datavorous/entrance-exam-dataset huggingface dataset:
A system consists of three masses m1, m2 and m3 connected by a string passing over a pulley P. The mass m3 hangs freely and m2 and m1 are on a rough horizontal table (the coefficient of friction =μ). The pulley is frictionless and of negligible mass. The downward acceleration of mass m3 is: (Assume m1=m2=m3=m)

Options:
A g1-μ9
B 2gÎ¼3
C g1-2Î¼3
D g1-2Î¼2

The correct answer is:
g1-2μ3

Solution:
Let the tension force between m1 and m2 is T1 and m2 and m3 is T2

T1-μmg=ma ...1

T2-T1-μmg=ma ...2

mg-T2=ma ...3

From equation 1,2 & 3,

3amg=mg-2μmg

⇒a=mg-2μmg3mg
⇒a=g-2μmg3=g1-2μ3

Topic: Physics, Laws of Motion, NEET, NEET 2014
A system goes from A to B via two processes I and II as shown in figure. If âU1 and âU2 are the changes in internal energies in the processes I and II respectively, then

Options:
A ∆U1= ∆U2
B Relation between ∆U1and ∆U2 cannot be determined
C ∆U2> ∆U1
D ∆U2< ∆U1

The correct answer is:
∆U1= ∆U2

Solution:
The change in internal energy does not depend upon path followed by the process. It only depends on initial and final states.
Hence, ∆U1=∆U2

Topic: Physics, Thermodynamics, NEET
A system goes from P to Q by two different paths in the P-V diagram as shown in Fig. 12.8. Heat given to the system in path 1 is 1000 J. The work done by the system along path 1 is more than path 2 by 100 J. What is the heat exchanged by the system in path 2?

Options:
A 600 J
B 700 J
C 800 J
D 900 J

The correct answer is:
900 J

Solution:
Correct Option is : (D)
900 J

Topic: Chemistry, Thermodynamics, NEET
A system is taken from state a to state $c$ by two paths $a d c$ and $a b c$ as shown in the figure. The internal energy a is $U_2=10$ J. Along the path adc the amount of heat heat absorbed $\delta Q_1=50 \mathrm{~J}$ and the work obtained $\delta W_1=20 \mathrm{~J}$ whereas along the path $a b c$ the heat absorbed $\delta Q_2=36 \mathrm{~J}$. The amount of work along the path $a b c$ is:

Options:
A $10 \mathrm{~J}$
B $12 \mathrm{~J}$
C $38 \mathrm{~J}$
D $6 \mathrm{~J}$

The correct answer is:
$6 \mathrm{~J}$

Solution:
According to $1^{\text {st }} 1$ aw of thermodynamics
$\delta Q=\delta V+\delta W$
Along the path adc
$\delta U_1=\delta Q_1-\delta W_1=50-20=30 \mathrm{~J}$
then path abc

Since change in Internal energy in path in dependence
$\begin{aligned}
\text {So, } \delta U_1 & =\delta U_2 \\
30 \mathrm{~J} & =36 \mathrm{~J}-\delta W_2 \\
\Rightarrow \delta W_2 & =36 \mathrm{~J}-30 \mathrm{~J} \\
& =6 \mathrm{~J}
\end{aligned}$

Topic: Physics, Thermodynamics, NEET, NEET 2013 (Karnataka)
A system is taken through a cyclic process represented by a circle as shown. The heat absorbed by the system is

Options:
A ÏÃ103Â J
B Ï2 J
C 4ÏÃ102Â J
D ÏÂ J

The correct answer is:
Ï2 J

Solution:
In the cyclic process
ΔQ= Work done = Area inside the closed curve. Treat the circle as an ellipse of area
ΔQ=π4P2-P1 V2-V1
⇒ ΔQ=π4150-50×10-3.40-20×10-6
⇒ ΔQ=12 J

Topic: Physics, Thermodynamics, NEET
A system of classification in which a large number of traits are considered is

Options:
A Artifical system
B Synthetic system
C Natural system
D Phylogenetic system

The correct answer is:
Natural system

Solution:
The natural system of classification considers comparable study of a number of characters so as to bring out natural similarities and dissimilarities and hence natural relationships among the organisms.

The system employs characters which are relatively constant. They include morphological characters, anatomical characters, cytological characters, physiology, ontogeny or development, reproduction, cytochemistry and bio­chemistry, experimental taxonomy, etc.

