datavorous/entrance-exam-dataset huggingface dataset:
The positive integer value of $n>3$ satisfying the equation
$$
\frac{1}{\sin \left(\frac{\pi}{n}\right)}=\frac{1}{\sin \left(\frac{2 \pi}{n}\right)}+\frac{1}{\sin \left(\frac{3 \pi}{n}\right)} \text { is }
$$

Options:

The correct answer is:
7

Solution:
Given, $n>3 \in$ Integer
$$
\begin{aligned}
& \text { and } \frac{1}{\sin \left(\frac{\pi}{n}\right)}=\frac{1}{\sin \left(\frac{2 \pi}{n}\right)}+\frac{1}{\sin \left(\frac{3 \pi}{n}\right)} \\
& \Rightarrow \frac{1}{\sin \frac{\pi}{n}}-\frac{1}{\sin \frac{3 \pi}{n}}=\frac{1}{\sin \frac{2 \pi}{n}} \\
& \Rightarrow \quad \frac{\sin \frac{3 \pi}{n}-\sin \frac{\pi}{n}}{\sin \frac{\pi}{n} \cdot \sin \frac{3 \pi}{n}}=\frac{1}{\sin \frac{2 \pi}{n}}
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad 2 \cos \left(\frac{2 \pi}{n}\right) \cdot \sin \frac{\pi}{n}=\frac{\sin \frac{\pi}{n} \cdot \sin \frac{3 \pi}{n}}{\sin \frac{2 \pi}{n}} \\
& \Rightarrow \quad 2 \sin \frac{2 \pi}{n} \cdot \cos \frac{2 \pi}{n}=\sin \frac{3 \pi}{n} \\
& \Rightarrow \quad \quad \sin \frac{4 \pi}{n}=\sin \frac{3 \pi}{n} \\
& \Rightarrow \frac{4 \pi}{n}=\pi-\frac{3 \pi}{n} \Rightarrow \frac{7 \pi}{n}=\pi \Rightarrow n=7
\end{aligned}
$$

Topic: Mathematics, Trigonometric Equations, JEE Main
The positive root of $x^{2}-78.8=0$ after first approximation by Newton Raphson method assuming initial approximation to the root is 14, is

Options:
A $9.821$
B $9.814$
C $9.715$
D $9.915$

The correct answer is:
$9.814$

Solution:
Here, $\quad x_{0}=14, f(x)=x^{2}-78.8$
and $f^{\prime}(x)=2 x$
$$
\begin{aligned}
\therefore \quad x_{1}=x_{0}-& \frac{f\left(x_{0}\right)}{f^{\prime}\left(x_{0}\right)} \\
&=14-\frac{(14)^{2}-(78.8)}{2 \times 14}=9.814
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Main
The positive value of λ for which the co-efficient of x2 in the expansion x2x+λx210 is 720, is

Options:
A 5
B 3
C 4
D 22

The correct answer is:
4

Solution:
Coefficient of x2 in x2x+λx210
= coefficient of x0 in x+λx210
A general term in the expansion x+λx210 is 10Crx10-rλx2r=10Crλrx5-5r2
⇒ coefficient of x0= 10C2λ2 (for r=2 )
=45λ2
Now 45λ2=720
⇒λ=±4

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2019 (10 Jan Shift 2)
The positive value of parameter a for which the area bounded by parabolas y=x-ax2 and ay=x2 attains the maximum value is

Options:

The correct answer is:
1

Solution:
Points of intersection are at x=0, a1+a2
Area=∫0a1+a2x-ax2-x2adx
=16.1a+1a2
Hence, area is maximum when a=1 ∵ a+1a≥2

Topic: Mathematics, Area Under Curves, JEE Main
The positive value of the determinant of the matrix A, whose AdjAdjA=1428-14-14142828-1414, is ______.

