datavorous/entrance-exam-dataset huggingface dataset:
A galvanometer having a resistance of $20 \Omega$ and 30 division 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 volt, is:

Options:
A $100 \Omega$
B $120 \Omega$
C $80 \Omega$
D $125 \Omega$

The correct answer is:
$80 \Omega$

Solution:
Deflection current

$=\operatorname{Ig}_{\max }=n x k=0.005 \times 30$

Where, $\mathrm{n}=$ Number of divisions $=30$ and $\mathrm{k}=0.005 \mathrm{amp} /$

division

$\begin{array}{l}

=15 \times 10^{-2}=0.15 \\

\mathrm{v}=\mathrm{I}_{\mathrm{g}}[20+\mathrm{R}] \\

15=0.15[20+\mathrm{R}] \\

100=20+\mathrm{R} \\

\mathrm{R}=80 \Omega

\end{array}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (11 Jan Shift 2)
A galvanometer having coil resistance 10 Ω shows a full scale deflection for a current of 3 mA. For it to measure a current of 8 A, the value of the shunt should be:

Options:
A 3×10-3Ω
B 4.85×10-3Ω
C 3.75×10-3Ω
D 2.75×10-3Ω

The correct answer is:
3.75×10-3Ω

Solution:
Given G=10 Ω, Ig=3 mA and I=8 A.

Let's consider the following figure:

From the figure, we have

IgG=I-IgS⇒S=IgGI-Ig⇒S=3×10-3108-0.003=3.75×10-3 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (29 Jan Shift 1)
A galvanometer having full scale deflection current of 10 mA and resistance 8 Î© is joined in circuit as shown. When the terminals A and B are used and C remains open, the range of the ammeter is I. When the terminals A and C are used and B remains open, then the range of the ammeter is

Options:
A I
B I2
C I3
D 2I3

The correct answer is:
I2

Solution:
When A and B is used
Ig=2⋅I2+10
When A and C is used
Ig=4⋅I′4+8
2I12=4I′12
I′=I2

Topic: Physics, Magnetic Properties of Matter, JEE Main
A galvanometer is used in laboratory for detecting the null point in electrical experiments. If, on passing a current of 6mA it produces a deflection of 2°, its figure of merit is close to :

Options:
A 333° A/div
B 6×10-3A/div
C 666° A/div.
D 3×10-3A/div

The correct answer is:
3×10-3A/div

Solution:
Current sensitivity =1θ=6×10-32=3×10-3

Topic: Physics, Current Electricity, JEE Main, JEE Main 2020 (05 Sep Shift 2)
A galvanometer of $10 \mathrm{ohm}$ resistance gives full scale deflection with 0.01 ampere of current. It is to be converted into an ammeter for measuring 10 ampere current. The value of shunt resistance required will be

Options:
A $\frac{10}{999} \mathrm{ohm}$
B $0.1 \mathrm{ohm}$
C $0.5 \mathrm{ohm}$
D $1.0 \mathrm{ohm}$

The correct answer is:
$\frac{10}{999} \mathrm{ohm}$

Solution:
$S=\frac{i_g G}{i-i_g}=\frac{10 \times 0.01}{10-0.01}=\frac{10}{999} \mathrm{ohm}$

Topic: Physics, Current Electricity, JEE Main
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, JEE Main
A galvanometer of resistance 25 Î© is connected to a battery of 2 V along with a resistance of 3000 Î©. In this case, a full-scale deflection of 30 units is obtained in the galvanometer. In order to reduce this deflection to 10 units, how much more resistance (in Î©) should be added to the circuit in series?

Options:

The correct answer is:
6050

Solution:
The deflection in a galvanometer is directly proportional to the current.

