datavorous/entrance-exam-dataset huggingface dataset:
An air bubble of radius $0.1 \mathrm{~cm}$ is in a liquid having surface tension $0.06 \mathrm{~N} / \mathrm{m}$ and density $10^3 \mathrm{~kg} / \mathrm{m}^3$. The pressure inside the bubble is $1100 \mathrm{Nm}^{-2}$ greater than the atmospheric pressure. At what depth is the bubble below the surface of the liquid? $\left(\mathrm{g}=9.8 \mathrm{~ms}^{-2}\right)$

Options:
A
$0.1 \mathrm{~m}$

B
$0.15 \mathrm{~m}$

C
$0.20 \mathrm{~m}$

D
$0.25 \mathrm{~m}$

The correct answer is:
$0.1 \mathrm{~m}$

Solution:
Given: Radius of air bubble,
$$
r=0.1 \mathrm{~cm}=10^{-3} \mathrm{~m}
$$
Surface tension of liquid,
$$
S=0.06 \mathrm{~N} / \mathrm{m}=6 \times 10^{-2} \mathrm{~N} / \mathrm{m}
$$
Density of liquid, $\rho=10^3 \mathrm{~kg} / \mathrm{m}^3$
Excess pressure inside the bubble,
$$
\rho_{\text {exe }}=1100 \mathrm{Nm}^{-2}
$$
Depth of bubble below the liquid surface, $h=$ ?
As we know,
$$
\begin{aligned}
&\rho_{\text {Excess }}=h \rho g+\frac{2 s}{r} \\
&\Rightarrow \quad 1100=h \times 10^3 \times 9.8+\frac{2 \times 6 \times 10^{-2}}{10^{-3}} \\
&\Rightarrow \quad 1100=9800 h+120 \\
&\Rightarrow \quad 9800 h=1100-120 \\
&\Rightarrow \quad h=\frac{980}{9800}=0.1 \mathrm{~m}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2014 (11 Apr Online)
An air column closed in a tube sealed at one end by a Hg column having height h . When the tube is placed with open end down, the height of the air column is l1. If the tube is turned so that its open end is at the top, the height of the air column is l 2 . What is the atmospheric pressure P0-

Options:
A P0=hl1+l2l2-l1Â cmÂ ofÂ Hg
B P0=hl1-l2l1+l2Â cmÂ ofÂ Hg
C 76Â cmÂ ofÂ Hg
D P0=hl1+l2l1-l2cm of Hg .

The correct answer is:
P0=hl1+l2l1-l2cm of Hg .

Solution:
Let P 0 be the value at atmospheric pressure
In case I:
Pgas+h cm of Hg=P0 cm of Hg
In case ii
Pgas=P0 cm of Hg+h cm of Hg
By applying Boyle's law between I & II
P0 â hÃl1=P0+hÃl2
or P0l1- hl1=P0l2+hl2
or P0 l1-l2=h l1+l2
or P0=hl1+l2l1-l2

Topic: Chemistry, States of Matter, JEE Main
An air column in a pipe, which is closed at one end, will be in resonance with a vibrating body of frequency $166 \mathrm{~Hz}$, if the length of the air column is

Options:
A 2.00 m
B 1.50 m
C 1.00 m
D 0.50 m

The correct answer is:
0.50 m

Solution:
For closed pipe $\mathrm{n} 1=\mathrm{v} 41 \Rightarrow 1=\mathrm{v} 4 \mathrm{n}=3324 \times 166=0.5 \mathrm{~m}$.

