datavorous/entrance-exam-dataset huggingface dataset:
The trajectory of projectile is

Options:
A semicircle
B an ellipse
C a parabola always
D a parabola in the absence of air resistance

The correct answer is:
a parabola in the absence of air resistance

Solution:
The trajectory of a projectile is a parabola in the absence of air resistance.
In the presence of air resistance, range and maximum height of projectile, both will decrease, hence path of projectile will not remain parabolic.

Topic: Physics, Motion In Two Dimensions, KCET, KCET 2021
The trans-alkenes are formed by the reduction of alkynes with

Options:
A Sn-HCl
B H2-Pd/C, BaSO4
C NaBH4
D Na/liq.NH3

The correct answer is:
Na/liq.NH3

Solution:
Birch reduction is anti-addition. So trans alkene will be produced.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2018 (08 Apr)
The transfer characteristic curve of a transistor, having input and output resistance 100 Î© and 100 kÎ© respectively, is shown in the figure. The voltage and power gain, are respectively:

Options:
A 5Ã104, 2.5Ã106
B 2.5Ã104,Â 2.5Ã106
C 5Ã104,Â 5Ã106
D 5Ã104,Â 5Ã105

The correct answer is:
5Ã104, 2.5Ã106

Solution:
Current gain Î²=ÎICÎIB= 5 Ã10-3 100 Ã10-6=50
Voltage gain (AV)=Î²RoutRin=50Ã100Ã103100=5Ã104
Power gain =AVÃÎ²=50Ã5Ã104=2.5Ã106

Topic: Physics, Semiconductors, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The transfer characteristic curve of a transistor, having input and output resistance 100 Î© and 100 kÎ© respectively, is shown in the figure. The voltage and power gain, are respectively:

Options:
A 5Ã104, 2.5Ã106
B 2.5Ã104,Â 2.5Ã106
C 5Ã104,Â 5Ã106
D 5Ã104,Â 5Ã105

The correct answer is:
5Ã104, 2.5Ã106

Solution:
Current gain Î²=ÎICÎIB= 5 Ã10-3 100 Ã10-6=50
Voltage gain (AV)=Î²RoutRin=50Ã100Ã103100=5Ã104
Power gain =AVÃÎ²=50Ã5Ã104=2.5Ã106

Topic: Physics, Semiconductors, JEE Main
The transfer RNA anticodon for the messenger RNA codon G-C-A is

Options:
A C-G-U
B C-C-U
C U-C-C
D G-U-C

The correct answer is:
C-G-U

Solution:
The transfer RNA anticodon for messanger RNA codon G-C-A is

C-G-U, because in anticodon of RNA,

G replaced by C
C replaced by G
A replaced by U

Here, C(cytosine) forms triple bond with G (guanine) and A (adenine) forms double bond with U (uracil).

Topic: Chemistry, Biomolecules, JEE Main
The transformation given below can be performed by

Options:
A (i) PCC (ii) MeMgBr (iii) POCl3
B TsCl/MeMgBr
C NaOH/MeI
D (i) HBr (ii) Mg (iii) MeI

The correct answer is:
(i) PCC (ii) MeMgBr (iii) POCl3

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The transformation occurring in Duma's method is given below
C2H7 N+2x+y2CuO→xCO2+y2H2O+z2 N2+2x+y2Cu
The value of y is (Integer answer)

Options:

The correct answer is:
7

Solution:
Dumas method formula:

CxHy Nz+2x+y2CuO→xCO2+y2H2O+z2 N2+2x+y2Cu

Comparing it with the given compound C2H7 N we get y=7

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2021 (31 Aug Shift 2)
The transition elements are more metallic than the representative elements because they have

Options:
A Electron pairs in d-orbitals
B Availability of d-orbitals for bonding
C The electron in p-orbitals
D Unpaired electron in metallic orbitals

The correct answer is:
Availability of d-orbitals for bonding

Solution:
Transition elements are more metallic than representative elements due to the availability of d-orbitals for bonding

