datavorous/entrance-exam-dataset huggingface dataset:
An ideal gas heat engine operates in carnot cycle between $227^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$. It absorbs $6 \times 10^4$ cal of heat at higher temperature. Amount of heat converted into work, is:

Options:
A $4.8 \times 10^4 \mathrm{cals}$
B $6 \times 10^4 \mathrm{cals}$
C $2.4 \times 10^4$ cals
D $1.2 \times 10^4$ cals

The correct answer is:
$1.2 \times 10^4$ cals

Solution:
For a carnot engine
$$
\text { efficiency } \begin{aligned}
e & =1-\frac{T_2}{T_1} \\
& =1-\frac{127+273}{227+273} \\
& =1-\frac{400}{500}=\frac{1}{5}
\end{aligned}
$$
Now, $\quad e=\frac{\text { Work Output }}{\text { Heat Output }}$
$$
=\frac{\mathrm{W}}{6 \times 10^4}
$$
$$
\begin{aligned}
\mathrm{W} & =e \times 6 \times 10^4 \\
& =\frac{1}{5} \times 6 \times 10^4 \\
& =1.2 \times 10^4 \mathrm{cal} .
\end{aligned}
$$

Topic: Physics, Thermodynamics, NEET, NEET 2005
An ideal gas in the state P1, V1 is isothermally expanded to state P2, V2. Then it is adiabatically compressed to initial volume V1. The pressure being P3 in the final state. If W is work done by the gas in the whole process then

Options:
A P3>P1 & W<0
B P3<P1 & W>0
C P3<P1 & W=0
D P3>P1 & W>0

The correct answer is:
P3>P1 & W<0

Solution:
P-V diagram of the process is,
From the diagram, P3>P1
W1-2<W2-3
∴ Overall W<0

Topic: Physics, Thermodynamics, NEET
An ideal gas is allowed to expand against a constant pressure of 2 bar from 10 L to 50 L in one step. Calculate the amount of work done by the gas. If the same expansion were carried out reversibly, will the work done be higher or lower than the earlier case? (Given that 1 L bar = 100 J)

Options:
A w = – 10 kJ and work done will be more in the reversible expansion
B w = – 8 kJ and work done will be less in the reversible expansion
C w = – 9 kJ and work done will be more in the reversible expansion
D w = – 8 kJ and work done will be more in the reversible expansion

The correct answer is:
w = – 8 kJ and work done will be more in the reversible expansion

Solution:
Correct Option is : (D)
w = – 8 kJ and work done will be more in the reversible expansion

Topic: Chemistry, Thermodynamics (C), NEET
An ideal gas is at temperture 300 K. The final temperature of the gas when its volume changes from V to 2V in the process P=Î±V (here Î± is a positive constant) is

Options:
A 900 K
B 1200 K
C 600 K
D 300 K

The correct answer is:
1200 K

Solution:
The given relation is pâV
Therefore, pâV
When V chnages from V to 2V, pressure p is also doubled
For an ideal gas, pVT=constant
â´ Tâpv. Hence, T becomes 2Ã2=4 times
i.e, 4Ã300 K=1200 K

Topic: Physics, Thermodynamics, NEET
An ideal gas is compressed to half of its initial volume by means of several processes. Which of the process results in the maximum work done on the gas?

Options:
A Isobaric
B Isochoric
C Isothermal
D Adiabatic

The correct answer is:
Adiabatic

Solution:
Since area under the curve is maximum for the adiabatic process so work done on the gas will be maximum for the adiabatic process.

Topic: Physics, Thermodynamics, NEET, NEET 2015 (Phase 2)
An ideal gas is expanded so that the amount of heat given is equal to the decrease in internal energy of the gas. The gas undergoes the process PV65=constant. The gas may be

Options:
A He
B O2
C Ar
D all of the above

The correct answer is:
O2

Solution:
Given that amount of heat given is equal to the decrease in internal energy,
dQ=-dU
C=-CV
Rγ-l+Rl-x=-Rγ-l
Rγ-l+Rl-65=-Rγ-l
52=lγ-l
⇒γ-l=25
⇒γ=75
The gas may be diatomic or polyatomic linear.

