datavorous/entrance-exam-dataset huggingface dataset:
A coil is placed perpendicular to a magnetic field of 5000 T. When the field is changed to 3000 T in 2 s, an induced emf of 22 V is produced in the coil. If the diameter of the coil is 0.02 m, then the number of turns in the coil is:

Options:
A 7
B 70
C 35
D 140

The correct answer is:
70

Solution:
Average induced EMF is given by, ε=N∆ϕt

Also, change in flux ∆ϕ=(∆B)A

Given: Bi=5000 T,

Bf=3000 T

d=0.02 m ⇒r=d2=0.01 m

Hence,

∆ϕ=(∆B)A

=2000π0.012=0.2π

EMF, ε=N∆ϕt⇒22=N0.2π2

⇒N=70

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2024 (31 Jan Shift 1)
A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating; it is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to :

Options:
A
development of air current when the plate is placed.

B
induction of electrical charge on the plate

C
shielding of magnetic lines of force as aluminium is a paramagnetic material.

D
electromagnetic induction in the aluminium plate giving rise to electromagnetic damping.

The correct answer is:
electromagnetic induction in the aluminium plate giving rise to electromagnetic damping.

Solution:
Oscillating coil produces time variable magnetic field. It cause eddy current in the aluminium plate which causes anti-torque on the coil, due to which is stops.

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2012 (Offline)
A coil of 40 henry inductance is connected in series with a resistance of $8 \mathrm{ohm}$ and the combination is joined to the terminals of a 2 volt battery. The time constant of the circuit is:

Options:
A 5 seconds
B 1/5 seconds
C 40 seconds
D 20 seconds

The correct answer is:
5 seconds

Solution:
According to the question Time constant $L R$ circuit is $T=\frac{L}{P}$
$$
\therefore \quad T=5 \mathrm{sec}
$$

Topic: Physics, Electromagnetic Induction, JEE Main
A coil of 100 turns and area 5 square centimetre is placed in a magnetic field $B=0.2 T$. The normal to the plane of the coil makes an angle of $60^{\circ}$ with the direction of the magnetic field. The magnetic flux linked with the coil is

Options:
A ${ }^{5 \times 10^{-3} \mathrm{~Wb}}$
B ${ }^{5 \times 10^{-5} \mathrm{~Wb}}$
C $0^{10^{-2} \mathrm{~Wb}}$
D $10^{-4} \mathrm{~Wb}$

The correct answer is:
${ }^{5 \times 10^{-3} \mathrm{~Wb}}$

Solution:
$\phi=N B A \cos \theta=100 \times 0.2 \times 5 \times 10^{-4} \cos 60^{\circ}$
$=5 \times 10^{-3} \mathrm{~Wb}$

Topic: Physics, Electromagnetic Induction, JEE Main
A coil of 200 turns and area 0.20 m2 is rotated at half a revolution per second and is placed in uniform magnetic field of 0.01 T perpendicular to axis of rotation of the coil. The maximum voltage generated in the coil is 2πβ volt. The value of β is ______.

Options:

The correct answer is:
5

Solution:
Time taken for one complete revolution = time period =2 s.

Therefore, ω=2πT=π rad s-1

Now flux, ϕ=NABcos(ωt)

Therefore, induced EMF ε=-dϕdt=NABωsin(ωt)

The maximum voltage generated, εmax=NABω

=200×0.2×0.01×π

=4π10=2π5 volt

Hence, β=5.

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2024 (01 Feb Shift 2)
A coil of area $100 \mathrm{~cm}^2$ has 500 turns. Magnetic field of 0.1 weber $/$ metre $^2$ is perpendicular to the coil. The field is reduced to zero in 0.1 second. The induced e.m.f. in the coil is

Options:
A $1 \mathrm{~V}$
B $5 \mathrm{~V}$
C $50 \mathrm{~V}$
D Zero

The correct answer is:
$5 \mathrm{~V}$

Solution:
$e=-\frac{N\left(B_2-B_1\right) A \cos \theta}{\Delta t}$
$=-\frac{500 \times(0-0.1) \times 100 \times 10^{-4} \cos 0}{0.1}=5 \mathrm{~V}$