First case:

30x=225+3000

Second case:

10x=225+3000+R

â3=25+3000+R25+3000

On solving, we get

R=6050 Î©

Topic: Physics, Experimental Physics, JEE Main
A galvanometer of resistance 30 Î© is connected to a battery of emf 2 V with 1970 Î© resistance in series. A full-scale deflection of 20 divisions is obtained in the galvanometer. To reduce the deflection to 10 divisions, the resistance value (in ohm) in series required is

Options:

The correct answer is:
3970

Solution:
i=VReff=V1970+30=22000=1Ã10-3A=1 mA
This current provides full scale deflection (i.e. 20 division) In order to limit the deflection 10 divisions,
the resistances needed to connect can be obtained as
Î¸=niABk â´ Î¸1Î¸2=i1i2
â´ Î¸1 :Î¸2=2
â´ i1:i2=2:1
â´ i2=i12=1Ã10-32=5Ã10-4mA
â´ i=VReff+RsâRs=Vi-Reff
=25Ã10-4-2000
=25Ã104-2000
=4Ã103-2000
=2000 Î©
â´ The resistance of 1970 Î© is to be replaced by 1970+2000=3970 Î©

Topic: Physics, Current Electricity, JEE Main
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 $4450 \Omega$
B $5050 \Omega$
C $5550 \Omega$
D $6050 \Omega$

The correct answer is:
$4450 \Omega$

Solution:
$$
\begin{gathered}
30 i_0=\frac{V}{R_g+2950} ; R_g=50 \Omega \\
20 i_0=\frac{V}{R_g+R} \Rightarrow R=4450 \Omega
\end{gathered}
$$

Topic: Physics, Current Electricity, NEET, NEET 2008 (Mains)
A galvanometer of resistance 50 Î© is connected to a battery of 3 V along with a resistance of 2950 Î© 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 Ω
B 5550 Ω
C 6050 Ω
D 4450 Ω

The correct answer is:
4450 Ω

Solution:
Current through the galvanometer
I=3(50+2950)=10-3 A
Current for 30 divisions =10-3 A
Current for 20 divisions
=10-330Ã20=23Ã10-3 A

For the same deflection to obtain for 20 divisions, let resistance added be R
â´ 23Ã10-3=350+R
or R=4450 Î©

Topic: Physics, Experimental Physics, JEE Main
A galvanometer of resistance 50 Î© is converted into an ammeter by connecting a low resistance (shunt) of value 1 Î© in parallel to the galvanometer, S. If full-scale deflection current of the galvanometer is 10 mA, then the maximum current that can be measured by the ammeter is -

Options:
A 500Â mA
B 510 mA
C 610Â mA
D 490Â mA

The correct answer is:
510 mA

Solution:
S=GIgIâIg
Shunt and Galvnometer are in parallel combination
1=50Ã10Iâ10
I-10=500
I=510 mA

Topic: Physics, Current Electricity, JEE Main
A galvanometer of resistance $100 \Omega$ gives full-scale defection for $10 \mathrm{~mA}$ current. What should be the value of the shunt so that it can measure a current of $100 \mathrm{~mA}$ ?

Options:
A $11.11 \Omega$
B $9.9 \Omega$
C $1.1 \Omega$
D $4.4 \Omega$

The correct answer is:
$11.11 \Omega$

Solution:
Given,
Resistance of a galvanometer, $G=100 \Omega$,
Current at full-scale deflection, $\mathrm{i}_{\mathrm{g}}=10 \mathrm{~mA}$
Measured current, $\mathrm{i}=100 \mathrm{~mA}$
The value of the shunt is calculated as, $\frac{R}{Q}=\frac{i_3}{i-i_g}$
$\begin{aligned}
& \Rightarrow R=\frac{G_y}{i-i_y} \\
& R=\frac{100 \times 10 \times 10^{-3}}{(100-10) \times 10^{-3}}=11.11 \Omega
\end{aligned}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
A galvanometer of resistance 100 Ω has 50 divisions on its scale and has sensitivity of 20μA/ division . It is to be converted to voltmeter with three ranges, of 0-2 V,0-10 V and 0-20 V. The appropriate circuit to do so is:

Options:
A
B
C
D

The correct answer is:

Solution:
Given, the resistance of the galvanometer G=100 Ω
No. of divisions =50
Current Sensitivity =20 μA/division

