datavorous/entrance-exam-dataset huggingface dataset:
For a body moving with relativistic speed, if the velocity is doubled, then

Options:
A Its linear momentum is doubled
B Its linear momentum will be less than double
C Its linear momentum will be more than double
D Its linear momentum remains unchanged

The correct answer is:
Its linear momentum will be more than double

Solution:
Relativistic momentum $=\frac{m_0 v}{\sqrt{1-v^2 / c^2}}$
If velocity is doubled then the relativistic mass also increases. Thus value of linear momentum will be more than double.

Topic: Physics, Motion In Two Dimensions, JEE Main
For a body projected at an angle with the horizontal from the ground, choose the correct statement

Options:
A Gravitational potential energy is maximum at the highest point.
B The horizontal component of velocity is zero at highest point.
C The vertical component of momentum is maximum at the highest point.
D The kinetic energy (K.E.) is zero at the highest point of projectile motion.

The correct answer is:
Gravitational potential energy is maximum at the highest point.

Solution:
In a projectile motion, a particle has two components of velocity. The horizontal component of velocity remains constant over time, while the vertical component of velocity changes.

At highest point, the vertical component of velocity vy becomes zero, but the horizontal component is non-zero. So the kinetic energy of the particle is not zero.

The formula to calculate the potential energy U of the particle is given by

U= mgh...................(1)

where, m is the mass, g is the acceleration due to gravity and h is the vertical height attained by the object.

Hence, the potential energy is maximum when the projectile attains the maximum height.

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (01 Feb Shift 2)
For a cell reaction involving a two-electron change, the standard e.m.f. of the cell is found to be $0.295 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction at $25^{\circ} \mathrm{C}$ will be

Options:
A
$29.5 \times 10^{-2}$

B
10

C
$1 \times 10^{10}$

D
$1 \times 10^{-10}$

The correct answer is:
$1 \times 10^{-10}$

Solution:
The equilibrium constant is related to the standard emf of cell by the expression
$\log K=E_{\text {cell }}^0 \times \frac{n}{0.059}=0.295 \times \frac{2}{0.059}$
$\log \mathrm{K}=\frac{590}{59}=10$ or $\mathrm{K}=1 \times 10^{-10}$

Topic: Chemistry, Electrochemistry, JEE Main
For a cell reaction involving a two-electron change, the standard e.m.f. of the cell is found to be $0.295 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction at $25^{\circ} \mathrm{C}$ will be

Options:
A
$29.5 \times 10^{-2}$

B
10

C
$1 \times 10^{10}$

D
$1 \times 10^{-10}$

The correct answer is:
$1 \times 10^{-10}$

Solution:
The equilibrium constant is related to the standard emf of cell by the expression
$\log K=E_{\text {cell }}^0 \times \frac{n}{0.059}=0.295 \times \frac{2}{0.059}$
$\log \mathrm{K}=\frac{590}{59}=10$ or $\mathrm{K}=1 \times 10^{-10}$

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2003
For a cell, terminal potential difference is $2.2 \mathrm{~V}$ when circuit is open. If it reduces to $1.8 \mathrm{~V}$ when the cell is connected to a resistance of $5 \Omega$. The internal resistance of cell $(r)$ is then:

Options:
A $\frac{10}{9} \Omega$
B $\frac{9}{10} \Omega$
C $\frac{11}{9} \Omega$
D $\frac{5}{9} \Omega$

The correct answer is:
$\frac{10}{9} \Omega$

Solution:
Terminal potential difference
$$
\begin{aligned}
& V=E-I r \\
& \therefore V=E-\left[\frac{E}{R-r}\right] r=\frac{E R}{R+r}
\end{aligned}
$$
From given condition $=E=2.2$ and where $R=5$ then potential $V=1.8 \mathrm{~V}$ Therefore $1.8=\frac{2.2 \times 5}{5+r}$
$\Rightarrow r=\frac{10}{9} \Omega$

Topic: Physics, Current Electricity, JEE Main
For a certain first order reaction 32% of the reactant is left after 570 s. The rate constant of this reaction is ×10-3 s-1. (Round off to the Nearest Integer).