Topic: Biology, The Living World, NEET
A system of two identical, uniform discs with identical circular cavities, is shown in the figure. Different relevant coordinates are given in the figure. The coordinates of the centre of mass of the system are

Options:
A 3R2, 5R4
B 19R6, 5R2
C R2, R4
D 20R6, 5R2

The correct answer is:
19R6, 5R2

Solution:
Mass of the removed portion
=MπR2×πR22
=M4
xcm=M·0-M4×R+M×7R-M4×8RM-M4+M-M4
=19R6
ycm=M·0-M40+M5R-M45RM-M4+M-M4
=5R2

Topic: Physics, Center of Mass Momentum and Collision, NEET
A system shown in the figure consists of a massless pulley, a spring of force constant k and a block of mass m. If the block is slightly displaced vertically downwards from its equilibrium position and released, then the period of vertical oscillations is

Options:
A T=πm⁡4k
B T=2πm⁡4k
C T=2πm⁡2k
D T=2πm⁡3k

The correct answer is:
T=2πm⁡4k

Solution:
In this situation, if mass m moves downwards a distance x from the equilibrium position, the pulley will also move by x and so, the spring will stretch by 2x.
Therefore, the spring force will be 2kx. The restoring force on the block will be 4 kx.
Hence, F=- 4 kx

⇒ mg−4kx=ma
⇒ ma=−4kx+mg
⇒ a = - 4 k x m⁡ +g⇒ ω 2 =| a x |
T=2π|xa|
T=2π(m⁡4k)

Topic: Physics, Oscillations, NEET
A table fan, rotating at a speed of 2400  rpm is switched off and the resulting variation of the rpm with time is shown in the figure. The total number of revolutions of the fan before it comes to rest is

Options:
A 420
B 280
C 240
D 380

The correct answer is:
280

Solution:
Number of revolutions = area under the curve

no. of revolution = 1 2 Ã 2400 60 + 600 60 Ã8+ 1 2 Ã 600 60 Ã16
no. of revolution =â280

Topic: Physics, Laws of Motion, NEET
A tactic taken up by the prey to escape from Predators is a method called ___.

Options:
A Commensalism
B Mutualism
C Camouflage
D Predation

The correct answer is:
Camouflage

Solution:
A tactic taken up by the prey to escape from Predators is a method called camouflage

Topic: Biology, Organisms and Populations, NEET
A tall true breeding garden pea plant is crossed with a dwarf true breeding garden pea plant. When the F1 plants were selfed the resulting genotypes were in the ratio

Options:
A 1 : 2 : 1 :: Tall homozygous : Tall heterozygous : Dwarf
B 1 : 2 : 1 :: Tall heterozygous : Tall homozygous : Dwarf
C 3 : 1 :: Tall : Dwarf
D 3 : 1 :: Dwarf : Tall

The correct answer is:
1 : 2 : 1 :: Tall homozygous : Tall heterozygous : Dwarf

Solution:
Mendel did the crossing of pure tall plant represented by (TT) with a pure dwarf plant represented by (tt). In the F1 generation of this cross, plants formed are tall (Tt). When this plant from F1 generation undergoes self-pollination, then in the F2 generation, three tall plants and one dwarf plants are seen. In this 1 homozygous tall plant (TT), two heterozygous tall plants(Tt), 1 homozygous dwarf plant(tt) are formed. Phenotypic ratio is 3:1::Tall:Dwarf, while the genotypic ratio becomes 1:2:1::Tall homozygous : Tall heterozygous : Dwarf

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2016 (Phase 1)
A tangential force F acts at the top of a thin spherical shell of mass m and radius R. The acceleration of the shell, if it rolls without slipping is

Options:
A 5F6m
B 6F5m
C 7m2F
D 2m7F

The correct answer is:
6F5m

Solution:
torque about 'O' τ=Iα
F( 2R )=α 5 3 m R 2
α= 6F 5mR
Linear acceleration a=Rα= 6F 5m

Topic: Physics, Rotational Motion, NEET
A tank is filled with water of density 1 g cm-3 and oil of density 0.9 g cm-3. The height of the water layer is 100 cm and of the oil layer is 400 cm. If g=980 cm s-2, then the velocity of efflux from an opening in the bottom of the tank is

