datavorous/entrance-exam-dataset huggingface dataset:
A functional mammary gland is characteristic of all female mammals. The glandular tissue of each breast is divided into ____A____containing clusters of cells called alveoli. The cells of alveoli secrete milk, which is stored in the cavities (lumens) of alveoli. The alveoli open into____B____ . The ____B____of each lobe join to form a ____C____. Several ____C____join to form a wider____D____ which is connected to ____E____through which milk is sucked out.

Options:
A

A	B	C	D	E
7-8	mammary duct	mammary ampulla	mammary tubules	lactiferous duct

B

A	B	C	D	E

15-20	mammary ampulla	mammary duct	lactiferous duct	mammary tubules

C

A	B	C	D	E

15-20	mammary tubules	mammary duct	mammary ampulla	lactiferous duct

D

A	B	C	D	E
7-8	mammary duct	mammary ampulla	lactiferous duct	mammary tubules

The correct answer is:

A	B	C	D	E

15-20	mammary tubules	mammary duct	mammary ampulla	lactiferous duct

Solution:
Mammary glands are modified sweat glands and characteristics of mammalian females. In human males it is a vestigial organ. The mammary glands are paired structures (breasts) that contain glandular tissue and a variable amount of fat. The glandular tissue of each breast is divided into 15-20 mammary lobes containing clusters of cells called alveoli. The cells of alveoli secrete milk, which is stored in the cavities (lumens) of alveoli. The alveoli open into mammary tubules. The tubules of each lobe join to form a mammary duct. Several mammary ducts join to form a wider mammary ampulla which is connected to lactiferous duct through which milk is sucked out.

Topic: Biology, Human Reproduction, NEET
A fungus contains cells with two nuclei from different genomes. The nuclei do not fuse but divide independently and simultaneously as new cells are formed. It belongs to

Options:
A Phycomycetes
B Zygomycetes
C Deuteromycetes
D Basidiomycetes

The correct answer is:
Basidiomycetes

Solution:
In basidiomycetes, dikaryotisation occurs. In this process, two nuclei from different genomes are present in a cell; they do not fuse but divide independently and simultaneously as new cells are formed.

Topic: Biology, Biological Classification, NEET
Ag2S+NaCN→A→ZnB ,Hence A and B are -

Options:
A Na2ZnCN4,Zn
B NaAgCN2,Ag
C Na2AgCN4,Ag
D Na2AgCN4,Ag

The correct answer is:
NaAgCN2,Ag

Solution:
It is Mac-Arthur cyanide process is a hydrometallurgical process for extraction of gold and silver. In this ore is converted to water-soluble coordination complex.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
A galvanic cell has electrical potential of 1.1 V. If an opposing potential of 1.1 V is applied to this cell, what will happen to the cell reaction and current flowing through the cell?

Options:
A The cell reaction will continue as before, and the current will flow in the same direction.
B The cell reaction will stop, and no current will flow through the cell.
C The cell reaction will reverse, and the current will flow in the opposite direction.
D The cell reaction will become faster, and the current will increase.

The correct answer is:
The cell reaction will stop, and no current will flow through the cell.

Solution:
Correct Option is : (B)
The cell reaction will stop, and no current will flow through the cell.

Topic: Chemistry, Electrochemistry, NEET
A galvanometer acting as a voltmeter will have:

Options:
A a high resistance in parallel with its coil
B a high resistance in series with its coil
C a low resistance in parallel with its coil
D a coil resistance in series with its coil

The correct answer is:
a high resistance in series with its coil

Solution:
A galvanometer acting as a voltmeter will have a high resistance in series with its coil.

