datavorous/entrance-exam-dataset huggingface dataset:
Two mercury drops (each of radius r) merge to form a bigger drop. The surface energy of the bigger drop, if T is the surface tension, is

Options:
A 253Â Ïr2T
B 4Â Ï2T
C 2Â Ïr2T
D 283 Ïr2T

The correct answer is:
283 Ïr2T

Solution:
Let R be the radius of the bigger drop, then Volume of bigger drop =2× volume of small drop
43πR3=2×43πr3⇒R=21/3r
Surface energy of bigger drop,
E=4πR2T=4×22/3πr2T=28/3πr2T

Topic: Physics, Mechanical Properties of Fluids, NEET
Two metal rods 1 and 2 of same lengths have same temperature difference between their ends. Their thermal conductivities are $K_1$ and $K_2$ and crosssectional areas $A_1$ and $A_2$ respectively. If the rate of heat conduction in 1 is four times that in 2 , then:

Options:
A $K_1 A_1=4 K_2 A_2$
B $K_1 A_1=2 K_2 A_2$
C $4 K_1 A_1=K_2 A_2$
D $K_1 A_1=K_2 A_2$

The correct answer is:
$K_1 A_1=4 K_2 A_2$

Solution:
Let $L=$ length of each rod Temperature difference $=\Delta T$
$\therefore$ Rate of flow of rod (i) is $H_1=$ $\frac{K_1 A_1 \Delta T}{L}$ and Rate of flow of rod (2) is $H_2=\frac{K_2 A_2 \Delta T}{L}$ from the equation
$$
\begin{array}{rlrl}
H_1 & =4 H_2 \\
\Rightarrow & \frac{K_1 A_1 \Delta \mathrm{T}}{L} & =4 \frac{K_2 A_2 \Delta T}{L} \\
\Rightarrow & K_1 A_1 & =4 K_2 A_2
\end{array}
$$

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2013 (Karnataka)
Two metal spheres, one of radius $\mathrm{R}$ and the other of radius $2 \mathrm{R}$ respectively have the same surface charge density $\sigma$. They are brought in contact and separated. What will be the new surface charge densities on them?

Options:
A $\sigma_1=\frac{5}{6} \sigma, \quad \sigma_2=\frac{5}{2} \sigma$
B $\sigma_1=\frac{5}{2} \sigma, \quad \sigma_2=\frac{5}{6} \sigma$
C $\sigma_1=\frac{5}{2} \sigma, \quad \sigma_2=\frac{5}{3} \sigma$
D $\sigma_1=\frac{5}{3} \sigma, \quad \sigma_2=\frac{5}{6} \sigma$

The correct answer is:
$\sigma_1=\frac{5}{3} \sigma, \quad \sigma_2=\frac{5}{6} \sigma$