Options:

The correct answer is:
14

Solution:
Given,

1428-14-14142828-1414=adjadjA

Now we know that adjA=A2 if it is 3×3 matrix

So adjadjA=adjA2

⇒A2×2=A4

Now, A4=1428-14-14142828-1414

⇒ A4=143 12-1-1122-11

⇒A4=143×14

A4=144 ⇒ A=14

Topic: Mathematics, Matrices, JEE Main, JEE Main 2022 (27 Jun Shift 1)
The possibility of photochemical smog formation is more at

Options:
A

Marshy lands

B

Industrial areas

C

Himalayan villages in winter

D

The places with healthy vegetation

The correct answer is:

Industrial areas

Solution:
Photochemical smog is a mixture of pollutants that are formed when nitrogen oxides and volatile organic compounds (VOCs) react to sunlight, creating a brown haze above cities. It tends to occur more often in summer, because that is when we have the most sunlight.

Photochemical smog formation will be more at Industrial areas.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2023 (06 Apr Shift 1)
The possibility of photochemical smog formation will be minimum at

Options:
A Srinagar, Jammu and Kashmir in January
B Kolkata in October
C Mumbai in May
D New-Delhi in August (Summer)

The correct answer is:
Srinagar, Jammu and Kashmir in January

Solution:
Photochemical smog is a mixture of pollutants that are formed when nitrogen oxides and volatile organic compounds (VOCs) react to sunlight, creating a brown haze above cities. It tends to occur more often in summer, because that is when we have the most sunlight. The average temperature of Kashmir in January is estimated to be -3.6°C in Srinagar, -11.8°C in Pahalgam, and -10.4°C in Gulmarg as their minimum temperatures.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2023 (15 Apr Shift 1)
The possible final products R and S of the following reaction sequence are

Options:
A

R	S

B

R	S

C

R	S

D

R	S

The correct answer is:

R	S

Solution:
When Bromine react with toluene (in which methyl group is ring activating and para directing group according to resonance) in the presence of Fe/dark then product P which is obtained is o−bromotoluene.and product Q is p−bromotoluene and this reaction is called electrophilic substitution reaction.

Now, when o−bromotoluene react with Mg/ether in the presence of water this will lead to formation of phenol and when it is further treated with HCHO then we will be left with o-hydroxymethylphenol and p-hydroxymethylphenol that willbe our final product R and S.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The possible number of structural and stereo isomers for the complex MBr2Cl2SO4 is

Options:

The correct answer is:
5

Solution:
The compound also shows ionization isomerism. The ionization isomers are
(i) MBr2SO4Cl2
(ii) MCl2SO4Br2
(iii) MBrClSO4BrCl
The complex ion [MBr2Cl2]2+ shows geometrical isomerism.
The geometrical isomers are
(i) cis- MBr2Cl2SO4
(ii) trans- MBr2Cl2SO4
Thus, a total of five (3 structural and 2 geometrical) isomers are formed.

Topic: Chemistry, Coordination Compounds, JEE Main
The possible number of values of a for which the common chord of the circles x2+y2=8 and x-a2+y2=8 subtends a right angle at the origin is

Options:
A 0
B 2
C 5
D 3

The correct answer is:
2

Solution:
Equation of the common chord is x-a2+y2-8-x2-y2+8=0
⇒-2ax+a2=0⇒x=a2
If the chord x=a2 subtends a right angle at 0,0 then the distance of x=a2 from the origin is equal to 22cos⁡45o=2
⇒a=4 or -4

Topic: Mathematics, Circle, JEE Main
The possible product obtained from the reaction of cyclobutyl amine with HNO2 followed by treating with water is

Options:
A
B
C
D Both A and C

The correct answer is:
Both A and C

Solution:

Topic: Chemistry, Amines, JEE Main
The possible value of the ordered triplet a,b,c such that the function fx=x3+ax2+bx+c is a monotonic function is

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

The correct answer is:
2,2,4

Solution:
fx=x3+ax2+bx+c
f'x=3x2+2ax+b≥0, ∀x∈R
D≤0
4a2-12b≤0
a2≤3b
From the given options, 2,2,4 satisfies the above condition

Topic: Mathematics, Application of Derivatives, JEE Main
The possible value of x if SD of the numbers 2, 3, 2x and 11 is 3.5 is equal to

