datavorous/entrance-exam-dataset huggingface dataset:
Trichloroacetaldehyde, $\mathrm{CCl}_3 \mathrm{CHO}$ reacts with chlorobenzene in presence of sulphuric acid and produces -

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Trichloroacetaldehyde, $\mathrm{CCl}_3 \mathrm{CHO}$ reacts with chlorobenzene in presence of sulphuric acid and produces -

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, NEET, NEET 2009 (Mains)
Trichloroacetaldehyde was subjected to Cannizzaro's reaction by using $\mathrm{NaOH}$. The mixture of the products contains sodium trichloroacetate and another compound. The other compound is :

Options:
A
Trichloromethanol

B
2, 2, 2-Trichloropropanol

C
Chloroform

D
2, 2, 2-Trichloroethanol

The correct answer is:
2, 2, 2-Trichloroethanol

Solution:
$2 \mathrm{CCl}_3 \mathrm{CHO} \stackrel{\mathrm{OH}^{(-)}}{\longrightarrow} \mathrm{CCl}_3 \mathrm{COONa}+\mathrm{CCl}_3 \mathrm{CH}_2 \mathrm{OH}$
Cannizaro reaction is a disproportionation reaction
One aldehyde molecule is oxidized to salt of the carboxylic Acid, other one is reduced to Alcohol. So the compound is $\mathrm{CCl}_3 \mathrm{CH}_2 \mathrm{OH}$
IUPAC Name is $2,2,2,-$ Trichloro ethanol

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2011
Trichoderma harzianum has proved a useful microorganism for

Options:
A Biological control of soi-borne plant pathogens
B Bioremediation of contaminated soils
C Reclamation of wastelands
D Gene transfer in higher plants

The correct answer is:
Biological control of soi-borne plant pathogens

Solution:
It secretes chitinase enzyme which is responsible for antifungal activities.

Topic: Biology, Microbes in Human Welfare, NEET, NEET 2008 (Mains)
Trigonal bipyramidal geometry is shown by:

Options:
A
$\mathrm{XeOF}_2$

B
$\mathrm{XeO}_3 \mathrm{F}_2$

C
$\mathrm{FXeOSO}_2 \mathrm{F}$

D
$\left[\mathrm{XeF}_8\right]^{2-}$

The correct answer is:
$\mathrm{XeO}_3 \mathrm{F}_2$

Solution:
The hybridization of $\mathrm{XeO}_3 \mathrm{F}_2$ is $s p^3 d$ and its structure is trigonal bipyramidal in which oxygen atoms are situated on the plane and the fluoride atoms are on the top and bottom.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Trigonal bipyramidal geometry is shown by:

Options:
A
$\mathrm{XeOF}_2$

B
$\mathrm{XeO}_3 \mathrm{F}_2$

C
$\mathrm{FXeOSO}_2 \mathrm{F}$

D
$\left[\mathrm{XeF}_8\right]^{2-}$

The correct answer is:
$\mathrm{XeO}_3 \mathrm{F}_2$

Solution:
The hybridization of $\mathrm{XeO}_3 \mathrm{F}_2$ is $s p^3 d$ and its structure is trigonal bipyramidal in which oxygen atoms are situated on the plane and the fluoride atoms are on the top and bottom.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2013 (09 Apr Online)
Trigonometric and inverse trigonometric functions are differentiable in their respective domain.

Options:
A
True

B
False

The correct answer is:
True

Solution:
No Solution Available

Topic: Mathematics, Continuity and Differentiability
Trimolecular reactions are uncommon because

Options:
A the probability of three molecules colliding at an instant is low.
B the probability of three molecules colliding at an instant is high.
C the probability of three molecules colliding at an instant is zero.
D the probability of many molecules colliding at an instant is high.

The correct answer is:
the probability of three molecules colliding at an instant is low.

Solution:
According to collision theory reaction occurs mostly due to the effective collision of two molecules. The chance of three molecules to collide simultaneously is low. In three or more molecules due to the loss of active species on collision probability of reactants conversion into the product will be less.