Topic: Physics, Thermodynamics, NEET
An ideal gas is expanding such that PT2 = constant. The coefficient of volume expansion of the gas is

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

The correct answer is:
3 T

Solution:
P T 2 = constant
∴ n RT V T 2 = constant
or, T 3 V - 1 = constant
Differentiating the equation, we get
3 T 2 V · dT - T 3 V 2 dV = 0
or, 3 dT = T V · dV ..........(i)
From the equation, dV = V γ dT
γ = Coefficient of volume expansion of gas = dV V · dT
From Eq.(i) γ = dV V · dT = 3 T

Topic: Physics, Thermodynamics, NEET
An ideal gas is expanding such that PT2 = constant. The coefficient of volume expansion of the gas is

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

The correct answer is:
3T

Solution:
PT2=constant
∴ nRTVT2=constant
or,T3V-1=constant
Differentiating the equation, we get
3T2V·dT-T3V2dV=0
or, 3dT=TV·dV ..........(i)
From the equation,dV=VγdT
γ=Coefficient of volume expansion of gas=dVV·dT
From Eq.(i) γ=dVV·dT=3T

Topic: Physics, Thermodynamics, NEET
An ideal gas is expanding such that pT2= constant. The coefficient of volume expansion of the gas is

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

The correct answer is:
3T

Solution:
From the ideal gas equation, pV=μRT
⇒p=μRTV
Given, pT2= constant
μRTV.T2= constant
T3=kV ... (i)
Where, k= constant.
On differentiating both sides, we get
3T2dT=kdV...(ii)
From Equations (i) and (ii), we get
3dTT=dVV⇒dVV dT=3T
∴ Coefficient of volume expansion, dVVdT=3T.

Topic: Physics, Thermodynamics, NEET
An ideal gas is found to obey an additional law VP2= constant. The gas is initially at temperature T and volume V. When it expands to a volume 2 V, the temperature becomes

Options:
A 2 T
B 2T
C 22 T
D 4T

The correct answer is:
2 T

Solution:
VP 2 = constant
V RT V 2 = constant from ideal gas equation
or T 2 V = constant
Let the new temperature become T â² T â² T 2 = 2V V = 2 or T â² = (2) T

Topic: Physics, Thermodynamics, NEET
An ideal gas is heated at constant pressure and in this process, it absorbs Q amount of heat. If the adiabatic exponent is Î³, then the fraction of heat absorbed in performing the work is

Options:
A 1-1γ
B 1+1γ
C 1-2γ
D 1+2γ

The correct answer is:
1-1γ

Solution:
Heat absorbed by the system at constant pressure
Q=ncp∆T
Change in internal energy ∆U=ncv∆T
W=Q-∆U
∴ WQ=Q-∆UQ=1-∆UQ
=1-ncv∆Tncp∆T=1-cvcp
=1-1γ

Topic: Physics, Thermodynamics, NEET
An ideal gas is taken around the cycle ABCA as shown in the P-V diagram. The total work done by the gas during the cycle is

Options:
A PV
B 2PV
C 4PV
D 3PV

The correct answer is:
3PV

Solution:
Area of triangle ABC will give the total work done
=12×2V⁡×3P⁡=3P⁡V⁡

Topic: Physics, Thermodynamics, NEET
An ideal gas is taken through a cyclic thermodynamical process through four steps. The amounts of heat involved in these steps are Q1=5960âJ,âQ2=-5585âJ,âQ3=-2980âJ and Q4=3645âJ, respectively. The corresponding works involved are W1=2200 J, W2=-825 J, W3=-1100 J and W4, respectively. The value of W4 is

Options:
A 1315 J

B 275 J

C 765 J

D 675 J

The correct answer is:
765 J

Solution:
As, ΔQ=Q1+Q2+Q3+Q4
=5960-5585-2980+3645
=1040 J
Also, ΔW=W1+W2+W3+W4
=2200-825-1100+W4
=275+W4
For a cyclic process, ΔU=0
i.e., Uf-Ui=0,
From first law of thermodynamics
ΔQ=ΔU+ΔW
1040=0+275+W4
Or W4=765 J

