datavorous/entrance-exam-dataset huggingface dataset:
the curvature radii of a concavo-convex glass lens are 20 cm and 60 cm. The convex surface of the lens is silvered. With the lens horizontal, the concave surface is filled with water. The focal length of the effective mirror is (Î¼ of glass = 1.5, Î¼ of water = 43)

Options:
A 9 0 1 3 cm
B 8 0 1 3 cm
C 2 0 3 cm
D 4 5 8 cm

The correct answer is:
9 0 1 3 cm

Solution:
R=60 cm, R2=20 cm

Let P1 be power of lens of water ,P2 be power of glass lens, and P3 be power of concave mirror
Peq= 2P1 + 2P2 + P3
1-feq=2f1+2f2-1f3 ....(1)
for the lens of water
1 f 1 = 4 3 - 1 1 ∞ - 1 - 6 0
1 f 1 = 1 1 8 0
for glass lens
1 f 2 = ( 3 2 - 1 ) ( 1 - 6 0 - 1 - 2 0 )
1 f 2 = 1 6 0
For concave mirror f 3 = R 2
f3=-10 cm
1 - f eq = 2 180 + 2 60 - 1 - 10

1-feq=26180⇒feq=-9013cm

Topic: Physics, Ray Optics, JEE Main
The curve having differential equation, xcos⁡ydydx+sin⁡y=x and passing through the origin, also passes through

Options:
A 2,Ï2
B -2,π2
C 4,3Ï2
D -8,3Ï2

The correct answer is:
2,π2

Solution:
Given equation is xcos⁡ydy+sin⁡ydx=xdx
Or dxsin⁡y=xdx
Integrating, we get ∫dxsin⁡y=∫xdx
⇒xsin⁡y=x22+C

Since, it passes through 0,0

C=0

⇒xsin⁡y=x22

Topic: Mathematics, Differential Equations, JEE Main
The curve passing through P2,72 and having slope 1-1x2 at Px,y also passes through

Options:
A -2,23
B -2,-32
C -2,1
D -2,6

The correct answer is:
-2,-32

Solution:
As dydx=1-1x2
⇒dy=1-1x2dx
Integrating, we get
∫dy=∫1-1x2dx
⇒y=x+1x+C
As it passes through 2,72
∴72=2+12+C
⇒C=1
Thus, y=x+1x+1
which also passes through -2,-32

Topic: Mathematics, Differential Equations, JEE Main
The curve passing through PÏ2,Ï is such that for a tangent drawn to it at a point Q, the ratio of the y-intercept and the ordinate of Q is 1:2. Then, the equation of the curve is

Options:
A y=πx2
B y=πx
C y=x
D y=π2x

The correct answer is:
y=x

Solution:
General equation of the tangent at Qx,y is Y-y=mX-x
Its y-intercept =y-mx
â´y-mxy=12
â2y-2mx=y
â2mx=y
âm=dydx=y2x
or â«dyy=â«dx2x
â´lnâ¡y=12lnâ¡x+lnâ¡C
or y=Cx
As it passes through Ï2,ÏâC=1
â´ The curve is y=x

Topic: Mathematics, Differential Equations, JEE Main
The curve satisfying the differential equation dxdy=x+2yx2y-2x3 and passing through (1, 0) is given by

Options:
A x2+y2=1
B x2+y2+yx=1
C y2-yx-x2=-1
D x2-y2=1

The correct answer is:
x2+y2+yx=1

Solution:
Given equation is
2xdx+2ydy+dyx=0
Integrating, we get,

x2+y2+yx=c
As (1, 0) lies on it, hence, c=1

Topic: Mathematics, Differential Equations, JEE Main
The curve satisfying the differential equation sin⁡x3eydy+3x2cos⁡x3eydx=xsin⁡x2dx (where, C is the constant of integration) is λsin⁡x3ey+cos⁡x2=C. Then, the value of λ is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
Given equation is dsin⁡x3ey=xsin⁡x2dx
On integrating, we get,

∫dsin⁡x3ey=∫xsin⁡x2dx
⇒sin⁡x3ey=-12cos⁡x2+C
or 2sin⁡x3ey+cos⁡x2=C
∴λ=2

Topic: Mathematics, Differential Equations, JEE Main
The curve satisfying the differential equation, $\left(x^2-y^2\right) d x+2 x y d y=0$ and passing through the point $(1,1)$ is

Options:
A
a circle of radius two

B
a circle of radius one

C
a hyperbola

D
an ellipse

The correct answer is:
a circle of radius one

Solution:
$\left(x^2-y^2\right) d x+2 x y d y=0$
$$
\Rightarrow \frac{d y}{d x}=\frac{y^2-x^2}{2 x y}
$$
Let $y=v x$
$$
\begin{aligned}
& \frac{d y}{d x}=v+x \frac{d v}{d x} \\
\Rightarrow & v+x \frac{d v}{d x}=\frac{v^2 x^2-x^2}{2 v x^2} \\
\Rightarrow & v+x \frac{d v}{d x}=\frac{v^2-1}{2 v} \\
\Rightarrow & x \frac{d v}{d x}=\frac{-v^2-1}{2 v} \\
\Rightarrow & \frac{2 v d v}{v^2+1}=-\frac{d x}{x}
\end{aligned}
$$
After integrating, we get
$$
\begin{aligned}
&\ln \left|v^2+1\right|=-\ln |x|+\ln c \\
&\frac{y^2}{x^2}+1=\frac{c}{x}
\end{aligned}
$$
As curve passes through the point $(1,1)$, so $1+1=c \Rightarrow c=2$
$x^2+y^2-2 x=0$, which is a circle of radius one.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
The curve satisfying the differential equation xln⁡xdy-ydx=xln⁡x2dx is (where, C is an arbitrary constant)

Options:
A y=ln⁡x2+C
B y=x2+C
C y=xln⁡x+C
D y=ln⁡xx+C

The correct answer is:
y=ln⁡xx+C

Solution:
Given equation is
lnx dy−y⋅ 1 x ⋅dx lnx 2 =dx
or ∫dyln⁡x=∫dx
⇒yln⁡x=x+C
⇒y=ln⁡xx+C

Topic: Mathematics, Differential Equations, JEE Main
The curve satisfying the differential equation, ydx-x+3y2dy=0 and passing through the point (1,1) also passes through the point

Options:
A 14,-12
B -13,13
C 14,12
D 13,-13

The correct answer is:
-13,13

Solution:
Given differential equation, ydx-xdy-3y2 dy=0
⇒ dxdy=xy+3y
dxdy-xy=3y
I.F. = e-∫1y dy =e-ln⁡y=1y
∴ solution is
xy= ∫3y .1ydy
⇒ xy=3y+c
Passes through (1,1),
∴ 1=3+c ;c=-2
Equation of required curve, x=3y2-2y
Clearly, it passes through −13,13.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2017 (08 Apr Online)
The curve $y=a x^2+b x$ passes through the point $(1,2)$ and lies above the $X$-axis for $0 \leq x \leq 8$. If the area enclosed by this curve, the $X$-axis and the line $x=6$ is 108 square units, then $2 b-a=$

Options:
A 2
B 0
C 1
D -1

The correct answer is:
0

Solution:
Given, curve $y=a x^2+b x$ passes through point $(1,2)$.
$$
\therefore \quad 2=a(1)^2+b(1)
$$
$\begin{aligned} & \text { Given, area under curve }=\int_0^6\left(a x^2+b x\right) d x=108 \\ & \Rightarrow \quad\left[\frac{a x^3}{3}+\frac{b x^2}{2}\right]_0^6=108 \\ & \Rightarrow \quad 72 a+18 b=108\end{aligned}$
By solving Eqs. (i) and (ii), we get
$$
\begin{aligned}
& a=\frac{4}{3} \text { and } b=\frac{2}{3} \\
\therefore \quad & 2 b-a=2 \times \frac{2}{3}-\frac{4}{3}=\frac{4}{3}-\frac{4}{3}=0
\end{aligned}
$$

Topic: Mathematics, Area Under Curves, JEE Main
The curve y=fx in the first quadrant is such that the y -intercept of the tangent drawn at any point P is equal to twice the ordinate of P. If y=fx passes through Q2,3, then the equation of the curve is

Options:
A x2y=12
B xy=6
C xy2=18
D x+y2=11

The correct answer is:
xy=6

Solution:
General equation of the tangent at Px,y is
Y-y=mX-x
∴ y-intercept is y-mx put X=0
⇒y-mx=2y
⇒mx=-y
or m=dydx=-yx
⇒dyy=-dxx

On integrating, we get,
lny=-lnx+lnc
⇒xy=c
As it passes through 2,3⇒c=6
⇒y=fx is xy=6

Topic: Mathematics, Differential Equations, JEE Main
The curve yx=ax3+bx2+cx+5 touches the x-axis at the point P-2,0 and cuts the y-axis at the point $\mathrm{Q}$, where y' is equal to 3. Then the local maximum value of yx is

Options:
A 274
B 294
C 374
D 92

The correct answer is:
274

Solution:
Given that yx=ax3+bx2+cx+5 pass through -2,0

so 8a-4b+2c=5 ...i

Since the curve touches x-axis at -2,0, so its slope would be 0

i.e y'-2=0⇒3ax2+2bx+cx=-2=0

12a-4b+c=0 ...ii

Also given, for x=0, y'(x)=3

c=3 ...iii

Solving eq. i, ii & iii, we get,

a=-12, b=-34

For local maxima y'x=-32x2-32x+3=0

⇒x2+x-2=0⇒x-1x+2=0

⇒x=1 and y"1<0

So yx has local maxima at x=1

Hence, y1=274

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2022 (25 Jul Shift 1)
The curves $\frac{x^2}{a^2}+\frac{y^2}{4}=4$ and $y^3=16 x$ intersect each other orthogonally, then $\mathrm{a}^2=$

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

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

Solution:
$$
\begin{aligned}
& \frac{x^2}{a^2}+\frac{y^2}{4}=1 \\
& \therefore \frac{1}{a^2} 2 x+\frac{1}{4} 2 y \frac{d y}{d x}=0 \quad \Rightarrow \frac{d y}{d x}=\left(\frac{-x}{a^2}\right)\left(\frac{4}{y}\right)
\end{aligned}
$$
Also $\mathrm{y}^3=16 \mathrm{x}$
$$
\therefore 3 y^2 \frac{d y}{d x}=16 \Rightarrow \frac{d y}{d x}=\frac{16}{3 y^2}
$$
Since curves intersect orthogonally, from (1) and (2), we write
$$
\left(\frac{-x}{a^2}\right)\left(\frac{4}{y}\right)\left(\frac{16}{3 y^2}\right)=-1
$$
$\therefore \frac{64 x}{3 a^2 y^3}=1$ and we have $y^3=16 x$
$$
\therefore \frac{64 \mathrm{x}}{3 \mathrm{a}^2(16 \mathrm{x})}=1 \Rightarrow \mathrm{a}^2=\frac{4}{3}
$$

Topic: Mathematics, Application of Derivatives, JEE Main
The cut-off voltage of the diodes (shown in figure) in forward bias is 0.6 V. The current through the resister of 40Ω is _____ mA.

Options:

The correct answer is:
4

Solution:
Here diode D1 is forward biased and diode D2 is reverse biased.

So, no current will flow through diode D2 resistor.

Current through the resistor of 40 Ω is i=1-0.660+40=0.4100=4 mA

Topic: Physics, Semiconductors, JEE Main, JEE Main 2022 (27 Jun Shift 2)
The cyanide process is used for obtaining

Options:
A $\mathrm{Na}$
B $A g$
C $\mathrm{Cu}$
D $Z n$

The correct answer is:
$A g$

Solution:
Correct Option is 2 $\mathrm{Ag}$
cyanide process is manly method of extracting silver and gold from their ores. It is obtained by dissolving them in a dilute solution of sodium cyanide or potassium cyanide.
Hence, the correct answer is option 2.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
The cylindrical tube of a spray pump has a cross-section of 8 cm2, one end of which has 40 fine holes each of area 10-8 m2. If liquid flows inside the tube with a speed of 0.15 m min-1, the speed with which the liquid is ejected through the hole is

Options:
A 50 m s-1
B 5 m s-1
C 0.05 m s-1
D 0.5 m s-1

The correct answer is:
5 m s-1

Solution:
According to equation of continuity,
av=constant
â´ For tube, 8Ã10-4Ã0.1560=a1v1
For holes 40Ã10-8Ãv=a2v2
â´ a1v1=a2v2
â´ 40Ã10-8Ãv=8Ã10-4Ã0.1560
â¹ v=8Ã10-4Ã0.1540Ã10-8Ã60=5 m s-1

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The $d$-block elements consists mostly of

Options:
A Monovalent metals
B All non-metals
C Elements which generally form stoichiometric metal oxide
D Many metals with catalytic properties

The correct answer is:
Many metals with catalytic properties

Solution:
Many metals with catalytic properties because
(i) They provide surface area for reaction to occur
(ii) They decreases the ionisation energy.
(iii) They have vacant $d$-orbitals.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The $d$-electron configurations of $\mathrm{Cr}^{2+}, \mathrm{Mn}^{2+}, \mathrm{Fe}^{2+}$ and $\mathrm{Co}^{2+}$ are $d^4, d^5, d^6$ and $d^7$ respectively. Which one of the following will exhibit the lowest paramagnetic behaviour?
(Atomic no. $\mathrm{Cr}=24, \mathrm{Mn}=25, \mathrm{Fe}=26, \mathrm{Co}=27$ ).

Options:
A
$\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

B
$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

C
$\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

D
$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

The correct answer is:
$\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

Solution:
$\begin{array}{cc}\text { electronic } & \text { no. of unpaired } \\ \text { configuration } & \text { electrons }\end{array}$
$\therefore \quad$ Since $\mathrm{Co}^{2+}$ has lowest no. of unpaired electrons hence lowest paramagnetic behaviour is shown by $\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2012 (07 May Online)
The damping force on an oscillator is directly proportional to the velocity. The units of the constant of proportionality are

Options:
A $\mathrm{kg} \mathrm{ms}^{-1}$
B $\mathrm{kg} \mathrm{ms}^{-2}$
C $\mathrm{kg} \mathrm{s}^{-1}$
D $\mathrm{kg} \mathrm{s}$

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

Solution:
Given, Damping force $\propto$ velocity
$\begin{aligned}
& F \propto y \\
& F=k v \\
& k=\frac{F}{v}
\end{aligned}$
$\begin{aligned}
\text {Unit of } k & =\frac{\text { unit of } F}{\text { unit of } v} \\
& =\frac{\mathrm{kg}-\mathrm{ms}^{-2}}{\mathrm{~ms}^{-1}}=\mathrm{kgs}^{-1}
\end{aligned}$

Topic: Physics, Units and Dimensions, JEE Main
The dark purple colour of KMnO4 disappears in the titration with oxalic acid in acidic medium. The overall change in the oxidation number of manganese in the reaction is

Options:
A 5
B 1
C 7
D 2

The correct answer is:
5

Solution:
The overall reaction takes place in the process is

2KMnO4+3H2SO4+5COOH2→K2SO4+2MnSO4+8H2O+10CO2↑

Potassium permanganate is a strong oxidising agent and in the presence of sulfuric acid it acts as a powerful oxidising agent. In acidic medium the oxidising ability of KMnO4 is represented by the following equation.

MnO4-+8H++5e-→Mn2++4H2O

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2022 (26 Jul Shift 1)
The DC common-emitter current gain of an n-p-n transistor is 50. The potential difference applied across the collector and emitter of a transistor used in CE configuration is, VCE=2 V. If the collector resistance RC=4 kÎ©, the base current IB and the collector current IC are

Options:
A IB=10 Î¼A,IC=0.5 mA
B IB=0.5 Î¼A,IC=10 mA
C IB=5 μA, IC=1 mA
D IB=1 Î¼A,IC=0.5 ;mA

The correct answer is:
IB=10 Î¼A,IC=0.5 mA

Solution:
Given, Î²=50, VCE=2 V, Rc=4 kÎ©
IC=VCERC=24Ã103=5Ã10-4 A
IC=0.5 mA
DC current gain, Î²=ICIB
IB=ICÎ²=0.5Ã10-350=10-5 A
Hence, IB=10 Î¼A

Topic: Physics, Semiconductors, JEE Main
The de Broglie wavelength $(\lambda)$ associated with a photoelectron varies with the frequency $(v)$ of the incident radiation as,$\left[v_{0}\right.$ is threshold frequency]:

Options:
A $\lambda \propto \frac{1}{\left(v-v_{0}\right)}$
B $\lambda \propto \frac{1}{\left(v-v_{0}\right)^{\frac{1}{4}}}$
C $\lambda \propto \frac{1}{\left(v-v_{0}\right)^{\frac{3}{2}}}$
D $\lambda \propto \frac{1}{\left(v-v_{0}\right)^{\frac{1}{2}}}$

The correct answer is:
$\lambda \propto \frac{1}{\left(v-v_{0}\right)^{\frac{1}{2}}}$

Solution:
According to de-Broglie wavelength equation,

$$

\lambda=\frac{h}{m v} \Rightarrow \lambda \propto \frac{1}{v}

$$

According to photoelectric effect,

$$

\begin{array}{l}

\mathrm{hv}-\mathrm{hv}_{0}=\frac{1}{2} \mathrm{mv}^{2} ; \mathrm{v}-\mathrm{v}_{0}=\frac{1}{2}=\frac{\mathrm{mv}^{2}}{\mathrm{~h}} \\

\mathrm{v}-\mathrm{v}_{0} \propto \mathrm{v}^{2} \\

\mathrm{v} \propto\left(\mathrm{v}-\mathrm{v}_{0}\right)^{1 / 2} \\

\therefore \quad \lambda \propto \frac{1}{\left(\mathrm{v}-\mathrm{v}_{0}\right)^{1 / 2}}

\end{array}

$$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2019 (11 Jan Shift 2)
The de Broglie wavelength of a car of mass 1000 $\mathrm{kg}$ and velocity $36 \mathrm{~km} / \mathrm{hr}$ is :

Options:
A
$6.626 \times 10^{-34} \mathrm{~m}$

B
$6.626 \times 10^{-38} \mathrm{~m}$

C
$6.626 \times 10^{-31} \mathrm{~m}$

D
$6.626 \times 10^{-30} \mathrm{~m}$

The correct answer is:
$6.626 \times 10^{-38} \mathrm{~m}$

Solution:
$$
\begin{aligned}
& \text { } \lambda=\frac{h}{m v} \\
& h=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s} \\
& m=1000 \mathrm{~kg} \\
& v=36 \mathrm{~km} / \mathrm{hr}=\frac{36 \times 10^3}{60 \times 60} \mathrm{~m} / \mathrm{sec}=10 \mathrm{~m} / \mathrm{sec} \\
& \therefore \quad \lambda=\frac{6.6 \times 10^{-34}}{10^3 \times 10}=6.6 \times 10^{-38} \mathrm{~m} \\
&
\end{aligned}
$$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2013 (23 Apr Online)
The de Broglie wavelength of a molecule in a gas at room temperature 300 K is λ1. If the temperature of the gas is increased to 600 K, then the de Broglie wavelength of the same gas molecule becomes

Options:
A 12λ1
B 2λ1
C 12λ1
D 2λ1

The correct answer is:
12λ1

Solution:
The root mean squared velocity of a gas is given by

v=3RTM

⇒v∝T

Let

T1=300 KT2=600 K

Taking velocity ratios at the given temperatures,

v1v2=T1T2=300600=12

The de Broglie wavelength is given by λ=hmv. So,

λ∝1v

The ratio of the wavelengths is

λ1λ2=v2v1=21

⇒λ2=12λ1

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2023 (10 Apr Shift 1)
The de Broglie wavelength of a proton and α-particle are equal. The ratio of their velocities is

Options:
A 4:1
B 4:3
C 1:4
D 4:2

The correct answer is:
4:1

Solution:
λ=hmv

λp=λα

mPVP=mαvα

mPvP=4 mpvα mα=4 mP

vpvα=4

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2021 (24 Feb Shift 2)
The de Broglie wavelength of a tennis ball of mass $60 \mathrm{~g}$ moving with a velocity of 10 metres per second is approximately

Options:
A
$10^{-31}$ metres

B
$10^{-16}$ metres

C
$10^{-25}$ metres

D
$10^{-33}$ metres Planck's constant, $\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}$.

The correct answer is:
$10^{-33}$ metres Planck's constant, $\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}$.

Solution:
$\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.6 \times 10^{-34}}{60 \times 10^{-3} \times 10}=10^{-33} \mathrm{~m}$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2003
The de Broglie wavelength of an electron accelerated between two plates having a potential difference of 900 $\mathrm{V}$ is neariy.

Options:
A $0.015nm$
B $0.01nm$
C $0.02mm$
D $0.04nm$

The correct answer is:
$0.04nm$

Solution:
de Broglie wavelength is given by
$\begin{aligned}
& \lambda=\frac{12.27 \times 10^{-10}}{\sqrt{\mathrm{V}}} \\
& =\frac{12.27 \times 10^{-10}}{\sqrt{900}} \\
& =\frac{12.27 \times 10^{-10}}{30} \\
& =0.04 \mathrm{~nm}
\end{aligned}$

Topic: Physics, Dual Nature of Matter, JEE Main
The de Broglie wavelength of an electron having kinetic energy E is λ. If the kinetic energy of electron becomes E4, then its de-Broglie wavelength will be:

Options:
A 2λ
B λ2
C λ2
D 2λ

The correct answer is:
2λ

Solution:
The de-Broglie wavelength in the first case is

λ=h2mE

In the second case the kinetic energy is K=E4.

Let the de-Broglie wavelength be λ'.

The value of wavelength is

λ'=h2mE4=2h2mE=2λ

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2023 (15 Apr Shift 1)
The de Broglie wavelength of an electron in the 4th Bohr orbit is:

Options:
A 2πa0
B 4πa0
C 6πa0
D 8πa0

The correct answer is:
8πa0

Solution:
2πr=nλ
2π×n2Za0=nλ
2π×421a0=nλ
λ=8πa0

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2020 (09 Jan Shift 1)
The de Broglie wavelengths of a proton and an α particle are λ and 2λ respectively. The ratio of the velocities of proton and α particle will be :

Options:
A 1 : 8
B 1 : 2
C 4 : 1
D 8 : 1

The correct answer is:
8 : 1

Solution:
As we know, de Broglie wavelength is given by, λ=hp=hmv

⇒v=hmλ

Therefore,

vpvα=mαmp×λαλp

=4×2=8

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2024 (01 Feb Shift 1)
The de-Broglie wavelength associated with a material particle is

Options:
A Directly proportional to its energy
B Directly proportional to momentum
C Inversely proportional to its energy
D Inversely proportional to momentum

The correct answer is:
Inversely proportional to momentum

Solution:
de-Broglie equation is $\lambda=\frac{h}{p}$.

Topic: Chemistry, Structure of Atom, JEE Main
The de-Broglie wavelength associated with a particle of mass $1 \mathrm{~kg}$ moving with a velocity of $10 \mathrm{~m} \mathrm{~s}^{-1}$ is:

Options:
A $6.63 \times 10^{-35} \mathrm{~m}$
B $6.63 \times 10^{-34} \mathrm{~m}$
C $6.63 \times 10^{-32} \mathrm{~m}$
D $6.63 \times 10^{-33} \mathrm{~m}$

The correct answer is:
$6.63 \times 10^{-35} \mathrm{~m}$

Solution:
$m=1 \mathrm{~kg}, v=10 \mathrm{~m} / \mathrm{s}$
Formula
$\lambda=h/mv$
$\begin{aligned}& \lambda=\frac{6.63 \times 10^{-34}}{1 \times 10} \\& \lambda=6.63 \times 10^{-35} \mathrm{~m}\end{aligned}$

Topic: Physics, Dual Nature of Matter, JEE Main
The de-Broglie wavelength associated with a particle of mass $m$ and energy $E$ is $h / \sqrt{2 m E}$. The dimensional formula for Planck's constant is :

Options:
A $\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]$
B $\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$
C $\left[\mathrm{MLT}^{-2}\right]$
D $\left[\mathrm{M}^2 \mathrm{~L}^2 \mathrm{~T}^{-2}\right]$

The correct answer is:
$\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]$

Solution:
$\begin{aligned} & \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}} \text { or } \mathrm{E}=\mathrm{h} \nu \\ & {\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]=\mathrm{h}\left[\mathrm{T}^{-1}\right]} \\ & \mathrm{h}=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]\end{aligned}$

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2024 (09 Apr Shift 2)
The deBroglie wavelength associated with a steel ball of mass $1000 \mathrm{gm}$ moving at a speed of $1 \mathrm{~ms}^{-1}$ is $\left[\mathrm{h}=6.626 \times 10^{-34} \mathrm{Js}\right]$

Options:
A $6.626 \times 10^{-31} \mathrm{~m}$
B $6.626 \times 10^{-37} \mathrm{~m}$
C $6.626 \times 10^{-34} \mathrm{~m}$
D $6.626 \times 10^{34} \mathrm{~m}$

The correct answer is:
$6.626 \times 10^{-34} \mathrm{~m}$

Solution:
de Broglie wavelength is given by
$\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.626 \times 10^{-34}}{1000 \times 10^{-3} \times 1}=6.626 \times 10^{-34} \mathrm{~m}$

Topic: Physics, Dual Nature of Matter, JEE Main
The de-Broglie wavelength associated with an electron and a proton were calculated by accelerating them through same potential of 100 V. What should nearly be the ratio of their wavelengths ? mp=1.00727 u,me=0.00055 u

Options:
A 1860 : 1
B 18602 : 1
C 41.4 : 1
D 43 : 1

The correct answer is:
43 : 1

Solution:
λ=hmv=h2mK=h2mqV

λ1λ2=m2 m1

λeλp=mpme=1831.4=42.79

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2021 (16 Mar Shift 2)
The de-Broglie wavelength associated with the electron in the n=4 level is:

Options:
A Half of the de-Broglie wavelength of the electron in the ground state
B Four times the de-Broglie wavelength of the electron in the ground state
C 14th of the de-Broglie wavelength of the electron in the ground state
D Two times the de-Broglie wavelength of the electron in the ground state

The correct answer is:
Four times the de-Broglie wavelength of the electron in the ground state

Solution:
De-Broglie wavelength is given as
λ=hp
∵p22m=k⇒p=2mk
⇒λ=h2mk
At ground level energy =E0 so
At n=4 level energy =E0n2=E016
∴ at ground level
λ0=h2m E0
At n = 4 level
λ′=h2mE016=4h2m E0
⇒λ′=4λ0

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2015 (11 Apr Online)
The de-Broglie wavelength associated with the particle of mass $m$ moving with velocity $v$ is

Options:
A $\mathrm{h} / \mathrm{mv}$
B $\mathrm{mv} / \mathrm{h}$
C $\mathrm{mh} / \mathrm{v}$
D $\mathrm{m} / \mathrm{hv}$

The correct answer is:
$\mathrm{h} / \mathrm{mv}$

Solution:
Correct option is $\mathrm{(1)} : h / m v$
The de-Broglie wavelength associated with with the particle of mass $m$ moving with velocity $v$ is
$\lambda=\frac{h}{p}=\frac{h}{m v}$

Topic: Physics, Dual Nature of Matter, JEE Main
The de-Broglie wavelength λB associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state λG by:

Options:
A Î»B= 3Î»G
B Î»B=2Î»G
C Î»B=Î»G3
D Î»B=Î»G2

The correct answer is:
Î»B= 3Î»G

Solution:
We know that wavelength is inversely proportional to momentum, let λB is de-Broglie Wavelength corresponds to momentum PB, similarly let λG is Wavelength at ground state corresponds to momentumPG,
λBλG= PGPB= mvGmvB, and velocity, V∝zn, where Z is the atomic number, and n is a number of orbits, here, nG=1 (ground state), nB=3 (second excited state). So, λBλG= nBnG= 31⇒λB=3λG

Topic: Physics, Dual Nature of Matter, JEE Main
The de-Broglie wavelength λB associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state λG by:

Options:
A Î»B= 3Î»G
B Î»B=2Î»G
C Î»B=Î»G3
D Î»B=Î»G2

The correct answer is:
Î»B= 3Î»G

Solution:
We know that wavelength is inversely proportional to momentum, let λB is de-Broglie Wavelength corresponds to momentum PB, similarly let λG is Wavelength at ground state corresponds to momentumPG,
λBλG= PGPB= mvGmvB, and velocity, V∝zn, where Z is the atomic number, and n is a number of orbits, here, nG=1 (ground state), nB=3 (second excited state). So, λBλG= nBnG= 31⇒λB=3λG

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2018 (16 Apr Online)
The de-Broglie wavelength λn of the electron in the nth orbit of hydrogen atom is

Options:
A Inversely proportional to n
B Proportional to n2
C Proportional to n
D Inversely proportional to n2

The correct answer is:
Inversely proportional to n

Solution:
âµ de-Broglie wavelength Î»n of the electron in the nth orbit of hydrogen atom
Î»=2ÏnhÃhr
â Î»n=2Ïrn
As, Î»=hp=hmvor mvr=hrÎ»=nh2Ï,Î»=2ÏnhÃhr
Î»nâân
Hence, Î»n is directly proportional to n.

Topic: Physics, Dual Nature of Matter, JEE Main