Topic: Chemistry, Chemical Kinetics, JEE Main
Triple bond of ethyne is made of or Cylindrical shape of an alkyne is due to

Options:
A Three $\sigma$-bonds
B Three $\pi$-bonds
C Two $\sigma$ and one $\pi$-bond
D Two $\pi$ and one $\sigma$-bond

The correct answer is:
Two $\pi$ and one $\sigma$-bond

Solution:
In $\mathrm{CH} \equiv \mathrm{CH}$ triple bond consists of one $\sigma$ and two $\pi$ bonds.

Topic: Chemistry, Hydrocarbons, JEE Main
Tritium is an isotope of hydrogen whose nucleus triton contains neutrons and 1 proton. Free neutrons decay into $\mathrm{p}+\overline{\mathrm{e}}+\overline{\mathrm{n}}$. If one of the neutrons is Triton decays, it would transform into $\mathrm{He}^3$ nucleus. This does not happen. This is because

Options:
A Triton energy is less than that of a $\mathrm{He}^3$ nucleus

B
The electron created in the beta decay process cannot remain in the nucleus

C both the neutrons in triton have to decay simultaneously resulting in a nucleus with 3 protons, which is not a $\mathrm{He}^3$ nucleus.

D free neutrons decay due to external perturbations which is absent in triton nucleus

The correct answer is:
Triton energy is less than that of a $\mathrm{He}^3$ nucleus

Solution:
Tritium $\left({ }_1 \mathrm{H}^3\right)$ has proton and 2 neutrons. If a neutron decays as $\mathrm{n} \rightarrow \mathrm{P}+\overline{\mathrm{e}}+\overline{\mathrm{v}}$, then the nucleus may have 2 protons and neutron, i.e., tritium atom will convert into ${ }_2 \mathrm{He}^3(2$ protons and 1 neutron).
Binding energy of ${ }_1 \mathrm{H}^3$ is less than that of ${ }_2 \mathrm{He}^3$ nucleus, So, transformation is not possible energetically.

Topic: Physics, Nuclear Physics
Tritium with a half-life of 12.5 years undergoing β -decay. What fraction of a sample of pure tritium will remain undecayed after 25 years?

Options:
A One half
B One fourth
C One third
D Three fourth

The correct answer is:
One fourth

Solution:
Given that, half-life of tritium
t1/2=12.5 years
T=25 years
â´ Number of half-life, n=Tt1/2=2512.5=2
â´ Fraction of sample remain undecayed
N=N012n, where, N0= initial sample
=N0122=N04

Topic: Physics, Atomic Physics, NEET
Tropical rain forests once covered more than ____a____ percent of the Earth surface, now covers not more than ___b___percent.

Options:
A a-30, b-19.8
B a-8.l, b-2.4
C a-l4,b-6
D a-19.8,b-8.l

The correct answer is:
a-l4,b-6

Solution:
The region near the middle of the Earth between the Tropic of Cancer and Tropic of Capricon is considered as the Tropic region. This region receives the most amount of solar energy which leads to better productivity and higher species diversity. Also, the seasonal variations in such regions are low and environment mainly remains the same supporting high species richness and promoting niche specialization. Tropical rain forests used to cover more than 14 per cent of the Earth surface. Now due to human activities leading to destruction of habitat and pushing many species to the brink of extinction has tropical rainforests cover no more than 6 percent.

Topic: Biology, Biodiversity and its Conservation, NEET
Tropical regions show greatest level of species richness because
A. Tropical latitudes have remained relatively undisturbed for millions of years, hence more time was available for species diversification.
B. Tropical environments are more seasonal.
C. More solar energy is available in tropics.
D. Constant environments promote niche specialization.
E. Tropical environments are constant and predictable.
Choose the correct answer from the options given below.

Options:
A $A$ and $B$ only
B A, B and E only
C A, B and D only
D A, C, D and E only

The correct answer is:
A, C, D and E only

Solution:
Only statement B is incorrect because tropical environments unlike temperate ones, are less seasonal, relatively more constant and predictable.
Thus statements $\mathrm{A}, \mathrm{C}, \mathrm{D}$ and E are correct.

