datavorous/entrance-exam-dataset huggingface dataset:
A cell of emf $4 \mathrm{~V}$ and internal resistance $0.5 \Omega$ is connected to a $7.5 \Omega$ external resistance. The terminal potential difference of the cell is:

Options:
A $3.75 \mathrm{~V}$
B $4.25 \mathrm{~V}$
C $4 \mathrm{~V}$
D $0.375 \mathrm{~V}$

The correct answer is:
$3.75 \mathrm{~V}$

Solution:
As, $i=\frac{\mathrm{E}}{\mathrm{R}+r}$
Hence, $\quad i=\frac{4}{7.5+0.5}=\frac{4}{8}=\frac{1}{2}$
And, $\mathrm{V}=\mathrm{E}-i r$
(where, the symbols have their usual meanings)
$\begin{aligned}
\mathrm{V} & =4-0.5 \times 0.5 \\
\mathrm{~V} & =3.75 \text { volt }
\end{aligned}$

Topic: Physics, Current Electricity, NEET, NEET 2022 (Phase 2)
A cell organelle containing hydrolytic enzymes is:

Options:
A Lysosome
B Golgi bodies
C Ribosome
D Mesosome

The correct answer is:
Lysosome

Solution:
These are membrane-bound vesicular structures formed by the process of packaging in the Golgi apparatus. The isolated lysosomal vesicles have been found to be very rich in almost all types of hydrolytic enzymes (hydrolases - lipases, proteases, carbohydrases) optimally active at the acidic pH. These enzymes are capable of digesting carbohydrates, proteins, lipids and nucleic acids.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2016 (Phase 2)
A cell organelle containing hydrolytic enzymes is:

Options:
A Lysosome
B Golgi bodies
C Ribosome
D Mesosome

The correct answer is:
Lysosome

Solution:
These are membrane-bound vesicular structures formed by the process of packaging in the Golgi apparatus. The isolated lysosomal vesicles have been found to be very rich in almost all types of hydrolytic enzymes (hydrolases – lipases, proteases, carbohydrases) optimally active at the acidic pH. These enzymes are capable of digesting carbohydrates, proteins, lipids and nucleic acids.

Topic: Biology, Cell - The Unit of Life, NEET
A cell swells up when kept in

Options:
A Hypotonic solution
B Hypertonic solution
C Isotonic solution
D All of the above

The correct answer is:
Hypotonic solution

Solution:
A cell swells up when kept in a hypotonic solution due to the process of endosmosis. Endosmosis is the movement of water into the cell and such a plant cell is referred to as turgid.

Topic: Biology, Transport in Plants, NEET
A cell undergoes mitosis in 30 minutes. Find the number of cells produced after 24 hours if initially 10⁵ cells are present.

Options:
A 224×105
B 248×105
C 248+105
D 48×105-

The correct answer is:
248×105

Solution:
A cell undergoes one round of mitosis in 30 minutes; thus, it undergoes 48 divisions in 24 hours. Hence, the number of cells produced is 248. Each cell undergoes division, so the total number of cells produced is 248 Ã 105.

Topic: Biology, Cell Cycle and Cell Division, NEET
A cell with DPD = 5 is surrounded by cells A, B, C, and D having OP and TP respectively 5 and 4, 7 and 5, 3 and 2, and zero and zero. From which cell it will draw water faster?

Options:
A D
B C
C B
D A

The correct answer is:
D

Solution:
DPD of the given cell = 5
DPD = OP - TP
DPD for A = 5 - 4 = 1
DPD for B = 7 - 5 = 2
DPD for C = 3 - 2 = 1
DPD for D = 0 - 0 = 0
Since cell D has the lowest DPD value, the pressure difference developed will be higher and hence, water will be drawn fastest from cell D.

