datavorous/entrance-exam-dataset huggingface dataset:
A lead bullet strikes a target with a velocity of 480 m s-1. If the bullet falls dead, the rise in temperature of the bullet (c = 0.03 cal g-1 oC-1 ), assuming that heat produced is equally shared between the bullet and target is

Options:
A 557 oC
B 457 oC
C 857 oC
D 754 oC

The correct answer is:
457 oC

Solution:
mc Î T = 1 2 1 2 mv 2
Î T = v 2 4c = 4 8 0 Ã 4 8 0 4 Ã 0 Â· 0 3 Ã 4 Â· 2 Ã 1 0 3 = 4 5 7 o C

Topic: Physics, Thermal Properties of Matter, NEET
A leading strand is

Options:
A Chain of DNA which is continuously formed
B Formation of chain which polymerises in 3' to 5' direction
C Chain of DNA which is discontinuously formed
D Chain of DNA contains Okazaki fragments.

The correct answer is:
Chain of DNA which is continuously formed

Solution:
Replication of DNA is semi discontinuous process in which one strand with polarity 3' to 5' is synthesized continuously and is known as leading strand and the other strand with polarity 5' to 3' is discontinuously synthesised in the form of small fragments known as Okazaki fragments.

Topic: Biology, Molecular Basis of Inheritance, NEET
A leaf peeling of Tradescantia is kept in a medium having 10% NaCl. After a few minutes, if we observe the leaf peel under the microscope, we are likely to see

Options:
A Entry of water into the cell
B The cells bursting out
C Diffusion of NaCl into the cell
D Exit of water from the cell

The correct answer is:
Exit of water from the cell

Solution:
When a leaf peeling of Tradescantia is kept in a medium having 10% NaCl solution, the cells shrink in size, this is followed by separation of protoplast from cell wall due to exosmosis. This phenomenon is called plasmolysis.

Topic: Biology, Transport in Plants, NEET
A leaf peeling of Tradescantia is kept in a medium having 10% NaCl. After a few minutes, if we observe the leaf peel under the microscope, we are likely to see

Options:
A Entry of water into the cell
B The cells bursting out
C Diffusion of NaCl into the cell
D Exit of water from the cell

The correct answer is:
Exit of water from the cell

Solution:
When a cell is placed in a hypertonic solution then exosmosis occurs. Due to exosmosis, the protoplasm shrinks and leaves the cell wall and thus becomes flaccid, which is called plasmolysis. Thus, exosmosis leads to plasmolysis. Since 10% NaCl solution is a hypertonic solution as compared to cell sap exosmosis, i.e., movement of water from inside to outside the cell will occur.

Topic: Biology, Transport in Plants, NEET
A length l of wire carries a steady current i. It is bent first to form a circular plane coil of one turn. The same length is now bent more sharply to give three loops of smaller radius. The magnetic field at the centre caused by the same current is

Options:
A One-third of its first value
B Unaltered
C Three times of its initial value
D Nine times of its initial value

The correct answer is:
Nine times of its initial value

Solution:
B â n2 (n is number of turns)

Topic: Physics, Magnetic Effects of Current, NEET
A lens is made up of 3 different transparent media as shown in figure. A point object $O$ is placed on its axis beyond $2 f$. How many real images will be obtained on the other side?

Options:
A No image will be formed
B $3$
C $2$
D $1$

The correct answer is:
$3$

Solution:
Since all three medium will form individual real images at different points on the other side of lens (as object distance is more than $2 f$ )
$\therefore$ Number of real images formed $=3$.

Topic: Physics, Ray Optics, NEET, NEET 2023 (Manipur)
A lens of large focal length and large aperture is best suited as an objective of an astronomical telescope since:

Options:
A a large aperture contributes to the quality and visibility of the images.
B a large area of the objective ensures better light gathering power.
C a large aperture provides a better resolution.

D all of the above.

The correct answer is:
all of the above.

Solution:
The formula of magnification power and resolving power of a telescope is, M.P. = f0fe & R.P.=a 1.22 λ. A large aperture(a) of the objective lens provides better resolution so, the good quality of the image is formed, and also it gathers more light. The larger the objective, the lighter the telescope collects and increases the brightness of the image and large focal length enhances the magnifying power of the telescope.

