datavorous/entrance-exam-dataset huggingface dataset:
In a potentiometer experiment, the balancing length obtained with a cell is 240 cm. On shunting the cell with a resistance of 2 Î©, the balancing length becomes 120 cm. The internal resistance of the cell is

Options:
A 4 Ω
B 1 Ω
C 0.5 Ω
D 2 Î©

The correct answer is:
2 Î©

Solution:
In the first case, the emf of the battery is

E=Kl1, where K is the potential gradient and l1 is the first balancing length.

In the second case the potential difference across the battery is

ERr+R=Kl2, where R is the shunt resistance used.

Dividing the above two equations, we get

rR+1=l1l2

⇒r=2240-120120=2 Ω

Topic: Physics, Current Electricity, JEE Main
In a potentiometer experiment the balancing with a cell is at length $240 \mathrm{~cm}$. On shunting the cell with a resistance of $2 \Omega$ the balancing length becomes $120 \mathrm{~cm}$. The internal resistance of the cell is

Options:
A
$1 \Omega$

B
$0.5 \Omega$

C
$4 \Omega$

D
$2 \Omega$

The correct answer is:
$2 \Omega$

Solution:
$\mathrm{r}=\mathrm{R}\left[\frac{\ell_1}{\ell_2}-1\right]=2\left[\frac{240}{120}-1\right]=2 \Omega$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2005
In a potentiometer experiment, the null point is obtained on $7^{\text {th }}$ wire for given cell. To shift the null point on $9^{\text {th }}$ wire for the same cell what should we do?

Options:
A Decrease applied e.m.f.
B Increase resistance in main circuit
C Attach resistance in series with a cell
D Decrease resistance in main circuit

The correct answer is:
Increase resistance in main circuit

Solution:
Potential gradient,
$\left(\frac{\Delta V}{L}\right)=\frac{\left[\frac{E R}{\left(R+R^{\prime}\right)}\right]}{L}$
To increase the balancing length, we need to decrease the potential gradient. Hence, we have to increase the resistance in main circuit.

Topic: Physics, Current Electricity, MHT CET, MHT CET 2022 (11 Aug Shift 1)
In a potentiometer of 10 wires, the balance point is obtained on the 6th wire. To shift the balance point to 8th wire, we should

Options:
A Increase resistance in the main circuit
B

Decrease resistance in the main circuit

C

Increase resistance in series with the cell whose emf is to be measured

D Decrease resistance in series with the cell whose emf is to be measured

The correct answer is:
Increase resistance in the main circuit

Solution:
By increasing resistance in the primary circuit the P.G of the wire decreases thereby it requires more length to measure the emf of secondary cell.

∴By increasing the resistance in series circuit we can move balancing point from sixth to eight wire.

Topic: Physics, Current Electricity, AP EAMCET, AP EAMCET 2021 (19 Aug Shift 2)
In a $\Delta \mathrm{PQR}$, if $3 \sin \mathrm{P}+4 \cos \mathrm{Q}=6$ and $4 \sin \mathrm{Q}+3 \cos \mathrm{P}=1$, then the angle $\mathrm{R}$ is equal to

Options:
A
$\frac{5 \pi}{6}$

B
$\frac{\pi}{6}$

C
$\frac{\pi}{4}$

D
$\frac{3 \pi}{4}$

The correct answer is:
$\frac{\pi}{6}$

Solution:
$3 \sin P+4 \cos Q=6 \quad\dots(1)$
$4 \sin Q+3 \cos P=1\quad\dots(2)$
From $(1)$ and $(2) \angle P$ is obtuse.
$(3 \sin P+4 \cos Q)^2+(4 \sin Q+3 \cos P)^2=37$
$\Rightarrow 9+16+24(\sin P \cos Q+\cos P \sin Q)=37$
$\Rightarrow 24 \sin (P+Q)=12$
$\Rightarrow \sin (P+Q)=\frac{1}{2} \quad \Rightarrow P+Q=\frac{5 \pi}{6} \quad \Rightarrow R=\frac{\pi}{6}$

Topic: Mathematics, Properties of Triangles, JEE Main, JEE Main 2012 (Offline)
In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy (in J) for the process?

