datavorous/entrance-exam-dataset huggingface dataset:
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 Main
Statement 1: $y=m x-\frac{1}{m}$ is always a tangent to the parabola, $y^2=-4 x$ for all non-zero values of $m$.
Statement 2: Every tangent to the parabola, $y^2=-4 x$ will meet its axis at a point whose abscissa is non-negative.

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

B
Statement 1 is false, Statement 2 is true.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1.

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

Solution:
Both the given statements are true. Statement $-2$ is not the correct explanation for statement $-1$.

Topic: Mathematics, Parabola, JEE Main, JEE Main 2012 (07 May Online)
Statement (A) For an ideal liquid, the bulk modulus is infinite and the shear modulus is zero.
Statement (B) The volume contraction of a metal cube of bulk modulus $140 \mathrm{GPa} .10 \mathrm{~cm}$ on a edge, when subjected to a hydraulic pressure of $7 \times 10^6 \mathrm{~Pa}$ is $-0.05 \mathrm{~m}^3$.
Statement (C) A spiral spring is stretched by a weight attached to it. The strain is tensile.

Options:
A A, B and C are true.
B $A, B$ are true $C$ is false.
C $A, C$ are true $B$ is false.
D $\mathrm{B}$ and $\mathrm{C}$ are true $\mathrm{A}$ is false.

The correct answer is:
$A, C$ are true $B$ is false.

Solution:
For an ideal liquid, the bulk modulus, is infinite and shear modulus is zero.
Hence, statement (A) is correct.
In statement (B), Bulk modulus
$$
\begin{aligned}
B & =140 \mathrm{GPa} \\
& =140 \times 10^9 \mathrm{~Pa} \\
& =1.4 \times 10^{11} \mathrm{~Pa} \\
p & =7 \times 10^6 \mathrm{~Pa} \\
V & =a^3=(0.1)^3 \\
& =0.001 \mathrm{~m}^3
\end{aligned}
$$
$$
[a=10 \mathrm{~cm}=0.1 \mathrm{~m}]
$$
since, $\quad B=\frac{p}{\left(\frac{-\Delta V}{V}\right)}$
$$
\begin{aligned}
& \Rightarrow 1.4 \times 10^{11}=\frac{7 \times 10^6}{\frac{-\Delta V}{0.001}} \\
& \Rightarrow 1.4 \times 10^{11}=\frac{7 \times 10^3}{-\Delta V}
\end{aligned}
$$
$$
\Rightarrow 1.4 \times 10^{11}=\frac{7 \times 10^3}{-\Delta V}
$$
$$
\begin{aligned}
\Rightarrow \quad \Delta V & =-\frac{7 \times 10^3}{1.4 \times 10^{11}} \\
& =-5 \times 10^{-8} \mathrm{~m}^3
\end{aligned}
$$
Hence, statement (B) is incorrect.
Strain produced in spiral spring stretched by a weight attached to it, is tensile.
Hence, statement $(C)$ is correct.

Topic: Physics, Mechanical Properties of Solids, JEE Main
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 Main
Statement I: A cyclist is moving on an unbanked road with a speed of 7 km h-1 and takes a sharp circular turn along a path of the radius of 2 m without reducing the speed. The static friction coefficient is 0.2. The cyclist will not slip and pass the curve g=9.8 m s-2

Statement II : If the road is banked at an angle of 45°, cyclist can cross the curve of 2 m radius with the speed of 18.5 km h-1 without slipping. In the light of the above statements, choose the correct answer from the options given below.

Options:
A Statement I is incorrect and statement II is correct
B Statement I is correct and statement II is
incorrect
C Both statement I and statement II are false
D Both statement I and statement II are true

The correct answer is:
Both statement I and statement II are true

Solution:
Statement Ivmax=μRg=0.2×2×9.8

vmax=1.97 m s-1

7 km h-1=1.944 m s-1

Speed is lower than vmax, hence it can take safe
turn.

Statement II

vmax=Rgtanθ+μ1-μtanθ

=2×9.81+0.21-0.2=5.42 m s-1

18.5 km h-1=5.14 m s-1

Speed is lower than vmax, hence it can take safe turn.

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2021 (16 Mar Shift 2)
Statement (I) : A device in which heat measurement can
be made is called calorimeter.
Statement (II) : Skating is possible on snow due to the
formation of water below the skates. Water is formed due
to the increase of temperature and ice melts.
Statement (III) : Two bodies at different temperature are
mixed in a calorimeter. Total internal energy of the two
bodies remains conserved.
Which of the following is correct?