[Given: log10 2=0.301, ln10=2.303]

Options:

The correct answer is:
2

Solution:
For 1st order reaction,

K=2.303t·logA0At=2.303570sec·log10032

=1.999×10-3sec-1≈2×10-3sec-1

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2021 (17 Mar Shift 1)
For a certain lens, the magnification of an object when placed at a distance of 0.15 m is twice of the magnification produced, when the distance was 0.2 m. If in both the situations a real image is formed, then what is the focal length (in cm) of the lens?

Options:

The correct answer is:
10

Solution:
Let as shown, 1 and 2 are positions of objects and images in two different situations.
Object
It is given
vâuâ=2vâuâ
Here, uâ= -15 cm, uâ= -20 cm
â´vâ=2vâ Ãuâuâ=2vâ Ã1520=32vâ
now, 1f=1v-1u
â´1f=1vâ-1uâ and1f=1vâ-1uâ
so, 1v1-1u1=1v2-1u2
â¹22vâ+115=1vâ+120
â¹v=20 cm
â´vâuâ=2vâuâ=2 Ã2020=2
â¹vâ=2uâ=2Ã15=30 cm
Therefore,1f=1vâ-1u1=115+130=330
â´f=10 cm

Topic: Physics, Ray Optics, JEE Main
For a certain organ pipe, the first three resonance frequencies are in the ratio of 1:3:5 respectively. If the frequency of fifth harmonic is 405 Hz and the speed of sound in air is 324 m s–1 the length of the organ pipe is _____ m.

Options:

The correct answer is:
1

Solution:
It is given,

f5=405 Hz.

As the first three resonance frequencies are in the ratio of 1:3:5, therefore it must be a closed organ pipe.

The frequency formula in a closed organ pipe is given by f=nv4l. So,

5v4I=405 Hz

⇒I=405×45×324=1 m

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2023 (12 Apr Shift 1)
For a certain reaction at 300 K, K=10, then ΔG° for the same reaction is – ________ ×10-1 kJ mol-1. (Given R=8.314JK-1 mol-1)

Options:

The correct answer is:
57

Solution:
The standard Gibbs free energy of the formation of a compound is basically the change of Gibbs free energy that is followed by the formation of one mole of that substance from its component element available at their standard states or the most stable form of the element which is at 25 °C and 100 kPa. Its symbol is ΔfG˚.

ΔG°=-RTℓnK

=-8.314×300ln(10)

=5744.14 J/mole

=57.44×10-1 kJ/mole

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2024 (01 Feb Shift 2)
For a certain reaction the variation of rate constant with temperature is given by the equation

ln kt=lnk0+ln3t10t â¥ 0Â°C where k0=k at zero degree centigrade

The value of temperature coefficient of the reaction is -

Options:
A 0.1
B 1
C 10
D 3

The correct answer is:
3

Solution:
The variation of 'k' with 'T' is given by
k T  =  k 0 μ t 10
where Î¼ is temperature coefficient
∴ ln  k T  = ln  k 0 + t 10 ln μ
comparing with given eqn we get μ = 3

Topic: Chemistry, Chemical Kinetics, JEE Main
For a certain thermochemical reaction M→N at T=400 K,ΔHo=77.2 kJ mol-1, ΔSo=122JK-1, log equilibrium constant (logK) is - _____×10-1.

Options:

The correct answer is:
37

Solution:
Given,

T=400 K,ΔHo=77.2 kJ mol-1,ΔSo=122JK-1

ΔGo=ΔHo-TΔSo

=77.2×103-400×122=28400 J

ΔG°=-2.303 R T log K

Where K is the equilibrium constant.

By putting the values in above equation,

⇒28400=-2.303×8.314×400×log K

⇒log K=-3.708=-37.08×10-1

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2024 (27 Jan Shift 2)
For a charged spherical ball, electrostatic potential inside the ball varies with r as V=2ar2+b.

