datavorous/entrance-exam-dataset huggingface dataset:
When a ceiling fan is switched off, its angular velocity falls to half while it makes 36 rotations. How many rotations will it make before coming to rest?

Options:
A 24
B 36
C 18
D 12

The correct answer is:
12

Solution:
From the third equation of angular motion,
ω2=ω02-2αθ (Here,ω=ω02,θ=36×2π)
∴ ω022=ω02-2α×36×2π
or 4×36πα=ω02-ω024
or 4×36πα=3ω024
or α=ω0216×12π ...(i)
According to question again applying the third equation of angular motion
ω2=ω02-2αθ (Here, ω=0)
∴ 0=ω022-2×ω02.θ16×12π
or θ=24π or θ=12×2π
But 2π=1 cycle
So, θ=12 cycle

Topic: Physics, Rotational Motion, JEE Main
When a ceiling fan is switched off, its angular velocity reduces to 50% of its initial value while it makes 36 rotations. If angular retardation of the fan is uniform, then how many more rotations will it make before coming to rest?

Options:
A 18
B 12
C 36
D 48

The correct answer is:
12

Solution:
Ï022=Ï02- 2Î±Î¸1 â 2Î± Î¸1=34 Ï02 ..........(i)
0=Ï022- 2Î±Î¸2 â 2Î±Î¸2=Ï024 .............(ii)
from (i) & (ii) Î¸1Î¸2=3
â Î¸2=Î¸13=363=12 rotation

Topic: Physics, Laws of Motion, JEE Main
When a certain amount of solid A is dissolved in 100 g of water at 25 °C to make a dilute solution, the vapour pressure of the solution is reduced to one-half of that of pure water. The vapour pressure of pure water is 23.76mmHg. The number of moles of solute A added is

Options:

The correct answer is:
5.55

Solution:
We know that relative lowering in vapour pressure is given as

P0-PSPo=solutennSolute+solventn

P0-PSPS=solutensolventn

P0-P02P02=solutensolventn

soluten=solventn=10018=5.55 mol

Topic: Chemistry, Solutions, JEE Main, JEE Main 2022 (27 Jul Shift 2)
When a certain metal surface is illuminated with light of frequency $v$, the stopping potential for photoelectric current is $V_{0}$. When the same surface is illuminated by light of frequency $\frac{v}{2}$ the stopping potential is $\frac{V_{0}}{4} .$ The threshold
frequency for photoelectric, emission is

Options:
A $\frac{v}{6}$
B $\frac{v}{3}$
C $\frac{2 v}{3}$
D $\frac{4 v}{3}$

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

Solution:
Ist case
$$
\begin{array}{l}
h v=h v_{0}+e V_{0} \\
\frac{h v}{2}=h v_{0}+\frac{e V_{0}}{4}
\end{array}
$$
Solving Eqs. (i) and (ii), we get
$$
v_{0}=\frac{v}{3}
$$

Topic: Physics, Dual Nature of Matter, JEE Main
When a certain metallic surface is illuminated with monochromatic light of wavelength λ, the stopping potential for photoelectric current is 3V0 and when the same surface is illuminated with light of wavelength 2λ, the stopping potential is V0. The threshold wavelength of this surface for photoelectric effect is

Options:
A 6λ
B 4λ
C λ4
D λ6

The correct answer is:
4λ

Solution:
Using photoelectric equation, hf-hf0=12mv2=eVs
Or hcλ-hcλ0=eVs
For the first case, hcλ-hcλ0=e3V0 ...(i)
For the second case, hc2λ-hcλ0=eV0 ...(ii)
Solving λ0=4λ

Topic: Physics, Dual Nature of Matter, JEE Main
When a certain photosensitive surface is illuminated with a monochromatic light of frequency ν, the stopping potential of the photo current is -V02. When the surface is illuminated by monochromatic light of frequency ν2, the stopping potential is -V0. The threshold frequency for photoelectric emission is

