datavorous/entrance-exam-dataset huggingface dataset:
Statement 1 For practical purposes, the earth is used as a reference at zero potential in electrical circuits.
and Statement 2 The electrical potential of a sphere of radius $R$ with charge $Q$ uniformly distributed on the surface is given by $\frac{Q}{4 \pi \varepsilon_0 R}$.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

Solution:
No solution is required.

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2008 (Paper 2)
Statement $1 \mathrm{In}$ a Meter Bridge experiment, null point for an unknown resistance is measured. Now, the unknown resistance is put inside an enclosure maintained at a higher temperature. The null point can be obtained at the same point as before by decreasing the value of the standard resistance.
and Statement 2 Resistance of a metal increase with increase in temperature.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is false, Statement 2 is true

Solution:
With increase in temperature, the value of unknown resistance will increase.
In balanced Wheatstone bridge condition, $\frac{R}{X}=\frac{l_1}{l_2}$
Here, $R=$ value of standard resistance,
$X=$ value of unknown resistance .
To take null point at same point or $\frac{l_1}{l_2}$ to remain unchanged, $\frac{R}{X}$ should also remain unchanged. Therefore, if $X$ is increasing $R$ should also increase.
$\therefore$ correct option is (d).

Topic: Physics, Current Electricity, JEE Advanced, JEE Advanced 2008 (Paper 1)
Statement 1 It is easier to pull a heavy object than to push it on a level ground.
and Statement 2 The magnitude of frictional force depends on the nature of the two surfaces in contact.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

Solution:
Both statements are correct. But statement-2, does not explain correctly, statement- 1 . Correct explanation is : "There is increase in nomal reaction when the object is pushed and there is decrease in normal reaction when the object is pulled (but strictly not horizontally).

Topic: Physics, Laws of Motion, JEE Advanced, JEE Advanced 2008 (Paper 2)
Statement $1 \mathrm{~Pb}^{4+}$ compounds are stronger oxidising agents than $\mathrm{Sn}^{4+}$ compounds.
Statement 2 The higher oxidation states for the group 14 elements are more stable for the heavier members of the group due to 'inert pair effect'.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is true, Statement 2 is false.

Solution:
In early p-block, on moving top to bottom, the stability of lower oxidation state increases due to inert pair effect. $\mathrm{Pb}^{4+}$ is less stable than $\mathrm{Sn}^{4+}$, making it a better oxidising agent The lower oxidation state for the group 14 elements are more stable for the heavier members.

Topic: Chemistry, Chemical Kinetics, JEE Advanced, JEE Advanced 2008 (Paper 1)
Statement 1 The geometrical isomer of the complex $\left.\left[\mathrm{M}_{\left(\mathrm{NH}_3\right.}\right)_4 \mathrm{Cl}_2\right]$ are optically inactive.
Statement 2 Both geometrical isomers of the complex $\left[\mathrm{M}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]$ possess axis of symmetry.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

Solution:
$$
\text { The cis and trans both form of complex }\left[M\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \text { are optically inactive }
$$
There are plane of symmetry in addition to that there is alternate axis of symmetry $\left(C_n\right)$ which also causes their optical inactivity.
A molecule possesses $n$ fold alternating axis of symmetry, if when rotated through an angle of $360^{\circ} / \mathrm{n}$ about the axis and then followed by reflection in a plane perpendicular to the axis, the molecule is indistinguishable from the original molecule.

Topic: Chemistry, Coordination Compounds, JEE Main
Statement 1 The geometrical isomer of the complex $\left.\left[\mathrm{M}_{\left(\mathrm{NH}_3\right.}\right)_4 \mathrm{Cl}_2\right]$ are optically inactive.
Statement 2 Both geometrical isomers of the complex $\left[\mathrm{M}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]$ possess axis of symmetry.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

Solution:
$$
\text { The cis and trans both form of complex }\left[M\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \text { are optically inactive }
$$
There are plane of symmetry in addition to that there is alternate axis of symmetry $\left(C_n\right)$ which also causes their optical inactivity.
A molecule possesses $n$ fold alternating axis of symmetry, if when rotated through an angle of $360^{\circ} / \mathrm{n}$ about the axis and then followed by reflection in a plane perpendicular to the axis, the molecule is indistinguishable from the original molecule.

