datavorous/entrance-exam-dataset huggingface dataset:
The parabola $y^2=4 x$ divides the area of the circle $x^2+y^2=5$ in two parts. The area of the smaller part is equal to:

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

The correct answer is:
$\frac{2}{3}+5 \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)$

Solution:
$\begin{aligned} & y^2=4 x \\ & x^2+y^2=5\end{aligned}$
$\therefore$ Area of shaded region as shown in the figure will be
$\begin{aligned} & \mathrm{A}_1=\int_0^1 \sqrt{4 \mathrm{x}} \mathrm{dx}+\int_1^{\sqrt{5}} \sqrt{5-\mathrm{x}^2} \mathrm{dx} \\ & =\frac{4}{3} \cdot\left[\mathrm{x}^{\frac{3}{2}}\right]_0^1+\left[\frac{\mathrm{x}}{2} \sqrt{5-\mathrm{x}^2}+\frac{5}{2} \sin ^{-1} \frac{\mathrm{x}}{\sqrt{5}}\right]_1^{\sqrt{5}} \\ & =\frac{1}{3}+\frac{5 \pi}{4}-\frac{5}{2} \sin ^{-1}\left(\frac{1}{\sqrt{5}}\right) \\ & \therefore \text { Required Area }=2 \mathrm{~A}_1 \\ & =\frac{2}{3}+\frac{5 \pi}{2}-5 \sin ^{-1}\left(\frac{1}{\sqrt{5}}\right) \\ & =\frac{2}{3}+5\left(\frac{\pi}{2}-\sin ^{-1} \frac{1}{\sqrt{5}}\right) \\ & =\frac{2}{3}+5 \cos ^{-1} \frac{1}{\sqrt{5}} \\ & =\frac{2}{3}+5 \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2024 (09 Apr Shift 1)
The parabola $y^2=x$ divides the circle $x^2+y^2=2$ into two parts whose areas are in the ratio

Options:
A
$9 \pi+2: 3 \pi-2$

B
$9 \pi-2: 3 \pi+2$

C
$7 \pi-2: 2 \pi-3$

D
$7 \pi+2: 3 \pi+2$

The correct answer is:
$9 \pi-2: 3 \pi+2$

Solution:
$$
\begin{aligned}
& \text { Area of circle }=\pi(\sqrt{2})^2=2 \pi \\
& \text { Area of } O C A D O=2\{\text { Area of } O C A O\} \\
& \quad=2 \text { area of } O C B+\text { area of } B C A\} \\
& =2 \int_0^1 y_p d x+2 \int_1^{\sqrt{2}} y_c d x \\
& \text { where } y_p=\sqrt{x} \text { and } y_c=\sqrt{2-x^2} \\
& \therefore \text { Required Area } \\
& =2 \int \frac{1}{x} d x+2 \int_0^{\sqrt{2}} \sqrt{2-x^2} d x \\
& =2\left[\frac{2}{3} \cdot 1-0\right]+2\left[\frac{x \sqrt{2-x^2}}{2}+\sin ^{-1} \frac{x}{\sqrt{2}}\right]_1^{\sqrt{2}}
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{4}{3}+2\left\{\frac{\pi}{2}-\frac{\pi}{4}-\frac{1}{2}\right\}=\frac{4}{3}+2\left\{\frac{\pi}{4}-\frac{1}{2}\right\}=\frac{3 \pi+2}{6} \\
& \text { Bigger area }=2 \pi-\frac{3 \pi+2}{6}=\frac{9 \pi-2}{6} \\
& \therefore \text { Required Ratio }=\frac{9 \pi-2}{3 \pi+2} \text { i.e., } 9 \pi-2: 3 \pi+2
\end{aligned}
$$

Topic: Mathematics, Area Under Curves, JEE Main
The parabola $y^2=x$ divides the circle $x^2+y^2=2$ into two parts whose areas are in the ratio