Topic: Biology, Transport in Plants, NEET
A certain amount of an ideal gas is contained in a closed vessel. The vessel is moving with a constant velocity v. The molecular mass of gas is M. The rise in temperature of the gas when the vessel is suddenly stopped is γ = C P / C v

Options:
A Mv2γ-12Rγ+1
B Mv2γ-12R
C Mv22Rγ+1
D Mv22RÎ³-1

The correct answer is:
Mv2γ-12R

Solution:
If m is the total mass of the gas, then its kinetic energy = 12mv2
When the vessel is suddenly stopped, disordered kinetic energy will increase the temperature of the gas (because the process will be adiabatic), i.e.,
12mv2=μCvΔT
=mMCvΔT
⇒ mMRγ-1ΔT=12mv2 AsCv=Rγ-1
⇒ ΔT=Mv2γ-12R

Topic: Physics, Thermodynamics, NEET
A certain amount of an ideal gas is contained in a closed vessel. The vessel is moving with a constant velocity v. The rise in the temperature of the gas, when the vessel is suddenly stopped, is (M is the molecular mass) Î³ =CPCV

Options:
A Mv2(Î³-1)2R
B Mv2γ+12R
C Mv22Rγ
D Mv22Rγ+1

The correct answer is:
Mv2(γ-1)2R

Solution:
12Mv2=msΔT
Sgm=CmolarM
CV=Rγ-1

Topic: Physics, Thermodynamics, NEET
A certain buffer solution contains equal concentration of X- and HX. The Kb for X- is 10-10. The pH of the buffer is

Options:
A 4
B 7
C 10
D 14

The correct answer is:
4

Solution:
X −  +  H 2 O ⇌  OH −  + HX
Kb=[OH-][HX][X-]
HX ⇌  H +  +  X −
Ka=[H+][X-][HX]
∴Ka×Kb=[H+][OH-]=Kw=10-14
Hence Ka=10-4
Now as [X-]=[HX],pH=pKa=4 .

Topic: Chemistry, Ionic Equilibrium, NEET
A certain current liberated 0.504 g of hydrogen in 2 h. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution? Given Atomic mass of Cu = 63.5

Options:
A 12.9 g
B 15.9 g
C 31.7 g
D 36.9 g

The correct answer is:
15.9 g

Solution:
Weight of CuWeight of H2=Eq.wt.of CuEq.wt.of H2
Weightm of Cu 0.504 = 63.6/2 1
∴ Weight of Cu =15.9 g

Topic: Chemistry, Electrochemistry, NEET
A certain ideal gas undergoes a polytropic process PVn= constant such that the molar specific heat during the process is negative. If the ratio of the specific heats of the gas be Î³ , then the range of values of n will be

Options:
A 0 < n < γ
B 1 < n < Î³
C n = γ
D n > γ

The correct answer is:
1 < n < γ

Solution:
Since PVn= constant and also PV=RT, taking 1 mol of the gas for simplicity,
dU=CvdT
where Cv→ molar specific heat at constant volume.
Now the molar specific heat in a polytropic process PVn= constant is given by
C=Rγ-1-Rn-1=n-γRn-1γ-1 ...(i)
From this equation, we see that C will be negative when n < γ and n > 1, simultaneously, i.e., 1 < n < γ . Since γ for all ideal gases is greater than 1, if n > γ or n < 1, then Cv will be positive.

Topic: Physics, Thermal Properties of Matter, NEET
A certain mass of gas occupies a volume of 300 cc at 27℃ and 620 mm pressure. The volume of this gas at 47℃ and 640 mm pressure will be

Options:
A 400 cc
B 510 cc
C 310 cc
D 350 cc

The correct answer is:
310 cc

Solution:
Given initial volume V1=300 cc; initial temperature T1=27â=300 K, initial pressure p1=620 mm, final temperature T2=47â=320 K and final pressure p2=640 mm. We know from the general gas equation
p1V1T1=p2V2T2
=620Ã300300=640ÃV2320
=310 cc
where, V2 is the final volume of the gas

Topic: Chemistry, States of Matter, NEET
A certain mass of hydrogen is changed to helium by the process of fusion. The mass defect in fusion reaction is $0.02866 \mathrm{u}$. The energy liberated per $u$ is (given $1 \mathrm{u}=931 \mathrm{MeV}$ )

Options:
A $2.67 \mathrm{MeV}$
B $26.7 \mathrm{MeV}$
C $6.675 \mathrm{MeV}$
D $13.35 \mathrm{MeV}$

The correct answer is:
$6.675 \mathrm{MeV}$

Solution:
Here, $\Delta m=0.02866 U$
$$
\begin{aligned}
& \therefore \text { Energy liberated } \\
& =\frac{0.02866 \times 931}{4}=\frac{26.7}{4} \mathrm{MeV}=6.675 \mathrm{MeV}
\end{aligned}
$$

