datavorous/entrance-exam-dataset huggingface dataset:
A number $\mathrm{x}$ is chosen at random from the set $\{1$, $2,3,4, \ldots ., 100\}$. Define the event: $\mathrm{A}=$ the chosen number $x$ satisfies $\frac{(x-10)(x-50)}{(x-30)} \geq 0$
Then $\mathrm{P}(\mathrm{A})$ is:

Options:
A
$0.71$

B
$0.70$

C
$0.51$

D
$0.20$

The correct answer is:
$0.71$

Solution:
Given $\frac{(x-10)(x-50)}{(x-30)} \geq 0$
Let $x \geq 10, x \geq 50$ equation will be true $\forall x \geq 50$ as $\left(\frac{x-50}{x-30}\right) \geq 0, \forall x \in[10,30)$
$$
\frac{(x-10)(x-50)}{x-30} \geq 0 \quad \forall x \in[10,30)
$$
Total value of $x$ between 10 to 30 is 20 . Total values of $x$ between 50 to 100 including 50 and 100 is 51 .
Total values of $x=51+20=71$
$$
P(\mathrm{~A})=\frac{71}{100}=0.71
$$

Topic: Mathematics, Probability, JEE Main, JEE Main 2014 (12 Apr Online)
A one metre long (both ends open) organ pipe is kept in a gas that has double the density of air at STP. Assuming the speed of sound in air at STP is 300 m/s, the frequency difference between the fundamental and second harmonic of this pipe is __________ Hz.

Options:

The correct answer is:
106

Solution:
V=Bρ
VpipeVair=B2ρBρ=12
Vpipe=Vair2
fn=(n+1)Vpipe2l
f1-f0=Vpipe2l=30022=105.75 Hz  if 2=1.41
=106.05 Hz  if 2=1.414

Topic: Physics, Waves and Sound, JEE Main
A one metre long (both ends open) organ pipe is kept in a gas that has double the density of air at STP. Assuming the speed of sound in air at STP is 300 m/s, the frequency difference between the fundamental and second harmonic of this pipe is __________ Hz.

Options:

The correct answer is:
106

Solution:
V=Bρ
VpipeVair=B2ρBρ=12
Vpipe=Vair2
fn=(n+1)Vpipe2l
f1-f0=Vpipe2l=30022=105.75 Hz  if 2=1.41
=106.05 Hz  if 2=1.414

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2020 (08 Jan Shift 1)
A one metre steel wire of negligible mass and area of cross-section $0.01 \mathrm{~cm}^2$ is kept on a smooth horizontal table with one end fixed. A ball of mass $1 \mathrm{~kg}$ is attached to the other end. The ball and the wire are rotating with an angular velocity of $\omega$. If the elongation of the wire is $2 \mathrm{~mm}$, then $\omega$ is (Young's modulus of steel $=2 \times 10^{11} \mathrm{Nm}^{-2}$ )

Options:
A $5 \mathrm{rad} \mathrm{s}^{-1}$
B $10 \mathrm{rad} \mathrm{s}^{-1}$
C $15 \mathrm{rad} \mathrm{s}^{-1}$
D $20 \mathrm{rad} \mathrm{s}^{-1}$

The correct answer is:
$20 \mathrm{rad} \mathrm{s}^{-1}$

Solution:
Given, elongation of the wire, $\Delta l=2 \mathrm{~mm}$
$$
=2 \times 10^{-3} \mathrm{~m}
$$
Mass of the ball, $m=1 \mathrm{~kg}$
Length of wire, $l=1 \mathrm{~m}$
Area of cross-sectional of wire,
$$
A=0.01 \mathrm{~cm}^2=0.01 \times 10^{-4} \mathrm{~m}
$$
Young's modulus of steel, $Y=2 \times 10^{11} \mathrm{Nm}^{-2}$
$\because$ Tension force in wire, $T=m \omega^2 l$
$\because$ Stress $=\frac{\text { Tension }}{\text { Area }}=\frac{m \omega^2 l}{A}$
Strain $=\frac{\Delta l}{l}=\frac{\text { stress }}{\text { Young's modulus }}$
or $\quad \Delta l=\frac{m \omega^2 l^2}{Y A}$
or $\quad \omega=\sqrt{\frac{Y A \Delta l}{m l^2}}$
Putting the given values, we get
$$
\begin{aligned}
& =\sqrt{\frac{2 \times 10^{11} \times 0.01 \times 10^{-4} \times 2 \times 10^{-3}}{1 \times(1)^2}} \\
\omega & =20 \mathrm{rad} / \mathrm{sec}^{-1}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A one molal solution of sodium chloride in water has the same boiling point as