The maximum current through galvanometer will be
Imax=50×20×10-6=10-3 A
For V1 = 2 V

using Ohm's law V1= Imax G+R1
2=10-3 ×100+R1
∴R1=1900 Ω

For V2 = 10 V

using Ohm's law V2 =ImaxG+R1+R2

10=10-3 ×100 + 1900+R2
∴R2=8000 Ω

For V3 =20 V

using Ohm's law V3 =Imax G+R1+R2+R3
20=10-3 100 +1900+8000+R3
∴R3=10×103 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (12 Apr Shift 1)
A galvanometer of resistance $200 \Omega$ is to be converted into an ammeter. The value of shunt resistance which allows $3 \%$ of the main current through the galvanometer is equal to (nearly)

Options:
A $7 \Omega$
B $5 \Omega$
C $10 \Omega$
D $6 \Omega$

The correct answer is:
$6 \Omega$

Solution:
Given:
Resistance of galvanometer is $G=200 \Omega$.
Current through the galvanometer is $i_g=3 \%$ of $i$.
To convert into an ammeter, the galvanometer with resistance $G$ has to be shunted with $S$. The potential drop across the galvanometer and shunt is the same:
$\begin{aligned} & i_g G=\left(i-i_g\right) S \\ & \Rightarrow \frac{i}{i_g}=1+\frac{G}{S}\end{aligned}$
Given, $\frac{i_g}{i}=\frac{3}{100}$
$\begin{aligned} & \therefore \frac{100}{3}=1+\frac{200}{S} \\ & \Rightarrow S=\frac{200 \times 3}{97} \approx 6 \Omega\end{aligned}$

Topic: Physics, Current Electricity, MHT CET, MHT CET 2022 (10 Aug Shift 2)
A galvanometer of resistance $240 \Omega$ allows only $4 \%$ of the main current after connecting a shunt resistance. The value of the shunt resistance is

Options:
A $10 \Omega$
B $20 \Omega$
C $8 \Omega$
D $5 \Omega$

The correct answer is:
$10 \Omega$

Solution:
Given, galvanometer resistance $G=240 \Omega$
Shunt resistance $S=$ ?
$$
\mathrm{I}_{\mathrm{G}}=\frac{4}{100} \mathrm{I}
$$
From figure voltage through the circuit.
$$
\left(\mathrm{I}-\mathrm{I}_{\mathrm{G}}\right) \mathrm{S}=\mathrm{I}_{\mathrm{G}} \mathrm{G}
$$
or $\quad\left(1-\frac{4 \mathrm{I}}{100}\right) \mathrm{S}=\frac{4 \mathrm{I}}{100} \times 240$
or
$$
S=\frac{4 \times 240}{96}=10 \Omega
$$

Topic: Physics, Current Electricity, KCET, KCET 2009
A galvanometer of resistance G is converted into a voltameter of range 0-1 V by connecting a resistance R in series with it. The additional resistance that should be connected in series with R1 to increase the range of the voltmeter to 0-2V will be :

Options:
A G
B R1
C R1-G
D R1+G

The correct answer is:
R1+G

Solution:
igR1+G=1

igR1+G+R2=2

1R1+GR1+G+R2=2

R1+G+R2=2R1+2G

R2=R1+G

Topic: Physics, Current Electricity, JEE Main, JEE Main 2020 (05 Sep Shift 1)
A galvanometer of resistance ' $G$ ' is shunted by a resistance of $5 \Omega$. To keep the main current in the circuit unchanged, the resistance to be put in series with the same galvanometer is

Options:
A $\frac{\mathrm{G}^2}{\mathrm{~S}+\mathrm{G}}$
B $\frac{5 G}{S+G}$
C $\frac{G}{S+G}$
D $\frac{S^2}{S+G}$

