω
B 2ω
C 4ω
D ω2

The correct answer is:
2ω

Solution:
x=sin2⁡ωt-2cos2⁡ωt
=1-3cos2⁡ωt
=1-31+cos⁡2ωt2
= -12-32cos⁡2ωt
Which is a periodic function with angular frequency of 2ω .

Topic: Physics, Oscillations, NEET
A simple harmonic oscillator has amplitude A angular frequency ω and mass m. Then average kinetic energy in one time period is

Options:
A 12mω2A2
B 14mω2A2
C mω2A2
D zero

The correct answer is:
14mω2A2

Solution:
Let displacement be x=Asinωt+ϕ
Kinetic energy, K=12mω2A2cos2ωt+ϕ
As the average value of cos2ωt+ϕ over a cycle is 12 .
Therefore average kinetic energy is Kav=14mω2A2

Topic: Physics, Oscillations, NEET
A simple pendulum 4 m long swings with an amplitude of 0.2 m. The acceleration at the ends of its path is g=10 m s-2

Options:
A zero
B 10 m s-2
C 0.5 m s-2
D 2.5 m s-2

The correct answer is:
0.5 m s-2

Solution:
T=mgcos⁡θ
∴Fnet=mgsin⁡θ
and acceleration=gsin⁡θ=gtan⁡θ sin⁡θ≅tan⁡θ, for small value of θ
=100.24=0.5 m s-2

Topic: Physics, Oscillations, NEET
A simple pendulum 4 m long swings with an amplitude of 0.2 m . What is its acceleration at the ends of its path ? g=10 m s-2

Options:
A zero
B 10 m s-2
C 0.5m s-2
D 2.5m s-2

The correct answer is:
0.5m s-2

Solution:
T=mgcosâ¡Î¸
â´Fnet=mg sin Î¸
and acceleration =gsinâ¡Î¸=gtanâ¡Î¸ as angle is very small sinâ¡Î¸=tanâ¡Î¸
∴anet=10×0.24=0.5 m s-2

Topic: Physics, Oscillations, NEET
A simple pendulum consisting of a mass M attached to a string of length L is released from rest at an angle Î±. A pin is located at a distance l below the pivot point. When the pendulum swings down, the string hits the pin as shown in the figure. If the particle doesn't undergo a complete vertical circle, then the maximum angle Î¸ <Ï2 through which the string swings after hitting the pin is

Options:
A cos-1â¡Lcosâ¡Î±+l L+l
B cos-1â¡Lcosâ¡Î±-l L-l
C cos-1⁡Lcos⁡α+l L-l
D cos-1⁡Lcos⁡α-l L+l

The correct answer is:
cos-1â¡Lcosâ¡Î±-l L-l

Solution:
At the bottom most point, square of speed of bob,
v2=2gL1 â cos Î±
It will rise further to a height,
h=v22g= L1-cosâ¡Î±
or L-l 1-cosâ¡Î¸=L1-cosâ¡Î±
â´ Î¸=cos-1â¡Lcosâ¡Î±-lL-l

Topic: Physics, Laws of Motion, NEET
A simple pendulum has time period T1. The point of suspension is now moved upward according to the relation y=kt2, (k=1 m s-2), where y is the vertical displacement. The time period now becomes T2. The ratio of T12T22 is (Take g=10 m s-2)

Options:
A 65
B 56
C 1
D 45

The correct answer is:
65

Solution:
y=kt2

d2ydt2=2k

or ay = 2 m s-2 (as k=1 m s-2)
T1=2Ïlg

and T2=2Ïlâ¡g+ay
â´ geff=g+ay

T12T22=g+ayg=10+210=65
Hence, the correct answer is 65

Topic: Physics, Oscillations, NEET
A simple pendulum is oscillating with an angular amplitude of 90Â°. If the direction of resultant acceleration of the bob is horizontal at a point then angle made by the string with vertical is

Options:
A sin-1⁡13
B cos-1⁡13
C sin-1⁡13
D cos-1⁡13

The correct answer is:
cos-1⁡13

Solution:
tan⁡θ=v2rgsin⁡θ
But v2=2gh ;
h=rcos⁡θ
cos⁡θ=13⇒θ=cos-1⁡13

Topic: Physics, Laws of Motion, NEET
A simple pendulum is oscillating without damping. When the displacement of the bob is less than maximum, its acceleration vector a → is correctly shown in

Options:
A
B
C
D

The correct answer is:

