datavorous/entrance-exam-dataset huggingface dataset:
Suppose the initial charge on the capacitor in $\mathrm{q}_{\mathrm{n}} \cdot 7.7$ is $6 \mathrm{mC}$. What is the total energy stored in the circuit initially? What is the total energy at later time?

Options:

None

Solution:
Given: Capacity $=30 \mu \mathrm{F}$, charge $\mathrm{q}=6 \mathrm{mC}$
To find: Total energy U
Formula: $U=\frac{q^2}{2 C}$
For a capacitor, energy stored
$$
\begin{aligned}
&U=\frac{q^2}{2 C}=\frac{\left(6 \times 10^{-3}\right)^2}{2 \times 30 \times 10^{-6}} \\
&U=\frac{36 \times 10^{-6}}{60 \times 10^{-6}}=0.6 \mathrm{~J}
\end{aligned}
$$
This initial energy of capacitor is shared by inductance during oscillation (capacitor charges and discharges through the inductor) but the total energy of the LC tank circuit remains same.

Topic: Physics, Alternating Current
Suppose the law of gravitational attraction suddenly changes and becomes an inverse cube law i.e. $F \propto 1 / r^3$, but still remaining a central force. Then

Options:
A Keplers law of areas still holds
B Keplers law of period still holds
C Keplers law of areas and period still hold
D Neither the law of areas, nor the law of period still holds

The correct answer is:
Neither the law of areas, nor the law of period still holds

Solution:
Neither the law of areas, nor the law of period still holds.
If the law of gravitational attraction becomes cube low then,
$F=\frac{G M m}{r^3}=\frac{m v^2}{r}$ or, $v=\sqrt{\frac{G M}{r^2}}=\frac{\sqrt{G M}}{r}$
$\therefore$ Time period, $T=\frac{2 \pi r}{\nu}=\frac{2 \pi r^2}{\sqrt{G m}}$
$T^2 \propto r^4 \rightarrow$ law of period doesn't hold.
It means a planet will not have an elliptical orbit. and circular orbit also may not be possible as gravitational force obey inverse square law not cube law. So, Kepler's and law (law of areas) will also not be valid.

Topic: Physics, Gravitation, JEE Main
Suppose the limit L=limn→∞n∫0111+x2ndx exists and is larger than 12. Then,

Options:
A 12<L<2
B 2<L<3
C 3<L<3
D L≥4

The correct answer is:
12<L<2

Solution:
We have,

L=limn→∞n∫0111+x2ndx

1+x2n>1+nx2

⇒11+x2n<11+nx2

⇒∫01dx1+x2n<∫01dx1+nx2

⇒∫01dx1+x2n<1ntan-1nx01

⇒∫01dx1+x2n<1ntan-1n

⇒L=limn→∞n∫0111+x2ndx<limn→∞nntan-1n

⇒L<tan-1∞=π2

∴12<L<2 ∵π2<2

Topic: Mathematics, Limits, JEE Main
Suppose the line x-2α=y-2-5=z+22 lies on the plane x+3y-2z+β=0. Then (α+β) is equal to

.

Options:

The correct answer is:
7

Solution:
Given the line x-2α=y-2-5=z+22

2, 2, -2 Lies in plane

Equation of plane x+3y-2z+β=0

2+3×2-2×-2+β=0

2+6+4+β=0

β=-12

The direction ratio of the line are α,-5,2

The line is in the plane

α×(1)-5×(3)+2×(-2)=0

α=19

Hence, α+β=19-12

=7

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (31 Aug Shift 2)
Suppose the pairs of straight lines $2 x^2+a x y+3 y^2=0$ and $2 x^2+b x y-3 y^2=0$ are such that they have one common line with the other two remaining perpendicular. Then the values of $a$ and $b$ respectively are

