datavorous/entrance-exam-dataset huggingface dataset:
Starting from the origin, a body oscillates simple harmonically with a period of $2 \mathrm{~s}$. After what time will its kinetic energy be $75 \%$ of the total energy?

Options:
A
$\frac{1}{12} \mathrm{~s}$

B
$\frac{1}{6} \mathrm{~s}$

C
$\frac{1}{4} \mathrm{~s}$

D
$\frac{1}{3} s$

The correct answer is:
$\frac{1}{6} \mathrm{~s}$

Solution:
$\frac{1}{2} m v^2=\frac{3}{4}\left(\frac{1}{2} m v_{\max }^2\right)$
$\mathrm{A}^2 \omega^2 \cos ^2 \omega \mathrm{t} \Rightarrow \frac{3}{4} \mathrm{~A}^2 \omega^2$
$\cos \omega \mathrm{t}=\frac{\sqrt{3}}{2}$
$\omega t=\frac{\pi}{6} \Rightarrow t=\frac{1}{6}$ sec

Topic: Physics, Oscillations, JEE Main, JEE Main 2006
Starting from the origin at time t=0, with initial velocity 5 j⏜ ms−1, a particle moves in the x -y plane with a constant acceleration of 10i⏜+4j⏜ms−2. At time t, its coordinates are 20 m, y0 m. The values of t and y0 are, respectively:

Options:
A 2 s and 18 m
B 4 s and 52 m
C 2 s and 24 m
D 5 s and 25 m

The correct answer is:
2 s and 18 m

Solution:
Equation 1

Sx=12axt2

20=12×10×t2

t=2

Equation 2

Sy=uyt+12ayt2

y=52+12422

y=18

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2020 (04 Sep Shift 1)
Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after some time. Has equilibrium been reached?

Options:
A No, equilibrium has not been reached.
B Yes, equilibrium has been reached.
C Cannot be determined with the given information.
D Invalid data

The correct answer is:
Yes, equilibrium has been reached.

Solution:
Correct Option is : (B)
Yes, equilibrium has been reached.

Topic: Chemistry, Ionic Equilibrium, NEET
Starting with a sample of pure ${ }^{66} \mathrm{Cu}, 7 / 8$ of it decays into $\mathrm{Zn}$ in 15 minutes. The corresponding half-life is

Options:
A
10 minutes

B
15 minutes

C
5 minutes

D
$7 \frac{1}{2}$ minutes

The correct answer is:
5 minutes

Solution:
$\mathrm{N}_0 \stackrel{\mathrm{t}_{1 / 2}}{\longrightarrow} \frac{\mathrm{N}_0}{2} \stackrel{\mathrm{t}_{1 / 2}}{\longrightarrow} \frac{\mathrm{N}_0}{4} \stackrel{\mathrm{t}_{1 / 2}}{\longrightarrow} \frac{\mathrm{N}_0}{8}$
$3 \mathrm{t}_{1 / 2}=15 \therefore \mathrm{t}_{1 / 2}=5$

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2005
Starting with the same initial conditions, an ideal gas expands from volume V1 to V2 in three different ways. The work done by the gas is W1 if the process is purely isothermal, W2, if the process is purely adiabatic and W3 if the process is purely isobaric. Then, choose the correct option

Options:
A W2<W1<W3
B W1<W2<W3
C W2<W3<W1
D W3<W1<W2

The correct answer is:
W2<W1<W3

Solution:
As we know, in isobaric process pressure remains constant and magnitude of slope of adiabatic curve will be more than the isothermal curve at the common point, therefore diagram for process will be as given below.

Also, W= area under P-V curve.

Hence,

∴ W2<W1<W3

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2022 (29 Jun Shift 2)
State a condition under which a biomolecular reaction is kinetically first order reaction.

Options:

None

Solution:
A biomolecular reaction is kinetically first order when one of the reactant is in excess such that its concentration remains constant and rate of such reaction depends upon concentration of one reactant only. Such reaction is called pseudo first order reaction.

Topic: Chemistry, Chemical Kinetics
State reflection property of a parabola:

Options:
A The tangent line at a point $\mathrm{P}$ on a parabola makes $90^{\circ}$ angle with the line through $\mathrm{P}$ parallel to the axis of symmetry and the line through $\mathrm{P}$ and the focus.
B The tangent line at a point $\mathrm{P}$ on a parabola makes equal angle with the line through $\mathrm{P}$ parallel to the axis of symmetry and the line through $\mathrm{P}$ and the Pole.
C The tangent line at a point $P$ on a parabola makes equal angle with the line through $\mathrm{P}$ parallel to the axis of symmetry and the line through $\mathrm{P}$ and the focus.
D None of the above

The correct answer is:
The tangent line at a point $\mathrm{P}$ on a parabola makes equal angle with the line through $\mathrm{P}$ parallel to the axis of symmetry and the line through $\mathrm{P}$ and the Pole.

