datavorous/entrance-exam-dataset huggingface dataset:
In a chemical reaction, a catalyst used for

Options:
A Decreases the energy of activation
B Increases the energy of activation
C Does not change energy of activation
D Decreases or increase the energy of activation

The correct answer is:
Decreases or increase the energy of activation

Solution:
In a chemical reaction, a catalyst can increase or decreases the activation energy of reaction and hence, increases or decrease the rate of reaction.

Topic: Chemistry, Surface Chemistry, NEET
In a chloroplast the highest number of protons are found in:

Options:
A Stroma
B Lumen of thylakoids
C Inter membrane space
D Antennae complex

The correct answer is:
Lumen of thylakoids

Solution:
During photophosphorylation process, PS-II helps in the photolysis of water molecules, and the PS-II are located towards the lumen of the thylakoid. The photolysis yields number of protons. For the photolysis of the water several cations and anions are needed, like Magnesium, Manganese and Chloride. Thus, the lumen of thylakoid has a maximum number of protons.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2016 (Phase 1)
In a chromosome, there is a specific DNA sequence, responsible for initiating replication. It is called as:

Options:
A Recognition sequence
B Cloning site
C Restriction site
D $ori$ site

The correct answer is:
$ori$ site

Solution:
Option (4) is the correct answer because origin of replication (ori) is a sequence from where replication starts and any piece of DNA when linked to this sequence can be made to replicate, within the host cells.
Option (1) is incorrect because recognition sequence is a specific palindromic nucleotide sequence in DNA recognised by restriction endonuclease.
Options (2) and (3) are not the answers because cloning site/restriction site is the site on the vector for the use of restriction enzyme and it is similar to recognition site.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2024 (Re-NEET)
In a circuit $\mathrm{L}, \mathrm{C}$ and $\mathrm{R}$ are connected in series with an alternating voltage source of frequency $f$. The current leads the voltage by $45^{\circ}$. The value of $\mathrm{C}$ is:

Options:
A $\frac{1}{\pi f(2 \pi f(L-R)}$
B $\frac{1}{2 \pi f(2 \pi f L-R)}$
C $\frac{1}{\pi f(2 \pi f L+R)}$
D $\frac{1}{2 \pi f(2 \pi f L+R)}$

The correct answer is:
$\frac{1}{2 \pi f(2 \pi f L+R)}$

Solution:
$$
\begin{aligned}
\tan \phi & =\frac{X_C-X_L}{R} \\
\tan \left(\frac{\pi}{4}\right) & =\frac{\frac{1}{\omega C}-\omega L}{R} \\
\Rightarrow \quad R & =\frac{1}{\omega C}-\omega L \\
(R+2 \pi f L) & =\frac{2}{2 \pi f C} \\
C & =\frac{1}{2 \pi f(R+2 \pi f L)}
\end{aligned}
$$

Topic: Physics, Alternating Current, NEET, NEET 2005
In a circuit (refer to Fig 3.7) with two cells of the same electromotive force (EMF) E, but different internal resistances r1 and r2, connected in series to an external resistor R, what value should R hold to ensure that the potential difference across the terminals of the first cell becomes zero?

Options:
A R = r1 + r2
B R = r1 - r2
C R = r1 * r2
D R = r1 / r2

The correct answer is:
R = r1 - r2

Solution:
Correct Option is : (B)
R = r1 - r2

Topic: Chemistry, Current Electricity, NEET
In a circuit represented in Fig. if each diode has a forward bias resistance of 25Ω and infinite resistance in reverse bias, what will be the value of the current I1?

Options:
A 0A
B 25A
C 0.025A
D 1A

The correct answer is:
0A

Solution:
Correct Option is : (A)
0A

Topic: Chemistry, Semiconductors, NEET
In a circuit represented in Fig. if each diode has a forward bias resistance of 25Ω and infinite resistance in reverse bias, what will be the value of the current I2?

Options:
A 0A
B 25A
C 0.025A
D 1A

The correct answer is:
0.025A

Solution:
Correct Option is : (C)
0.025A

Topic: Chemistry, Semiconductors, NEET
In a circular motion of a particle, the tangential acceleration of the particle is given by at=9 m s-2. The radius of the circle is 4 m. The particle was initially at rest. Time after which total acceleration of the particle makes an angle of 45Â° with the radial acceleration is

Options:
A 13 s
B 23 s
C 53 s
D 43 s

The correct answer is:
23 s

Solution:
At 45°, at=ar=9
v 2 r = v 2 4 = 9
v=6 m s-1
6=0+9t
t=23 s.