Options:

The correct answer is:
307.00

Solution:
Given, q = + 701 J (heat is absorbed, heance q is positive.)
W = - 394 J (work is done by the system, hence W is negative.)
By first law of thermodynamics;
Internal energy change, Î U = q + w
= + 701 J + (-394 J ) = + 307 J.
Hence, internal energy of the system increases by 307 J.

Topic: Chemistry, Thermodynamics (C), JEE Main
In a p-type semiconductor, the acceptor level is situated \( 57 \mathrm{meV} \) above the valence band. The maximum wavelength of light required to produce a hole will be

Options:
A \( 57 \mathrm{~A} \)
B \( 57 \times 10^{-3} \dot{A} \)
C \( 217100 Å \)
D \( 11.61 \times 10^{-33} \mathrm{~m} \)

The correct answer is:
\( 217100 Å \)

Solution:
Given that the level of energy of the acceptor band is above the valence band by,
E=57 meV
The energy of the photon having just enough energy to overcome the gap is given by,
E=hcλ
⇒λ=hcE

Speed of light c=3×108 ms
Planck's constant h=6.63×10-34 Js,

Substituting values,
λ=6.6×10-34×3×10857×10-3×1.6×10-19λ=21700A∘

The maximum wavelength of light required to produce a hole isλ=21700 A∘ .

Topic: Physics, Semiconductors, JEE Main
In a pure inductive circuit or In an ac circuit containing inductance only, the current

Options:
A Leads the e.m.f. by $90^{\circ}$
B Lags behind the e.m.f. by $90^{\circ}$
C Sometimes leads and sometime lags behind the e.m.f.
D Is in phase with the e.m.f.

The correct answer is:
Lags behind the e.m.f. by $90^{\circ}$

Solution:
As we know,
In a pure inductor (zero resistance), voltage leads the current by $90^{\circ}$ i.e., $\frac{\pi}{2}$

Topic: Physics, Alternating Current, JEE Main
In a quadrilateral $A B C D$, the point $P$ divides $D C$ in the ratio $1: 2$ and $Q$ is the mid point of $A C$. If $\overrightarrow{\mathbf{A B}}+2 \overrightarrow{\mathbf{A D}}+\overrightarrow{\mathbf{B C}}-2 \overrightarrow{\mathbf{D C}}=k \overrightarrow{\mathbf{P Q}}$, then $k$ is equal to

Options:
A -6
B -4
C 6
D 4

The correct answer is:
-6

Solution:
Now, $\overrightarrow{\mathbf{A B}}+2 \overrightarrow{\mathbf{A D}}+\overrightarrow{\mathbf{B C}}-2 \overrightarrow{\mathbf{D C}}$
$=\overrightarrow{\mathbf{A C}}+2 \overrightarrow{\mathbf{A D}}-2 \overrightarrow{\mathbf{D C}}$
$=\overrightarrow{\mathbf{A C}}+2(\overrightarrow{\mathbf{A C}}+\overrightarrow{\mathbf{C D}})-2 \overrightarrow{\mathbf{D C}}$
$\begin{aligned}
& =3 \overrightarrow{\mathbf{A C}}-4 \overrightarrow{\mathbf{D C}} \\
& =3(2 \overrightarrow{\mathbf{Q C}})-4\left(\frac{3}{2} \overrightarrow{\mathbf{P C}}\right) \\
& =6 \overrightarrow{\mathbf{Q C}}-6 \overrightarrow{\mathbf{P C}}=6(\overrightarrow{\mathbf{Q C}}+\overrightarrow{\mathbf{C P}})
\end{aligned}$
$\Rightarrow \quad k \overrightarrow{\mathbf{P Q}}=6 \overrightarrow{\mathbf{Q P}}=-6 \overrightarrow{\mathbf{P Q}} \quad$ (given)
$\Rightarrow \quad k=-6$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
In a radioactive disintegration, the ratio of initial number of atoms to the number of atoms present at time $t=\frac{1}{2 \lambda}$ is $[\lambda=$ decay constant $]$

Options:
A $\frac{1}{\mathrm{e}}$
B $\sqrt{\mathrm{e}}$
C $\mathrm{e}$
D $2 \mathrm{e}$

