datavorous/entrance-exam-dataset huggingface dataset:
For a system to follow the law of conservation of linear momentum during a collision, the condition is :
(1) total external force acting on the system is zero.
(2) total external force acting on the system is finite and time of collision is negligible.
(3) total internal force acting on the system is zero.

Options:
A (1) only
B (2) only
C (3) only
D (1) or (2)

The correct answer is:
(1) only

Solution:
From Newton's second law,
$F=\frac{d p}{d t}$
If $F=0$, then $\frac{d p}{d t}=0$
$\Rightarrow \quad p=$ constant
Thus, if total external force acting on the system is zero, then linear momentum of the system remains conserved.

Topic: Physics, Rotational Motion, AP EAMCET, AP EAMCET 2006
For a system to follow the law of conservation of linear momentum during a collision, the condition is :
(1) total external force acting on the system is zero.
(2) total external force acting on the system is finite and time of collision is negligible.
(3) total internal force acting on the system is zero.

Options:
A (1) only
B (2) only
C (3) only
D (1) or (2)

The correct answer is:
(1) only

Solution:
From Newton's second law,
$F=\frac{d p}{d t}$
If $F=0$, then $\frac{d p}{d t}=0$
$\Rightarrow \quad p=$ constant
Thus, if total external force acting on the system is zero, then linear momentum of the system remains conserved.

Topic: Physics, Rotational Motion, TS EAMCET, TS EAMCET 2006
For a transistor amplifier in common emitter configuration having load impedance of $1 \mathrm{k} \Omega\left(\mathrm{h}_{\mathrm{fe}}\right.$ $=50$ and $h_{o e}=25$ ) the current gain is

Options:
A
$-5.2$

B
$-15.7$

C
$-24.8$

D
$-48.78$

The correct answer is:
$-48.78$

Solution:
In CE configuration, $A_{\mathrm{i}}=\frac{-\mathrm{h}_{\mathrm{fe}}}{1+\mathrm{h}_{0 \mathrm{e}} R_{\mathrm{L}}}$
$=\frac{-50}{1+25 \times 10^{-6} \times 1 \times 10^3}=-48.78$

Topic: Physics, Semiconductors, JEE Main, JEE Main 2004
For a transistor amplifier, the voltage gain
(a) remains constant for all frequencies.
(b) is high at high and low frequencies and constant in the middle frequency range.
(c) is low at high and low frequencies and constant at mid frequencies.
(d) none of the above.

Options:

None

Solution:
(c)

Topic: Physics, Semiconductors
For a transistor amplifier, the voltage gain

Options:
A is low at high and low frequencies and constant at mid frequencies
B remains constant for all frequencies
C constant at high frequencies and low at low frequencies
D is high at high and low frequencies and constant in the middle frequency range

The correct answer is:
is low at high and low frequencies and constant at mid frequencies

Solution:
(A)

Topic: Physics, Semiconductors, KCET, KCET 2019
For a transistor α and β are given as α=ICIE and β=ICIB. Then the correct relation between α and β will be:

Options:
A α=1-ββ
B αβ=1
C β=α1-α
D α=β1-β

The correct answer is:
β=α1-α

Solution:
In the case of the transistor the relation between emitter current, base current and, the collector current is, iE=iB+iC, ⇒iEiC=iBiC+1,

⇒1α=1β+1

⇒β=α1-α

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (27 Aug Shift 2)
For a transistor in CE mode to be used as an amplifier, it must be operated in:

Options:
A Cut-off region only
B Saturation region only
C The active region only
D Both cut-off and Saturation

The correct answer is:
The active region only

Solution:
Active region of the CE transistor is linear region and is best suited for its use as an amplifier.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (27 Aug Shift 1)
For a transistor, the current amplification factor Î± is 0.8. If it is connected in common emitter configuration, the change in collector current when the base current changes by 6 mA is

Options:
A 6 mA
B 4.8 mA
C 24 mA
D 8 mA

The correct answer is:
24 mA

Solution:
Ic=α1-αIb=0.80.2Ib=4Ib
Ic=collector current
Ib=base current
Ic=24 mA

Topic: Physics, Semiconductors, JEE Main
For a transistor to act as a switch, it must be operated in