Topic: Chemistry, d and f Block Elements, JEE Main
The transition elements exhibit variable oxidation states because

Options:
A they form large number of complexes
B \( (\mathrm{n}-1) \mathrm{d} \) and \( \mathrm{ns} \) electrons are of nearly the same energy level
C they have high density
D they have high melting and boiling points

The correct answer is:
\( (\mathrm{n}-1) \mathrm{d} \) and \( \mathrm{ns} \) electrons are of nearly the same energy level

Solution:
Transition elements are those elements in which, partially filled d-orbitals are present, either in their atomic state or in ionic state. The electrons present in (n –1)d and ns orbitals are of nearly the same energy level, due to which, the transition elements exhibit variable oxidation states. Both the energy levels can be used for bond formation.

Topic: Chemistry, d and f Block Elements, JEE Main
The transition elements that has the lowest enthalpy of atomisation is:

Options:
A V
B Fe
C Zn
D Cu

The correct answer is:
Zn

Solution:
Zn is usually not considered as a transition element as the d-orbital electrons do not take part in reactions. This is why Zn exhibit only +2 oxidation state.
But since, it is placed in the d-block along with other transition metals, it is technically a d-block element and a transition metal. Zinc has the lowest enthalpy of atomization among the first row transition metals.

Elements	V	Cu	Zn	Fe

Enthalpy of atomization	515 KJ/mol	339 KJ/mol	216 KJ/mol	416 KJ/mol

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2019 (09 Jan Shift 2)
The transition from the state $n=3$ to $n=1$ in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from

Options:
A $2 \rightarrow 1$
B $3 \rightarrow 2$
C $4 \rightarrow 2$
D $4 \rightarrow 3$

The correct answer is:
$4 \rightarrow 3$

Solution:
Infrared radiation is found in Paschan, Brackett and pfund series and it is obtain when electron transition occur from high energy level to minimum third level.

Topic: Physics, Atomic Physics, NEET, NEET 2012 (Mains)
The transition from the state $n=4$ to $n=3$ in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from

Options:
A
$2 \rightarrow 1$

B
$3 \rightarrow 2$

C
$4 \rightarrow 2$

D
$5 \rightarrow 4$

The correct answer is:
$5 \rightarrow 4$

Solution:
IR corresponds to least value of $\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
i.e. from Paschen, Bracket and Pfund series. Thus the transition corresponds to $5 \rightarrow 3$.

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2009
The transition from the third shell to any other shell is known as

Options:
A Balmer series
B Paschen series
C Pfund series
D None of the above

The correct answer is:
Paschen series

Solution:
The transition from the third shell to any other shell is known as Paschen series.

Topic: Physics, Atomic Physics, JEE Main
The transition metal having highest 3rd ionisation enthalpy is:

Options:
A Cr
B Mn
C V
D Fe

The correct answer is:
Mn

Solution:
Ionization energy, in chemistry and physics, the amount of energy required to remove an electron from an isolated atom or molecule.

3rd Ionisation energy:

V : 2833 kJ/mol

Cr: 2990 kJ/mol

Mn:3260 kJ/mol

Fe:2962 kJ/mol

alternative

Mn:3 d54 s2

Fe:3 d64 s2

Cr:3 d54 s1

V:3 d34 s2

So Mn has highest 3rd IE among all the given elements due to d5configuration

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (01 Feb Shift 2)
The transition of an electrom from $n_{2}=5,6, \ldots \ldots \ldots$. to $n_{1}=4$ gives rise to

Options:
A Pfund series
B Lyman series
C Paschen series
D Brackett series

The correct answer is:
Brackett series

Solution:
According to Bohr's, Brackett series is obtained when an electron jumps to $4^{\text {th }}$ orbit from any other outer orbit

Topic: Physics, Atomic Physics, JEE Main
The translational degrees of freedom $\left(f_t\right)$ and rotational degrees of freedom $\left(f_r\right)$ of $\mathrm{CH}_4$ molecule are:

Options:
A $f_t=2$ and $f_r=2$
B $f_t=3$ and $f_r=3$
C $f_t=3$ and $f_r=2$
D $f_t=2$ and $f_r=3$

The correct answer is:
$f_t=3$ and $f_r=3$

Solution:
Since $\mathrm{CH}_4$ is polyatomic Non-Linear
D.O.F of $\mathrm{CH}_4$
T. DOF $=3$
$\mathrm{RDOF}=3$

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2024 (04 Apr Shift 2)
The transpiration driven ascent of xylem sap depends mainly on?