Options:
A 5v3
B 3v2
C 43v
D 2v

The correct answer is:
3v2

Solution:
hÎ½â=W+V02e
hÎ½â2=W+V0e
On solving, we get, W=32h Î½â
âhÎ½â0=32h Î½â
âÎ½â0=32Î½â

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2019 (12 Jan Shift 2)
When a certain photosensitive surface is illuminated with a monochromatic light of frequency ν, the stopping potential of the photo current is -V02. When the surface is illuminated by monochromatic light of frequency ν2, the stopping potential is -V0. The threshold frequency for photoelectric emission is

Options:
A 5v3
B 3v2
C 43v
D 2v

The correct answer is:
3v2

Solution:
hÎ½â=W+V02e
hÎ½â2=W+V0e
On solving, we get, W=32h Î½â
âhÎ½â0=32h Î½â
âÎ½â0=32Î½â

Topic: Physics, Dual Nature of Matter, JEE Main
When a certain weight is suspended from a long uniform wire, its length increases by one $\mathrm{cm}$. If the same weight is suspended from another wire of the same material and length but having a diameter half of the first one then the increase in length will be

Options:
A $0.5 \mathrm{~cm}$
B $2 \mathrm{~cm}$
C $4 \mathrm{~cm}$
D $8 \mathrm{~cm}$

The correct answer is:
$4 \mathrm{~cm}$

Solution:
$l=\frac{F L}{A Y} \Rightarrow l \propto \frac{1}{r^2}(F, L$ and $Y$ are constant $)$
$\frac{l_2}{l_1}=\left(\frac{r_1}{r_2}\right)^2=(2)^2=4 \Rightarrow l_2=4 l_1=4 \mathrm{~cm}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
When a charged particle moves perpendicular to a uniform magnetic field, then

Options:
A its momentum changes and total energy is same
B Both momentum and its total energy remain the same
C Both momentum and its total energy will change
D total energy changes but momentum remains same

The correct answer is:
its momentum changes and total energy is same

Solution:
When a charged particle moves in a uniform magnetic field normally, then a magnetic force acts on it. Due to this force, it will move in a circular path.
At every point of circle, its velocity changes direction. So, its mumentun also changes bul the Lotal energy will remain same.

Topic: Physics, Magnetic Effects of Current, COMEDK, COMEDK 2016
When a charged particle moving with velocity $\vec{v}$ is subjected to a magnetic field of induction $\vec{B}$, the force on it is non-zero. This implies that:

Options:
A angle between is either zero or $180^{\circ}$
B angle between is necessarily $90^{\circ}$
C angle between can have any value other than $90^{\circ}$
D angle between can have any value other than zero and $180^{\circ}$

The correct answer is:
angle between can have any value other than zero and $180^{\circ}$

Solution:
$$
\begin{aligned}
\vec{F} & =q(\vec{V} \times \vec{B}) \\
& =q v B \sin \theta
\end{aligned}
$$
Therefore when $\theta=0^{\circ}$ or $\theta=180^{\circ}$
$$
F=0
$$

Topic: Physics, Magnetic Effects of Current, JEE Main
When a chlorine gas is treated with barium hydroxide it produces [X], barium chloride and water, when sulphuric acid is
added to \( [\mathrm{X}] \), it produced \( [\mathrm{Y}] \) and barium sulphate. When \( \mathrm{Y} \) is heated above \( 365 \mathrm{~K} \), it produces \( \mathrm{Z} \) and \( \mathrm{H} 2 \mathrm{O} \). [ \( \mathrm{Y} \) ] and [ \( \mathrm{Z} \) ]
refers to

Options:
A \( \mathrm{HCIO}_{4}, \mathrm{CIO}_{2} \)
B \( \mathrm{HClO}_{3}, \mathrm{CIO}_{2} \)
C \( \mathrm{HCIO}_{3}, \mathrm{CIO}_{6} \)
D \( \mathrm{HCIO}_{4}, \mathrm{CI}_{2} \mathrm{O}_{7} \)

The correct answer is:
\( \mathrm{HClO}_{3}, \mathrm{CIO}_{2} \)