Solution:
Refer the above image which shows that the tangent line at a point $\mathrm{P}$ on a parabola makes equal angle with the line through $\mathrm{P}$ parallel to the axis of symmetry and the line through $\mathrm{P}$ and the focus which is known as reflective property of the parabola.

Topic: Mathematics, Parabola, JEE Main
State the different modes of $\mathrm{CO}_2$ transport in blood.

Options:

None

Solution:
The blood carries carbon dioxide in three forms.
(i) In dissolved State About $7 \%$ of $\mathrm{CO}_2$ is carried by physical solution. Under normal temperature and pressure.
(ii) As carbamino Compounds-Carbon dioxide binds directly with $\mathrm{Hb}$ to form an unstable compound carbamino compounds $\left(\mathrm{CO}_2 \mathrm{Hb}\right)$. About $23 \% \mathrm{CO}_2$ is transported in this form. When $\mathrm{pCO}_2$ is high and $\mathrm{pO}_2$ is low as in the tissues, more binding of $\mathrm{CO}_2$ occurs whereas, when $\mathrm{pCO}_2$ is low and $\mathrm{pO}_2$ is high as in alveoli as tissue dissociation of $\mathrm{CO}_2$ from carbamino-haemoglobin takes place.
$\mathrm{HbO}_2+\mathrm{CO}_2 \rightleftharpoons \mathrm{HbCO}_2+\mathrm{H}^{+}+\mathrm{O}_2$
(iii) As bicarbonate lons $\mathrm{CO} 2$ reacts with water in the presence of carbonic anhydrase to form carbonic acid $\left(\mathrm{H}_2 \mathrm{CO}_3\right)$ in $\mathrm{RBC} \mathrm{H}_2 \mathrm{CO}_3$ dissociates into hydrogen and bicarbonate ions $\left(\mathrm{HCO}_3^{-}\right)$.
The whole reaction proceeds as follows
The carbonic anhydrase reaction mainly occur in RBC as it contain high concentration of enzyme carbonic anhydrase and minute quantity of it is present in plasma too.

Topic: Biology, Breathing and Exchange of Gases
State the volume of air remaining in the lungs after a normal breathing.

Options:

None

Solution:
Volume of air remaining in lungs after a normal respiration is called functional residual capacity. It includes $E R V+R V$ $=$ Expiratory reserve volume $+$ residual volume

Topic: Biology, Breathing and Exchange of Gases
Statement 1: A function $f: R \rightarrow R$ is continuous at $x_0$ if and only if $\lim _{x \rightarrow x_0} f(x)$ exists and $\lim _{x \rightarrow x_0} f(x)=f\left(x_0 \cdot\right)$
Statement 2: A function $f: R \rightarrow R$ is discontinuous at $x_0$ if and only if, $\lim _{x \rightarrow x_0} f(x)$ exists and $\lim _{x \rightarrow x_0} f(x) \neq f\left(x_0.\right)$

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1.

B
Statement 1 is false, Statement 2 is true.

C
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation of Statement 1.

D
Statement 1 is true, Statement 2 is false.

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

Solution:
Statement-1 is true.
It is the definition of continuity.
Statement $-2$ is false.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2012 (12 May Online)
Statement $1$: An equation of a common tangent to the parabola $y^2=16 \sqrt{3} x$ and the ellipse $2 x^2+y^2=4$ is $y=2 x+2 \sqrt{3}$.
Statement $2$: If the line $y=m x+\frac{4 \sqrt{3}}{m},(m \neq 0)$ is a common tangent to the parabola $y^2=16 \sqrt{3} x$ and the ellipse $2 x^2+y^2=4$, then $m$ satisfies $m^4+2 m^2=24$.