Topic: Physics, Ray Optics, NEET, NEET 2021
A lift of mass $1000 \mathrm{~kg}$ which is moving with an acceleration of $1 \mathrm{~ms}^{-2}$ in the upward direction, then the tension developed in the string which is connected to lift is:

Options:
A 9800 N
B 10,800 N
C 11000 N
D 10,000 N

The correct answer is:
10,800 N

Solution:
Using the equation
$\begin{aligned}
& T=m(g+a)=1000(9.8+1) \\
& =10800 \mathrm{~N}
\end{aligned}$

Topic: Physics, Laws of Motion, NEET, NEET 2002
A light and heavy body have equal momentum, which one has greater kinetic energy?

Options:
A A light body
B A heavy body
C Both have equal kinetic energies
D Data given is incomplete

The correct answer is:
A light body

Solution:
Let m1 and m2 be the mass of light and heavy bodies respectively, i.e. m2>m1
m2m1>1 ...(i)
K1 and K2 be the kinetic energy of light and heavy bodies respectively, then
K1=p22m1 and K2=p22m2 , [where p= momentum]
∴ K1K2=p2/2m1p2/2m2=m2m1
K1K2=m2m1
But m2m1>1 , From Equation. (i),
∴ K1K2>1
K1>K2
i.e. Light body has greater kinetic energy.

Topic: Physics, Work Power Energy, NEET
A light beam consists of two types of photons. In one type, each photon has the energy 2 eV and in another type, each photon has energy 3 eV. The light beam is incident on a photoelectric material of work function 1 eV. The maximum kinetic energy of emitted photoelectron is

Options:
A 2 eV
B 3 eV
C 4 eV
D 1 eV

The correct answer is:
2 eV

Solution:
Given,
Work function of photoelectric material Ï=1 eV
We know that,
KEmax=hv-Ï
=3-1=2 eV

Topic: Physics, Dual Nature of Matter, NEET
A light of wavelength 4000 Aâ is allowed to fall on a metal surface having work function 2 eV. The maximum velocity of the emitted electrons is (h=6.6Ã10-34 J s)

Options:
A 1.35Ã105 m s-1
B 2.7Ã105 m s-1
C 6.2Ã105 m s-1
D 8.1Ã105 m s-1

The correct answer is:
6.2Ã105 m s-1

Solution:
12mv2=hcÎ»-Ï (in eV)
=6.6Ã10-34Ã3Ã1084000Ã10-10Ã1.6Ã10-19-2
=3.1-2=1.1 eV=1.1Ã1.6Ã10-19 J
=1.76Ã10-19 J
v=1.76Ã10-19Ã29Ã10-13
=6.2Ã105 m s-1

Topic: Physics, Dual Nature of Matter, NEET
A light of wavelength 6000 â« is incident on a single slit. First minimum is obtained at a distance of 0.4 cm from the centre. If width of the slit is 0.3 mm, then the distance between slit and screen will be

Options:
A 1.0 m
B 1.5 m
C 2.0 m
D 2.3 m

The correct answer is:
2.0 m

Solution:
Using asinâ¡Î¸=nÎ»
Or aÃxD=nÎ» or a=nÎ»ÃDx
Or D=aÃxnÎ»
=3Ã10-4Ã4Ã10-31Ã6000Ã10-10 m=2.0 m

Topic: Physics, Wave Optics, NEET
A light of wavelength 6328 â« is incident normally on a slit of width 0.2 mm. The angular width of the central maximum on the screen will be

Options:
A 0.9°
B 0.18°
C 0.54°
D 0.36Â°

The correct answer is:
0.36°

Solution:
asinâ¡Î¸=nÎ»
sinâ¡Î¸=nÎ»a=6328Ã10-10Ã10.2Ã10-3
=3164Ã10-6 rad=0.003164 rad
â´ Angular width =2Î¸
=2Ã0.003164Ã180Â°Ï=0.36Â°

Topic: Physics, Wave Optics, NEET
A light ray enters through a right angled prism at point $P$ with the angle of incidence $30^{\circ}$ as shown in figure. It travels through the prism parallel to its base $B C$ and emerges along the face $A C$. The refractive index of the prism is:

Options:
A $\frac{\sqrt{5}}{2}$
B $\frac{\sqrt{3}}{4}$
C $\frac{\sqrt{3}}{2}$
D $\frac{\sqrt{5}}{4}$