Solution:
a → = a c → + a t →

Topic: Physics, Laws of Motion, NEET
A simple pendulum is released from A as shown. If m and l represent the mass of the bob and length of the pendulum, the gain in kinetic energy at B is

Options:
A mgl2
B mgl2
C 32mgl
D 23mgl

The correct answer is:
32mgl

Solution:
Applying the principle of conservation of energy,
âK+âU=0 ; K-0+-mgh=0
K=mgh=mgl cos30Â°
â´K=mgl cosâ¡30Â°=32 mgl

Topic: Physics, Work Power Energy, NEET
A simple pendulum is released from A as shown. If m and l represent the mass of the bob and length of the pendulum, the gain in kinetic energy at B is

Options:
A mgl2
B mgl2
C 32mgl
D 23mgl

The correct answer is:
32mgl

Solution:
Applying principle of conservation of enegy,
∆K+∆U=0 ; K-0+-mgh=0
K=mgh=mgl cos30°
∴K= mg l cos⁡30°=32 mgl

Topic: Physics, Work Power Energy, NEET
A simple pendulum is suspended from the ceiling of a lift. When the lift is at rest its time period is T. With what acceleration should the lift be accelerated upwards in order to reduce its period to T/2? (g is the acceleration due to gravity)

Options:
A 2g
B 3g
C 4g
D g

The correct answer is:
3g

Solution:
Time period of a simple pendulum is given by,
T=2πlg ...(i)
When the lift is moving up with an acceleration a, then time period becomes
T′=2πlg+a
Here, T′=T2
⇒ T2=2πlg+a ...(ii)
Dividing Eq.(ii) by Eq. (i), we get
a=3g

Topic: Physics, Oscillations, NEET
A simple pendulum is taken to 64 km above the earth's surface. Its new time period will

Options:
A increase by 1%
B decrease by 1%
C increase by 2%
D decrease by 2%

The correct answer is:
increase by 1%

Solution:
T=2Ïlg=2Ïâ(R+h)2GM
Tâ(R+h)
or TT=R+hR=1+hR=1+1100â´T'T-1=1100
â´ T'-TTÃ100=1
Time period will increase by 1%

Topic: Physics, Gravitation, NEET
A simple pendulum is taken to 64 km above the earth's surface. It's time period will be

Options:
A Increase by 1%
B Decrease by 1%
C Increase by 2%
D Decrease by 2%

The correct answer is:
Increase by 1%

Solution:
g eff at height of h
g eff = g ( 1+ h R e ) 2 = g R e 2 ( R e +h ) 2
âT=2 T 1 l g =2 T 1 l ( R e +h ) 2 R e 2
â T 2 Î± T 1 â R e +h R e = 6400+64 6400
â T 2 T 1 =( 1+ 1 100 )
â´ Time period will increase by 1%.

Topic: Physics, Oscillations, NEET
A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad s-1 The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad s-1 and an amplitude of 10-2 m. The relative change in the angular frequency of the pendulum is best given by:

Options:
A 10-1 rad s-1
B 1 rad s-1
C 10-5 rad s-1
D 10-3 rad s-1

The correct answer is:
10-3 rad s-1

Solution:
Angular frequency of simple pendulum,
ω=gl
As the support oscillates up and down, effective g changes
∆g=2.ω02A=2×12×10-2 m s-2
=0.02 m s-2
Change in angular frequency dω of the pendulum is given by
dωω=-12⋅dgg
∆ωω=12∆gg
=12×0.0210
=0.001 rad s-1

Topic: Physics, Oscillations, NEET
A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period ?

Options:
A $T=\sqrt{\frac{l}{\sqrt{{{g}^{2}}+\frac{{{v}^{4}}}{{{R}^{2}}}}}}$
B $T=\sqrt{\frac{l}{\sqrt{{{g}^{2}}+\frac{{{v}^{4}}}{2{{R}^{2}}}}}}$
C $T=\sqrt{\frac{2l}{\sqrt{{{g}^{2}}+\frac{2{{v}^{4}}}{{{R}^{2}}}}}}$
D $T=\sqrt{\frac{2l}{\sqrt{{{g}^{2}}+\frac{{{v}^{4}}}{{{R}^{2}}}}}}$

The correct answer is:
$T=\sqrt{\frac{l}{\sqrt{{{g}^{2}}+\frac{{{v}^{4}}}{{{R}^{2}}}}}}$

Solution:
Correct Option is : (A)
$T=\sqrt{\frac{l}{\sqrt{{{g}^{2}}+\frac{{{v}^{4}}}{{{R}^{2}}}}}}$