Options:
A 920×980 cm s-1
B 900×980 cm s-1
C 1000×980 cm s-1
D 92×980 cm s-1

The correct answer is:
920×980 cm s-1

Solution:
Pressures at the bottom of the tank must be equal to the pressures due to water of height h.
If dw and d0 be the densities of water and oil respectively, Pressure at the bottom of the tank
=hwdwg+h0d0g
If this pressure be equivalent to pressure due to water of height h.
Then, hdwg=hwdwg+h0d0g
h=hw+h0d0dw=100+400×0.91
=100+360=460
According to Torricelli's theorem
v=2gh=2×980×460
=920×980 cm s-1

Topic: Physics, Mechanical Properties of Fluids, NEET
A tank of height H is fully filled with water. If the water rushing from a hole made in the tank below the free surface, strikes the floor at a maximum horizontal distance, then the depth of the hole from the free surface must be

Options:
A 34H
B 23H
C 14H
D 12H

The correct answer is:
12H

Solution:
As hole is made in the tank below the free surface of water, so water rushing from this hole follows a parabolic path.
The velocity of efflux liquid,
v=2gh
Time, t=2H-hg
Horizontal range, R=vt
R=2gh×2H-hg12
ie, R2=4hH-h=4Hh-h2
The range is maximum if
dRdh=0
or 2RdRdh=4H-2h
or 0=H-2h
or h=H2

Topic: Physics, Mechanical Properties of Fluids, NEET
A tap root is always

Options:
A +vely geotropic and +vely hydrotropic
B +vely phototropic and +vely hydrotropic
C -vely geotropic and +vely hydrotropic
D -vely hydrotropic and -vely geotropic

The correct answer is:
+vely geotropic and +vely hydrotropic

Solution:
Tap root always grow towards the soil and bends towards the source of water.

Topic: Biology, Morphology of Flowering Plants, NEET
A tautonym is

Options:
A non Latinised name
B same name for genus and species
C common name used as scientific name
D unscientific explanation of a phenomenon

The correct answer is:
same name for genus and species

Solution:
A scientific name having the same generic name and specific name is called tautonyms. Tautonyms are accepted for animals but not accepted for plants. Example Ratus ratus, Naja naja, Catla catla.

Topic: Biology, The Living World, NEET
A taxon is

Options:
A a group of related families
B a group of related species
C a type of living organisms
D a taxonomic group of any ranking.

The correct answer is:
a taxonomic group of any ranking.

Solution:
The word taxon signifies a taxonomic group of any rank that represents the real biological organisms included in a category. The term taxon was coined by Adolf Meyer (1926). Basic unit or smallest taxon of taxonomy/classification is species.

Topic: Biology, The Living World, NEET
A taxon is

Options:
A A Group of related families
B A group of related species
C A type of living organisms
D A taxonomic group of any ranking

The correct answer is:
A taxonomic group of any ranking

Solution:
The term taxon has been introduced in 1948 which indicates a taxonomic group of any rank. It was first used in 1926 by ‘Adolf Meyer-Abich for animal groups, as a back-formation from the word taxonomy. Herman Johannes Lam in 1948 used it for plants and it was adopted at VII International Botanical Congress, held in 1950.

Topic: Biology, The Living World, NEET
A technique in which sperm are stored for a long duration is

Options:
A incubator
B cryopreservation
C deep freezing
D all of these

The correct answer is:
cryopreservation

Solution:
The sperm from a Donor is preserved in liquid nitrogen and can be used when it is required and this process of storage at -196oC is called cryopreservation.

Topic: Biology, Reproductive Health, NEET
A telephone wire of length 200 km has a capacitance of 0.014 Î¼F per km. If it carries an AC frequency 5âkHz, what should be the value of an inductor required to be connected in series so that the impedance of the circuit is minimum?

Options:
A 0.35âmH
B 3.5 mH
C 2.5 mH
D zero

The correct answer is:
0.35 mH

Solution:
Capacitance of wire
C=0.014×10-6×200
=2.8×10-6 F=2.8 μF
For impedance of the circuit to be minimum
XL=XC
⇒2πfL=12πfC
L=14π2f2C
=143.142×5×1032×2.8×10-6
=0.35×10-3 H=0.35 mH

Topic: Physics, Alternating Current, NEET
A telescope has an objective lens of 10 $\mathrm{cm}$ diameter and is situated at a distance of one kilometers from two objects. The minimum distance between these two objects, which can be resolved by the telescope, when the mean wavelength of light is $5000 Å$, is of the order of:

Options:
A $0.5 \mathrm{~m}$
B $5 \mathrm{~m}$
C $5 \mathrm{~mm}$
D $5 \mathrm{~cm}$