The correct answer is:
$\frac{\sqrt{5}}{2}$

Solution:
In prism, $r_1+c=A$
$\begin{gathered}
r_1=90^{\circ}-c \ldots(1)\\
\sin c=\frac{1}{\mu} \Rightarrow \cos c=\frac{\sqrt{\mu^2-1}}{\mu}
\end{gathered}$
$\Rightarrow$ Apply Snell's law, on incidence surface
$\begin{aligned}
1 \cdot \sin 30^{\circ}=\mu \sin \left(r_1\right) \Rightarrow 1 \times \frac{1}{2} & =\mu \times \sin \left(90^{\circ}-c\right) \\
\frac{1}{2} & =\mu \times \frac{\sqrt{\mu^2-1}}{\mu}
\end{aligned}$
On squaring $\frac{1}{4}=\mu^2-1$
$\Rightarrow \mu^2=\frac{5}{4} \Rightarrow \mu=\frac{\sqrt{5}}{2}$

Topic: Physics, Ray Optics, NEET, NEET 2024
A light ray falls on a glass surface of refractive index 3, at an angle 60°. The angle between the refracted and reflected rays would be:

Options:
A 60°
B 90°
C 120°
D 30°

The correct answer is:
90°

Solution:
From Snell's law,

μ1sini=μ2sinr

⇒1×sin60°=3sinr

⇒ r=30°

Angle between the refracted say and reflected say is

θ=180°-i+r=90°

Topic: Physics, Ray Optics, NEET, NEET 2022 (Phase 1)
A light ray is incident perpendicular to one face of a 90Â° prism and is totally internally reflected at the glass-air interface. If the angle of reflection is 45Â°, we conclude that the refractive index n

Options:
A n<12
B n>12
C n<2
D n>2

The correct answer is:
n>2

Solution:
Total internal reflection occurs in denser medium when light is incident at surface of separation at an angle exceeding critical angle of the medium.
Given i=45° in the medium and total internal reflection occurs at the glassair interface
∴ n>1sin C>1sin45°=2

n>2

Topic: Physics, Ray Optics, NEET
A light ray moving in medium-I (of refractive index n1) is incident on interface of two media and it is totally internally reflected at the interface. Now, refractive index n2 of medium-II is decreased, then

Options:
A ray will move completely parallel to the interface
B ray will be still totally internally reflected at interface
C ray will be totally transmitted into medium-II only, if angle of incidence is increased
D ray will be totally transmitted in medium-II

The correct answer is:
ray will be still totally internally reflected at interface

Solution:
As n2 decreases, iC=sin-1⁡n2n1 also decreases.
So, condition i>iC is still satisfied and there will be still total internal reflection at interface. If angle of incidence is increased, then ray will be still totally internally reflected at interface because i>iC.

Topic: Physics, Ray Optics, NEET
A light rod of length, l has two masses, m1 and m2 attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is

Options:
A m1m2m1+m2l2
B m1+m2m1m2l2
C m1+m2l2
D m1m2l2

The correct answer is:
m1m2m1+m2l2

Solution:
x1=m2lm1+m2 and x2=m1lm1+m2.

I=m1x12+m2x22=m1m2m1+m2l2.

Topic: Physics, Rotational Motion, NEET, NEET 2016 (Phase 2)
A light string passing over a smooth light pulley connects two blocks of masses m1 and m2 (vertically). If the acceleration of the system is g8, then the ratio of masses is

Options:
A 8:1
B 9:7
C 4:3
D 5:3

The correct answer is:
9:7

Solution:
In the given system,
a=m1-m2m1+m2=g8
â´ m1-m2m1+m2=18
8m1-8m2=m1+m2
7m1=9m2
m1m2=97

Topic: Physics, Laws of Motion, NEET
A light string passing over a smooth light pulley connects two blocks of masses m1 and m2 (vertically). If the acceleration of the system is (g/8), then the ratio of masses is

Options:
A 8 : 1
B 9 : 7
C 4 : 3
D 5 : 3

The correct answer is:
9 : 7

Solution:
In the given system,
a=m1-m2m1+m2=g8
∴ m1-m2m1+m2=18
8m1-8m2=m1+m2
7m1=9m2
m1m2=97

