datavorous/entrance-exam-dataset huggingface dataset:
Which of the following statement(s) is/are correct?

Options:
A
If the electric field due to a point charge varies as $r^{-2.5}$ instead of $r^{-2}$, then the Gauss's law will still be valid

B
The Gauss's law can be used to calculate the field distribution around an electric dipole

C
If the electric field between two point charges is zero somewhere, then the sign of the two charges is the same

D
The work done by the external force in moving a unit positive charge from point $A$ at potential $V_A$ to point $B$ at potential $V_B$ is $\left(V_B-V_A\right)$

The correct answers are:
If the electric field between two point charges is zero somewhere, then the sign of the two charges is the same
,
The work done by the external force in moving a unit positive charge from point $A$ at potential $V_A$ to point $B$ at potential $V_B$ is $\left(V_B-V_A\right)$

Solution:
If charges are of opposite signs, then the two fields are along the same direction. So, they cannot be zero. Hence, the charges should be of same sign.
Therefore, option (c) is correct.
Further, work done by external force = change in potential energy
$$
\begin{aligned}
& \therefore \quad W_{A \rightarrow B}=q(\Delta V) \\
& =(+1)\left(V_B-V_A\right) \\
& \text { or } \quad W_{A \rightarrow B}=V_B-V_A \\
&
\end{aligned}
$$
Therefore, option (d) is also correct.
$\therefore$ Correct options are (c) and (d).
Analysis of Question
(i) Question is simple.
(ii) $W=\Delta U$ (by external force)
If charge is just moved, without changing the kinetic energy. If nothing is mentioned in the question regarding the kinetic energy then we take $W=\Delta U$.

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2011 (Paper 2)
Which of the given statement(s) about $\mathbf{N}, \mathbf{O}, \mathbf{P}$ and $\mathbf{Q}$ with respect to $\mathbf{M}$ is (are) correct?

Options:
A $\mathbf{M}$ and $\mathbf{N}$ are non-mirror image stereoisomers
B $\mathbf{M}$ and $\mathbf{O}$ are identical
C $\mathbf{M}$ and $\mathbf{P}$ are enantiomers
D $\mathbf{M}$ and $\mathbf{Q}$ are identical

The correct answers are:
$\mathbf{M}$ and $\mathbf{N}$ are non-mirror image stereoisomers, $\mathbf{M}$ and $\mathbf{O}$ are identical, $\mathbf{M}$ and $\mathbf{P}$ are enantiomers

Solution:
The given conformation are of the following compound

Note : Fisher conformation always show the eclipsed form of molecule. Thus, we have to convert the given staggered conformation of 'O' to an eclipsed conformation.

Based on the above structures, we can say that

(a) $\mathrm{M}$ and $\mathrm{N}$ are diastereomers or non-mirror image stereoisomers.

(b) $\mathrm{M}$ and $\mathrm{O}$ are identical.

(c) $\mathrm{M}$ and $\mathrm{P}$ are enantiomers.

(d) $\mathrm{M}$ and $\mathrm{Q}$ are diastereomers and therefore not identical.

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2012 (Paper 2)
Which one of the following options represents the magnetic field B→ at O due to the current flowing in the given wire segments lying on the xy plane?

Options:
A B→=-μ0IL32+142πk^
B B→=-μ0IL32+122πk^
C B→=-μ0IL1+142πk^
D B→=-μ0IL1+14πk^

The correct answer is:
B→=-μ0IL1+142πk^

Solution:
Magnetic field due to a straight wire will be zero if the line of the wire passes through the point of interest. Therefore, In the above figure,

BBC+BDE+BFG=0

Therefore,

B→net =B→AB+B→CD +B→EF

Magnetic field due to a finite wire is given by,

B=μ0I4πdsinα+sinβ

Using above formula,

B→AB=μ0I4πL×12-k^

Magnetic field due to a circular arc is,

B=μ0I2R×θ2π

Therefore,

B→CD=μ0I4L2-k^ and B→EF=μ0I8×L4-k^

Therefore, net magnetic field

B→net=μ0I4π×L12+μ0I4L2+μ0I8×L4-k^

⇒B→net=μ0I2L+μ0I2L+μ0I42πL-k^

⇒B→net=μ0IL1+142π-k^

Topic: Physics, Magnetic Effects of Current, JEE Advanced, JEE Advanced 2022 (Paper 2)
Which one of the following statement is WRONG in the context of X-rays generated from X-ray tube?