Topic: Physics, Waves and Sound, JEE Main
An air column in a pipe, which is closed at one end, will be in resonance wtih a vibrating tuning fork of frequency $264 \mathrm{~Hz}$ if the length of the column in $\mathrm{cm}$ is (velocity of sound $=330 \mathrm{~m} / \mathrm{s}$ )

Options:
A
$125.00$

B
$93.75$

C
$62.50$

D
$187.50$

The correct answer is:
$93.75$

Solution:
Given : Frequency of tuning fork, $n=264 \mathrm{~Hz}$
Length of column $L=$ ?
For closed organ pipe
$$
\begin{aligned}
& n=\frac{v}{4 l} \\
& \Rightarrow l=\frac{v}{4 n}=\frac{330}{4 \times 264}=0.3125 \\
& \text { or, } l=0.3125 \times 100=31.25 \mathrm{~cm}
\end{aligned}
$$
In case of closed organ pipe only odd harmonics are possible.
Therefore value of $l$ will be $(2 n-1) l$ Hence option (b) i.e. $3 \times 31.25=93.75 \mathrm{~cm}$ is correct.

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2012 (26 May Online)
An air column in a tube $32 \mathrm{~cm}$ long, closed at one end, is in resonance with a tuning fork. The air column in another tube, open at both ends, of length $66 \mathrm{~cm}$ is in resonance with
another tuning fork. When these two tuning forks are sounded together, they produce 8 beats per second. Then the frequencies of the two tuning forks are, (Consider fundamental frequencies only)

Options:
A $250 \mathrm{~Hz}, 258 \mathrm{~Hz}$
B $240 \mathrm{~Hz}, 248 \mathrm{~Hz}$
C $264 \mathrm{~Hz}, 256 \mathrm{~Hz}$
D $280 \mathrm{~Hz}, 272 \mathrm{~Hz}$

The correct answer is:
$264 \mathrm{~Hz}, 256 \mathrm{~Hz}$

Solution:
We knows frequency of a closed end an column
$$
n_1=\frac{v}{4 / 1}
$$
We knows frequency of a open end an column
$$
n_2=\frac{v}{2 l_2}
$$
Given, $l_1=32 \mathrm{~cm}, I_2=66 \mathrm{~cm}$
and $\quad n_1-n_2=8$ heat $/ \mathrm{s}$
So, $\quad n_1=\frac{v}{4 \times 32}=\frac{v}{128}$
and $\quad n_2=\frac{v}{2 \times 66}=\frac{v}{132}$
In given condition,
$$
\begin{aligned}
& \frac{v}{128}-\frac{v}{132}=8 \\
& v=8448 \times 4 \\
& v=33792 \\
& \text { Hence, } \quad n_1=\frac{33792}{128} \\
& n_1=264 \mathrm{~Hz} \\
& \text { and } \\
& n_2=\frac{33792}{132} \\
& n_2=256 \mathrm{~Hz} \\
&
\end{aligned}
$$

Topic: Physics, Waves and Sound, JEE Main
An aircraft has three engines $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$. The aircraft crashes if all the three engines fail. The probabilities of failure are $0.03,0.02$ and $0.05$ for engines $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ respectively. What is the probability that the aircraft will not crash?

Options:
A $0.00003$
B $0.90$
C $0.99997$
D $0.90307$

The correct answer is:
$0.99997$

Solution:
Since, probabilities of failure for engines A, B and C $\mathrm{P}(\mathrm{A}), \mathrm{P}(\mathrm{B})$ and $\mathrm{P}(\mathrm{C})$ are $0.03,0.02$ and $0.05$ respectively.
The aircraft will crash only when all the three engine fail. So, probability that it crashes $=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B}) \cdot \mathrm{P}(\mathrm{C})$
$=0.03 \times 0.02 \times 0.05$
$=0.00003$
Hence, the probability that the aircraft will not crash, $=1-0.00003$
$=0.99997$

Topic: Mathematics, Probability, JEE Main
An aircraft is moving with uniform velocity 150 m/s in the space. If all forces acting on it are balanced, then it will

Options:
A Keep moving with same velocity
B Remain floating at its place
C Escape in space
D Fall down on earth

The correct answer is:
Keep moving with same velocity

Solution:
As all the forces acting on the aircraft are balanced, so the net force on it will be zero, i.e., no external force act on it. Thus, the aircraft will keep moving with the same velocity of 150 m/s in the space.