Options:
A Cohesion
B Adhesion
C Surface tension
D All of the above

The correct answer is:
All of the above

Solution:
The transpiration driven ascent of xylem sap depends mainly on the

Topic: Biology, Transport in Plants, NEET
The transverse axis of a hyperbola is along the $x$ -axis and its length is $2 a$. The vertex of the hyperbola bisects the line segment joining the centre and the focus. The equation of the hyperbola is

Options:
A $6 x^{2}-y^{2}=3 a^{2}$
B $x^{2}-3 y^{2}=3 a^{2}$
C $x^{2}-6 y^{2}=3 a^{2}$
D $3 x^{2}-y^{2}=3 a^{2}$

The correct answer is:
$3 x^{2}-y^{2}=3 a^{2}$

Solution:
Let e be the eccentricity of hyperbola and length of conjugate axis be $2 h$. since, vertex $(a, 0)$ bisects the join of centre (0,0) and focus $(a e, 0)$
$$
\begin{array}{l}
a=\frac{a e+0}{2} \Rightarrow e=2 \\
b^{2}=a^{2}\left(c^{2}-1\right)=3 a^{2}
\end{array}
$$
$\therefore$ Required hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
$\Rightarrow \quad \frac{x^{2}}{a^{2}}-\frac{y^{2}}{3 a^{2}}=1 \Rightarrow 3 x^{2}-y^{2}=3 a^{2}$

Topic: Mathematics, Hyperbola, JEE Main
The transverse displacement of a wave on string is given by yx, t=e-ax2+bt2+2ab xt, this represents a _______

Options:
A Wave moving along + x-axis with speed ab
B Wave moving along - x-axis with speed ba
C Standing wave of frequency b
D Standing wave of frequency 1b

The correct answer is:
Wave moving along - x-axis with speed ba

Solution:
General equation of travelling wave is
y=fax ± bt=fx ± vt
y=e-ax2+bt2+2ab xt=e-ax +b t 2
So, the wave will be travelling along - x-axis with speed ba.

Topic: Physics, Waves and Sound, JEE Main
The transverse displacement $y(x, t)$ of a wave on a string is given by $y(x, t)=e^{-\left(a x^2+b t^2+2 \sqrt{a b} x t\right)}$. This represents a

Options:
A
wave moving in $-x$ direction with speed $\sqrt{\frac{b}{a}}$

B
standing wave of frequency $\sqrt{b}$

C
standing wave of frequency $\frac{1}{\sqrt{b}}$

D
wave moving in $+x$ direction with $\sqrt{\frac{a}{b}}$

The correct answer is:
wave moving in $-x$ direction with speed $\sqrt{\frac{b}{a}}$

Solution:
$$
y_{(x, t)}=e^{-}(\sqrt{a} x+\sqrt{b} t)^2 v=\sqrt{\frac{b}{a}}
$$
Wave moving in - ve $\mathrm{x}$-direction.

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2011
The treatment of $\mathrm{CH}_3 \mathrm{MgX}$ with $\mathrm{CH}_3 \mathrm{C} \equiv \mathrm{C}-\mathrm{H}$ produces

Options:
A
$$
\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2
$$

B

C

D

The correct answer is:

Solution:
$\mathrm{CH}_3-\mathrm{MgX}+\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{C}-\mathrm{H} \longrightarrow \mathrm{CH}_4$

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2008
The triangle of maximum area that can be inscribed in a given circle of radius 'r' is :

Options:
A An equilateral triangle having each of its side of length 3r.
B An isosceles triangle with base equal to 2r.
C An equilateral triangle of height 2r3.
D A right angle triangle having two of its sides of length 2r and r.