Topic: Chemistry, Coordination Compounds, JEE Advanced, JEE Advanced 2008 (Paper 2)
Statement 1 The plot of atomic number ( $Y$-axis) versus number of neutrons ( $x$-axis) for stable nuclei shows a curvature towards $X$-axis from the line of $45^{\circ}$ slope as the atomic number is increased.
Statement 2 Proton-proton electrostatic repulsions begin to overcome attractive forces involving protons and neutrons in heavier nuclides.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

Solution:
The stability relationship can be represented by a line with a slope of $45^{\circ}$, i.e. the maximum stability is attained, when $N=Z$.
Right or the curve a radioactive nuclide would be neutron rich and would decay be $\beta$-emission to produce a daughter nucleus with a lower $\frac{n}{p}$ ratio.
For heavier nuclides, $p-p$ repulsions start to offset the attractive forces and an excess of neutrons over protons, is required for stability

Topic: Chemistry, Structure of Atom, JEE Main
Statement 1 The plot of atomic number ( $Y$-axis) versus number of neutrons ( $x$-axis) for stable nuclei shows a curvature towards $X$-axis from the line of $45^{\circ}$ slope as the atomic number is increased.
Statement 2 Proton-proton electrostatic repulsions begin to overcome attractive forces involving protons and neutrons in heavier nuclides.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

Solution:
The stability relationship can be represented by a line with a slope of $45^{\circ}$, i.e. the maximum stability is attained, when $N=Z$.
Right or the curve a radioactive nuclide would be neutron rich and would decay be $\beta$-emission to produce a daughter nucleus with a lower $\frac{n}{p}$ ratio.
For heavier nuclides, $p-p$ repulsions start to offset the attractive forces and an excess of neutrons over protons, is required for stability

Topic: Chemistry, Structure of Atom, JEE Advanced, JEE Advanced 2008 (Paper 1)
Statement 1 The sensitivity of a moving coil galvanometer is increased by placing a suitable magnetic material as a core inside the coil.
and Statement 2 Soft iron has a high magnetic permeability and cannot be easily magnetized or demagnetized.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is true, Statement 2 is false.

Solution:
$c \phi=B I N A$
$$
\therefore \phi=\left(\frac{B N A}{c}\right) I
$$
Using iron core, value of magnetic field increases. So, deflection increases for same current. Hence, sensitivity increases. Soft iron can be easily magnetized or demagnetized.
$\therefore$ correct option is (c).

Topic: Physics, Magnetic Effects of Current, JEE Advanced, JEE Advanced 2008 (Paper 2)
Statement 1 The stream of water flowing at high speed from a garden hose pipe tends to spread like a fountain when held vertically up, but tends to narrow down when held vertically down.
and Statement 2 In any steady flow of an incompressible fluid, the volume flow rate of the fluid remains constant.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

Solution:
From continuity equation,
$$
A v=\text { constant or } A \propto \frac{1}{v}
$$
At lower heights, speed will be more. Therefore, area of cross-section will be less. $\therefore$ correct option is (a).

Topic: Physics, Mechanical Properties of Fluids, JEE Advanced, JEE Advanced 2008 (Paper 1)
Statement 1 There is a natural asymmetry between converting work to heat and converting heat to work.
Statement 2 No process is possible in which the sole result in the absorption of heat from a reservoir and its complete conversion into work.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

Solution:
There is natural asymmetry between converting work to heat and converting heat to work because of the lapse of energy in this process which has been generalised in the form of second law of thermodynamics (Statement II).