Options:
A $-5,1$
B $5,-1$
C 5,1
D $5, \frac{1}{5}$

The correct answer is:
5,1

Solution:
For $a x^2+2 h x y+b y^2=0$ ...(i)
If the slope of 2 lines representing ...(i)
are $m_1$ and $m_2$ then
$\begin{aligned} & m_1+m_2=-2 h / b \\ & m_1 \cdot m_2=a / b\end{aligned}$
For $2 x^2+a x y+3 y^2=0$ ...(ii)
If slope of common line is $m$ and slope of other 2 are $m_1$ and $m_2$ respectively
$\begin{aligned} & m+m_1=-a / 3, m . m_1=2 / 3 \\ & m+m_2=\mathrm{b} / 3, m \cdot m_2=-2 / 3 \\ & m m_1=-m m_2 \\ & m_1=-m_2\end{aligned}$
other 2 lines are perpendicular
$\begin{aligned} & m_1 \cdot m_2=-1 \\ & m_2=-1 / m_1 \\ & m_1=-m_2=-\left(\frac{-1}{m_1}\right) \\ & m_1^2=1 \\ & m_1= \pm 1 \\ & \therefore m_2=\mp 1\end{aligned}$
For $m_1=1$ and $m_2=-1$
$\begin{aligned} & m=2 / 3 \\ & m+m_1=-\frac{a}{3}=\frac{2}{3}+1=\frac{5}{3} \\ & a=-5 \\ & m+m_2=\frac{b}{3}=\frac{2}{3}-1=\frac{-1}{3} \\ & b=-1\end{aligned}$
$(a, b) \equiv(-5,-1)$
for $m_1=-1, m_2=1$
$\begin{aligned} & m=-2 / 3 \\ & m+m_1=\frac{-2}{3}-1=-\frac{5}{3}=-\frac{a}{3} \Rightarrow a=5 \\ & m+m_2=-\frac{2}{3}+1=\frac{1}{3}=\frac{b}{3} \Rightarrow b=1 \\ & (a, b) \equiv(5,1)\end{aligned}$

Topic: Mathematics, Straight Lines, AP EAMCET, AP EAMCET 2022 (05 Jul Shift 2)
Suppose the parabola $(\mathrm{y}-\mathrm{k})^{2}=4(\mathrm{x}-\mathrm{h})$, with vertex $\mathrm{A}$, passes through $\mathrm{O}=(0,0)$ and $\mathrm{L}=(0,2)$. Let $\mathrm{D}$ be an end point of the latus rectum. Let the y-axis intersect the axis of the parabola at $\mathrm{P}$. Then $\angle \mathrm{PDA}$ is equal to

Options:
A $\tan ^{-1} \frac{1}{19}$
B $\tan ^{-1} \frac{2}{19}$
C $\tan ^{-1} \frac{4}{19}$
D $\tan ^{-1} \frac{8}{19}$

The correct answer is:
$\tan ^{-1} \frac{2}{19}$

Solution:
Curve, $S:(y-k)^{2}=4(x-h)$
LLR $=4 ;$ Clearly $\mathrm{k}=1 ; \Rightarrow \mathrm{A}(\mathrm{h}, 1) \& ' \mathrm{M}^{\prime}$ is focus $(\mathrm{h}+1,1)$
So D $(\mathrm{h}+1,3)$
$\mathrm{S}_{(0,0)}=0 \Rightarrow \mathrm{k}^{2}=-4 \mathrm{~h}$ $\Rightarrow \mathrm{h}=\frac{-1}{4} \quad \Rightarrow \mathrm{D}\left(\frac{3}{4}, 3\right)$
Now $; \tan \alpha=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|=\left|\frac{\frac{8}{3}-2}{1+\frac{8}{3} \times 2}\right|=\frac{2}{19}$
where, $m_{1}=\frac{3-1}{\frac{3}{4}-0}=\frac{2}{\frac{3}{4}}=\frac{8}{3}$
$$
\mathrm{m}_{2}=\frac{3-1}{1}=2
$$

Topic: Mathematics, Parabola, JEE Main
Suppose the quadratic polynomial $\mathrm{P}(\mathrm{x})=a x^{2}+\mathrm{b} \mathrm{x}+\mathrm{c}$ has positive coefficients $\mathrm{a}, \mathrm{b}, \mathrm{c}$ in arithmetic progression in that order. If $\mathrm{P}(\mathrm{x})=0$ has integer roots $\alpha$ and $\beta$, then $\alpha+\beta+\alpha \beta$ equals