The correct answer is:
$\sqrt{\mathrm{e}}$

Solution:
According to radioactive disintegration law,
$$
\begin{aligned}
& \mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}} \\
& \frac{\mathrm{N}}{\mathrm{N}_0}=\mathrm{e}^{-\lambda t} \\
& \frac{\mathrm{N}}{\mathrm{N}_0}=\mathrm{e}^{-\lambda \times \frac{1}{2 \lambda}} \\
& \left(\because t=\frac{1}{2 \lambda}\right) \\
& \frac{\mathrm{N}}{\mathrm{N}_0}=\mathrm{e}^{-\frac{1}{2}} \\
& \frac{\mathrm{N}_0}{\mathrm{~N}}=\mathrm{e}^{\frac{1}{2}} \\
& \frac{\mathrm{N}_0}{\mathrm{~N}}=\sqrt{\mathrm{e}} \\
&
\end{aligned}
$$

Topic: Physics, Nuclear Physics, JEE Main
In a radioactive material, a fraction of active material remaining after the time t is 916 . The fraction that was remaining after the time t2 is :

Options:
A 45
B 35
C 34
D 78

The correct answer is:
34

Solution:
N=N0e-λt ...1

N'=N0e-λt2….2

from (1) & (2)

N'N0=NN012=91612=34.

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2020 (03 Sep Shift 1)
In a radioactive material the activity at time $t_1$, is $A_1$ and at a later time $t_2$, it is $A_2$. If the decay constant of the material is $\lambda$, then

Options:
A
$A_1=A_2 e^{-\lambda\left(t_1-t_2\right)}$

B
$A_1=A_2 e^{\lambda\left(t_1-t_2\right)}$

C
$A_1=A_2\left(t_2 / t_1\right)$

D
$A_1=A_2$

The correct answer is:

$A_1=A_2 e^{-\lambda\left(t_1-t_2\right)}$

Solution:
From radioactive decay law,
$$
-\frac{d N}{d t} \propto N \text { or }-\frac{d N}{d t}=\lambda N
$$
Thus,
$$
A=-\frac{d N}{d t} \text { or } A=\lambda N
$$
$$
\Rightarrow \quad A=\lambda N_0 e^{-\lambda t}
$$
Where, $A_0=\lambda N_0$ is the activity of the radioactive material at time, $t=0$
At time, $t_1 \quad A_1=A_0 e^{-\lambda t_1}$
At time, $t_2 \quad A_2=A_0 e^{-\lambda t_2}$
On dividing Eq. (ii) by Eq. (iii), we have
$$
\begin{aligned}
\quad \frac{A_1}{A_2} & =\frac{e^{-\lambda t_1}}{e^{-\lambda t_2}}=e^{-\lambda\left(t_1-t_2\right)} \\
\Rightarrow \quad \quad A_1 & =A_2 e^{-\lambda\left(t_1-t_2\right)}
\end{aligned}
$$

Topic: Physics, Nuclear Physics, JEE Main
In a random experiment, a fair die is rolled until two fours are obtained in succession. The probability that the experiment will end in the fifth throw of the die is equal to :

Options:
A 15065
B 17565
C 22565
D 20065

The correct answer is:
17565

Solution:
The probability of getting a 4 in a throw of a die is 16.

The probability of getting other than 4 i.e. 4 i.e. numbers 1, 2, 3, 5& 6 in a single throw of a die is 56.

Total throws done =5.

For the event to be completed in 5th throw, 4th and 5th throw must be 4 . Also, 3rd throw must be other than 4 i.e. 4. Hence, cases for the first two position can be :

Case I : 1st throw must be 4 and 2nd throw must not be 4 i.e. 4

Thus, P44-4-44=16.56.56.16.16=2565.

Case II : 1st throw must not be 4 i.e. 4 and 2nd throw must not be 4 i.e. 4

Thus, P4-4-4-44=56.56.56.16.16=12565.

Case III : 1st throw must not be 4 i.e. 4 and 2nd throw must be 4

Thus, P4-44-44=56.16.56.16.16=2565.

Therefore, probability that experiment will end in the fifth throw given that two fours occurred in succession will be

P44-4-44+P4-44-44+P4-4-4-44

=2565+2565+12565

=17565.