Topic: Physics, Laws of Motion, NEET
In a cis isomer of a square planar complex, how are the ligands positioned?

Options:
A Adjacent to each other
B Opposite to each other
C In a square-shaped arrangement
D In a triangular arrangement

The correct answer is:
Adjacent to each other

Solution:
In a cis isomer of a square planar complex, the ligands are positioned adjacent to each other.

Topic: Chemistry, Coordination Compounds, NEET
In a closed insulated container a liquid is stirred with a paddle to increase the temperature, which of the following is true?

Options:
A $\Delta \mathrm{U}=w \neq 0, q=0$
B $\Delta \mathrm{U}=w=q \neq 0$
C $\Delta \mathrm{U}=0, w=q \neq 0$
D $w=0, \Delta \mathrm{U}=q \neq 0$

The correct answer is:
$\Delta \mathrm{U}=w \neq 0, q=0$

Solution:
System is in a closed container. So there will be no heat transfer. Therefore the heat is $q=0$. Now liquid is stirred with a paddle to increase the temperature which increases the internal energy $\Delta E$ of the system.
$$
\begin{aligned}
& \Delta \mathrm{E}=n \mathrm{C} \Delta \mathrm{T} \\
& \Delta \mathrm{E} \neq 0
\end{aligned}
$$
Work is also done on the system when started with a paddle.
$$
\begin{gathered}
\mathrm{W} \neq 0 \\
\Delta \mathrm{E}=q+\mathrm{W}=0+\mathrm{W}, \Delta \mathrm{E}=\mathrm{W} \\
\Delta \mathrm{E}=\Delta \mathrm{W} \neq 0, q=0
\end{gathered}
$$

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2002
In a closed system : A (s)âââ2B (g) + 3C(g) , if the partial pressure of C is doubled, then partial pressure of B will be

Options:
A two times the original pressure
B one half of its original value
C 122 times to the original value
D 22 times to the original value

The correct answer is:
122 times to the original value

Solution:
K P = P A 2 P B 3
Kp=2p2 Ã 3p3=27 Ã 4 p5 where p is the original press.
A s â â â2B g + 3C
pâ²âââ6p
â´pâ²2 Ã 6p3=27 Ã 4p5
or pâ²2 =27Ã4 p533Ã23p3= p22
â´pâ²=p2= final partial press.of Bâ´pâ²2p =final partial press of Binitial partial press of B= p22p
=122

Topic: Chemistry, Chemical Equilibrium, NEET
In a closed vessel a mixture of two moles of carbon monoxide and one mole of oxygen is ignited to convert the carbon monoxide to carbon dioxide. If ΔH is the enthalpy change and ΔE is the change in internal energy, then

Options:
A ΔH=ΔE
B ΔH>ΔE
C ΔH<ΔE
D The relationship depends on the capacity of the vessel

The correct answer is:
ΔH<ΔE

Solution:
Formation of CO2 from CO is an exothermic reaction; heat is evolved from the system, i.e., energy is lowered. Thus, exothermic reactions occur spontaneously on account of decrease in enthapy of system. Thus, ΔE>ΔH .

Topic: Chemistry, Thermodynamics (C), NEET
In a closed vessel containing a binary solution of two volatile liquids, how is the equilibrium between the vapor phase and the liquid phase established?

Options:
A Through condensation
B Through dissolution
C Through evaporation
D Through sublimation

The correct answer is:
Through evaporation

Solution:
The paragraph mentions that in a closed vessel, an equilibrium is established between the vapor phase and the liquid phase through evaporation.

Topic: Chemistry, Solutions, NEET
In a coal-fired power plant, electrostatic precipitators are installed to control emission of

Options:
A SO2
B NOx
C SPM
D CO

The correct answer is:
SPM

Solution:
The electrostatic precipitation controls SPM (suspended particulate matter) in cocal fire power plant.

Topic: Biology, Environmental Issues, NEET
In a coil of area 10 cm2 and 10 turns, magnetic field is directed perpendicular to the plane and is changing at a rate of 104 T s-1. The resistance of the coil is 20 Ω. The current in the coil will be

Options:
A 0.5Â A
B 5 A
C 50Â A
D 5Ã108Â A

The correct answer is:
5 A

Solution:
Induced current is,
i=1RNdϕdt=1RNAdBdt
i=12010 10×10-4104=5 A
i=5 A

Topic: Physics, Electromagnetic Induction, NEET
In a common base transistor circuit, IC is the Output current and IE is the input current. The current gain α is

Options:
A Greater than one
B Less than one
C Equal to one
D None of these

The correct answer is:
Less than one

Solution:
The ratio of collector current to emitter current is called as current gain alpha.

