datavorous/entrance-exam-dataset huggingface dataset:
The fraction of atoms of a radioactive element that decays in 6 days is 78 . The fraction that decays in 10 days will be

Options:
A 7780
B 7180
C 3132
D 1516

The correct answer is:
3132

Solution:
78 fraction decays in 6 days
18 fraction is active after 6 days
By using, N=N012 t/T1/2
N=N02n
N08=N02n
n=3

6 days = half lives
t12=2 days
10 days=5 t1/2
N=N025=N032
So decay=N0-N032=3132N0

Topic: Physics, Nuclear Physics, JEE Main
The fraction of the initial number of radioactive nuclei which remain undecayed after half of a half-life of the radioactive sample is

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

The correct answer is:
$\frac{1}{\sqrt{2}}$

Solution:
$N_{0}=2^{n} N$
$N=\frac{N_{0}}{2^{n}}=\frac{N_{0}}{2^{4 / 2}}=\frac{N_{0}}{\sqrt{2}}$

Topic: Physics, Nuclear Physics, KCET, KCET 2011
The fractional part of 27831 is

Options:
A 231
B 431
C 631
D 831

The correct answer is:
831

Solution:
278=23⋅275=82515=831+115
=815C03115+15C13114+........+15C1431+15C15
278=831k+1
27831=8k+831

Topic: Mathematics, Binomial Theorem, JEE Main
The fractional part of a real number $x$ is $x-[x]$, where $[\mathrm{x}]$ is the greatest integer less than or equal to $\mathrm{x}$. Let $\mathrm{F}_{1}$ and $\mathrm{F}_{2}$ be the fractional parts of $(44-\sqrt{2017})^{2017}$ and $(44+\sqrt{2017})^{2017}$ respectively. Then $\mathrm{F}_{1}+\mathrm{F}_{2}$ lies between the numbers

Options:
A 0 and $0.45$
B $0.45$ and $0.9$
C $0.9$ and $1.35$
D $1.35$ and $1.8$

The correct answer is:
$0.9$ and $1.35$

Solution:
$\mathrm{I}+\mathrm{F}_{2}=(\sqrt{2017}+44)^{2017}$
$\quad \mathrm{~F}_{2}^{\prime}=(\sqrt{2017}-44)^{2017} ; 0 < \mathrm{F}_{2}^{\prime} < 1$
$\mathrm{I}+\mathrm{F}_{2}-\mathrm{F}_{2}^{\prime}=2\left[{ }^{2017} \mathrm{C}_{1}(\sqrt{2017})^{2016}(44)+\ldots .\right]$
$\mathrm{F}_{2}=\mathrm{F}_{2}{ }^{\prime}$
$\mathrm{F}_{2}=(0.911)^{2017}$
Now, $\mathrm{F}_{1}=(44-\sqrt{2017})^{2017}=-(0.911)^{2017}$
Fractional part can not $-\mathrm{ve}$.
So, $F_{1}=1-(0.911)^{2017}$
So, $F_{1}+F_{2}=1$
1 lie Between $0.9 \& 1.35$

Topic: Mathematics, Functions, JEE Main
The fragrance of flowers is due to the presence of some steam volatile organic compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapour in the vapour phase. A suitable method for the extraction of these oils from the flowers is:

Options:
A crystallisation
B distillation under reduced pressure
C distillation
D steam distillation

The correct answer is:
steam distillation

Solution:
Steam distillation technique is applied to separate substances which are steam volatile and are immiscible with water.

As the organic compounds steam volatile and are miscible with water vapour in the vapour phase, steam distillation will be the most suitable method for extraction of oils from flowers. Steam distillation is carried out by passing dry steam through the plant material whereby the steam volatile compounds are volatilised, condensed and collected in receivers. The most common use of steam distillation is the extraction of natural products from plant materials. It has been in use for essential oil extraction for many years.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (31 Jan Shift 2)
The free end of a thread wound on a bobbin is passed round a nail A hammered into the wall. The thread is pulled at a constant velocity. Assuming pure rolling of bobbin, find the velocity $v_0 \mid$ of the centre of the bobbin at the instant when the thread forms an angle a with the vertical.

Options:
A $\mid\frac{v \bar{R}}{R \sin \alpha-r}\mid$
B $\mid\frac{v \bar{R}}{R \sin \alpha+r}\mid$
C $\mid\frac{2v \bar{R}}{R \sin \alpha+r}\mid$
D $\mid\frac{v \bar{R}}{R \sin \alpha+r}\mid$

The correct answer is:
$\mid\frac{v \bar{R}}{R \sin \alpha-r}\mid$

Solution:
When the thread is pulled, the bobbin rolls to the right. Resultant velocity of point $B$ along the thread is $\mid v=v_0 \sin \alpha-\omega r \mid$, where $\mid v_0 \sin \alpha \mid$ is the component of translational velocity along the thread and $\mid \omega r|$ linear velocity due to rotation. As the bobbin rolls without slipping, $\mid v_0=\omega R \mid$. Solving the obtained equations, we get $\mid \left.v_0=\frac{v R}{R \sin \alpha-r} \right\rvert\,$

Topic: Physics, Rotational Motion, JEE Main
The free energy change for a reversible reaction at equilibrium is

Options:
A Large positive
B Small negative
C Small positive
D 0

The correct answer is:
0

Solution:
$\Delta G$ at equilibrium $=0$.