Topic: Physics, Nuclear Physics, NEET, NEET 2013 (All India)
A certain number of spherical drops of a liquid of radius r coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid then:

Options:
A Energy 4VT1r-1R is released
B Energy =3VT1r+1R is released
C Energy =3VT1r-1R is released
D Energy is neither released nor absorbed

The correct answer is:
Energy =3VT1r-1R is released

Solution:
∆U=ST∆A
Ainitial=4πr2n
Afinal=4πR2
∆A=4πr2n-4πR2
43πr3n=43πR3
n=R3r3
∆A=4πR3r3.r2-R2=4πR3r-R3R=4πR3331r-1R
∆A=3V1r-1R
∆U=3VT1r-1R

Topic: Physics, Mechanical Properties of Fluids, NEET, NEET 2014
A certain particle carries 2.5 × 10–16C of static electric charge.What are the number of electrons present in it?

Options:
A 1422
B 1682
C 1560
D 1398

The correct answer is:
1560

Solution:
Correct Option is : (C)
1560

Topic: Chemistry, Structure of Atom, NEET
A certain plant homozygous for yellow seeds and red flowers was crossed with a plant homozygous for green seeds and white flowers. The $F_1$ plants had yellow seeds and pink flowers. The $F_1$ plants were selfed to get $F_2$ progeny. Assuming independent assortment of the two characters, how many phenotypic categories are expected for these characters in the $F_2$ generation?

Options:
A $4$
B $6$
C $9$
D $16$

The correct answer is:
$6$

Solution:
As six different, phenotypic categories are visible, thus option (2) is correct.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2023 (Manipur)
A certain quantity of electricity deposits 0.54 g of silver from AgNO3 solution. What volume of hydrogen will the same quantity of electricity liberate at 27°C and 750 mm Hg pressure?

Options:
A 62.34 ml
B 6.234 ml
C 623.5 ml
D None of these

The correct answer is:
62.34 ml

Solution:
WAgEAg=WH2EH2
or 0.54108=W1121
WH2=0.005 g
P = 750 mm = 750/760 atm
Using ideal gas equation
V=WRTMP=0.005Ã0.082Ã300Ã7602Ã750
= 62.34 ml .

Topic: Chemistry, Electrochemistry, NEET
A certain road accident patient with unknown blood group needs immediate blood transfusion. His one doctor friend at once offers his blood. What was the blood group of the donor?

Options:
A Blood group B
B Blood group $\mathrm{AB}$
C Blood group $\mathrm{O}$
D Blood group $\mathrm{A}$

The correct answer is:
Blood group $\mathrm{O}$

Solution:
Blood group is tested by two types of sera i.e., anti-A (antibody-A) and anti-B (antibody-B). Persons with blood group $\mathrm{O}$ possess both antibodies in their plasma but have no antigens in their RBCs. So, RBCs of blood group ' $\mathrm{O}$ ' do not show clumping in any of the two sera. That's why, persons with blood group ' $\mathrm{O}$ ' are called universal donor and they can donate blood to a person with any type of blood group.

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2012 (Screening)
A certain salt X, gives the following results. (i) Its aqueous solution is alkaline to litmus. (ii) It swells up to a glassy material Y on strong heating. (iii) When conc. H2SO4 is added to a hot solution of X, white crystal of an acid Z separates out.

Options:
A 1. \(\mathrm{XOH} \rightarrow \mathrm{X}^{+}+\mathrm{OH}^{-}\)
2. \(X+\) Heat \(\rightarrow Y\)
3. \(\mathrm{X}+\mathrm{H} 2 \mathrm{SO} 4 \rightarrow \mathrm{Z}+\mathrm{HXSO} 4\)
B 1. \(\mathrm{X}+\mathrm{H} 2 \mathrm{O} \rightarrow \mathrm{XO}^{-}+\mathrm{H}^{+}\)
2. \(X+\) Heat \(\rightarrow X O\)
3. \(\mathrm{X}+\mathrm{H} 2 \mathrm{SO} 4 \rightarrow \mathrm{HXSO} 4+Z\)
C 1. \(\mathrm{XOH} \rightarrow \mathrm{X}^{+}+\mathrm{OH}^{-}\)
2. \(Y+\) Heat \(\rightarrow X\)
3. \(\mathrm{X}+\mathrm{H}_2 \mathrm{SO} 4 \rightarrow \mathrm{HXSO} 4+\mathrm{Z}\)
D 1. \(\mathrm{X}+\mathrm{H} 2 \mathrm{O} \rightarrow \mathrm{XO}^{-}+\mathrm{H}^{+}\)
2. \(Y+\) Heat \(\rightarrow X O\)
3. \(\mathrm{X}+\mathrm{H} 2 \mathrm{SO} 4 \rightarrow \mathrm{Z}+\mathrm{HXSO4}\)