Solution:
The surface charge density of a closed surface area having charge $\mathrm{Q}$ is given by
$$
\sigma=\frac{\text { Charge }}{\text { Area }}=\frac{\mathrm{Q}}{\mathrm{A}} \text { or } \mathrm{Q}=\sigma \mathrm{A}
$$
Thus, the charges on sphere P and Q having same charge density as shown in the figure below is given by
$$
\begin{gathered}
\mathrm{Q}_{\mathrm{P}}=\sigma \times 4 \pi \mathrm{R}^2=4 \pi \sigma \mathrm{R}^2 \\
\text { and } \mathrm{Q}_{\mathrm{Q}}=\sigma \times 4 \pi(2 \mathrm{R})^2=16 \pi \sigma \mathrm{R}^2
\end{gathered}
$$
when they are brought in contact with each other, the total charge will be
$$
\begin{aligned}
\mathrm{Q}_{\mathrm{t}} & =\mathrm{Q}_{\mathrm{P}}+\mathrm{Q}_{\mathrm{Q}} \\
= & 4 \pi \sigma \mathrm{R}^2+16 \pi \sigma \mathrm{R}^2 \text { [From Eq. (i) and (ii)] } \\
& =20 \pi \sigma \mathrm{R}^2
\end{aligned}
$$
In connection of two charged conducting bodies, the potential will become same on both, i.e.
$$
\begin{aligned}
\frac{\mathrm{Q}_{\mathrm{P}}}{4 \pi \in_0 \mathrm{R}}=\frac{\mathrm{Q}_{\mathrm{Q}}}{4 \pi \epsilon_0 2 \mathrm{R}} \\
\Rightarrow \quad \frac{\mathrm{Q}_{\mathrm{P}}}{\mathrm{R}}=\frac{\mathrm{Q}_{\mathrm{Q}}}{2 \mathrm{R}} \Rightarrow \frac{\mathrm{Q}_{\mathrm{P}}}{\mathrm{Q}_{\mathrm{Q}}}=\frac{1}{2}
\end{aligned}
$$
So, the charges on the sphere $\mathrm{P}$ and $\mathrm{Q}$ after separation will be distributed as
$$
\Rightarrow \quad Q_P^{\prime}=\frac{1}{3} Q_{\mathrm{t}} \text { and } \mathrm{Q}_{\mathrm{Q}}^{\prime}=\frac{2}{3} \mathrm{Q}_{\mathrm{t}}
$$
After separation, the new surface charge densities on $\mathrm{P}$ and $\mathrm{Q}$ will be
$$
\begin{aligned}
\sigma_{\mathrm{P}}=\frac{\mathrm{Q}_{\mathrm{P}}^{\prime}}{\text { Area }} & =\frac{1}{3} \frac{\mathrm{Q}_{\mathrm{t}}}{\text { Area }}=\frac{1}{3} \frac{20 \pi \sigma \mathrm{R}^2}{4 \pi \mathrm{R}^2}=\frac{5}{3} \sigma \\
\text { and } \sigma_{\mathrm{Q}}=\frac{\mathrm{Q}_{\mathrm{Q}}^{\prime}}{\text { Area }} & =\frac{2}{3} \frac{\mathrm{Q}_{\mathrm{t}}}{\text { Area }}=\frac{2}{3} \times \frac{20 \pi \sigma \mathrm{R}^2}{4 \pi(2 \mathrm{R})^2} \\
& =\frac{2}{3} \times \frac{5}{4} \sigma=\frac{5}{6} \sigma
\end{aligned}
$$

Topic: Physics, Electrostatics, NEET, NEET 2019 (Odisha)
Two metal wires of identical dimensions are connected in series. If v1 and v2 are the conductivities of the metal wires respectively, the effective conductivity if the combination is:

Options:
A Ï1+Ï22Â Ï1Â Ï2
B Ï1+Ï2Ï1Â Ï2
C Ï1Â Ï2Ï1+Ï2
D 2 σ1 σ2σ1+σ2

The correct answer is:
2 σ1 σ2σ1+σ2

Solution:
Rec=lσ1A+lσ1A=leqσeqAeq
2lσeqA=lAσ1+σ2σ1σ2
σeq=2σ1σ2σ1+σ2

Topic: Physics, Current Electricity, NEET, NEET 2015 (Phase 2)
Two metal wires P and Q of same length and material are stretched by the same load. Their masses are in the ratio m1:m2. The ratio of elongations of wire is

Options:
A m12:m22
B m22:m12
C m2 :m1
D m1:m2

The correct answer is:
m2 :m1

Solution:
m 1 : m 2
Y= Fl AÎl
Îl= Fl YA
m=ÏV=ÏÃAÃl
A Î± m
Î l 1 Î l 2 = A 2 A 1 = m 2 m 1

Topic: Physics, Mechanical Properties of Solids, NEET
Two metallic spheres each of mass M are suspended by two strings each of length L. The distance between the upper ends of strings is L. The angle which the strings will make with the vertical due to mutual attraction of the spheres is

Options:
A tan-1GMgL
B tan-1GM2gL
C tan-1GMgL2
D tan-12GMgL2

The correct answer is:
tan-1GMgL2

Solution:
The metallic spheres will be at positions as shown in the figure.
Tsinθ=F= GM×M L 2 = G M 2 L 2 ..............( 1 )
Tcosθ=Mg .................( 2 )
divide equations ( 1 ) & ( 2 )
∴           tanθ= GM g L 2
or          θ= tan −1 ( GM g L 2 )

Topic: Physics, Gravitation, NEET
Two microbes found to be very useful in genetic engineering are:

Options:
A crown gall bacterium and Caenorhabditis elegans
B Escherichia coli and Agrobacterium tumefaciens
C Vibrio cholerae and a tailed bacteriophage
D Diplococcus sp. and Pseudomonas sp.