Topic: Physics, Thermodynamics, NEET
An ideal gas is taken through a series of changes, from P1, V1â4P1, V1âP1, 3V1 represented in the figure. The net work done by the gas at the end of the cycle is equal to

Options:
A P1V1
B 3P1V1
C 6P1V1
D 12P1V1

The correct answer is:
3P1V1

Solution:
Work done =12×2V1×3P1=3P1V1

Topic: Physics, Thermodynamics, NEET
An ideal gas is taken through a series of changes, from P1,V1â4P1,V1âP1,3V1 represented in the figure. The net work done by the gas at the end of the cycle is equal to

Options:
A P1V1
B 3P1V1
C 6P1V1
D 12P1V1

The correct answer is:
3P1V1

Solution:
Work done =12×2V1×3P1=3P1V1

Topic: Physics, Thermodynamics, NEET
An ideal gas is taken through the cycle AâBâCâA, as shown in the figure. If the net heat supplied to the gas in the cycle is 5 J , the work done by the gas in the process CâA is

Options:
A -5 J
B -10 J
C -15 J
D -20 J

The correct answer is:
-5 J

Solution:
For cyclic process,
ΔU=0

⇒Q=W
Q=WAB+WBC+WCA
5=10×1+0+WCA

⇒ WCA=-5 J

Topic: Physics, Thermodynamics, NEET
An ideal gas is taken through the cycle AâBâCâA, as shown in the figure. If the net heat supplied to the gas in the cycle is 5 J, the work done by the gas in the process CâA is

Options:
A â5 J
B −10 J
C −15 J
D −20 J

The correct answer is:
â5 J

Solution:
Δ W AB = p Δ V = 1 0 2 - 1 = 10 J
ΔWBC=0 as V = constant
From first law of thermodynamics,
Δ Q = Δ W + Δ U
Δ U = 0
Process ABCA is cyclic,
∴ Δ Q = Δ W AB + Δ W BC + Δ W CA
∴ Δ W CA = Δ Q - Δ W AB - Δ W BC
= 5 - 1 0 - 0 = - 5 J

Topic: Physics, Thermodynamics, NEET
An ideal gas mixture filled inside a balloon expands according to the relation PV2/3= constant. The temperature inside the balloon is

Options:
A Increasing
B Decreasing
C Constant
D Can't be said

The correct answer is:
Increasing

Solution:
For the process with ideal gas PVn= constant. If expansion happens with n > 1 then only the temperature decreases. But here n=23 , so here n<1 so temperature must increase.

Topic: Physics, Thermodynamics, NEET
An ideal gas, obeying kinetic theory of gases can not be liquefied, because

Options:
A it solidifies before becoming a liquid
B forces acting between its molecules are negligible
C its critical temperature is above 0°C
D its molecules are relatively small in size

The correct answer is:
forces acting between its molecules are negligible

Solution:
A gas can only be liquified, if some forces of attraction are acting in its molecules. Since, an ideal gas is devoid of force of attraction in its molecules, therefore it can not be liquefied.

Topic: Chemistry, States of Matter, NEET
An ideal gas undergoes four different processes from the same initial state as shown in the figure below. Those processes are adiabatic, isothermal, isobaric and isochoric. The curve which represents the adiabatic process among 1, 2,3 and 4 is:

Options:
A 2
B 3
C 4
D 1

The correct answer is:
2

Solution:
For adiabatic process,

P∝1Vγ

For isothermal process,

P∝1V

For isobaric process, P=constant

For isochoric process, V=constant

∴ Graphs for different processes

Adiabatic ⇒2

Isothermal ⇒3

Isobaric ⇒1

∴The graph for adiabatic process is 2.