Options:
A Active region
B Saturation and cut-off region
C Cutoff region only
D Saturated region only

The correct answer is:
Saturation and cut-off region

Solution:
The saturation zone and cut-off area are known as the transistor switch’s working regions. This implies that, by switching between its “top-off” (saturation) and “absolute OFF,” the transistor is used as a switch.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2022 (27 Jun Shift 2)
For a transvers wave travelling, along a straight line, the distance between two peaks (crests) is 5 m, while the distance between one crest and one trough is 1.5 m. The possible wavelengths (in m) of the waves are:

Options:
A 1, 3, 5
B 11,13,15, . . . . .
C 1, 2, 3, . . . .
D 12,14,16, . . . . .

The correct answer is:
11,13,15, . . . . .

Solution:
Trough to crest distance

1.5=2n1+1λ2 ......(1)

Trough to trough distance

5=n2λ .....(2)

from (1) and (2)

1.55=2n1+12n2

3n2=10n1+5

n1 and n2 are integer

(1) n1=1, n2=5, λ=1

(2) n1=4, n2=15, λ=1/3

(3) n1=7, n2=25, λ=1/5

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2020 (04 Sep Shift 1)
For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through O (the centre of mass) and O' (corner point) is:

Options:
A 2/3
B 1/4
C 1/8
D 1/2

The correct answer is:
1/4

Solution:
We will use Parallel Axes Theorem here,

Distance between OO' is,

OO'=12a2+b2=a24+b24

So, according to parallel axis theorem we have,

IOIO'=M12a2+b2M12a2+b2+Ma24+b24=M12a2+b2M3a2+b2=14

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2020 (04 Sep Shift 2)
For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance h from its centre. Then value of h is:

Options:
A R2
B R5
C R2
D R

The correct answer is:
R2

Solution:
Electric field at a distance h from centre on axis
E=14πε0QhR2+h232
To get maximum E, dEdh=0
0=14πε0QR2+h232-Qh32R2+h2122hR2+h23
⇒h=±R2

Topic: Physics, Electrostatics, JEE Main, JEE Main 2019 (09 Jan Shift 1)
For a uniformly charged thin spherical shell, the electric potential V radially away from the centre O of shell can be graphically represented as

Options:
A
B
C
D

The correct answer is:

Solution:
The potential is constant inside the sphere as the electric field is zero inside the sphere. So, the value is given by

V=Q4πε0R for r<R

The potential at the surface is

V=Q4πε0R

The potential outside the sphere starts decreasing and is given by the formula

V=Q4πε0r for r>R

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (06 Apr Shift 1)
For a weak acid $H A$, Ostwald's dilution law is represented by the equation

Options:
A $K_a=\frac{\alpha c}{1-\alpha^2}$
B $K_a=\frac{\alpha^2 c}{1-\alpha}$
C $\alpha=\frac{K_0 c}{1-c}$
D $K_a=\frac{\alpha^2 c}{1-\alpha^2}$

The correct answer is:
$K_a=\frac{\alpha^2 c}{1-\alpha}$

Solution:

Topic: Chemistry, Ionic Equilibrium, JEE Main
For a zero order reaction

Options:
A the concentration of the reactant remains constant during the reaction
B the concentration changes only when the temperature changes
C the rate remains constant throughout the reaction
D the rate of the reaction is proportional to the concentration of reactants

The correct answer is:
the rate remains constant throughout the reaction

Solution:
Rate law expression for zero order reaction is given by:

r=kA0r=k

So, as we can see, for zero order reaction the rate remains constant throughout the reaction, as it does not depend on concentration.