Here, a and b are constant and r is the distance from the center. The volume charge density inside the ball is -λaε. The value of λ is ______.

ε= permittivity of medium.

Options:

The correct answer is:
12

Solution:
As we know, E=-dVdr=-4ar

Now, electric field inside the charged sphere is given by E=ρr3ε.

Comparing both, we get ρ=-12aε.

Therefore, λ=12.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (29 Jan Shift 2)
For a chemical reaction A+B=C+D ΔrHΘ=80 kJ mol-1 the entropy change ΔrSΘ depends on the temperature T (in K) as ΔrSΘ=2TJK-1mol-1. Minimum temperature at which it will become spontaneous is

K. (Integer)

Options:

The correct answer is:
200

Solution:
ΔG°=ΔH°-T×ΔS°

ΔG°=ΔH°-T×2T

T=200 K

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2021 (26 Feb Shift 1)
For a chemical reaction A→B, it was found that concentration of B is increased by 0.2 mol L-1 in 30 min. The average rate of the reaction is ________ ×10-1 mol L-1 h-1. (in nearest integer)

Options:

The correct answer is:
4

Solution:
A⟶B

t=0 0

t=30 min 0.2M

Av. rate of reaction =-Δ[A]Δt=Δ[B]Δt=(0.2-0)12

=0.4=4×10-1 mol/L×hr

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2021 (25 Jul Shift 2)
For a chemical reaction A+B→ Product, the order is 1 with respect to A and B.

Rate
mol L-1 S-1	[A]
mol L-1	[B]
mol L-1

0.10	20	0.5

0.40	x	0.5

0.80	40	y

What is the value of x and y?

Options:
A 160 and 4
B 80 and 4
C 80 and 2
D 40 and 4

The correct answer is:
80 and 2

Solution:
To determine the values of x and y, we can use the information provided and the rate expression for the given reaction, which is first order with respect to both A and B:

Rate r = K[A]1 [B]1

0.1 = K(20)1 (0.5)1 …(i) 0.40 = K(x)1 (0.5)1 …(ii) 0.80 = K(40)1 (y)1 …(iii)

From (i) and (ii)

x = 80

From (i) and (iii)

y = 2

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2023 (11 Apr Shift 2)
For a chemical reaction $\mathrm{A} \rightarrow \mathrm{B}$, the rate of the reaction is $2 \times 10^{-3} \mathrm{~mol} \mathrm{} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$, when the initial concentration is $0.05 \mathrm{~mol} \mathrm{dm}^{-3}$. The rate of the same reaction is $1.6 \times 10^{-2} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$ when the initial concentration is $0.1 \mathrm{~mol} \mathrm{} \mathrm{dm}^{-3}$. The order of the reaction is

Options:
A 2
B 0
C 3
D 1

The correct answer is:
3

Solution:
$\mathrm{A} \longrightarrow \mathrm{B}$
Rate $=k[A]^{n}$
$(\text { Rate })_{1}=\mathrm{k}(0.05)^{\mathrm{n}}=2 \times 10^{-3} \quad \text{...(i)}$
$(\text { Rate })_{2}=\mathrm{k}(0.1)^{\mathrm{n}}=1.6 \times 10^{-2} \quad \text{...(ii)}$
Dividing the Eq. (ii) by Eq. (i).
$$
\begin{gathered}
\frac{(\text { Rate })_{2}}{(\text { Rate })_{1}}=\frac{\mathrm{k}(0.1)^{\mathrm{n}}}{\mathrm{k}(0.05)^{\mathrm{n}}}=\frac{1.6 \times 10^{-2}}{2 \times 10^{-3}} \\
\quad(2)^{\mathrm{n}}=8 \quad \text { or } \quad(2)^{\mathrm{n}}=2^{3} \\
\therefore \quad \mathrm{n}=3
\end{gathered}
$$

Topic: Chemistry, Chemical Kinetics, JEE Main
For a chemical reaction, the free energy change (ΔG) is negative. The reaction is

Options:
A A spontaneous reaction
B An equilibrium reaction
C A non-spontaneous reaction
D Characherised by r f  =  r b (where, r f snd r b are rates of forward and backward reactions respectively)

The correct answer is:
A spontaneous reaction

Solution:
For spontaneous reaction, the free energy change (ΔG) should be negative.