Topic: Physics, Thermodynamics, NEET, NEET 2022 (Phase 1)
An ideal gas with pressure P, volume V and temperature T is expanded isothermally to a volume 2V and final pressure P I . The same gas is expanded adiabatically to a volume 2V and final pressure in this case is P A . In terms of the ratio of the two specific heats for the gas Î³ , the ratio P I / P A is

Options:
A 2 γ - 1
B 2 1 - γ
C 2 γ
D 2 γ

The correct answer is:
2 γ - 1

Solution:
For an isothermal process
PV = constant ∴ PV = P I 2 V
Pâ¡ Iâ¡ = Pâ¡ 2 … i
For an adiabatic process
PV γ = constant ∴ PV γ = P A 2 V γ
or Pâ¡ Aâ¡ = Pâ¡ 2 γ … i i
Divide (i) by (ii), we get
Pâ¡ Iâ¡ Pâ¡ Aâ¡ = 2 γ 2 = 2 γ - 1

Topic: Physics, Thermodynamics, NEET
An ideal mixture of two liquids A and B is put in a cylinder containing piston. Piston is pulled out isothermally so that the volume of liquid decreases but that of vapours increases. Negligibly small amount of liquid was left and mole-fraction of A in vapour is 0.4. If P A â = 0.4 and P B â = 1.2 atm at the experimental temperature, which of the following is the total pressure at which the liquid is almost evaporated?

Options:
A 0.22 atm
B 0.431 atm
C 0.667 atm
D 1 atm

The correct answer is:
0.667 atm

Solution:
Mole-fraction in vapour phase, Y for A
Y A = 0.4 , Y B = 0.6
P A â = 0.4 P B â = 1.2 } given
P T = P A â x A + P B â x B ...(i)
Y A = 0.4 = P A â x A P T â 0.4 x A P T = 0.4
or P T = x A ...(ii)
Y B = 0.6 = P B â x B P T â 1.2 x B P T = 0.6
or P T = 2 x B ...(iii)
Comparing (ii) and (iii),
x A = 2 x B
we know, x A + x B = 1
Hence, x B = 1 3 and x A = 2 3
Putting in Equation (i)
P T = 0.4 Ã 2 3 + 1.2 Ã 1 3 = 0.667 atm

Topic: Chemistry, Solutions, NEET
An ideal mono-atomic gas is taken around the cycle ABCDA as shown in the P-V diagram. The work done during the cycle is given by

Options:
A 12PV
B PV

C 2PV

D 4PV

The correct answer is:
PV

Solution:
The area of P-V diagram gives work done
∴ W=(2V-V) (2P-P)=PV

Topic: Physics, Thermodynamics, NEET
An ideal monoatomic gas is taken round the cycle. ABCDA as shown in figure. The work done during the cycle is

Options:
A -PV
B -2PV
C -12PV
D 0

The correct answer is:
-PV

Solution:
As net work done during a complete cycle is equal to area under the cycle so here work done is equal to -PV .

Topic: Chemistry, Thermodynamics (C), NEET
An ideal monoatomic gas undergoes process PV1.25= constant. Then

Options:
A upon increase in pressure temperature decreases
B upon increase in pressure heat is absorbed by gas
C if heat is given to gas its internal energy increases
D heat is absorbed by gas if volume increases

The correct answer is:
heat is absorbed by gas if volume increases

Solution:
Specific heat of an ideal gas for process PVx= constant is,
C=Rγ-1-Rx-1
C=R153-1-11.25-1<0
⇒ If heat is given to gas then temperature decreases and hence internal energy also decreases.
Also, PV1.25= constant
PV=constantV0.25
nRT=constantV0.25
⇒ Upon increase in pressure, volume decreases and Temperature increases.
If volume increases then temperature decreases and heat is absorbed by the gas.