Topic: Physics, Electromagnetic Induction, JEE Main
A coil of circular cross-section having 1000 turns and $4 \mathrm{~cm}^2$ face area is placed with its axis parallel to a magnetic field which decreases by $10^{-2} \mathrm{~Wb}$ $\mathrm{m}^{-2}$ in $0.01 \mathrm{~s}$. The e.m.f. induced in the coil is:

Options:
A
$400 \mathrm{mV}$

B
$200 \mathrm{mV}$

C
$4 \mathrm{mV}$

D
$0.4 \mathrm{mV}$

The correct answer is:
$400 \mathrm{mV}$

Solution:
Given: No. of turns $N=1000$
Face area, $A=4 \mathrm{~cm}^2=4 \times 10^{-4} \mathrm{~m}^2$
Change in magnetic field,
$$
\Delta \mathrm{B}=10^{-2} \mathrm{wbm}^{-2}
$$
Time taken, $t=0.01 \mathrm{~s}=10^{-2} \mathrm{sec}$
Emf induced in the coil $e=$ ?
Applying formula,
Induced emf, $e=\frac{-d \phi}{d t}$
$$
\begin{aligned}
&=N\left(\frac{\Delta B}{\Delta t}\right) A \cos \theta \\
&=\frac{1000 \times 10^{-2} \times 4 \times 10^{-4}}{10^{-2}}=400 \mathrm{mV}
\end{aligned}
$$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2014 (11 Apr Online)
A coil of cross-sectional area A having n turns is placed in a uniform magnetic field B. When it is rotated with an angular velocity ω, the maximum e.m.f. induced in the coil will be:

Options:
A 32nBAω
B 3nBAω
C nBAÏ
D 12nBAω

The correct answer is:
nBAω

Solution:
Magnetic flux through a coil of area A of n turns at an angle θ to the magnetic field of magnitude B is, φ=nBA cosθ=nBAcosωt. The induced emf,
ε=dϕdt=nBAωsin⁡ωt. Since, the maximum value of the sine function is 1,
εmax=nBAω

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2018 (16 Apr Online)
A coil of inductance 1 H and resistance 100 Ω is connected to a battery of 6 V. Determine approximately :

(a) The time elapsed before the current acquires half of its steady-state value
(b) The energy stored in the magnetic field associated with the coil at an instant 15 ms after the circuit is switched on. (Given ln2=0.693, e-32=0.25)

Options:
A t=10 ms;U=2 mJ
B t=10 ms;U=1 mJ
C t=7 ms;U=1 mJ
D t=7 ms;U=2 mJ

The correct answer is:
t=7 ms;U=1 mJ

Solution:
Given circuit is R-L growth circuit

The current is the circuit is given by,

i=ER1-e-tτ

For current to be half of the peak value,

i=E2R=ER1-e-tτ

Solving t=τln2

t=LRln2=11000.693=0.00693≃7 ms

At time t=15 ms,

i15 ms=ER1-e-1510

⇒i=61001-14=34×6100=0.045 A

⇒U=12Li2

by solving we get U=1 mJ.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2022 (29 Jul Shift 1)
A coil of inductance 2 H having negligible resistance is connected to a source of supply whose voltage is given by V=3t volt. (where t is in second). If the voltage is applied when t=0, then the energy stored in the coil after 4 s in J,

Options:

The correct answer is:
144

Solution:
ε=LdIdt

3∫04tdt=2∫01dI

32×16=2I

I=12

V=12LI2=12×2122=144 J

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2021 (25 Feb Shift 1)
A coil of inductance $300 \mathrm{mH}$ and resistance $2 \Omega$ is connected to a source of voltage 2V. The current reaches half of its steady state value in