The correct answer is:
1. \(\mathrm{X}+\mathrm{H} 2 \mathrm{O} \rightarrow \mathrm{XO}^{-}+\mathrm{H}^{+}\)
2. \(X+\) Heat \(\rightarrow X O\)
3. \(\mathrm{X}+\mathrm{H} 2 \mathrm{SO} 4 \rightarrow \mathrm{HXSO} 4+Z\)

Solution:
Correct Option is : (B)
1. \(\mathrm{X}+\mathrm{H} 2 \mathrm{O} \rightarrow \mathrm{XO}^{-}+\mathrm{H}^{+}\)
2. \(X+\) Heat \(\rightarrow X O\)
3. \(\mathrm{X}+\mathrm{H} 2 \mathrm{SO} 4 \rightarrow \mathrm{HXSO} 4+Z\)

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
A certain weak acid has a dissociation constant 1.0×10-4 . The equilibrium constant for its reaction with a strong base is

Options:
A 1.0×10-4
B 1.0×10-10
C 1×1010
D 1.0×10-14

The correct answer is:
1×1010

Solution:
HA⇌H++A-
Ka=[H+][A-][HA] …(i)
Also, HA+B++OH-→B++A-+H2O
Keq=[H2O][A-][HA][OH-] …(ii)
By Eqs. (i) and (ii),
KeqKa=1[H+][OH-]=1Kw
So Keq=KaKw=10-410-14=1010 .

Topic: Chemistry, Ionic Equilibrium, NEET
A certain weak acid has a dissociation constant 10 × 10 −4 . The equilibrium constant for its reaction with a strong base is:

Options:
A 1 . 0 × 1 0 - 4
B 1 . 0 × 1 0 - 10
C 1 . 0 × 1 0 - 14
D 1 Ã 1 0 10

The correct answer is:
1 × 1 0 10

Solution:
HA + aq â H aq + + A - aq
K a = H + A - HA
Also HA + B + + OH - â B + + H 2 O + A -
HA + OH - â H 2 O + A -
K eq = A - HA OH - = A - H + HA OH - H + = K a K wâ¡
â´ K eq = K a K w = 10 −4 10 −14 = 10 10

Topic: Chemistry, Ionic Equilibrium, NEET
A certain wire $A$ has resistance $81 \Omega$. The resistance of another wire $B$ of same material and equal length but of diameter thrice the diameter of $A$ will be

Options:
A $729 \Omega$
B $243 \Omega$
C $81 \Omega$
D $9 \Omega$

The correct answer is:
$9 \Omega$

Solution:
We know that, $R=\frac{\rho l}{A}$, where $A=\frac{\pi}{4} d^2$
$\begin{aligned} & \Rightarrow 81=\frac{\rho \times I}{\frac{\pi}{4} d^2}, \text { also, } R^{\prime}=\frac{\rho \times I}{\frac{\pi}{4} \times(3 d)^2}=\frac{\rho l}{\frac{\pi}{4} \times 9 d^2} \\ & \Rightarrow R^{\prime}=\frac{R}{9}=\frac{81}{9}=9 \Omega\end{aligned}$

Topic: Physics, Current Electricity, NEET, NEET 2023 (Manipur)
A chain of 125 links is 1.25 m long and has mass of 2 kg with the ends fastened together. It is set for rotating at 50 revolutions per second. The centripetal force on each link is

Options:
A 3.14 N
B 0.314 N
C 314 N
D None of these

The correct answer is:
314 N

Solution:
circumference =2πr=1.25 m
so radius r=1.25m2π=0.199 m
Mass of each link is m =2 kg 125 =0.016  kg
angular velocity is w=50 revolutions per second =50×2π rad s-1=314 rad s-1
The centripetal force on each link is f=m×r×w2=0.016×0.199×3142 N
f=314 N

Topic: Physics, Laws of Motion, NEET
A chain of length L and of mass m is placed upon a smooth surface. The length of BA is L-b . Now, the chain is released and it slides down. Calculate the speed of the chain when its end reaches B.