Topic: Chemistry, Thermodynamics (C), JEE Main
For a circle of diameter $R$, touching $x^2+y^2-4 y=0$ and passing through $(4,5)$, which of the following is correct?

Options:
A $3 \leq \mathrm{R} \leq 7$
B $0 < \mathrm{R} < 3$
C $\mathrm{R}>7$
D $\frac{3}{2} \leq R \leq \frac{7}{2}$

The correct answer is:
$\frac{3}{2} \leq R \leq \frac{7}{2}$

Solution:
The distance between centres of circles will be R/2 + 2
Put (4,5) in general equation of circle

Topic: Mathematics, Circle, JEE Main
For a closed system consisting of a reaction $\mathrm{N}_{2} \mathrm{O}_{4}$ (g) $\rightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})$, the pressure

Options:
A remains constant
B decreases
C increases
D becomes zero

The correct answer is:
decreases

Solution:
As the system is closed, hence the reaction will be reversible, hence according to Le-chatelier principle pressure decreases since the volume is increasing.

Topic: Chemistry, Chemical Equilibrium, VITEEE, VITEEE 2006
For a colour of light the wavelength for air is $6000 Å$ and in water the wavelength is $4500 Å$. Then the speed of light in water will be

Options:
A $5 . \times 10^{14} \mathrm{~m} / \mathrm{s}$
B $2.25 \times 10^8 \mathrm{~m} / \mathrm{s}$
C $4.0 \times 10^8 \mathrm{~m} / \mathrm{s}$
D Zero

The correct answer is:
$2.25 \times 10^8 \mathrm{~m} / \mathrm{s}$

Solution:
$\begin{aligned} & v \propto \lambda \Rightarrow \frac{v_1}{v_2}=\frac{\lambda_1}{\lambda_2} \\ & \quad \therefore \quad v_2=\frac{v_1}{\lambda_1} \times \lambda_2=3 \times 10^8 \times \frac{4500}{6000}=2.25 \times 10^8 \mathrm{~m} / \mathrm{s}\end{aligned}$

Topic: Physics, Ray Optics, JEE Main
For a common emitter configuration, if α and β have their usual meanings, the correct relationship between α and β is:

Options:
A α=β1+β
B Î± =Î²21+Î²2
C 1Î±=1Î²-1
D α=β1-β

The correct answer is:
α=β1+β

Solution:
IE=IC+IB
IEIC=1+IBIC
1∝=1+1β

1∝=β+1β
∝=β1+β

Topic: Physics, Semiconductors, JEE Main, JEE Main 2016 (03 Apr)
For a common-emitter transistor amplifier, the current gain is 60 . If the emitter current is 6.6mA then its base current is

Options:
A 6.492mA
B 0.108mA
C 0.11mA
D 0.343mA

The correct answer is:
0.108mA

Solution:
For a common-emitter transistor amplifier, the current gain β is 60
The emitter current IE is 6.6mA
IEIB=1+β⇒IB=IE1+β

Topic: Physics, Semiconductors, JEE Main
For a complex number Z, if all the roots of the equation Z3+aZ2+bZ+c=0 are unimodular, then

Options:
A a>3 and c=1
B aâ¤3 and c=3
C a>3 and c=13
D aâ¤3 & c=1

The correct answer is:
a≤3 & c=1

Solution:
Given, Z1=Z2=Z3=1
Now, Z1+Z2+Z3≤Z1+Z2+Z3
⇒-a≤1+1+1
⇒a≤3
Also, Z1Z2Z3=Z1×Z2×Z3
⇒-c=1×1×1
⇒c=1