Topic: Physics, Current Electricity, NEET, NEET 2004
A galvanometer coil has a resistance of 50 Î© and the meter shows full-scale deflection for a current of 5 mA. This galvanometer is converted into a voltmeter of range 0-20 V by connecting

Options:
A 3950 Î© in series with galvanometer
B 4050Â Î© in series with galvanometer
C 3950Â Î© in parallel with galvanometer
D 4050Â Î© in parallel with galvanometer

The correct answer is:
3950 Î© in series with galvanometer

Solution:
Here,
R= ?
G=50 Î©
V=20 V
Ig=5Ã10-3 A
Where G: Resistance of the galvanometer
V: Required range of the voltmeter
Ig: Current in galvanometer which given full deflection
R: High resistance in series with galvanometer
Applying the formula,
R=VIg-G
=205Ã10-4-50
=4000-50
=3950 Î©

Topic: Physics, Magnetic Effects of Current, NEET
A galvanometer has a coil of resistance $100 \Omega$ and gives a full scale deflection for $30 \mathrm{~mA}$ current. If it is to work as a voltmeter of $30 \mathrm{~V}$ range, the resistance required to be added will be

Options:
A $900 \Omega$
B $1800 \Omega$
C $500 \Omega$
D $1000 \Omega$

The correct answer is:
$900 \Omega$

Solution:
Required resistance $R=\frac{V}{i_g}-G$
$$
=\frac{30}{30 \times 10^{-3}}-100=900 \Omega
$$

Topic: Physics, Current Electricity, NEET, NEET 2010 (Screening)
A galvanometer having a coil resistance of $60 \Omega$ shows full scale deflection when a current of 1.0 A passes through it. It can be converted into an ammeter to read currents upto $5.0 \mathrm{~A}$ by

Options:
A putting in parallel a resistance of $240 \Omega$
B putting in series a resistance of $15 \Omega$
C putting in series a resistance of $240 \Omega$
D putting in parallel a resistance of $15 \Omega$

The correct answer is:
putting in parallel a resistance of $15 \Omega$

Solution:
Key Idea To convert a galvanometer to ammeter a small resistance is connected in parallel to the coil of the galvanometer.
$$
\text { Here, } \begin{aligned}
& \mathrm{G}_1=60 \Omega, \mathrm{I}_{\mathrm{g}}=1.0 \mathrm{~A}, \mathrm{I}=5 \mathrm{~A} \\
& \mathrm{I}_{\mathrm{g}} \mathrm{G}_1=\left(\mathrm{I}-\mathrm{I}_{\mathrm{g}}\right) \mathrm{S} \\
& \mathrm{S}=\frac{\mathrm{I}_{\mathrm{g}} \mathrm{G}_1}{\mathrm{I}-\mathrm{I}_{\mathrm{g}}}=\frac{1}{5-1} \times 60=15 \Omega
\end{aligned}
$$
Putting $15 \Omega$ resistance in parallel.

Topic: Physics, Current Electricity, NEET, NEET 2009 (Screening)
A galvanometer having a resistance of 20 Î© and 30 divisions on both sides has figure of merit 0.005 ampere/division. The resistance that should be connected in series such that it can be used as a voltmeter upto 15 V, is:

Options:
A 120Â Î©
B 100Â Î©
C 80 Î©
D 125Â Î©

The correct answer is:
80 Î©

Solution:
Current for full scale deflection =0.005×30=0.150 A
Voltage for full scale deflection =0.150×20 V
=3.0 V
Resistance required to be connected in series =153-1 20 Ω
=80 Ω

Topic: Physics, Current Electricity, NEET
A galvanometer of $50 \mathrm{ohm}$ resistance has 25 divisions. A current $4 \times 10^{-4}$ ampere gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of 25 volts, it should be connected with a resistance of:

Options:
A $2500 \Omega$ as a shunt
B $2450 \Omega$ as a shunt
C $2550 \Omega$ in series
D $2450 \Omega$ in series

The correct answer is:
$2450 \Omega$ in series

Solution:
Total current in the galvanometer $=$
$$
\begin{aligned}
& 25 \times 4 \times 10^{-4} \mathrm{~A} \\
& \mathrm{I}_g=10^{-2} \mathrm{~A}
\end{aligned}
$$
The value of resistance connected in series to convert galvanometer into voltmeter of $25 \mathrm{~V}$ is
$$
\begin{aligned}
R & =\frac{V}{I_g}-G \\
& =\frac{25}{10^{-2}}-50 \\
& =2450 \Omega
\end{aligned}
$$

Topic: Physics, Current Electricity, NEET, NEET 2004
A galvanometer of resistance 22.8 Î© measures 1 A. How much shunt should be used, so that it can be used to measure 20 A?