The correct answer is:
$\frac{\mathrm{G}^2}{\mathrm{~S}+\mathrm{G}}$

Solution:
Current will be unchanged if resistance remains same to
$\begin{aligned} & G=\frac{G S}{G+S}+R \\ & \Rightarrow R=G-\frac{G S}{G+S}=\frac{G^2}{G+S}\end{aligned}$

Topic: Physics, Current Electricity, MHT CET, MHT CET 2022 (07 Aug Shift 1)
A galvanometer of resistance, $G$ is shunted by a resistance $S \mathrm{~ohm}$. To keep the main current in the circuit unchanged, the resistance to be put in series with the galvanometer is

Options:
A $\frac{S^2}{(S+G)}$
B $\frac{S G}{(S+G)}$
C $\frac{G^2}{(S+G)}$
D $\frac{G}{(S+G)}$

The correct answer is:
$\frac{G^2}{(S+G)}$

Solution:
Current will be unchanged if resistance remains same so
$$
\begin{aligned}
& G=\frac{G S}{G+S}+R \\
& R=G-\frac{G S}{G+S} \\
& R=\frac{G^2}{G+S}
\end{aligned}
$$

Topic: Physics, Current Electricity, NEET, NEET 2011 (Mains)
A galvanometer of resistance, $\mathrm{G}$ is shunted by a resistance S ohm. To keep the main current in the circuit unchanged, the resistance to be put in series with the galvanometer is

Options:
A $\frac{\mathrm{S}^{2}}{(\mathrm{~S}+\mathrm{G})}$
B $\frac{\mathrm{SG}}{(\mathrm{S}+\mathrm{G})}$
C $\frac{\mathrm{G}^{2}}{(\mathrm{~S}+\mathrm{G})}$
D $\frac{G}{(S+G)}$

The correct answer is:
$\frac{\mathrm{G}^{2}}{(\mathrm{~S}+\mathrm{G})}$

Solution:
If resistance remains same so current will be unchanged.
$$
\begin{array}{l}
\mathrm{G}=\frac{\mathrm{GS}}{\mathrm{G}+\mathrm{S}}+\mathrm{R} \Rightarrow \mathrm{R}=\mathrm{G}-\frac{\mathrm{GS}}{\mathrm{G}+\mathrm{S}} \\
\text { or, } \mathrm{R}=\frac{\mathrm{G}^{2}}{\mathrm{G}+\mathrm{S}}
\end{array}
$$

Topic: Physics, Current Electricity, VITEEE, VITEEE 2012
A galvanometer of resistance $G \Omega$, is shunted by a resistance $S \Omega$. To keep the main current in the circuit unchanged, the resistance to be connected in series with the galvanometer is

Options:
A $\frac{G^2}{S+G}$
B $\frac{S}{S+G}$
C $\frac{S^2}{S+G}$
D $\frac{S G}{S+G}$

The correct answer is:
$\frac{G^2}{S+G}$

Solution:
Given, galvanometer resistance, $R_G=G$ and shunt resistance, $R_S=S$,
Let the series resistance is $R$, so the net resistance of circuit after connected $R$,
Since, it is given that the current should not change. So, it should be $G$,
From Eq. (i) and (ii), we get
$$
\begin{aligned}
G & =R+\frac{G \times S}{G+S} \\
\Rightarrow \quad R & =G-\frac{G S}{G+S} \Rightarrow R=\frac{G^2}{G+S}
\end{aligned}
$$
Hence, the correct option is (a).

Topic: Physics, Current Electricity, AP EAMCET, AP EAMCET 2019 (21 Apr Shift 1)
A galvanometer with its coil resistance 25 Ω requires a current of 1 mA for its full deflection. In order to construct an ammeter to read up to a current of 2 A the approximate value of the shunt resistance should be:

Options:
A 1.25 ×10-2 Ω
B 2.5 Ã10-3 Î©
C 2.5 Ã10-2 Î©
D 1.25 Ã10-3 Î©

The correct answer is:
1.25 ×10-2 Ω

Solution:
From the above figure, it is clear that, IgRg=(I- Ig)S
S ≃10-3×252⇒S ≃12.5 ×10-3
⇒1.25 ×10-2Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2018 (16 Apr Online)
A gas absorbs a photon of $355 \mathrm{~nm}$ and emits at two wavelengths. If one of the emissions is at $680 \mathrm{~nm}$, the other is at:

Options:
A
$325 \mathrm{~nm}$

B
$743 \mathrm{~nm}$

C
$518 \mathrm{~nm}$

D
$1035 \mathrm{~nm}$

The correct answer is:
$743 \mathrm{~nm}$

Solution:
$\frac{1}{\lambda_{\text {absorted }}}=\frac{1}{\lambda_1}+\frac{1}{\lambda_2}$
$\Rightarrow \frac{1}{355}=\frac{1}{680}+\frac{1}{\lambda_2}$
$\Rightarrow \lambda_2=742.8 \cong 743 \mathrm{~nm}$

Topic: Chemistry, Structure of Atom, JEE Main
A gas absorbs a photon of $355 \mathrm{~nm}$ and emits at two wavelengths. If one of the emissions is at $680 \mathrm{~nm}$, the other is at:

Options:
A
$325 \mathrm{~nm}$

B
$743 \mathrm{~nm}$

C
$518 \mathrm{~nm}$

D
$1035 \mathrm{~nm}$

The correct answer is:
$743 \mathrm{~nm}$

Solution:
$\frac{1}{\lambda_{\text {absorted }}}=\frac{1}{\lambda_1}+\frac{1}{\lambda_2}$
$\Rightarrow \frac{1}{355}=\frac{1}{680}+\frac{1}{\lambda_2}$
$\Rightarrow \lambda_2=742.8 \cong 743 \mathrm{~nm}$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2011
A gas at NTP is suddenly compressed to one-fourth of its original volume. If $\gamma$ is supposed to be $\frac{3}{2}$, then the final pressure is

Options:
A 4 atmosphere
B $\frac{3}{2}$ atmosphere
C 8 atmosphere
D $\frac{1}{4}$ atmosphere

The correct answer is:
8 atmosphere

Solution:
$\begin{aligned} & P V^\gamma=\text { constant } \Rightarrow \frac{P_2}{P_1}=\left(\frac{V_1}{V_2}\right)^\gamma \Rightarrow \frac{P_2}{1}=\left(\frac{V_1}{V_1 / 4}\right)^{3 / 2}=8 \\ & \Rightarrow P_2=8 \mathrm{~atm} .\end{aligned}$

Topic: Physics, Thermodynamics, JEE Main
A gas can be liquefied

Options:
A Above its critical temperature
B At its critical temperature
C Below its critical temperature
D At any 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, JEE Main
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 used then 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, JEE Main, JEE Main 2019 (09 Jan Shift 1)
A gas expands adiabatically at constant pressure such that
T∝1V
The value of γ , i.e., CPCV of the gas will be

Options:
A 1.30
B 1.50
C 1.70
D 2

The correct answer is:
1.50

Solution:
T∝1V
TV12= constant
For adiabatic process, TV12=TVγ-1= constant
So γ-1=12
γ=32 .

Topic: Chemistry, Thermodynamics (C), JEE Main
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), JEE Main
A gas formed by the action of alcoholic $\mathrm{KOH}$ on ethyl iodide, decolourises alkaline $\mathrm{KMnO}_{4}$. The gas is

Options:
A $\mathrm{C}_{2} \mathrm{H}_{6}$
B $\mathrm{CH}_{4}$
C $\mathrm{C}_{2} \mathrm{H}_{2}$
D $\mathrm{C}_{2} \mathrm{H}_{4}$

The correct answer is:
$\mathrm{C}_{2} \mathrm{H}_{4}$

Solution:
By the action of alc. $\mathrm{KOH}$ on ethyl iodide, ethylene gas is formed by elimination reaction. Ethylene decolourises alkaline $\mathrm{KMnO}_{4}$.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
A gas has molar heat capacity C=37.55 J mol-1 K-1, in this process pT= constant. The number of degree of freedom of the molecule of gas is