Options:
A $1 \mathrm{~m}$ solution of magnesium sulphate
B $1 \mathrm{~m}$ solution of magnesium chloride
C $1 \mathrm{~m}$ solution of aluminium sulphate
D $1 \mathrm{~m}$ solution of aluminium chloride

The correct answer is:
$1 \mathrm{~m}$ solution of magnesium sulphate

Solution:
$\Delta T_{b}=i K_{b} m$
Two solutions having same molality and same van' $t$ Hoff factors will have same elevation in boiling points $\left(\Delta T_{b}\right)$ and thus, have same boiling points $\left(T_{b}\right)$.
For NaCl,$i=2 ; \mathrm{MgSO}_{4}, i=2 ; \mathrm{MgCl}_{2}, i=3$;
$\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}, i=5$ and $\mathrm{AlCl}_{3}, i=4$.
Thus, one molal solution of sodium chloride in water has the same boiling point as one molal solution of $\mathrm{MgSO}_{4}$.

Topic: Chemistry, Solutions, JEE Main
A P32 radionuclide with half-life T = 14.3 days is produced in a reactor at a constant rate q= 2.7 Ã 109 nuclei per second. How soon after the beginning of production of that nuclide will its activity be equal to A = 1.0 Ã 109 dis./s?

Options:
A 9.5 days

B 8 days

C 7.5 days

D 6 days

The correct answer is:
9.5 days

Solution:
According to the problem, the radio nuclide is being produced at a constant rate as well as it decay with rate - d⁡ N⁡ d⁡ t = λ N⁡ . Hence net rate of accumulation of radio nuclide is given by
d⁡ N⁡ d⁡ t = q⁡ - λ N⁡ … 1
or ∫ 0 N⁡ d⁡ N⁡ q⁡ - λ N⁡ = ∫ 0 t d⁡ t
or - 1 λ log q⁡ - λ N⁡ 0 N = t
or - 1 λ ln q⁡ - λ N⁡ q⁡ = t
or t = T⁡ 1 / 2 ln 2 ln 1 - A⁡ q⁡
where λ = ln 2 T⁡ 1 / 2 and A⁡ = λ N⁡
On substituting numerical values, we get
t = 9.5 days

Topic: Physics, Nuclear Physics, JEE Main
A $p$ - $n$ junction diode can withstand upto $20 \mathrm{~mA}$ current under forward bias. The diode has a potential difference of $0.5 \mathrm{~V}$ across it, which is assumed to be independent of current. What is the maximum voltage of the battery used to forward bias the diode when a resistance of $125 \Omega$ is connected in series with it?

Options:
A $3.0 \mathrm{~V}$
B $2.5 \mathrm{~V}$
C $3.2 \mathrm{~V}$
D $2.0 \mathrm{~V}$

The correct answer is:
$3.0 \mathrm{~V}$

Solution:
Given, potential difference across diode,
$V_D=0.5 \mathrm{~V}$
Maximum current in forward bias,
$i=20 \mathrm{~mA}=20 \times 10^{-3} \mathrm{~A}$
$\Rightarrow$ Resistance of diode,
$R_D=\frac{V_D}{i}=\frac{0.5}{20 \times 10^{-3}}=25 \Omega$
Now, another resistance of $125 \Omega$ is connected with this diode in series.
So, effective resistance,
$R=25+125=150 \Omega$
Clearly, maximum voltage to be used during forward bias,
$V=i R=20 \mathrm{~mA} \times 150 \Omega$
$=3000 \mathrm{mV}=3.0 \mathrm{~V}$