Topic: Physics, Laws of Motion, NEET
A light whose frequency is equal to 6Ã1014 Hz is incident on a metal whose work function is 2 eV. [h=6.63Ã10-34 J s, 1eV=1.6Ã10-19 J]. The maximum energy of the electrons emitted will be

Options:
A 2.49 eV

B 4.49 eV

C 0.49 eV

D 5.49 eV

The correct answer is:
0.49 eV

Solution:
KEmax=hν-ϕ
Where hν=energy of incident photon,
ϕ=work function
KEmax = 6.6×10-34×6×1014-2×1.6×10-19
= 3.96×10-19-3.2×10-19
= 0.76×10-191.6×10-19 eV=0.475 eV

Topic: Physics, Dual Nature of Matter, NEET
A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5Ã10-5 cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is

Options:
A 0.10  cm
B 0.25  cm
C 0.20  cm
D 0.15  cm

The correct answer is:
0.15  cm

Solution:
By the principle of single slit diffraction pattern
θ=λa=yf
y=fλa=60×5×10-50.02=0.15 cm

Topic: Physics, Wave Optics, NEET, NEET 2016 (Phase 2)
A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5Ã10-5 cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is

Options:
A 0.10  cm
B 0.25  cm
C 0.20  cm
D 0.15  cm

The correct answer is:
0.15  cm

Solution:
By the principle of single slit diffraction pattern
Î¸=Î»a=yf
y=fÎ»a=60Ã5Ã10-50.02=0.15 cm

Topic: Physics, Wave Optics, NEET
A linear graphic representation of the sequence and relative distances of the various genes present in a chromosome is

Options:
A linkage map or chromosomal map
B genetic map or linkage map
C chromosomal map or genetic map
D chromosomal map or genetic map or linkage map

The correct answer is:
chromosomal map or genetic map or linkage map

Solution:
A linkage or genetic chromosome map is a linear graphic representation of the sequence and relative distances of the various genes present in a chromosome. A chromosome map is also called a linkage map or genetic map.
The percentage of crossing over between two genes is directly proportional to their distance. The unit of crossing over has been termed as by Haldane as centiMorgan (cM). One unit of map distance (cM) is, therefore, equivalent to 1% crossing over.

Topic: Biology, Principles of Inheritance and Variation, NEET
A linear object of height 10 cm is kept in front of a concave mirror of radius of curvature 15 cm, at a distance of 10 cm. The image formed is

Options:
A Magnified and erect
B Magnified and inverted
C Diminished and erect
D Diminished and inverted

The correct answer is:
Magnified and inverted

Solution:
Given,
Height of object h0=10 cm
Radius of curvature R=15 cm
Distance of object u=-10 cm
We know that,
f=R2=152 cm
According to mirror formula,
1f=1u+1v
1-152=1-10+1v
â 1v=-215+110
â 1v=-4+330
â 1v=-130
â v=-30 cm
Magnification m=hiho=-vu
=hi10=--30-10
â hi=-3
Hence, image will be magnified and inverted.

Topic: Physics, Ray Optics, NEET
A liquid can easily change its shape but a solid can not because

Options:
A the density of a liquid is smaller than that of a solid
B the forces between the molecules is stronger in solid than in liquids
C the atoms combine to form bigger molecules in a solid
D the average separation between the molecules is larger in solids

The correct answer is:
the forces between the molecules is stronger in solid than in liquids

Solution:
Molecular force between molecules of solid is stronger than that of the liquid. Therefore, liquids have no fixed shape, they take the shape of the container, while solids have fixed shape

Topic: Physics, Mechanical Properties of Solids, NEET
A liquid compound (x) can be purified by steam distillation only if it is

Options:
A Steam volatile, immiscible with water
B Not steam volatile, miscible with water
C Steam volatile, miscible with water
D Not steam volatile, immiscible with water

The correct answer is:
Steam volatile, immiscible with water

Solution:
Compounds purified by steam distillation which are immiscible in water but steam volatile.

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2020 (Phase 2)
A liquid does not wet the solid surface if angle of contact is:

Options:
A equal to 45°
B equal to 60°
C greater than 90°
D zero

The correct answer is:
greater than 90°

Solution:
A liquid does not wet the solid surface if the angle of contact is obtuse ie, greater than 90°. In this case cohesive forces will be greater than adhesive forces and so, the liquid does not wet the surface of solid.