Options:
A 2
B 3
C 5
D 7

The correct answer is:
5

Solution:
Given, C=37.55 J mol-1 K-1

Also, PT= constant K

According to ideal gas equation, PV=RT

⇒ p=RTV .....i

Putting the value of p in Eq. (i), we get

RTV×T=K ⇒ V=RT2K .....ii

On differentiating above equation both sides, we get

dVdT=2RTK....iii
But TK=1p fromEq. i
Hence, Eq. iiibecomes

dVdT=2 Rp
So, C=CV+pdVdT
or C=CV+p×2 Rp=CV+2 R
or CV=C-2 R ...iv

As CV=nR2, where n= number of degrees of freedom.
Putting the value of CV in Eq. (iv), we get

nR2=C-2 R

n=2 C-2 RR=2 37.55-2×8.38.3=5.048≈5

Topic: Physics, Kinetic Theory of Gases, JEE Main
A gas in an airtight container is heated from 25 Â°C to 90 Â°C. The density of the gas will

Options:
A increase slightly
B increase considerably
C remain the same
D decrease slightly

The correct answer is:
remain the same

Solution:
As mass and volume of the gas in an airtight container will remain same, so density will also remain same.

Topic: Physics, Thermodynamics, NEET
A gas is allowed to expand at constant temperature from a volume of 1.0 L to 10.1 L against an external pressure of 0.50 atm. If the gas absorbs 250 J of heat from the surroundings, what are the values of q, w and ΔE? (Given 1 L atm = 101.0 J)

Options:
A

q	w	ΔE

250 J	- 460 J	- 210 J

B

q	w	ΔE
-250 J	- 460 J	- 710 J

C

q	w	ΔE
250 J	460 J	710 J

D

q	w	ΔE
-250 J	460 J	210 J

The correct answer is:

q	w	ΔE

250 J	- 460 J	- 210 J

Solution:
Since heat is absorbed, q=+250 J
Work done =-pV2-V1
=-50 atmÃ10.1 L-1.0 L
=-0.50 atmÃ9.1 L
=-0.50Ã9.1 L atmÃ101.0 J1 L atm=-460.0 J
Also ÎE=q+w=250 J-460 J=-210 J

Topic: Chemistry, Thermodynamics (C), JEE Main
A gas is compressed adiabatically, which one of the following statement is NOT true?

Options:
A There is no heat supplied to the system
B There is no change in the internal energy
C

The temperature of the gas increases

D The change in the internal energy is equal to the work done on the gas

The correct answer is:
There is no change in the internal energy

Solution:
In adiabatic process,

∆Q=0 and ∆U≠0

Using the first law of thermodynamics, ∆Q=∆U+W

⇒∆U=-W W is the work done by the gas, so -W will be the work done on the gas

As the gas is compressed

W <0

Therefore, ∆U >0 and hence ∆T >0.

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2023 (10 Apr Shift 2)
A gas is compressed at a constant pressure of $50 \mathrm{~N} / \mathrm{m}^2$ from a volume of $10 \mathrm{~m}^3$ to a volume of $4 \mathrm{~m}^3$. Energy of $100 \mathrm{~J}$ then added to the gas by heating. Its internal energy is

Options:
A Increased by $400 \mathrm{~J}$
B Increased by $200 \mathrm{~J}$
C Increased by $100 \mathrm{~J}$
D Decreased by $200 \mathrm{~J}$

The correct answer is:
Increased by $400 \mathrm{~J}$

Solution:
$\begin{aligned} & \text { From FLOT } \Delta Q=\Delta U+\Delta W=\Delta U+P \Delta V \\ & \qquad \Rightarrow 100=\Delta U+50 \times(4-10) \Rightarrow \Delta U=400 \mathrm{~J}\end{aligned}$