Topic: Physics, Laws of Motion, MHT CET, MHT CET 2019 (Shift 2)
An aircraft loops a vertical loop of radius R = 500 m with a constant velocity v=360 kmh-1. Find the weight of the flier of mass m=70 kg in the lower, upper, and middle points of the loop.

Options:
A 2.1 kN, 0.7 kN, 1.5 kN
B 0.7 kN, 1.5 kN, 2.1 kN
C 1.5 kN, 2.1 kN, 0.7 kN
D none of these

The correct answer is:
2.1 kN, 0.7 kN, 1.5 kN

Solution:
Forces on the flier in the frame of aircaraft (non inertial reference frame) is:
(i) Normal reaction (N)
(ii)Pseudo force m vâ¡ 2 Râ¡
(iii) Gravitational force d(mg)
According to problem,
vT = vb = vm = v
The net force on flier at every point should be zero in the frame of aircraft. For top point T,
N T + mg = mvâ¡ T 2 Râ¡ â´ N T = mvâ¡ 2 Râ¡ - mg
= 0.7 kN
For bottom point, Nâ¡ Bâ¡ = mg + mvâ¡ b 2 Râ¡ = mg + mvâ¡ 2 Râ¡
= 2.1 kN.
For middle point, N m = mv 2 R 2 + mg 2 = 1 . 5 k Nâ¡

Topic: Physics, Laws of Motion, JEE Main
An air-filled parallel plate capacitor has capacitance C. The capacitor is connected through a resistor to a voltage source providing a constant potential difference V

A dielectric plate with a dielectric constant K is inserted into the capacitor, filling it completely. After the equilibrium is established plate is quickly removed. Find the amount of heat generated in the resistor by the time, the equilibrium is re-established.

Options:
A CV2 K- 1
B 12CV2 K- 1
C CV2 K- 12
D 12CV2K2- 1

The correct answer is:
12CV2 K- 1

Solution:
After the insertion of the slab and steady-state is reached, the energy stored in the capacitor is

Ui=12KCV2

After the removal of the lab and steady-state is reached, the energy stored in the capacitor is

Uf=12CV2

Heat=12CK- 1V2

Topic: Physics, Capacitance, JEE Main
An air-filled parallel plate capacitor has capacitance C. The capacitor is connected through a resistor to a voltage source providing a constant potential difference V

A dielectric plate with a dielectric constant K is inserted into the capacitor, filling it completely. After the equilibrium is established plate is quickly removed. Find the amount of heat generated in the resistor by the time, the equilibrium is re-established.

Options:
A CV2 K- 1
B 12CV2 K- 1
C CV2 K- 12
D 12CV2K2- 1

The correct answer is:
12CV2 K- 1

Solution:
After the insertion of the slab and steady-state is reached, the energy stored in the capacitor is

Ui=12KCV2

After the removal of the lab and steady-state is reached, the energy stored in the capacitor is

Uf=12CV2

Heat=12CK- 1V2

Topic: Physics, Current Electricity, JEE Main
An alcohol on oxidation is found to give $\mathrm{CH}_3 \mathrm{COOH}$ and $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}$. The structure of the alcohol is [BIT 1990]

Options:
A $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}$
B $\left(\mathrm{CH}_3\right)_2 \mathrm{C}(\mathrm{OH}) \mathrm{CH}_2 \mathrm{CH}_3$
C $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHOHCH}_3$
D $\mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$

The correct answer is:
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHOHCH}_3$

Solution:
\(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \xrightarrow[\text { Iodofomm test }]{\mathrm{NaOH} / \mathrm{L}_2} \mathrm{CHI}_3\) (yellow ppt) \(\mathrm{CH}_3 \mathrm{OH} \xrightarrow{\mathrm{NcOH} / \mathrm{I}_2} \mathrm{No~ppt}\)