Topic: Biology, Biodiversity and its Conservation, NEET, NEET 2024
Tropolone is

Options:
A benzenoid and aromatic
B non-benzenoid and not aromatic
C non-benzenoid and aromatic
D non-benzenoid and anti-aromatic

The correct answer is:
non-benzenoid and aromatic

Solution:
Tropolone is
$$
\text { Tropolone is non-benzenoid quasi aromatics. }
$$

Topic: Chemistry, General Organic Chemistry, JEE Main
Truth table for system of four NAND gates as shown in figure is

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Physics, Semiconductors, JEE Main, JEE Main 2012 (Offline)
Truth table for the given circuit will be

Options:
A
$$
\begin{array}{cc|c}
\mathrm{x} & \mathrm{y} & \mathrm{z} \\
\hline 0 & 0 & 1 \\
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}
$$

B
$$
\begin{array}{cc|c}
\mathrm{x} & \mathrm{y} & \mathrm{z} \\
\hline 0 & 0 & 0 \\
0 & 1 & 0 \\
1 & 0 & 0 \\
1 & 1 & 1
\end{array}
$$

C
$$
\begin{array}{cc|c}
x & y & z \\
\hline 0 & 0 & 1 \\
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 1
\end{array}
$$

D
$$
\begin{array}{cc|c}
\mathrm{x} & \mathrm{y} & \mathrm{z} \\
\hline 0 & 0 & 0 \\
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 1
\end{array}
$$

The correct answer is:
$$
\begin{array}{cc|c}
x & y & z \\
\hline 0 & 0 & 1 \\
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 1
\end{array}
$$

Solution:
Truth table of the circuit is as follows
$$
\begin{array}{|c|c|c|c|c|c|}
\hline x & y & \bar{x} & a=x \cdot y & b=\bar{x} \cdot y & z=\overline{a \cdot b} \\
\hline 0 & 0 & 1 & 0 & 0 & 1 \\
\hline 0 & 1 & 1 & 0 & 1 & 1 \\
\hline 1 & 0 & 0 & 0 & 0 & 1 \\
\hline 1 & 1 & 0 & 1 & 0 & 1 \\
\hline
\end{array}
$$

Topic: Physics, Semiconductors, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
Tube $A$ has both ends open while tube $B$ has one end closed, otherwise they are identical. The ratio of fundamental frequency of tube $A$ and $B$ is

Options:
A $1: 2$
B $1: 4$
C $2: 1$
D $4: 1$

The correct answer is:
$2: 1$

Solution:
Fundamental freq. for,
Tube $\mathrm{A} \Rightarrow \mathrm{f}_{\mathrm{A}}=\frac{\mathrm{v}}{2 \mathrm{~L}}$
Tube $\mathrm{B} \Rightarrow \mathrm{f}_{\mathrm{B}}=\frac{\mathrm{v}}{4 \mathrm{~L}}$
Now,
$$
\begin{array}{l}
\frac{\mathrm{f}_{\mathrm{A}}}{\mathrm{f}_{\mathrm{B}}}=\frac{\mathrm{v}}{2 \mathrm{~L}} \times \frac{4 \mathrm{~L}}{\mathrm{v}}=\frac{2}{1} \\
\mathrm{f}_{\mathrm{A}}: \mathrm{f}_{\mathrm{B}}=2: 1
\end{array}
$$

Topic: Physics, Waves and Sound, JEE Main
Tube A has both ends open while tube B has one end closed, otherwise they are identical. The ratio of fundamental frequency of tube $A$ and $B$ is

Options:
A
1 : 2

B
1 : 4

C
2 : 1

D
4 : 1

The correct answer is:
2 : 1

Solution:
$\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{1}{2} \Rightarrow \frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{n}_{\mathrm{B}}}=\frac{2}{1}$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2002
Twelve resistors each of resistance 1 Î© are connected as shown in the figure. Net resistance between points A and H would be

Options:
A 56 Î©
B 1 Ω
C 34 Ω
D 712Â Î©

The correct answer is:
34 Ω

Solution:
Let current i enters from point A and leaves at H. Then current can be distributed as follows :

from the above diagram we can see that no current will flow in wires CE and FD. So, the simple circuit is as follows:

â´ R AH = 3 4 Î©

Topic: Physics, Current Electricity, JEE Main
Twelve tickets are numbered 1 to 12 . One ticket is drawn at random, then the probability of the number to be divisible by 2 or 3 , is

Options:
A $\frac{2}{3}$
B $\frac{7}{12}$
C $\frac{5}{6}$
D $\frac{3}{4}$

The correct answer is:
$\frac{2}{3}$

Solution:
$P(2 \cup 3)=P(2)+P(3)-P(2 \cap 3)$
$=\frac{6}{12}+\frac{4}{12}-\frac{2}{12}=\frac{2}{3}$.

Topic: Mathematics, Probability, JEE Main
Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is:

Options:
A 12.5
B 10
C 25
D 30

The correct answer is:
25

Solution:
Perimeter =2r+rθ=20
θ=20-2rr
Now, Area A=r2θ2
⇒A=r22 20-2rr=10r-r2
⇒dAdr=10-2r=0
⇒r=5
d2Adr2r=5=-2<0⇒r=5 is the point of maxima
Amax=10×5-25=25

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2017 (02 Apr)
Two 5 molal solutions are prepared by dissolving a non-electrolyte, non-volatile solute separately in the solvents $\mathrm{X}$ and $\mathrm{Y}$. The molecular weights of the solvents are $\mathrm{M}_{\mathrm{X}}$ and $\mathrm{M}_{\mathrm{Y}}$, respectively where $\mathrm{M}_X=\frac{3}{4} \mathrm{M}_{\mathrm{Y}}$. The relative lowering of vapour pressure of the solution in $\mathrm{X}$ is " $\mathrm{m}$ " times that of the solution in Y. Given that the number of moles of solute is very small in comparison to that of solvent, the value of "m" is:

Options:
A
$\frac{3}{4}$

B
$\frac{1}{2}$

C
$\frac{1}{4}$

D
$\frac{4}{3}$

The correct answer is:
$\frac{3}{4}$

Solution:
The relationship between molar masses of the two solvents is
$$
\mathrm{M}_{\mathrm{X}}=\frac{3}{4} \mathrm{M}_{\mathrm{Y}}
$$
The relative lowering of vapour pressure of the two solutions is
$$
\left(\frac{\Delta \mathrm{P}}{\mathrm{P}}\right)_{\mathrm{X}}=\mathrm{m}\left(\frac{\Delta \mathrm{P}}{\mathrm{P}}\right)_{\mathrm{Y}}
$$
But, the relative lowering of vapour pressure of solutions is directly proportional to the mole fraction of solute.
Given 5 molal solution, means 5 moles of solute are dissolved in $1 \mathrm{~kg}$ ( or $1000 \mathrm{~g}$ ) of solvent.
The number of moles of solvent $=\frac{1000 \mathrm{~g}}{\mathrm{M}}$
The mole fraction of solute $=\frac{5}{1000 / \mathrm{M}}$
$$
=\mathrm{M} \times \frac{5}{1000}
$$
hence $\mathrm{M}_{\mathrm{X}} \times \frac{5}{1000}=\mathrm{m} \times \mathrm{M}_{\mathrm{Y}} \times \frac{5}{1000}$.
Substitute equation (i) in equation (ii)
$$
\begin{aligned}
&\frac{3}{4} \times \mathrm{M}_{\mathrm{Y}} \times \frac{5}{1000}=\mathrm{m} \times \mathrm{M}_{\mathrm{Y}} \times \frac{5}{1000} \\
&\mathrm{~m}=\frac{3}{4}
\end{aligned}
$$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
Two 10 cm long, straight wires, each carrying a current of 5 A are kept parallel to each other. If each wire experienced a force of 10-5 N, then separation between the wires is _____ cm.

Options:

The correct answer is:
5

Solution:
Given: F=10-5 N

Force between wires Fl=μ0i1i22πd, where, d is separation between them.