Options:
A 1 Î©
B 2 Î©
C 1.2 Î©
D 2.2 Î©

The correct answer is:
1.2 Î©

Solution:
Shunt is a low resistance used in parallel with the galvanometer to make it ammeter.
The voltage across galvanometer = voltage across the shunt
Given, G = 22.8 Î© , i = 20 A , iG = 1 A
â´S=1Ã22.820â1=22.819=1.2Î©

Topic: Physics, Current Electricity, NEET
A galvanometer of resistance 25 Î© measures 10-3 A. shunt required to increase range up to 2 A is

Options:
A 12.5 mÎ©
B 0.125 mΩ
C 0.125 Ω
D 1.25 mΩ

The correct answer is:
12.5 mÎ©

Solution:
i g G = i - i g S ; 1 0 - 3 - 2 5 â 2 Ã S
S = 2 5 Ã 1 0 - 3 2 = 0 . 0 1 2 5 mÎ©

Topic: Physics, Current Electricity, NEET
A galvanometer of resistance 50 Î© giving full-scale deflection for a current of 10 mA is to be changed into a voltmeter of range 100 V. What should be the value of resistance to be added in series with this galvanometer?

Options:
A 9950 Ω
B 10025 Ω
C 10000 Ω
D 9975 Ω

The correct answer is:
9950 Ω

Solution:
V=IG+R
100=R+50×10×10-3
R+50=10000
⇒ R=9950 Ω

Topic: Physics, Magnetic Effects of Current, NEET
A galvanometer of resistance $50 \Omega$ is connected to a battery of $3 \mathrm{~V}$ along with a resistance of $2950 \Omega$ in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be

Options:
A $5050 \Omega$
B $5550 \Omega$
C $6050 \Omega$
D $4450 \Omega$

The correct answer is:
$4450 \Omega$

Solution:
Current through the galvanometer
\(I=\frac{3}{(50+2950)}=10^{-3} \mathrm{~A}\)
Current for 30 divisions \(=10^{-3} \mathrm{~A}\)
Current for 20 divisions \(=\frac{10^{-3}}{30} \times 20\)
\(=\frac{2}{3} \times 10^{-3} \mathrm{~A}\)
For the same deflection to obtain for 20 divisions, let resistance added be R
\(\begin{aligned}
& \therefore \frac{2}{3} \times 10^{-3}=\frac{3}{(50+1 \mathrm{R})} \\
& \text {or } \mathrm{R}=4450 \Omega
\end{aligned}\)

Topic: Chemistry, Ionic Equilibrium, NEET, NEET 2008 (Screening)
A galvanometer of resistance 50 Î© is connected to a battery of 3 V along with a resistance of 2950 Î© in series shows full-scale deflection of 30 divisions. The additional series resistance required to reduce the deï¬ection to 20 divisions is

Options:
A 2950Â Î©
B 1500 Î©
C 4440 Î©
D 7400Â Î©

The correct answer is:
1500 Î©

Solution:
Current flowing in galvanometer,
I=350+2950
I=10-3A
Current for 30 division =10-3A
Current for 20 division =10-330Ã20=23Ã10-3A
Let the additional series resistance =R
â´ 23Ã10-3=350+2950+R
R=1500 Î©

Topic: Physics, Current Electricity, NEET
A galvanometer with a scale divided into 100 equal divisions has a current sensitivity of 10 divisions per milliampere and a voltage sensitivity of 2 divisions per millivolt. The galvanometer resistance will be