This ratio is less than one.

Topic: Physics, Semiconductors, NEET
In a common base transistor circuit, the current gain is 0.98. On changing emitter current by 5.00 mA, the change in collector current is

Options:
A 0.196 mA
B 2.45 mA
C 4.9 mA
D 5.1 mA

The correct answer is:
4.9 mA

Solution:
Δic=αΔie
⇒Δic=0.98×5 mA
⇒Δic=4.9 mA

Topic: Physics, Semiconductors, NEET
In a common emitter amplifier, using output resistance of 5000 Ω and input resistance of 2000 Ω, if the peak value of input signal voltage is 10 mV and β=50 is given. Find peak value of output voltage -

Options:
A 5×10-6 V
B 2.5×10-4 V
C 1.25 V
D 125 V

The correct answer is:
1.25 V

Solution:
for CE amplifier
Voltage gain A=βRoRi
A=50×50002000 ;
A=125=voutputVinput ;
125=V010mV ;
V0=1.25V

Topic: Physics, Semiconductors, NEET
In a common emitter (CE) amplifier having a voltage gain $G$, the transistor used has transconductance 0.03 mho and current gain 25. If the above transistor is replaced with another one with transconductance $0.02 \mathrm{mho}$ and current gain 20 , the voltage gain will

Options:
A $\frac{2}{3} G$
B $1.5 \mathrm{G}$
C $\frac{1}{3} G$
D $\frac{5}{4} G$

The correct answer is:
$\frac{2}{3} G$

Solution:
$\mathrm{As} A_v=\beta \frac{R_L}{R_j} \quad\left[\because g_m=\frac{\Delta_C}{\Delta V_B}=\frac{\Delta_C}{\Delta /_B R_j}\right]$
or
$$
G=\left(\frac{\beta}{R_i}\right) R_L \quad\left[\because g_m=\frac{\beta}{R_i}\right]
$$
$$
\begin{array}{ll}
\Rightarrow & G=g_m R_i \\
\Rightarrow & G \propto g_m
\end{array}
$$
$\begin{aligned} & \therefore \quad \frac{G_2}{G_1}=\frac{g_{m_1}}{g_{m_2}} \\ & \Rightarrow \quad G_2=\frac{0.02}{0.03} \times G \\ & \therefore \text { Voltage gain } G_2=\frac{2}{3} G\end{aligned}$

Topic: Physics, Semiconductors, NEET, NEET 2013 (All India)
In a common emitter transistor amplifier, Î²=60, R0=5000 Î© and internal resistance of a transistor is 500 Î©. The voltage amplification of the amplifier will be

Options:
A 500
B 460
C 600
D 560

The correct answer is:
600

Solution:
Voltage amplification
Av=βRoRi=60×5000500=600

Topic: Physics, Semiconductors, NEET
In a common-emitter transistor amplifier, the audio signal voltage across the collector is 3 V. The resistance of collector is 3 kÎ©. If current gain is 100 and the base resistance is 2 kÎ©, the voltage and power gain of the amplifier is

Options:
A 200 and 1000
B 15 and 200
C 150 and 15000
D 20 and 2000

The correct answer is:
150 and 15000

Solution:
Current gain Î²=100
Voltage gain AV=Î²RcRb =100 32 =150
Power gain =AVÎ² =150 100 =15000

Topic: Physics, Semiconductors, NEET, NEET 2017
In a common-emitter transistor amplifier, the audio signal voltage across the collector is 3 V. The resistance of collector is 3 kÎ©. If current gain is 100 and the base resistance is 2 kÎ©, the voltage and power gain of the amplifier is

Options:
A 200 and 1000
B 15 and 200
C 150 and 15000
D 20 and 2000

The correct answer is:
150 and 15000

Solution:
Current gain Î²=100
Voltage gain AV=Î²RcRb =100 32 =150
Power gain =AVÎ² =150 100 =15000

Topic: Physics, Semiconductors, NEET
In a common-emitter transistor amplifier, the output resistance is 500 kÎ© and the current gain Î²=49. If the power gain of the amplifier is 5Ã106, the input resistance is

Options:
A 165 Ω
B 198 Ω
C 225 Ω
D 240 Ω

The correct answer is:
240 Ω

Solution:
Given, Ro= 500 kΩ, β =49 and P = 5×106
We have P=β 2 RoRi
5×106=492×500×103Ri
Ri=49×49×500×1035×106=240 Ω

Topic: Physics, Semiconductors, NEET
In a comparative analogy between the aeroplane and ecosystem (the rivet popper hypothesis) used by Paul Ehrlich, find out the correct matching.