⇒10-50.1=2×10-7×25d

⇒d=50×10-3 m=5 cm

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (26 Jun Shift 2)
Two adjacent piano keys are struck simultaneously. The notes emitted by them have frequencies $n_1$ and $n_2$. The number of beats heard per second is

Options:
A $\frac{1}{2}\left(n_1-n_2\right)$
B $\frac{1}{2}\left(n_1+n_2\right)$
C $n_1 \sim n_2$
D $2\left(n_1-n_2\right)$

The correct answer is:
$n_1 \sim n_2$

Solution:
The number of beats heard is given by difference in frequencies A the notes
$\therefore B=n_1 \sim n_2$

Topic: Physics, Waves and Sound, JEE Main
Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are $0.3$ and $0.2$, respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is

Options:
A
$0.06$

B
$0.14$

C
$0.2$

D
None of these

The correct answer is:
None of these

Solution:
The desired probability
$\begin{aligned}
& =0.7 \times 0.2+(0.7)(0.8)(0.7)(0.2)+(0.7)(0.8)(0.7)(0.8)(0.7)(0.2)+\ldots . \\
& =0.14\left[1+(0.56)+(0.56)^2+\ldots\right] \\
& =0.14\left[\frac{1}{1-0.56}\right]=\frac{0.14}{0.44}=\frac{7}{22}.
\end{aligned}$

Topic: Mathematics, Probability, JEE Main, JEE Main 2007
Two angles of a triangle are $\frac{\pi}{6}$ and $\frac{\pi}{4}$ and the length of the included side is $(\sqrt{3}+1) \mathrm{cm}$. The area of the triangle is

Options:
A $\frac{\sqrt{3}-1}{2} \mathrm{~cm}^2$
B $\frac{\sqrt{3}}{2} \mathrm{~cm}^2$
C $\frac{\sqrt{3}+1}{2} \mathrm{~cm}^2$
D None of these

The correct answer is:
$\frac{\sqrt{3}+1}{2} \mathrm{~cm}^2$

Solution:
$A=\pi-\frac{\pi}{4}-\frac{\pi}{6}=180^{\circ}-45^{\circ}-30^{\circ}=105^{\circ}$
$\therefore \frac{\sqrt{3}+1}{\sin 105^{\circ}}=\frac{c}{\sin \frac{\pi}{6}}$ i.e. $\frac{C}{\sin 30^{\circ}}$
$\therefore C=\frac{1}{2} \cdot \frac{\sqrt{3}+1}{\sin \left(60^{\circ}+45^{\circ}\right)}$
$=\frac{\sqrt{3}+1}{2\left(\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}}+\frac{1}{2} \cdot \frac{1}{\sqrt{2}}\right)}$
$=\frac{(\sqrt{3}+1)}{2 \cdot(\sqrt{3}+1)} \cdot 2 \sqrt{2}=\sqrt{2}$
$\therefore$ reqd. area $=\frac{1}{2}$ ca $\sin B$
$=\frac{1}{2} \sqrt{2}(\sqrt{3}+1) \sin \frac{\pi}{4}$
$=\frac{\sqrt{2}}{2}(\sqrt{3}+1) \cdot \frac{1}{\sqrt{2}}$
$=\frac{\sqrt{3}+1}{2} \mathrm{~cm}^2$

Topic: Mathematics, Properties of Triangles, JEE Main
Two aromatic compounds having formula \(\mathrm{C}_7 \mathrm{H}_8 \mathrm{O}\) which are easily identifiable by \(\mathrm{FeCl}_3\) solution test (violet colouration) are

Options:
A o- cresol and benzyl alcohol
B m-cresol and p-cresol
C o-cresol and p-cresol
D methyl phenyl ether and benzyl alcohol

The correct answer is:
o- cresol and benzyl alcohol

Solution:
Hints: \(\mathrm{O}\) - cresol contains phenolic group, thus it gives violet coloration with \(\mathrm{FeCl}_3\) where as benzylalchol donot contains phenolic group, hence no coloration with \(\mathrm{FeCl}_3\). Hence Identifiable

Topic: Chemistry, Hydrocarbons, JEE Main
Two astronauts in space are traveling directly towards each other. Astronaut A has a mass of $70 \mathrm{~kg}$ and a velocity of $15 \frac{\mathrm{m}}{\mathrm{s}}$ and Astronaut $B$ has a mass of $60 \mathrm{~kg}$ and a velocity of $5 \frac{\mathrm{m}}{\mathrm{s}}$. When the astronauts collide, they grab onto each other. What is the velocity of the two astronauts after the collision as they continue to grab onto each other?