Topic: Chemistry, Thermodynamics (C), JEE Advanced, JEE Advanced 2008 (Paper 2)
Statement 1 Two cylinders, one hollow (metal) and the other solid (wood) with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first.
and Statement 2 By the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is false, Statement 2 is true

Solution:
In case of pure rolling on inclined plane,
$$
\begin{aligned}
& a=\frac{g \sin \theta}{1+I / m R^2} \\
& I_{\text {solid }} < I_{\text {hollow }} \\
\therefore \quad & a_{\text {solid }}>a_{\text {hollow }}
\end{aligned}
$$
$\therefore$ solid cylinder will reach the bottom first.
Further, in case of pure rolling on stationary ground, work done by friction is zero. Therefore, mechanical energy of both the cylinders will remain constant.
$$
\begin{aligned}
& \therefore(\mathrm{KE})_{\text {Hollow }}=(\mathrm{KE})_{\text {solid }}=\text { decrease in } \mathrm{PE}=m g h \\
& \therefore \text { correct option is }(\mathrm{d})
\end{aligned}
$$

Topic: Physics, Rotational Motion, JEE Advanced, JEE Advanced 2008 (Paper 1)
Statement I A block of mass $m$ starts moving on a rough horizontal surface with a velocity $v$. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of $30^{\circ}$ with the horizontal and the same block is made to go up on the surface with the same initial velocity $v$. The decrease in the mechanical energy in the second situation is smaller than that in the first situation.
Statement II The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination.

Options:
A
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

B
Statement I is true, Statement II is true; Statement II is NOT a correct explanation for Statement I

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is true, Statement II is false

Solution:
In statement I Decrease in mechanical energy in case I will be
$$
\Delta U_1=\frac{1}{2} m v^2
$$
But decrease in mechanical energy in case II will be
$$
\begin{aligned}
& \Delta U_2=\frac{1}{2} m v^2-m g h \\
& \Delta U_2 < \Delta U_1
\end{aligned}
$$
$$
\therefore \quad \Delta U_2 < \Delta U_1
$$
or Statement $\mathrm{I}$ is correct.
In Statement II Coefficient of friction will not change or this Statement is wrong.
$\therefore$ Option (c) is correct.

Topic: Physics, Work Power Energy, JEE Advanced, JEE Advanced 2007 (Paper 1)
Statement I A cloth covers a table. Some dishes are kept on it. The cloth can be pulled out without dislodging the dishes from the table.
Statement II For every action there is an equal and opposite reaction.

Options:
A
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement.

B
Statement I is true, Statement II is true; Statement II is NOT a correct explanation for Statement I

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is true, Statement II is true; Statement II is NOT a correct explanation for Statement I

Solution:
The cloth can be pulled out without dislodging the dishes from the table due to law of inertia, which is Newton's first law. While the Statement II is true, but it is Newton's third law.
$\therefore$ Correct option is (b).

Topic: Physics, Laws of Motion, JEE Advanced, JEE Advanced 2007 (Paper 2)
Statement I A vertical iron rod has a coil of wire wound over it at the bottom end. An alternating current flows in the coil. The rod goes through a conducting ring as shown in the figure. The ring can float at a certain height above the coil.
Statement II In the above situation, a current is induced in the ring which interacts with the horizontal component of the magnetic field to produce an average force in the the upward direction.

Options:
A
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement

B
Statement I is true, Statement II is true; Statement II is NOT a correct explanation for Statement I

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement

Solution:
The induced current in the ring will interact with horizontal component of magnetic field and both will repel each other.
This repulsion will balance the weight of ring.
$$
\text { Hence, option (a) is correct. }
$$

Topic: Physics, Magnetic Effects of Current, JEE Main
Statement I A vertical iron rod has a coil of wire wound over it at the bottom end. An alternating current flows in the coil. The rod goes through a conducting ring as shown in the figure. The ring can float at a certain height above the coil.
Statement II In the above situation, a current is induced in the ring which interacts with the horizontal component of the magnetic field to produce an average force in the the upward direction.

Options:
A
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement

B
Statement I is true, Statement II is true; Statement II is NOT a correct explanation for Statement I

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement

Solution:
The induced current in the ring will interact with horizontal component of magnetic field and both will repel each other.
This repulsion will balance the weight of ring.
$$
\text { Hence, option (a) is correct. }
$$

Topic: Physics, Magnetic Effects of Current, JEE Advanced, JEE Advanced 2007 (Paper 2)
Statement I Alkali metals dissolve in liquid ammonia to give blue solutions. Statement II Alkali metals in liquid ammonia give solvated species of the type $\left(M\left(\mathrm{NH}_3\right)_n\right]^{+}(M=$ alkali metals $)$.