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
An alcohol X on heating with concentrated H2SO4 gives an alkene Y which can show geometrical isomerism. The alcohol X is

Options:
A CH3CH2CHOHCH3
B CH32COHCHCH32
C CH33COH
D CH32COHCH2CH3

The correct answer is:
CH3CH2CHOHCH3

Solution:
CH3CH2CHOHCH3X→conc.H2SO4CH3CH=CHCH3Y

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
An aldohexose (e.g., glucose) and 2-oxohexose (e.g., fructose) can be distinguished with the help of

Options:
A Tollen's reagent
B Fehling's solution
C Benedict solution
D Br 2  +  H 2 O

The correct answer is:
Br 2  +  H 2 O

Solution:
an aldohexose on reaction with Br 2  +  H 2 O forms glyconic acids, HOCH2CHOH4COOH , with open-chain glucose.

Topic: Chemistry, Biomolecules, JEE Main
An aliphatic amine with molar mass \( 73 \mathrm{~g} \) on heating with excess of \( \mathrm{CH}_{3} \mathrm{I} \) gave a quaternary salt. Which of the following amine follow this data?

Options:
A N,N-diethylethanamine
B N-methylpropanamine
C 2-Propanamine
D Neopentylamine

The correct answer is:
N-methylpropanamine

Solution:
C H 3 C H 2 C H 2 NHC H 3 â Excess C H 3 I [ C H 3 âC H 2 C H 2 N ( C H 3 ) 3 ] + I -
Salt

Topic: Chemistry, Amines, JEE Main
An alkali is titrated against acid with methyl orange as an indicator, which of the following is a correct combination?

Options:
A BaseAcidEnd pointStrongStrongPink to colourless
B BaseAcidEndÂ pointWeakSrtongColourlessÂ toÂ pink
C BaseAcidEndÂ pointStrongstrongPinkishÂ redÂ toÂ yellow
D BaseAcidEnd pointWeakStrongYellow to pinkish red

The correct answer is:
BaseAcidEnd pointWeakStrongYellow to pinkish red

Solution:
The pH range between 3.1 (red) and 4.4 (yellow) is the colour-change interval of methyl orange.

Weak base has pH > 7. So, when methyl orange indicator is added the solution turns yellow & if the solution is titrated against strong acid it turns pinkish red at the end point i.e., pH <3.1.

Topic: Chemistry, Redox Reactions, JEE Main
An alkali is titrated against acid with methyl orange as an indicator, which of the following is a correct combination?

Options:
A BaseAcidEnd pointStrongStrongPink to colourless
B BaseAcidEndÂ pointWeakSrtongColourlessÂ toÂ pink
C BaseAcidEndÂ pointStrongstrongPinkishÂ redÂ toÂ yellow
D BaseAcidEnd pointWeakStrongYellow to pinkish red

The correct answer is:
BaseAcidEnd pointWeakStrongYellow to pinkish red

Solution:
The pH range between 3.1 (red) and 4.4 (yellow) is the colour-change interval of methyl orange.

Weak base has pH > 7. So, when methyl orange indicator is added the solution turns yellow & if the solution is titrated against strong acid it turns pinkish red at the end point i.e., pH <3.1.

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2018 (08 Apr)
An alkali is titrated against an acid with methyl orange as indicator, which of the following is a correct combination?

Options:
A

Base	Acid	End point
Strong	Strong	Pink to colourless

B

Base	Acid	End point
Weak	Strong	Colourless to pink

C

Base	Acid	End point
Strong	Strong	Pinkish red to yellow

D

Base	Acid	End point
Weak	Strong	Yellow to pinkish red

The correct answer is:

Base	Acid	End point
Weak	Strong	Yellow to pinkish red

Solution:
Methyl orange is weak organic base. It is used in the titration of weak base vs strong acid.
MeOHUnionized form (Yellow)⇌Me++OH-Ionized form(Red)
In basic medium, equilibrium lies in backward direction and therefore it shows yellow colour.
In acidic medium, equilibrium shifts in forward direction and therefore, colour changes from yellow to red.