Options:
A gsin⁡θL2-b2L
B 2gsin⁡θL-bL
C 2gsin⁡θL-b
D gsin⁡θL-b

The correct answer is:
2gsin⁡θL-b

Solution:
Given, u=0
Applying, v2-u2=2as
a=gcos⁡90-θ
=gsin⁡θ
v2=2gsin⁡θL-b
∴v=2gsin⁡θL-b

Topic: Physics, Motion In Two Dimensions, NEET
A chain of length â < Ï R 2 is placed on a smooth surface whose some part is horizontal and some part is on quarter circular of radius R in the vertical plane as shown. Initially the whole part of chain lies in the circular part with one end at topmost point of circular surface. If the mass of chain is m, then work required to pull very slowly the whole chain on horizontal part is â

Options:
A m â gR 2 sin â R
B m â gR 2 cos â R
C m â gR 2 â R - sin â R
D m â gR 2 â R - cos â R

The correct answer is:
m ℓ gR 2 ℓ R - sin ℓ R

Solution:
Taking flat surface as reference,
dU i = - m â Rd Î¸ Ã g Ã R 1 - cos Î¸
dU i = - mgR 2 â 1 - cos Î¸ d Î¸
â´ U i = - mgR 2 â â R - sin â R
and U f = 0
â´ W ext = - Î U

Topic: Physics, Work Power Energy, NEET
A change of 8.0 mA in the emitter current brings a change of 7.9 mA in the collector current. The values of Î± and Î² are

Options:
A 0.99, 90

B 0.96, 79

C 0.97, 99

D 0.99, 79

The correct answer is:
0.99, 79

Solution:
Given that, change in emitter current,
∆IE=8 mA
And the change in collector current,
∆IC=7.9 mA
We know that,
α=∆IC∆IE⇒α=7.98⇒α≈0.99
Also, we know that
β=α1-α⇒β=7.981-7.98
⇒ β=7.98-7.9⇒β=7.90.1=79
Hence, α=0.99 and β=79

Topic: Physics, Semiconductors, NEET
A change of 8.0 mA in the emitter current brings a change of 7.9 mA in the collector current. The values of α and β are

Options:
A 0.99, 90
B 0.96, 79
C 0.97, 99
D 0.99, 79

The correct answer is:
0.99, 79

Solution:
Given that, change in emitter current, ∆IE=8 mA and change in collector current, ∆IC=7.9 mA.
We know that,
α=∆IC∆IE
⇒α=7.98
⇒α≈0.99
Also, we know that
β=α1-α
⇒β=7.981-7.98=7.98-7.9
⇒β=7.90.1=79
Hence, the required answer is α=0.99 and β=79.

Topic: Physics, Semiconductors, NEET
A change of 8.0 mA in the emitter current brings a change of 7.9 mA in the collector current. The values of Î± and Î² are

Options:
A 0.99, 90

B 0.96, 79

C 0.97, 99

D 0.99, 79

The correct answer is:
0.99, 79

Solution:
Given that, change in emitter current, ∆IE= 8 mA
and change in collector current, ∆IC=7.9 mA
We know that,
α=∆IC∆IE
⇒=7.98⇒ α≈0.99
Also, we know that
β=α1-α
⇒=7.981-7.98=7.98-7.9
Or β=7.90.1 = 79
Hence, the required answer is α=0.99 and β=79.

Topic: Physics, Semiconductors, NEET
A change of 8.0 mA in the emitter current brings a change of 7.9 mA in the collector current. The values of Î± and Î² are

Options:
A 0.99, 90
B 0.96, 79
C 0.97, 99
D 0.99, 79

The correct answer is:
0.99, 79

Solution:
Given that, change in emitter current, ∆IE=8 mA and change in collector current, ∆IC=7.9 mA.
We know that,
α=∆IC∆IE
⇒α=7.98
⇒α≈0.99
Also, we know that
β=α1-α
⇒β=7.981-7.98=7.98-7.9
⇒β=7.90.1=79
Hence, the required answer is α=0.99 and β=79.