Options:
A
$0.05 \mathrm{~s}$

B
$0.1 \mathrm{~s}$

C
$0.15 \mathrm{~s}$

D
$0.3 \mathrm{~s}$

The correct answer is:
$0.1 \mathrm{~s}$

Solution:
$\mathrm{I}=\mathrm{I}_{\circ}\left(1-\mathrm{e}^{-\frac{\mathrm{R}}{\mathrm{L}}\mathrm{t}}\right)$
$0.693=\frac{\mathrm{R}}{\mathrm{L}} \mathrm{t}$
$\mathrm{t}=\frac{.3 \times 0.693}{2}=0.1 \mathrm{sec}$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2005
A coil of inductive reactance 3 1 Î© has a resistance of 8 Î© . It is placed in series with a condenser of capacitive reactance 2 5 Î© . The combination is connected to an ac source of 110âV . The power factor of the circuit is

Options:
A 0.33
B 0.56
C 0.64
D 0.80

The correct answer is:
0.80

Solution:
X L = 3 1 Î© , X C = 2 5 Î© , R = 8 Î©
Impedance of series LCR is
Z = R 2 + X L - X C 2
= 8 2 + 3 1 - 2 5 2 = 6 4 + 3 6 = 1 0 Î©
Power factor, cos Ï = R Z = 8 1 0 = 0.8

Topic: Physics, Electromagnetic Induction, JEE Main
A coil of ' $n$ ' turns and radius ' $R$ ' carries a current ' $I$ '. It is unwound and rewound again to make another coil of radius $\left(\frac{\mathrm{R}}{3}\right)$, current remaining the same. The ratio of magnetic moment of the new coil to that of original coil is

Options:
A $3: 1$
B $1: 3$
C $9: 1$
D $1: 9$

The correct answer is:
$1: 3$

Solution:
As the length of the wire remains the same, we can write
$$
\begin{aligned}
& \mathrm{N}_1 2 \pi \mathrm{R}=\mathrm{N}_2 \times 2 \pi \frac{\mathrm{R}}{3} \\
& \mathrm{~N}_1=\frac{\mathrm{N}_2}{3} \\
& \mathrm{~N}_2=3 \mathrm{~N}_1
\end{aligned}
$$
$\begin{array}{ll} & \text { Magnetic moment of coil } \mu=\text { NIA } \\ \therefore & \mu_1=N_1 I A_1=N_1 I \pi R^2 \\ \therefore & \mu_2=N_2 I A_2=3 \frac{N_1 I \pi R^2}{9} \\ \therefore & \frac{\mu_2}{\mu_1}=3 \frac{N_1 I \pi R^2}{9} \times \frac{1}{N_1 I \pi R^2}=\frac{1}{3}\end{array}$

Topic: Physics, Magnetic Effects of Current, MHT CET, MHT CET 2023 (12 May Shift 2)
A coil of resistance $10 \Omega$ and an inductance $5 \mathrm{H}$ is connected to a 100 volt battery. Then energy stored in the coil is

Options:
A $125 \mathrm{erg}$
B $125 \mathrm{~J}$
C $250 \mathrm{erg} \quad$
D $250 \mathrm{~J}$

The correct answer is:
$250 \mathrm{~J}$

Solution:
$U=\frac{1}{2} L i^2 \Rightarrow$ $U=\frac{1}{2} \times 5 \times\left(\frac{100}{10}\right)^2=250 \mathrm{~J}$

Topic: Physics, Electromagnetic Induction, JEE Main
A coil of self inductance 10 mH and resistance of 0.1 Ω is connected through a switch to a battery of internal resistance 0.9 Ω . After the switch is closed, the time taken for the current to attain 80% of the saturation value is: [ ln5=1.6 ]

Options:
A 0.103 s
B 0.002 s
C 0.324 s
D 0.016 s

The correct answer is:
0.016 s

Solution:
The instantaneous current in an LR circuit is given by the equation :
(instantaneous current) I=Is1-e-RtL

Where I, Is, R, t, L are the instantaneous current, saturation current, resistance, time and inductance respectively.