The correct answer is:
Escherichia coli and Agrobacterium tumefaciens

Solution:
Escherichia coli and Agrobacterium tumefaciens are the microbes found to be very useful in genetic engineering. E. coli is a motile, gram negative, rod shaped bacterium which is a normal inhabitant of human colon. It is most extensively used in bacterial genetics and molecular biology. Agrobacterium tumefaciens is a soil bacterium. It has Ti plasmid (Tumour inducing plasmid) and it can be used for the transfer of a desired gene in dicot plants.
Related Theory
Using Agrobacterium vectors, nematode specific genes are introduced into host plants. The introduction of DNA produces both sense and antisense RNA in host cells. These two RNAs being complementary form a double stranded RNA (dsRNA) that initiates the RNAs and silence the specific $m R N A$ of the nematode. As a result of this, the parasite cannot survive in the transgenic host expressing specific interfering $R N A$.

Topic: Biology, Biotechnology - Principles and Processes, NEET, NEET 2006
Two mirrors are inclined at an angle Î¸ as shown in the diagram. Light ray is incident parallel to one of the mirrors. The ray will start retracting its path after third reflection if :

Options:
A θ = 4 5 â
B Î¸ = 3 0 â
C θ = 6 0 â
D All three

The correct answer is:
θ = 3 0 ∘

Solution:
α = 9 0 - θ 2
α 1 = 9 0 - θ = 9 0 - 9 0 - θ 2
9 0 - θ = 4 5 + θ 2
4 5 = 3 θ 2
θ = 3 0 ∘

Topic: Physics, Ray Optics, NEET
Two molecules of a gas have speeds of 9×106 m/s and 1×106 m/s, respectively. What is the root mean square speed of these molecules?

Options:
A √41×103m/s
B 6.1×104m/s
C 6.1×105m/s
D √41×106m/s

The correct answer is:
√41×106m/s

Solution:
Correct Option is : (D)
√41×106m/s

Topic: Chemistry, Kinetic Theory of Gases, NEET
Two molecules of the primary acceptor of carbon dioxide in Calvin cycle undergoes photorespiration. How many glycine molecules will be formed from it?

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
At high temperature and low concentration of carbon dioxide, C3 plants undergo photorespiration. In photorespiration, instead of carbon dioxide, atmospheric oxygen combines with RuBP (5C) to form One molecule of 3-PGA (3C) and One molecule of phospho-glycolate (2C) in the chloroplast. 3-PGA enters the Calvin cycle while Phospho- glycolate gets dephosphorylated to form glycolate (2C) which is then transported to the peroxisomes. In the peroxisomes, glycolate (2C) gets oxidised to form glyoxylate(2C) and hydrogen peroxide. Glyoxylate then gets converted to amino acid glycine (2C) which is transported to mitochondria. Thus one molecule of RUBP gives rise to one molecule of glycine.

Topic: Biology, Photosynthesis in Higher Plants, NEET
Two moles of a certain ideal gas at a temperature 300 K were cooled isochorically so that the gas pressure is reduced Î· (= 2) times. The gas is expanded isobarically afterwards till its temperature got back to original value. The total amount of heat absorbed by gas in this process is Cv=1.5R

Options:
A 2000Â J
B 2500 J
C 3000 J
D 4000Â J

The correct answer is:
2500 J

Solution:
ÎQV=constant=Î¼CV dT
=2Ã3R2Ã(-150) .... (1)
ÎQP=constant=2Ã5R2Ã150 ..... (2)
Total heat absorbed
Q=ÎQP=constant+ÎQV=constant
=750-4508.3=300Ã8.3=830Ã3
=2490 J=2500 J

Topic: Physics, Thermodynamics, NEET
two moles of an ideal gas are allowed to expand from a volume of 10 dm3 to 2 m3 at 300 K against a pressure of 101.325 kPa. Calculate the work done