Options:
A 3
B 5
C 7
D 14

The correct answer is:
7

Solution:
$\begin{array}{l}
\mathrm{P}(\mathrm{x})=\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=\mathrm{a}(\mathrm{x}-\alpha)(\mathrm{x}-\beta) \\
\text { and } \alpha+\beta+\alpha \beta+1-1=(\alpha+1)(\beta+1)-1 \\
=\frac{(\mathrm{a}-\mathrm{b}+\mathrm{c})}{\mathrm{a}}-1 \\
\Rightarrow \alpha+\beta+\alpha \beta=\frac{\mathrm{b}}{\mathrm{a}}-1=\lambda_{1}-1
\end{array}$
i.e., $\frac{\mathrm{b}}{\mathrm{a}}$ is integer $=\lambda_{1}$
If $\mathrm{b}=\mathrm{a} \lambda_{1}$
then, $\mathrm{c}=\mathrm{a}\left(2 \lambda_{1}-1\right)$ \{because $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in A.P.\}
$\begin{aligned}
\therefore \mathrm{P}(\mathrm{x}) &=\mathrm{ax}^{2}+\mathrm{a} \lambda_{1} \mathrm{x}+\mathrm{a}\left(2 \lambda_{1}-1\right) \\
&=\mathrm{a}\left[\mathrm{x}^{2}+\lambda_{1} \mathrm{x}+\left(2 \lambda_{1}-1\right)\right]
\end{aligned}$
$\mathrm{D}=\lambda_{1}^{2}-4\left(2 \lambda_{1}-1\right)$ is perfect square for integral roots $\Rightarrow \mathrm{D}=\lambda_{1}{ }^{2}-8 \lambda_{1}+4$ is perfect square
Let $\mathrm{D}=\left(\lambda_{1}-4\right)^{2}-12=\mathrm{k}^{2}\{$ where $\mathrm{k} \in \mathrm{I}\}$
$\Rightarrow\left(\lambda_{1}-4-\mathrm{k}\right)\left(\lambda_{1}-4+\mathrm{k}\right)=12$
This gives $\lambda_{1}-4-\mathrm{k}=2$
$\begin{array}{c}
\Rightarrow \quad \lambda_{1}-4=4 & \& k=1 \\
\Rightarrow \alpha_{1}=8 \\
\therefore \alpha+\beta+\alpha \beta=8-1=7
\end{array}$

Topic: Mathematics, Quadratic Equation, JEE Main
Suppose the sides of a triangle form a geometric progression with common ratio r. Then $r$ lies in the interval-

Options:
A $\left(0, \frac{-1+\sqrt{5}}{2}\right]$
B $\left(\frac{1+\sqrt{5}}{2}, \frac{2+\sqrt{5}}{2}\right]$
C $\left(\frac{-1+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right]$
D $\left(\frac{2+\sqrt{5}}{2}, \infty\right)$

The correct answer is:
$\left(\frac{-1+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right]$

Solution:
$\mathrm{}$
$$
\begin{array}{l}
a+a r>a r^{2} \\
r^{2}-r-1 < 0 \\
r \in\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)...(1) \\
a r^{2}+a r>a \\
r^{2}+r-1>0 \\
r>\frac{-1+\sqrt{5}}{2}, r < \frac{-1-\sqrt{5}}{2}...(2)
\end{array}
$$
$\mathrm{ar}^{2}+\mathrm{a}>$ ar ; $\quad \mathrm{r}^{2}-\mathrm{r}+1>0$ always true
ving (1) \& (2)
$$
\mathrm{r} \in\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)
$$

Topic: Mathematics, Sequences and Series, JEE Main
Suppose there is a circuit consisting of only resistances and batteries. Suppose one is to double (or increase it to n-times) all voltages and all resistances. Show that currents are unaltered. Do this for circuit of ExampLes 3,7 in the NCERT Text Book for Class XII.

Options:

None

Solution:
Consider the effective internal resistance of the battery is $R_{\text {eff }}$ the effective external resistance $R$ and the effective voltage of the battery is $\mathrm{V}_{\text {eff }}$.
Applying Ohm's law, $(V=I R)$
Then current through $R$ is
If all the resistances and the effective voltage are increased n-times, then we get
and
$$
\begin{aligned}
&\mathrm{V}_{\text {eff }}^{\text {new }}=\mathrm{nV}_{\text {eff }} \mathrm{R}_{\text {eff }}^{\text {new }}=\mathrm{nR}_{\mathrm{eff}} \\
&\mathrm{R}^{\text {new }}=\mathrm{nR}
\end{aligned}
$$
Then, the new current in circuit is :
$$
\begin{aligned}
\mathrm{I}_2 &=\frac{\mathrm{nV}_{\text {eff }}}{n R_{\text {eff }}+\mathrm{nR}} \\
&=\frac{\mathrm{n}\left(\mathrm{V}_{\mathrm{eff}}\right)}{\mathrm{n}\left(\mathrm{R}_{\text {eff }}+\mathrm{R}\right)} \\
&=\frac{\left(\mathrm{V}_{\mathrm{eff}}\right)}{\left(\mathrm{R}_{\mathrm{eff}}+\mathrm{R}\right)}=\mathrm{I}_1
\end{aligned}
$$
Hence, current remains the same.

Topic: Physics, Current Electricity
Suppose we consider a large number of containers each containing initially 10000 atoms of a radioactive material with a half life of 1 yr. After 1 year,

Options:
A all the containers will have 5000 atoms of the material

B
all the contains will conatin the same number of atoms of the material but that number will only be approximately 5000

C the containers will in general have different numbers of the atoms of the material but their average will be close to 5000

D none of the containers can have more than 5000 atoms.

The correct answer is:
the containers will in general have different numbers of the atoms of the material but their average will be close to 5000

Solution:
Radioactivity is a process due to which a radiactive material spontaneously decays. Half-life time for a radioactive remain average half of its or of radioactive atoms will decay. So, the containers will in general have different number of atoms of the material,after one year means one half life i.e., average atom of radioactive subtance remain after 1 year, in each container is equal to $\frac{1}{2}$ of $10000=5000$ atoms.

Topic: Physics, Nuclear Physics
Suppose you have taken a dilute solution of oleic acid in such a way that its concentration becomes 0.01 cm3 of oleic acid per cm3 of the solution. Then you make a thin film of this solution (monomolecular thickness) of area 4 cm2 by considering 100 spherical drops of radius 340π13×10-3 cm. Then the thickness of oleic acid layer will be x×10-14 m. Where x is ________ .

Options:

The correct answer is:
25

Solution:
4tT=100×43πr3

=100×4π3×340π×10-9=10-8 cm3

tT=25×10-10 cm=25×10-12 m

t0=0.01tT=25×10-14 m⇒x=25.

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2021 (17 Mar Shift 2)
Supposing 4 letters are placed in 4 different envelopes. In how many ways can they be taken out from their original envelopes and distributed among the 4 different envelopes so that no letter remains in its original envelope?

Options:
A $9$
B $10$
C $12$
D $15$

The correct answer is:
$9$

Solution:
This is clearly a case of derangement. Using the formula for the number of derangements that are possible out of 4 letters in 4 envelopes, we get the number of ways as:
$4!(1-1+1 / 2!-1 / 3!+1 / 4!)$
$=24(1-1+1 / 2-1 / 6+1 / 24)$
$=9$.

Topic: Mathematics, Permutation Combination, JEE Main
Supposing the distance between the atoms of a diatomic gas to be constant, its specific heat at constant volume per mole (gram mole) is

Options:
A $\frac{5}{2} R$
B $\frac{3}{2} R$
C $R$
D $\frac{1}{2} R$

The correct answer is:
$\frac{5}{2} R$

Solution:
$C_V=\frac{f}{2} R$; for diatomic gas $f=5 \Rightarrow C_V=\frac{5}{2} R$

Topic: Physics, Kinetic Theory of Gases, JEE Main
Surface tension of a soap bubble is 2.0×10-2 N m-1 . Work done to increase the radius of soap bubble from 3.5 cm to 7 cm will be : [Take π=227 ]

Options:
A 0.72×10-4 J
B 5.76×10-4 J
C 18.48×10-4 J
D 9.24×10-4 J

The correct answer is:
18.48×10-4 J

Solution:
As we know,

Work done = change in surface energy of soap bubble

Therefore,

Work done =2T4πr22-4πr12

=8πTr22-r12

=8×227×0.0272-3.52×10-4 J

=18.48×10-4J

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Surface tension of the liquid is S. Work done in increasing the radius of soap bubble from S. to 3R at given temperature will be