Topic: Physics, Semiconductors, NEET
A character in human beings is being controlled by 5 polygenes. The possible phenotypes are

Options:
A 11
B 10
C 9
D 25

The correct answer is:
11

Solution:
When traits are controlled by three or more genes, they are called polygenic traits, and their inheritance is called polygenic inheritance. Human skin color is a classic example following polygenic inheritance as the skin colour is regulated by three pairs of genes. Polygenic inheritance is influenced by the environment as well and not just the polygenes.
If 5 polygenes are there to control a character the number of phenotypes will be 2×5+1=11 .

Topic: Biology, Principles of Inheritance and Variation, NEET
A character is controlled by 5 alleles. How many genotypes are possible for this character?

Options:
A 15
B 6
C 12
D 25

The correct answer is:
15

Solution:
Irrespective of the phenotypes, genotypes are fixed for a given number of alleles in the population. The number of possible genotypes, in this case, would be 15.n2 X (n+1) = number of genotypes in a population. Where n is the number of alleles of that gene.

Topic: Biology, Principles of Inheritance and Variation, NEET
A character is controlled by 5 alleles of the same gene. It will be considered as the case of

Options:
A Multiple allelism - Deviation from the law of unit factors
B Multiple allelism - Deviation from the law of segregation
C Multiple allelism - Deviation from the law of independent assortment
D Polygenic inheritance - Deviation from the law of dominance

The correct answer is:
Multiple allelism - Deviation from the law of unit factors

Solution:
Three or more alternative forms of a gene (alleles) can occupy the same locus. However, only two of the alleles can be present in a single organism. For example, the ABO system of blood groups is controlled by three alleles, only two of which are present in an individual.

Topic: Biology, Principles of Inheritance and Variation, NEET
A characteristic of drupe is

Options:
A stony mesocarp.
B stony endocarp.
C fleshy seed coat.
D stony pericarp.

The correct answer is:
stony endocarp.

Solution:
A drupe is a simple indehiscent fruit derived from a single ovary. A drupe fruit mostly contains only one seed which is enclosed by stony endocarp. Hardened endocarp of drupe adheres to the seed coat. A drupe has thin exocarp and fleshly and edible mesocarp. Apricots, peaches, plums, mangoes, olives, cherries and Indian gooseberries are some example of typical drupe

Topic: Biology, Morphology of Flowering Plants, NEET
A charge flows through a 200 Î© galvanometer connected to a 400 Î© circular coil of 1000 turns wound on a wooden stick 20 mm in diameter. If a magnetic field B=0.012âT parallel to the axis of the stick is decreased suddenly to zero, the charge is

Options:
A 6.3 Î¼C
B 63 Î¼C
C 0.63 Î¼C
D 630 Î¼C

The correct answer is:
6.3 Î¼C

Solution:
q= ÎÏ R
q=NB2-B1AR=NÏr2B2-B1R
= 1000ÃÏÃ10-4Ã0.012-0200+400
=6.3 Ã10-6 C=6.3 Î¼C

Topic: Physics, Electromagnetic Induction, NEET
A charge of 1 Î¼C is divided into two parts such that their charges are in the ratio of 2:3. These two charges are kept at a distance 1 m apart in vacuum. Then, the electric force between them (in N) is -

Options:
A 0.216
B 0.00216
C 0.0216
D 2.16

The correct answer is:
0.00216

Solution:
q1q2=23
q1 + q2 = 1 μC
⇒ q1 = 25μC , q2= 35 μC
F = Kq1q2r2
F = 9 × 1 0 9 × 2 5 × 3 5 × 1 0 - 1 2 1 2
=0.00216 N

Topic: Physics, Electrostatics, NEET
A charge of magnitude 1.6Ã10-19 C is moving in a circle of radius 5 cm in a uniform magnetic field 0.5 T. After applying E=0.15 V m-1 charge starts moving in a straight line mass of charge will be

Options:
A 23×10-20 kg
B 43Ã10-20 kg
C 63×10-20 kg
D 13×10-20 kg

The correct answer is:
43Ã10-20 kg

Solution:
For a particle to move in a circle under magnetic field
qVB=mv2R ...(i)
For a particle to move in a straight line
In both magnetic field and electric field
qVB=2E
⇒V=EB ...(ii)
From (i) to (ii)
qB=mvR
=m.EBR
⇒m=qE2RE=1.6×10-13×0.52×10-20.15 kg=43×10−20 kg