Options:
A
Statement $1$ is false, statement $2$ is true

B
Statement $1$ is true, statement $2$ is true; statement $2$ is a correct explanation for statement $1$

C
Statement $1$ is true, statement $2$ is true; statement $2$ is not a correct explanation for statement $1$

D
Statement $1$ is true, statement $2$ is false

The correct answer is:
Statement $1$ is true, statement $2$ is true; statement $2$ is a correct explanation for statement $1$

Solution:
$y^2=16 \sqrt{3} x$
$\frac{x^2}{2}+\frac{y^2}{4}=1$
$y=m x+\frac{4 \sqrt{3}}{m}$ is tangent to parabola
which is tangent to ellipse
$\Rightarrow c^2=a^2 m^2+b^2$
$\Rightarrow \frac{48}{m^2}=2 m^2+4 \quad \Rightarrow m^4+2 m^2=24 \quad \Rightarrow m^2=4$

Topic: Mathematics, Parabola, JEE Main, JEE Main 2012 (Offline)
Statement 1 Aniline on reaction with $\mathrm{NaNO}_2 / \mathrm{HCl}$ at $0^{\circ} \mathrm{C}$ followed by coupling with $\beta$-naphthol gives a dark coloured precipitate.
Statement 2 The colour of the compound formed in the reaction of aniline with $\mathrm{NaNO}_2 / \mathrm{HCl}$ at $0^{\circ} \mathrm{C}$ followed by coupling with $\beta$-naphthol is due to the extended conjugation.

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:
Statement II is correct. Statement I appears correct in the light of Statement II. What is the colour of a dye that dye is not in included in JEE syllabus but what is the cause and understanding behind a colour could be essentially expected.
Concept Usually a certain proportion of the flight is absorbed and the rest reflected, the substance has the colour of the reflected light.
If only a single band is absorbed, the substance has the complementary colour.
- The wavelength of photon absorbed depends on the energy difference between the excited and ground states. The smaller the difference, the longer being the wavelength.
- The larger the number of electrons involved in resonance, the smaller is the energy difference between the ground and excited state. Hence, the more extended the conjugation in a molecule and the greater the contribution of charged structures, the longer is the wavelength of the photon required to excite the molecule.
- Especially, the presence of $+I /+M$ and $-I /-M$ groups in the conjugated system will both extend the conjugation and increase the contributions of charged structures to the resonance hybrid. Therefore, in these circumstances, the wavelength should become progressively longer.
- Depending on this discussion it is apparent that the concerned resultant azodye will absorb higher wavelength.
$\begin{array}{ccc}\text { Wavelength } & \text { Colour absorbed } & \text { Complimentary colour } \\ 490-500 & \text { Blue-green } & \text { Red } \\ 580-595 & \text { Yellow } & \text { Blue } \\ 605-750 & \text { Red } & \text { Blue-green }\end{array}$

Topic: Chemistry, Amines, JEE Main
Statement 1: Cryopreservation is one of the best methods of germplasm storage.

Statement 2: In cryopreservation, the cells remain in a state of suspended metabolism.

Options:
A Statement 1 is True, Statement 2 is True; Statement 2 is the correct explanation for Statement 1
B Statement 1 is True, Statement 2 is True; Statement 2 is not the 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 the correct explanation for Statement 1

Solution:
Although some microorganisms and higher plant pollen can be freeze dried, suspension of growth in higher plant somatic tissues can be achieved only by transfer to ultra low temperatures such as that of liquid nitrogen. At these temperatures, metabolism is suspended, and free radical damage caused by ionizing background radiation is the only factor potentially causing deterioration.

Topic: Biology, Biodiversity and its Conservation, NEET
Statement - 1: For every natural number $n \geq 2, \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{n}}>\sqrt{n}$.
Statement $-2$ : For every natural number $n \geq 2, \sqrt{n(n+1)} < n+1$.

Options:
A
Statement $-1$ is false, Statement $-2$ is true

B
Statement $-1$ is true, Statement $-2$ is true, Statement $-2$ is a correct explanation for Statement $-1$

C
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement $-1$.

D
Statement $-1$ is true, Statement $-2$ is false.

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

Solution:
$$
\begin{aligned}
& P(n)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{n}} \\
& P(2)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}>\sqrt{2}
\end{aligned}
$$
Let us assume that $P(k)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{k}}>\sqrt{k}$ is true $\therefore P(k+1)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}>\sqrt{k+1}$ has to be true. L.H.S. $>\sqrt{\mathrm{k}}+\frac{1}{\sqrt{\mathrm{k}+1}}=\frac{\sqrt{\mathrm{k}(\mathrm{k}+1)}+1}{\sqrt{\mathrm{k}+1}}$
Since $\sqrt{k(k+1)}>k \quad(\forall k \geq 0)$
$$
\therefore \frac{\sqrt{\mathrm{k}(\mathrm{k}+1)}+1}{\sqrt{\mathrm{k}+1}}>\frac{\mathrm{k}+1}{\sqrt{\mathrm{k}+1}}=\sqrt{\mathrm{k}+1}
$$
Let $P(n)=\sqrt{n(n+1)} < n+1$
Statement $-1$ is correct.
$$
P(2)=\sqrt{2 \times 3} < 3
$$
If $P(k)=\sqrt{k(k+1)} < (k+1)$ is true
Now $P(k+1)=\sqrt{(k+1)(k+2)} < k+2$ has to be true
Since $(k+1) < k+2$
$$
\therefore \sqrt{(\mathrm{k}+1)(\mathrm{k}+2)} < (\mathrm{k}+2)
$$
Hence Statement $-2$ is not a correct explanation of Statement $-1$.