Given that the current becomes so percent of saturation current.
⇒0.8 Is=Is1-e-RtL

R=0.1Ω+0.9Ω=1Ω
⇒0.8=1-e-1×t10×10-3
⇒0.8=1-e-100t
Therefore, the time is:
t=ln⁡11-0.8100=ln⁡5100=0.016 s

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2019 (10 Apr Shift 2)
A coil of self inductance $L$ is connected at one end of two rails as shown in figure. A connector of length $l$, mass $m$ can slide freely over the two parallel rails. The entire set up is placed in a magnetic field of induction $B$ going into the page. At an instant $t=0$ an initial velocity $v_0$ is imparted to it and as a result of that it starts moving along $x$-axis. The displacement of the connector is represented by the figure.

Options:
A

B

C

D

The correct answer is:

Solution:
No Solution Available

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2012 (19 May Online)
A coil of self-inductance $L$ is connected in series with a bulb $B$ and an $A C$ source. Brightness of the bulb decreases when

Options:
A frequency of the $\mathrm{AC}$ source is decreased
B number of turns in the coil is reduced
C a capacitance of reactance $X_C=X_L$ is included in the same circuit
D an iron rod is inserted in the coil

The correct answer is:
an iron rod is inserted in the coil

Solution:
As $Z=\sqrt{R^2+X_L^2}=\sqrt{R^2+(2 \pi f V L)^2}$
As $/=\frac{V}{Z}, P=/^2 R$
i.e., $V \uparrow, L \uparrow \Rightarrow Z \uparrow, / \downarrow$ and $P \downarrow$

Topic: Physics, Alternating Current, JEE Main
A coil resistance $20 \Omega$ and inductance $5 \mathrm{H}$ is connected with a $100 \mathrm{~V}$ battery. Energy stored in the coil will be

Options:
A $41.5 \mathrm{~J}$
B $62.50 \mathrm{~J}$
C $125 \mathrm{~J}$
D $250 \mathrm{~J}$

The correct answer is:
$62.50 \mathrm{~J}$

Solution:
$U=\frac{1}{2} L i^2=\frac{1}{2} L\left(\frac{E}{R}\right)^2$ $=\frac{1}{2} \times 5 \times\left(\frac{100}{20}\right)^2=62.50 \mathrm{~J}$

Topic: Physics, Electromagnetic Induction, JEE Main
A coil takes 15 minutes to boil a certain amount of water. Another coil takes 20 minutes for the same process. Time taken to boil the same amount of water when both coils are connected in series is

Options:
A 5 minutes
B 8.6 minutes
C 35 minutes
D 30 minutes

The correct answer is:
35 minutes

Solution:
From Joule's law,
H=V2tRJ cal
Where V is potential difference, t is the time and R is the resistance.
R=R1+R2
Let tR=HJV2= constant=k
⇒R1=kt1 and R2=kt2
∴ kt=kt1+kt2
t=15+20
t=35 minutes

Topic: Physics, Current Electricity, JEE Main
A coin is biased so that the head is 3 times as likely to occur as tail. This coin is tossed until a head or three tails occur. If X denotes the number of tosses of the coin, then the mean of X is

Options:
A 3816
B 1516
C 2116
D 8164

The correct answer is:
2116

Solution:
It is given that P(H)=3P(T)

∴P(H)=34,P(T)=14

Since coin is tossed till either 1H or 3T occurs. So, process will end in the maximum of 3 throws.

Xi	1	2	3

PXi	34	14·34	14×14×14+14×14×34

Now we know that,

Mean is given by ∑i=1nXiPXi

Mean=1×34+2×316+3×116=2116

Hence, this is the correct option.

Topic: Mathematics, Probability, JEE Main, JEE Main 2023 (13 Apr Shift 1)
A coin is dropped in a lift. It takes time $t_1$ to reach the floor when lift is stationary. It takes time ${ }^{t_2}$ when lift is moving up with constant acceleration. Then

Options:
A $t_1\gtt_2$
B $t_2\gtt_1$
C $t_1=t_2$
D $t_1 \gg t_2$

The correct answer is:
$t_1\gtt_2$

Solution:
For stationary lift $\left.t_1=\sqrt{\frac{2 h}{g}} \right\rvert\,$ and when the lift is moving up with constant acceleration $t_2=\sqrt{\frac{2 h}{g+a}} \quad \therefore t_1\gtt_2$

Topic: Physics, Laws of Motion, JEE Main
A coin is placed on a disc. The coefficient of friction between the coin and the disc is μ. If the distance of the coin from the center of the disc is r, the maximum angular velocity which can be given to the disc, so that the coin does not slip away, is :

Options:
A μgr
B rμg
C μgr
D μrg

The correct answer is:
μgr

Solution:
Normal force acting on the coin is, N=mg.