Options:
A Statements I, II and III are true
B Statement I is true, but statements II and III are false
C Both statements I and II are true, but statement III is false
D Both statements I, III are true, but statement II is false

The correct answer is:
Both statements I, III are true, but statement II is false

Solution:
While skating a person pressure the snow downwards
with his weight due to which melting point of ice lowers
and ice melt into water, which acts as a lubricant for skates.
So, (II) is false.
And, statement (I) and (II) are correct

Topic: Physics, Thermal Properties of Matter, TS EAMCET, TS EAMCET 2022 (20 Jul Shift 2)
Statement I: An alloy of lithium and magnesium is used to make aircraft plates.

Statement II : The magnesium ions are important for cell-membrane integrity.

In the light the above statements, choose the correct answer from the options given below

Options:
A Both Statement I and Statement II are true
B Both Statement I and Statement II are false
C Statement I is true but Statement II is false
D Statement I is false but Statement II is true

The correct answer is:
Both Statement I and Statement II are false

Solution:
Lithium metal is used to make useful alloys, for example with lead to make ‘white metal’ bearings for motor engines, with aluminium to make aircraft parts, and with magnesium to make armour plates. So, Alloy of Li and Mg is used to make armour plates and not aircraft plates.

Calcium plays important roles in neuromuscular function, interneuronal transmission and cell membrane integrity

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2022 (28 Jul Shift 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 Main
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 Main
Statement I: By doping silicon semiconductors with pentavalent material, the electrons density increases.

Statement II: The n-type of semiconductor has a net negative charge.

In the above statements, choose the most appropriate answer from the options given below:

Options:
A Both Statement I and Statement II are false.
B Statement I is true but Statement II is false.
C Statement I is false but Statement II is true.
D Both Statement I and Statement II are true.

The correct answer is:
Statement I is true but Statement II is false.

Solution:
Pentavalent material has excess free e- because we mix the impurities like phosphorus and arsenic having five electrons in the outermost shell. So e-density increases, but overall semiconductor is neutral. So we can say that Statement I is true but, Statement II is false.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (26 Aug Shift 1)
Statement I:- Dipole moment is a vector quantity and by convention it is depicted by a small arrow with tail on the negative centre and head pointing towards the positive centre.

Statement II:- The crossed arrow of the dipole moment symbolizes the direction of the shift of charges in the molecules.

In the light of the above statements, choose the most appropriate answer from the options given below:-

Options:
A Both Statement I and Statement II are correct.
B Statement I is incorrect but Statement II is correct.
C Both Statement I and Statement II are incorrect.
D Statement I is correct but Statement II is incorrect.

The correct answer is:
Statement I is incorrect but Statement II is correct.

Solution:
Dipole moment is a vector quantity and by convention it is depicted by a small arrow with tail on the negative centre and head pointing towards the positive centre. But in chemistry presence of dipole moment is represented by the crossed arrow put on Lewis structure of the molecule. The cross is on positive end and arrow head is on negative end. So, statement I is incorrect.

The crossed arrow of the dipole moment symbolises the direction of the shift of electron density in the molecules. So, statement II is correct.

Hence the correct option is B

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Statement I: For colloidal particles, the values of colligative properties are of small order as compared to values shown by true solutions at same concentration.

Statement II: For colloidal particles, the potential difference between the fixed layer and the diffused layer of same charges is called the electrokinetic potential or zeta potential.

In the light of the above statements, chooses the correct answer from the options given below.

Options:
A Statement I is true but statement II is false
B Statement I is false but Statement II is true
C Both Statement I and Statement II are true
D Both Statement I and Statement II are false

The correct answer is:
Both Statement I and Statement II are true

Solution:
The value of colligative properties of colloidal solution are of small order in comparison to those of true solution of same concentration because of colloidal particles are comparatively less in number.