Topic: Chemistry, Chemical Kinetics, JEE Main
For above chemical reactions, identify the correct statement from the following:

Options:
A Both compound 'A' and compound 'B' are dicarboxylic acids
B Both compound 'A' and compound 'B' are diols
C Compound 'A' is diol and compound 'B' is dicarboxylic acid
D Compound 'A' is dicarboxylic acid and compound 'B' is diol

The correct answer is:
Compound 'A' is dicarboxylic acid and compound 'B' is diol

Solution:
KMnO4 in acidic medium is very strong oxidising agent due to which carboxylic acid is produced from alkene. KMnO4 in neutral medium is not very strong oxidising agent therefore, syn hydroxylation occurs and diol is produced from alkene.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2021 (20 Jul Shift 1)
For adsorption of a gas on a solid, the plot of log⁡xm vs log P is linear with slope equal to (n being whole number)

Options:
A K
B log K
C n
D 1n

The correct answer is:
1n

Solution:
log⁡xm=log⁡K+1nlog⁡P
It is of the form y = mx + C.
Therefore, slope of the line will be 1n .

Topic: Chemistry, Surface Chemistry, JEE Main
For Ag 2 CO 3 , K sp =6.2× 10 −12 . For AgCl , K sp =2.8× 10 −10 . Solid Ag 2 CO 3 and solid AgCl are added to a beaker containing Na 2 CO 3 ( aq ) . Under these conditions the [ CO 3 2− ]=1.00 M . Calculate the [ Cl − ] in solution when equilibrium is established.

Options:
A 1 . 1 Ã 1 0 - 4
B 1 . 2 6 × 1 0 - 8
C 0.15
D 2 . 8 × 1 0 - 6

The correct answer is:
1 . 1 × 1 0 - 4

Solution:
[Ag+ ] to be calculated from Ksp(Ag2CO3) with lower
Ksp but with greater solubility
Ag + = K sp Ag 2 CO 3 CO 3 2 - = 6 . 2 Ã 1 0 - 1 2 1
= 2.49 x 10- 6 M
Cl - = K sp Ag Cl Ag + = 2 . 8 Ã 1 0 - 1 0 2 . 4 9 Ã 1 0 - 6
= 1.12 x 10- 4 M

Topic: Chemistry, Ionic Equilibrium, JEE Main
For air at room temperature, the atmospheric pressure is $1.0 \times 10^{5} \mathrm{Nm}^{-2}$ and density of ait is $1.2 \mathrm{kgm}^{-3}$. For a tube of length $1.0 \mathrm{m}$, closed at one end, the lowest frequency $\begin{array}{lllllll}\text { generated is } & 84 & \text { Hz. } & \text { The } & \text { value } & \text { of } \boldsymbol{\gamma}\end{array}$ (ratio of two specific heats) for air is

Options:
A 2.1
B 1.5
C 1.8
D 1.4

The correct answer is:
1.4

Solution:
The tube is closed at one end. So, it will act like a closed organ pipe. The lowest frequency produced in the tube

$$
\begin{array}{l}
l=\frac{\lambda}{4}, \lambda=4 \\
l=1 \mathrm{m} \\
f=84 \mathrm{Hz} \\
V=n \lambda=84 \times 4
\end{array}
$$
When wave propagate in air, the velocity
Here,
$$
\begin{array}{l}
v=\sqrt{\frac{\gamma p}{\rho}} \\
p=1.0 \times 10^{5} \mathrm{N} / \mathrm{m}^{2}
\end{array}
$$
$$
\begin{aligned}
\rho &=1.2 \mathrm{kg} / \mathrm{m}^{3} \\
V^{2} &=\frac{\gamma P}{\rho} \\
\gamma &=\frac{\mathrm{v}^{2} \cdot \rho}{\mathrm{P}} \\
&=\frac{(84 \times 4)^{2} \times 1.2}{1.0 \times 10^{5}} \\
&=\frac{135472.2}{10^{5}}=1.354 \\
&=1.4
\end{aligned}
$$

Topic: Physics, Waves and Sound, JEE Main
For all complex numbers z of the form 1+iα, α∈R, if z2=x+iy, then

Options:
A y2-4x+4=0
B y2+4x-4=0
C y2-4x+2=0
D y2+4x+2=0

The correct answer is:
y2+4x-4=0

Solution:
Given z=1+iα, z2=x+iy

⇒1+iα2=x+iy

⇒12+i2α2+2iα=x+iy

Using i2=-1,

⇒1-α2+2iα=x+iy

Equating the real and imaginary parts, we get

x=1-α2 and y=2α

⇒α2=1-x and y2=4α2

⇒y2=41-x

⇒y2=4-4x

⇒y2+4x-4=0.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2014 (19 Apr Online)
For all real values of $x$, increasing function $f(x)$ is

Options:
A $x^{-1}$
B $x^2$
C $x^3$
D $x^4$

The correct answer is:
$x^3$

Solution:
Since $f(x)=x^3 \Rightarrow f^{\prime}(x)=3 x^2$, which is non-negative for all real values of $x$.