Topic: Mathematics, Probability, JEE Main, JEE Main 2019 (12 Jan Shift 1)
In a reaction, $A+B \rightleftharpoons C+D, 40 \%$ of $B$ has reacted at equilibrium, when 1 mole of $A$ was heated with 1 mole of $B$ in a $10 \mathrm{~L}$ closed vessel. The value of $K_C$ is

Options:
A $0.44$
B $0.18$
C $0.22$
D $0.36$

The correct answer is:
$0.44$

Solution:
$\begin{aligned} & A+B \rightleftharpoons C \quad+D \\ & \text { At } t=0,1 \quad 1 \quad 0 \quad 0 \\ & \text { At equil. }(1-0.4)(1-0.4) 0.40 .4 \\ & =0.6 \quad=0.6 \\ & K_c=\frac{[C][D]}{[A][B]} \\ & =\frac{0.4 \times 0.4}{0.6 \times 0.6}=0.44 \\ & \end{aligned}$

Topic: Chemistry, Chemical Equilibrium, JEE Main
$$
\text { In a reaction of ani line a coloured product } \mathrm{C} \text { was obtained. }
$$
$$
\text { The structure of } \mathrm{C} \text { would be }
$$

Options:
A

B

C

D

The correct answer is:

Solution:
Diazotisation followed electrophillic substitution at para position.

Topic: Chemistry, Amines, JEE Main
$$
\text { In a reaction of ani line a coloured product } \mathrm{C} \text { was obtained. }
$$
$$
\text { The structure of } \mathrm{C} \text { would be }
$$

Options:
A

B

C

D

The correct answer is:

Solution:
Diazotisation followed electrophillic substitution at para position.

Topic: Chemistry, Amines, NEET, NEET 2008 (Mains)
In a reaction of aniline a coloured product $C$ was obtained.
$\	ext {The structure of } C \text { would be}$

Options:
A

B

C

D

The correct answer is:

Solution:
Key Idea : $1^{\circ}$ amines form diazonium salts at low temperature and diazonium salts are weak electrophiles, so react with strongly activated ring. (takes $\mathrm{H}$ from benzene nucleus).
Note : The $-\mathrm{N}=\mathrm{N}-$ linkage is called -azo linkage, and the reactions in which azo or diazo compounds are formed, are called diazo coupling reactions.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In a reaction

reagents ‘X’ and ‘Y’ respectively are:

Options:
A CH3CO2O/H+ and CH3OH/H+, ∆
B

CH3CO2O/H+ and CH3CO2O/H+

C CH3OH/H+, ∆ and CH3OH/H+, ∆
D CH3OH/H+∆ and CH3CO2O/H+

The correct answer is:
CH3CO2O/H+ and CH3OH/H+, ∆

Solution:
When salicylic acid is heated with methyl alcohol, the carboxyl group of salicylic acid is esterified producing a strong-smelling liquid ester (methyl salicylate). Aspirin is prepared by chemical synthesis from salicylic acid, through acetylation with acetic anhydride.

Hence, reagent 'Y' and 'X' respectively are: CH3OH/H+, ∆ and CH3CO2O/H+.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2023 (01 Feb Shift 2)
In a reaction, the time required to complete half of the reaction was found to increase 16 times when the initial concentration of the reactant was reduced to 1/4th. What is the order of the reaction?

Options:

The correct answer is:
3

Solution:
t2t1=a1a2n-1
16=11/4n-1
So n = 3
i.e., third order reaction.

Topic: Chemistry, Chemical Kinetics, JEE Main
In a reactor, 2 kg of U23592 fuel is fully used up in 30 days. The energy released fission is 200 MeV. Given that the Avogadro number, N=6.023×1026 per kilo mole and 1 eV=1.6×10-19 J. The power output of the reactor is close to:

Options:
A 35 MW
B 60 MW
C 125 MW
D 54 MW

The correct answer is:
60 MW

Solution:
P=Et

=2235×6.023×1026×200×1.6×10-1930×24×60×60=60 MW

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2020 (02 Sep Shift 1)
In a rectangle $A B C D$, the coordinates of $A$ and $B$ are $(1,2)$ and $(3,6)$ respectively and some diameter of the circumscribing circle of $A B C D$ has equation $2 x-y+4=0$. Then the area of the rectangle is