Topic: Chemistry, Oscillations, NEET
A simple pendulum of length L has mass M and it oscillates freely with amplitude A. At the extreme position, its potential energy is (g = acceleration due to gravity)

Options:
A MgA22L
B MgA2L
C MgA2L
D 2MgA2L

The correct answer is:
MgA22L

Solution:
Potential energy =12MÏ2A2
=12M.gL.A2 â´ Ï=gL

Topic: Physics, Oscillations, NEET
A simple pendulum of length l has maximum angular displacement Î¸. The maximum kinetic energy of the bob of mass m is ( g= acceleration due to gravity)

Options:
A mgl (1+cosâ¡Î¸)
B mglÂ (1+cos2â¡Î¸)
C mgl (1-cosâ¡Î¸)
D mglÂ (cosâ¡Î¸-1)

The correct answer is:
mgl (1-cosâ¡Î¸)

Solution:
When bob is at rest, the pendulum has only potential energy which is given as P.E=mgl
When the bob is released from maximum displaced position, it will lose P.E. and gain K.E
Thus loss in P.E = gain in K.E
â´ââââK.E=mglâmglcosÎ¸=mgl( 1âcosÎ¸ )

Topic: Physics, Oscillations, NEET
A simple pendulum of length l is suspended from the ceiling of a cart which is sliding without friction on an inclined plane of inclination Î¸. What will be the time period of the pendulum?

Options:
A 2 Ï lâ¡ g cos Î¸
B - 2 π 3 l&ApplyFunction; g cos θ
C 4 π 2 l&ApplyFunction; g cos θ
D - 3 π 4 l&ApplyFunction; 2g cos θ

The correct answer is:
2 π l⁡ g cos θ

Solution:
Here, the point of suspension has an acceleration. a → =gsin θ (down the plane). Further, g→ can be resolved into two components g sin θ (along the plane) and g cos θ (perpendicular to the plane).
g→eff=g→-a→
=gcosθ (perpendicular to plane)
T=2πl⁡|g→eff|
=2πl⁡gcosθ

Topic: Physics, Oscillations, NEET
A simple pendulum oscillating in air has a period of $\sqrt{3} \mathrm{~s}$. If it is completely immersed in non-viscous liquid, having density $\left(\frac{1}{4}\right)^{\text {th }}$ of the material of the bob, the new period will be

Options:
A $2 \mathrm{~s}$
B $\frac{\sqrt{3}}{2} s$
C $2 \sqrt{3} s$
D $\frac{2}{\sqrt{3}} \mathrm{~s}$

The correct answer is:
$2 \mathrm{~s}$

Solution:
Time period of simple pendulum
$\begin{aligned} & T=2 \pi \sqrt{\frac{l}{g_{\mathrm{eff}}}} \\ & \sqrt{3}=2 \pi \sqrt{\frac{l}{g}}\end{aligned}$
After immersed in liquid
$\begin{aligned} & M g_{\text {eff }}=M g-\frac{M g}{\frac{\rho_s}{\rho_l}} \\ & M g_{\text {eff }}=M g-\frac{M g}{4}=\frac{3 M g}{4} \\ & g_{\text {eff }}=\frac{3 g}{4} \\ & T_1=2 \pi \sqrt{\frac{l}{3 g}}=\left(2 \pi \sqrt{\frac{I}{g}}\right) \times \frac{2}{\sqrt{3}} \\ & T_1=\frac{2 T}{\sqrt{3}}=2 \mathrm{~s}\end{aligned}$

Topic: Physics, Oscillations, NEET, NEET 2023 (Manipur)
A simple pendulum performs simple harmonic motion about $\mathrm{x}=0$ with an amplitude a and time period $T$. The speed of the pendulum at $\mathrm{x}=\frac{\mathrm{a}}{2}$ will be

Options:
A $\frac{\pi \mathrm{a} \sqrt{3}}{2 \mathrm{~T}}$
B $\frac{\pi \mathrm{a}}{\mathrm{T}}$
C $\frac{3 \pi^2 a}{T}$
D $\frac{\pi \mathrm{a} \sqrt{3}}{\mathrm{~T}}$