Topic: Mathematics, Complex Number, JEE Main
For a complex number Z, if arg⁡Z=π4 and Z+1Z=4, then the value of Z-1Z is equal to

Options:
A 14
B 18
C 4
D 12

The correct answer is:
14

Solution:
Let, Z=Zeiπ4 then,

1Z=1Ze-iπ4
⇒Z+1Z=Zcosπ4+isinπ4+1Zcosπ4-isinπ4=4
⇒12Z+iZ+1Z-iZ=4
⇒Z+1Z+Z-1Zi=42
⇒Z+1Z2+Z-1Z2=32
⇒2Z2+1Z2=32⇒Z2+1Z2=16
⇒Z-1Z2=14⇒Z-1Z=14

Topic: Mathematics, Complex Number, JEE Main
For a complex number z, If argz∈-π,π, then arg1+cos⁡6π5+isin⁡6π5 is (here i2=-1)

Options:
A 3π5
B 2π5
C -2π5
D -3π5

The correct answer is:
-2π5

Solution:
Let, z=1+cos⁡6π5+isin⁡6π5=2cos⁡3π5cos⁡3π5+isin⁡3π5
=-2cos⁡3π5-cos⁡3π5-isin⁡3π5
=-2cos⁡3π5cos⁡3π5-π+isin⁡3π5-π

i.e. ω=-2cos⁡3π5cos⁡-2π5+isin⁡-2π5
Hence, argz=-2π5

Topic: Mathematics, Complex Number, JEE Main
For a complex number Z, if argZ∈-π,π, then arg1+cos⁡6π7+isin⁡6π7 is (here i2=-1)

Options:
A 3π7
B 2π7
C -2π7
D -3π7

The correct answer is:
3π7

Solution:
Let W=1+cos⁡6π7+isin⁡6π7=2cos⁡3π7cos⁡3π7+isin⁡3π7
Hence, argW=3π7

Topic: Mathematics, Complex Number, JEE Main
For a complex number Z, if one root of the equation Z2-aZ+a=0 is 1+i and its other root is Î±, then the value of aÎ±4 is equal to

Options:
A 4
B -12
C 2
D -2

The correct answer is:
-12

Solution:
Î±+1+i=a
Î±1+i=a
Î±+Î±i=Î±+1+i
⇒Î±=1+ii=1-i
a=1-i+1+i=2
Î±4=Î±22=1+-1-2i2
=-2i2=4i2=-4

Topic: Mathematics, Complex Number, JEE Main
For a complex number Z, if the argument of 3+3i and Z-2Z--1 are equal, then the maximum distance of Z from the x-axis is equal to (where, i2=-1)

Options:
A 1+22 units
B 2 units
C 32 units
D 2+22 units

The correct answer is:
1+22 units

Solution:
Let, Z=x+iy
Now, Z-2Z--1=ZZ--2Z--Z+2
=x2+y2-3x+2+iy
arg of 3+3i=π4
⇒x2+y2-3x+2=y where y>0
⇒x2+y2-3x-y+2=0
Centre of the circle is 32,12 and radius =12

⇒ required value =12+12 units

Topic: Mathematics, Complex Number, JEE Main
For a complex number Z, if the argument of Z-aZ¯-b is π4 or -3π4 (where a, b are two real numbers), then the value of ab such that the locus of Z represents a circle with centre 32+i2 is (where, i2=-1)

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
Let, Z=x+iy, then
Z-aZÂ¯-b=ZZÂ¯-aZÂ¯-bZ+ab
=x2+y2-ax-iy-bx+iy+ab
=x2+y2-a+bx+ia-by+ab
If the argument of the above complex number is Ï4 or -3Ï4, then
x2+y2-a+bx+ab=a-by
â Centre of the circle is a+b2,a-b2
âa+b2=32 & a-b2=12
âa=2, b=1âab=2