Topic: Physics, Thermodynamics, NEET
An ideal refrigerator has a freezer at a temperature of -13 oC. The coefficient of performance of the engine is 5. The temperature of the air at which heat is rejected will be

Options:
A 325 oC
B 325 K
C 39 oC
D 320 oC

The correct answer is:
39 oC

Solution:
Coefficient of performance α=TCTH-TC
TC=273-13=260 K
5=260TH-260
TH=312 K
∴ TH=312-273=39 oC

Topic: Physics, Thermodynamics, NEET
An ideal solution contains two volatile liquids A (po = 100 torr) and B (po = 200 torr). If mixture contains 1 mole of A and 4 moles of B, then total vapour pressure of the distillate is

Options:
A 150 torr
B 180 torr
C 188.88 torr
D 198.88 torr

The correct answer is:
188.88 torr

Solution:
p Total = p A â X A + p B â X B
= 1 0 0 × 1 5 + 2 0 0 × 4 5
= 2 0 + 1 6 0 = 1 8 0
(Vapours are condensed and again brought in equilibrium with vapours).
i.e for distillate
yA → to be replaced by xA
yB → to be replaced by xB.
y A = p A o x A p T = 2 0 1 8 0
y B = p B o x B p T = 1 6 0 1 8 0
P distillate = 1 0 0 × 2 0 1 8 0 + 2 0 0 × 1 6 0 1 8 0
= 188.88

Topic: Chemistry, Solutions, NEET
An ideal spring supports a disc of mass M. A body of mass m is released from a certain height from where it falls to hit M. The two masses stick together at the moment they touch and move together from then on. The oscillations reach to a height a above the original level of the disc and depth b below it. The constant of the force of the spring is

Options:
A 2mgb-a
B mgb-a
C 2mga-b
D mg2a-b

The correct answer is:
2mga-b

Solution:
Amplitude of vibration =b+a2
Mean position =a-b+a2=a-b2
∴ ka-b2=mg⇒k=2mga-b

Topic: Physics, Oscillations, NEET
An ideal spring with spring constant k is hung from the ceiling and a block of mass M is attached to its lower end. If the mass is released with the spring initially unstretched, then the maximum extension in the spring is

Options:
A 4Mgk
B 2Mgk
C Mgk
D Mg2k

The correct answer is:
2Mgk

Solution:
Let x be the maximum extension of the spring. From conservation of mechanical energy
Decrease in gravitational potential energy = increase in elastic potential energy
â´ Mg x = 1 2 kx 2
or x = 2 Mg k

Topic: Physics, Work Power Energy, NEET
An illuminated object and a screen are placed 90 cm apart. The focal length and nature of the lens required to produce a clear image on the screen, twice the size of the object would be:

Options:
A 10 cm
B 20 cm
C 15 cm
D 30 cm

The correct answer is:
20 cm

Solution:
â´u+v=90 ...(i)

m= Ï u â2= Ï u âv=2u ...(ii)

Putting the value of v in (i), we get

u+2u=90âu= 90 30 =30

â´v=2Ã30=60

Using lens formula, we get

1 f = 1 Ï â 1 u â 1 f = 1 60 â 1 â30 â 1 f = 1 60 + 1 30 â 1 f = 1+2 60 = 3 60 âf= 60 3 â´f=20 cm.

Topic: Physics, Ray Optics, NEET
An image is formed at a distance of 100 cm from a glass surface with refractive index 1.5 when a point object is placed in the air at a distance of 100 cm from the glass surface. What is the radius of curvature is of the surface?

Options:
A 20 cm
B 40 cm
C 30 cm
D 50 cm

The correct answer is:
20 cm

Solution:
Î¼2V-Î¼1U=Î¼2-Î¼1R
1.5100+1100=1.5-1R
R=20 cm

Topic: Physics, Ray Optics, NEET
An image is formed at a distance of 100 cm from the glass surface with refractive index 1.5, when a point object is placed in the air at a distance of 100 cm from the glass surface. The radius of curvature is of the surface is

Options:
A 20 cm
B 40 cm
C 30 cm
D 50 cm

The correct answer is:
20 cm

Solution:
Î¼2V-Î¼1U=Î¼2-Î¼1R
1.5100+1100=1.5-1R
R=20 cm

Topic: Physics, Ray Optics, NEET
An immersion heater with electrical resistance 7 Î© is immersed in 0.1 kg of water at 20 â for 3 min. If the flow of current is 4 A, what is the final temperature of the water assuming whole of heat produced is consumed in water. Specific heat capacity of water=4.2Ã103 J kg-1K-1