The correct answer is:
$\frac{\pi \mathrm{a} \sqrt{3}}{\mathrm{~T}}$

Solution:
$$
\begin{aligned}
v=\frac{d y}{d t}=A \omega \cos \omega t & =A \omega \sqrt{1-\sin ^2 \omega t} \\
& =\omega \sqrt{A^2-y^2}
\end{aligned}
$$
Here, $\mathrm{y}=\frac{\mathrm{a}}{2}$
$$
\therefore \mathrm{v}=\omega \sqrt{\mathrm{a}^2-\frac{\mathrm{a}^2}{4}}=\omega \sqrt{\frac{3 \mathrm{a}^2}{4}}=\frac{2 \pi}{\mathrm{T}} \frac{\mathrm{a} \sqrt{3}}{2}=\frac{\pi \mathrm{a} \sqrt{3}}{\mathrm{~T}}
$$

Topic: Physics, Oscillations, NEET, NEET 2009 (Screening)
A simple stirred tank bioreactor has all of the following properties, except

Options:
A It is usually cylindrical or with a curved base to facilitate the mixing of the reactor contents
B The stirrer facilitates even mixing and oxygen availability throughout the bioreactor
C Foam control system is present
D Sterile air bubbles are sparged into the simple stirred tank bioreactor

The correct answer is:
Sterile air bubbles are sparged into the simple stirred tank bioreactor

Solution:
Sterile air bubbles are sparged and mixed in sparged stirred-tank bioreactor, which increases the surface area for oxygen transfer. Sparger is not found in the simple stirred tank bioreactor.

Topic: Biology, Biotechnology - Principles and Processes, NEET
A simple telescope consisting of an objective lens of focal length 60 cm and an eye lens of focal length 5 cm, is focused on a distant object. If the object subtends an angle of 2Â° at the objective lens, the angular width of the image is

Options:
A 50°
B 16°
C 10°
D 24Â°

The correct answer is:
24°

Solution:
Magnifying power for normal adjustment (distant object) is given by,
M.P.=βα=fofe
⇒β2=605
⇒β=24°

Topic: Physics, Ray Optics, NEET
A simple wave motion is represented by y=5sin⁡4πt+3cos⁡4πt. Its amplitude is

Options:
A 5 units
B 53 units
C 103 units
D 10 units

The correct answer is:
10 units

Solution:
y=5sin⁡4πt+3cos⁡4πt
y=5sin⁡4πt+53cos⁡4πt
A=A12+A22
A=52+532
=25+75=100
A=10

Topic: Physics, Waves and Sound, NEET
A single alkene is produced when an alkyl bromide reacts with sodium ethoxide and ethanol. This alkene undergoes hydrogenation and produces 2-methyl butane. What is the identity of the alkyl bromide?

Options:
A 1-bromo-2, 2-dimethylpropane
B 1-bromobutane
C 1-bromo-2-methylbutane
D 2-bromo-2-methylbutane

The correct answer is:
1-bromo-2-methylbutane

Solution:

Topic: Chemistry, Hydrocarbons, NEET
A single compound of the structure
is obtainable from ozonolysis of which of the following cyclic compounds?

Options:
A
B
C
D

The correct answer is:

Solution:
All these three options will give the different products, only the option A will give the desired product by reductive ozonolysis.

Topic: Chemistry, Aldehydes and Ketones, NEET, NEET 2015 (Phase 1)
A single compound of the structure

is obtained from ozonolysis of which of the following cyclic compounds?

Options:
A
B
C
D

The correct answer is:

Solution:
From the ozonolysis of rest other compounds this compound can not be obtained.
For example

Topic: Chemistry, Aldehydes and Ketones, NEET
A single slit of width 0.1 mm is illuminated by a parallel beam of light of wavelength 6000 â« and diffraction bands are observed on a screen 0.5 m from the slit. The distance of the third dark band from the central bright band is:

Options:
A 9 mm
B 3 mm
C 4.5 mm
D 1.5 mm

The correct answer is:
9 mm

Solution:
a=0.1 mm=10-4m
Î»=6000Ã10-10=6Ã10-7m
D=0.5 m
For 3rd dark
asinâ¡Î¸=3Î»
sinâ¡Î¸=3Î»a=xD
x=3Î»Da
=3Ã6Ã10-7Ã0.510-4
=9 mm

Topic: Physics, Wave Optics, NEET
A single species might show high diversity over its distributional range at the

Options:
A Genetic level
B Species level
C Ecosystem level
D Both a and b

The correct answer is:
Genetic level

Solution:
The variety of flora and fauna in a particular habitat or the globe is called biodiversity. When the diversity in organisms at the species level is termed as species diversity.
If a particular area/region shows the variation in the ecosystem will be included in ecological diversity. India has more ecological diversity than a northern nations like Finland, Norway.
A particular species displaying variations at the genetic level at different regions are distribution comes under genetic diversity. India has mango varieties which are approximately 1000 varieties and for rice approximately 50000 varieties.