Options:
A – 201.6 kJ
B 13.22 kJ
C – 810 kJ
D – 18.96 kJ

The correct answer is:
– 201.6 kJ

Solution:
V1=10 dm3=10-2m3,V2=2m3
P=101.325×103Pa
w=-PΔV
w=-101.325×1031.99=-201.6 kJ

Topic: Chemistry, Thermodynamics (C), NEET
Two moles of an ideal gas at 300 K were cooled at a constant volume so that the pressure is reduced to half the initial value. Then as a result of heating at constant pressure, the gas expands till it attains the original temperature. The total heat absorbed by the gas, if R is the gas constant, is

Options:
A 150 R
B 300 R
C 75 R
D 100 R

The correct answer is:
300 R

Solution:
For 1 mol of gas,
ÎQ=ÂµCvÎT+PÎV
At constant volume, ÎV=0
For 2 moles of gas,
ÎâQ=2CvÎT
From PV = nRT = 2R Ã 300
and P2V=2RTfâ¡
â´ Tfâ¡=150K
â´ ÎQ=2CvTfâ¡-Ti=2Cv150-300
=-300Cv
In the next process,
ÎQ=2CpÎT=2Cp=2Cp300-150
=300Cp
â´ Net heat absorbed =-300Cv+300Cp
=300Cp-Cv=300 R

Topic: Physics, Thermodynamics, NEET
Two moles of an ideal gas have undergone a cyclic process 1 â 2 â 3 â 1 . If net heat exchange in the process is -300 J , then work done by the gas in the process 2-3 is R=8.3 J mol-1 K-1

Options:
A -5000 J
B 5000 J
C -3000 J
D none of these

The correct answer is:
none of these

Solution:
The process 1-2 is isobaric therefore work done by the gas is
W1→2=nRΔT=2×R×300
W1→2=600R=4980 J
The process 3-1 is isochoric therefore work done by the gas is zero.
Therefore,
W1→2→3→1=W1→2+W2→3
Also for cyclic process,
ΔW=ΔQ
W1→2+W2→3+0=ΔQ
4980J⁡+W2→3=-300J
W2→3=-5280J

Topic: Physics, Thermodynamics, NEET
Two moles of an ideal gas is expanded isothermally and reversibly from 2 litre to 20 litre at 300 K. The enthalpy change (in kJ) for the process is

Options:
A 22.4 kJ
B –22.4 kJ
C 9.6 kJ
D 0 kJ

The correct answer is:
0 kJ

Solution:
For isothermal process for an ideal gas, ÎTâ=â0
Hence, ΔH =  nC p  ΔT = 0

Topic: Chemistry, Thermodynamics (C), NEET
Two moles of an ideal gas, with CPCV=53, are mixed with 3 moles of another ideal gas CPCV=43. The value of CPCV for the mixture is

Options:
A 1.45
B 1.50
C 1.47
D 1.42

The correct answer is:
1.42

Solution:
1.42

Topic: Physics, Thermodynamics, NEET
Two moles of helium gas expanded isothermally and irreversibly at 27 o C from volume 1 dm3 to 1 m3 at constant pressure of 100 k Pa . Calculate the work done.

Options:
A 99900 J
B â 99900 J
C 34464.65 J
D â34464.65 J

The correct answer is:
â 99900 J

Solution:
W=-PâV
Given, P = 1 0 0 k Pa = 1 0 5 Pa
V1=1 dm3=10-3m3,V2=1 m3
W=â 10 5 Ã 1â 10 â3 âJ
â´ W=â 99900âJ

Topic: Chemistry, Thermodynamics (C), NEET
Two moles of oxygen is mixed with eight moles of helium. The effective specific heat of the mixture at constant volume is

Options:
A 1.3Â R
B 1.4Â R
C 1.7 R
D 1.9Â R

The correct answer is:
1.7 R

Solution:
Let for mixture of gases, specific heat at constant volume be CV
C V = n 1 ( C V ) 1 + n 2 ( C V ) 2 n 1 + n 2
where for oxygen; ( C V ) 1 =5R2, n1=2 mol
For helium; ( C V ) 2 =3R2, n2=8 mol
Therefore, CV=2Ã5R2+8Ã3R22+8=17R10=1.7 R