Topic: Mathematics, Basic of Mathematics, JEE Main, JEE Main 2008
Statement 1: If $A$ and $B$ be two sets having $p$ and $q$ elements respectively, where $q>p$. Then the total number of functions from set $A$ to set $B$ is $q^p$
Statement 2: The total number of selections of $p$ different objects out of $q$ objects is ${ }^q \mathrm{C}_p$.

Options:
A
Statement 1 is true, Statement 2 is false.

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

C
Statement 1 is false, Statement 2 is true

D
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation of Statement 1.

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

Solution:
Statement-1: $n(A)=p, n(B)=q, q>p$
Total number of functions from $A \rightarrow B=q^p$ It is a true statement.
Statement - 2 : The total number of selections of $p$ different objects out of $q$ objects is ${ }^q \mathrm{C}_p$.
It is also a true statement and it is a correct explanation for statement $-1$ also.

Topic: Mathematics, Functions, JEE Main, JEE Main 2012 (12 May Online)
Statement 1: If the points $(1,2,2),(2,1,2)$ and $(2,2, z)$ and $(1,1,1)$ are coplanar, then $z=2$.
Statement 2: If the 4 points $P, Q, R$ and $S$ are coplanar, then the volume of the tetrahedron $P Q R S$ is 0.

Options:
A
Statement 1 is false,, Statement 2 is true.

B
Statement 1 is true, Statement 2 is false.

C
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation of Statement 1.

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 false,, Statement 2 is true.

Solution:
Statement-1
Points (1, 2, 2), (2, 1, 2), (2, 2, z) and (1, 1, 1) are coplanar then $z=2$ which is false.
$$
\begin{aligned}
& \because\left|\begin{array}{ccc}
1 & -1 & 0 \\
1 & 0 & z-2 \\
0 & -1 & -1
\end{array}\right|=0 \\
& \Rightarrow 1(z-2)+1(-1)=0 \Rightarrow z=3
\end{aligned}
$$
Statement $-2$ is the true statement.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2012 (12 May Online)
Statement 1: If the system of equations $x+k y+$ $3 z=0,3 x+k y-2 z=0,2 x+3 y-4 z=0$ has a nontrivial solution, then the value of $k$ is $\frac{31}{2}$.
Statement 2: A system of three homogeneous equations in three variables has a non trivial solution if the determinant of the coefficient matrix is zero.

Options:
A
Statement 1 is false, Statement 2 is true.

B
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

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

D
Statement 1 is true, Statement 2 is false.

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

Solution:
Given system of equations is
$$
\begin{aligned}
& x+k y+3 z=0 \\
& 3 x+k y-2 z=0 \\
& 2 x+3 y-4 z=0
\end{aligned}
$$
Since, system has non-trivial solution
$$
\begin{aligned}
& \therefore\left|\begin{array}{ccc}
1 & k & 3 \\
3 & k & -2 \\
2 & 3 & -4
\end{array}\right|=0 \\
\Rightarrow & 1(-4 k+6)-k(-12+4)+3(9-2 k)=0 \\
\Rightarrow & 4 k+33-6 k=0 \Rightarrow k=\frac{33}{2}
\end{aligned}
$$
Hence, statement - 1 is false.
Statement- 2 is the property.
It is a true statement.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2012 (26 May Online)
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 Main
Statement 1: The degrees of the differential equations $\frac{d y}{d x}+y^2=x$ and $\frac{d^2 y}{d x^2}+y=\sin x$ are equal.
Statement 2: The degree of a differential equation, when it is a polynomial equation in derivatives, is the highest positive integral power of the highest order derivative involved in the differential equation, otherwise degree is not defined.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is not 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 a correct explanation of Statement 1.