Options:
A
$9 \pi+2: 3 \pi-2$

B
$9 \pi-2: 3 \pi+2$

C
$7 \pi-2: 2 \pi-3$

D
$7 \pi+2: 3 \pi+2$

The correct answer is:
$9 \pi-2: 3 \pi+2$

Solution:
$$
\begin{aligned}
& \text { Area of circle }=\pi(\sqrt{2})^2=2 \pi \\
& \text { Area of } O C A D O=2\{\text { Area of } O C A O\} \\
& \quad=2 \text { area of } O C B+\text { area of } B C A\} \\
& =2 \int_0^1 y_p d x+2 \int_1^{\sqrt{2}} y_c d x \\
& \text { where } y_p=\sqrt{x} \text { and } y_c=\sqrt{2-x^2} \\
& \therefore \text { Required Area } \\
& =2 \int \frac{1}{x} d x+2 \int_0^{\sqrt{2}} \sqrt{2-x^2} d x \\
& =2\left[\frac{2}{3} \cdot 1-0\right]+2\left[\frac{x \sqrt{2-x^2}}{2}+\sin ^{-1} \frac{x}{\sqrt{2}}\right]_1^{\sqrt{2}}
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{4}{3}+2\left\{\frac{\pi}{2}-\frac{\pi}{4}-\frac{1}{2}\right\}=\frac{4}{3}+2\left\{\frac{\pi}{4}-\frac{1}{2}\right\}=\frac{3 \pi+2}{6} \\
& \text { Bigger area }=2 \pi-\frac{3 \pi+2}{6}=\frac{9 \pi-2}{6} \\
& \therefore \text { Required Ratio }=\frac{9 \pi-2}{3 \pi+2} \text { i.e., } 9 \pi-2: 3 \pi+2
\end{aligned}
$$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2012 (07 May Online)
The parabola \( y^{2}=8 x \) and the circle \( x^{2}+y^{2}=2 \)

Options:
A Have only two common tangents which are mutually perpendicular
B Have only two common tangents which are parallel to each other
C Have infinitely many common tangents
D Does not have any common tangent

The correct answer is:
Have only two common tangents which are mutually perpendicular

Solution:
Let the equation y=mx+c be the common tangents so the curve y2=8x and x2+y2=2
Then, c=2m and c2=2(1+m2)
If m2=t, then
4t=21+t⇒t2+t-2=0
⇒t+2t-1=0⇒t=1, -2
Thus, m=±1 (∵t≠-2)
Hence, tangents are y=x+c and y=-x+c which are perpendicular to each other

Topic: Mathematics, Parabola, JEE Main
The parabolas : ax2+2bx+cy=0 and d2+2ex+fy=0 intersect on the line y=1. If a, b, c, d, e, f are positive real numbers and a,b,c are in G.P., then

Options:
A d,e,f are in A.P.
B da,eb,fc are in G.P.
C da,eb,fc are in A.P.
D d,e,f are in G.P.

The correct answer is:
da,eb,fc are in A.P.

Solution:
Given,

ax2+2bx+cy=0 and d2+2ex+fy=0 intersect on the line y=1,

And a,b,c are in G.P.

So, b2=ac

Now putting the value of b in ax2+2bx+cy=0and taking y=1 we get,

ax2+2bx+c=0

⇒ax2+2acx+c=0 ∵b2=ac

⇒(xa+c)2=0

⇒x=-ca and x2=ca .......1

Now, putting the value of x & x2 in dx2+2ex+fy=0 and taking y=1 we get,

⇒dca+2e-ca+f=0

⇒dca+f=2eca

⇒da+fc=2e1ac

⇒da+fc=2eb as b=ac

∴da,eb,fcare in A.P.