Options:
A
Wavelength of characteristic X-rays decreases when the atomic number of the target increases

B
Cut-off wavelength of the continuous X-rays depends on the atomic number of the target

C
Intensity of the characteristics X-rays depends on the electrical power given to the X-ray tube

D
Cut-off wavelength of the continuous X-rays depends on the energy of the electrons in the X-ray tube

The correct answer is:
Cut-off wavelength of the continuous X-rays depends on the atomic number of the target

Solution:
Cut-off wavelength depends on the applied voltage not on the atomic number of the target. Characteristic wavelengths depend on the atomic number of target.

Topic: Physics, Electromagnetic Waves, JEE Advanced, JEE Advanced 2008 (Paper 1)
Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen?

Options:
A $\mathrm{HNO}_{3}, \mathrm{NO}, \mathrm{NH}_{4} \mathrm{Cl}, \mathrm{N}_{2}$
B $\mathrm{HNO}_{3}, \mathrm{NO}, \mathrm{N}_{2}, \mathrm{NH}_{4} \mathrm{Cl}$
C $\mathrm{HNO}_{3}, \mathrm{NH}_{4} \mathrm{Cl}, \mathrm{NO}, \mathrm{N}_{2}$
D $\mathrm{NO}, \mathrm{HNO}_{3}, \mathrm{NH}_{4} \mathrm{Cl}, \mathrm{N}_{2}$

The correct answer is:
$\mathrm{HNO}_{3}, \mathrm{NO}, \mathrm{N}_{2}, \mathrm{NH}_{4} \mathrm{Cl}$

Solution:

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Advanced, JEE Advanced 2012 (Paper 1)
Why does ozone hole forms over Antarctica? How will enhanced ultraviolet radiation affect us?

Options:

None

Solution:
A large amount of ODS (Ozone Depleting Sub-stances) like $\mathrm{CFCs}, \mathrm{N}_2 \mathrm{O}$, halons, $\mathrm{SO}_2, \mathrm{CH}_4, \mathrm{Cl}^{-}$are released by advanced countries like USA, Japan, European countries. These are released in stratosphere, drift towards poles and reach there before the coming of winter. During winter (temp. $85^{\circ} \mathrm{C}$ ) ice clouds are formed over Antarctica and no sunrise is received in polar areas. It catalyses release of $\mathrm{Cl}$ from CFCs. With the coming of spring season, $\mathrm{Cl}$ reacts with ozone in the presence of sunlight and converts $\mathrm{O}_3$ into $\mathrm{O}_2$ causing ozone depletion/thining of ozone shield in stratosphere called ozone hole. This hole disappears in summer due to free mixing of air of Antarctica with the rest of the global air.
Effect of Enhanced UV Radiation. (i) Snow blindness or inflammation of cornea (ii) Damage of skin cells and development of skin cancer. (iii) Damage to nucleic acids and proteins. (iv) Reduced immunity (v) Higher number of cataracts in humans.

Topic: Biology, Environmental Issues
With reference to the scheme given below, which of the given statement(s) about $\mathrm{T}, \mathrm{U}, \mathrm{V}$ and $\mathrm{W}$ is (are) correct?

Options:
A $\mathrm{T}$ is soluble in hot aqueous $\mathrm{NaOH}$
B $\mathrm{U}$ is optically active
C Molecular formula of $\mathrm{W}$ is $\mathrm{C}_{10} \mathrm{H}_{18} \mathrm{O}_{4}$
D V gives effervescence on treatment with aqueous $\mathrm{NaHCO}_{3}$.

The correct answers are:
$\mathrm{T}$ is soluble in hot aqueous $\mathrm{NaOH}$, Molecular formula of $\mathrm{W}$ is $\mathrm{C}_{10} \mathrm{H}_{18} \mathrm{O}_{4}$, V gives effervescence on treatment with aqueous $\mathrm{NaHCO}_{3}$.

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Advanced, JEE Advanced 2012 (Paper 2)
With respect to graphite and diamond, which of the statement(s) given below is (are) correct?

Options:
A Graphite is harder than diamond.
B Graphite has higher electrical conductivity than diamond.
C Graphite has higher thermal conductivity than diamond.
D Graphite has higher C-C bond order than diamond.

The correct answers are:
Graphite has higher electrical conductivity than diamond., Graphite has higher C-C bond order than diamond.

Solution:
(a) Diamond is harder than graphite.

(b) Graphite is good conductor of electricity as each carbon is attached to three C-atoms leaving one valency free, which is responsible for electrical conduction, while in diamond, all the four valencies of carbon are satisfied, hence insulator.

(c) Diamond is better thermal conductor than graphite. Whereas electrical conduction is due to availability of free electrons; thermal conduction is due to transfer of thermal vibrations from atom to atom. A compact and precisely aligned crystal like diamond thus facilitates fast movement of heat.

(d) In graphite, $\mathrm{C}-\mathrm{C}$ bond acquires double bond character, hence higher bond order than in diamond.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Advanced, JEE Advanced 2012 (Paper 2)