The correct answer is:
An equilateral triangle having each of its side of length 3r.

Solution:
h=rsinθ+r

base =BC=2rcosθ

θ∈0,π2

Area of ΔABC=12BC.h

Δ=122rcosθ.rsinθ+r

=r2cosθ.1+sinθ

dΔdθ=r2cosθddθ1+sinθ+1+sinθddθcosθ

dΔdθ=r2cosθ0+cosθ+1+sinθ-sinθ

dΔdθ=r2cos2θ-sinθ-sin2θ

dΔdθ=r21-sin2θ-sinθ-sin2θ

=r21-sinθ-2sin2θ

=r2-2sin2θ-sinθ+1

=r2-2sin2θ-2sinθ+sinθ+1

=r2-2sinθsinθ+1+1sinθ+1

=r21+sinθ⏟positive1-2sinθ=0

⇒θ=π6

⇒ Δ is maximum where θ=π6

Δmax=334r2= area of equilateral Δ with BC=3r.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (26 Feb Shift 2)
The trigger for activation of toxin of Bacillus thuringiensis is

Options:
A
acidic $\mathrm{pH}$ of stomach

B
high temperature

C
alkaline $\mathrm{pH}$ of gut

D
mechanical action in the insect gut

The correct answer is:
alkaline $\mathrm{pH}$ of gut

Solution:
$B t$ toxins are inactive protoxins but after ingested by the insect, these inactive toxin becomes active, due to the alkaline $\mathrm{pH}$ of the gut, that solublise the crystals. Hence, high $\mathrm{pH}$ value is required to make $B t$ toxins active. Under high temperature and acidic $\mathrm{pH}, \mathrm{Bt}$ toxins remain insoluble and inactive.

Topic: Biology, Biotechnology and Its Applications
The trigonometric equation $\sin ^{-1} x=2 \sin ^{-1}$ a has a solution for

Options:
A
$|\mathrm{a}| \geq \frac{1}{\sqrt{2}}$

B
$\frac{1}{2} < |\mathrm{a}| < \frac{1}{\sqrt{2}}$
C
all real values of a

D
$|a| < \frac{1}{2}$

The correct answer is:
all real values of a

Solution:
$\sin ^{-1} x=2 \sin ^{-1} a$
$-\frac{\pi}{2} \leq \sin ^{-1} \mathrm{x} \leq \frac{\pi}{2} ; \therefore-\frac{\pi}{2} \leq 2 \sin ^{-1} \mathrm{a} \leq \frac{\pi}{2} \quad-\frac{\pi}{4} \leq \sin ^{-1} \mathrm{a} \leq \frac{\pi}{4}$ or $\frac{-1}{\sqrt{2}} \leq \mathrm{a} \leq \frac{1}{\sqrt{2}}$
$\therefore|\mathrm{a}| \leq \frac{1}{\sqrt{2}}\left(\right.$ As $\frac{1}{\sqrt{2}}>\frac{1}{2}$ ). Out of given four option no one is absolutely correct but (c) could be taken into consideration. $\rightarrow|\mathrm{a}| \leq \frac{1}{\sqrt{2}}$ is correct, if $\mathrm{a} < \frac{1}{\sqrt{2}}$ is taken as correct then it domain satisfy for $\mathrm{a}=\frac{1}{\sqrt{3}}$ but equation is satisfied. $\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{3}}>\frac{1}{2}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2003
The true statement amongst the following is:

Options:
A Both ΔS and S are functions of temperature.
B Both S and ΔS are not functions of temperature
C S is not a function of temperature but ΔS is a function of temperature
D S is a function of temperature but ΔS is not a function of temperature

The correct answer is:
Both ΔS and S are functions of temperature.