Options:
A 4 Î©
B 5 Î©
C 3 Î©
D 7 Î©

The correct answer is:
5 Î©

Solution:
Here, the current sensitivity is 10 divmA and there are 100 divisions on the scale. So, the current required for full-scale deflection is
Ig=110×100 mA=10 mA
= 0.01 A
Also, the voltage sensitivity is 2 divmV so the voltage required for full-scale deflection is
Vg=12×100 mV
= 0.05 V
Galvanometer resistance is given by
G=VgIg=0.050.01
=5 Ω

Topic: Physics, Magnetic Effects of Current, NEET
A galvanometer, with a scale divided into 150 equal divisions, has current sensitivity of 10 divisions per milliampere and voltage sensitivity of 2 divisions per millivolt. Find shunt resistance for making an ammeter of 6 A.

Options:
A 1.25Â ÃÂ 10â3Â Î©
B 12.5 Ã 10â3 Î©
C 125Â ÃÂ 10â3Â Î©
D 0.125Â ÃÂ 10â3Â Î©

The correct answer is:
12.5 Ã 10â3 Î©

Solution:
Ig=15 mAVg=75 mVI=6 A⇒ Rg=VgIg=5 Ω
S = IgRgI–Ig=12.5 × 10–3 Ω

Topic: Physics, Experimental Physics, NEET
A gas absorbs 200 J of heat and expands by 500  cm 3 against a constant pressure of 2×10 5   Nm −2 . The change in internal energy of the gas will be

Options:
A -200 J
B -100 J
C +100 J
D +200 J

The correct answer is:
+100 J

Solution:
ÎE=q+w
Here q=+200 J
w = −PΔV = −2 ×  10 5   Nm −2  × 500  cm 3  ×  1 m 100 cm 3
= −2 ×  10 5  × 500 ×  10 −6   Nm −2  ×  m 3
=−100 Nm=−100 J
So, ΔE=200 J−100 J
=+100 J

Topic: Chemistry, Thermodynamics (C), NEET
A gas at 350 K and 15 bar has a molar volume 20 per cent smaller than that for an ideal gas under the same conditions. The correct option about the gas, and its compressibility factor Z is

Options:
A Z>1 and repulsive forces are dominant.
B Z>1 and attractive forces are dominant.
C Z<1 and attractive forces are dominant.
D Z<1 and repulsive forces are dominant.

The correct answer is:
Z<1 and attractive forces are dominant.

Solution:
Z=PVnRT as V is 20% lesser than it would have been for an ideal gas.

∴PV will be lesser than expected.

∴Z<1

Since the volume is lesser, the attractive forces are dominant.

Topic: Chemistry, States of Matter, NEET, NEET 2019 (All India)
A gas can be compressed to a fraction of its volume. The same volume of a gas can be spread all over a room. The reason for this is that

Options:
A The volume occupied by molecules of a gas is negligible as compared to the total volume of the gas
B Gases consists of molecules which are in a state of random motion
C Gases consist of molecules having very large inter - molecular space which can be reduced or increased.
D None of these

The correct answer is:
Gases consist of molecules having very large inter - molecular space which can be reduced or increased.

Solution:
4FeO +  O 2 → 2Fe 2 O 2
FeO 3 +3CO → 400   o C− 700  o C 2Fe+3CO 2
Gases consist of molecules having very large inter - molecular space which can be reduced or increased.

Topic: Chemistry, States of Matter, NEET
A gas can be liquefied

Options:
A At any temperature
B Above its critical temperature
C At its critical temperature
D Below its critical temperature

The correct answer is:
Below its critical temperature

Solution:
The temperature below which the gas can be liquefied by the application of pressure alone is called critical temperature.

Topic: Chemistry, States of Matter, NEET
A gas can be taken from A to B via two different processes ACB and ADB.