Topic: Physics, Kinetic Theory of Gases, NEET
Two moles of PCl5 were heated in a closed vessel of 2 L. At equilibrium 40% of PCl5 is dissociated into PCl3 and Cl2. The value of equilibrium constant is

Options:
A 0.53
B 0.267
C 2.63
D 5.3

The correct answer is:
0.267

Solution:
Degree of dissociation=0.4
PCl 5 ââââââ PCl 3 + Cl 2 a(1âx)ââââââaxâââââââax
a=2, x=0.4, V=2
PCl5=2(1-0.4)2=0.6 mol/L
PCl3=2Ã0.42=0.4 mol/L
Cl2=2Ã0.42=0.4 mol/L
â´ Kc=PCl3Cl2PCl5
=0.4Ã0.40.6=0.267

Topic: Chemistry, Chemical Equilibrium, NEET
"Two monochromatic beams A and B of equal intensity I, hit a screen. The number of photons hitting the screen by beam A is twice that by beam B. What inference can you make about their frequencies?"

Options:
A The frequency of beam A is twice that of B.
B The frequency of beam B is twice that of A.
C The frequencies of both beams are equal.
D The frequencies of both beams cannot be determined with the given information.

The correct answer is:
The frequency of beam B is twice that of A.

Solution:
Correct Option is : (B)
The frequency of beam B is twice that of A.

Topic: Chemistry, Dual Nature of Matter, NEET
Two monomers in maltose are

Options:
A α -D-glucose and α -D-Fructose
B α -D-glucose and β -D-galactose
C α -D-glucose and α -D-glucose
D α -D-glucose and α -D-galactose

The correct answer is:
α -D-glucose and α -D-glucose

Solution:
Maltose on hydrolysis give 2 mole of α -D Glucose because in maltose glucosidic linkage is present in between C1 & C4 of α -D glucose.

Topic: Chemistry, Biomolecules, NEET
Two mya, ___________ probably lived in East African grasslands. Evidences show they hunted with stone weapons but essentially ate fruit.

Options:
A Australopithecus
B Neanderthal man
C Homo erectus
D Dryopithecus

The correct answer is:
Australopithecus

Solution:
Two mya, Australopithecines probably lived in East African grasslands. Evidence shows they hunted with stone weapons but essentially ate the fruit. Some of the bones among the bones discovered were different. This creature was called the first human-like being the hominid and was called Homo habilis. The brain capacities were between 650-800cc. They probably did not eat meat.

Topic: Biology, Evolution, NEET
Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities Ï 1 and Ï 2, respectively, touch each other. The net electric field at a distance 2R from the centre of the smaller sphere, along the line joining the centres of the spheres, is zero. The ratio Ï1Ï2 is

Options:
A 2
B 8
C 4
D 5

The correct answer is:
4

Solution:
Field due to small sphere
= K ρ 1 4 3 π R 3 2 R 2
Field due to bigger sphere
= K ρ 2 4 3 π 2 R 3 × R 2 R 3
Equating above two, we get ρ 1 ρ 2 = 4

Topic: Physics, Electrostatics, NEET
Two non-mixing liquids of densities ρ and nρn>1 are put in a container. The height of each liquid is h. A solid cylinder of length L and density d is put in this container. The cylinder floats with its axis vertical and length pLp<1 in the denser liquid. The density d is equal to:

Options:
A 1+n+1pρ
B 2+n+1pρ
C 2+n-1pρ
D 1+n-1pρ

The correct answer is:
1+n-1pρ

Solution:
f0=mg
ρA1-pLg+nρApLg=dALg
ρ1-p+nρp=d
1-p+npρ=d
1+n-1pρ=d

Topic: Physics, Mechanical Properties of Fluids, NEET, NEET 2016 (Phase 1)
Two north poles each of pole strength m and a south pole of pole strength m are placed at the three corners of an equilateral triangle of side a. The intensity of magnetic induction field strength at the centre of the triangle is