Column I	Column II

a. Rivets	1. Ecosystem

b. Aeroplane	2. Species

c. Popping a rivet	3. Key species

d. Rivets on the wings	4. Proper functioning of ecosystem

e. Flight safety	5. A species to become extinct

Options:
A aâ1,Â bâ2,Â câ3,Â dâ4,Â eâ5
B aâ2,Â bâ1,Â câ4,Â dâ5,Â eâ3
C aâ2,Â bâ1,Â câ5,Â dâ4,Â eâ3
D aâ2, bâ1, câ5, dâ3, eâ4

The correct answer is:
aâ2, bâ1, câ5, dâ3, eâ4

Solution:
Paul Enrlich gave the Rivet Popper hypothesis to explain the significance of the key species in functioning of an ecosystem. The hypothesis uses an aeroplane as an analogue to the ecosystem and the rivets of the airplane as species that join parts aeroplane together. Passengers are the human counter-part in anthropogenic activities. As per the hypothesis, if the passengers start removing the rivets, initially it will not have any impact of the flight capability of the plane but later, the plane will lose it flying capability. If the rivets of the wings are removed it directly affect the functioning of the plane. In the same analogy, if the key species are exploited and they get extinct, it will not harm the ecosystem initially but, eventually, it will lead to collapse of the ecosystem. The Rivet Popper hypothesis compares the rivets of the wings to the key species that helps in proper functioning and stabilising of the ecosystem.

Topic: Biology, Biodiversity and its Conservation, NEET
In a compound AB, electro negativity difference between A and B is 1.9. Atomic radius A and B are 4A° and 2A° . The distance between A and B atoms means dA−B is

Options:
A

6.72A°

B 5.82A°
C 6.9A°
D 7.5A°

The correct answer is:
5.82A°

Solution:
dA−B=rA+rB−0.09(xA−xB)=6−0.09(1.9)≈5.829

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
In a compound C , H and N atoms are present in 9 :1 :3.5 by weight. Molecular weight of compound is 108. Molecular formula of compound is

Options:
A C2H6N2
B C3H4N
C C6H8N2
D C9H12N3

The correct answer is:
C6H8N2

Solution:
Molecular weight of compound=108
C12→12×6=72
H1→1×8=8
N14→14×2=28
Total molecular weight =108
∴ Molecular formula =C6H8N2
atom ratio 72 8 : 8 8 : 28 8
=9:1:3.5

Topic: Chemistry, General Organic Chemistry, NEET
In a compound microscope, the focal lengths of two lenses are 1.5 cm and 6.25 cm and an object is placed at 2 cm from objective and the final image is formed at 25 cm from eye lens. The distance between the two lenses is

Options:
A 6.00Â cm
B 7.75Â cm
C 9.25Â cm
D 11.00 cm

The correct answer is:
11.00 cm

Solution:
f0=1.5 cm, fe=6.25 cm, u0=-2 cm, ve=-D=-25 cm
By objective lens 1f0=1v0-1u0
11.5=1v0-1-2
⇒ 1v0=11.5-12 or v0=6 cm
By eyepiece 1fe=1ve-1ue
16.25=1-25-1-ue
⇒ 1ue=16.25+125=425+125=15
ue=5 cm

Length of tube =L=v0+ue=6.0cm⁡+5.0cm

L=11 cm

Topic: Physics, Ray Optics, NEET
In a compound microscope, the focal lengths of two lenses are 1.5 cm and 6.25 cm. An object is placed at 2 cm from objective and the final image is formed at 25 cm from eye lens. The distance between the two lenses is

Options:
A 6.00 cm
B 7.75 cm
C 9.25 cm
D 11.00 cm

The correct answer is:
11.00 cm

Solution:
f0=1.5cm,⁡fe=6.25cm,⁡u0=-D=-25 cm
By objective lens 1f0=1v0-1u0
11.5=1v0-1-2
⇒ 1v0=11.5-12
or v0=6 cm
By eyepiece
1fe=1ve-1ue
16.25=1-25-1-ue
⇒1ue=16.25+125
⇒1ue=425+125=15
⇒1ue=15
ue=5 cm , Length of tube =L=vo+ue=6.0+5.0=11 cm

Topic: Physics, Ray Optics, NEET
In a compound, nitrogen atoms (N) make cubic close packed lattice and metal atoms (M) occupy one-third of the tetrahedral voids present. Determine the formula of the compound formed by M and N?