Topic: Mathematics, Complex Number, JEE Main
For a complex number Z, if Z-1+i+Z+i=1, then the range of the principle argument of Z is (where principle argZâ-Ï,Ï )

Options:
A -π4,π4
B π4,π2
C -π2,-π4
D -π2,π2

The correct answer is:
-π2,-π4

Solution:
Z-1+i+Z+i=1-i--i
â locus of Z is the line segment joining -i with 1-i
â minimum arg Z=-Ï2 & maximum arg Z=-Ï4
â principal arg Zâ-Ï2,-Ï4

Topic: Mathematics, Complex Number, JEE Main
For a complex number z, if z2+z--z=4i and z does not lie in the first quadrant, then (where i2=-1)

Options:
A z=2
B z=22
C arg⁡z=-π4
D arg⁡z=3π4

The correct answer is:
z=2

Solution:
Put z=x+iy
⇒x2-y2+2ixy-2iy=4i
⇒x2-y2=0⇒x=y or -yxy-y=2
Case i if x=y, then x2-x-2=0
⇒x=-1,2
⇒z1=-1-iz2=2+2i rejected
Case ii if x=-y, then x2-x+2=0
⇒x∈ϕ

Hence, z=-1-i=2

Topic: Mathematics, Complex Number, JEE Main
For a complex number Z, if Z=Ï41+i41-ÏiÏ+i+Ï-i1+Ïi, then the value of ZampZ is equal to (where ampZâ-Ï,Ï)

Options:
A π
B 4
C 1
D 3π

The correct answer is:
4

Solution:
Z=π41+i221-πi1+πi+π-iπ+iπ+πi+i-π
⇒Z=π41-1+2i21+π+π+1π+1i
⇒Z=π44-12i=-2πi=2πi
⇒Z=2π and ampZ=π2
⇒ZampZ=4

Topic: Mathematics, Complex Number, JEE Main
For a complex number Z, if Z-iâ¤2 and Z1=5+3i, then the maximum value of iZ+Z1 is (where, i2=-1)

Options:
A 7+13
B 7+12
C 7
D 34-2

The correct answer is:
7

Solution:
iZ+Z1=iZ-i+Z1-1
≤iZ-i+Z1-1
≤iZ-i+4+3i
≤2+42+32=7

Topic: Mathematics, Complex Number, JEE Main
For a complex number Z, the equation of the line of common chord of the circles Z-3=2 and Z=2 is

Options:
A Z+Z¯=3
B Z-Z¯=3
C Z¯-Z=3
D Z+Z¯+3=0

The correct answer is:
Z+Z¯=3

Solution:
Z-3=2⇒Z-3Z¯-3¯=4
⇒ZZ¯-3Z-3Z¯+9=4 … 1
Z=2⇒ZZ¯=4 … 2
For the equation of common chord, Subtracting 2 from 1, we get,
-3Z-3Z¯+9=0
⇒Z+Z¯=3

Topic: Mathematics, Complex Number, JEE Main
For a complex number z, the equation z2+p+iqz+r+is=0 has a real root (where p,q,r,s are non-zero real numbers and i2=-1), then

Options:
A pqr=r2+p2s
B prs=q2+r2p
C qrs=p2+s2q
D pqs=s2+q2r

The correct answer is:
pqs=s2+q2r

Solution:
Since, the equation has a real root
â´ z=x+iy=x
âx2+p+iqx+r+is=0
â´x2+px+r=0 and qx+s=0
Putting x=-sq in the first equation, we get,
pqs=s2+q2r which is the required condition.

Topic: Mathematics, Complex Number, JEE Main
For a complex number \( z \), the equation \( z^{2}+(p+i q) z+r+i s=0 \) has a real root (where \( p, q, r, s \) are non-zero real numbers and \( i^{2}=-1 \) ), then

Options:
A \( p q r=r^{2}+p^{2} s \)
B \( p r s=q^{2}+r^{2} p \)
C \( q r s=p^{2}+s^{2} q \)
D \( p q s=s^{2}+q^{2} r \)

The correct answer is:
\( p q s=s^{2}+q^{2} r \)

Solution:
Since, the equation has a real root
â´ z=x+iy=x
âx2+p+iqx+r+is=0
â´x2+px+r=0 and qx+s=0
Putting x=-sq in the first equation, we get,
pqs=s2+q2r which is the required condition.