Topic: Chemistry, Redox Reactions, JEE Main
An alkaline earth metal 'M' readily forms water soluble sulphate and water insoluble hydroxide. Its oxide MO is very stable to heat and does not have rock-salt structure. M is

Options:
A Sr
B Ca
C Mg
D Be

The correct answer is:
Be

Solution:
BeSO4 Soluble in water

Be(OH)2 Insoluble in water

Structure of BeO is Hexagonal Wurtzite.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2020 (04 Sep Shift 2)
An alkane of molecular weight 86 g/mol on monochlorination gives two products. The alkane is

Options:
A 2-Methylbutane
B n-butane
C 2, 2-Dimethyl propane
D 2, 3-Dimethyl butane

The correct answer is:
2, 3-Dimethyl butane

Solution:
CnH2n+2= 86
12n+ 2n+2=86
14n= 84
n=6
molecular formula =C6H14
for two monochloro products alkane must have 2-type H

Topic: Chemistry, Hydrocarbons, JEE Main
An alkyl bromide reacts with sodium to form 4, 5-diethyloctane. The alkyl bromide is

Options:
A CH3CH22CH2Br
B CH3CH24CH2Br
C CH3CH23CHBrCH3
D CH3CH22CHBrCH2CH3

The correct answer is:
CH3CH22CHBrCH2CH3

Solution:
R-X→Na/etherR-R
so given alkane divided by half and gets corresponding alkyl halide as following:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
An alloy is composed of two radioactive materials $A$ and $B$ having equal weight. The half life of $A$ and $B$ are 10 yrs and $20 \mathrm{yrs}$ respectively. After time $t$, the alloy was found to consist of $\left(\frac{1}{e}\right) \mathrm{kg}$ of $A$ and $1 \mathrm{~kg}$ of $B$. . If the atomic weight of $A$ and $B$ are same, then the value of $t$ is (Assume, $\ln 2=0.7$ )

Options:
A $\left(\frac{200}{7}\right) \mathrm{yrs}$
B $\left(\frac{10}{7}\right) \mathrm{yrs}$
C 7 yrs
D 70 yrs

The correct answer is:
$\left(\frac{200}{7}\right) \mathrm{yrs}$

Solution:
Given, initial amount of radioactive element, $A=$ initial amount of radioactive element, $B$ i.e., $\left(N_0\right)_A=\left(N_0\right)_B$
half life of element $A,\left(t_2 / 2\right)_A=10 \mathrm{Yr}$
and half life of element $B,\left(t_{1 / 2}\right)_B=20 \mathrm{yr}$
After time $t$, remaining the amount of element $A$,
$N_A=\frac{1}{e} \mathrm{~kg}$
and remaining amount of element $B$,
$N_{\mathcal{B}}=1 \mathrm{~kg}$
For element $A$, after time $t$, amount of element lef undecayed is given as
$\begin{aligned}
N_A & =\left(N_0\right)_A\left(\frac{1}{2}\right)^{\frac{t}{\left(t_{1 / 2}\right)_A}} \\
\frac{1}{e} & =\left(N_0\right)_A\left(\frac{1}{2}\right)^{\frac{t}{10}}
\end{aligned}$
Similarly, for element $B$,
$\begin{gathered}
N_B=\left(N_0\right)_B\left(\frac{1}{2}\right)^{\frac{t}{20}} \\
1=\left(N_0\right)_A\left(\frac{1}{2}\right)^{\frac{t}{20}}
\end{gathered}$
Dividing Eqs. (ii) and (iii), and using Eq. (i), we get
$\frac{1}{e}=\frac{\left(\frac{1}{2}\right)^{\frac{t}{10}}}{(1)^{t / 20}}=\left(\frac{1}{2}\right)^{\frac{t}{20}}$
Taking log on the both sides, we get
$\begin{aligned}
\log \frac{1}{e} & =\frac{t}{20} \log \frac{1}{2} \\
t & =\frac{20}{\log 2} \\
& =\frac{20}{0.7}=\frac{200}{7} y \mathrm{r} \quad[\because \log 2=0.7]
\end{aligned}$