Options:
A
Statement I is true, Statement II is true, Statement II is a correct explanation for Statement I

B
Statement I is true, Statement II is true, Statement II is not a correct explanation for Statement I

C
Statement I is true, Statement II is false

D
Statement $\mathrm{I}$ is false, Statement $\mathrm{II}$ is true

The correct answer is:
Statement I is true, Statement II is true, Statement II is not a correct explanation for Statement I

Solution:
Alkali metals in ammonia gives blue colour due to ammoniated $e^{-}$

Topic: Chemistry, s Block Elements, JEE Advanced, JEE Advanced 2007 (Paper 2)
Statement I Band gap in germanium is small.
Statement II The energy spread of each germanium atomic energy level is infinitesimally small.

Options:
A
Statement I is true, Statement II is true, Statement II is a correct explanation for Statement I

B
Statement I is true, Statement II is true, Statement II is not a correct explanation for Statement I

C
Statement-I is true, Statement II is false

D
Statement-I is talse, Statement II is true

The correct answer is:
Statement-I is true, Statement II is false

Solution:
Germanium, semi-conductor substance, has small band gap in comparison to insulator (non-metal).
On moving top to bottom in the group 14 band gap decreases.

Topic: Chemistry, Structure of Atom, JEE Advanced, JEE Advanced 2007 (Paper 2)
Statement I Boron always forms covalent bond
Statement II The small size of $\mathrm{B}^{3+}$ favours formation of covalent bond.

Options:
A
Statement I is true, Statement II is true, Statement II is a correct explanation for Statement I

B
Statement I is true, Statement II is true, Statement II is not a correct explanation for Statement II

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is true, Statement II is true, Statement II is a correct explanation for Statement I

Solution:
Boron always form covalent bond because boron requires very high energy to form $\mathrm{B}^{3+}$ and again $\mathrm{B}^{3+}$ due to its very small size having high polarising power thus cause greater polarisation and eventually significant covalent characteristics-Fajans rule.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Advanced, JEE Advanced 2007 (Paper 1)
Statement I Glucose gives a reddish-brown precipitate with Fehling's solution.
Statement II Reaction of glucose with Fehling's solution gives $\mathrm{CuO}$ and gluconic acid.

Options:
A
Statement I is true, Statement II is true, Statement II is a correct explanation for Statement I

B
Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is true, Statement II is false

Solution:
Glucose gives a reddish brown precipitate with Fehling's solution. Solution consists of $\mathrm{Cu}^{2+}$ which gets reduced to $\mathrm{Cu}^{+}\left(\mathrm{Cu}_2 \mathrm{O}\right)$.

Topic: Chemistry, Biomolecules, JEE Advanced, JEE Advanced 2007 (Paper 2)
Statement I If the accelerating potential in an X-ray tube is increased, the wavelengths of the characteristic $X$-rays do not change.
Statement II When an electron beam strikes the target in an X-ray tube, part of the kinetic energy is converted into X-ray energy.

Options:
A
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

B
Statement I is true, Statement II is true; Statement II is NOT a correct explanation for Statement I

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is true, Statement II is true; Statement II is NOT a correct explanation for Statement I

Solution:
Cut-off wavelength depends on the accelerating voltage, not the characteristic $\mathrm{X}$-rays. Further approximately $2 \%$ kinetic energy of the electrons is utilised in producing X-rays. Rest $98 \%$ is lost in heat.
$\therefore$ Option (b) is correct.

Topic: Physics, Electromagnetic Waves, JEE Advanced, JEE Advanced 2007 (Paper 1)
Statement I If there is no external torque on a body about its centre of mass, then the velocity of the centre of mass remains constant.
Statement II The linear momentum of an isolated system remains constant.

Options:
A
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement

B
Statement I is true, Statement II is true; Statement II is NOT a correct explanation for Statement I

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is false, Statement II is true

Solution:
If a force is applied at centre of mass of a rigid body, its torque about centre of mass will be zero, but acceleration will be non-zero. Hence, velocity will change.
$\therefore$ Correct option is $(\mathrm{d})$.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main