Options:
A $5.77 \frac{m}{s}$
B $4.34 \frac{m}{s}$
C $10 \frac{m}{s}$
D $5 \frac{m}{s}$

The correct answer is:
$5.77 \frac{m}{s}$

Solution:
Momentum is always conserved. Equation for conservation of momentum:
$m_1 v_1+m_2 v_2=\left(m_1+m_2\right) v_{\text {total }}$
There is only one velocity on the right since the two astronauts grab onto each other, thus they move together at the same velocity. Solve.
$\begin{aligned} & \left(70 \mathrm{~kg} \cdot 15 \frac{\mathrm{m}}{\mathrm{s}}\right)+\left(60 \mathrm{~kg}\left(-5 \frac{\mathrm{m}}{\mathrm{s}}\right)\right)=(70 \mathrm{~kg}+60 \mathrm{~kg}) v_{\text {total }} \\ & v_{\text {total }}=5.77 \frac{\mathrm{m}}{\mathrm{s}}\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two atoms are said to be isotopes if:

Options:
A They have same atomic number but different mass numbers
B They have same number of neutrons but different mass number
C The sum of the number of protons and neutrons is same but number of protons are different
D They have same number of neutrons but different modes of radioactive decay

The correct answer is:
They have same atomic number but different mass numbers

Solution:
Any of the two or more species of atoms of a chemical element, with the same atomic number and a nearly identical chemical behaviour, but with different atomic mass or mass number and different physical properties are called isotopes.

Topic: Chemistry, Structure of Atom, JEE Main
Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at t=0 s. Ball B is thrown vertically down with an initial velocity u at t=2 s. After a certain time, both balls meet 100 m above the ground. Find the value of u in m s-1. [use g=10 m s-2]

Options:
A 10
B 15
C 20
D 30

The correct answer is:
30

Solution:
Displacement covered by first ball,180-100=0+12×10×t2 ⇒t=4 s

Now, the second body gets only, t2=4-2=2 s.

Displacement covered by second will be same,

80=u×2+12×10×22

⇒u=80-202=30 m s-1

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2022 (29 Jun Shift 1)
Two balls A and B are projected from the same position with horizontal velocity 20 m s-1 and 5 m s-1 as shown. The separation between these two bodies at the moment their velocities become perpendicular to each other is x m. The value of x is

Options:

The correct answer is:
25

Solution:
v1→=5i^-gtj^ ⇒u1=5

v2→=20i^-gtj^ ⇒u2=20

As v1→ and v2→ are perpendicular,

v→1·v→2=0
t=u1 u2g
s→rel=u→rel×t
= (u1+u2) t
=u1+u2u1 u2g
=25×1010=25 m

Topic: Physics, Motion In Two Dimensions, JEE Main
Two balls are thrown in air. The second ball is thrown \( 2 \) seconds after the first ball. The acceleration of the center of mass of the two balls while in air:

Options:
A Depends on the direction of the motion of the balls
B Depends on the masses of the balls
C Depends on the speed of two balls
D is equal to ' \( g \) '

The correct answer is:
is equal to ' \( g \) '

Solution:
aCM=fnet m=mg+mg2m=g

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two balls of masses 2 g and 6 g are moving with kinetic energy in the ratio of 3:1. What is the ratio of their linear momentum?