Topic: Physics, Mechanical Properties of Fluids, NEET, NEET 2020 (Phase 2)
A liquid drop having surface energy E is spread into 512 droplets of the same size. The final surface energy of the droplets is

Options:
A 2E
B 4E
C 8E
D 12E

The correct answer is:
8E

Solution:
The surface area of the liquid drop is A=4πR2
Its surface energy is E
When the drop splits in 512 droplets, the surface area of each droplet is 4πr2
∴ Total surface area A2 =512×4πr2
The volume of the bigger drop is 43πR3 and the volume of small droplets is 512×43πr3
∴ 43πR3=512×43πr3⇒r=R8
∴ A2=512×4πr2=512×4πR82=8A1
Surface energy E=A.T (T is surface tension and A is area)
∴ EnE0=A2⋅TA1⋅T=8⋅A1A1=8
∴ En=8E

Topic: Physics, Mechanical Properties of Fluids, NEET
A liquid is stirred in thermally insulated container, for about 2 hrs. Which of the following is correct?

Options:
A w > 0, q = 0, ΔU>0
B w < 0, q = 0, ΔU=0
C w > 0, q > 0, ΔU=0
D w < 0, q < 0, ΔU=0

The correct answer is:
w > 0, q = 0, ΔU>0

Solution:
A liquid is stirred in thermally insulated container, for about 2 hrs. Work is done on the system (w > 0), internal energy will increase ΔU>0 and as system is insulated so q = 0.

Topic: Chemistry, Thermodynamics (C), NEET
A liquid is stirred with a paddle to increase its temperature in a closed insulated container. Which of the following is true in this process?

Options:
A ΔE=W=Q=0
B ΔE=W≠0,Q=0
C ΔE≠0,Q=W=0
D ΔE=Q≠0,W=0

The correct answer is:
ΔE=W≠0,Q=0

Solution:
As the system is closed and insulated no heat enter or leave the system, i.e. q=0 ; ∴ΔE=Q+W=W .

Topic: Chemistry, Thermodynamics (C), NEET
A liquid of density 0.85 g cm-3 flows through a calorimeter at the rate of 8.0 cm3 s-1. Heat is added by means of a 250 W electric heating coil and a temperature difference of 15Â° C is established in steady-state conditions between the inflow and the outflow points of the liquid. The specific heat for the liquid will be (1 kcal =4186 J )

Options:
A 0.6 kcal kg-1K-1
B 0.3 kcal kg-1K-1
C 0.5 kcal kg-1K-1
D 0.4 kcal kg-1K-1

The correct answer is:
0.6 kcal kg-1K-1

Solution:
This is a problem on 'flow calorimeter' used to measure the specific heat of the liquid.
Amount of heat supplied to the water per second by the heating coil = Qs = 250 J = 2 5 0 4 1 8 6 kcal
The volume of liquid flowing out per second = 8.0 cm3 = 8 × 10-6 m3

Mass of this liquid = (0.85) × 1000 × 8 × 10-6 kg
The temperature rise of this mass of liquid = 15° C

Hence, = 2 5 0 4 1 8 6 = mst = 0.85 × 8 × 1 0 - 3 × s × 1 5
Hence, s=250×1034186×0.85×8×15=0.6 kcal kg-1K-1

Topic: Physics, Thermal Properties of Matter, NEET
A liquid of density ρ is coming out of a hose pipe of radius a with horizontal speed v and hits a mesh. 50% of the liquid passes through the mesh unaffected. 25% looses all of its momentum and 25% comes back with the same speed. The resultant pressure on the mesh will be:

Options:
A 12ρv2
B ρv2
C 34ρv2
D 14ρv2

The correct answer is:
34ρv2

Solution:
Mass per unit time =ρAv
Force due to momentum loss =14ρAv×v
Force due to bounce back =14ρAv×2v
Pressure =ρAv24+ρAv22A=34ρv2

Topic: Physics, Laws of Motion, NEET
A liquid of density Ï is filled in a U-tube is accelerated with an acceleration a so that the height of liquid in its two vertical arms are h1 and h2 as shown in the figure. If l is the length of horizontal arm of the tube, the acceleration a is

Options:
A g(h1-h2)2l towards right
B g(h1-h2)2l towards left
C g(h1-h2)l towards right
D g(h1-h2)l towards left