Solution:
The change in entropy and entropy at the absolute scale of the temperature are the function of the temperature. The entropy change is defined as the heat change at the constant temeprature.

ΔS=∫dqT
ST=∫0TncdTT

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2020 (09 Jan Shift 2)
The truth table for the circuit given in the figure is:

Options:
A ABY000010101111
B ABY001010100110
C ABY001011101111
D ABY001011100110

The correct answer is:
ABY001011100110

Solution:
C=A+B
and y=A.C¯

ABC=(A+B)ACy=A.C¯00001011011011011110

Topic: Physics, Semiconductors, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The truth table for the following logic circuit is :

Options:
A
B
C
D

The correct answer is:

Solution:
y=(AB¯+A¯B)¯

y=AB¯·A¯B¯

y=(A¯+B)·(A+B¯)

y=A¯·A+A¯B¯+A·B+BB¯

y=AB+A¯B¯

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (25 Feb Shift 2)
The truth table for this given circuit is:

Options:
A
B
C
D

The correct answer is:

Solution:
With respect to the above diagram, the final output can be calculated as follows:

Y=A·B+A¯·B=(A+A¯)·B⇒Y=1.B⇒Y=B

Hence, the required truth table can be written as follows:

Topic: Physics, Semiconductors, JEE Main, JEE Main 2024 (29 Jan Shift 2)
The truth table given in fig. represents:

A	B	Y

0	0	0

0	1	1

1	0	1

1	1	1

Options:
A OR-Gate
B NAND-Gate
C AND-Gate
D NOR-Gate

The correct answer is:
OR-Gate

Solution:
The given truth table represents OR gate.

Its boolean algebra is Y=A+B, where A and B are input and Y is output.

OR gate gives output 1 if any one of its input is 1

Topic: Physics, Semiconductors, JEE Main, JEE Main 2016 (09 Apr Online)
The truth table of the given circuit diagram is :

Options:
A A B Y 001010100111
B A B Y 000011101110
C A B Y 000010100111
D A B Y 001011101110

The correct answer is:
A B Y 000011101110

Solution:
From the diagram, it is clear that the first set of gates are NOT gates, the second set of gates are AND gates and the third and final gate is an OR gate.

The final output from all the combination of the gates can be written as

Y=A·B¯+A¯·B

This indicates a XOR GATE. Hence, this option indicates the correct truth table.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2024 (27 Jan Shift 2)
The truth table shown in figure is for:
$\begin{array}{|l|l|l|l|l|} \hline A & 0 & 0 & 1 & 1 \\ \hline B & 0 & 1 & 0 & 1 \\ \hline Y & 1 & 0 & 0 & 1 \\ \hline \end{array}$

Options:
A XOR
B AND
C XNOR
D OR

The correct answer is:
XNOR

Solution:
For 'XNOR' gate $Y=\bar{A} \bar{B}+A B$
$\begin{aligned}
& \text { i.e. } \overline{0} \cdot \overline{0}+0.0=1.1+0.0=1+0=1 \\
& \overline{0} \cdot \overline{1}+0.1=1 \cdot 0+0.1=0+0=0 \\
& \overline{1} \cdot \overline{0}+1 \cdot 0=0.1+1.0=0+0=0 \\
& \overline{1} \cdot \overline{1}+1 \cdot 1=0.0+1 \cdot 1=0+1=1
\end{aligned}$

Topic: Physics, Semiconductors, JEE Main
The truth-table given below is for which gate
$\begin{array}{l|l|l|l|l}
A & 0 & 0 & 1 & 1 \\
\hline B & 0 & 1 & 0 & 1 \\
\hline C & 1 & 1 & 1 & 0
\end{array}$

Options:
A XOR
B OR
C AND
D NAND

The correct answer is:
NAND

Solution:
For 'NAND' gate $C=\overline{A \cdot B}$ i.e. $\overline{0.0}=\overline{0}=1, \overline{0.1}=\overline{0}=1$ $\overline{1.0}=\overline{0}=1, \overline{1.1}=\overline{1}=0$