Options:
A M2N5
B M2N
C M2N3
D MN3

The correct answer is:
M2N3

Solution:
Correct Option is : (C)
M2N3

Topic: Chemistry, Solid State, NEET
In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298 K to 298 K due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K-1 , the value for the enthalpy of combustion of the gas is

Options:
A 90 kJ mol-1
B 9 kJ mol-1
C 18 kJ mol-1
D 6 kJ mol-1

The correct answer is:
9 kJ mol-1

Solution:
Energy released by combustion of 3.5 g gas
=2.5Ã(298.45-298) kJ
Energy released by 1 mole of gas =2.5Ã0.453.5/28=9 kJmol-1

Topic: Chemistry, Thermodynamics (C), NEET
In a cool fired power plant electrostatic precipitators are installed to control emission of:

Options:
A $\mathrm{NO}_x$
B $\mathrm{SPM}$
C $\mathrm{CO}$
D $\mathrm{SO}_2$.

The correct answer is:
$\mathrm{SPM}$

Solution:
Electrostatic precipitators are widely used in power plants to remove particulate matter. ESPs work by inducing an electrostatic charge on the particulate matter in the flowing gas and using the forces that arise due to these electrostatic charges to filter out the particulate matter from the gas. The performance of an electrostatic precipitator is dependent on the electrical resistivity and the particle-size distribution.
Related Theory
Recent years, the increases in toxicsubstances of $\mathrm{NO}_x$ and/or SPM (SuspendedParticulate Matter) from vehicles come tobe serious problem. About the correlation between the traffic volume in our model andthe real volume, we obtained the correlation coefficient of 0.74. Simultaneously, we gotthe result that theconcentration of $\mathrm{NO}_x$ was approximately 70 to $230 \mathrm{ppb}$. That of SPM was approximately 40 to $100 \mu \mathrm{g} / \mathrm{m}^3$.
Caution
Since the energy is only applied to the particulate matter in the gas, electrostatic precipitators are believed to be more efficient than wet scrubbers (in terms of energy consumption).

Topic: Biology, Environmental Issues, NEET, NEET 2007
In a coordination compound, what is collectively termed the coordination sphere?

Options:
A The central atom/ion and the ligands attached to it
B The ionisable groups
C The counter ions
D The central atom/ion alone

The correct answer is:
The central atom/ion and the ligands attached to it

Solution:
The coordination sphere in a coordination compound includes the central atom/ion and the ligands attached to it.

Topic: Chemistry, Coordination Compounds, NEET
In a couple, both the man and the wife have normal vision, although their fathers were colourblind and mothers do not have any allele for colourblindness. In their progeny, the probability of colourblind daughters is

Options:
A 0%
B 15%
C 25%
D 50%

The correct answer is:
0%

Solution:
- Daughters are completely normal.

Topic: Biology, Principles of Inheritance and Variation, NEET
In a cross between a male and female, both heterozygous for sickle cell anaemia gene, what percentage of the progeny will be diseased?

Options:
A 50 %
B 75 %
C 25 %
D 100 %

The correct answer is:
25 %

Solution:
Since both the parents are heterozygous for sickle-cell anaemia, i.e., shows sickle-cell trait, they are represented by the symbol HbAHbS. Each parent thus will produce two types of gametes; one having a composition HbA and the other having a composition for sickle cell anaemia as HbS. The progeny cross will be as follows:

Progeny:

HbA	HbS

HbA	HbAHbA
(Normal)	HbAHbS
(Sickle-cell trait)

HbS	HbAHbS
(Sickle-cell trait)	HbSHbS
(Diseased)

Phenotypic ratio: Normal: Sickle-cell trait: Diseased: 1: 2: 1
Thus, 25% of the progeny will be affected.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2021
In a cross between AABB × aabb, the ratio of F2 genotypes between AABB, AaBB, Aabb and aabb would be

Options:
A 9 : 3 : 3 : 1
B 2 : 1 : 1 : 2
C 1 : 2 : 2 : 1
D 7 : 5 : 3 : 1

The correct answer is:
1 : 2 : 2 : 1

Solution:
Genotype is the genetic make up of an individual irrespective of the mendelian characters of genes impressing.