Solution:
This problem is based on the concept of preparation and properties of oxo acid of chlorine. to solve this problem, draw the road map of problem given followed by choosing the correct option out of given four.
6 Cl 2 + 2 Ba OH 2 → Ba ClO 3 2 X + 5 BaCl 2 + 6 H 2 O
Ba ClO 3 2 + H 2 SO 4 → 2 HClO 3 X + BaSO 4
2 HClO 3 → Δ > 3 6 5 K Δ 2 ClO 2 Z + H 2 O + 1 2 O 2

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
When a cluster of genes show linkage behaviour they:

Options:
A do not show a chromosome map
B show recombination during meiosis
C donot show independent assortment
D induce cell division

The correct answer is:
donot show independent assortment

Solution:
Genes present near to each other have less probability of being separated onto different chromatids during chromosomal crossover. Therefore, when a cluster of genes show linkage behaviour, they do not show independent assortment.
Related Theory
The strength of linkage between two genes is inversely proportional to the distance between them in the chromosome i.e., it decreases with increase in distance between the genes. It ensures the maintenance of parental trait in the offspring and reduces the chance of creation of variability with sexual reproduction.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2003
When a conducting slab fills the space between the two plates of a capacitor, its capacitance

Options:
A becomes infinite.
B becomes four times the original one.
C remains same.
D becomes zero.

The correct answer is:
becomes infinite.

Solution:
Capacitance is nothing but charge to potential difference ratio. Capacitance C is given by

C=QV

where Q and V are charge and voltage, respectively.

When a conducting slab is put in between, then potential difference V between the plates of the capacitor becomes zero. So, the capacitance becomes infinite.

Topic: Physics, Capacitance, JEE Main
When a conservative force does positive work on a body, the potential energy of the body

Options:
A increases
B decreases
C remains unaltered.
D None

The correct answer is:
decreases

Solution:
Correct Option is : (B)
decreases

Topic: Chemistry, Work Power Energy, NEET
When a current of 5 mA is passed through a galvanometer having a coil of resistance 15 Ω, it shows full-scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0-10 V is:

Options:
A 4.005×103 Ω
B 1.985×103 Ω
C 2.045×103 Ω
D 2.535×103 Ω

The correct answer is:
1.985×103 Ω

Solution:
∆V=IgRg+RH
10=5×10-3x+15
x=1.985×103 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2017 (02 Apr)
When a Daniel cell is connected in the secondary circuit of a potentiometer, the balancing length is found to be 540 cm. If the balancing length becomes 500 cm when the cell is short-circuited with 1 Î©, the internal resistance of the cell (in Î©) is

Options:

The correct answer is:
0.08

Solution:
Internal resistance,
r=Rl1-l2l2=1540-500500=40500=0.08 Î©

Topic: Physics, Current Electricity, JEE Main
When a DC voltage is applied at the two ends of a circuit kept in closed box, it is observed that the current gradually increases from zero to a certain value and then remains constant. What do you think that the circuit contains?

Options:
A A resistor alone
B A capacitor alone
C A resistor and an inductor in series
D A resistor and a capacitor in series

The correct answer is:
A resistor and an inductor in series

Solution:
Hint:
LR circuit transient

Topic: Physics, Electromagnetic Induction, WBJEE, WBJEE 2020
When a diode is forward biased, it has a voltage drop of 0.5 V . the safe limit of current through the diode is 10 mA. If a battery of emf 1.5 V is used in the circuit, the value of minimum resistance to be connected in series with the diode so that the current does not exceed the safe limit is :

Options:
A 300 Ω
B 50 Ω
C 100 Ω
D 200 Ω

The correct answer is:
100 Ω

Solution:
Vdiode=0.5 volt

VR=1.5-0.5=1 volt

iR=1

R=1i=110-2=100 Ω

Topic: Physics, Semiconductors, JEE Main, JEE Main 2020 (03 Sep Shift 1)
When a disc rotates with uniform angular velocity, which of the following is not true?