Topic: Biology, Biodiversity and its Conservation, NEET
A single step large mutation is called

Options:
A speciation
B saltation
C genetic drift
D Sewall Wright effect

The correct answer is:
saltation

Solution:
Sudden and large mutation occurring between generation and leading to a single step speciation is referred to as saltation.
Speciation the process of formation of new species due to the course of evolution.
Change in frequency of the gene type in a population due to random selection is referred to as genetic drift. It is also termed as Sewall wright effect.

Topic: Biology, Evolution, NEET
A single strand of nucleic acid tagged with a radioactive molecule is called

Options:
A vector
B selectable marker
C plasmid
D probe

The correct answer is:
probe

Solution:
Probes are 15-30 bases long radioactive labelled oligonucleotides (RNA or DNA) used to detect complementary nucleotide sequences, used for disease diagnosis, etc.

Topic: Biology, Biotechnology - Principles and Processes, NEET, NEET 2012 (Screening)
A single turn of the citric acid cycle yields

Options:
A 2 FADH2, 2 NADH2, 2 GTP
B 1 FADH2, 2 NADH2, 1 GTP
C 1 FADH2, 3 NADH2, 1 GTP
D 1 FADH2, 4 NADH2, 1 GTP

The correct answer is:
1 FADH2, 3 NADH2, 1 GTP

Solution:
Citric acid cycle or Kreb's cycle occurs in the mitochondrial matrix and comprises eight cyclic steps. During its single turn, it produces CO2, one of the end products of respiration, one molecule of GTP, three NADH2 and one FADH2 as a reduced coenzyme are formed.

Topic: Biology, Respiration in Plants, NEET
A single U-shaped bone at the base of the buccal cavity is called:

Options:
A Hyoid
B Thyroid
C Cricoid
D Mandible

The correct answer is:
Hyoid

Solution:
The hyoid is a U shaped bone which is situated at the root of the tongue and between lower jaw and voice box. In the act of swallowing the hyoid bone, larynx and tongue all move upwards. It primarily serves as a structure for connecting the tongue and muscles of the oral cavity. Its functions include protection of oesophagus and enabling movements involved in speaking and swallowing.

Topic: Biology, Locomotion and Movement, NEET
A sinusoidal wave with amplitude y is travelling with speed v on a string with linear density Ï . The angular frequency of the wave is Ï . The following conclusions are drawn. Mark the one which is correct. (in particular option, if we are changing one quantity, assume others are kept constant)

Options:
A Doubling the frequency doubles the rate at which energy is carried along the string
B If the amplitude were doubled, the rate at which energy is carried would be halved
C If the amplitude were doubled, the rate at which energy is carried would be doubled
D The rate at which energy is carried is directly proportional to the velocity of the wave

The correct answer is:
The rate at which energy is carried is directly proportional to the velocity of the wave

Solution:
Par=12μvω2A2
So P∝v
∝ ω 2
∝ A 2
So if frequency or amplitude is doubled power will be quadrupled, but if velocity is doubled power will also be doubled.

Topic: Physics, Waves and Sound, NEET
A siren emitting a sound of frequency 800 Hz moves away from an observer towards a cliff at a speed of 15 m s-1. Then, the frequency of sound that the observer hears in the echo reflected from the cliff is
(Take velocity of sound in air =330 m s-1)

Options:
A 765  Hz
B 800  Hz
C 838ââHz
D 885  Hz

The correct answer is:
838  Hz

Solution:
Given
fÎ¿=800 Hz
Vsource=15 m s-1
fa=f0(V+V0VâVs)â
âfa=330330-15Ã800=330315Ã800
fa=838 Hz

Topic: Physics, Waves and Sound, NEET
A siren emitting a sound of frequency 800 Hz moves away from an observer towards a cliff at a speed of 15 m s-1. Then, the frequency of sound that the observer hears in the echo reflected from the cliff is
(Take velocity of sound in air =330 m s-1)

Options:
A 765  Hz
B 800  Hz
C 838ââHz
D 885  Hz

The correct answer is:
838  Hz

Solution:
Given here,
fo=800 Hz
Vs=15 m s-1
Using Doppler's effect, the frequency of sound that the observer hears in the echo reflected from the cliff is fa=foV+VoV−Vs
⇒fa=330330-15×800=330315×800
fa=838 Hz

Topic: Physics, Waves and Sound, NEET, NEET 2016 (Phase 1)