Topic: Biology, Principles of Inheritance and Variation, NEET
In a cross between true:breeding red:flowered plants and true:breeding white:flowered plants in the dog flower, the F₁ generation produced pink flowers due to incomplete dominance. If two F₁ plants are crossed, what would be the expected phenotype ratio of the F₂ generation?

Options:
A 1 red: 2 pink: 1 white
B 1 red: 1 pink: 1 white
C 3 red: 1 white
D 1 red: 1 white

The correct answer is:
1 red: 2 pink: 1 white

Solution:
When two F₁ plants are crossed, the genotypic ratio of the F₂ generation would be 1 RR : 2 Rr : 1 rr, following Mendelian monohybrid cross rules. However, the phenotype of the F₂ generation would be different from the classic 3:1 ratio of a dominant/recessive relationship, due to incomplete dominance. The expected phenotype ratio of the F₂ generation would be 1 red : 2 pink : 1 white.

Topic: Biology, Principles of Inheritance and Variation, NEET
In a cross in Drosophila, the heterozygous member with grey body (b+) and long wings (vg+) was crossed with black body and vestigial wings, the progeny had the following ratio grey vestigial 24: grey long 126: black long 26: black vestigial 124. What is the frequency of recombinants in the population?

Options:
A 15.8 %
B 16.7 %
C 17.5 %
D 14.5 %

The correct answer is:
16.7 %

Solution:
In a cross in Drosophila, the heterozygous member with grey body (b+) and long wings (vg+) was crossed with black body and vestigial wings, the progeny had the following ratio grey vestigial 24: grey long 126: black long 26: black vestigial 124.
P1 = grey and long
P2 = black and vestigial
Their recombination results in grey vestigial and black long variants.
Total population: 300
Recombinant population (26 grey vestigeal + 24 black long): 50
Frequency of recombinants in the population: 50/300 * 100 = 16.7 %

Topic: Biology, Principles of Inheritance and Variation, NEET
In a cubic close packed structure, how many tetrahedral voids are present per unit cell?

Options:
A 11
B 8
C 9
D 7

The correct answer is:
8

Solution:
Correct Option is : (B)
8

Topic: Chemistry, Solid State, NEET
In a cyclic process, work done by the system is

Options:
A Zero
B More than the heat given to the system
C Equal to heat given to the system
D Independent of heat given to the system

The correct answer is:
Equal to heat given to the system

Solution:
The first law of thermodynamics states that heat supplied in a cyclic process. Δ Q = Δ U + Δ W .
Since in a cyclic process. the system returns to its original state, therefore Δ U = 0 . Hence Δ Q = Δ W .

Topic: Physics, Thermodynamics, NEET
In a cyclotron, if a deuteron can gain an energy of 40 MeV , then a proton can gain an energy of

Options:
A 40Â MeV
B 60 MeV
C 20Â MeV
D 80 MeV

The correct answer is:
80 MeV

Solution:
Radius of the path
R=2mKqB
âµRd=Rp
â´2mdKdqdB=2mpKpqpB
or 2(2mp)KdeB=2mpKpeB
∴ Kp=2Kd=2×40 MeV⁡=80 MeV

Topic: Physics, Magnetic Effects of Current, NEET
In a cylindrical vessel containing a liquid of density Ï, there are two holes in the side walls at heights of h1 and h2 respectively such that the range of efflux at the bottom of the vessel is same. The height of a hole for which the range of efflux would be maximum will be

Options:
A h2-h1
B h2+h1
C h2-h12
D h2+h12

The correct answer is:
h2+h12

Solution:
If H is the height of the liquid surface then for same range
h2=H-h1
and for maximum range
h=H2=h1+h22

Topic: Physics, Mechanical Properties of Fluids, NEET
In a diabetic patient, glucose appears in the urine. Which of the following statements is the best explanation for this phenomenon?

Options:
A Low plasma levels of glucose causes glucose to be secreted by the proximal tubule.
B The kidney is a significant site for glucose synthesis, which causes glucose to occasionally appear in the urine.
C The high concentration of glucose saturates the reabsorption pumps, causing glucose to appear in the urine.
D The vasa recta transport glucose into the urine.

The correct answer is:
The high concentration of glucose saturates the reabsorption pumps, causing glucose to appear in the urine.

Solution:
Diabetes mellitus is a condition wherein there is an excess of glucose produced in the body due to the ineffective functioning of the pancreas. The diabetic patients often have glucosuria (glucose in the urine) due to the malfunctioning of the renal tubules to reabsorb all the filtered glucose back into the blood.

Topic: Biology, Excretory Products and their Elimination, NEET