Options:
A 64πSR2
B 18πSR23
C 8πSR2
D 16πSR2

The correct answer is:
64πSR2

Solution:
Given, surface tension of the liquid is S.
Work done in blowing a soap bubble, so as to increase its radius from R1=R to R2=3R is given by
W=S⋅8Ï [R22-R12]
where S= surface tension
W=S⋅8Ï[(3R)2-(R)2]
=S⋅8Ï [9R2-R2]
=S⋅8Ï⋅8R2=64 ÏR2S

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Sustainable agriculture does not include

Options:
A Ecofriendly cropping
B Green revolution
C IPM
D Biological control

The correct answer is:
Green revolution

Solution:
The Green Revolution in India refers to a period when Indian agriculture was converted into an industrial system due to the adoption of modern methods and technology such as the use of high yielding variety (HYV) seeds, tractors, irrigation facilities, pesticides, and fertilizers. It was mainly found by M.S. Swaminathan. This was part of the larger Green revolution endeavor initiated by Norman Borlaug, which leveraged agricultural research and technology to increase agricultural productivity in the developing world.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
Sweet potato is a modified root. Potato is a modified stem. Both these plants are classified in order

Options:
A Polymoniales
B Sapindales
C Poales
D Asparagales

The correct answer is:
Polymoniales

Solution:
Ipomoea batatas or sweet potato is a plant which belongs to the family Convolvulaceae. It is also called the Morning glory Family. Potato belongs to the family Solanaceae. Both are included in the order Polymoniales.
Generally, order and other higher taxonomic categories are identified based on the aggregates of characters. Order is an assemblage of families that exhibit a few similar characters. The similar characters are less in number as compared to different genera included in a family. Plant families like Convolvulaceae, Solanaceae are included in the order polymoniales mainly based on the floral characters.

Topic: Biology, The Living World, NEET
Sweet potato is homolgous to

Options:
A potato
B colocasia
C ginger
D turnip

The correct answer is:
turnip

Solution:
Sweet potato (Ipomoea batatas) and turnip (Brassica rapa) both are roots. However, sweet potato is tuberous adventitious root while turnip is napiform tap root.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2011 (Mains)
\( S(x, y)=0 \) represent a circle. The equation \( S(x, 2)=0 \) gives two identical solutions \( x=1 \) and the equation \( S(1, y)=0 \) gives two distinct solutions \( y=0,2 \). If the equation of the circle is \( x^{2}+y^{2}-\lambda(x+y)+m=0 \) find \( \lambda+3 m \).

Options:

The correct answer is:
5

Solution:
S(x, 2) = 0 given two identical solutions x = 1
⇒ line y = 2 is a tangent to the circle S(x, y) = 0 at the point (1, 2) and S(1, y) = 0 gives two distinct solution y = 0, 2
⇒Line x = 1 cut the circle S(x, y) = 0 at points (1, 0) A(1, 2) and B(1, 0) are diametrically opposite points.
∴equation of the circle is (x – 1)2 + y (y – 2) = 0
x2 + y2 – 2x – 2y + 1 = 0

Topic: Mathematics, Circle, JEE Main
Synapsis occurs between :

Options:
A two homolog chromosomes
B a male and a female gamete
C mRNA and ribosomes
D spindle fibres and centromere

The correct answer is:
two homolog chromosomes

Solution:
No Solution Available

Topic: Biology, Cell Cycle and Cell Division, NEET, NEET 2009 (Mains)
Synthesis of each molecule of glucose in photosynthesis involves

Options:
A 8 molecules of ATP
B 6 molecules of ATP
C 18 molecules of ATP
D 10 molecules of ATP

The correct answer is:
18 molecules of ATP

Solution:
Synthesis of glucose molecule in photosynthesis involves following reaction:

1 2 H 2 O + 1 2 NADP + 18 ADP ⟶ Light reaction 6 O 2 + 1 8 ATP + 1 2 NADPH
6 CO 2 + 1 2 NADPH + 1 8 ATP ⟶ Dark reaction C 6 H 1 2 O 6 + 1 2 NADP + 1 8 ADP + 6 H 2 O
6 CO 2 + 6 H 2 O ⟶ C 6 H 1 2 O 6 + 6 O 2

Topic: Chemistry, Biomolecules, JEE Main
Synthesis of each molecule of glucose in photosynthesis involves