Options:
A $4, \frac{5}{3}$
B $5, \frac{7}{3}$
C $3, \frac{7}{3}$
D $2, \frac{5}{3}$

The correct answer is:
$3, \frac{7}{3}$

Solution:
Given Standard deviation of 2, 3, 2x and 11 is 3.5.
$\begin{aligned} & \because \text { Mean }(\bar{x})=\frac{2+3+2 x+11}{4}=\frac{8+x}{2} \\ & \text { Now, } \sum_{i=1}^4\left(x_i-\bar{x}\right)^2\left\{2-\left(\frac{8+x}{2}\right)\right\}^2+\left\{3-\left(\frac{8+x}{2}\right)\right\}^2 \\ &+\left\{2 x-\left(\frac{8+x}{2}\right)\right\}^2+\left\{11-\left(\frac{8+x}{2}\right)\right\}^2 \\ &=\left(\frac{4+x}{2}\right)^2+\left(\frac{2+x}{2}\right)^2+\left(\frac{3 x-8}{2}\right)^2+\left(\frac{14-x}{2}\right)^2\end{aligned}$
$\begin{aligned} & \because \mathrm{SD}=\sqrt{\left[\begin{array}{l}{\left[\frac{4+x}{4}\right)^2+\left(\frac{2+x}{2}\right)^2+\left(\frac{3 x-8}{2}\right)^2} \\ +\left(\frac{14-x}{2}\right)^2\end{array}\right]} \\ & (3 \cdot 5)^2 \times 4=\frac{1}{4}\end{aligned}$
$\begin{aligned} & {\left[16+x^2+8 x+4+x^2+4 x+9 x^2+64-48 x\right.} \\ & \left.\quad+196+x^2-28 x\right] \\ & \\ & \Rightarrow 12 x^2-64 x+84=0 \\ & \Rightarrow \quad x=3,7 / 3\end{aligned}$

Topic: Mathematics, Statistics, JEE Main
The possible values of n for which the equation nx2+2nâ1x+nâ1=0 has roots of opposite sign is/are given by

Options:
A no values of n
B all values of n
C –1<n<0
D 0<n<1

The correct answer is:
0<n<1

Solution:
D=2nâ12â4nnâ1=4n2+1â4nâ4n2+4n=1>0
Product of roots =n-1n<0
⇒nnâ1<0
⇒nâ0,1

Topic: Mathematics, Quadratic Equation, JEE Main
The possible values of scalar k such that the matrix A - 1 - kI is singular where A= 1 0 2 0 2 1 1 0 0 , are

Options:
A −1 2 , 2
B −1,  1 2
C 1 2 ,  −1 2
D −1, 1

The correct answer is:
−1,  1 2

Solution:
A - 1 - kI = 0
A A - 1 - kI = 0 A â 0
I - kA = 0
Ik-A=0 â A-1k. I=0
âA-Î»I=0, where Î»=1k
â 1 - Î» 0 2 0 2 - Î» 1 1 0 - Î» = 0
â 1 - Î» - Î» 2 - Î» + 2 0 - 2 - Î» = 0
â - Î» 3 + 3 Î» 2 - 2 Î» - 4 + 2 Î» = 0
â Î» 3 â3 Î» 2 +4=0âââââÎ»=2, 2,â1âk=â1,â 1 2

Topic: Mathematics, Determinants, JEE Main
The posterior lobe of pituitary gland secretes:

Options:
A FSH, GH and LH
B Somatostatin, GH and TSH
C TSH, ADH and prolactin
D ADH and oxytocin

The correct answer is:
ADH and oxytocin

Solution:
Neurohypophysis (pars nervosa) also known as posterior pituitary, stores and releases two hormones called oxytocin and vasopressin, which are actually synthesised by the hypothalamus and are transported axonally to neurohypophysis. Oxytocin acts on the smooth muscles of our body and stimulates their contraction. In females, it stimulates a vigorous contraction of uterus at the time of child birth, and milk ejection from the mammary gland. Vasopressin acts mainly at the kidney and stimulates resorption of water and electrolytes by the distal tubules and thereby reduces loss of water through urine (diuresis). Hence, it is also called as anti-diuretic hormone (ADH). An impairment affecting synthesis or release of ADH results in a diminished ability of the kidney to conserve water leading to water loss and dehydration. This
condition is known as diabetes insipidus.

Topic: Biology, Chemical Coordination and Integration, NEET
The potassium ferrocyanide solution gives a Prussian blue colour, when added to :

Options:
A CoCl3
B CoCl2
C FeCl2
D FeCl3

The correct answer is:
FeCl3

Solution:
6NaCN+FeSO4⟶Na4Fe(CN)6

Na4Fe(CN)6+4Fe3+⟶xH2OFe4Fe(CN)63.xH2OFerric ferrocyanide (Prussian blue complex)

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2021 (01 Sep Shift 2)
The potential at a point $x$ (measured in $\mu \mathrm{m}$ ) due to some charges situated on the $\mathrm{x}$-axis is given by $\mathrm{V(x)=20 /\left(x^2-4\right)} \mathrm{~Volts}$. The electric field $\mathrm{E}$ at $\mathrm{x=4 ~\mu m}$ is given by

Options:
A
$5 / 3 \mathrm{~Volt}/$ $\mu \mathrm{m}$ and in the $-ve$ $x$ direction

B
$5 / 3 \mathrm{~Volt}/$ $\mu \mathrm{m}$ and in the $+ve$ $x$ direction.

C
$10 / 9 \mathrm{~Volt} / \mu \mathrm{m}$ and in the $-ve$ $x$ direction

D
$10 / 9 \mathrm{~Volt} / \mu \mathrm{m}$ and in the $+ve$ $x$ direction.

The correct answer is:
$10 / 9 \mathrm{~Volt} / \mu \mathrm{m}$ and in the $+ve$ $x$ direction.

Solution:
$\mathrm{V_x=\frac{20}{x^2-4}}$
$\mathrm{E=-\frac{d V}{d x}=\frac{20}{\left(x^2-4\right)^2}(2 x-0)=\frac{160}{144}=\frac{10}{9}}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2007
The potential at point A, in the circuit, is (Point N is grounded, i.e. the potential of that point is zero.)

Options:
A 10Â V
B 7.5 V
C 5Â V
D 2.5Â V

The correct answer is:
7.5 V

Solution:
VAâVN=10ÃCC+C3=30C4C=7.5
VAâ0=7.5
VA=7.5 V

Topic: Physics, Electrostatics, JEE Main
The potential difference across a Coolidge tube is 20 kV and 10 mA current flows through the voltage supply. Only 0.5% of the energy, carried by electrons striking the target, is converted into X-ray. Then the power carried by X-ray beam (in W) is

Options:

The correct answer is:
1

Solution:
P=VI
Total power drawn by the tube =PT=VI=200 W
As 0.5% of energy is carried by electrons, the power of X-ray will be 0.5%ÃPT=1 W

Topic: Physics, Atomic Physics, JEE Main
The potential difference across a resistor is measured to be V=(8±0.5) V and the current through it is measured to be I=(2±0.2) A, then the value of the resistance R is

Options:
A 4±16.25%
B 4±6.25%
C 4±10%
D 4±8%

The correct answer is:
4±16.25%

Solution:
V=(8+0.5)
I=(2+0.2)
R=82=4
∆RR%=∆VV+∆II
=0.58+0.22×100=16.25%
∴ R=(4±16.25%)

Topic: Physics, Units and Dimensions, JEE Main
The potential difference between point $A ~\&~ B$ is

Options:
A $\frac{20}{7} \mathrm{~V}$
B $\frac{40}{7} \mathrm{~V}$
C $\frac{10}{7} \mathrm{~V}$
D 0

The correct answer is:
0

Solution:
The given circuit is a balanced Wheatstone bridge circuit. Hence potential difference between $A$ and $B$ is zero.

Topic: Physics, Current Electricity, JEE Main
The potential difference between the points A and B in the given circuit is

Options:
A 1 V
B 3 V
C 6 V
D 2 V

The correct answer is:
2 V

Solution:
VAB=11+21+3111+11+11=1+2+33=63=2volt

Topic: Physics, Current Electricity, JEE Main
The potential difference between two parallel plates is $10^4 \mathrm{~V}$. If the plates are separated by $0.5 \mathrm{~cm}$, the force on an electron between the plates is

Options:
A $32 \times 10^{-13} \mathrm{~N}$
B $0.32 \times 10^{-13} \mathrm{~N}$
C $0.032 \times 10^{-13} \mathrm{~N}$
D $3.2 \times 10^{-13} \mathrm{~N}$

The correct answer is:
$3.2 \times 10^{-13} \mathrm{~N}$

Solution:
The force $F=\frac{e V}{d}$
$=\frac{1.6 \times 10^{-19} \times 10^4}{0.5 \times 10^{-2}}$
$=3.2 \times 10^{-13} \mathrm{~N}$

Topic: Physics, Capacitance, JEE Main
The potential energy for a force field $\overrightarrow{\mathrm{F}}$ is given by $U(x, y)=\cos (x+y)$. The force acting on a particle at position given by coordinates $(0, \pi / 4)$ is $-$

Options:
A $-\frac{1}{\sqrt{2}}(\hat{\mathrm{i}}+\hat{\mathrm{j}})$
B $\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$
C $\left(\frac{1}{2} \hat{\mathrm{i}}+\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right)$
D $\left(\frac{1}{2} \hat{\mathrm{i}}-\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right)$

The correct answer is:
$\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$

Solution:
$\quad \mathrm{F}_{\mathrm{x}}=-\frac{\partial \mathrm{U}}{\partial \mathrm{x}}=\sin (\mathrm{x}+\mathrm{y})$
$$
\begin{array}{l}
\mathrm{F}_{\mathrm{y}}=-\frac{\partial \mathrm{U}}{\partial \mathrm{x}}=\sin (\mathrm{x}+\mathrm{y}) \\
\left.\mathrm{F}_{\mathrm{x}}=\sin (\mathrm{x}+\mathrm{y})\right]_{(0, \pi / 4)}=\frac{1}{\sqrt{2}} \\
\left.\mathrm{~F}_{\mathrm{y}}=\sin (\mathrm{x}+\mathrm{y})\right]_{(0, \pi / 4)}=\frac{1}{\sqrt{2}} \\
\therefore \mathrm{F}=\frac{1}{\sqrt{2}}[\hat{\mathrm{i}}+\hat{\mathrm{j}}]
\end{array}
$$

Topic: Physics, Work Power Energy, JEE Main
The potential energy for a force field Fâ is given by U(x, y)=cos (x + y). The force acting on a particle at the position given by coordinates 0 , Ï / 4 is

Options:
A - 1 2 i ^ + j ^
B 1 2 i ^ + j ^
C 1 2 i ^ + 3 2 j ^
D 1 2 i ^ - 3 2 j ^

The correct answer is:
1 2 i ^ + j ^

Solution:
F x = - ∂ U ∂ x = sin x + y , F y = - ∂ U ∂ x = sin x + y
F x = sin x + y ] 0 , π / 4 = 1 2 ,
F y = sin x + y ] 0 , π / 4 = 1 2
∴ F = 1 2 i ^ + j ^

Topic: Physics, Work Power Energy, JEE Main
The potential energy function for the force between two atoms in a diatomic molecule is approximately given by $U(x)=\frac{a}{x^{12}}-\frac{b}{x^6}$, where a and $b$ are constants and $x$ is the distance between the atoms. If the dissociation energy of the molecule is $D=\left[U(x=\infty)-U_{\text {at equilbrium }}\right], D$ is

Options:
A
$\frac{b^2}{2 a}$

B
$\frac{b^2}{12 a}$

C
$\frac{b^2}{4 a}$

D
$\frac{b^2}{6 a}$

The correct answer is:
$\frac{b^2}{4 a}$

Solution:
$$
\begin{aligned}
& U(x)=\frac{a}{x^{12}}-\frac{b}{x^6} \\
& U(x=\infty)=0
\end{aligned}
$$
as, $\quad \mathrm{F}=-\frac{\mathrm{dU}}{\mathrm{dx}}=-\left[\frac{12 \mathrm{a}}{\mathrm{x}^{13}}+\frac{6 \mathrm{~b}}{\mathrm{x}^7}\right]$ at equilibrium, $\quad F=0$
$\begin{array}{ll}\therefore & x^6=\frac{2 a}{b} \\ \therefore & U_{\text {at equiltrium }}=\frac{a}{\left(\frac{2 a}{b}\right)^2}-\frac{b}{\left(\frac{2 a}{b}\right)}=\frac{-b^2}{4 a} \\ \therefore & D=\left[U(x=\infty)-U_{\text {at equilibium }}\right]=\frac{b^2}{4 a}\end{array}$

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2010
The potential energy function (in J ) of a particle in a region of space is given as U=2x2+3y3+2z. Here x, y and z are in meter. The magnitude of x - component of force (in N ) acting on the particle at point P(1, 2, 3) m is:

Options:
A 2
B 6
C 4
D 8

The correct answer is:
4

Solution:
Given U=2x2+3y3+2z

The expression for the x-component of the force can be calculated as follows:

Fx=-∂U∂x=-4x

Hence, at the point 1, 2, 3 m, the value of the force is given by

Fx=-4×1 N=4 N

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2024 (29 Jan Shift 1)
The potential energy of 1 kg particle free to move along the X-axis is given by U=x44-x22 J. The total mechanical energy of the particle is 2 J. Maximum speed of the particle is

Options:
A 42
B 12
C 32
D 2

The correct answer is:
32

Solution:
U=x44-x22 (given)
For maxima or minima of PE =dUdx=(x3-x)
∴xx2-1=0⇒x=0 or ±1
d2Udx2=3x2-1
at x=±1
d2Udx=+ ve  i.e.,  at  x=±1,
P.E. is minimum
U=-14
E=Kmax+Vmin⇒2=Kmax-14
∴12mvmax2=94
∵m=1 kg (given) ⇒vmax=32 m s-1

Topic: Physics, Work Power Energy, JEE Main
The potential energy of a $1 \mathrm{~kg}$ particle free move along the $\mathrm{x}$-axis is given by
$$
V(x)=\left(\frac{x^4}{4}-\frac{x^2}{2}\right) J
$$
The total mechanical energy of the particle $2 \mathrm{~J}$. Then, the maximum speed (in $\mathrm{m} / \mathrm{s}$ ) is

Options:
A
2

B
$3 / \sqrt{2}$

C
$\sqrt{2}$

D
$1 / \sqrt{2}$

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

Solution:
$\mathrm{k} \mathrm{E}_{\max }=\mathrm{E}_{\mathrm{T}}-\mathrm{U}_{\min }$
$\mathrm{U}_{\min }(\pm 1)=-1 / 4 \mathrm{~J}$
$\mathrm{KE}_{\max }=9 / 4 \mathrm{~J} \Rightarrow \mathrm{U}=\frac{3}{\sqrt{2}} \mathrm{~J}$

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2006
The potential energy of a 1 kg particle, free to move along the x-axis, is given by Vâ¡(x)=(x44-x22) J. The total mechanical energy of the particle is 2 J then, the maximum speed (in m s-1) is

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

The correct answer is:
3 / 2

Solution:
Total energy ET =2 J It is fixed. For maximum speed. kinetic energy is maximum The potential energy should therefore be minimum
∵ V⁡ ( x ) = x 4 4 - x 2 2
or d⁡ V⁡ d⁡ x = 4 x 3 4 - 2 x 2 = x ( x 2 - 1 )
For V to be minimum, d⁡ V⁡ d⁡ x = 0
∴ x ( x 2 - 1 ) = 0 ,⁡ or ⁡ x = 0 ,⁡ ± 1
At x=0, V(x)=0
At x = ± 1 ,⁡ V⁡ ( x ) = - 1 4 J
∴ ( Kinetic energy ) max = E⁡ T⁡ - V⁡ min
or ∴(Kinetic energy)max=(14)=94 J
or 12m⁡vm⁡2=94
vm2=9×2m×4
or vm⁡2=9×2m⁡×4=9×21×4
∴vm⁡=32 m s-1

Topic: Physics, Work Power Energy, JEE Main
The potential energy of a 2 kg particle, free to move along the x-axis is given by Vâ¡(x)=(x44-x22) J. The total mechanical energy of the particle is 2 J then, the maximum speed (in m s-1) is

Options:

The correct answer is:
1.5

Solution:
Total energy ET= 2 J It is fixed. For maximum speed, kinetic energy is maximum The potential energy should, therefore, be minimum
âµVâ¡(x)=x44-x22
or dâ¡Vâ¡dâ¡x=4x34-2x2=x(x2-1)
For V to be minimum,dâ¡Vâ¡dâ¡x=0
â´x(x2-1)=0,â¡or â¡x=0,â¡Â±1

at x=0, V(x)=0

At x=Â±1,â¡Vâ¡(x)=-14 J
â´(Kinetic energy)max=Eâ¡Tâ¡-Vâ¡min
or â´(Kinetic energy)max=2+14=94 J
or 12mâ¡Î½mâ¡2=94
Î½m2=9Ã2mÃ4
or Î½mâ¡2=9Ã2mâ¡Ã4=9Ã22Ã4
â´Î½mâ¡=1.5 m s-1

Topic: Physics, Work Power Energy, JEE Main
The potential energy of a body is given by, $\quad U=A-B x^2$ (Where $x$ is the displacement). The magnitude of force acting on the particle is

Options:
A Constant
B Proportional to $x$
C Proportional to $x^2$
D Inversely proportional to $x$

The correct answer is:
Proportional to $x$

Solution:
$U=A-B x^2 \Rightarrow F=-\frac{d U}{d x}=2 B x \Rightarrow F \propto x$

Topic: Physics, Work Power Energy, JEE Main
The potential energy of a certain spring when stretched through a distance \( S \) is \( 10 \) joule. The amount of work (in joule) that must be done on this spring to stretch it through additional distance \( S \) will be

Options:
A \( 30 \)
B \( 40 \)
C \( 10 \)
D \( 20 \)

The correct answer is:
\( 30 \)

Solution:
U⁡ = 1 2 K⁡ x 2 = 1 2 KS 2 = 1 0 J ...(i)
U⁡ ′ = 1 2 K⁡ x + x 2 = 1 2 K⁡ S⁡ + S⁡ 2
= 1 2 K⁡ 4S 2 = 4 0 J ...(ii)
∴ W=U'–U=40 J –10 J=30 J

Topic: Physics, Work Power Energy, JEE Main
The potential energy of a long spring when stretched by $2 \mathrm{~cm}$ is $U$. If the spring is stretched by $8 \mathrm{~cm}$ the potential energy stored in it is:

Options:
A $U / 4$
B $4 U$
C $8 U$
D $16 U$

The correct answer is:
$16 U$

Solution:
Potential energy of a spring
$\begin{aligned}
& \begin{aligned}
&= \frac{1}{2} \times \text { force constant } \\
& \times(\text { extension })^2 \\
& \Rightarrow \text { Potential energy } \propto(\text { extension })^2
\end{aligned} \\
& \text {Hence } \frac{U_1}{V_2}=\left(\frac{x_1}{x_2}\right)^2 \\
& \Rightarrow \frac{V_1}{V_2}=\left(\frac{2}{8}\right)^2 \\
& \Rightarrow \frac{U_1}{U_2}- \frac{1}{16}\left(\because U_1=V\right)
\end{aligned}$

Topic: Physics, Work Power Energy, JEE Main
The potential energy of a particle executing S.H.M. is $2.5 \mathrm{~J}$, when its displacement is half of amplitude. The total energy of the particle be

Options:
A $18 \mathrm{~J}$
B $10 \mathrm{~J}$
C $12 \mathrm{~J}$
D $2.5 \mathrm{~J}$

The correct answer is:
$10 \mathrm{~J}$

Solution:
$\frac{\text { Potential energy }(U)}{\text { Total energy }(E)}=\frac{\frac{1}{2} m \omega^2 y^2}{\frac{1}{2} m \omega^2 a^2}=\frac{y^2}{a^2}$
So $\frac{2.5}{E}=\frac{\left(\frac{a}{2}\right)^2}{a^2} \Rightarrow E=10 J$

Topic: Physics, Oscillations, JEE Main
The potential energy of a particle of mass 5 kg moving in the x - y plane is given by U=-7x+24y J. x and y are in meter. If the particle starts from rest from origin then speed of the particle at t=2 s is

Options:
A 5 m s-1
B 14 m s-1
C 17.5 m s-1
D 10 m s-1

The correct answer is:
10 m s-1

Solution:
F= -∂U∂x i^-∂U∂y j^=7i^-24j^
∴ ax=Fxm=75=1.4 m s-2
ay=Fym= -245
=4.8 m s-2 along negative y-axis
∴ vx=axt=1.4 ×2
=2.8 m s-1 along negative y-axis
And vy=4.8 ×2=9.6 m s-1
∴ v= vx2+vy2=10 m s-1

Topic: Physics, Work Power Energy, JEE Main
The potential energy of a particle of mass 5 kg moving in the x-y plane is given by U=(-7x+24y) J, x and y being in metre. If the particle starts from rest from origin, then what is the speed (in m s-1) of particle at t=2 s?

Options:

The correct answer is:
10

Solution:
F→=-∂U∂xi^-∂U∂yj^=7i^-24j^
∴ ax=Fxm=75=1.4 ms-2 along positive x-axis
ay=Fym=-245
=4.8ms-2 along negative y-axis
∴ vx=axt=1.4×2
=2.8 ms-2
and vy=4.8×2=9.6 ms-1
∴ v=vx2+vy2=10 ms-1

Topic: Physics, Work Power Energy, JEE Main
The potential energy of a particle of mass \( m \) is given by \( U=\frac{1}{2} k x^{2} \) for \( x < 0 \) and \( U=0 \) for \( x \geq 0 \). If total mechanical energy of the particle is \( E \). Then its speed at \( x=\sqrt{\frac{2 E}{k}} \) is

Options:
A Zero
B \( \sqrt{\frac{2 \mathrm{E}}{\mathrm{m}}} \)
C \( \sqrt{\frac{\mathrm{E}}{\mathrm{m}}} \)
D \( \sqrt{\frac{E}{2 m}} \)

The correct answer is:
\( \sqrt{\frac{2 \mathrm{E}}{\mathrm{m}}} \)

Solution:
Potential energy of particle at x = 2 E k is zero
∴ KE = E
or 1 2 mv 2 = E or v = 2 E m

Topic: Physics, Work Power Energy, JEE Main
The potential energy of a particle under a conservative force is given by Ux=x2-3x J. The equilibrium position of the particle is at x m. The value of 10x will be

Options:

The correct answer is:
15

Solution:
Ux=x2â3xJ
For a conservative field, Force, Fâ¡=-dâ¡Uâ¡dâ¡x
â´Fâ¡=-dâ¡dâ¡xx2-3x=-2x-3=-2x+3
At equilibrium position, F=0
-2x+3=0âx=32mâ¡=1Â·5m

Topic: Physics, Work Power Energy, JEE Main
The potential energy of a particle varies as

Ux=E0 for 0â¤xâ¤1
=0 for x>1

For 0â¤xâ¤1 , the de-Broglie wavelength is Î»1 and for x>1 the de-Broglie wavelength is Î»2. The total energy of the particle is 2E0. Find Î» 1 Î» 2

Options:
A 3
B 7
C 2
D 5

The correct answer is:
2

Solution:
For 0≤x≤1 , U=E0
∴ Kinetic energy, K1 = Total energy - U
=2E0-E0=E0
∴λ1=h2mE0
For x>1, U=0
∴ Kinetic energy K2 = Total energy -U

=2E0
∴ λ 2 = h⁡ 2 m⁡ ( 2E⁡ 0 ) ...(ii)
From Eqs. (i) and (ii), we have
λ 1 λ 2 = 2

Topic: Physics, Dual Nature of Matter, JEE Main