Topic: Mathematics, Parabola, JEE Main, JEE Main 2023 (30 Jan Shift 2)
The parabolas C1:y2=4ax-a and C2:y2=-4ax-k intersect at two distinct points A and B. If the slope of the tangent at A on C1 is same as the slope of the normal at B on C2 , then the value of k is equal to

Options:
A 3a
B 2a
C a
D 0

The correct answer is:
3a

Solution:
Let, A=p,q & B=p,-q
Now q2=4ap-a & q2=-4ap-k …(i)
⇒p-a=-p+k⇒p=a+k2 …(ii)
Now for C1, dydx=2ay
dydxp,q=2aq & -dxdyp,-q=q2a
⇒ 2aq=q2a⇒q2=4a2 … (iii)
Putting values from (ii) & (iii) in equation (i), we get,
4a2=4aa+k2-a ⇒a=k-a2⇒k=3a

Topic: Mathematics, Parabola, JEE Main
The parabolas $\mathrm{y}^2=4 \mathrm{x}$ and $\mathrm{x}^2=4 \mathrm{y}$ divide the square region bounded by the lines $\mathrm{x}=$ $4, y=4$ and the coordinate axes. If $S_1, S_2, S_3$ are respectively the areas of these parts numbered from top to bottom; then $\mathrm{S}_1: \mathrm{S}_2: \mathrm{S}_3$ is

Options:
A
$1: 2: 1$

B
$1: 2: 3$

C
$2: 1: 2$

D
$1: 1: 1$

The correct answer is:
$1: 1: 1$

Solution:
$\mathrm{y}^2=4 \mathrm{x}$ and $\mathrm{x}^2=4 \mathrm{y}$ are symmetric about line $\mathrm{y}=\mathrm{x}$
$\Rightarrow$ area bounded between $y^2=4 x$ and $y=x$ is $\int_0^4(2 \sqrt{x}-x) d x=\frac{8}{3}$
$\Rightarrow \mathrm{A}_{\mathrm{s}_2}=\frac{16}{3}$ and $\mathrm{A}_{\mathrm{s}_1}=\mathrm{A}_{\mathrm{s}_3}=\frac{16}{3}$
$\Rightarrow A_{s_1}: A_{s_2}: A_{s_3}:: 1: 1: 1$.

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2005
The parabolas y2=4x and x2=4y divide the square region bounded by the lines x=4, y=4 and the coordinate axes. If S1,S2,S3 are respectively the areas of these parts numbered from top to bottom of the square region, then S1:S2:S3 is

Options:
A 2Â :1Â :2
B 1 :1 :1
C 1Â :2Â :1
D 1Â :2Â :3

The correct answer is:
1 :1 :1

Solution:
y2=4x and x2=4y are symmetric about line y=x
⇒ Area bounded between y2=4x and y=x is ∫042x-xdx=83
⇒ A S 2 =163 and AS1=AS3=163 { Area of rectangle− S 2 2 }
⇒ AS1:AS2:AS2::1 :1 :1 .

Topic: Mathematics, Area Under Curves, JEE Main
The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a battery of voltage, V. When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant K is placed the two plates of the first capacitor. The new potential difference of the combined system is:

Options:
A n+1VK+n
B nVK+n
C VK+n
D V

The correct answer is:
n+1VK+n

Solution:
i)
ii) Battery disconnected after full charge,
Total charge of the system,
=CV+nCV
=n+1CV
iii) Inserting dielectric of constant K
Common potential, VC=total chargetotal capacitance
=n+1CVKC+nC=n+1VK+n

Topic: Physics, Capacitance, JEE Main, JEE Main 2019 (09 Apr Shift 2)
The parameter that remains the same for molecules of all gases at a given temperature is :

Options:
A kinetic energy
B momentum
C mass
D speed

The correct answer is:
kinetic energy

Solution:
From the kinetic interpretation of temperature, we know, KE=32kBT. Clearly, it depends only on temperature.

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2024 (31 Jan Shift 1)
The parametric representation of a point on the ellipse whose foci are \( (-1,0) \) and \( (7,0) \) and eccentricity \( \frac{1}{2} \) is

Options:
A \( (3+8 \cos \theta, 4 \sqrt{3} \sin \theta) \)
B \( (8 \cos \theta, 4 \sqrt{3} \sin \theta) \)
C \( (3+4 \sqrt{3} \cos \theta, 8 \sin \theta) \)
D none of these

The correct answer is:
\( (3+8 \cos \theta, 4 \sqrt{3} \sin \theta) \)

Solution:
Distance between two foci, 2ae=7+1=8
∴ ae=4⇒a=8, ∴e=12, given
∵b2=a21-e2=64 1-14
⇒b=43
Since, the centre of the ellipse is the mid point of the line joining two foci, therefore the coordinates of the centre will be 3,0.
Then, its equation is
x-3282+y-02432=1 ........1
Hence, the parametric coordinates of a point on ellipse are (3+8cos⁡θ, 43sin⁡θ )

Topic: Mathematics, Ellipse, JEE Main
The part of a circuit shown in the figure consists of two ideal diodes and a few resistors. The equivalent resistance of the circuit between A and B is

Options:
A 4Â Î©
B 13Â Î©
C 4 Î© or 13 Î©
D 4 Î© or zero

The correct answer is:
4 Î© or 13 Î©

Solution:
If VA>VB, both the junction diodes are forward biased and the given circuit diagram becomes a balanced Wheatstone bridge. The equivalent resistance, in this case, becomes 4 Ω.
If VB>VA, the diodes are reverse biased. In that case, 4 Ω, 5 Ω and 4 Ω are in series.

Topic: Physics, Semiconductors, JEE Main
The part of a transistor which is most heavily doped to produce large number of majority carriers is

Options:
A
emmiter

B
base

C
collector

D
can be any of the above three

The correct answer is:
emmiter

Solution:
No Solution Available

Topic: Physics, Semiconductors, JEE Main, JEE Main 2002
The part of internal ear responsible for dynamic balancing is

Options:
A cochlea
B semicircular canal
C utriculus
D Sacculus

The correct answer is:
semicircular canal

Solution:
The inner ear is fluid-filled and divisible mainly into 2 parts:
1. Bony labyrinth: It is a series of bony channels inside which the membranous labyrinth lies. It is filled with perilymph.
2. Membranous labyrinth: It is filled with a fluid called endolymph.
- Membranous labyrinth = Cochlea + Vestibular Apparatus
- The cochlea is meant for hearing and while the vestibular apparatus (Saccule and Utricle for static balance and 3 semicircular canals for dynamic balance) are meant for body balancing.

Topic: Biology, Neural Control and Coordination, NEET
The part of nephron involved in conditional reabsorption of sodium is

Options:
A Distal convoluted tubule
B Proximal convoluted tubule
C Bowman's capsule
D Descending limb of Henle's loop

The correct answer is:
Proximal convoluted tubule

Solution:
Conditional reabsorption of Na+ and water takes place in the DCT segment. DCT is also capable of reabsorption of HCO3- and selective secretion of hydrogen and potassium ions and NH3 to maintain the pH and sodium-potassium balance in the blood.

Topic: Biology, Excretory Products and their Elimination, NEET, NEET 2016 (Phase 2)
The part of nephron involved in conditional reabsorption of sodium is

Options:
A Distal convoluted tubule
B Proximal convoluted tubule
C Bowman's capsule
D Descending limb of Henle's loop

The correct answer is:
Distal convoluted tubule

Solution:
Conditional reabsorption of Na+ and water takes place in the DCT segment. DCT is also capable of reabsorption of HCO3â and selective secretion of hydrogen and potassium ions and NH3 to maintain the pH and sodium-potassium balance in the blood.

Topic: Biology, Excretory Products and their Elimination, NEET
The part of the embryonal axis that remains above the level of attachment of scutellum is

Options:
A plumule.
B coleoptile.
C hypocotyl.
D epicotyl.

The correct answer is:
epicotyl.

Solution:
The portion of embryonal axis above the level of the cotyledons is the epicotyls, which terminates with the plumule or stem tip. The cylindrical portion below the level of the cotyledon is the hypocotyl that terminates at the lower end in the radical or root tip.

Topic: Biology, Morphology of Flowering Plants, NEET
The particle A is converted into C via the following reaction.
AâB+ 2He4
Bâ C + 2e-
Then

Options:
A A and C are isobars
B A and C are isotopes
C A and B are isobars
D A and B are isotopes

The correct answer is:
A and C are isotopes

Solution:
Let x be the mass number of A and y the atomic number. Then , since atomic number and mass number remain conserved, we have
yAx→ y-2Bx-4 + 2He4
y-2Bx-4→ yCx-4 + 2 -1e0
Hence, we observe that A and C are isotopes as their atomic numbers are same but mass numbers are different.

Topic: Physics, Nuclear Physics, JEE Main
The particular solution of the differential equation
$\sin ^{2} y \frac{d x}{d y}+x=\cot y$ when $x=0$ and $y=\frac{3 \pi}{4}$ is

Options:
A $x=1+\cot y$
B $x y=\cot (x+y)$
C $x y=\cot (x-y)$
D $y=1+\cot x$

The correct answer is:
$x=1+\cot y$

Solution:
We have,
$\sin ^{2} \mathrm{y} \frac{\mathrm{d} x}{\mathrm{dy}}+\mathrm{x}=\cot \mathrm{y}$
$\therefore \frac{\mathrm{d} \mathrm{x}}{\mathrm{dy}}+\left(\operatorname{cosec}^{2} \mathrm{y}\right) \mathrm{x}=\cot \mathrm{y} \cdot \operatorname{cosec}^{2} \mathrm{y}$
$\therefore$ I.F. $=\mathrm{e}^{\int \operatorname{cosec}^{2} \mathrm{ydy}}=\mathrm{e}^{-\cot y}$
$\therefore \mathrm{xe}^{-\cot y}=\int \mathrm{e}^{-\cot y} \cot \mathrm{y} \cos \mathrm{ec}^{2} \mathrm{y} \mathrm{dy}$
In RHS, put $-\cot \mathrm{y}=\mathrm{t} \Rightarrow \operatorname{cosec}^{2} \mathrm{y} \mathrm{dy}=\mathrm{dt}$
$\therefore \quad \mathrm{xe}^{-\cot y}=\int \mathrm{e}^{t}(-\mathrm{t}) \mathrm{dt}=-\int \mathrm{t} \mathrm{e}^{t} \mathrm{dt}$
$=-\left[\mathrm{t} \mathrm{e}^{t}-\int \mathrm{e}^{t} \mathrm{dt}\right]=-\mathrm{t} \mathrm{e}^{t}+\mathrm{e}^{t}=\mathrm{e}^{t}(1-\mathrm{t})$
$\quad x \mathrm{e}^{-\cot y}=\mathrm{e}^{-\cot y}(1+\cot y)+C$...(2)
At $x=0, y=\frac{3 \pi}{4}$, we get $0=\mathrm{e}(1-1)+C$
From equation $(2)$, required solution is $x \mathrm{e}^{-\cot y}=\mathrm{e}^{-\cot y}(1+\cot y)$
$\quad x \quad=1+\cot y$ $x=1+\cot y$

Topic: Mathematics, Differential Equations, JEE Main
The pass-axes of two polarisers were kept such that the incident unpolarised beam of intensity $I_0$, gets completely blocked. Another polariser was introduced in between these two polarisers with its pass-axis $60^{\circ}$ with respect to the pass-axis of the first one. The output intensity would then become

Options:
A 0
B $\frac{3}{32} I_0$
C $\frac{3}{16} I_0$
D $\frac{3}{8} I_0$

The correct answer is:
$\frac{3}{32} I_0$

Solution:
Since, the pass axes of two polarisers $P_1$ and $P_2$ were kept such that the incident unpolarised beam of intensity $I_0$ gets completely blocked.
This means that both polarisers were kept cross (perpendicular of their pass axes) to each other. When third polariser $P_3$ is introduced between $P_1$ and $P_2$ such that the angle between the pass axes of $P_1$ and $P_3$ is $60^{\circ}$, i.e. $\theta_1=60^{\circ}$.
intensity of polarised light emerging from first polaroid, $I_1=\frac{I_0}{2}$
Intensity of polarised light $\left(I_3\right)$ emerging from polaroid $P_3$ is given by law of Malus, i.e.,
$$
\begin{aligned}
I_3 & =I_1 \cos ^2 \theta, \\
& =\frac{I_0}{2} \cos ^2 60=\frac{I_0}{2} \frac{1}{4} \\
I_3 & =\frac{I_0}{8}
\end{aligned}
$$
Now, the angle between pass axis of $P_3$ and $P_2$, $\theta_2=90^{\circ}-60^{\circ}=30^{\circ}$
$\therefore$ Intensity of polarised light emerging from lost (second) polaroid $P_2$.
$$
I_2=I_3 \cos ^2 30^{\circ}=\frac{I_0}{8} \cdot \frac{3}{4}=\frac{3 I_0}{32}
$$

Topic: Physics, Wave Optics, JEE Main
The passive movement of mineral ions into the apoplast occurs

Options:
A at a rapid pace selectively through ion-channels.
B at the expenditure of metabolic energy.
C at a slow pace freely through the ion-channels.
D independently of the gradient.

The correct answer is:
at a rapid pace selectively through ion-channels.

Solution:
Studies revealed that the process of absorption of minerals can be demarcated into two main phases. In the first phase, an initial rapid uptake of ions into the ‘free space’ or ‘outer space’ of cells – the apoplast, is passive. In the second phase of uptake, the ions are taken in slowly into the ‘inner space’ – the symplast of the cells. The passive movement of ions into the apoplast usually occurs through ion-channels, the trans-membrane proteins that function as selective pores.

Topic: Biology, Mineral Nutrition, NEET
The path executed by a charged particle whose motion is perpendicular to magnetic field is

Options:
A A straight line
B An ellipse
C A circle
D A helix

The correct answer is:
A circle

Solution:
A charged particle describes a circle when it moves perpendicular to magnetic field.

Topic: Physics, Magnetic Effects of Current, JEE Main
The path of a positively charged particle \( 1 \) through a rectangular region of a uniform electric field is as shown in the figure. What are the directions of the electric fields and the directions of deflection of particles \( 2,3 \) and \( 4 \) ?

Options:
A Top; down; top; down
B Top; down; down; top
C Down; top; top; down
D Down; top; down; down

The correct answer is:
Top; down; top; down

Solution:
The positive charged particle moves as a parabolic path in electric field.

Positive charge diverts in the direction of the field.
∴The field must be towards the top.
Particle 2 deflects down, 3 top.
Particle 4 deflects down, 3 top.

Topic: Physics, Electrostatics, JEE Main
The pathogen Microsporum responsible for ringworm disease in humans belongs to the same kingdom of organisms as that of

Options:
A Taenia, a tapeworm
B Wuchereria, a filarial worm
C Rhizopus, a mould
D Ascaris, a round worm

The correct answer is:
Rhizopus, a mould

Solution:
Many fungi belonging to the generaMicrosporum, Trichophyton and Epidermophyton are responsible for ringworms which is one of the most common infectious disease in man. Microsporum is a member of class-Deuteromycetes (Fungi Imperfecti) of Kingdom-fungi. Rhizopus (bread mould) is a member of class-Zygomycetes (Conjugation fungi) of Kingdom-Fungi.

Topic: Biology, Human Health and Diseases, NEET, NEET 2011 (Mains)
The pdf of a random variable X is fx=31-2x2,0<x<1= 0 otherwise. The P14<X<13=…

Options:
A 179864
B 159864
C 169864
D 189864

The correct answer is:
179864

Solution:
We have, p.d.f of a random varlabe
X is fx=31-2x2,0<x<1
=0, otherwise
∴p14<X<13=∫1/41/3fxdx
=∫1/41/331-2x2dx
=3x-23x31/41/3
=313-23133-14-23143
=313-14-23133-143
=3112-23×371728
=3×1792592
=179864

Topic: Mathematics, Probability, JEE Main
The P.E. ' $U$ ' of a moving particle of mass ' $m$ ' varies with ' $\mathrm{x}$ ' is shown in the figure. The de-Broglie wavelength of the particle in the regions $0 \leq x \leq 1$ and $x>1$ are $\lambda_1$ and $\lambda_2$ respectively. If the total energy of the particle is ' $\mathrm{nE}$ ', then the ratio $\lambda_1 / \lambda_2$ is

Options:
A $\sqrt{\frac{n^2}{n-1}}$
B $\sqrt{\frac{n-1}{n}}$
C $\sqrt{\frac{n}{n-1}}$
D $\sqrt{\frac{n(n-1)}{n}}$

The correct answer is:
$\sqrt{\frac{n}{n-1}}$

Solution:
In the region $0 \leq x \leq 1$, the potential energy of the particle is $E$. Total energy is $\mathrm{nE}$.
Hence, kinetic energy, $\mathrm{K}=\mathrm{nE}-\mathrm{E}=(\mathrm{n}-1) \mathrm{E}$
Its momentum, $\mathrm{p}_1=\sqrt{2 \mathrm{mK}}=\sqrt{2 \mathrm{~m}(\mathrm{n}-1) \mathrm{E}}$
$$
\therefore \lambda_1=\frac{\mathrm{h}}{\mathrm{p}_1}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}(\mathrm{n}-1) \mathrm{E}}}
$$
In the region $\mathrm{x}>1, \mathrm{PE}$ is zero
Hence, total is kinetic energy.
$$
\begin{aligned}
& \therefore \mathrm{K}=\mathrm{nE} \\
& \therefore \lambda_2=\frac{\mathrm{h}}{\mathrm{p}_2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mnE}}} \\
& \therefore \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{\mathrm{n}}{\mathrm{n}-1}}
\end{aligned}
$$

Topic: Physics, Dual Nature of Matter, JEE Main
The peak electric field produced by the radiation coming from the 8 W bulb at a distance of 10 m is x10μ0cπ V m-1. The efficiency of the bulb is 10% and it is a point source. The value of x is,

Options:

The correct answer is:
2

Solution:
I=12cεE020

84π×102×12=14×c×1μ0c2×E02

E0=210×μ0cπ⇒x=2

Topic: Physics, Electrostatics, JEE Main, JEE Main 2021 (25 Feb Shift 2)
The peak value of an Alternating current is $6 \mathrm{amp}$, then r.m.s. value of current will be

Options:
A $3 \mathrm{~A}$
B $3 \sqrt{3} \mathrm{~A}$
C $3 \sqrt{2} A$
D $2 \sqrt{3} \mathrm{~A}$

The correct answer is:
$3 \sqrt{2} A$

Solution:
$i_{r . m . s}=\frac{6}{\sqrt{2}}=3 \sqrt{2} \mathrm{~A}$

Topic: Physics, Alternating Current, JEE Main
The peak value of an alternating emf ' $\mathrm{e}$ ' given by $\mathrm{e}=\mathrm{e}_0 \cos \omega \mathrm{t}$ is 10 volt and its frequency is $50 \mathrm{~Hz}$. At time $\mathrm{t}=\frac{1}{600} \mathrm{~s}$, the instantaneous e.m.f is

Options:
A 10 V
B $\frac{10}{\sqrt{3}} \mathrm{~V}$
C 5 V
D $5 \sqrt{3} \mathrm{~V}$

The correct answer is:
$5 \sqrt{3} \mathrm{~V}$

Solution:
$\begin{aligned} & \mathrm{e}=\mathrm{e}_0 \cos \omega \mathrm{t}=10 \cos 2 \pi \mathrm{ft} \\ & \mathrm{f}=50 \mathrm{~Hz}, \mathrm{t}=\frac{1}{600} \mathrm{~s} \\ & \therefore \mathrm{e}=10 \cos 100 \pi \times \frac{1}{600}=10 \cos \frac{\pi}{6}=5 \sqrt{3} \mathrm{~V}\end{aligned}$

Topic: Physics, Alternating Current, JEE Main
The pendulum suspended from the ceiling of a train has a period T when the train is at rest. When the train is accelerating with a uniform acceleration, the period of oscillation will

Options:
A increase
B decrease
C remain unaffected
D become infinite

The correct answer is:
decrease

Solution:
Comparing with T=2π L g eff ,
⇒geff=g2+a2
⇒T′=2π1g2+a2
Clearly, T′<T.

Topic: Physics, Oscillations, NEET
The penetration of light into the region of geometrical shadow is called

Options:
A Polarisation
B Interference
C Diffraction
D Refraction

The correct answer is:
Diffraction

Solution:
It is caused due to turning of light around corners.

Topic: Physics, Wave Optics, JEE Main
The peptide that gives positive ceric ammonium nitrate and carbylamine tests is:

Options:
A Gln-Asp
B Asp-Gln
C Lys-Asp
D Ser-Lys

The correct answer is:
Ser-Lys

Solution:
Serine ⇒
Lysine ⇒
Lysine has -NH2 group hence gives ⊕ ve carbyl amine test and serine has -OH group hence gives ⊕ ceric ammonium nitrate test.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2019 (09 Apr Shift 2)
The percentage composition of carbon by mole in methane is:

Options:
A 75 %
B 20 %
C 80 %
D 25 %

The correct answer is:
20 %

Solution:
Chemical Formula of Methane is CH4.

The mole percentage of carbon =moles of  carbon  Total  moles in Methane×100

=11+4×100

=20%

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2019 (08 Apr Shift 2)
The percentage decrease in the weight of a rocket, when taken to a height of 32 km above the surface of earth will, be
(Radius of earth =6400 km)

Options:
A 1%
B 3%
C 4%
D 0.5%

The correct answer is:
1%

Solution:
Acceleration due to gravity at a height h<<R is

g'=g1-2hR

Now, change in g is

∴Δgg=2hR

Percentage change is

⇒Δgg×100=2hR×100

=2×326400×100=1%

Topic: Physics, Gravitation, JEE Main, JEE Main 2022 (26 Jul Shift 1)
The percentage errors in quantities P, Q, R and S are 0.5%, 1%, 3% and 1.5% respectively in the measurement of a physical quantity A= P 3 Q 2 R S . The maximum percentage error in the value of A will be:

Options:
A 6.5 %
B 7.5Â %
C 6.0Â %
D Â 8.5Â %

The correct answer is:
6.5 %

Solution:
100×∆AA=3∆PP+2∆QQ+12∆RR+∆SS ×100
max percentage error, A =3 ×0.5+2 ×1+12 ×3+1.5
=1.5+2+1.5+1.5
= 6.5 %

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2018 (16 Apr Online)
The percentage errors in quantities P, Q, R and S are 0.5%, 1%, 3% and 1.5% respectively in the measurement of a physical quantity A= P 3 Q 2 R S . The maximum percentage error in the value of A will be:

Options:
A 6.5 %
B 7.5Â %
C 6.0Â %
D Â 8.5Â %

The correct answer is:
6.5 %

Solution:
100×∆AA=3∆PP+2∆QQ+12∆RR+∆SS ×100
max percentage error, A =3 ×0.5+2 ×1+12 ×3+1.5
=1.5+2+1.5+1.5
= 6.5 %

Topic: Physics, Mathematics in Physics, JEE Main