Required friction force to provide centripetal force for the coin is, f=mω2r and maximum value of static friction force available is, f=μN=μmg.

For limiting condition, we can write

⇒μmg=mrω2

⇒ω=μgr

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2024 (31 Jan Shift 1)
A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency $\omega$. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time

Options:
A
at the highest position of the platform

B
at the mean position of the platform

C
for an amplitude of $\frac{\mathrm{g}}{\omega^2}$

D
for an amplitude of $\frac{\mathrm{g}^2}{\omega^2}$

The correct answer is:
for an amplitude of $\frac{\mathrm{g}}{\omega^2}$

Solution:
$A \omega^2=g$
$\Rightarrow A=g / \omega^2$

Topic: Physics, Oscillations, JEE Main, JEE Main 2006
A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency $\omega$. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time

Options:
A
at the highest position of the platform

B
at the mean position of the platform

C
for an amplitude of $\frac{\mathrm{g}}{\omega^2}$

D
for an amplitude of $\frac{\mathrm{g}^2}{\omega^2}$

The correct answer is:
for an amplitude of $\frac{\mathrm{g}}{\omega^2}$

Solution:
$A \omega^2=g$
$\Rightarrow A=g / \omega^2$

Topic: Physics, Oscillations, JEE Main
A coin is tossed. If head turns up then pair of dice is thrown and sum of numbers is noted. If tail turns up then a card numbered from \( 1 \) to \( 9 \) is drawn and number is noted. The probability that the sum of numbers on dice or the number on card is \( 7 \) or \( 8 \), is:

Options:
A \( \frac{13}{36} \)
B \( \frac{15}{72} \)
C \( \frac{19}{72} \)
D \( \frac{11}{36} \)

The correct answer is:
\( \frac{19}{72} \)

Solution:
P7or 8

=PH P7 or 8+PT P7or 8

=12×1136+12×29=1172+19=1972

Topic: Mathematics, Probability, JEE Main
A coin is tossed three times in succession. If $E$ is the event that there are at least two heads and $F$ is the event in which first throw is a head, then $P\left(\frac{E}{F}\right)=$

Options:
A $\frac{3}{4}$
B $\frac{3}{8}$
C $\frac{1}{2}$
D $\frac{1}{8}$

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

Solution:
$S=\{H H H, H H T, H T H, T H H, H T T, T H T, T T H, T T T$
$n(E)=4, n(F)=4$ and $n(E \cap F)=3$
$\therefore P\left(\frac{E}{F}\right)$ $=\frac{P(E \cap F)}{P(F)}=\frac{3 / 8}{4 / 8}=\frac{3}{4}$.

Topic: Mathematics, Probability, JEE Main
A coin is tossed three times. What is the probability of getting head and tail (HTH) or tail and head (THT) alternatively?

Options:
A $1 / 4$
B $1 / 5$
C $1 / 6$
D $1 / 8$

The correct answer is:
$1 / 4$

Solution:
Total possible outcomes, $\mathrm{S}=\{\mathrm{HHH}, \mathrm{HHT}, \mathrm{HTH}, \mathrm{THT}$.
TTH. THH, TTT, HTT $\}$ and desired outcomes E $=\{\mathrm{HTH}, \mathrm{THT}\}$
$\Rightarrow \mathrm{n}(\mathrm{E})=2$ and $\mathrm{n}(\mathrm{S})=8$
Hence, required probability $=\mathrm{P}(\mathrm{E})=\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{2}{8}=\frac{1}{4}$

Topic: Mathematics, Probability, JEE Main
A coin placed on a rotating turn table just slips if it is placed at a distance of $4 \mathrm{~cm}$ from the centre. If the angular velocity of the turn table is doubled it will just slip at a distance of