Topic: Physics, Nuclear Physics, JEE Main
An alloy is composed of two radioactive materials $A$ and $B$ having equal weight. The half life of $A$ and $B$ are 10 yrs and $20 \mathrm{yrs}$ respectively. After time $t$, the alloy was found to consist of $\left(\frac{1}{e}\right) \mathrm{kg}$ of $A$ and $1 \mathrm{~kg}$ of $B$. . If the atomic weight of $A$ and $B$ are same, then the value of $t$ is (Assume, $\ln 2=0.7$ )

Options:
A $\left(\frac{200}{7}\right) \mathrm{yrs}$
B $\left(\frac{10}{7}\right) \mathrm{yrs}$
C 7 yrs
D 70 yrs

The correct answer is:
$\left(\frac{200}{7}\right) \mathrm{yrs}$

Solution:
Given, initial amount of radioactive element, $A=$ initial amount of radioactive element, $B$ i.e., $\left(N_0\right)_A=\left(N_0\right)_B$
half life of element $A,\left(t_2 / 2\right)_A=10 \mathrm{Yr}$
and half life of element $B,\left(t_{1 / 2}\right)_B=20 \mathrm{yr}$
After time $t$, remaining the amount of element $A$,
$N_A=\frac{1}{e} \mathrm{~kg}$
and remaining amount of element $B$,
$N_{\mathcal{B}}=1 \mathrm{~kg}$
For element $A$, after time $t$, amount of element lef undecayed is given as
$\begin{aligned}
N_A & =\left(N_0\right)_A\left(\frac{1}{2}\right)^{\frac{t}{\left(t_{1 / 2}\right)_A}} \\
\frac{1}{e} & =\left(N_0\right)_A\left(\frac{1}{2}\right)^{\frac{t}{10}}
\end{aligned}$
Similarly, for element $B$,
$\begin{gathered}
N_B=\left(N_0\right)_B\left(\frac{1}{2}\right)^{\frac{t}{20}} \\
1=\left(N_0\right)_A\left(\frac{1}{2}\right)^{\frac{t}{20}}
\end{gathered}$
Dividing Eqs. (ii) and (iii), and using Eq. (i), we get
$\frac{1}{e}=\frac{\left(\frac{1}{2}\right)^{\frac{t}{10}}}{(1)^{t / 20}}=\left(\frac{1}{2}\right)^{\frac{t}{20}}$
Taking log on the both sides, we get
$\begin{aligned}
\log \frac{1}{e} & =\frac{t}{20} \log \frac{1}{2} \\
t & =\frac{20}{\log 2} \\
& =\frac{20}{0.7}=\frac{200}{7} y \mathrm{r} \quad[\because \log 2=0.7]
\end{aligned}$

Topic: Physics, Nuclear Physics, TS EAMCET, TS EAMCET 2019 (03 May Shift 1)
An alloy of Pb-Ag weighing 1.08 g was dissolved in dilute HNO3 and the volume made to 100 mL. A silver electrode was dipped in the solution and the emf of the cell set-up, Pt(s), H2(g)|H+(1M)||Ag+(aq)Ag(s) was 0.62 V. If Ecello is 0.80 V, what is the percentage of Ag in the alloy? (At 25°C , RT/F = 0.06)

Options:

The correct answer is:
50

Solution:
Overall cell reaction is
H2(g)+2Ag+â2Ag(s)+2H+(aq)
E=EÂ°-0.06Ã2.3032logâ¡[H+]2[Ag+]2pH2
0.62=0.80+2Ã0.06Ã2.3032logâ¡[Ag+]
[Ag+]=0.05 M
Number of moles of Ag+ in 100 mL =MV1000=0.05Ã1001000=0.005
Mass of silver =0.005Ã108 g
Percentage of Ag in 1.08 g of alloy =0.005Ã108Ã1001.08=50%

Topic: Chemistry, Electrochemistry, JEE Main
An alpha nucleus of energy 12mv2 bombards on a heavy nuclear target of the charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to

Options:
A Z2v2
B v2Z2
C Zv2
D v2Z

The correct answer is:
Zv2

Solution:
For the closest approach, kinetic energy is converted into potential energy
12mv2=14πε0q1q2r0=14πε0Ze2er0
r0∝Zv2

Topic: Physics, Atomic Physics, JEE Main
An alpha nucleus of energy $\frac{1}{2} m v^2$ bombards a heavy nuclear target of charge $Z e$. Then the distance of closest approach for the alpha nucleus will be proportional to

Options:
A $v^2$
B $1 / m$
C $1 / v^4$
D $1 / Z e$

The correct answer is:
$1 / m$

Solution:
Since, here nuclear target is heavy it can be assumed safely that it will remain stationary and will not move due to the Coulombic interaction force.
At distance of closest approach relative velocity of two particles is $v$. Here target is considered as stationary, so $\alpha$-particle comes to rest instantaneously at distance of closest approach. Let required distance is $r$, then from work energy-theorem.
$\begin{aligned} 0-\frac{m v^2}{2} & =-\frac{1}{4 \pi \varepsilon_0} \frac{Z e \times 2 e}{r} \\ \Rightarrow \quad r & \propto \frac{1}{m} \\ & \propto \frac{1}{v^2} \\ & \propto Z e^2\end{aligned}$

Topic: Physics, Work Power Energy, JIPMER, JIPMER 2009
An alpha nucleus of energy $\frac{1}{2} m v^2$ bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to

Options:
A
$\frac{1}{\mathrm{Ze}}$

B
$v^2$

C
$\frac{1}{m}$

D
$\frac{1}{v^4}$

The correct answer is:
$\frac{1}{m}$

Solution:
$\frac{1}{m}$

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2006
An alpha particle $\left({ }^{4} \mathrm{He}\right)$ has a mass of 4.00300 amu. A proton has a mass of 1.00783 amu and a neutron has a mass of 1.00867 amu respectively. The binding energy of alpha estimated from these data is the closest to

Options:
A $27.9 \mathrm{MeV}$
B $22.3 \mathrm{MeV}$
C $35.0 \mathrm{MeV}$
D $20.4 \mathrm{MeV}$

The correct answer is:
$27.9 \mathrm{MeV}$

Solution:
The mass defect, $\Delta m=2\left(m_{p}+m_{n}\right)-m_{\text {He }}$
$=2(1.00783+1.00867)-4.00300$
$=0.0300$ amu
So, the binding energy, $E=\Delta m c^{2}$
$=0.03 \times 931 \mathrm{MeV}=27.9 \mathrm{MeV}$

Topic: Physics, Nuclear Physics, JEE Main
An \( \alpha \) - particle and a proton are subjected to the same electric field, then the ratio of the forces acting on them is

Options:
A \( 2: 3 \)
B \( 1: 2 \)
C \( 3: 2 \)
D \( 2: 1 \)

The correct answer is:
\( 1: 2 \)

Solution:
F=Ee

Topic: Physics, Electrostatics, JEE Main
An alpha- particle of mass m suffers 1- dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing 64% of its initial kinetic energy. The mass of the nucleus is