Options:
A
The sense of rotation remains same

B
The orientation of the axis of rotation remains same

C
The speed of rotation is non-zero and remains same

D
The angular acceleration is non-zero and remains same

The correct answer is:
The angular acceleration is non-zero and remains same

Solution:
Angular acceleration $\alpha=\frac{d \omega}{d t}$ where $\omega$ is angular velocity of the disc and is uniform or constant.
$$
\alpha=\frac{d \omega}{d t}=0
$$
Hence, angular acceleration is zero.

Topic: Physics, System of Particles and Rotational Motion
When a fluoride is heated with concentration H2SO4 in a glass tube and if a drop of water is held at the mouth of the glass tube, a white deposit formed is of

Options:
A H2SiF6
B SiO2
C H2SiO3
D SiF4+H2F2

The correct answer is:
H2SiO3

Solution:
CaF2+H2SO4conc.âCaSO4+H2F2
SiO2+2H2F2âSiF4+2H2O
3SiF 4  +  3H 2 O → H 2 SiO 3 Silicic acid (Gelatinous mass)  +   2H 2 SiF 6 (Hydrofluoro silicic acid)

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
When a foreign DNA is introduced into an organism, how is it maintained in the host and how is it transferred to the progeny of the organisms?

Options:

None

Solution:
Foreign DNA is usually ligated to a plasmid vector and then introduced in the host. As plasmid replicates, and makes multiple copies of itself, the foreign gene gets replicated receives as well, and several copies of it are made. When the host organism divides, its progeny also receives the plasmid. DNA containing the foreign gene.

Topic: Biology, Biotechnology and Its Applications
When a freshly precipitated substance is converted into a colloidal solution with the help of a third substance, the process is known as

Options:
A Coagulation
B Peptization
C Electrodispersion
D Dialysis

The correct answer is:
Peptization

Solution:
Correct option is 2. Peptization
Peptization or Deflocculation is the process responsible for the formation of converting precipitate into colloid by shaking with it an electrolyte.

Topic: Chemistry, Surface Chemistry, JEE Main
When AgNO3 solution is added in excess to 1M solution of CoCl3.XNH3 one mole of AgCl is formed? What is the value of 'X'? (Assume that the co-ordination number is 6)

Options:
A 1
B 4
C 3
D 2

The correct answer is:
4

Solution:
CoCl3.XNH3 gives 1 mole of AgCl. So one Clâ ion is outside the complex. So 2Clâ and 4NH3 satisfy coordination number 6.

From the given data the possible complex is
[ Co ( NH 3 ) 4 Cl 2 ]Cl
So x is 4

Topic: Chemistry, Coordination Compounds, JEE Main
When a gas filled in a closed vessel is heated by raising the temperature by 1°C, its pressure increases by 0.4% The initial temperature of the gas is _____ K.

Options:

The correct answer is:
250

Solution:
From ideal gas equation, PV=nRT, where, n is amount of substance, R is ideal gas constant and V is volume.

In closed vessel volume remains constant.

So, dPP×100=dTT×100

0.4=1T×100

⇒T=250 K

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2022 (25 Jun Shift 2)
When a gas is bubbled through water at 298 K, a very dilute solution of gas is obtained. Henry's law constant for the gas is 100 kbar. If gas exerts a pressure of 1 bar, the number of moles of gas dissolved in 1 litre of water is x ×  10 −5 . Find the value of x here?
Report your answer upto two decimal places.

Options:

The correct answer is:
55.55

Solution:
p=KHÃx
x=pKH=1100Ã103=1Ã10-5
Mole of fraction =Mole of gasTotal moles
Moles of H 2 O =  1000 18  = 55.55  ∴1 L = 1000 g  ×  10 −5
Mole fraction =xx+55.55 55.55>>>x
So 10-5=x55.55
or x=55.55Ã10-5

Topic: Chemistry, Solutions, JEE Main
When a glass prism of refracting angle $60^{\circ}$ is immersed in a liquid its angle of minimum deviation is $30^{\circ}$. The critical angle of glass with respect to the liquid medium is