Options:
A 28 â
B 48 â
C 52 â
D 68 â

The correct answer is:
68 â

Solution:
i2Rt=ms âT
â 42Ã7Ã3Ã60
=0.1Ã4.2Ã103ÃT-20
â´ T=68 â

Topic: Physics, Current Electricity, NEET
An important characteristic that Hemichordates share with Chordates is

Options:
A Absence of notochord
B Ventral tubular nerve cord
C Pharynx with gill slits
D Pharynx without gill slits

The correct answer is:
Pharynx with gill slits

Solution:
Pharyngeal gill slits are present in Hemichordates as well as in Chordates. The notochord is present in Chordates only. Ventral tubular nerve cord is a characteristic feature of non-chordates.

Topic: Biology, Animal Kingdom, NEET, NEET 2017
An important step in the manufacture of pulp for paper industry from the woody tissues of plants is the:

Options:
A preparation of pure cellulose by removing lignin
B treatment of wood with chemicals that break down cellulose
C removal of oils present io the wood by treatment with suitable chemicals
D removal of water from the wood by prolonged heating at approximately $50^{\circ} \mathrm{C}$.

The correct answer is:
preparation of pure cellulose by removing lignin

Solution:
Paper production is basically a two-step process in which a fibrous raw material is first converted into pulp, and then the pulp is converted into paper. The harvested wood is first processed so that the fibres are separated from the unusable fraction of the wood, the lignin.
Related Theory
The pulping process is aimed at removing lignin without loosed fibre strength, thereby freeing the fibres and removing impurities that cause discoloration and possible future disintegration of the paper. Hemicellulose plays an important role in fibre-to-fibre bonding in papermaking. Several extractives such as waxes, resins are contained in wood but they do not contribute to its strength properties; these too are removed during the pulping process. In general, the softwoods (e.g., pines, firs, and spruces) yield long and strong fibres and hardwoods produce $a$ weaker paper as they contain shorter fibres.

Topic: Biology, Biodiversity and its Conservation, NEET, NEET 2005
An improved variety of transgenic basmati rice

Options:
A does not require chemical fertilizers and growth hormones
B gives high yield and is rich in vitamin-A
C is completely resistant to all insect pests and diseases of paddy
D gives high yield but has no characteristic aroma

The correct answer is:
gives high yield and is rich in vitamin-A

Solution:
Golden rice is a variety of Oryza sativa produced through genetic engineering to biosynthesize beta-carotene, a precursors of pro-vitamin-A in the edible parts of rice. The research that led to golden rice was conducted with the goal of helping children who suffer from vitamin-A deficiency in poor countries. Golden rice has been bred to be especially disease-resistant, resulting in better crop yields.

Topic: Biology, Biotechnology and Its Applications, NEET, NEET 2010 (Screening)
An improved variety of transgenic basmati rice:

Options:
A Gives high yield and is rich in Vitamin A
B Is completely resistant to all insect pests and diseases of paddy
C Gives high yield but has no characteristic aroma
D Does not require chemical fertilizers and growth hormones

The correct answer is:
Gives high yield and is rich in Vitamin A

Solution:
Golden Rice is a new type of rice that contains beta carotene, a source of vitamin A.

Topic: Biology, Biotechnology - Principles and Processes, NEET
An inclined plane making an angle β with horizontal. A projectile projected from the bottom of the plane with a speed u at angle α with horizontal, then its maximum range Rmax is