The correct answer is:
g(h1-h2)l towards right

Solution:
Pressure on left end of horizontal tube,
p1=p0+h1Ïg
Pressure on right end of horizontal tube,
p2=p0+h2Ï g
As p1>p2, so acceleration should be towards right hand side. If A is the area of cross-section of the tube in the horizontal portion of U-tube, then
p1 A-p2 A=(l A Ï)a
Or h1-h2Ï g A=l A Ï a
a=g ( h 1 â h 2 ) l

Topic: Physics, Mechanical Properties of Fluids, NEET
A liquid was mixed with ethanol and a drop of concentrated H2SO4 was added, a compound with a fruity smell was formed. The liquid was:

Options:
A HCHO
B CH3COCH3
C CH3COOH
D CH3OH

The correct answer is:
CH3COOH

Solution:
Here esterification reaction is involved to give ethyl acetate which has fruity smell.
CH3COOHI+C2H5OH→H+CH3COOC2H5I+H2O

Topic: Chemistry, Carboxylic Acid Derivatives, NEET
A liquid with high boiling point decomposes on simple distillation but it can be steam distilled for its purification. Explain how is it possible?

Options:
A Steam distillation allows for higher temperatures than simple distillation.
B Steam distillation is a special type of separation process for temperature sensitive materials like natural organic compounds. Some organic compounds tend to decompose at higher temperature and normal distillation does not suit this purpose. So, steam/water is added to the apparatus and the temperature of the compounds are depressed, evaporating them at lower temperature. Once the distillation is accomplishing, the vapours are condensed and hence there is the separation of the constituents at ease.
C Steam distillation uses a lower pressure than simple distillation.
D Steam distillation involves the use of specialized equipment.

The correct answer is:
Steam distillation is a special type of separation process for temperature sensitive materials like natural organic compounds. Some organic compounds tend to decompose at higher temperature and normal distillation does not suit this purpose. So, steam/water is added to the apparatus and the temperature of the compounds are depressed, evaporating them at lower temperature. Once the distillation is accomplishing, the vapours are condensed and hence there is the separation of the constituents at ease.

Solution:
Correct Option is : (B)
Steam distillation is a special type of separation process for temperature sensitive materials like natural organic compounds. Some organic compounds tend to decompose at higher temperature and normal distillation does not suit this purpose. So, steam/water is added to the apparatus and the temperature of the compounds are depressed, evaporating them at lower temperature. Once the distillation is accomplishing, the vapours are condensed and hence there is the separation of the constituents at ease.

Topic: Chemistry, General Organic Chemistry, NEET
A litre of solution is saturated with AgCl . To this solution if 1.0×10-4 mole of solid NaCl is added, what will be the [Ag+] , assuming no volume change

Options:
A increases
B decreases
C does not change
D zero

The correct answer is:
decreases

Solution:
Due to common ion effect of Cl- equilibrium shift to beckward direction, that is why [Ag+] will decrease.

Topic: Chemistry, Ionic Equilibrium, NEET
A load of mass m falls from a height h on to the scale pan hung from a spring, as shown in the figure. If the spring constant is k, mass of the scale pan is zero and the mass m does not bounce, relative to the pan, then the amplitude of vibration is

Options:
A mgk
B mgk 1+2hkmg
C mgk+mgkÂ 1+2hkmg
D mgkÂ 1+2hkmg-mgk

The correct answer is:
mgk 1+2hkmg

Solution:
According to conservation principle,
mgh=12 kx02-mgx0
where x0 is maximum elongation in spring (when particle is in its lowest extreme position)
or 12kx02-mgx0-mgh=0
or x02-2mgk x0-2mgk h=0
â´ x0=2mgk Â± 2mgk2+ 4 Ã 2mgkh2
âµ Amplitude = elongation in spring for lowest extreme position - elongation in spring for equilibrium position =x0-x1
=mgk 1+2hkmg

Topic: Physics, Oscillations, NEET
A location with luxuriant growth of lichens on the trees indicates that the:

Options:
A Trees are very healthy
B Trees are heavily infested
C Location is highly polluted
D Location is not polluted

The correct answer is:
Location is not polluted

Solution:
Lichens are indicators of SO2 pollution. If the location is not polluted by SO2 then the growth of lichen will enhance. Lichens are also useful in the detection of toxic elemental pollutants and radioactive metals.

Topic: Biology, Biological Classification, NEET, NEET 2014