Options:
A $42^{\circ}$
B $45^{\circ}$
C $50^{\circ}$
D $52^{\circ}$

The correct answer is:
$45^{\circ}$

Solution:
$A=60^{\circ}, \delta=30^{\circ}$
Refractive index of prism with respect to liquid,
$\begin{aligned} \mu & =\frac{\sin \frac{A+\delta_n}{2}}{\sin \frac{A}{2}}=\frac{\sin \frac{\left(60^{\circ}+30^{\circ}\right)}{2}}{\sin \frac{60^{\circ}}{2}} \\ & =\frac{\sin 45^{\circ}}{\sin 30^{\circ}}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}=\sqrt{2}\end{aligned}$
$\therefore$ Critical angle $=\sin ^{-1}\left(\frac{1}{\mu}\right)=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$
$=\sin ^{-1}\left(\sin 45^{\circ}\right)=45^{\circ}$

Topic: Physics, Ray Optics, JEE Main
When a hydrocarbon A undergoes combustion in the presence of air, it requires 9.5 equivalents of oxygen and produces 3 equivalents of water. What is the molecular formula of A?

Options:
A C8H6
B C9H9
C C6H6
D C9H6

The correct answer is:
C8H6

Solution:
CxHy+x+y4O2→xCO2+y2H2O

Number of equivalents of O2 = Number of equivalents of H2O

Number of equivalents of H2O = y2=3 ⇒ y = 6

Number of equivalents of O2 = x+y4= 9.5

⇒x + 64 = 9.5

⇒ x + 1.5 = 9.5

⇒ x = 9.5 - 1.5 = 8

⇒x=8, y=6.

So, the formula is C8H6.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2023 (29 Jan Shift 2)
When a hydrocarbon A undergoes complete combustion it requires 11 equivalents of oxygen and produces 4 equivalents of water. What is the molecular formula of A?

Options:
A C11H8
B C11H4
C C5H8
D C9H8

The correct answer is:
C9H8

Solution:
By using the formula of combustion analysis:

CxHy+x+y4O2→CO2+y2H2O

Given, x+y4a=11a

x+y4=11

y2a=4a

y=8,

then x+84=11

x = 11-84

x=9

Formula =C9H8

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2023 (31 Jan Shift 2)
When a hydrogen atom emits a photon in going from \( \mathrm{n}=5 \) to \( \mathrm{n}=1 \), its recoil speed is almost

Options:
A \( 10^{-4} \mathrm{~m} / \mathrm{s} \)
B \( 2 \times 10^{-2} \mathrm{~m} / \mathrm{s} \)
C \( 4 \mathrm{~m} / \mathrm{s} \)
D \( 8 \times 10^{-2} \mathrm{~m} / \mathrm{s} \)

The correct answer is:
\( 4 \mathrm{~m} / \mathrm{s} \)

Solution:
Energy of photon emitted \(=13.6\left(\frac{1}{1}-\frac{1}{25}\right) \mathrm{eV}=13.056 \mathrm{eV}\)
Now, momentum of the photon \(=\frac{13.056 \times 1.6 \times 10^{19}}{3 \times 10^{8}}\)
Momentum of the hydrogen \(=\) same as (1)
Let recoil speed be V.
\(\mathrm{V}\left(1841 \times 9.1 \times 10^{-31}\right)=\frac{13.056 \times 1.6 \times 10^{19}}{3 \times 10^{8}}\)
\(\Rightarrow \mathrm{V} \approx 4 \mathrm{~m} / \mathrm{s}\)

Topic: Physics, Atomic Physics, JEE Main
When a hydrogen atom going from n=2 to n=1 emits a photon, its recoil speed is x5 m s-1. Where x=_______. (Use: mass of hydrogen atom =1.6×10-27 kg, charge of electron e=1.6×10-19 C)

Options:

The correct answer is:
17

Solution:
According to the above diagram, the difference in energy between the two states is given by

ΔE=-3.4 eV--13.6 eV=10.2 eV

The formula to calculate the recoil speed is given by

v=∆Emc ...1

From equation (1), it follows that

v=10.2 eV×1.6×10-19 J1 eV1.6×10-27 kg×3×108 m s-1=3.4 m s-1=175 m s-1

Therefore, x=17.