Options:
A 16
B $2 \sqrt{10}$
C $2 \sqrt{5}$
D 20

The correct answer is:
16

Solution:
Slope of $\mathrm{AB}=\frac{4}{2}=2$
slope of $\mathrm{BC}=-\frac{1}{2}$
$\ell(\mathrm{AB})=\sqrt{4+16}=2 \cdot 5$
distance between $2 \mathrm{x}-\mathrm{y}+4=0 \quad \& \quad 2 \mathrm{x}-\mathrm{y}=0 \Rightarrow \frac{4}{\sqrt{5}}$
Area $=2 \sqrt{5} \cdot \frac{8}{\sqrt{5}}=16$

Topic: Mathematics, Straight Lines, JEE Main
In a reflecting telescope, a secondary mirror is used to:

Options:
A reduce the problem of mechanical support
B make chromatic aberration zero
C move the eyepiece outside the telescopic tube
D remove spherical aberration

The correct answer is:
move the eyepiece outside the telescopic tube

Solution:
Mirrors are used in reflecting telescopes, which allow astronomers to see distant objects in space more clearly.

A mirror forms an image by gathering light from spatial objects. A second mirror receives the picture that was reflected by the first mirror, which may be rather wide. The light is reflected by this tiny mirror onto an eyepiece lens, which magnifies or enlarges the image of the object.

Hence, the secondary mirror is used to move the eyepiece outside the telescopic tube.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (08 Apr Shift 1)
In a region, steady and uniform electric and magnetic fields are present. These two fields are parallel to each other. A charged particle is released from rest in this region. The path of the particle will be a

Options:
A
circle

B
helix

C
straight line

D
ellipse

The correct answer is:
straight line

Solution:
Straight line

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2006
In a region, the potential is represented by $\mathrm{V}(\mathrm{x}, \mathrm{y}, \mathrm{z})=6 \mathrm{x}-8 \mathrm{xy}-8 \mathrm{y}+6 \mathrm{yz},$ where $\mathrm{V}$ is in volts and $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are in metres. The electric force experienced by a charge of 2 coulomb situated at point (1,1,1) is :

Options:
A $6 \sqrt{5} \mathrm{~N}$
B $30 \mathrm{~N}$
C $24 \mathrm{~N}$
D $4 \sqrt{35} \mathrm{~N}$

The correct answer is:
$4 \sqrt{35} \mathrm{~N}$

Solution:
$\overrightarrow{\mathrm{E}}=-\frac{\partial \mathrm{V}}{\partial \mathrm{x}} \hat{\mathrm{i}}-\frac{\partial \mathrm{V}}{\partial \mathrm{y}} \hat{\mathrm{j}}-\frac{\partial \mathrm{V}}{\partial \mathrm{z}} \hat{\mathrm{k}}$
$=-[(6-8 \mathrm{y}) \hat{\mathrm{i}}+(-8 \mathrm{x}-8+6 \mathrm{z}) \hat{\mathrm{j}}+(6 \mathrm{y}) \hat{\mathrm{k}}]$
$\mathrm{At}(1,1,1), \overrightarrow{\mathrm{E}}=2 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}$
$\Rightarrow \quad(\overrightarrow{\mathrm{E}})=\sqrt{2^{2}+10^{2}+6^{2}}=\sqrt{140}=2 \sqrt{35}$
$\therefore \quad \mathrm{F}=\mathrm{q} \overrightarrow{\mathrm{E}}=2 \times 2 \sqrt{35}=4 \sqrt{35}$

Topic: Physics, Electrostatics, JEE Main
In a regular hexagon $A B C D E F, \overrightarrow{A E}=$

Options:
A $\overrightarrow{A C}+\overrightarrow{A F}+\overrightarrow{A B}$
B $\overrightarrow{A C}+\overrightarrow{A F}-\overrightarrow{A B}$
C $\overrightarrow{A C}+\overrightarrow{A B}-\overrightarrow{A F}$
D None of these

The correct answer is:
$\overrightarrow{A C}+\overrightarrow{A F}-\overrightarrow{A B}$

Solution:
Obviously, $\overrightarrow{A E}=\overrightarrow{A C}+\overrightarrow{C D}+\overrightarrow{D E}$
$=\overrightarrow{A C}+\overrightarrow{A F}-\overrightarrow{A B},\{\overrightarrow{C D}=\overrightarrow{A F}$ and $\overrightarrow{D E}=-\overrightarrow{A B}\}$