Options:
A $1 \mathrm{~cm}$
B $2 \mathrm{~cm}$
C $4 \mathrm{~cm}$
D $8 \mathrm{~cm}$

The correct answer is:
$1 \mathrm{~cm}$

Solution:
When the coin placed on a rotating table is first about to slips, then till the time the coin rotates along with the table, necessary centripetal force for rotation will be provided by the frictional force,
i.e., $\quad m r \omega^{2}=\mu m g$
where,
$m=$ mass of the coin,
$r=$ distance of the coin from the centre of the table,
$\omega=$ angular velocity,
$\mu=$ coefficient of friction
and $g=$ acceleration due to gravity.
Since, $m, \mu$ and $g$ remains constant, so
$r \omega^{2}=$ constant
$\Rightarrow \quad r \propto \frac{1}{\omega^{2}}$
or $\quad \frac{r_{2}}{r_{1}}=\left(\frac{\omega_{1}}{\omega_{2}}\right)^{2}$
Given, $\omega_{2}=2 \omega_{1}, r_{1}=4 \mathrm{~cm}$
$\Rightarrow \quad \frac{r_{2}}{r_{1}}=\left(\frac{\omega_{1}}{2 \omega_{1}}\right)^{2}$
or
$r_{2}=\frac{r_{1}}{4}=\frac{4}{4}=1 \mathrm{~cm}$

Topic: Physics, Rotational Motion, KCET, KCET 2021
A collection of all the alleles of all the genes of a crop plant is called

Options:
A
germplasm collection

B
protoplasm collection

C
herbarium

D
somaclonal collection

The correct answer is:
germplasm collection

Solution:
The entire collection of plants/seeds having all the diverse alleles for all genes in a given crop is germplasm collection. The given definition does not fit in other options because protoplasm collection deals mainly with collection of protoplast, i.e. plant cell without cell wall.
A herbarium is museum of preserved plant that are used for botanical research, mainly in identification and classification of plants.
Somaclonal collection involves plants produced from a single cell which are gentically variable from their parents.

Topic: Biology, Strategies for Enhancement in Food Production
A collection of all the alleles of all the genes of a crop plant is called

Options:
A Germplasm collection
B Protoplasm collection
C Somaclonal collection
D Herbarium

The correct answer is:
Germplasm collection

Solution:
Germplasm are living genetic resources such as seeds or tissues that are maintained for the purpose of animal and plant breeding, preservation, and other research uses. These resources may take the form of seed collections stored in seed banks, trees growing in nurseries, animal breeding lines maintained in animal breeding programs or gene banks, etc.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
A colloid is formed by adding FeCl3in excess of hot water. What will happen if excess sodium chloride is added to this colloid?

Options:
A The colloid will become more stable.
B The colloid will coagulate.
C The colloid will disintegrate.
D The colloid will remain unchanged.

The correct answer is:
The colloid will coagulate.

Solution:
Correct Option is : (B)
The colloid will coagulate.

Topic: Chemistry, Surface Chemistry, NEET
A colloidal solution can be purified by

Options:
A Filtration
B Peptization
C Coagulation
D Dialysis

The correct answer is:
Dialysis

Solution:
Dialysis: Process of removing a dissolved substance from a colloidal solution by means of diffusion through a suitable membrane.

Topic: Chemistry, Surface Chemistry, JEE Main
A coloured compound is formed when MnO2 is fused with KOH , the compound and its colour is

Options:
A K2MnO4 , purple green
B KMnO4 , purple
C Mn2O3 , brown
D Mn3O4 black

The correct answer is:
K2MnO4 , purple green

Solution:
Potassium maganate is formed when mangnese dioxide is fused with KOH, which is purple green coloured compound.
MnO2+KOH+O2→K2MnO4+H2O .