Topic: Physics, Semiconductors, JEE Main
A p-n junction has acceptor impurity concentration of 1017 cm-3 in the p side and donor impurity concentration of 1016 cm-3 in the n side. The contact potential at the junction is (kT= thermal energy, intrinsic carrier concentration ni=1.4Ã 1010 cm-3 )

Options:
A kTe ln (4×1012)
B kTe ln (2.5×1023)
C kTe ln (1023)
D kTe ln (109)

The correct answer is:
kTe ln (4×1012)

Solution:
Constant potential at the junction
Vconstant=kTe ln nandni2
Vconstant=kTe ln 1017×10161.4×10102=kTe ln (4×1012)

Topic: Physics, Semiconductors, JEE Main
A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade, is :

Options:
A 34
B 52867
C 3950
D 22425

The correct answer is:
3950

Solution:
E1 : Event denotes spade is missing

PE1=14;PE¯1=34

A : Event drawn two cards are spade

PAE1=C212C251, PAE1¯=C213C251

PE1A=34×C213C25114×C212C251+34×C213C251=3950

Topic: Mathematics, Probability, JEE Main, JEE Main 2021 (16 Mar Shift 1)
A pair has two children. If one of them is boy, then the probability that other is also a boy, is

Options:
A $\frac{1}{2}$
B $\frac{1}{4}$
C $\frac{1}{3}$
D None of these

The correct answer is:
$\frac{1}{3}$

Solution:
Let $P(A)=$ probability of a boy in two children $=\frac{3}{4}$
Because cases are $B B, B G, G B, G G=4$
Favourable cases are $B B, B G, G B=3$
The probability that the second child is also boy is $P(A \cap B)=\frac{1}{4}$
We have to find $P(B / A)=\frac{P(A \cap B)}{P(A)}=\frac{1 / 4}{3 / 4}=\frac{1}{3}$.

Topic: Mathematics, Probability, JEE Main
A pair of 12-sided fair dice with faces numbered $1,2,3, \ldots \ldots, 12$ is rolled. The probability that the sum of the numbers appearing has remainder 2 when divided by 9 is

Options:
A $\frac{7}{72}$
B $\frac{5}{48}$
C $\frac{11}{144}$
D $\frac{1}{9}$

The correct answer is:
$\frac{1}{9}$

Solution:
\begin{array}{l|c}
\mathrm{x}_{1}+\mathrm{x}_{2}=11 or & \mathrm{x}_{1}+\mathrm{x}_{2}=20 \\
possible cases & \\
\qquad(1,10) & (8,12) \\
(2,9) & (9,11) \\
(3,8) & (10,10) \\
(4,7) & \\
(5,6) & \\
\end{array}
$\frac{10}{144}$ $+\frac{6}{144}=\frac{16}{144}=\frac{1}{9}$

Topic: Mathematics, Probability, KVPY, KVPY 2017 (19 Nov SB/SX)
A pair of dice is thrown 5 times. For each throw, a total of 5 is considered a success. If the probability of at least 4 successes is k311, then k is equal to

Options:
A 82
B 75
C 164
D 123

The correct answer is:
123

Solution:
Given that,

A pair of dice is thrown 5 times.

Required Sample space is 1,1,1,2,1,3...........6,6

Total number of observations =36

Now, a sum of 5 is observed in 1,4,4,1,2,3,3,2

No of favourable outcomes =4

P(success) p=436=19

P(failure) q=1-19=89

We know that for a binomial distribution, Bn,p

PX=k=Cknpkqn-k where k≥0 and p+q=1

∴Required probability

⇒PX≥4=PX=4+PX=5

=C4519489+C55195

=5·895+195=4195

=41×3310×3=123311

But given that PX≥4=k311

⇒k311=123311

⇒k=123

Hence, this is the correct option.