The correct answer is:
$5 \mathrm{~mm}$

Solution:
According to the question:

$\begin{array}{ll}
\theta =\frac{1.22}{a} \lambda \\
\therefore \frac{d}{D} =\frac{\lambda}{\mathrm{a}} \\
\Rightarrow a^{\prime}=\frac{\lambda D}{a} \\
a =\frac{5000 \times 10^{-10} \times 10^3}{10 \times 10^{-2}}=5 \mathrm{~mm}
\end{array}$

Topic: Physics, Ray Optics, NEET, NEET 2004
A telescope of objective lens diameter 2 m uses light of wavelength 5000 â« for viewing stars. The minimum angular separation between two stars whose image is just resolved by the telescope is

Options:
A 4Ã10-4Â rad
B 0.25Ã10-6Â rad
C 0.31Ã10-6 rad
D 5Ã10-3Â rad

The correct answer is:
0.31Ã10-6 rad

Solution:
Minimum angular separation between two stars whose image is just resolved by telescope is
Δθ=1.22λa
∴Δθ=1.22×5000×10-102
Δθ=3050×10-10=3.05×10-7 rad
Δθ≈0.31×10-6 rad

Topic: Physics, Wave Optics, NEET
A temporary endocrine gland in the human body is

Options:
A Pineal gland
B Corpus cardiacum
C Corpus luteum
D Corpus allatum

The correct answer is:
Corpus luteum

Solution:
The corpus luteum is the temporary endocrine structure formed in the ovary after ovulation. It is responsible for the release of hormones like progesterone, estrogen, etc.
The pineal gland is a small endocrine gland in the brain of animals. The pineal gland produces melatonin that modulates sleep patterns in both circadian and seasonal cycles.

Corpus cardiacum: In insects, it is located posterior to the brain and function in the storage and secretion of a brain hormone.

The corpus allatum (plural: corpora allata) is an endocrine gland that generates juvenile hormones.

Topic: Biology, Chemical Coordination and Integration, NEET, NEET 2017
A test cross is carried out to

Options:
A find the number of alleles of a gene
B determine two species will breed successfully
C predict whether two traits are linked
D determine the genotype of a plant in F2 generation.

The correct answer is:
determine the genotype of a plant in F2 generation.

Solution:
Test crosses are used to test an individual's genotype by crossing it with an individual of a known genotype. Individuals that show the recessive phenotype are known to have a homozygous recessive genotype. Individuals that show the dominant phenotype, however, may either be homozygous dominant or heterozygous.

Topic: Biology, Principles of Inheritance and Variation, NEET
A test cross is performed to distinguish between

Options:
A two homozygous forms
B two heterozygous forms
C a homozygous dominant and a heterozygous form
D a homozygous recessive and the heterozygous form

The correct answer is:
a homozygous dominant and a heterozygous form

Solution:
Test cross is a process to examine and identify whether the organism is heterozygous or not. In it, the organisms is crossed with the recessive homozygous as it will ensure the expression of the dominant character in heterozygous form for F1 generation enabling us to know the zygosity.

Topic: Biology, Principles of Inheritance and Variation, NEET
A tetra-atomic molecule X on reaction with nitrogen oxide (Oxidation State = 1) produces two substances Y and Z. Y is a dehydrating agent while compound Z is a diatomic gas which shows almost inert gas behavior. The substances X, Y and Z are

Options:
A P4, N2O5, O2
B P4, P4O10, Ar
C P4, P2O3, O2
D P4, P4O10, N2

The correct answer is:
P4, P4O10, N2

Solution:
When P4 is treated with N2O (O.S = 1) it produces P4O10 and N2 (diatomic molecule).
P4O10 used as dehydrating agent
P 4 + 1 0 N 2 O → P 4 O 1 0 + 1 0 N 2 X Y Z

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
A tetrapeptide has -COOH group on alanine. This produces glycine (Gly), valine (Val), phenyl alanine (Phe) and alanine (Ala), on complete hydrolysis. For this tetrapeptide, the number of possible sequences (primary structures) with -NH2 group attached to a chiral center is

Options:
A 2
B 3
C 4
D 5

The correct answer is:
4

Solution:
Hence, number of possible sequences (primary structures) with -NH2 group attached to a chiral center is 4.