The surface of a colloidal particle acquires a positive or a negative charge by selective adsorption of ions carrying +ve or – ve charges respectively. Thus, an electrical double layer is formed on the surface of the particles i.e., one due to adsorbed ions and the other due to oppositely charged ions forming a diffused layer. This potential difference between the fixed charge layer and diffused layer of opposite charge is called electro kinetic potential or zeta potential

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2023 (24 Jan Shift 1)
Statement I : If \( \int_{0}^{1} e^{\sin x} d x=\lambda \), then \( \int_{0}^{200} e^{\sin x} d x=200 \lambda \)
Statement II : \( \int_{0}^{n a} f x d x=n \int_{0}^{a} f(x) d x, n \in I \) given \( \mathrm{f}(\mathrm{a}+\mathrm{x})=\mathrm{f}(\mathrm{x}) \)

Options:
A Both Statement I and Statement II are true and the Statement II is the correct explanation of the Statement I
B Both Statement I and Statement II are true but the Statement II is not the correct explanation of the Statement I
C Statement I is true but Statement II is false
D Statement I is false but Statement II is true

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

Solution:
We have given that ∫01esinxdx=λ
f(x)=esinx
f(x) is a periodic function, its value will repeat as sinx repeat its value.
∴Period of f(x)=2π
∫0200esinxdx=∫0100π·2πesinxdx
But 1 0 0 π is not an integer
∴ Statement I is wrong.

and ∫0nafxdx=n∫0afxdx, n∈I
So Statement II is true.

Topic: Mathematics, Definite Integration, JEE Main
Statement I In an elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision.
Statement II In an elastic collision, the linear momentum of the system is conserved.

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:
In case of elastic collision, coefficient of restitution $e=1$ or relative speed of separation.
$\therefore$ Option (b) is correct.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Statement I: Leaching of gold with cyanide ion in absence of air / O2 leads to cyano complex of Au(III).

Statement II: Zinc is oxidized during the displacement reaction carried out for gold extraction.

In the light of the above statements, choose the correct answer from the options given below.

Options:
A Both Statement 1 and statement 2 are correct
B Both Statement 1 and statement 2 are incorrect
C Statement 1 is correct but statement 2 is incorrect
D Statement 1 is incorrect but statement 2 is correct

The correct answer is:
Statement 1 is incorrect but statement 2 is correct

Solution:
4 Aus+8CN-aq+2H2Oaq+O2g→4AuCN2-aq+4OH-aq

The oxidation state of gold in the complex is +1.

2AuCN2-aq+Zns→2Aus+ZnCN42-aq

The oxidation state of zinc in the complex is +2.

Oxidation state of Zn in ZnCN42- +2.

As, Oxidation state of zinc is converted from zero(elemental state) to +2.It is undergoing oxidation.

Hence,Statement 1 is incorrect but statement 2 is correct

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2022 (27 Jun Shift 2)
Statement I: Lysosomes are membrane-bound structures.

Statement II: These are golgian vacuoles.

Options:
A Both the statements are false
B Both the statements are true
C Statement I is true and statement II is false
D Statement I is flase and statement II is true

The correct answer is:
Both the statements are true

Solution:
Lysosomes are membrane-bound organelles that contain digestive enzymes that break down proteins, lipids, carbohydrates, and nucleic acids. They are important in processing the contents of vesicles taken in from outside the cell. It is crucial to maintain the integrity of the lysosomal membranes because the enzymes they contain can digest cellular components as well.
Golgian vacuoles are expanded part of the cisternae which have become modified to form vacuoles. The vacuoles develop from the concave or maturing face. Golgian vacuoles contain amorphous or granular substance. Some of the golgian vacuoles function as a lysosomes.

Topic: Biology, Cell - The Unit of Life, NEET
Statement I : Sodium hydride can be used as an oxidising agent.

Statement II : The lone pair of electrons on nitrogen in pyridine makes it basic.

Choose the CORRECT answer from the options given below :

Options:
A Both statement I and statement II are false
B Statement I is true but statement II is false
C Statement I is false but statement II is true
D Both statement I and statement II are true

The correct answer is:
Statement I is false but statement II is true

Solution:
(1) NaH (sodium Hydride) is used as a reducing reagent.

(2) In pyridine, due to free electron on N atom, it is basic in nature.

Hence, statement I is false and II is true.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2021 (16 Mar Shift 2)
Statement I : The \( 3 \)-chloro-3, 4-dimethyl hex-1-ene undergo reaction with base to give butadiene.
Statement II : This is due to Sytzeff rule which states that more substituted product will be the more stable product.

Options:
A Both Statement I and Statement II are true and the Statement II is the correct explanation of the Statement I.
B Both Statement I and Statement II are true and the Statement II is not the correct explanation of the Statement I.
C Statement I is false but Statement II is true.
D Both Statement I and Statement II are false.

The correct answer is:
Both Statement I and Statement II are true and the Statement II is not the correct explanation of the Statement I.