Topic: Mathematics, Vector Algebra, JEE Main
In a regular octahedral molecule, $\mathrm{MX}_6$ the number of $\mathrm{X}-\mathrm{M}-\mathrm{X}$ bonds at $180^{\circ}$ is:

Options:
A three
B two
C $\operatorname{six}$
D four

The correct answer is:
three

Solution:
In octahedral structure $\mathrm{MX}_6$, the six hybrid orbitals $\left(s p^3 d^2\right)$ are directed towards the corners of a regular octahedral with an angle of $90^{\circ}$. According to following structure of $\mathrm{MX}_6$, the number of $\mathrm{X}-\mathrm{M}-\mathrm{X}$ bonds at $180^{\circ}$ must be three.
Related Theory
In octahedral, two orbitals contain lone pairs of electrons on opposite sides of the central atom. The remaining four atoms connected to the central atom give the molecule a square planar shape.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2004
In a resonance column first and second resonance are obtained at depths 22.7 cm and 70.2 cm. The third resonance will be obtained at a depth

Options:
A 117.7 cm

B 92.9 cm

C 115.5 cm

D 113.5 cm

The correct answer is:
117.7 cm

Solution:
l1+x=22.7 + x=λ4

l2+x=70.2 + x=3λ4

l3+x=5λ4
x=l2-3l12=70.2-68.12=2.12=1.05 cm
l3+xl1+x=5
l3=5l1+4x=5×22.7+4×1.05=117.7 cm

Topic: Physics, Waves and Sound, JEE Main
In a resonance column, two successive resonances are obtained at depths \( 30 \mathrm{~cm} \) and \( 50 \mathrm{~cm} \), respectively. The next
resonance will be obtained at a depth,

Options:
A \( 60 \mathrm{~cm} \)
B \( 80 \mathrm{~cm} \)
C \( 70 \mathrm{~cm} \)
D \( 90 \mathrm{~cm} \)

The correct answer is:
\( 70 \mathrm{~cm} \)

Solution:
In a resonance column,

difference in depth of ;two successive resonances = distance between two consecutive nodes

=λ2

λ2=50-30cm=20 cm

Next resonance will be obtained at 50+λ2=70 cm.

Topic: Physics, Waves and Sound, JEE Main
In a resonance tube experiment, to determine the speed of sound in air, a pipe of diameter 5 cm is used. The air column in the pipe resonates with a tuning fork of frequency 480 Hz when the minimum length of the air column is 16 cm. Find the speed of sound ( in m s-1 ) in the air at room temperature.

Options:

The correct answer is:
336

Solution:
v4L+0.6r=480
â v = 480 Ã 4 Ã [L + 0.6r]

âv=336 m s-1

Topic: Physics, Waves and Sound, JEE Main
In a reverse-biased diode, when the applied voltage changes by 1 V, the current is found to change by 0.5 Î¼A. The reverse bias resistance of the diode is

Options:
A 2Ã105Â Î©
B 2Ã106Â Î©
C 200 Î©
D 2 Î©

The correct answer is:
2×106 Ω

Solution:
Reverse resistance = ∆V∆I=10.5×10-6=2×106 Ω

Topic: Physics, Semiconductors, JEE Main
In a reversible chemical reaction at equilibrium, if the concentration of any one of the reactants is doubled, then the equilibrium constant will

Options:
A also be doubled
B be halved
C remains the same
D becomes one-fourth

The correct answer is:
remains the same

Solution:
Equilibrium constant is independent of the concentration of reactants.