Topic: Mathematics, Probability, JEE Main, JEE Main 2023 (06 Apr Shift 1)
A pair of fair dice is thrown independently three times. The probability of getting a score of exactly $9$ twice is

Options:
A
$1 / 729$

B
$8 / 9$

C
$8 / 729$

D
$8 / 243$

The correct answer is:
$8 / 243$

Solution:
Probability of getting score $9$ in a single throw $=\frac{4}{36}=\frac{1}{9}$
Probability of getting score $9$ exactly twice $={ }^3 \mathrm{C}_2 \times\left(\frac{1}{9}\right)^2 \times \frac{8}{9}=\frac{8}{243}$.

Topic: Mathematics, Probability, JEE Main, JEE Main 2007
A pair of tangents are drawn from a point P to the circle x2+y2=1. If the tangents make an intercept of 2 units on the line x=1, then the locus of the point P is

Options:
A y2=2x+1
B 2y2=-x+1
C y2=2x-1
D y2=-2x+1

The correct answer is:
y2=2x+1

Solution:
Let P(h,k) be the point from which tangents are drawn.
So, combined equation of PA and PB is
xh+yk-12=x2+y2-1h2+k2-1 ∵T2=SS1
Putting x=1, we get,

h-12+2kyh-1=y2h2-1
⇒h+1y2-2ky-h-1=0
where roots are y1 & y2 and y1-y2=2
⇒2k2+4h+1h-1h+1=2
⇒y2+x2-1=x2+2x+1
⇒y2=2x+1

Topic: Mathematics, Circle, JEE Main
A parabola has the origin as its focus and the line $x=2$ as the directrix. Then the vertex of the parabola is at

Options:
A
$(0,2)$

B
$(1,0)$

C
$(0,1)$

D
$(2,0)$

The correct answer is:
$(1,0)$

Solution:
Vertex is $(1,0)$

Topic: Mathematics, Parabola, JEE Main, JEE Main 2008
A parabola is drawn through two given points \( A(2,0) \) and \( B(-2,0) \) such that its directrix always touch the circle \( x^{2}+y^{2}=16 \), then locus of focus of the parabola is

Options:
A \( 3 x^{2}+4 y^{2}=48 \)
B \( 4 x^{2}+3 y^{2}=48 \)
C \( 3 x^{2}+4 y^{2}=60 \)
D \( 4 x^{2}+3 y^{2}=60 \)

The correct answer is:
\( 3 x^{2}+4 y^{2}=48 \)

Solution:
Let focus be Sh,k, then

h-22+k2=4cosθ-22 …i

h+22+k2=4cosθ+22 …ii

⇒cosθ=h4

⇒h-22+k2=4h4-22⇒h216+k212=1

Topic: Mathematics, Parabola, JEE Main
A parachutist after bailing out falls $50 \mathrm{~m}$ without friction. When parachute opens, it decelerates at $2 \mathrm{~m} / \mathrm{s}^2$. He reaches the ground with a speed of $3 \mathrm{~m} / \mathrm{s}$. At what height, did he bail out?

Options:
A
$91 \mathrm{~m}$

B
$182 \mathrm{~m}$

C
$293 \mathrm{~m}$

D
$111 \mathrm{~m}$

The correct answer is:
$293 \mathrm{~m}$

Solution:
$s=50+\left(\frac{3^2-(2 \times 10 \times 50)}{2(-2)}\right)$
$=293 \mathrm{~m}$.

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2005
A parallel beam of electrons travelling in x - direction falls on a slit of width d (see figure). If after passing the slit, an electron acquires momentum py in the y - direction then for a majority of electrons passing through the slit (h is Planck's constant):

Options:
A | p y |d<h
B | p y |d>h
C | p y |d≃h
D | p y |d>>h

The correct answer is:
| p y |d<h

Solution:
The electron beam will be diffractive at an angle θ
For Ist minima
dsin⁡θ= λ
dsin⁡θ= hp
Also psin⁡θ=py
⇒ d py=h
∴ majority of electrons passing through the slit are lying in the central maxima d py <h

Topic: Physics, Wave Optics, JEE Main, JEE Main 2015 (10 Apr Online)
A parallel beam of fast moving electrons is incident normally on a narrow slit. A screen is placed at a large distance from the slit. If the speed of the electrons is increased, which of the following statement is correct?