Topic: Physics, Semiconductors, JEE Main
The turn ratio of transformers is given as $2: 3 .$ If the current through the primary coil is $3 \mathrm{~A}$, thus calculate the current through load resistance

Options:
A $1 \mathrm{~A}$
B $4.5 \mathrm{~A}$
C $2 \mathrm{~A}$
D $1.5 \mathrm{~A}$

The correct answer is:
$2 \mathrm{~A}$

Solution:
Transformation ratio, $\frac{\mathrm{I}_{\mathrm{P}}}{\mathrm{I}_{\mathrm{S}}}=\frac{\mathrm{n}_{\mathrm{S}}}{\mathrm{n}_{\mathrm{S}}}$
i.e. $\frac{3}{\mathrm{I}_{\mathrm{S}}}=\frac{3}{2}$ or, $\mathrm{I}_{\mathrm{S}}=2 \mathrm{~A}$

Topic: Physics, Alternating Current, JEE Main
The TV transmission tower at a particular station has a height of 125 m. For doubling the coverage of its range, the height of the tower should be increased by

Options:
A 500 m
B 375 m
C 250 m
D 125 m

The correct answer is:
375 m

Solution:
Maximum distance covered is given by d=2Rh, here R is the radius.

Now, to increase the length of radio transmission to 2 times,2d=2h'R

Height of tower is h'=4h=4×125=500 m

\(\begin{aligned} \Delta H=h^{\prime}-h &=500-125 \\ &=375 \end{aligned}\)

Topic: Physics, Communication System, JEE Main, JEE Main 2022 (29 Jun Shift 2)
The two circles x2+y2=ax and x2+y2=c2 c>0 touch each other, if ca is equal to

Options:
A 2
B 12
C 1
D None of these

The correct answer is:
1

Solution:
The two circles touch each other internally as shown in the figure below

For the circle x2+y2-2ax=0, the centre is at a2,0 and radius is a2 units.

Also, the circle is passing through the origin.

For the circle x2+y2=a2, the centre is at 0,0 and radius is c units

From the figure,

c=2a2=a

Hence, ca=1

Topic: Mathematics, Circle, JEE Main
The two circles $x^2+y^2=a x$ and $x^2+y^2=c^2(c>0)$ touch each other if

Options:
A
$|a|=c$

B
$a=2 c$

C
$|a|=2 c$

D
$2|a|=c$

The correct answer is:
$|a|=c$

Solution:
$\mathrm{c}_1=\left(\frac{\mathrm{a}}{2}, 0\right) ; \mathrm{c}_2=(0,0)$
$\mathrm{r}_1=\frac{\mathrm{a}}{2} ; \mathrm{r}_2=\mathrm{c}$
$\mathrm{c}_1 \mathrm{c}_2=\mathrm{r}_1-\mathrm{r}_2 \Rightarrow \frac{\mathrm{a}}{2}=\mathrm{c}-\frac{\mathrm{a}}{2} \Rightarrow \mathrm{c}=\mathrm{a}$

Topic: Mathematics, Circle, JEE Main, JEE Main 2011
The two electrons have the following set of quantum numbers :
\[
\begin{array}{l}
P=3,2,-2,+\frac{1}{2} \\
Q=3,0,0,+\frac{1}{2}
\end{array}
\]
Which of the following statement is true?

Options:
A \( P \) and \( Q \) have same energy
B \( P \) has greater energy than \( Q \)
C \( P \) has lesser energy than \( Q \)
D \( \mathrm{P} \) and \( \mathrm{Q} \) represent same electron

The correct answer is:
\( P \) has greater energy than \( Q \)

Solution:
For electron $P$, the values of quantum numbers are $n=3, I=2, m l=-2$ and $s=+1 / 2$. For electron $Q$ quantum numbers are $n=3, I=0, m l=0$ and $s=+1 / 2$. Therefore, from the quantum number values, electron $P$ belongs to $3 d-$ orbital and electron $Q$ to $3 s$ - orbital. Energy of the orbital is $s < p < d < f$.
Hence, energy of electron $P$ is higher than electron $Q$.