Topic: Chemistry, Biomolecules, NEET
A tetraploid female flowering plant and a hexaploid male flowering plant are crossed, endosperm will be

Options:
A heptaploid.
B triploid.
C hexaploid.
D pentaploid.

The correct answer is:
heptaploid.

Solution:
Double fertilization is a unique characteristic observed in the Angiosperm group of plants. In this, two fertilizations are observed - 1st between the female and male gametic cells forming zygote and 2nd is between the second male gamete and polar nuclei forming the primary endosperm nucleus. The ploidy of PEN is 3n (male gamete 1n + polar nuclei 2n). If the female plant is tetraploid then polar nuclei will be tetraploid as well. The male plant is hexaploid thus, its male gamete will be triploid. Therefore, the ploidy level of endosperm will 4n+3n=7nâ heptaploid.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
A tetravalent element forms monoxide and dioxide with oxygen. When air is passed over heated element (1273 K), producer gas is obtained. Monoxide of the element is a powerful reducing agent and reduces ferric oxide to iron. Identify the element and write formulas of its monoxide and dioxide.

Options:
A Carbon, CO, CO2
B Silicon, SiO, SiO2
C Boron, BO, BO2
D Phosphorus, PO, PO2

The correct answer is:
Carbon, CO, CO2

Solution:
Correct Option is : (A)
Carbon, CO, CO2

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
A. The conditions primitive on earth (before the origin of life) were: low temperature, volcanic storms, reducing atmosphere containing CH4, NH3, etc.
B. Cheetah and humans share similarities in the pattern of bones of forelimbs.
C. In a collection of moths made in the 1850s, i.e., before industrialization set in, it was observed that there were more dark-winged moths on trees than white-winged moths.
D. Branching descent and natural selection are the two key concepts of Darwinian theory of evolution.
E. The Neanderthal man with a brain size of 1400cc lived in east and central Asia 25,000–10,000 years ago.

Which set of given statements are correct?

Options:
A A, B, E
B B, D
C A, C, E
D B, C, D

The correct answer is:
B, D

Solution:
A. The conditions primitive on earth (before the origin of life) were - high temperature, volcanic storms, reducing atmosphere containing CH4, NH3, etc.
C. In a collection of moths made in the 1850s, i.e., before industrialization set in, it was observed that there were more white-winged moths on trees than dark-winged moths.
D. The Neanderthal man with a brain size of 1400cc lived in Neanderthal valley Germany, between 1,00,000-40,000 years back.

Topic: Biology, Evolution, NEET
a. The passage through which movement of RNA and protein molecules takes place in both directions in a nuclear pore.
b. A disc-shaped structure is kinetochore is present as a lateral side of centromere.
c. DNA moves easily in both directions through the nuclear pore
Read the following and select the correct option

Options:
A a and b are true.
B b and c are true.
C c and a are true.
D a, b and c are true.

The correct answer is:
a and b are true.

Solution:
The materials like RNA and ribosomal proteins moving from the nucleus to the cytoplasm and proteins like DNA polymerase, carbohydrates, signaling molecules, and lipids move into the nucleus through the nuclear pore. Kinetochores are a proteinaceous structure present at the side of the centromere for the attachment of spindle fiber during cell division. There is very little access to the nuclear interior that is seen through the nuclear pores for DNA (mitosis/gene transfer).

Topic: Biology, Cell - The Unit of Life, NEET
A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig
Its volume is then reduced to the original value from E to F by an isobaric process. What is the total work done by the gas from D to E to F

Options:
A 220J
B 400J
C 120J
D 450J

The correct answer is:
450J

Solution:
Correct Option is : (D)
450J

Topic: Chemistry, Thermodynamics, NEET
A thermodynamic system is taken through the cycle $A B C D$ as shown in figure. Heat rejected by the gas during the cycle is

Options:
A $2 \mathrm{pV}$
B $4 p V$
C $\frac{1}{2} p V$
D $p V$

The correct answer is:
$2 \mathrm{pV}$

Solution:
For given cyclic process, $\Delta U=0$
$\therefore Q=W$
Also, $W=-$ area enclosed by the curve
$\begin{aligned}
& =A B \times A D \\
& =(2 p-p)(3 V-V) \\
& =-p \times 2 V
\end{aligned}$
$\therefore$ Heat rejected $=2 p_V$

Topic: Physics, Thermodynamics, NEET, NEET 2012 (Screening)
A thermodynamic system is taken through the cycle $a b c d a$. The work done by the gas along the path $b c$ is:

Options:
A 30 J
B -90 J
C -60 J
D Zero

The correct answer is:
Zero

Solution:
Path $b c$ is an isochoric process.
$\therefore \quad$ Work done by gas along path $b c$ is zero.