Topic: Physics, Thermodynamics, JEE Main
A gas is expanded from an initial state to a final state along a path that consists of (a) an isothermal expansion doing $40 \mathrm{~J}$ work, (b) an adiabatic expansion doing $\mathrm{W}$ work, (c) an isothermal expansion doing $30 \mathrm{~J}$ work. If the total change in the internal energy of the gas is $-20 \mathrm{~J}$, the work done by the gas during the adiabatic expansion $\mathrm{W}=$

Options:
A $50 \mathrm{~J}$
B $90 \mathrm{~J}$
C $70 \mathrm{~J}$
D $20 \mathrm{~J}$

The correct answer is:
$20 \mathrm{~J}$

Solution:
Isothermal Expansion,
In this process temperature remains constant.
$\therefore \Delta \mathrm{V}=0$
Using 1st law of thermodynamics
$\Delta \mathrm{Q}_1=\Delta \mathrm{V}_1+\Delta \mathrm{W}_1$
$\Delta \mathrm{Q}_1=\Delta \mathrm{W}_1=40 \mathrm{~J}$ ... (1)
Adiabatic expansion,
In this process heat remains constant,
$\Delta Q_2=0$
$\Delta \mathrm{U}_2=-\mathrm{W}$ ... (2)
Isothermal expansion,
$\begin{aligned} & \Delta \mathrm{W}_3=30 \mathrm{~J} \\ & \Delta \mathrm{U}_3=0\end{aligned}$
$\Delta \mathrm{Q}_3=\Delta \mathrm{W}_3=30 \mathrm{~J}$ ...(3)
Change in internal energy, $\mathrm{V}=-20 \mathrm{~J}$
It can be written as
$\begin{aligned} & \mathrm{U}_1+\mathrm{U}_2+\mathrm{U}_3=-20 \mathrm{~J} \\ & 0+\mathrm{U}_2+0=-20 \quad \mathrm{U}_2=-20 \mathrm{~J}\end{aligned}$
Now, using equation (2),
$\begin{aligned} & \mathrm{W}=-\mathrm{V}_2=20 \mathrm{~J} \\ & \mathrm{~W}=20 \mathrm{~J}\end{aligned}$

Topic: Physics, Thermodynamics, AP EAMCET, AP EAMCET 2023 (18 May Shift 1)
A gas is filled in a cylinder. Its temperature is increased by 20% on the Kelvin scale and volume is reduced by 10%. How much percentage of the gas has to leak for pressure to remain constant?

Options:
A 30%
B 40%
C 15%
D 25%

The correct answer is:
25%

Solution:
V ∝μT, 90100V∝μ′6T5
μ′μ×65=910 or μ′μ=910×56=34
Percentage of gas leaked out
=μ-μ′μ×100=4-34×100=25%

Topic: Physics, Kinetic Theory of Gases, NEET
A gas is filled in a cylinder. Its temperature is increased by 20% on the Kelvin scale and volume is reduced by 10%. How much percentage of the gas has to leak for pressure to remain constant?

Options:
A 30%
B 40%
C 15%
D 25%

The correct answer is:
25%

Solution:
V ∝μT, 90100V∝μ′6T5
μ′μ×65=910 or μ′μ=910×56=34
Percentage of gas leaked out
=μ-μ′μ×100=4-34×100=25%

Topic: Physics, Thermodynamics, JEE Main
A gas is heated in such a way that its pressure and volume both becomes double. Then by decreasing temperature, some of the molecules of air has been added in container to maintain the doubled volume and pressure. Assuming 14th of initial number of moles has been added in for the purpose. The fraction by which the temperature must have been raised finally of initial absolute temperature?

Options:
A 4Â times
B 165 times
C 45times
D 15Â times

The correct answer is:
165 times

Solution:
I initially : Pressure P volume V temp T moles x.
II on 2P, 2V then temperature heating becomes 4T After decreasing temperature :
2P Ã 2V = n Ã R Ã 4T
III Pressure volume Temperature moles 2P, 2V, Tâ² (assume) n+n4=5n4
III case
2P Ã 2V=5n4 Ã R Ã Tâ² â Tâ²=165 T

Topic: Chemistry, States of Matter, NEET