Topic: Mathematics, Application of Derivatives, JEE Main
For all real $x$, the minimum value of $\frac{1-x+x^{2}}{1+x+x^{2}}$ is

Options:
A 0
B $1 / 3$
C 1
D 3

The correct answer is:
$1 / 3$

Solution:
Let $y=\frac{1-x+x^{2}}{1+x+x^{2}}=1-\frac{2 x}{1+x+x^{2}}$
$$
\begin{aligned}
&=1-\frac{2}{\frac{1}{x}+1+x} \\
\Rightarrow \quad y &=1-\frac{2}{t}
\end{aligned}
$$
where
$$
t=\frac{1}{x}+1+x
$$
Now, $y$ is minimum, when $\frac{2}{t}$ is $\max \Rightarrow t$ is $\mathrm{min}$.
$$
\begin{array}{lc}
\therefore & \frac{d t}{d x}=-\frac{1}{x^{2}}+1=0 \\
\Rightarrow \quad x=\pm 1 \\
\frac{d^{2} t}{d x^{2}}=\frac{2}{x^{3}}>0, \text { for } x=1
\end{array}
$$
$\therefore$ Minimum value of $y$ is
$$
\begin{aligned}
1-\frac{2}{1+1+1} &=1-\frac{2}{3} \\
&=\frac{1}{3}
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Main
For all twice differentiable functions f : R→R, with f0=f1=f'0=0,

Options:
A f"x≠0 at every point xε0,1
B f"x=0, for some x ε 0,1
C f"0=0
D f"x=0, at every point x ε0,1

The correct answer is:
f"x=0, at every point x ε0,1

Solution:
Applying Rolle's theorem in [0,1] for function f(x)

f'c=0, c ε0,1

again applying Rolle's theorem in [0, c] for function f'(x),

f''x=0, c∈0,1

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (06 Sep Shift 2)
For $\alpha-$ amino acids having the structure
Which of the following statements are true
(A) Water solubility is maximum at a $\mathrm{pH}$ when concentrations of anions and cations are equal
(B) They give ninhydrin test
(C) On reacting with nitrous acid give off $\mathrm{N}_2$

Options:
A A
B $B$ and $C$
C $A$ and $B$
D A

The correct answer is:
A

Solution:
Amino acids can form zwitterions, that is a molecule or ion having separate positively and negatively charged groups. Amino acids form due to the presence of an amine group that can ionize to form $\mathrm{NH}^{+3}$ and a carboxyl group that can form $\mathrm{COO}-$. When they are in the zwitterion form, they do not migrate in an electric field and they precipitate out of the solution. Therefore amino acids are least soluble at a $\mathrm{pH}$ when they exist in the zwitterion form, that is when the concentration of anions and cations are equal.The other two statements are true:The amine group of amino acids reacts with ninhydrin reagent to give a purple color complex.
And amino acids react with nitrous acid releasing $\mathrm{N}_2$.