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

Solution:
Statement-1
Given differential equations are $\frac{d y}{d x}+y^2=x$ and $\frac{d^2 y}{d x^2}+y=\sin x$
Their degrees are 1 .
Both have equal degree.
Also, Statement $-2$ is the correct explanation for Statement - 1.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2012 (12 May Online)
Statement 1: The only circle having radius $\sqrt{10}$ and a diameter along line $2 x+y=5$ is $x^2+y^2-6 x$ $+2 y=0$.
Statement $2: 2 x+y=5$ is a normal to the circle $x^2+y^2-6 x+2 y=0$.

Options:
A
Statement 1 is false; Statement 2 is true.

B
Statement 1 is true; Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

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 for Statement 1.

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

Solution:
Circle: $x^2+y^2-6 x+2 y=0$
Line: $2 x+y=5$
Centre $=(3,-1)$
Now, $2 \times 3-1=5$, hence centre lies on the given line. Therefore line passes through the centre. The given line is normal to the circle.
Thus statement- 2 is true, but statement- 1 is not true as there are infinite circle according to the given conditions.

Topic: Mathematics, Circle, JEE Main, JEE Main 2013 (25 Apr Online)
Statement 1: The shortest distance between the lines $\frac{x}{2}=\frac{y}{-1}=\frac{z}{2}$ and $\frac{x-1}{4}=\frac{y-1}{-2}=\frac{z-1}{4}$ is $\sqrt{2}$.
Statement 2: The shortest distance between two parallel lines is the perpendicular distance from any point on one of the lines to the other line.

Options:
A
Statement 1 is true, Statement 2 is false.

B
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

C
Statement 1 is false, Statement 2 is true.

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

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

Solution:
On solving we will get shortest distance
$$
\neq \sqrt{2}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2012 (19 May Online)
Statement $1$: The sum of the series $1+(1+2+4)+(4+6+9)+(9+12+16)+\ldots \ldots+(361+380+ 400)$ is $8000$.
Statement $2$: $\sum_{k=1}^n\left(k^3-(k-1)^3\right)=n^3$ for any natural number $n$.

Options:
A
Statement $1$ is false, statement $2$ is true.

B
Statement $1$ is true, statement $2$ is true; statement $2$ is a correct explanation for statement $1$

C
Statement $1$ is true, statement $2$ is true; statement $2$ is not a correct explanation for statement $1$

D
Statement $1$ is true, statement $2$ is false

The correct answer is:
Statement $1$ is true, statement $2$ is true; statement $2$ is a correct explanation for statement $1$

Solution:
Statement $1$ has $20$ terms whose sum is $8000$
And statement $2$ is true and supporting statement $1$.
$\because k^{\text {th }}$ bracket is $(k-1)^2+k(k-1)+k^2=3 k^2-3 k+1$.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2012 (Offline)
Statement 1: The variance of first $n$ odd natural numbers is $\frac{n^2-1}{3}$
Statement 2: The sum of first $\mathrm{n}$ odd natural number is $n^2$ and the sum of square of first $n$ odd natural numbers is $\frac{n\left(4 n^2+1\right)}{3}$.

Options:
A
Statement 1 is true, Statement 2 is false.

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

C
Statement 1 is false, Statement 2 is true.

D
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

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

Solution:
Statement 2 : Sum of first $n$ odd natural numbers is not equal to $n^2$.
So, statement - 2 is false.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2012 (26 May Online)
Statement 1: The vectors $\vec{a}, \vec{b}$ and $\vec{c}$ lie in the same plane if and only if $\vec{a} \cdot(\vec{b} \times \vec{c})=0$
Statement 2: The vectors $\vec{u}$ and $\vec{v}$ are perpendicular if and only if $\vec{u} \cdot \vec{v}=0$ where $\vec{u} \times \vec{v}$ is a vector perpendicular to the plane of $\vec{u}$ and $\vec{v}$

Options:
A
Statement 1 is false, Statement 2 is true.

B
Statement 1 is true, Statement 2 is true, Statement 2 is correct explanation for Statement 1.

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 for Statement 1.

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

Solution:
Statement - 1
The vectors $\vec{a}, \vec{b}$ and $\vec{c}$ lie in the same plane.
$\Rightarrow \vec{a}, \vec{b}$ and $\vec{c}$ are coplanar.
We know, the necessary and sufficient conditions for three vectors to be coplanar is that $[\vec{a} \vec{b} \vec{c}]=0$
i.e. $\vec{a} \cdot(\vec{b} \times \vec{c})=0$
Hence, statement-1 is true.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2012 (26 May Online)