Options:
A 8 molecules of ATP
B 6 molecules of ATP
C 18 molecules of ATP
D 10 molecules of ATP

The correct answer is:
18 molecules of ATP

Solution:
Synthesis of glucose molecule in photosynthesis involves following reaction:

1 2 H 2 O + 1 2 NADP + 18 ADP ⟶ Light reaction 6 O 2 + 1 8 ATP + 1 2 NADPH
6 CO 2 + 1 2 NADPH + 1 8 ATP ⟶ Dark reaction C 6 H 1 2 O 6 + 1 2 NADP + 1 8 ADP + 6 H 2 O
6 CO 2 + 6 H 2 O ⟶ C 6 H 1 2 O 6 + 6 O 2

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2013 (07 Apr)
Synthetic polymer bakelite can be prepared from following compounds

Options:
A
Styrene and vinyl chloride

B
Acrylonitrile and vinyl chloride

C
Adipic acid and ethylene glycol

D
Phenol and formaldehyde

The correct answer is:
Phenol and formaldehyde

Solution:
Phenol and formaldehyde undergo condensation polymerisation under two different conditions to give a cross linked polymer called bakelite.

Topic: Chemistry, Polymers, JEE Main, JEE Main 2012 (12 May Online)
System is shown in the figure. Assume that cylinder remains in contact with the two wedges. The velocity of the cylinder is

Options:
A 19-43u2 m s-1
B 13u2 m s-1
C 3u m s-1
D 7u m s-1

The correct answer is:
7u m s-1

Solution:
In the frame of A
3usin⁡30=Vycos⁡30 . . . (i)
Vx=2u . . . (ii)
V= V x 2 + V y 2 = ( 2u ) 2 + ( 3 u ) 2 =u 7

Topic: Physics, Laws of Motion, NEET
T0 is the time period of a simple pendulum at a place. If the length of the pendulum is reduced to 116 times of its initial value, the modified time period is

Options:
A T0
B 8πT0
C 4T0
D 14T0

The correct answer is:
14T0

Solution:
T0=2πℓg
New time period, T==2πℓ16 g=2π4ℓg
T=T04

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (22 Jul Shift 1)
$\mathrm{t}_{1 / 4}$ can be taken as the time taken for the concentration of a reactant to drop to $\frac{3}{4}$ of its initial value. If the rate constant for a first order reaction is $\mathrm{K}$, the $\mathrm{t}_{1 / 4}$ can be written as

Options:
A
$0.10 / \mathrm{K}$

B
$0.29 / \mathrm{K}$

C
$0.69 / \mathrm{K}$

D
$0.75 / \mathrm{K}$

The correct answer is:
$0.29 / \mathrm{K}$

Solution:
$$
t_{1 / 4}=\frac{2.303}{K} \log \frac{1}{1-\frac{1}{4}}=\frac{0.29}{K}
$$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2005
t87.5 is the time required for the reaction to undergo 87.5% completion and t50 is the time required for the reaction to undergo 50% completion. The relation between t87.5 and t50 for a first order reaction is t87.5=x×t50
The value of x is _________. (Nearest integer)

Options:

The correct answer is:
3

Solution:
The time for completion of the reaction and concentration of reactant relation is as follows,

At=A0×12.5100=A08[87.5%completed]A0→t1/2A02→t1/2A04→t1/2A08t87.5=3t1/2So, x=3

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2023 (13 Apr Shift 1)
$\frac{1}{\tan 3 A-\tan A}-\frac{1}{\cot 3 A-\cot A}=$

Options:
A $\tan A$
B $\tan 2 A$
C $\cot A$
D $\cot 2 A$

The correct answer is:
$\cot 2 A$

Solution:
$\frac{1}{\tan 3 A-\tan A}-\frac{1}{\cot 3 A-\cot A}$
$=\frac{1}{\tan 3 A-\tan A}+\frac{\tan A \tan 3 A}{\tan 3 A-\tan A}=\frac{1}{\tan 2 A}=\cot 2 A$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
tan620o-33tan420o+27tan220o+4=

Options:

The correct answer is:
7

Solution:
Let, tan 20o=k
then, tanâ¡60o=3tanâ¡20o-tan320o1-3tan220o
3=3k-k31-3k2
31-3k22=3k-k32
31+9k4-6k2=9k2+k6-6k4
3+27k4-18k2=9k2+k6-6k4
k6-33k4+27k2-3=0
k6-33k4+27k2+4=7
Hence, tan620o-33tan420o+27tan220o+4=7