Options:
A diffraction pattern is not observed on the screen in the case of electrons
B the angular width of the central maximum of the diffraction pattern will increase
C the angular width of the central maximum will decrease
D the angular width of the central maximum will remain the same

The correct answer is:
the angular width of the central maximum will decrease

Solution:
Fast moving electrons create electromagnetic waves. So, the diffraction pattern will be observed. Also, the angular width of the central maximum is given by
$\mathrm{W}=\frac{2 \lambda}{\mathrm{a}}$, where $\mathrm{a}=$ width of the slit
$\lambda=$ wavelength of the light .
When the speed of electrons will increase, the frequency will increase which will result in decreasing the wavelength as the speed of light is a constant. Therefore, the angular width will decrease.

Topic: Physics, Current Electricity, VITEEE, VITEEE 2008
A parallel beam of light is allowed to fall on a transparent spherical globe of diameter 30 cm and refractive index 1.5. The distance from the centre of the globe at which beam of light can converge is _____ mm.

Options:

The correct answer is:
225

Solution:
For the refraction at the left surface, u=-∞, R=+15 cm.

μ2v-μ1u=μ2-μ1R

⇒1.5v1-1-∞=1.5-1+15

⇒v1=45 cm

For the refraction at the right surface, u=45-30=15 cm, R=-15 cm

μ2v-μ1u=μ2-μ1R

⇒1v2-1.515=1-1.5-15

⇒v2=7.5 cm

Therefore, the distance from the centre where the beam will converge is,

15+7.5=22.5 cm=225 mm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2022 (29 Jun Shift 1)
A parallel beam of light is incident normally on a plane surface absorbing 40 % of the light and reflecting the rest. If the incident beam carries 60 W of power, the force exerted by it on the surface is

Options:
A 3.2Ã10-8 N
B 3.2Ã10-7 N
C 5.12Ã10-7 N
D 5.12Ã10-8 N

The correct answer is:
3.2Ã10-7 N

Solution:
Momentum of incident light per second
p1=Ec=603Ã108=2Ã10-7
Momentum of reflected light per second
p2=60100ÃEc=603Ã108=1.2Ã10-7
Force on the surface = change in momentum per second
=p2--p1=p2+p1=2+1.2Ã10-7=3.2Ã10-7 N

Topic: Physics, Dual Nature of Matter, JEE Main
A parallel beam of light of intensity $I_0$ is incident on a coated glass plate. If $25 \%$ of the incident light is reflected from the upper surface and $50 \%$ of light is reflected from the lower surface of the glass plate, the ratio of maximum to minimum intensity in the interference region of the reflected light is

Options:
A $\left(\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)^2$
B $\left(\frac{\frac{1}{4}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)^2$
C $\frac{5}{8}$
D $\frac{8}{5}$

The correct answer is:
$\left(\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)^2$

Solution:
According to question, we can draw the following diagram
$$
I_1=\frac{I_0}{4} \Rightarrow I_2=\frac{3}{8} I_0
$$
We know that,
$$
\frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2}{\left(\sqrt{I_1}-\sqrt{I_2}\right)^2}=\left(\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)^2
$$

Topic: Physics, Wave Optics, JEE Main
A parallel beam of light of wavelength $\lambda$ is incident normally on a narrow slit. A diffraction pattern formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is:

Options:
A $2 \pi$
B $3 \pi$
C $4 \pi$
D $\pi \pi$

The correct answer is:
$4 \pi$

Solution:
We know that path difference for second minimum $=2 \lambda$
$$
\begin{aligned}
\therefore \text { Phase difference } & =\frac{2 \pi}{\lambda} \times 2 x \\
& =4 \pi
\end{aligned}
$$

Topic: Physics, Wave Optics, JEE Main
A parallel beam of monochromatic light falls normally on a single narrow slit. The angular width of the central maximum in the resulting diffraction pattern

Options:
A decreases with increase of slitwidth.
B increases with increase of slitwidth.
C decreases with decrease of slitwidth.
D may increase or decrease.