Topic: Physics, Magnetic Effects of Current, NEET
A charge $Q \mu C$ is placed at the centre of a cube. The flux coming out from any one of its faces will be (in SI unit)

Options:
A $\frac{Q}{6 \varepsilon_0} \times 10^{-3}$
B $\frac{Q}{6 \varepsilon_0} \times 10^{-6}$
C $\frac{Q}{\varepsilon_0} \times 10^{-6}$
D $\frac{2 Q}{3 \varepsilon_0} \times 10^{-3}$

The correct answer is:
$\frac{Q}{6 \varepsilon_0} \times 10^{-6}$

Solution:
According to Gauss law, total electric flux through a closed surface is
$\phi_{\text {net }}=\frac{q}{\varepsilon_0}$
$\therefore$ Total electric flux through the cube is
$\phi_T=\frac{Q \times 10^{-6}}{\varepsilon_0}$
Since charge is placed at the centre of the cube, therefore flux through each face is same.
Hence, $\phi=\frac{1}{6} \phi_T=\frac{1}{6} \times \frac{Q \times 10^{-6}}{\varepsilon_0}$

Topic: Physics, Electrostatics, NEET, NEET 2023 (Manipur)
A charge q is accelerated through a potential difference V. It is then passed normally through a uniform magnetic field, where it moves in a circle of radius r. The potential difference required to move it in a circle of radius 2 r is

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

The correct answer is:
4 V

Solution:
r=mvqB=2m.KEqB
=2mqVâqB
2r=2mq.Vâ'qB
âVâ'=4 V

Topic: Physics, Magnetic Effects of Current, NEET
A charge Q is distributed over two concentric hollow spheres of radii r and RR>r such that the surface densities are equal. Find the potential at the common center.

Options:
A 14πε0 QR+rR2+r2
B 14πε0QR-rR2+r2
C 14πε0QR+r2R2+r2
D 14πε0QR+r2R2+2r2

The correct answer is:
14πε0 QR+rR2+r2

Solution:
Since surface charge densities are equal, Q is divided in the ratio of surface areas (i.e., r2. R2 )
Qr= r2Qr2+R2 QR=R2Qr2+R2
∴ V=14πε0 Q2r+QRR=14πε0 . Qr2+R2 r+R

Topic: Physics, Electrostatics, NEET
A charge q is distributed uniformly on a ring of radius r . A sphere of equal radius r is constructed with its centre at the periphery of the ring. The electric flux through the surface of the sphere is

Options:
A qε0
B 2qε0
C q2ε0
D q3ε0

The correct answer is:
q3ε0

Solution:
Q =qeqÎµ0=q3.Îµ0
If angle at centre is 360Â° then charge is q
If angle at centre is 120Â° then charge is q360Ã120

Topic: Physics, Electrostatics, NEET
A charge Q is divided into two charges q and Q-q. The value of q such that the force between them is maximum is

Options:
A Q
B 3Q4
C Q2
D Q3

The correct answer is:
Q2

Solution:
By Coulomb's law,
When charge Q is divided into two charges q and Q - q
F=14πε0 .q.Q-qr2
The value of q
Fmax=14πε0 .Q2Q-Q2r2
(putting q=Q2 for the maximum force)
=14πε0 .Q2Q2r2
=14πε0.Q24r2
=14πε0 .Q24r2
=14πε0 .Q22r2
So, option Q2 is correct.

Topic: Physics, Electrostatics, NEET
A charge $q$ is located at the centre of a cube. The electric flux through any face is:

Options:
A \(\frac{2 \pi q}{6\left(4 \pi \varepsilon_0\right)}\)
B \(\frac{4 \pi q}{6\left(4 \pi \varepsilon_0\right)}\)
C \(\frac{\pi q}{6\left(4 \pi \varepsilon_0\right)}\)
D \(\frac{\mathrm{q}}{6\left(4 \pi \varepsilon_0\right)}\)

The correct answer is:
\(\frac{4 \pi q}{6\left(4 \pi \varepsilon_0\right)}\)

Solution:
Electric flux through any face
$\begin{aligned}
& =\frac{\text { Total flux }}{\text { number of faces }}=\frac{\left(q / E_0\right)}{6} \\
& =\frac{4 \pi q}{6\left(4 \pi \varepsilon_0\right)}
\end{aligned}$

Topic: Physics, Electrostatics, NEET, NEET 2003