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2024 (29 Jan Shift 1)
When a large bubble rises from bottom of a water lake to its surface, then its radius doubles. If the atmospheric pressure is equal to the pressure of height H of a certain water column, then the depth of lake will be

Options:
A 2H
B H
C 7H
D 4H

The correct answer is:
7H

Solution:
When a large bubble rises from bottom of water lake to its surface, then its radius becomes double.
i.e., r2=2r1
Since, volume V∝r3
∴V2V1=r2r13=2r1r1=8
⇒V2=8V1 … (i)
From Boyle’s law,
p1V1=p2V2
Where, p2 = atmospheric pressure.
p1V1=p28V1 [from Eq. (ii)]
p2=p18
where, p2= atmospheric pressure
and p1 = pressure at depth d.
∴p1=patomspheric+pgd
p1=p2+pgd … (iii)
From Eqs. (ii) and (iii), we have
p2=p2+pgd8
7p2=pgd
7.pgH=pgd
7H=d⇒d=7H

Topic: Physics, Mechanical Properties of Fluids, JEE Main
When a large bubble rises from the bottom of a lake to the surface, its radius doubles. If atmospheric pressure is equal to that of a column of the water height H, then the depth of the lake is (assume temperature remains constant)

Options:
A H
B 2H
C 7H
D 8H

The correct answer is:
7H

Solution:
P 1 V 1 = P 2 V 2
⇒ (H+h)ρg× 4 3 π r 3 =H× 4 3 π (2r) 3
⇒H+h=8H
∴ h=7H

Topic: Physics, Mechanical Properties of Fluids, JEE Main
When a large bubble rises from the bottom of a lake to the surface, its radius doubles. If atmospheric pressure is equal to that of column of water height $H$, then the depth of lake is

Options:
A $\mathrm{H}$
B $2 \mathrm{H}$
C $7\mathrm{H}$
D $8 \mathrm{H}$

The correct answer is:
$7\mathrm{H}$

Solution:
$P_1 V_1=P_2 V_2 \Rightarrow(H+h) \rho g \times \frac{4}{3} \pi r^3=H \times \frac{4}{3} \pi(2 r)^3$
$\Rightarrow H+h=8 H . \therefore h=7 H$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
When a light of wavelength 300 nm falls on the photoelectric emitter, photoelectrons are just liberated. For another emitter, however light of 600 nm wavelength is sufficient for creating photoemission. What is the ratio of the work function Ï 0 of the two emitters?

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

The correct answer is:
2 : 1

Solution:
K.E max = hc λ - Ï
Ï 1 = 1240 ev-nm 300 nm = 4.13 eV
Ï 2 = 1240 ev-nm 600 nm = 2.06 eV
∴ Ï 1 Ï 2 = 4 2 = 2 1

Topic: Physics, Dual Nature of Matter, JEE Main
When a liquid boils, there is

Options:
A An increase in entropy
B A decrease in entropy
C An increase in heat of vaporization
D An increase in free energy

The correct answer is:
An increase in entropy

Solution:
Liquid $\longrightarrow$ Vapour, entropy increases.

Topic: Chemistry, Thermodynamics (C), JEE Main
When a liquid in a glass vessel is heated, its apparent expansion coefficient is 10.30Ã10-4 â-1. When the same liquid is heated in a metal vessel, its apparent expansion coefficient is 10.06Ã10-4 â-1. If the coefficient of linear expansion of glass = 9Ã10-6 â-1, what is the coefficient of linear expansion of metal?