Topic: Chemistry, Biomolecules, JEE Main
For an adiabatic expansion of an ideal gas, the fractional change in its pressure is equal to (where γ is the ratio of specific heats):

Options:
A -γdVV
B -γVdV
C -1γdVV
D dVV

The correct answer is:
-γdVV

Solution:
⇒PVγ= constant

Differentiating,

⇒dPdV=-γPV

⇒dPP=-γdVV

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2021 (18 Mar Shift 2)
For an amplitude modulated wave, the maximum amplitude is found to be 10 V while the minimum amplitude is found to be 2 V. The modulation index is

Options:
A 23
B 13
C 43
D 12

The correct answer is:
23

Solution:
Given, maximum amplitude Amax = 10 V
Minimum amplitude Amin = 2 V
Let Ac and Am be the amplitudes of carrier wave and signal wave
â´ Amax = Ac + Am = 10 ...(i)
and Amin = Ac â Am = 2 ...(i)
Adding the Eqs. (i) and (ii), we get
2Ac = 12
or Ac = 6 V
and Am = 10 â 6 = 4 V

Î¼=AmAc=46=23

Topic: Physics, Semiconductors, JEE Main
For an amplitude modulated wave the minimum amplitude is 3 V, while the modulation index is 60%. The maximum amplitude of the modulated wave is:

Options:
A 5 V
B 15 V
C 12 V
D 10 V

The correct answer is:
12 V

Solution:
Modulatio index is the ratio of the difference in amplitudes to the sum of the amplitudes.

The formula to calculate the modulation index of a wave is given by

μ=Amax-AminAmax+Amin ...1

Substitute the values of the known parameters into equation (1) and solve to calculate the required value of maximum amplitude.

0.6=Amax-3Amax+3⇒0.6Amax+1.8=Amax-3⇒0.4Amax=4.8⇒Amax= 4.80.4=12 V

Topic: Physics, Communication System, JEE Main, JEE Main 2023 (06 Apr Shift 2)
For an aqueous solution, the characteristic species of acid is

Options:
A $\mathrm{H}^{+}$ion
B $\mathrm{H}_3 \mathrm{O}^{+}$ion
C $\mathrm{H}_2{ }^{+}$ion
D $\mathrm{H}_4 \mathrm{O}^{+}$ion

The correct answer is:
$\mathrm{H}_3 \mathrm{O}^{+}$ion

Solution:
In an aqueous solution, the characteristic species of an acid is the hydronium ion ( $\left.\mathrm{H}_3 \mathrm{O}^{+}\right)$. This ion is formed when an acid donates a proton $\left(\mathrm{H}^{+}\right)$to a water molecule1. The concentration of hydronium ions determines the acidity of the solution. So, the correct answer to your question is (2) $\mathrm{H}_3 \mathrm{O}^{+}$ion

Topic: Chemistry, Ionic Equilibrium, JEE Main
For an ecosystem, the two main structural features are -

(i) Species composition

(ii) Energy flow

(iii) Decomposition

(iv) Stratification

Options:
A i and iv only
B ii and iii only
C i only
D ii, iii and iv only

The correct answer is:
i and iv only

Solution:
The basic structural and functional unit of ecology is an ecosystem which is made up of abiotic and biotic components. There is a characteristic physical structure within each ecosystem due to the interaction of abiotic and biotic components. Species composition and stratification form the two main structural features of an ecosystem. For any ecosystem to function and sustain, solar energy supply is a basic requirement.

Topic: Biology, Ecosystem, NEET
For an electrochemical cell, SnsSn2+aq,1 MPb2+aq,1 MPbs , the ratio Sn2+Pb2+ when this cell attains equilibrium is

Given: ESn2+|Sn0=-0.14V, EPb2+|Pb0=-0.13V,2.303 RTF=0.06

Options:
A 4.3
B 1
C -2.15
D 2.14

The correct answer is:
2.14

Solution:
At equilibrium
Ecell=0
Ecell0=0.01 V
Sn+Pb2+âSn2++Pb
Ecell=Ecell0-0.06nlogâ¡Q
0=0.01-0.062logâ¡Sn2+Pb2+
0.01=0.062logâ¡Sn2+Pb2+
13=logâ¡Sn2+Pb2+
Sn 2+ Pb 2+ = 10 1 3 =2.14

Topic: Chemistry, Electrochemistry, JEE Main
For an electrochemical cell SnsSn2+aq, 1MPb2+aq, 1MPbs the ratio Sn2+Pb2+ when this cell attains equilibrium is _______
(Given: ESn2+Sn0=-0.14 V,EPb2+Pb0=-0.13 V,  2.303RTF=0.06)

Options:

The correct answer is:
2.15

Solution:
Cell reaction is:
Sn(s)+ Pb +2 (aq)→ Sn +2 (aq)+Pb(s)
Apply Nernst equation:
E cell = E cell 0 − 0.06 2 log Sn +2 Pb +2 ..... 1
E cell 0 =−0.13+0.14=0.01V
At equilibrium : Ecell =0 Substituting in (1)
0=0.01− 0.06 2 log Sn +2 Pb +2
⇒  1 3 =log Sn +2 Pb +2
⇒  Sn +2 Pb +2 =2.15

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2020 (08 Jan Shift 2)
For an electron whose positional uncertainty is 1.0Ã10-10 m, the uncertainty in the component of the velocity in m s-1 will be

Options:
A 5.8× 10 5
B 109
C 102
D 1015

The correct answer is:
5.8× 10 5

Solution:
According to uncertainty principle
⇒δx × δp= h 4π
⇒ δp= h 4π×δx
⇒mδv= h 4πδx
δv=h4πmδx
= 6.625× 10 −34 9.1× 10 −31 × 10 −10 ×4×3.14 =5.8× 10 5

Topic: Chemistry, Structure of Atom, JEE Main
For an elementary chemical reaction, A2 ⇄k-1k12A , the expression for dAdt is:

Options:
A 2k1A2-2k-1A2
B 2k1A2-k-1A2
C k1A2+k-1A2
D k1A2-k-1A2

The correct answer is:
2k1A2-2k-1A2

Solution:
For the reaction, A2→K12A

rf=-dA2dt=12dAdt1=K1A2

dAdt1=2K1A2

For the reaction, 2A→k-1 A2

rb=-12dAdt2=K-1A2

dAdt2=-2K-1A2

Hence, the overall rate is given by:

dAdt=2k1A2-2k-1A2

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2019 (10 Jan Shift 2)
For an enclosure maintained at $1000 \mathrm{~K}$, the maximum radiation occurs at wavelength $\lambda_{\mathrm{m}}$. If the temperature is raised to $2000 \mathrm{~K}$, the peak will shift to

Options:
A $\frac{5}{2} \lambda_m$
B $\frac{1}{2} \lambda_m$
C $\frac{7}{2} \lambda_m$
D $\frac{3}{2} \lambda_m$.

The correct answer is:
$\frac{1}{2} \lambda_m$

Solution:
We know from Wien's displacement law,
$$
\begin{aligned}
& \lambda_m T=\text { constant } \\
\therefore \quad & \lambda_m T_l=\lambda_m^{\prime} T_2 \\
\Rightarrow & \lambda_m^{\prime}=\frac{\lambda_m T_1}{T_2}=\frac{\lambda_m \times 1000}{2000}=\frac{\lambda_m}{2}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, AIIMS, AIIMS 2001
For an exothermic reaction, following two steps are involved.

Step 1. A + B â I slow Step 2. I â AB fast

Which of the following graphs correctly represent this reaction ?

Options:
A
B
C
D

The correct answer is:

Solution:
In any reaction mechanism the no. of activated complexes showing maxima of potential energies is the no. of steps in that mechanism and in between these maxima is a valley where lies a more stable intermediate. There is one reaction intermediate lying at valley of two maxima representing two steps. Ea for first step should be higher than second step as first is slow step. These requirements fulfilled by choice (ii) & not choice (i) the choice (iv) is not possible because that is for endothermic reaction Δ H + ve & (ii) is single step equation so not possible.

Topic: Chemistry, Chemical Kinetics, JEE Main
For an ideal binary liquid mixture

Options:
A \( \Delta \mathrm{H}_{(\text {mix })}=0 ; \Delta \mathrm{S}_{(\text {mix })} < 0 \)
B \( \Delta \mathrm{S}_{(\mathrm{mix})}>0 ; \Delta \mathrm{G}_{(\mathrm{mix})} < 0 \)
C \( \Delta S_{(\mathrm{mix})}=0 ; \Delta G_{(\mathrm{mix})}=0 \)
D \( \Delta \mathrm{V}_{(\text {mix })}=0 ; \Delta \mathrm{G}_{(\text {mix })}>0 \)