When path ACB is used 60 J of heat flows into the system and 30 J of work is done by the system. If the path ADB is work done by the system is 10 J, the heat flows into the system in the path ADB is:

Options:
A 100 J
B 20 J
C 40 J
D 80Â J

The correct answer is:
40 J

Solution:
For path ACB,
ΔQ=ΔU+ΔW
60=ΔU+30
⇒ ΔU=30 J
For path ADB
ΔQ=ΔU+ΔW
As initial and find points are the same, change in internal energy is same in both the cases
ΔQ=30+10
=40 J

Topic: Physics, Thermodynamics, NEET
A gas cylinder containing cooking gas can withstand a pressure at 14.9 atm. The pressure gauge of the cylinder indicates 12 atm. at 27°C . Due to a sudden fire in the building its temperature starts rising. At what temperature will the cylinder explode?

Options:
A 372.5 K
B 3.725 K
C 37.25 K
D None of these

The correct answer is:
372.5 K

Solution:
T=273+27=300 K
P1T1=P2T2
Since volume of the cylinder is constant
T2=T1P2P1=300Ã14.912
=372.5 K

Topic: Chemistry, States of Matter, NEET
A gas does 4.5 J of external work during adiabatic expansion. If its temperature falls by 2 K, then its internal energy will be

Options:
A Increased by 4.5 J
B Decreased by 4.5 J
C Decreased by 2.25 J
D Increased by 9.0 J

The correct answer is:
Decreased by 4.5 J

Solution:
According to first law in thermodynamics dQ=dU+dW and for adiabatic dQ=0
dQ=0
dU=-dW
dU=-4.5 J

Topic: Physics, Thermodynamics, NEET
A gas expands adiabatically at constant pressure such that its temperature T∝1V . The value of CpCv of gas is

Options:
A 1.30
B 1.50
C 1.67
D 2.00

The correct answer is:
1.50

Solution:
Tâ1V
TV12= constant
For adiabatic process, TVÎ³-1= const.
Î³-1=12
â´ Î³=32=1.50

Topic: Physics, Thermodynamics, NEET
A gas expands from 3 dm3 to 5 dm3 against a constant pressure of 3 atm. The work done during expansion is used to heat 10 mole of water at temperature 290 K. Calculate final temperature of water. Specific heat of water = 4.184 J/ g/K.
[Given: 1 atm = 101.33 J]

Options:
A 290.81 K
B 290.61 K
C 290.41 K
D 290.21 K

The correct answer is:
290.81 K

Solution:
Work is done against constant P
ÎV=5-3=2 dm3=2 litre ;P=3 atm
â´W=-PÎV
=-3Ã2 litre atm
=−6×101.33=−608 J
Now, this work is used up in heating water
∴ W =n×C×ΔT
608=10Ã4.184Ã18ÎT
â´ÎT=0.81
â´Final temperature=T1+ÎT
=290+0.81=290.81 K

Topic: Chemistry, Thermodynamics (C), NEET
A gas expands with temperature according to the relation V=K T 2/3 , where K is a constant. Calculate work done when the temperature changes by 60 K?

Options:
A 10R
B 30R
C 40R
D 20R

The correct answer is:
40R

Solution:
dW⁡ = PdV = RT⁡ V⁡ dV⁡ … i
As V=K T 2/3 ∴ d V = K⁡ 2 3 T⁡ - 1 / 3 d⁡T⁡
∴ dV V⁡ = K⁡ 2 3 T⁡ - 1 / 3 d⁡ T⁡ KT 2 / 3 = 2 3 d⁡ T⁡ T⁡ … ii
From (i), W⁡ ∫ V⁡ 1 V⁡ 2 RT⁡ dV V⁡ = ∫ T⁡ 1 T⁡ 2 RT 2 3 d⁡ T⁡ T⁡ Using (ii)
W⁡ = 2 3 R⁡ T⁡ T⁡ 1 T⁡ 2
W⁡ = 2 3 R⁡ T⁡ 2 - T⁡ 1 = 2 3 R⁡ × 6 0 = 4 0R⁡

Topic: Physics, Thermodynamics, NEET
A gas for which γ=1.5 is suddenly compressed to the 14 th of the initial volume. Then the ratio of the final to the initial pressure is