Topic: Physics, Thermodynamics, NEET, NEET 2024
A thermodynamic system is taken through the cycle ABCD as shown in the figure. Heat rejected by the gas during the cycle is

Options:
A 2PV
B 4PV
C 12PV
D PV

The correct answer is:
2PV

Solution:
In a cyclic process Δ U = 0 . In a cyclic process work done is equal to the area under the cycle and is positive if the cycle is clockwise and negative if anticlockwise.

∴ΔW=-Area of rectangle ABCD =-P2V =-2PV
According to the first law of thermodynamics
Δ Q = Δ U + Δ W or Δ Q = Δ W As Δ U = 0
i.e., heat supplied to the system is equal to the work done
So heat absorbed, Δ Q = Δ W = - 2 PV
∴ Heat rejected by the gas = 2PV

Topic: Physics, Thermodynamics, NEET
A thermodynamic system is taken through the cycle PQRSP process. The net work done by the system is

Options:
A 20Â J
B -20 J
C 400Â J
D -374Â J

The correct answer is:
-20 J

Solution:
Work done by the system = Area of the loop on P-V diagram
W=-300-100×10-6×200-100×103
W=-20 J

(As cyclic process is in anti-clockwise sense, the work done is negative)

Topic: Physics, Thermodynamics, NEET
A thermodynamical system undergoes cyclic process ABCDA as shown in the figure. Work done by the system is

Options:
A Zero
B 2P0V0
C P0V0

D 32P0V0

The correct answer is:
Zero

Solution:
As can be seen from figure,

W BCOB = - area of Î BCO = - P 0 V 0 2
W AODA = + area of Î AOD = + P 0 V 0 2
â´ W net = W BCOB + W AODA = - P 0 V 0 2 + P 0 V 0 2 = zero

Topic: Physics, Thermodynamics, NEET
A thermodynamics system undergoes cyclic process ABCDA as shown in Figure. The work done by the system in the cycle is:

Options:
A P0V0
B 2P0V0
C P0V02
D Zero

The correct answer is:
Zero

Solution:
Work done in process ODA=12×P0×V0
While work done in the process OBC=-12 P0V0
Net work done =0

Topic: Physics, Thermodynamics, NEET, NEET 2014
A thick current carrying cable of radius R carries current I uniformly distributed across its cross-section. The variation of magnetic field B(r) due to the cable with the distance r from the axis of the cable is represented by:

Options:
A
B
C
D

The correct answer is:

Solution:
For inside part

J=IπR2

∫B→·dl→=μ0 Iinclosed …i

B 2πr=μ0IπR2×πr2

B 2πr=μ0Ir2R2

B∝r

For outside part

∫B→·dl→=μ0Iinclosed

B(2πr)=μ0I

∴ B∝1r

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2021
A thin biconvex lens of refractive index 32 and radius of curvature 30 cm is put in water (refractive index =43). Its focal length is

Options:
A 0.15 m
B 0.30 m
C 0.45 m
D 1.20 m

The correct answer is:
1.20 m

Solution:
Using lens-makers formula

1f=Î¼1Î¼2-11R1-1R2
1f=3/24/3-110.3+10.3
Or 1f=98-120.3
Or 1f=18Ã20.3 or f=1.20 m

Topic: Physics, Ray Optics, NEET
A thin circular disk is in the xy plane as shown in the figure. The ratio of its moment of inertia about z and z' axes will be:

Options:
A 1 :4
B 1:5
C 1:3
D 1:2

The correct answer is:
1:3

Solution:
IZ= mR22
(by parallel axis theorem)
Iz′=32mR2
IZIz′=13

Topic: Physics, Rotational Motion, NEET
A thin circular disk of radius R is uniformly charged with density ρ>0 per unit area. The disk rotates about its axis with a uniform angular speed ω. The magnetic moment of the disk is

Options:
A 2πR4ωρ
B ÏR4ÏÏ
C πR42ωρ
D ÏR44ÏÏ

The correct answer is:
πR44ωρ

Solution:
M L = q 2m
M= q 2m L
M= ρπ R 2 2m ×Iω
M=ρπR22m×mR22ω

M= 1 4 π R 4 ρω

Topic: Physics, Magnetic Effects of Current, NEET