Topic: Chemistry, Structure of Atom, JEE Main
The two electrons in $\mathrm{K}$ sub-shell will differ in

Options:
A Principal quantum number
B Azimuthal quantum number
C Magnetic quantum number
D Spin quantum number

The correct answer is:
Spin quantum number

Solution:
The two electrons will have opposite spins.

Topic: Chemistry, Structure of Atom, JEE Main
The two ends of a metal rod are maintained at temperatures 100 Â°C and 110 Â°C . The rate of heat flow in the rod is found to be 4.0 J s-1 . If the ends are maintained at temperatures 200 Â°C and 210 Â°C , the rate of heat flow will be:

Options:
A 44.0 J s-1
B 16.8 J s-1
C 8.0 J s-1
D 4.0 J s-1

The correct answer is:
4.0 J s-1

Solution:
In case I, temperature difference between the ends of rod is =110-100=10oC
In case II, temperature difference between the ends of rod is =210-200=10oC
Hence, in both the case temperature difference are same i.e. 10oC
∴ Rate of heat flow is also 4 J s-1 in both the cases.

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2015 (Phase 1)
The two ends of a metal rod are maintained at temperatures 100 Â°C and 110 Â°C . The rate of heat flow in the rod is found to be 4.0 J s-1 . If the ends are maintained at temperatures 200 Â°C and 210 Â°C , the rate of heat flow will be:

Options:
A 44.0 J s-1
B 16.8 J s-1
C 8.0 J s-1
D 4.0 J s-1

The correct answer is:
4.0 J s-1

Solution:
In case I, temperature difference between the ends of rod is =110-100=10oC
In case II, temperature difference between the ends of rod is =210-200=10oC
Hence, in both the case temperature difference are same i.e. 10oC
∴ Rate of heat flow is also 4 J s-1 in both the cases.

Topic: Physics, Thermal Properties of Matter, JEE Main
The two ends of a metal rod are maintained at temperatures 100 Â°C and 110 Â°C . The rate of heat flow in the rod is found to be 4.0 J s-1 . If the ends are maintained at temperatures 200 Â°C and 210 Â°C , the rate of heat flow will be:

Options:
A 44.0 J s-1
B 16.8 J s-1
C 8.0 J s-1
D 4.0 J s-1

The correct answer is:
4.0 J s-1

Solution:
In case I, temperature difference between the ends of rod is =110-100=10oC
In case II, temperature difference between the ends of rod is =210-200=10oC
Hence, in both the case temperature difference are same i.e. 10oC
∴ Rate of heat flow is also 4 J s-1 in both the cases.

Topic: Physics, Thermal Properties of Matter, NEET
The two ends of a rod of length $\mathrm{L}$ and a uniform cross-sectional area A are kept at two temperatures $T_1$ and $T_2\left(T_1>T_2\right)$. The rate of heat transfer, $\frac{\mathrm{dQ}}{\mathrm{dt}}$, through the rod in a steady state is given by :

Options:
A $\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{kA}\left(\mathrm{T}_1-\mathrm{T}_2\right)}{\mathrm{L}}$
B $\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{kL}\left(\mathrm{T}_1-\mathrm{T}_2\right)}{\mathrm{A}}$
C $\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{k}\left(\mathrm{T}_1-\mathrm{T}_2\right)}{\mathrm{LA}}$
D $\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{kLA}\left(\mathrm{T}_1-\mathrm{T}_2\right)$

The correct answer is:
$\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{kA}\left(\mathrm{T}_1-\mathrm{T}_2\right)}{\mathrm{L}}$

Solution:
For steady state $\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{kA}\left(\mathrm{T}_1-\mathrm{T}_2\right)}{\mathrm{L}}$

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2009 (Mains)