Options:
A 1:6
B 1:8
C 1:4
D 8:1

The correct answer is:
8:1

Solution:
Here, γ=1.5, V2=14V1; p2p1=?
As compression is sudden/adiabatic,
∴ p2V2γ=p1V1γ
p2p1=V1V2γ=41.5=8

Topic: Physics, Thermodynamics, NEET
A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy ΔU of the gas in joules will be

Options:
A – 500 J
B – 505 J
C + 505 J
D 1136.25 J

The correct answer is:
– 505 J

Solution:
As q = 0 so using I law of Thermodynamics.
ΔU=-p.ΔV
=-2.5(4.5-2.5)=-5 lit atm
=-5×101.3=-506.5 Joule
=-505 J

Topic: Chemistry, Thermodynamics (C), NEET
A gas is allowed to expand in a well-insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy ΔU of the gas in joules will be

Options:
A 1136.25 J.
B -500 J.
C -505 J.
D +505 J.

The correct answer is:
-505 J.

Solution:
ΔU=q+w
For adiabatic process, q=0
∴ ΔU=w
=-P⋅ΔV
= -2.5 atm×4.5-2.5L
= -2.5×2 L-atm
= -5×101.3 J
=-506.5 J
≈-505 J

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2017
A gas is at 1 atm pressure with a volume of 800 cm3. When 100 J of heat is supplied to the gas, it expands to 1 L, at a constant pressure. The change in its internal energy is

Options:
A 80 J
B -80 J
C 20 J
D -20 J

The correct answer is:
80 J

Solution:
Work done by the gas at constant pressure
ΔW=PΔV
=1×105 N m-21000-800×10-6 m3=20 J
ΔU=ΔQ-ΔW
ΔU=100J-20J=80J

Topic: Physics, Thermodynamics, NEET
A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then:

Options:
A Compressing the gas isothermally will require more work to be done.
B Compressing the gas through adiabatic process will require more work to be done.
C Compressing the gas isothermally or adiabatically will require the same amount of work.
D Which of the case (whether compression through isothermal or through adiabatic process) requires more work will depend upon the atomicity of the gas.

The correct answer is:
Compressing the gas through adiabatic process will require more work to be done.

Solution:
The slope of adiabatic curve is more hence it lie above the isothermal curve. So in adiabatic process more work to be done.

Topic: Physics, Thermodynamics, NEET, NEET 2016 (Phase 1)
A gas is expanded from volume V0 to 2V0 under three different processes. In figure process 1 is an isobaric process, process 2 is isothermal and process 3 is adiabatic. Let âU1, âU2 and âU3 be the change in internal energy of the gas in these three processes. Then

Options:
A ∆U1>∆U2>∆U3
B ∆U1<∆U2<∆U3
C ∆U2<∆U1<∆U3
D ∆U2<∆U3<∆U1

The correct answer is:
∆U1>∆U2>∆U3

Solution:
Process 1 is isobaric (p= constant) expansion
Hence, temperature of gas will increase
∴ âU1= positive
Process 2 is an isothermal expansion
∴ âU2=0
Process 3 is an adiabatic expansion
Hence, temperature of gas will fall
∴ âU3= negative
∴ âU1>âU2>âU3

Topic: Physics, Thermodynamics, NEET
A gas is present in a cylinder fitted with movable piston. Above and below the piston there is equal number of moles of gas. The volume above is two times the volume below at a temperature of 300 K. At what temperature will the volume above be four times the volume below-

Options:
A 600Â K
B 400Â K
C 200Â K
D 120 K

The correct answer is:
120 K

Solution:
Let pressure given by piston=P0
PA+P0=PB
or P0=PB- PA
or P0=nR Ã 300V/3 â nR Ã 3002V/3
Let the final temperature=T
PBâ²=P0+PAâ²
or P0=PBâ²-PAâ²
or P0=nRTV/5- nRT4V/5
â´nRTÃ5V Ã 34=nRÃ300V Ã 3 Ã 12
or T Ã 154=300 Ã 32
or T=415 Ã 300 Ã 32=120 K

Topic: Physics, Thermodynamics, NEET
A gas is taken through the cycle $A \rightarrow B \rightarrow C \rightarrow A$, as shown. What is the net work done by the gas?