Solution:
This problem includes conceptual mixing of elimination reaction, Sytzeff rule and conjugation.
Elimination reaction : In presence of base alkly halide undergo elimination as follows.
Product I is obtained accordingly to conjugation which deal about stability of product to a greater extent than product II obtained according to inductive effect only.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Statement I The curve $y=-\frac{x^2}{2}+x+1$ is symmetric with respect to the line $x=1$
Statement II A parabola is symmetric about its axis.

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 a correct explanation for Statement I

Solution:
$$
\begin{array}{rlrl}
y & =-\frac{x^2}{2}+x+1 \\
\Rightarrow & y-\frac{3}{2} & =-\frac{1}{2}(x-1)^2
\end{array}
$$
It is symmetric about $x=1$.

Topic: Mathematics, Parabola, JEE Main
Statement I The equation of the director circle to the ellipse $4 x^2+9 y^2=36$ is $x^2+y^2=13$
Statement II The locus of the point of intersection of perpendicular tangents to an ellipse is called the director circle.

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 |
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:
Given, ellipse is $4 x^2+9 y^2=36$
or
$\frac{x^2}{3^2}+\frac{y^2}{2^2}=1$
Equation of tangents in terms of slope $(\mathrm{m})$ are
$\begin{array}{ll}y=m x \pm \sqrt{\left(9 m^2+4\right)} \\ \Rightarrow (y-m x)^2=9 m^2+4 \\ \Rightarrow m^2\left(x^2-9\right)-2 m x y+y^2-4=0 \Rightarrow x^2+y^2=13 \text {, which is director circle of (i). } \\ \therefore m_1 m_2=-1 \Rightarrow \frac{y^2-4}{x^2-9}=-1\end{array}$
Hence, both statements are true and statement II is a correct explanation for statement I.

Topic: Mathematics, Ellipse, JEE Main
Statement I: The equation $\left(\sin ^{-1} \mathrm{x}\right)^3+$ $\left(\cos ^{-1} \mathrm{x}\right)^3-\mathrm{a} \pi^3=0$ has a solution for all $\mathrm{a} \geq \frac{1}{32}$.
Statement II: For any $\mathrm{x} \in \mathrm{R}$,
$\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$ and $0 \leq\left(\sin ^{-1} x-\frac{\pi}{4}\right)^2 \leq \frac{9 \pi^2}{16}$

Options:
A
Both statements I and II are true.

B
Both statements I and II are false.

C
Statement $\mathrm{I}$ is true and statement $\mathrm{II}$ is false.

D
Statement $\mathrm{I}$ is false and statement II is true.

The correct answer is:
Both statements I and II are true.

Solution:
$$
\begin{aligned}
&\sin ^{-1} x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\
&\Rightarrow-\frac{3 \pi}{4} \leq\left(\sin ^{-1} x-\frac{\pi}{4}\right) \leq \frac{\pi}{4} \\
&0 \leq\left(\sin ^{-1} x-\frac{\pi}{4}\right)^2 \leq \frac{9}{16} \pi^2
\end{aligned}
$$
Statement II is true
$$
\begin{aligned}
&\left(\sin ^{-1} x\right)^3+\left(\cos ^{-1} x\right)^3=a \pi^3 \\
\Rightarrow &\left(\sin ^{-1} x+\cos ^{-1} x\right)\left[\left(\sin ^{-1} x+\cos ^{-1} x\right)^2-\right.\\
&\left.3 \sin ^{-1} x \cos ^{-1} x\right]=a \pi^3 \\
\Rightarrow & \frac{\pi^2}{4}-3 \sin ^{-1} x \cos ^{-1} x=2 a \pi^2 \\
\Rightarrow & \sin ^{-1} x\left(\frac{\pi}{2}-\sin ^{-1} x\right)=\frac{\pi^2}{12}(1-8 a) \\
\Rightarrow &\left(\sin ^{-1} x-\frac{\pi}{4}\right)^2=\frac{\pi^2}{12}(8 a-1)+\frac{\pi^2}{16} \\
\Rightarrow &\left(\sin ^{-1} x-\frac{\pi}{4}\right)^2=\frac{\pi^2}{48}(32 a-1)
\end{aligned}
$$
Putting this value in equation (1)
$$
0 \leq \frac{\pi^2}{48}(32 a-1) \leq \frac{9}{16} \pi^2
$$
$$
\begin{aligned}
&\Rightarrow 0 \leq 32 a-1 \leq 27 \\
&\frac{1}{32} \leq a \leq \frac{7}{8}
\end{aligned}
$$
Statement-I is also true.

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2014 (12 Apr Online)