Topic: Mathematics, Complex Number, JEE Main
For a complex number z, the product of the real parts of the roots of the equation z2-z=5-5i is (where, i=-1)

Options:
A -25
B -6
C -5
D 25

The correct answer is:
-6

Solution:
Equation is z2-z-5+5i=0
⇒z=1±1+20-20i2
⇒z=1±5-2i22
⇒z=1±5-2i2=3-i,-2+i
The required value =-6

Topic: Mathematics, Complex Number, JEE Main
For a complex number Z, Z=1 and argZ=Î¸. If ZZ2Z3....Zn=1, then the value of Î¸ is

Options:
A 4mπnn+1,m∈Ι
B 2mπnn+1,m∈Ι
C mπnn+1,m∈Ι
D None of these

The correct answer is:
4mπnn+1,m∈Ι

Solution:
Let, Z=eiθ, Z2=ei2θ,..........,Zn=einθ
Given, ZZ2.......Zn=1
⇒eiθei2θ.......einθ=1
⇒eiθ1+2+....+n=ei2mπ m∈Ι
⇒θnn+12=2mπ, m∈Ι
⇒θ=4mπnn+1, m∈Ι

Topic: Mathematics, Complex Number, JEE Main
For a concentrated solution of a weak electrolyte (Keq=equilibrium constant) A2B3 of concentration ‘C’, the degree of dissociation ‘α’ is

Options:
A Keq5c415
B Keq108c415
C Keq25c215
D Keq6c515

The correct answer is:
Keq108c415

Solution:
Given that concentration of A2B3 is C and degree of dissociation is α

The equilibrium reaction of weak electrolyte A2B3 is

A2 B3⇌2 A3++3 B2-C--↓↓↓C(1-α)2Cα3Cα

Now, The equilibrium constant is the ratio of the concentrations of reactants and products raised to the stoichiometric coefficients.

Keq=(2Cα)2(3Cα)3C(1-α)=108C4α5

Keq108C4=α5⇒α=Keq108C415

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2023 (06 Apr Shift 1)
For a concrete sphere of radius R having a cavity of radius r packed with sawdust, to float with its entire volume submerged under water what will be the ratio of mass of concrete to mass of sawdust? Given the specific gravities of concrete and sawdust are respectively 2.4 and 0.3 .

Options:
A 8
B 4
C 3
D zero

The correct answer is:
4

Solution:
Let specific gravities of concrete and sawdust are Ï1 and Ï2 respectively.
According to principle of floatation, weight of whole sphere = upthrust on the sphere
4 3 Ï R 3 - r 3 Ï 1 g + 4 3 Ï r 3 Ï 2 g = 4 3 Ï R 3 Ã 1 Ã g
â R 3 Ï 1 - r 3 Ï 1 + r 3 Ï 2 = R 3
â R 3 Ï 1 - 1 = r 3 Ï 1 - Ï 2 â R 3 r 3 = Ï 1 - Ï 2 Ï 1 - 1
â R 3 - r 3 r 3 = Ï 1 - Ï 2 - Ï 1 + 1 Ï 1 - 1 â R 3 - r 3 Ï 1 r 3 Ï 2 = ( 1 - Ï 2 Ï 1 - 1 ) Ï 1 Ï 2
â Mass of concrete Mass of saw dust = ( 1 - 0 . 3 2 . 4 - 1 ) Ã 2 . 4 0 . 3 = 4

Topic: Physics, Mechanical Properties of Fluids, JEE Main
For a constant collector-emitter voltage of 8 V, the collector current of a transistor reached to the value of 6 mA from 4 mA, whereas base current changed from 20 μA to 25 μA value. If transistor is in active state, small signal current gain (current amplification factor) will be

Options:
A 240
B 400
C 0.0025
D 200

The correct answer is:
400

Solution:
Given here, VCE=8 V, IC=6 mA from 4 mA and IB=20 μA to 25 μA

Current gain is the ratio of change in output current divided by change in input current. For common CE (Common Emitter) configuration, it is the ratio of change in collector current divided by change in base current.