Options:
A $2000 \mathrm{~J}$
B $1000 \mathrm{~J}$
C Zero
D $-2000 \mathrm{~J}$

The correct answer is:
$1000 \mathrm{~J}$

Solution:
Net work done = Area enclased in $p V$ curve, i.e., $\triangle \mathrm{ABC}$
$\begin{aligned}
& =\frac{1}{2} \times 5 \times 10^{-3} \times 4 \times 10^5 \mathrm{~J} \\
& =10^3 \mathrm{~J} \\
& =1000 \mathrm{~J}
\end{aligned}$

Topic: Physics, Thermodynamics, NEET, NEET 2013 (All India)
A gas mixture consist of 2 moles of Oxygen and 4 moles of Argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system is

Options:
A 4RT
B 15RT
C 9RT
D 11RT

The correct answer is:
11RT

Solution:
Total internal energy of system
U=Uoxygen+Uargon
U=n1f12RT+n2f22RT
Degree of freedom is f1=5 (for Oxygen) and f2=3 (for Argon)
U=252RT+432RT
U=5RT+6RT
U=11RT

Topic: Physics, Kinetic Theory of Gases, NEET
A gas mixture consist of 2 mole of oxygen and 4 mole of argon at temperature T . Neglecting all vibrational modes, the total internal energy of the system is -

Options:
A 4 RT
B 15Â RT
C 9Â RT
D 11 RT

The correct answer is:
11 RT

Solution:
Total internal energy of system
=UOxygen+UArgon=μ1f12RT+μ2f22RT
=252RT+432RT=5RT+6RT=11RT
As f1 = 5 for Oxygen and f2 = 3 for Argon

Topic: Physics, Thermodynamics, NEET
A gas mixture consists of 2 moles of O2 and 4 moles of Ar at temperature T. Neglecting all vibrational modes, the total internal energy of the system is ( R is universal gas constant)

Options:
A 4Â RT
B 15Â RT
C 9Â RT
D 11 RT

The correct answer is:
11 RT

Solution:
For a gas having n moles of molecules with each f degrees of freedom at temperature T K, internal energy is U=12nfRT
â´ Total internal energy is
U=n1f12 RT+n2f22 RT
=2Ã52 RT+4Ã32RT
=5RT+6RT
U=11RT

Topic: Physics, Thermodynamics, NEET, NEET 2017
A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system is

Options:
A 4RT
B 15RT
C 9RT
D 11RT

The correct answer is:
11RT

Solution:
Internal energy of a gas is given by,
U=f2nRT
Total internal energy of the system is,
U=UO2+UAr
U=522RT+324RT
U=11RT

Topic: Physics, Thermodynamics, NEET
A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting all vibrational moles, the total internal energy of the system is

Options:
A 4 RT
B 15 RT
C 9 RT
D 11 RT

The correct answer is:
11 RT

Solution:
Oxygen being a diatomic gas possesses 5 degrees of freedom, 3 translational and 2 rotational.
Argon being monoatomic has 3 translational degrees of freedom.
Total energy of the system
=Eoxygen+Eargon
=n1f112RT+n2f2(12RT)
=2×5×12RT+4×3×12RT
=5RT+6RT=11RT

Topic: Physics, Kinetic Theory of Gases, NEET
A gas mixture used by deep sea divers to respirate is

Options:
A Oxygen and helium
B Oxygen and argon
C Oxygen and hydrogen
D Oxygen and nitrogen

The correct answer is:
Oxygen and helium

Solution:
An oxygen-helium mixture is used artificial respiration in deep sea diving instead of air because nitrogen present in air dissolves in blood under high pressure when sea diver goes into deep sea. When he comes to the surface, nitrogen bubbles out of the blood due to decrease in pressure, causing pains. This disease is called “bends”.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET