datavorous/entrance-exam-dataset huggingface dataset:
A substance breaks down by a stress of $10^6 \mathrm{~N} / \mathrm{m}^2$. If the density of the material of the wire is $3 \times 10^3$ $\mathrm{kg} / \mathrm{m}^3$, then the length of the wire of the substance which will break under its own weight when suspended vertically, is

Options:
A $66.6 \mathrm{~m}$
B $60.0 \mathrm{~m}$
C $33.3 \mathrm{~m}$
D $30.0 \mathrm{~m}$

The correct answer is:
$33.3 \mathrm{~m}$

Solution:
$L=\frac{p}{d g}=\frac{10^6}{3 \times 10^3 \times 10}=\frac{100}{3}=33.3 \mathrm{~m}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A substance $\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}$ yields on oxidation a compound, $\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}$ which gives an oxime and a positive iodoform test. The original substance on treatment with conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$ gives $\mathrm{C}_{4} \mathrm{H}_{8} .$ The structure of the compound is

Options:
A $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}$
B $\mathrm{CH}_{3} \mathrm{CHOHCH}_{2} \mathrm{CH}_{3}$
C $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}$
D $\mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{2} \mathrm{CH}_{3}$

The correct answer is:
$\mathrm{CH}_{3} \mathrm{CHOHCH}_{2} \mathrm{CH}_{3}$

Solution:
$$
\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}\left(\mathrm{R}-\mathrm{COCH}_{3}\right)
$$
Thus $\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}$ should be $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{3}$,
hence $\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}$ should be $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHOHCH}_{3}$

Topic: Chemistry, Aldehydes and Ketones, JEE Main
A substance containing hydrogen and releasing $\mathrm{H}^{+}$in aqueous medium is acid. Identify theory suggesting this concept, from following.

Options:
A Ostwald theory
B Bronsted-Lowry theory
C Arrhenius theory
D Lewis theory

The correct answer is:
Arrhenius theory

Solution:
According to Arrhenius theory an acid is a
Substance that contains hydrogen and gives $\mathrm{H}^{+}$ions in an aqueous solution.
- $\quad$ Lewis acid $\rightarrow$ lone pair acceptor
- $\quad$ Bronsted-Lowry acid is a substance which donates a proton or $\mathrm{H}^{+}$ion to the other compound and forms a conjugated base.

Topic: Chemistry, Ionic Equilibrium, JEE Main
A Substance undergoes first order decomposition. The decomposition follows two parallel first order reactions as

The percentage distribution of \(B\) and \(C\) are

Options:
A \(75 \% \mathrm{~B}\) and \(25 \% \mathrm{C}\)
B \(80 \% \mathrm{~B}\) and \(20 \% \mathrm{C}\)
C \(60 \% \mathrm{~B}\) and \(40 \% \mathrm{C}\)
D \(76.83 \% \mathrm{~B}\) and \(23.17 \% \mathrm{C}\)

The correct answer is:
\(76.83 \% \mathrm{~B}\) and \(23.17 \% \mathrm{C}\)

Solution:
\(\begin{aligned} & \text { % distribution of } \mathrm{B}=\frac{k_1}{k_1+k_2} \times 100 \\ & =\frac{1.26 \times 10^{-4}}{1.26 \times 10^{-4}+3.8 \times 10^{-4}} \times 100 \\ & \mathrm{~B} \%=76.83 \% \\ & \text { % Distribution of } \mathrm{C}=\frac{k_2}{k_1+k_2} \times 100 \\ & =\frac{3.8 \times 10^{-4}}{1.26 \times 10^{-4}+3.8 \times 10^{-4}} \times 100 \\ & \mathrm{C} \%=23.17 \%\end{aligned}\)

Topic: Chemistry, Chemical Kinetics, JEE Main
A substance which reacts with gangue to form fusible material is called

Options:
A Flux
B Catalyst
C Ore
D Slag

The correct answer is:
Flux

Solution:
Flux
Gangue $+\underset{\text { Infusible }}{\text { Flux }} \rightarrow \underset{\text { fusible }}{\text { Slag }}$

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
A substance $X$ is a compound of an element of group IA the substance $X$ gives a violet cololour in flame test, $X$ is

Options:
A $\mathrm{LiCl}$
B $\mathrm{NaCl}$
C $\mathrm{KCl}$
D None

The correct answer is:
$\mathrm{KCl}$

Solution:
Potassium react with halogens (chlorine) to gives violet cololour flame

Topic: Chemistry, s Block Elements, JEE Main
A sugar 'X' dehydrates very slowly under acidic condition to give furfural which on further reaction with resorcinol gives the coloured product after sometime. Sugar 'X' is

Options:
A Aldopentose
B Aldotetrose
C Oxalic acid
D Ketotetrose

The correct answer is:
Aldopentose

Solution:
Seliwanoff’s test is a chemical test which distinguishes between aldose and ketose sugars. If the sugar contains a ketone group, it is a ketose. If a sugar contains an aldehyde group, it is an aldose. This test relies on the principle that, when heated, ketoses are more rapidly dehydrated than aldoses. It is named after Theodor Seliwanoff, the chemist that devised the test. When added to a solution containing ketoses, a red color is formed rapidly indicating a positive test. When added to a solution containing aldoses, a slower forming light pink is observed instead. The acidic dehydration of the different aldopentoses gives furfural. This furfural combines with resorcinol to give cherry red product.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2022 (27 Jul Shift 1)
A suitcase is gently dropped on a conveyor belt moving at a velocity of 3 m s-1. If the coefficient of friction between the belt and the suitcase is 0.5, the displacement of the suitcase relative to conveyor belt before the slipping between the two is stopped, is (g=10 m s-2)

Options:
A 2.7 m

B 1.8 m
C 0.9 m
D 1.2 m

The correct answer is:
0.9 m

Solution:
Acceleration of the suitcase till the slipping continues is
a=fmaxm
a=μmgm=μg=0.5×10=5 m​s-2
Slipping will continue until its velocity also becomes 3 m s-1.
∴ v=u+at
or or t=0.6 s
In this time, the displacement of the suitcase
s1=12at2=12×5×0.62=0.9 m
and the displacement of the belt,
s2=vt=3×0.6=1.8 m
Displacement of the suitcase with respect to the belt
s1-s2=0.9 m

Topic: Physics, Laws of Motion, JEE Main
A sulphur containing amino acid is

Options:
A Methionine
B Valine
C Leucine
D Glutamic acid

The correct answer is:
Methionine

Solution:
Sulphur containing amino acids contribute substantially to the maintenance and integrity of cellular systems by influencing cellular redox state and cellular capacity to detoxify toxic compounds, free radicals, and reactive oxygen species. Methionine and cysteine are the two primary sulphur-containing amino acids in organisms.

Topic: Chemistry, Biomolecules, NEET
A superconducting loop of radius R has self-inductance L. A uniform and constant magnetic field B is applied perpendicular to the plane of the loop. Initially current in this loop is zero. The loop is rotated by 180°. The current in the loop after the rotation is

Options:
A zero
B BπR2L
C 2BπR2L
D BπR22L

The correct answer is:
2BπR2L

Solution:
Initially, the current was zero, so self-induced flux was also zero. Now when we rotate the loop, the change in flux is
Δϕ=2πR2×B
Finally, to maintain the same zero flux, the current in the superconducting loop will be such that the flux due to induced current will negative of the flux due to external field
L×i=2πR2×B.
i=2πR2×BL

Topic: Physics, Electromagnetic Induction, JEE Main
A superconducting loop of the radius R has self-inductance L. A uniform and the constant magnetic field B is applied perpendicular to the plane of the loop. Initially current in this loop is zero. The loop is rotated by 180°. The current in the loop after the rotation is

Options:
A zero
B BπR2L
C 2BπR2L
D BπR22L

The correct answer is:
2BπR2L

Solution:
Initially, the current was zero, so self-induced flux was also zero. Now when we rotate the loop, the change in flux is
Δϕ=2πR2×B
Finally, to maintain the same zero flux, the current in the superconducting loop will be such that the flux due to induced current will negative of the flux due to external field
L×i=2πR2×B.
i=2πR2×BL

Topic: Physics, Electromagnetic Induction, JEE Main
A survey shows that 63% of the people in a city read newspaper A whereas 76% read news paper B. If x% of the people read both the newspapers, then a possible value of x can be:

Options:
A 29
B 37
C 65
D 55

The correct answer is:
55

Solution:
nA=63

nB=76

nA∩B=x

Let n∪=100

nA∪B=nA+nB−nA∩B≤100

⇒63+76−x≤100

⇒x≥39

But nA∩B≤nA

∴ 39≤x≤63 .

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2020 (04 Sep Shift 1)
A survey shows that 69% students like mathematics, whereas 75% like chemistry. If x% students like both the subjects, then the maximum value of x is

Options:

The correct answer is:
69

Solution:
Let, total students in survey be 100

If mathematics is represented by m & chemistry by c, then
nm∩c=x
Given, nm=69, nc=75
Now, nm∪c=nm+nc-nm∩c
=69+75–x
=144–x
∵75≤nm∪c≤100
75≤144-x ≤100
∴ 44≤x≤69

Topic: Mathematics, Sets and Relations, JEE Main
A survey shows that 73% of the persons working in an office like coffee, whereas 65% like tea. If x denotes the percentage of them, who like both coffee and tea, then x cannot be:

Options:
A 63
B 36
C 54
D 38

The correct answer is:
36

Solution:
nC=73, nT=65, nC∩T=x,

Considering that value of nC∪T cannot exceed 100 in terms of percentage, we get

nC∪T≤100

⇒ nC+nT-nC∩T≤100

⇒ x≥38

Therefore, x≠36.

nC∩T≤minnC,nT ⇒ x≤65

⇒ 38≤x≤65

Hence x may assume values 63,54,38.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2020 (05 Sep Shift 1)
A symmetric double convex lens is cut in two equal parts by a plane perpendicular to the principal axis. If the power of the original lens is 4 D, the power of a cut lens will be

Options:
A 2 D

B 3 D

C 4 D

D 5 D

The correct answer is:
2 D

Solution:
Biconvex lens is cut perpendicularly to the principle axis, it will become a plano-convex lens.
Focal length of biconvex lens
1f=n-11R₁-1R₂
1f=n-12R (∵R1=R,R2=-R)
⟹f=R2n-1 ….…(i)
For plano-convex lens
1f₁=n-11R-1∞
f1=Rn-1 ………….(ii)
Comparing Eqs. (i) and (ii), we see that focal length becomes double.
Power of lens P ∝1focal length
Hence, power will become half.
New power = 42=2 D

Topic: Physics, Ray Optics, JEE Main
A symmetric star shaped conducting wire loop is carrying a steady state current I as shown in the figure. The distance between the diametrically opposite vertices of the star is 4a .

Options:
A μ0I4πa 3 3-1
B μ0I4πa63-1
C μ0I4πa6 3+1
D μ0I4πa32-3

The correct answer is:
μ0I4πa63-1

Solution:
The given points (1, 2, 3, 4, 5, 6) makes 360o angle at âOâ. Hence angle made by vertices 1 and 2 with âOâ is 60o.

Direction of magnetic field at âOâ due to each segment is same. Since it is symmetric star shape, magnitude will also be same.

Magnetic field due to section BC.

B1=kia sinâ¡60Â°-sinâ¡30Â°=ki2a 3-1

Bnet=12ÃB1=6kia 3-1 and k=Î¼04Ï

Topic: Physics, Magnetic Effects of Current, JEE Main
A symmetrical form of the line of intersection of the planes $x=a y+b$ and $z=c y+d$ is

Options:
A
$\frac{x-b}{a}=\frac{y-1}{1}=\frac{z-d}{c}$

B
$\frac{x-b-a}{a}=\frac{y-1}{1}=\frac{z-d-c}{c}$

C
$\frac{\mathrm{x}-\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{y}-0}{1}=\frac{\mathrm{z}-\mathrm{c}}{\mathrm{d}}$

D
$\frac{\mathrm{x}-\mathrm{b}-\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{y}-1}{0}=\frac{\mathrm{z}-\mathrm{d}-\mathrm{c}}{\mathrm{d}}$

The correct answer is:
$\frac{x-b-a}{a}=\frac{y-1}{1}=\frac{z-d-c}{c}$

Solution:
Given two planes:
$$
x-a y-b=0 \text { and } c y-z+d=0
$$
Let, $l, m, n$ be the direction ratio of the required line. Since the required line is perpendicular to normal of both the plane, therefore $l-a m=0$ and $c m-n=0$ $\Rightarrow l-a m+0 . n=0$ and $0 . l+c m-n=0$
$$
\therefore \quad \frac{l}{a-0}=\frac{m}{0+1}=\frac{n}{c-0}
$$
Hence, d.R of the required line are $a, 1$, $c$.
Hence, options (c) and (d) are rejected. Now, the point $(a+b, 1, c+d)$ satisfy the equation of the two given planes.
$\therefore$ Option (b) is correct.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A symmetrical form of the line of intersection of the planes $x=a y+b$ and $z=c y+d$ is

Options:
A
$\frac{x-b}{a}=\frac{y-1}{1}=\frac{z-d}{c}$

B
$\frac{x-b-a}{a}=\frac{y-1}{1}=\frac{z-d-c}{c}$

C
$\frac{\mathrm{x}-\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{y}-0}{1}=\frac{\mathrm{z}-\mathrm{c}}{\mathrm{d}}$

D
$\frac{\mathrm{x}-\mathrm{b}-\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{y}-1}{0}=\frac{\mathrm{z}-\mathrm{d}-\mathrm{c}}{\mathrm{d}}$

The correct answer is:
$\frac{x-b-a}{a}=\frac{y-1}{1}=\frac{z-d-c}{c}$

Solution:
Given two planes:
$$
x-a y-b=0 \text { and } c y-z+d=0
$$
Let, $l, m, n$ be the direction ratio of the required line. Since the required line is perpendicular to normal of both the plane, therefore $l-a m=0$ and $c m-n=0$ $\Rightarrow l-a m+0 . n=0$ and $0 . l+c m-n=0$
$$
\therefore \quad \frac{l}{a-0}=\frac{m}{0+1}=\frac{n}{c-0}
$$
Hence, d.R of the required line are $a, 1$, $c$.
Hence, options (c) and (d) are rejected. Now, the point $(a+b, 1, c+d)$ satisfy the equation of the two given planes.
$\therefore$ Option (b) is correct.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2014 (12 Apr Online)
A system changes from state X to Y with a change in internal energy measuring to 25 kJ mol-1 , by a reversible path and returns from Y to X by an irreversible path. What will be the net change in internal energy?

Options:

The correct answer is:
0

Solution:
For a cyclic process, net change in internal energy is zero. Change in internal energy does not depend on the path by which final stage is reached.

Topic: Chemistry, Thermodynamics (C), JEE Main
A system consist of a cylinder surrounded by a cylindrical shell. A cylinder is a radius $R$ and is made of material of thermal conductivity $K$, whereas a cylindrical shell has inner radius $R$ and outer radius $2 R$ and is made of material of thermal conductivity twice as that of cylinder. Assuming the system in steady state and negligible heat loss across the cylindrical surface, find the effective thermal conductivity of the system, if the two ends of the combined system are maintained at two different temperatures.

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

The correct answer is:
$\frac{7 K}{4}$

Solution:
$\therefore \quad R_1=\frac{L}{K_1 \cdot \pi R^2}$
and $\begin{aligned} R_2 & =\frac{L}{K_2\left(4 \pi R^2-\pi R^2\right)}=\frac{L}{3 K_2 \pi R^2} \\ R_{\text {eq }} & =\frac{L}{K_{\text {eq }} \cdot 4 \pi R^2}\end{aligned}$
$R_1$ and $R_2$ are in parallel.
$\therefore \quad \frac{1}{R_{\text {eq }}}=\frac{1}{R_1}+\frac{1}{R_2}$
$\begin{aligned} \Rightarrow & & \frac{4 \pi R^2 K_{\text {eq }}}{L} & =\frac{\pi R^2 K_1}{L}+\frac{3 \pi R^2 K_2}{L} \\ & \Rightarrow & 4 K_{\text {eq }} & =K_1+3 K_2\end{aligned}$
On putting values of $K_1$ and $K_2$, we get
$\begin{aligned} 4 K_{\text {eq }} & =K+3(2 K) \quad\left[\because K_1=K, K_2=2 K\right] \\ K_{\text {eq }} & =\frac{K+6 K}{4}=\frac{7 K}{4}\end{aligned}$

Topic: Physics, Thermal Properties of Matter, JIPMER, JIPMER 2016
A system consists of two identical spheres each of mass 1.5 kg and radius 50 cm at the ends of a light rod. The distance between the centres of the two spheres is 5 m. What will be the moment of inertia of the system about an axis perpendicular to the rod passing through its midpoint?

Options:
A 1.905×105 kg m2
B 18.75 kg m2
C 19.05 kg m2
D 1.875×105 kg m2

The correct answer is:
19.05 kg m2

Solution:
I=25MR2+Mx2×2=25×1.5×0.52+1.5×2.52×2=[0.15+9.375]×2=19.05kgm2

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (31 Aug Shift 2)
A system consists of two springs connected in series and each having the spring constant $10 \mathrm{Nm}^{-1}$. The minimum work required to stretch this system by $\mathrm{l} \mathrm{cm}$ in erg is

Options:
A $1500$
B $2000$
C $3000$
D $2500$

The correct answer is:
$2500$

Solution:
For series combination, effective force constant is given by
$\frac{1}{K_{\text {eff }}}=\frac{1}{K_1}+\frac{1}{K_2}$
Here, $\quad K_1=K_2=10 \mathrm{~N} / \mathrm{m}$
$\Rightarrow \quad \frac{1}{K_{\text {eff }}}=\frac{1}{10}+\frac{1}{10}=\frac{1}{5}$
or $\quad K_{\text {eff }}=5 \mathrm{~N} / \mathrm{m}=\frac{5 \times 10^5}{100} \mathrm{dyne} / \mathrm{cm}$
$\Rightarrow \quad K_{\text {eff }}=5 \times 10^3$ dyne $/ \mathrm{cm}$
Work done in stretching this spring,
$W=\frac{1}{2} K_{\mathrm{eff}} x^2=\frac{1}{2} \times 5 \times 10^3 \times 1^2=2500 \mathrm{erg}$

Topic: Physics, Laws of Motion, JEE Main
A system consists of two types of gas molecules A and B having the same number density 2×1025 m-3. The diameter of A and B are 10A and 5A respectively. They suffer collisions at room temperature. The ratio of average distance covered by the molecule A to that of B between two successive collisions is ___________×10-2

Options:

The correct answer is:
25

Solution:
∵ Mean free path

λ=12πd2n

λ1λ2=d22n2 d12n1

=5102=0.25=25×10-2

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2021 (25 Jul Shift 2)
A system containing masses and pulleys connected on an inclined plane is shown in the figure. If the system is in equilibrium then the value of \(m\) is

Options:
A \(1 \mathrm{~kg}\)
B \(0.5 \mathrm{~kg}\)
C \(0.75 \mathrm{~kg}\)
D \(0.25 \mathrm{~kg}\)

The correct answer is:
\(0.5 \mathrm{~kg}\)

Solution:
According to the question, the complete situation is shown in the following figure,
Since, system is in equilibrium, hence
\(\begin{aligned}
T & =m g \quad \ldots (i) \\
T & =\left(2 g-T_1\right) \sin 30^{\circ} \quad \ldots (ii) \\
T_1 & =1 g \quad \ldots (iii)
\end{aligned}\)
\(\therefore\) From Eqs. (ii) and (iii), we get
\(\begin{aligned}
& T=(2 g-\mathbf{l} g) \sin 30^{\circ}=g \sin 30^{\circ} \\
& T=\frac{g}{2} \quad \ldots (iv)
\end{aligned}\)
\(\therefore\) From Eqs. (i) and (iv), we get,
\(\begin{aligned}
m g & =\frac{g}{2} \\
\Rightarrow \quad m & =0.5 \mathrm{~kg}
\end{aligned}\)

Topic: Physics, Laws of Motion, JEE Main
A system does 200 J of work and at the same time absorbs 150 J of heat. The magnitude of the change in internal energy is ______J. (Nearest integer)

Options:

The correct answer is:
50

Solution:
w=-200 J,q=+150:ΔU=q+w

ΔU=150-200=-50 J: magnitude =50 J=|ΔU|

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2021 (25 Jul Shift 2)
A system goes from $A$ to $B$ via two processes I and II as shown in figure. If $\Delta U_1$ and $\Delta U_2$ are the changes in internal energies in the processes I and II respectively, then

Options:
A $\Delta U_{2}\gt\Delta U_1$
B $\Delta U_{2} \lt \Delta U_{\text {\u0964 }}$
C $\Delta U_1=\Delta U_{2}$
D Relation between $\Delta U_{\text {I }}$ and $\Delta U_{2}$ can not be determined

The correct answer is:
$\Delta U_1=\Delta U_{2}$

Solution:
As internal energy is a point function therefore change in internal energy does not depends upon the path followed i.e. $\Delta U_1=\Delta U_{2}$

Topic: Physics, Thermodynamics, JEE Main
A system goes from $A$ to $B$ via two processes I and II as shown in the figure. If $\Delta \mathrm{U}_1$ and $\Delta \mathrm{U}_2$ are the changes in internal energies in the processes I and II respectively, the

Options:
A
$\Delta \mathrm{U}_1=\Delta \mathrm{U}_2$

B
relation between $\Delta \mathrm{U}_1$ and $\Delta \mathrm{U}_2$ can not be determined

C
$\Delta \mathrm{U}_2>\Delta \mathrm{U}_1$

D
$\Delta \mathrm{U}_2 < \Delta \mathrm{U}_1$

The correct answer is:
$\Delta \mathrm{U}_1=\Delta \mathrm{U}_2$

Solution:
Internal energy is state function

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2005
A system of four gates is set up as shown. The 'truth table' corresponding to this system is :

Options:
A

B

C

D

The correct answer is:

Solution:
In the given system all four gate is NOR gate

Topic: Physics, Semiconductors, JEE Main, JEE Main 2013 (23 Apr Online)
A system of three charges are placed as shown in the figure:
If D >> d, the potential energy of the system is best given by:

Options:
A 14πϵ0-q2d-q Q dD2
B 14πϵ0+q2d+q Q dD2
C 14πϵ0-q2d-q Q d2D2
D 14πϵ0-q2d+2q Q dD2

The correct answer is:
14πϵ0-q2d-q Q dD2

Solution:
U=k-qqd+k-qQd2-D+kqQD+d2
=-kq2d-kQqD-d2+KQqD+d2
=-kq2d+kQq-D-d2+ D-d2D2-d24
=-kq2d-kQqdD2-d24
Since, D>>d
U=-kq2d-kQqdD2=14πϵ0-q2d-q Q dD2

Topic: Physics, Electrostatics, JEE Main, JEE Main 2019 (09 Apr Shift 1)
A system of two blocks of masses m=2 kg and M=8 kg is placed on a smooth table as shown in figure. The coefficient of static friction between two blocks is 0.5. The maximum horizontal force F that can be applied to the block of mass M so that the blocks move together will be g=9.8 m s-2

Options:
A 9.8 N
B 39.2 N
C 49 N
D 78.4 N

The correct answer is:
49 N

Solution:
Maximum acceleration of the upper block will be,

amax=μg=0.5×9.8=4.9 m s-2

Therefore, maximum force that can be applied on the lower block is,

Fmax=2+8×4.9=49 N

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2022 (27 Jun Shift 1)
A system of two charges separated by a certain distance apart stores electrical potential
energy. If the distance between them is increased, the potential energy of the system,

Options:
A may increase or decrease
B increases in any case
C remains the same
D decreases in any case

The correct answer is:
may increase or decrease

Solution:
(A)
Potential energy of a system of charges is $U=\left(\frac{1}{4 \Pi \varepsilon_{0}}\right) \frac{q_{1} q_{2}}{2}$
In case of like charges, $\left(q_{1} q_{2}\right)$ is positive
when $r$ increases, $U$ decreases
In case of unlike charges, $\left(q_{1} q_{2}\right)$ is negative
when $r$ increases, $|U|$ decreases
but because of negative sign, $U$ increases.

Topic: Physics, Electrostatics, KCET, KCET 2019
A system of two identical, uniform discs with identical circular cavities, is shown in the figure. Different relevant coordinates are given in the figure. The coordinates of the centre of mass of the system are

Options:
A 3R2, 5R4
B 19R6, 5R2
C R2, R4
D 20R6, 5R2

The correct answer is:
19R6, 5R2

Solution:
Mass of the removed portion
=MπR2×πR22
=M4
xcm=M·0-M4×R+M×7R-M4×8RM-M4+M-M4
=19R6
ycm=M·0-M40+M5R-M45RM-M4+M-M4
=5R2

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A system performs work $\Delta W$ when an amount of heat is $\Delta Q$ added to the system, the corresponding change in the internal energy is $\Delta U$. A unique function of the initial and final states (irrespective of the mode of change) is

Options:
A $\Delta Q$
B $\Delta W$
C $\Delta U$ and $\Delta Q$
D $\Delta U$

The correct answer is:
$\Delta U$

Solution:
the unique function of final and initial state, irrespective of mode of change is the internal energy.

Topic: Physics, Thermodynamics, JEE Main
A system undergoes a reversible adiabatic process. The entropy of the system

Options:
A Remains constant
B May increase or may decrease
C Increases
D Decreases

The correct answer is:
Remains constant

Solution:
In general, the temperature of the system may change during a process. If the process is reversible, the change in entropy is defined as
Sf-Si=â«ifâQT
In an adiabatic reversible process, no heat is given to the system. Therefore, entropy of the system remains constant.

Topic: Physics, Thermodynamics, JEE Main
A 'T' shaped object with dimensions shown in the figure, is lying on a smooth floor. A force $F$ is applied at the point $P$ parallel to $A B$, such that the object has only the translational motion without rotation. Find the location of $P$ with respect to $C$

Options:
A
$\frac{2}{3} \ell$

B
$\frac{3}{2} \ell$

C
$\frac{4}{3} \ell$

D
$\ell$

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

Solution:
P will be the centre of mass of system

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2005
A T shaped uniform density object with dimensions shown in the figure is lying on a smooth floor. A force Fâ is applied at the point P parallel to AB such that the object has only the translational motion without rotation. Find the location of P with respect to C.

Options:
A 43l
B l
C 34l
D 32l

The correct answer is:
43l

Solution:
It is a case of translation motion without rotation. The force should act at the centre of mass
Ycm=mÃ2l+2mÃlm+2m=4l3

( considering point C as origin )

Topic: Physics, Rotational Motion, JEE Main
A table is revolving on its axis at 5 revolutions per second. A sound source of frequency 1000 Hz is fixed on the table at 70 cm from the axis. The minimum frequency heard by a listener standing at a distance from the table will be (speed of sound = 352 m s-1)

Options:
A 1000 Hz

B 1066 Hz

C 941 Hz

D 352 Hz

The correct answer is:
941 Hz

Solution:
vs=rω=r×2πv
=70100×2×227×5=22 m s-1
Frequency is minimum when the source is moving away from the listener.
ν′=u×νu+vs=352×1000352+22=941 Hz

Topic: Physics, Waves and Sound, JEE Main