Options:
A μ04πmda2
B Î¼02Ï3ma2
C μ04π9ma2
D μ04πm2a2

The correct answer is:
μ02π3ma2

Solution:
Bres=2B

where, B=μ04πma32

⇒ Bres=2×μ04πma32⇒Bres=3μ02πma2

Topic: Physics, Magnetic Properties of Matter, NEET
Two nuclei A and B are isotonic with mass numbers 15 and 16 respectively. If A contains 7 protons, then the number of protons in B would be

Options:
A 7
B 8
C 9
D 10

The correct answer is:
8

Solution:
A1=15
A2=16
N1=N2=N
Z1=7
A1=Z1+N
A2=Z2+N
15=7+N .......(i)
16=Z2+N .......(ii)
Equation (ii), (i)
Z2-7=1
Z2=1+7
Z2=8

Topic: Physics, Nuclear Physics, NEET
Two nuclei have mass numbers in the ratio 1:8. The ratio of their nuclear radii would be:

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

The correct answer is:
1:2

Solution:
A 1 : A 2 =1:8 ⇒ A 1 A 2 = 1 8

Since ,

R= R 0 A 1/3

∴ R 1 R 2 = A 1 1/3 A 2 1/3 ⇒ ( A 1 A 2 ) 1/3 = ( 1 8 ) 1/3                          = 1 2

The ratio of their radii = 1 : 2.

Topic: Physics, Nuclear Physics, NEET
Two nuclei have mass numbers in the ratio 27:125. The ratio of their nuclear radii would be:

Options:
A 5:9
B 9:5
C 5:3
D 3:5

The correct answer is:
3:5

Solution:
Here ,

A 1 : A 2 =27:125 â A 1 A 2 = 27 125

Since, R= R 0 A 1/3

â´ R 1 R 2 = A 1 1 3 A 2 1 3 = ( A 1 A 2 ) 1 3 âââââââââââââââââââââââââ= ( 27 125 ) 1 3 âââââââââââââââââââââââââ= 3 5

â´ The ratio of their radii = 3 : 5

Topic: Physics, Nuclear Physics, NEET
Two nuclei have their mass numbers in the ratio of $1: 3$. The ratio of their nuclear densities would be

Options:
A $1: 3$
B $3: 1$
C $(3)^{1 / 3}: 1$
D $1: 1$

The correct answer is:
$1: 1$

Solution:
Density of nuclear matter is independent of mass number, so the required ratio is $1: 1$.
Alternative :
$A_1: A_2=1: 3$
Their radii will be in the ratio
$\begin{gathered}
R_0 A_1^{1 / 3}: R_0 A_2^{1 / 3}=1: 3^{1 / 3} \\
\text { Density }=\frac{A}{\frac{4}{3} \pi R^3} \\
\therefore \rho_{A_1}: \rho_{A_2}=\frac{1}{\frac{4}{3} \pi R_0^3 \cdot 1^3}: \frac{3}{\frac{4}{3} \pi R_0^3\left(3^{1 / 3}\right)^3}
\end{gathered}$
Their nuclear densities will be the same.

Topic: Physics, Nuclear Physics, NEET, NEET 2008 (Screening)
Two objects of mass 10 kg and 20 kg respectively are connected to the two ends of a rigid rod of length 10 m with negligible mass. The distance of the center of mass of the system from the 10 kg mass is:

Options:
A 203m
B 10 m
C 5 m
D 103 m

The correct answer is:
203m

Solution:
Let's consider the location of 10 kg mass to be origin.
Then location of COM will be,

xcom=m1x1+m2x2m1+m2=10×0+20×1010+20⇒xcom=203 m

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2022 (Phase 1)
Two opposite forces operate in the growth and development of every population. One of them relates to the ability to reproduce at a given rate. The force opposing it is called:

Options:
A morbidity
B fecundity
C biotic potential
D environmental resistance

The correct answer is:
environmental resistance

Solution:
Two opposite forces operating in growth and development of every populations. One of them has ability to reproduce at a given rate. The opposing force is environmental resistance.