The correct answer is:
decreases with increase of slitwidth.

Solution:
$X=\frac{2 \lambda D}{a}, X$ will decrease if a increases

Topic: Physics, Wave Optics, JEE Main
A parallel beam of monochromatic light of wavelength $5 \times 10^{-7} \mathrm{~m}$ is incident
normally on a single narrow slit of width $10^{-3} \mathrm{~mm}$. At what angle of diffraction,
the first minima is observed?

Options:
A $\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$
B $\sin ^{-1}(1)$
C $\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
D $\sin ^{-1}\left(\frac{1}{2}\right)$

The correct answer is:
$\sin ^{-1}\left(\frac{1}{2}\right)$

Solution:
$\sin \theta=\frac{\lambda}{a}=\frac{5 \times 10^{-7} \mathrm{~m}}{10^{-6} \mathrm{~m}}=0.5=\frac{1}{2}$
$\therefore \theta=\sin ^{-1}\left(\frac{1}{2}\right)$

Topic: Physics, Wave Optics, JEE Main
A parallel beam of monochromatic light of wavelength $600 \mathrm{~nm}$ passes through single slit of $0.4 \mathrm{~mm}$ width. Angular divergence corresponding to second order minima would be ______ $\times 10^{-3} \mathrm{rad}$.

Options:

The correct answer is:
6

Solution:
$\sin \theta \simeq \theta \simeq \frac{2 \lambda}{b}$
$=\frac{2 \times 600 \times 10^{-9}}{4 \times 10^{-4}}=3 \times 10^{-3} \mathrm{rad}$
Total divergence $=(3+3) \times 10^{-3}=6 \times 10^{-3} \mathrm{rad}$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2024 (08 Apr Shift 1)
A parallel beam of monochromatic light of wavelength 5000 Ao is incident normally on a single narrow slit of width 0.001 mm. The light is focused by convex lens on screen, placed on its focal plane. The first minima will be formed for the angle of diffraction of ______ (degree).

Options:

The correct answer is:
30

Solution:
The condition for the first minima in the single slit diffraction experiment can be written as

asinθ=λ ...1

Equation (1) implies that

sinθ=λa=5000×10-101×10-6=12

⇒θ=30°

Topic: Physics, Wave Optics, JEE Main, JEE Main 2024 (27 Jan Shift 2)
A parallel place capacitor is charged by connecting a $2 \mathrm{~V}$ battery across it. It is then disconnected from the battery and a glass slab is introduced between plates. Which of the following pairs of quantities decrease?

Options:
A Potential difference and energy stored
B Energy stored and capacitance
C Capacitance and charge
D Charge and potential difference

The correct answer is:
Potential difference and energy stored

Solution:
When a charged capacitor is disconnected from the battery and a glass slab (dielectric material) is introduced between the plates, then charge (Q), capacitance $(C)$, electric potential $(V)$ and stored energy $(U)$ is given as
$$
Q=Q_0, C=K C_0, V=\frac{V_0}{K} \text { and } U=\frac{U_0}{K}
$$
where, $K=$ dielectric constant
$\begin{aligned} V & =\text { electric potential } \\ U & =\text { potential energy } \\ V_0 & =\text { initial potential of charged capacitor } \\ U_0 & =\text { Initial energy of charged capacitor }\end{aligned}$
From above, we see that $V$ and $U$ decreases after introducing glass plate of dielectric constant $k$ whereas capacitance increases and charge remains same because battery is disconnected after introducing dielectric material.

Topic: Physics, Capacitance, KCET, KCET 2022
A parallel plate air capacitor has capacity C, distance of separation between plates is d and potential difference V is applied between the plates. The force of attraction between the plates of the capacitor is

Options:
A CV22d.
B CV2d.
C C2V22d2.
D C2V22d.

The correct answer is:
CV22d.