The correct answer is:
\( \Delta \mathrm{S}_{(\mathrm{mix})}>0 ; \Delta \mathrm{G}_{(\mathrm{mix})} < 0 \)

Solution:
For an ideal binary liquid mixture $\Delta S_{\operatorname{mix}}>0$ and $\Delta G_{\operatorname{mix}} < 0$
For ideal solution;
$\Delta H_{\operatorname{mix}}=0$ and $\Delta V_{\operatorname{mix}}=0$
$\Delta G=\Delta \mathrm{H}-\mathrm{T} \Delta S$
$\Delta G$ should be $-$ ve for reaction to be spontaneous, that is, $\Delta G < 0$.
When $\Delta G < 0$, then $\Delta S$ should have value greater than zero

Topic: Chemistry, Redox Reactions, JEE Main
For an ideal gas, number of moles per litre in terms of its pressure P, gas contant R and temperature T is

Options:
A
PT/R

B
PRT

C
P/RT

D
RT/P

The correct answer is:
P/RT

Solution:
PV = nRT (number of moles = n/V) $\therefore$ n/V = P/RT

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2002
For an ideal gas the instantaneous change in pressure P with volume V is given by the equation dPdV=-aP. If P=P0 at V=0 is the given boundary condition, then the maximum temperature one mole of gas can attain is: (Here R is the gas constant)

Options:
A aP0eR
B infinity
C 0°C
D P0aeR

The correct answer is:
P0aeR

Solution:
dPdV=-aP
∫P0PdPP=-∫0vadV⇒lnPP0=-aV
P=P0e-aV
Now, PV=RT(as n=1)
P+VdPdV=RdTdV
[ for maximum T, dTdV=0]
P-aPV=0⇒V=1a
Thus, T=PVR=P0e-1Ra=P0eRa

Topic: Physics, Kinetic Theory of Gases, JEE Main
For an ideal gas the instantaneous change in pressure P with volume V is given by the equation dPdV=-aP. If P=P0 at V=0 is the given boundary condition, then the maximum temperature one mole of gas can attain is: (Here R is the gas constant)

Options:
A aP0eR
B infinity
C 0°C
D P0aeR

The correct answer is:
P0aeR

Solution:
dPdV=-aP
∫P0PdPP=-∫0vadV⇒lnPP0=-aV
P=P0e-aV
Now, PV=RT(as n=1)
P+VdPdV=RdTdV
[ for maximum T, dTdV=0]
P-aPV=0⇒V=1a
Thus, T=PVR=P0e-1Ra=P0eRa

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2021 (31 Aug Shift 1)
For an ideal gas, the specific heat at constant pressure Cp is greater than the specific heat at constant volume Cv. This is because

Options:
A There is a finite work done by the gas on its environment when its temperature is increased while the pressure remains constant
B There is a finite work done by the gas on its environment when its temperature is increased while the volume remains constant
C The pressure of the gas remains constant when its temperature remains constant
D The internal energy of the gas at constant pressure is more than at constant volume

The correct answer is:
There is a finite work done by the gas on its environment when its temperature is increased while the pressure remains constant

Solution:
For an ideal gas, the specific heat at constant pressure Cp is greater than Cv . This is because some finite work has to be done by the gas on its environment when its temperature is increased while the pressure remains constant.

Topic: Physics, Thermodynamics, NEET
For an ideal gas, the specific heat capacity during an isentropic process is always

Options:
A zero
B infinite
C positive

D negative

The correct answer is:
zero

Solution:
A gas if fully insulated from the surroundings while undergoing adiabatic change. That is, no exchange of heat between the gas and surroundings. Hence, specific heat is zero.

Topic: Physics, Thermodynamics, JEE Main
For an ideal heat engine, the temperature of the source is 127°C. In order to have 60% efficiency the temperature of the sink should be __________C∘. (Round off to the nearest integer)

Options:

The correct answer is:
-113

Solution:
n=0.60=1-TLTH

TLTH=0.4⇒TL=0.4×400

=160 K

=-113°C

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2021 (16 Mar Shift 2)