Topic: Biology, Organisms and Populations, NEET, NEET 2003
Two organisms with different correlated morphological characters belongs to

Options:
A One biological species
B One taxonomic species
C Two biological species
D Two taxonomic species

The correct answer is:
Two taxonomic species

Solution:
Species is a group of similar individuals which produces offspring's that are fertile and capable of interbreeding.
Theodosius Dobzhansky gave the biological species concept defined as "a species as a group of individuals living in one or more populations that can potentially interbreed to produce healthy, fertile offspring."
Species sharing correlated morphological characters will not be considered as a biological species since they cannot produce fertile offsprings.
A natural taxon having pattern/morphological characters shared by all members of the group is considered as taxonomic species.

Topic: Biology, The Living World, NEET
Two organs which are similar in structure and origin but not necessarily in function are called:

Options:
A Homologous
B Analogous
C Apocrine
D None of these

The correct answer is:
Homologous

Solution:
Homologous organs are similar in structure and origin but not necessarily in function. Hence, organs from different species having a similar basic form, microscopic structures, body position and embryonic development are said to be homologous.

Topic: Biology, Evolution, NEET
Two parallel beams of electrons are moving with small velocities v<<c in the same direction, then there will be

Options:
A a net repulsion between them
B a net attraction between them
C the electric repulsion and magnetic attraction nullify each other
D the electric attraction and magnetic repulsion nullify each other

The correct answer is:
a net repulsion between them

Solution:
Due to electrostatic force, there will be repulsion between electron beams. Due to magnetic interaction electron beams will attract each other as they are moving in the same direction.
As the magnitude of electrostatic repulsion is more than magnetic attraction therefore there will be a net repulsion.

Topic: Physics, Magnetic Effects of Current, NEET
Two parallel infinite line charges with linear charge densities +λ C m-1 and -λ C m-1 are placed at a distance of 2R in free space. What is the electric field mid-way between the two-line charges?

Options:
A zero
B 2λπϵ0 R N C-1
C λπϵ0 R N C-1
D λ2πϵ0 R N C-1

The correct answer is:
λπϵ0 R N C-1

Solution:
E1=λ2πε0R
E2=λ2πε0R
Enet=E1+E2
=λπε0R N C-1

Topic: Physics, Electrostatics, NEET, NEET 2019 (All India)
Two parallel long wires carry currents i1 and i2 with i1>i2. When the currents are in the same direction, the magnetic field at a point midway between the wires is 10 Î¼T and when the direction of i2 is reversed, the field becomes 30 Î¼T. The ratio i1i2 is

Options:
A 4
B 3
C 2
D 1

The correct answer is:
2

Solution:
When currents are in the same direction,
B1â is into the plane and B2â is up the plane at point P.
10=B1-B2 ......(1)
When the direction is reversed,
Both B1â and B2â are into the plane.
30=B1-B2 ......(2)
From (1) and (2)
B1=20 T and B2=10 T
â´ B1B2=2010
â´ i1i2=21

Topic: Physics, Magnetic Effects of Current, NEET
Two parallel plate capacitors of capacitance C and 2C are connected in parallel and charged to a potential difference V0. The battery is then disconnected and the region between the plates of the capacitor C is completely filled with a material of dielectric constant 2. The potential difference across the capacitors now becomes

Options:
A V04
B V02
C 3V04
D V0

The correct answer is:
3V04

Solution:
Initial state
As both capacitors have the same capacitance total charge =C V 0 +2C V 0
=3C V 0
It will be divided equaly so q 1 = q 2 = 3C V 0 2 & potential = q C = ( 3C V 0 2 ) 2C = 3 V 0 4

Topic: Physics, Electrostatics, NEET
Two parallel plate of area A and separated by two different dielectric as shown in the figure. The net capacitance is

Options:
A 15ε0A4d
B 10ε0A7d
C 11ε0A5d
D 12ε0A3d

The correct answer is:
15ε0A4d

Solution:
Since, we know,
Capacitance of a parallel plate capacitance,
C=Kε0Ad
Where, symbols have their usual meaning.
In the given figure,
1Cnet=1C1+1C2=1K1ε0Ad/2+1K2ε0Ad/2
=13ε0Ad/2+15ε0Ad/2
=d6ε0A+d10ε0A=d2ε0A13+15
=8d15×2×ε0A=4d15ε0A
Thus, Cnet=15ε0A4d

Topic: Physics, Electrostatics, NEET