Options:
A Rmax=u2g1-sin⁡β
B Rmax=u2g1+sin⁡β
C Rmax=ug1-sin⁡β
D Rmax=ug1+sin⁡β

The correct answer is:
Rmax=u2g1+sin⁡β

Solution:
The expression of horizontal Range R=2u2sinâ¡Î±-Î²cosâ¡Î±gcos2Î²
From identity
2sinâ¡Acosâ¡B=sinâ¡A+B+sinâ¡A-B
2sinâ¡Î±-Î²cosâ¡Î±=sinâ¡2Î±-Î²+sinâ¡-Î²
R=u2gcos2â¡Î² sinâ¡2Î±-Î²-sinâ¡Î²
Range R along incline is maximum if 2Î±-Î²=Ï2 (as Î² is fixed)
So, Rmax=u2gcos2â¡Î² 1-sinâ¡Î²
= u2g1-sin2â¡Î² 1-sinâ¡Î²
= u 2 g 1 + sin Î² 1 - sin Î² 1 - sin Î²
= u2g1+sinâ¡Î²

Topic: Physics, Motion In Two Dimensions, NEET
An inclined track ends in a circular loop of the radius r. From what height on the track a particle should be released so that it completes the loop, assuming there is no friction?

Options:
A r2
B 3r2
C 2r
D 5r2

The correct answer is:
5r2

Solution:
For looping the loop v = 5 gr
By law of conservation of energy
Loss in potential energy = Gain in kinetic energy
â mgh = 1 2 mv 2 â h = 5 r 2

Topic: Physics, Work Power Energy, NEET
An incompressible liquid flows ina tube, as shown in the figure. The speed of the liquid in the lower branch will be

Options:
A 1.0 m s-1
B 1.5 m s-1
C 2.5 m s-1
D 3.0 m s-1

The correct answer is:
1.0 m s-1

Solution:
From equation of continuity
Av=A1v1+A2v2
0.123=0.121.5+0.18v
0.36=0.18+0.18v
0.18=0.18v
v=1 m/s

Topic: Physics, Mechanical Properties of Fluids, NEET
An incompressible liquid flows through a horizontal tube as shown in the following figure. Then the velocity v of the fluid is

​

Options:
A 3.0 m/s
B 1.5 m/s
C 1.0 m/s
D 2.25 m/s

The correct answer is:
1.0 m/s

Solution:
Volume flowing per second in left tube
= volume flowing per second 1st branch tube + volume flowing per second in 2nd branch tube
Av1=Av2+1.5 Av
AÃ3=AÃ1.5+1.5 Av
v=1 m/s

Topic: Physics, Mechanical Properties of Fluids, NEET
An incompressible non viscous fluid flows steadily through a cylindrical pipe which has radius 2R at point A and radius R at point B farther along the flow direction. If the velocity of the fluid at point A is V, its velocity at the point B will be

Options:
A V2
B 4V
C 2V
D V

The correct answer is:
4V

Solution:
We know that, from continuity equation,
A1V1=A2V2
(Symbols have usual meanings)
Ï(2R)2V=ÏR2VB
(VB= Velocity of the fluid at point B)
â Ï(2R)2V=ÏR2VB
â Ï4R2V=ÏR2VB
4V=VB
â´ VB=4V

Topic: Physics, Mechanical Properties of Fluids, NEET
An incorrect statement with respect to SN1 and SN2 mechanism for alkyl halide is

Options:
A A strong nucleophile is an aprotic solvent increases the rate or favours SN2 reaction.
B Competing reaction for SN2 reaction is rearrangement
C SN1 reactions can be catalysed by some Lewis acids
D A weak nucleophile and a protic solvent increases the rate of SN1 reaction

The correct answer is:
Competing reaction for SN2 reaction is rearrangement

Solution:
SN1 reaction occurs through the formation of carbocation as intermediate. Hence, rearrangement occurs in SN1 reaction and not in SN2 reaction. SN1 reaction is favoured in presence of a polar protic solvent and a weak nucleophite. These reactions can be catalysed by the some Lewis is acid. A strong nucleophile is an aprotic solvent increases the rate of SN2 reaction. Hence, option (B) in incorrect.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
An increase in the concentration of the reactants of a reaction leads to change in

Options:
A heat of reaction
B threshold energy
C collision frequency
D activation energy

The correct answer is:
collision frequency

Solution:
Collisional Frequency is the average rate in which two reactants collide for a given system and is used to express the average number of collisions per unit of time in a defined system<
Z∝ number of reactant molecules per unit volume.

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2020 (Phase 1)