Options:
A 1 : 1

B 2 : 1

C 1 : 2
D None of these

The correct answer is:
1 : 1

Solution:
K=12mv2=12Ãm(mv2)2=mv2m=mv22m or K=p22m
â´ K1K2=p122m1Ã2m2p22 or 31=p1p2Ã62
â´ p1: p2=1:1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two balls of masses $m_1$ and $m_2$ are separated from each other by a powder charge placed between them. The whole system is at rest on the ground. Suddenly the powder charge explodes and masses are pushed apart. The mass $m_1$ travels a distance $s_1$ and stops. If the coefficients of friction between the balls and ground are same, the mass $m_2$ stops after travelling the distance

Options:
A $s_2=\frac{m_1}{m_2} s_1$
B $s_2=\frac{m_2}{m_1} s_1$
C $s_2=\frac{m_1^2}{m_2^2} s_1$
D $s_2=\frac{m_2^2}{m_1^2} s_1$

The correct answer is:
$s_2=\frac{m_1^2}{m_2^2} s_1$

Solution:
We know that in the given condition $s \propto \frac{1}{m^2}$
$\therefore \frac{s_2}{s_1}=\left(\frac{m_1}{m_2}\right)^2 \Rightarrow s_2=\left(\frac{m_1}{m_2}\right)^2 \times s_1$

Topic: Physics, Laws of Motion, JEE Main
Two balls of same mass and carrying equal charge are hung from a fixed support of length $l$. At electrostatic equilibrium, assuming that angles made by each thread is small, the separation, $x$ between the balls is proportional to :

Options:
A
$l$

B
$l^2$

C
$l^{2 / 3}$

D
$l^{1 / 3}$

The correct answer is:
$l^{1 / 3}$

Solution:
In equilibrium, $\mathrm{F}_{\mathrm{e}}=\mathrm{T} \sin \theta$
$m g=T \cos \theta$
$\tan \theta=\frac{F_e}{m g}=\frac{q^2}{4 \pi \epsilon_0 x^2 \times m g}$
also $\tan \theta \approx \sin =\frac{x / 2}{\ell}$
Hence, $\frac{x}{2 \ell}=\frac{q^2}{4 \pi \epsilon_0 x^2 \times m g}$
$\Rightarrow x^3=\frac{2 q^2 \ell}{4 \pi \epsilon_0 m g}$
$\therefore x=\left(\frac{q^2 \ell}{2 \pi \epsilon_0 m g}\right)^{1 / 3}$
Therefore $\mathrm{x} \propto \ell^{1 / 3}$

Topic: Physics, Electrostatics, JEE Main
Two balls of same mass and carrying equal charge are hung from a fixed support of length $l$. At electrostatic equilibrium, assuming that angles made by each thread is small, the separation, $x$ between the balls is proportional to :

Options:
A
$l$

B
$l^2$

C
$l^{2 / 3}$

D
$l^{1 / 3}$

The correct answer is:
$l^{1 / 3}$

Solution:
In equilibrium, $\mathrm{F}_{\mathrm{e}}=\mathrm{T} \sin \theta$
$m g=T \cos \theta$
$\tan \theta=\frac{F_e}{m g}=\frac{q^2}{4 \pi \epsilon_0 x^2 \times m g}$
also $\tan \theta \approx \sin =\frac{x / 2}{\ell}$
Hence, $\frac{x}{2 \ell}=\frac{q^2}{4 \pi \epsilon_0 x^2 \times m g}$
$\Rightarrow x^3=\frac{2 q^2 \ell}{4 \pi \epsilon_0 m g}$
$\therefore x=\left(\frac{q^2 \ell}{2 \pi \epsilon_0 m g}\right)^{1 / 3}$
Therefore $\mathrm{x} \propto \ell^{1 / 3}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2013 (09 Apr Online)
Two balls with equal charges are in a vessel with ice at -10 Â°C at a distance of 25 cm from each other. On forming water at 0 Â°C, the balls are brought nearer to 5 cm for the interaction between them to be same. If the dielectric constant of water at 0 Â°C is 80, the dielectric constant of ice at -10 Â°C is

Options:
A 40
B 3.2
C 20
D 6.4

The correct answer is:
3.2

Solution:
(-10°C ice) (0°C water)
q1             k1            r1=25cmq2    q1         k2=80         r2=5cmq2
Given Fice=Fwater
14πε0  q1q2k1r12=14πε0  q1q2k2r22
∴   k1r12 = k2r22
∴k1=k2r2r12
= 80 × 5252=3.2

Topic: Physics, Electrostatics, JEE Main