Topic: Chemistry, Chemical Equilibrium, WBJEE, WBJEE 2012
In a right angle $\triangle \mathrm{ABC}, \angle \mathrm{A}=90^{\circ}$ and sides a, b, c are respectively, $5 \mathrm{~cm}, 4 \mathrm{~cm}$ and $3 \mathrm{~cm}$. If a force $\vec{F}$ has moments 0,9 and 16 in $N$ cm. units respectively about vertices $A, B$ and $C$, then magnitude of $\vec{F}$ is

Options:
A
3

B
4

C
5

D
9

The correct answer is:
5

Solution:
$\mathrm{F} \cdot 3 \sin \theta=9$
$\mathrm{~F} \cdot 4 \cos \theta=16$
$\Rightarrow \mathrm{F}=5$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2004
In a rocket of mass $1000 \mathrm{~kg}$ fuel is consumed at a rate of $40 \mathrm{~kg} / \mathrm{s}$. The velocity of the gases ejected from the rocket is $5 \times 10^4 \mathrm{~m} / \mathrm{s}$. The thrust on the rocket is

Options:
A $2 \times 10^3 \mathrm{~N}$
B $5 \times 10^4 \mathrm{~N}$
C $2 \times 10^6 \mathrm{~N}$
D $2 \times 10^9 \mathrm{~N}$

The correct answer is:
$2 \times 10^6 \mathrm{~N}$

Solution:
$2 x 10^6 \mathrm{~N}$
Step 1:Given
The rate of change of mass of the rocket due to the consumption of fuel is $\frac{d m}{d t}=40 \mathrm{~kg} / \mathrm{s}$
The velocity of the gases ejected from the rocket $v=5 x 10^4 \mathrm{~m} / \mathrm{s}$
Step 2: Formula used
We know that the accelerated mass in a particular direction causes an upthrust force that is equal in magnitude and opposite in direction.
For a changing mass, we get,
$\therefore F=m a=m \frac{d v}{d t}=v \frac{d m}{d t}$
Step 3:Calculate the thrust on the rocket:
Upon substituting the values we get,
$\begin{aligned}
& F=40 \times 5 \times 10^4 \\
& =200 \times 10^4 \\
& =2 \times 10^6 N
\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
In a rotor, a hollow vertical cylindrical structure rotates about its axis and a person rests against the inner wall. At a particular speed, the floor below the person is removed and the person hangs resting against the wall without any floor. If the radius of the rotor is 2 m and the coefficient of static friction between the wall and the person is 0.2, find the minimum speed (in m s-1) of any point the wall at which the floor may be removed. Take g = 10 m s-2.

Options:

The correct answer is:
10

Solution:
The situation is shown in the diagram

When the floor is removed, the forces on the person are
(1) weight mg downward
(2) normal force N due to the wall, towards the centre
(3) frictional force fs, parallel to the wall, upward.
The person is moving in a circle with a uniform speed, so its acceleration is v2r towards the centre.
Newton's law for the horizontal direction (2nd law) and for the vertical direction (1st law) give
N = mv2/r ...(i)
and fs = mg. ...(ii)
âFor the minimum speed when the floor may be removed, the friction is limiting one and so equals Î¼sN. This gives
Î¼sN = mg
or, Î¼ s mv 2 r = mg using (i)
or,v=rgÎ¼s=2 mÃ10 m/s20.2=10 m s-1

Topic: Physics, Laws of Motion, JEE Main
In a Rutherford scattering experiment when a projectile of charge $Z_1$ and mass $M_1$ approaches a target nucleus of charge $Z_2$ and mass $M_2$, the distance of closest approach is $r_0$. The energy of the projectile is :

Options:
A Directly proportional to mass $M_1$
B Directly proportional to $\mathrm{M}_1 \times \mathrm{M}_2$
C Directly proportional to $Z_1 Z_2$
D Inversely proportional to $Z_1$

The correct answer is:
Directly proportional to $Z_1 Z_2$

Solution:
$$
\text { } \text { Energy }=\frac{1}{4 \pi \in_0} \frac{Z_1 Z_2}{\mathrm{r}_0}
$$

Topic: Physics, Atomic Physics, NEET, NEET 2009 (Mains)
$$
\text { In a } \mathrm{S}_{\mathrm{N}} 2 \text { substitution reaction of the type }
$$
$$
\mathrm{R}-\mathrm{Br}+\mathrm{Cl}^{-} \stackrel{\mathrm{OMF}}{\longrightarrow} \mathrm{R}-\mathrm{Cl}+\mathrm{Br}^{-}
$$
Which one of the following has the highest relative rate?

Options:
A $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}$
B $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{Br}$
C

D

The correct answer is:
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}$

Solution:
For $\mathrm{S}_{\mathrm{N} 2}$ reaction, there should be low steric hinderance for better reactivity.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET, NEET 2008 (Mains)