Current gain βav=IC-'ICIB-'IB=6-4 mA25-20 μA=2 mA5 μA

βav=25×103=20005=400

Topic: Physics, Semiconductors, JEE Main, JEE Main 2022 (27 Jul Shift 2)
For a continuous series the mode is computed by the formula

Options:
A $l+\frac{f_{m-1}}{f_m-f_{m-1}-f_{m+1}} \times C$ or $l+\left(\frac{f_1}{f_m-f_1-f_2}\right) \times i$
B $l=\frac{f_m-f_{m-1}}{f_m-f_{m-1}-f_{m+1}} \times C$ or $l+\frac{f_m-f_1}{f_m-f_1-f_2} \times i$
C $l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times C$ or $l+\frac{f_m-f_1}{2 f_m-f_1-f_2} \times i$
D $l+\frac{2 f_m-f_{m-1}}{f_m-f_{m-1}-f_{m+1}} \times C$ or $l+\frac{2 f_m-f_1}{f_m-f_1-f_2} \times i$

The correct answer is:
$l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times C$ or $l+\frac{f_m-f_1}{2 f_m-f_1-f_2} \times i$

Solution:
$l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times C$ or $l+\left(\frac{f_m-f_1}{2 f_m-f_1-f_2}\right)$
$\mathrm{I}=$ lower boundary of the modal class
$\mathrm{C}$ or $\mathrm{i}=$ size of the modal class
$f_m=a$ frequency corresponding to the modal
class
$f_{m-1} / f_1=a$ frequency of preceding to the modal class
$f_{m+1} / f_2=$ a frequency next to the modal class

Topic: Mathematics, Statistics, JEE Main
For a diatomic ideal gas in a closed system, which of the following plots does not correctly describe the relation between various thermodynamic quantities?

Options:
A

B

C

D

The correct answer is:

Solution:
Cp is molar heat capacity at constant pressure. Since the quantity is defined at constant pressure, it does not vary with pressure.
At higher temperatures, the rotational degree of freedom becomes active.
$C_P=\frac{7}{2} R \quad \text { (Independent of } P \text { ) }$
$\mathrm{C}_{\mathrm{V}}=\frac{5}{2} \mathrm{R} \quad \text { (Independent of } \mathrm{V} \text { ) }$
$\mathrm{C}_{\mathrm{V}}$ depends on $\mathrm{T}$ and independent on $\mathrm{V}$.
$\mathrm{C}_{\mathrm{P}}$ depends on $\mathrm{T}$ and independent on $\mathrm{P}$.

Topic: Chemistry, Thermodynamics (C), JEE Main
For a diatomic ideal gas in a closed system, which of the following plots does not correctly describe the relation between various thermodynamic quantities?

Options:
A

B

C

D

The correct answer is:

Solution:
Cp is molar heat capacity at constant pressure. Since the quantity is defined at constant pressure, it does not vary with pressure.
At higher temperatures, the rotational degree of freedom becomes active.
$C_P=\frac{7}{2} R \quad \text { (Independent of } P \text { ) }$
$\mathrm{C}_{\mathrm{V}}=\frac{5}{2} \mathrm{R} \quad \text { (Independent of } \mathrm{V} \text { ) }$
$\mathrm{C}_{\mathrm{V}}$ depends on $\mathrm{T}$ and independent on $\mathrm{V}$.
$\mathrm{C}_{\mathrm{P}}$ depends on $\mathrm{T}$ and independent on $\mathrm{P}$.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2019 (12 Jan Shift 1)