Options:
A 1.5m
B 4m
C 3.5m
D 5m

The correct answer is:
4m

Solution:
From momentum conservation,

mv+0=-mv1+Mv2 ...(1)
Using kinetic energy conservation,
12mv2=12mv12+12Mv22⇒12mv2=0.36×12mv2+12Mv22⇒0.64mv2=Mv22⇒v2=0.8vmM
As it retain 36%,⇒0.36×12mv2=12mv12 ⇒0.36v2=v12 Hence, v1=0.6v
From (1), mv=-m0.6v+M0.8vmM

⇒1=-0.6+0.8Mm

On solving, we get M=4m

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2019 (12 Jan Shift 2)
An altermating current is flowing through a series $L-C - R$ circuit. It is found that the current reaches a value of $1 \mathrm{mA}$ at both $200 \mathrm{Hz}$ and $800 \mathrm{Hz}$ frequency. What is the resonance frequency of the circuit?

Options:
A $600 \mathrm{Hz}$
B $300 \mathrm{Hz}$
C $500 \mathrm{Hz}$
D $400 \mathrm{Hz}$

The correct answer is:
$400 \mathrm{Hz}$

Solution:
$\because$ Resonance frequency, $f_{0}=\sqrt{f_{1} f_{2}}$
$$
=\sqrt{200 \times 800}=400 \mathrm{Hz}
$$

Topic: Physics, Alternating Current, JEE Main
An alternating current in a circuit is given by $I=20 \sin (100 \pi t+0.05 \pi)$ A. The r.m.s. value and the frequency of current respectively are

Options:
A $10 \mathrm{A}$ and $100 \mathrm{Hz}$
B $10 \mathrm{A}$ and $50 \mathrm{Hz}$
C $10 \sqrt{2} \mathrm{A}$ and $50 \mathrm{Hz}$
D $10 \sqrt{2}$ A and $100 \mathrm{Hz}$

The correct answer is:
$10 \sqrt{2} \mathrm{A}$ and $50 \mathrm{Hz}$

Solution:
Given, $I=20 \sin (100 \pi t+0.05 \pi)$
The root mean square value of alternating
$$
\text { current }=l_{\mathrm{ms}}=\frac{l_{0}}{\sqrt{2}}=\frac{20}{\sqrt{2}}=10 \sqrt{2} \mathrm{A}
$$
Also, $\quad \omega=100 \pi$
$\Rightarrow \quad 2 \pi f=100 \pi$ or $f=50 \mathrm{Hz}$

Topic: Physics, Alternating Current, JEE Main
An alternating current is given by i=i1 cos Ït+i2 sin Ït. The value of rms current is

Options:
A i1+i22
B i1+i22
C i12+i222
D i12+i222

The correct answer is:
i12+i222

Solution:
i=i1cos Ït+i2sin Ït
i2mean=i12 cos2ÏtÂ¯+i22 sin2ÏtÂ¯+i1i2 cos Ït sin ÏtÂ¯
i12Ã12Ãi22Ã12+2i1i2Ã0
irms=i2mean=i12+i222

Topic: Physics, Alternating Current, JEE Main
An alternating current is given by the equation $i=i_1 \cos \omega t+i_2 \sin \omega t$. The r.m.s. current is given by

Options:
A $\frac{1}{\sqrt{2}}\left(i_1+i_2\right)$
B $\frac{1}{\sqrt{2}}\left(i_i+i_2\right)^2$
C $\frac{1}{\sqrt{2}}\left(i_1^2+i_2^2\right)^{1 / 2}$
D $\frac{1}{2}\left(i_1^2+i_2^2\right)^{1 / 2}$

The correct answer is:
$\frac{1}{\sqrt{2}}\left(i_1^2+i_2^2\right)^{1 / 2}$

Solution:
$i_{r m s}=\sqrt{\frac{i_1^2+i_2^2}{2}}=\frac{1}{\sqrt{2}}\left(i_1^2+i_2^2\right)^{1 / 2}$

Topic: Physics, Alternating Current, JEE Main