Solution:
Attraction force between the plates,
F=q22Aε0 where q=CV and C=ε0Ad
⇒F=C2V22Cd=CV22d

Topic: Physics, Capacitance, JEE Main
A parallel plate air capacitor has capacity \( C \), distance of separation between plates is \( d \) and potential difference \( V \) is applied between the plates. The force of attraction between the plates of the capacitor is

Options:
A \( \frac{C V^{2}}{2 d} \)
B \( \frac{C V^{2}}{d} \)
C \( \frac{C^{2} V^{2}}{2 d^{2}} \)
D \( \frac{C^{2} V^{2}}{2 d} \)

The correct answer is:
\( \frac{C V^{2}}{2 d} \)

Solution:
Attraction force between the plates,
F=q22Aε0 where q=CV and C=ε0Ad
⇒F=C2V22Cd=CV22d

Topic: Physics, Capacitance, JEE Main
A parallel plate air capacitor is charged up to $100 \mathrm{~V}$. A plate $2 \mathrm{~mm}$ thick is inserted between the plates. Then to maintain the same potential difference, the distance between the plates is increased by $1.6 \mathrm{~mm}$. The dielectric constant of the thick plate is

Options:
A 4
B 5
C 2
D 3

The correct answer is:
5

Solution:
If $d$ is the initial distance between the plates then capacitance is given by $\mathrm{C}=\frac{\mathrm{kA} \in_0}{\mathrm{~d}}$
When a plate of thickness $\mathrm{t}=2 \mathrm{~mm}$ is inserted between the plates the capacitance,
$$
C_1=\frac{k A \epsilon_0}{\left(d-t+\frac{t}{k}\right)}
$$
If distance between the plates is increased by $\mathrm{x}=1.6 \mathrm{~mm}$ Capacitance become, $\mathrm{C}_2=\frac{\mathrm{kA} \in_0}{\left(\mathrm{~d}+\mathrm{x}-\mathrm{t}+\frac{\mathrm{t}}{\mathrm{k}}\right)}$ But $\mathrm{C}_2=\mathrm{C}$
$\begin{aligned} & \therefore \mathrm{d}+\mathrm{x}-\mathrm{t}+\frac{\mathrm{t}}{\mathrm{k}}=\mathrm{d} \\ & \therefore \mathrm{t}-\mathrm{x}=\frac{\mathrm{t}}{\mathrm{k}} \\ & \therefore 2-1.6=\frac{2}{\mathrm{k}} \\ & \therefore 0.4=\frac{2}{\mathrm{k}} \quad \therefore \mathrm{k}=5\end{aligned}$

Topic: Physics, Capacitance, MHT CET, MHT CET 2021 (20 Sep Shift 1)
A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?

Options:
A The potential difference between the plates decreases K times.
B The energy stored in the capacitor decreases K times.
C The change in energy stored is 12CV21K-1.
D The charge on the capacitor is not conserved.

The correct answer is:
The charge on the capacitor is not conserved.

Solution:
For air K=1,
C1=C
q=C1V …i.
The charge remains constant.
U1=q22C
U2=q22CK
Change in energy stored in the capacitor,
U=U2-U1
=q22C 1K-1 =12CV2 1K-1
V'=qCK⇒V'=VK
Here,1- Potential difference between the plates decreases by K times.
2- The change in energy stored ∆U=12 CV2 1K-1.
3- Charge is conserved.
4- The energy stored in capacitor decrease by K, U2=12q2CK.

Topic: Physics, Capacitance, JEE Main
A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?

Options:
A The potential difference between the plates decreases K times.
B The energy stored in the capacitor decreases K times.
C The change in energy stored is 12CV21K-1.
D The charge on the capacitor is not conserved.

The correct answer is:
The charge on the capacitor is not conserved.

Solution:
For air K=1,
C1=C
q=C1V …i.
The charge remains constant.
U1=q22C
U2=q22CK
Change in energy stored in the capacitor,
U=U2-U1
=q22C 1K-1 =12CV2 1K-1
V'=qCK⇒V'=VK
Here,1- Potential difference between the plates decreases by K times.
2- The change in energy stored ∆U=12 CV2 1K-1.
3- Charge is conserved.
4- The energy stored in capacitor decrease by K, U2=12q2CK.

Topic: Physics, Capacitance, NEET, NEET 2015 (Phase 1)