Topic: Mathematics, Trigonometric Equations, JEE Main
tan620o-33tan420o+27tan220o+4=

Options:

The correct answer is:
7

Solution:
Let, tan 20o=k
then, tanâ¡60o=3tanâ¡20o-tan320o1-3tan220o
3=3k-k31-3k2
31-3k22=3k-k32
31+9k4-6k2=9k2+k6-6k4
3+27k4-18k2=9k2+k6-6k4
k6-33k4+27k2-3=0
k6-33k4+27k2+4=7
Hence, tan620o-33tan420o+27tan220o+4=7

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
T is the time period of simple pendulum on the earth's surface. Its time period becomes xT when taken to a height R (equal to earth's radius) above the earth's surface. Then, the value of x will be:

Options:
A 4
B 2
C 12
D 14

The correct answer is:
2

Solution:
Time period of a simple pendulum at Earth's surface is given by

T=2πlg.

When distance from centre of Earth, i.e., at h=R, acceleration due to gravity is g'=g1+hR2=g22=g4

Now, time period is xT=2πlg4

⇒xT=2×2πlg

⇒xT=2T⇒x=2

Thus, x=2.

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (25 Jan Shift 1)
TlAlSO42 . xH2O is bcc with edge length âaâ= 1.22 nm. If the density of the solid is 2.32 g/cc, then the value of x is
(Given: N A â=â6âÃâ 10 23 , atomtextc weight (in g/mol ) TIâ=â204,âAlâ=â27,âSâ=â32,âOâ=â16,âHâ=â1 )

Options:
A 2
B 4
C 47
D 70

The correct answer is:
47

Solution:
For any unit cell
d =  2 M a 3   N A (for BCC, Z = 2)
2.32=2 ÃM6Ã1023Ã1.223Ã10â21
â´ M=1264
So 4 2 3 + x 1 8 = 1 2 6 4
x = 4 7

Topic: Chemistry, Solid State, JEE Main
TlI3 is an ionic compound. In the aqueous solution it provides -

Options:
A Tl+ and I3-
B Tl3+ and I-
C Tl+,I- and I2
D Tl3+ and I3-

The correct answer is:
Tl+ and I3-

Solution:
Tl3+ does not exist because of inert pair effect.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
T O Diener discovered a

Options:
A free infectious RNA
B free infectious DNA
C infectious protein
D bacteriophage

The correct answer is:
free infectious RNA

Solution:
Viroids are small, circular, single stranded RNA molecules that are the smallest known pathogens. The extracellular form of the viroid is naked RNA, ie, there is no protein capsid of any kind. Even, the RNA molecule contains no protein encoding genes and therefore, the viroid is totally dependent on host function for its replication. No viroid diseases of animals are known and the precise mechanisms by which viroids cause plant diseases remain unclear.

Topic: Biology, Biological Classification, NEET, NEET 2009 (Screening)
\(t_{1 / 4}\) can be taken as the time taken for the concentration of a reactant to drop to \(3 / 4\) of its value. If the rate constant for a first order reaction is \(k\), the \(t_{1 / 4}\) can be written as \([\ln 2\) \(=0.695, \ln =1.1]\)

Options:
A \(0.69 / \mathrm{k}\)
B \(0.75 / \mathrm{k}\)
C \(0.10 / \mathrm{k}\)
D \(0.29 / \mathrm{k}\)

The correct answer is:
\(0.29 / \mathrm{k}\)

Solution:
\(\begin{aligned} & t_{\frac{1}{4}}=\frac{2.305}{K} \operatorname{fog} \frac{4}{3} \\ & =\frac{2.307}{K 0} \operatorname{Fog} 4-\log 3 \\ & =\frac{0.29}{K} \end{aligned}\)

Topic: Chemistry, Chemical Kinetics, JEE Main
Taj Mahal is being slowly disfigured and discoloured. This is primarily due to

Options:
A Global warming
B Acid rain
C Water pollution
D Soil pollution

The correct answer is:
Acid rain

Solution:
Acid rain.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2019 (11 Jan Shift 2)