Topic: Chemistry, d and f Block Elements, JEE Main
A colourless, neutral, paramagnetic oxide of nitrogen ' $P$ ' on oxidation gives reddish brown gas $Q$. $Q$ on cooling gives colourless gas $R . R$ on reaction with $P$ gives blue solid $S$. Identify $P, Q, R, S$ respectively

Options:
A $\mathrm{N}_{2} \mathrm{O}, \mathrm{NO}, \mathrm{NO}_{2}, \mathrm{~N}_{2} \mathrm{O}_{5}$
B $\mathrm{N}_{2} \mathrm{O}, \mathrm{NO}_{2}, \mathrm{~N}_{2} \mathrm{O}_{4}, \mathrm{~N}_{2} \mathrm{O}_{3}$
C $\mathrm{NO}_{2}, \mathrm{NO}_{2}, \mathrm{~N}_{2} \mathrm{O}_{4}, \mathrm{~N}_{2} \mathrm{O}_{3}$
D $\mathrm{NO}, \mathrm{NO}, \mathrm{N}_{2} \mathrm{O}_{4}, \mathrm{~N}_{2} \mathrm{O}_{5}$

The correct answer is:
$\mathrm{NO}_{2}, \mathrm{NO}_{2}, \mathrm{~N}_{2} \mathrm{O}_{4}, \mathrm{~N}_{2} \mathrm{O}_{3}$

Solution:
The complete given chemical reaction take place as follows
Note: Nitric oxide is paramagnetic as it contain unpaired electron as per molecular orbital theory.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the 4 μF and 9 μF capacitors), at a point distant 30 m from it, would equal:

Options:
A 420 N/C
B 480 N/C
C 240 N/C
D 360 N/C

The correct answer is:
420 N/C

Solution:
Ceq=5μF
Potential of 4 μF= 6 volt
∴charge on 4 μF q4=24 μC
Potential of 9 μF=2 volt

∴charge on 9 μF q9=18 μC
Total charge (q)=42 μC
E=kqr2=9×109×42×10-6900=420 N/C

Topic: Physics, Capacitance, JEE Main
A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the 4 μF and 9 μF capacitors), at a point distant 30 m from it, would equal:

Options:
A 420 N/C
B 480 N/C
C 240 N/C
D 360 N/C

The correct answer is:
420 N/C

Solution:
Ceq=5μF
Potential of 4 μF= 6 volt
∴charge on 4 μF q4=24 μC
Potential of 9 μF=2 volt

∴charge on 9 μF q9=18 μC
Total charge (q)=42 μC
E=kqr2=9×109×42×10-6900=420 N/C

Topic: Physics, Capacitance, JEE Main, JEE Main 2016 (03 Apr)
A combination of parallel plate capacitors is maintained at a certain potential difference.
When a 3 mm thick slab is introduced between all the plates, in order to maintain the same potential difference, the distance between the plates is increased by 2.4 mm. Find the dielectric constant of the slab.

Options:
A 6
B 4
C 3
D 5

The correct answer is:
5

Solution:
Assume initial separation as 3 mm
each capacitance C1=ε0A3 before
C 1 = k ε 0 A 3 . ε 0 A 2.4 k ε 0 A 3 + ε 0 A 2.4
to get same capacitance ε0A3=kε0A3 .ε0A2.4kε0A3+ε0A2.4
3k=2.4k+3
0.6k=3⇒k=30.6
k=306=5.

Topic: Physics, Capacitance, JEE Main
A combination of parallel plate capacitors is maintained at a certain potential difference.
When a 3 mm thick slab is introduced between all the plates, in order to maintain the same potential difference, the distance between the plates is increased by 2.4 mm. Find the dielectric constant of the slab.

Options:
A 6
B 4
C 3
D 5

The correct answer is:
5

Solution:
Assume initial separation as 3 mm
each capacitance C1=ε0A3 before
C 1 = k ε 0 A 3 . ε 0 A 2.4 k ε 0 A 3 + ε 0 A 2.4
to get same capacitance ε0A3=kε0A3 .ε0A2.4kε0A3+ε0A2.4
3k=2.4k+3
0.6k=3⇒k=30.6
k=306=5.

Topic: Physics, Capacitance, JEE Main, JEE Main 2017 (09 Apr Online)