Options:
A 51 Ã10-6 â-1
B 17 Ã10-6 â-1
C 25Â Ã10-6 â-1
D 43Â Ã10-6 â-1

The correct answer is:
17 Ã10-6 â-1

Solution:
Here, Î³ag=10.30 Ã10-4 â-1
Î³am=10.06Ã10-4 â-1
Î±g=9Ã10-6 â-1, Î±m= ?
Now, Î³r= Î³ag+Î³g= Î³am+Î³m
â´10.30Ã10-4+3Ã9Ã10-6=10.06Ã10-4+Î³m [âµÎ³g=3ÃÎ±a]
â´ Î³m=10.30+0.27-10.0610-4=0.51Ã10-4
Î±m=13Î³m= 0.51Ã10-43
= 0.17Ã10-4=17Ã10-6 â-1

Topic: Physics, Thermal Properties of Matter, JEE Main
When a long spring is stretched by $2 \mathrm{~cm}$, its potential energy is $U$. If the spring is stretched by $10 \mathrm{~cm}$, the potential energy stored it will be:

Options:
A $\frac{U}{5}$
B $5 U$
C $10 U$
D $25 U$

The correct answer is:
$25 U$

Solution:
$U=\frac{1}{2} K(2)^2$
$U^{\prime}=\frac{1}{2} K(10)^2$
From above equations
$U=25 \mathrm{~U}$

Topic: Physics, Work Power Energy, JEE Main
When a manganous salt is fused with a mixture of $\mathrm{KNO}_3$ and solid $\mathrm{NaOH}$ the oxidation number of Mn changes from +2 to

Options:
A +4
B +3
C +6
D +7

The correct answer is:
+6

Solution:
$$
\text { Hints : } \mathrm{Mn}^{+2}+\mathrm{NO}_3^{-}+\mathrm{O} \overline{\mathrm{H}} \rightarrow \stackrel{(+6)}{\mathrm{M} \mathrm{nO}_4^{-2}}+\mathrm{H}_2 \mathrm{O}
$$

Topic: Chemistry, d and f Block Elements, JEE Main
When a mass is rotated in a plane about a fixed point, its angular momentum is directed along the-

Options:
A radius
B tangent to orbit
C axis of rotation
D incline at an angle of 45° to plane of rotation

The correct answer is:
axis of rotation

Solution:
Angular momentum is directed along the axis of rotation.

Topic: Physics, Rotational Motion, JEE Main
When a mercury drop of radius 'R', breaks into 'n' droplets of equal size, the radius 'r' of each droplet is

Options:
A $\mathrm{r}=\frac{\mathrm{R}}{\sqrt{\mathrm{n}}}$
B $\mathrm{r}=\frac{\mathrm{R}}{\mathrm{n}}$
C $\mathrm{r}=\frac{\mathrm{R}}{\mathrm{n}^{\frac{1}{3}}}$
D $\mathrm{r}=\mathrm{R} \mathrm{n}^{\frac{1}{3}}$

The correct answer is:
$\mathrm{r}=\frac{\mathrm{R}}{\mathrm{n}^{\frac{1}{3}}}$

Solution:
Volume remains constant
$\therefore \mathrm{n} \times \frac{4}{3} \pi \mathrm{r}^{3}=\frac{4}{3} \pi \mathrm{R}^{3}$
$\therefore \mathrm{nr}^{3}=\mathrm{R}^{3}$
$\therefore \quad n^{\frac{1}{3}} r=R \quad \therefore r=\frac{R}{n^{\frac{1}{3}}}$

Topic: Physics, Mechanical Properties of Fluids, MHT CET, MHT CET 2020 (12 Oct Shift 2)
When a metal conductor connected to left gap of a meter bridge is heated, the balancing point

Options:
A shifts towards left
B remains unchanged
C shifts to the centre
D shifts towards right

The correct answer is:
shifts towards right

Solution:
We know that, resistance of unknown wire can be determined by using meter bridge as
$$
\frac{R}{S}=\frac{l}{100-l} \Rightarrow S=\frac{R(100-l)}{l}
$$
where $R=$ known resistance, $l=$ balancing length and $S=$ unknown resistance.
When a metal conductor of resistance $R$ connected to left gap is heated, then its resistance increases i.e. $R$ increases. Hence, from Eq. (i), we see that when $R$ will increase, then $(100-l)$ will decrease also, i.e $l$ will increase. Therefore, balancing point will shift towards right.

Topic: Physics, Current Electricity, KCET, KCET 2022