Topic: Chemistry, Thermodynamics (C), JEE Main
The free space inside a current carrying toroid is filled with a material of susceptibility $2 \times 10^{-2}$. The percentage increase in the value of magnetic field inside the toroid will be

Options:
A $2 \%$
B $0.2 \%$
C $0.1 \%$
D $1 \%$

The correct answer is:
$2 \%$

Solution:
Given,
Susceptibility of material, $\chi_m=2 \times 10^{-2}$
Using $\mu_{\mathrm{r}}=1+\chi_{\mathrm{m}}=1+0.02=1.02$
$\mathrm{B}_{\text {final }}=\mu_{\mathrm{r}} \mathrm{B}_0$ (here, $\mathrm{B}_0=$ initial magnetic field)
$\%$ increase in magnetic field
$=\frac{\mathrm{B}_{\text {final }}-\mathrm{B}_0}{\mathrm{~B}_0} \times 100=\frac{\mu_{\mathrm{r}} \mathrm{B}_0-\mathrm{B}_0 \times 100}{\mathrm{~B}_0}$
$=\frac{(\chi+1)-1 \times 100}{1}=0.02 \times 100=2 \%$

Topic: Physics, Magnetic Properties of Matter, JEE Main
The free space inside a current carrying toroid is filled with a material of susceptibility 2×10-2. The percentage increase in the value of magnetic field inside the toroid will be

Options:
A 0.2%
B 0.1%
C 2%
D 1%

The correct answer is:
2%

Solution:
It is given that

χm=2×10-2

The permeability is μr=1+χm=1+0.02=1.02.

The magnetic field is B=μrB0=1.02B0

Hence, percentage increase in magnetic field is

B-B0B0×100%=B01.02-1B0×100%=2%

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2023 (11 Apr Shift 1)
The freezing point (in ${ }^{\circ} \mathrm{C}$ ) of solution containing $0.1 \mathrm{~g}$ of $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ (molecular weight 329$)$ in $100 \mathrm{~g}$ of water $\left(K_f=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$ is

Options:
A
$-2.3 \times 10^{-2}$

B
$-5.7 \times 10^{-2}$

C
$-5.7 \times 10^{-3}$

D
$-1.2 \times 10^{-2}$

The correct answer is:
$-2.3 \times 10^{-2}$

Solution:
vant Hoff's factor $(i)=4\left\{3 \mathrm{~K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}\right\} \quad$ Molality $=\frac{0.1}{329} \times \frac{1000}{100}=\frac{1}{329}$
$$
\Rightarrow \quad-\Delta T_f=i K_f \cdot m=4 \times 1.86 \times \frac{1}{329}=2.3 \times 10^{-2} \Rightarrow T_f=-2.3 \times 10^{-2}{ }^{\circ} \mathrm{C}
$$
(As freezing point of water is $0^{\circ} \mathrm{C}$ )

Topic: Chemistry, Solutions, JEE Main
The freezing point of 1 molal NaCl solution assuming NaCl to be 100% dissociated in water is - Kf for H2O=1.860C m-1

Options:
A -1.860C
B -3.720C
C +1.860C
D +3.720C

The correct answer is:
-3.720C

Solution:
The freezing point of 1 molal NaCl solution assumtextng NaCl to be 100% dissociated in water is -
âTf=iÃKfÃm
=2Ã1.86Ã1=3.72oC
Freezing point = (0-3.72) = -3.72oC

Topic: Chemistry, Solutions, JEE Main
The freezing point of a 3% aqueous solution of 'A' is equal to the freezing point of 9% aqueous solution 'B'. If the molecular weight of 'A' is 60, then the molecular weight of 'B' will be

Options:
A 191.9
B 90
C 45
D 20

The correct answer is:
191.9

Solution:
(T f ) A  =  (T f ) B
â (ÎT f ) A â=â (ÎT f ) B
âiâÃâ K f âÃâ m A â=âiâÃâ K f âÃâ m B
â m A =â m B
â w A M A âÃâ 1000 w solventâ(g) â=â w B M B âÃâ 1000 w solventâ(g)
â 3 60 âÃâ 1000 97 â=â 9 M B âÃâ 1000 91
â M B = 9Ã60Ã97 3Ã91
â´ M B =191.9âg/mol

Topic: Chemistry, Solutions, JEE Main
The freezing point of a 4% aqueous solution of X is equal to the freezing point of a 12% aqueous solution of Y. If the molecular weight of X is A, then the molecular weight of Y will be

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

The correct answer is:
3A

Solution:
Freezing point depression refers to the lowering of the freezing point of solvent upon the addition of solute.

ΔTfX=ΔTfY
kfmx=kfmy
4×1000A×96=12×1000M×88
M=3.27A
≃3A

Topic: Chemistry, Solutions, JEE Main, JEE Main 2019 (12 Jan Shift 1)
The freezing point of a 0.08 molal aqueous solution of NaHSO4 is -0.372°C . The dissociation constant for the following reaction is ( Kf for H2O=1.86 K kg mol-1 )
HSO4-⇌H++SO42-

Options:
A 0.04
B 0.02
C 0.01
D 0.2

The correct answer is:
0.04

Solution:
NaHSO4→Na+0.08+HSO4-0.08
HSO4-0.08(1-α)⇌H+0.08α+SO42-0.08α
i=0.08+0.081-α+0.08α+0.08α0.08=2+α
ΔT=i×Kf×m
0.372=i×1.86×0.08
i=2.5
So 2+α=2.5
α=0.5
Dissociation constant, K=0.08α×0.08α0.081-α
=0.08×0.5×0.08×0.50.08×0.5=0.04

Topic: Chemistry, Solutions, JEE Main
The freezing point of a $1.00 \mathrm{~m}$ aqueous solution of $\mathrm{HF}$ is found to be $-1.91^{\circ} \mathrm{C}$. The freezing point constant of water, $K_f$ is $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. The percentage dissociation of $\mathrm{HF}$ at this concentration is

Options:
A
$30 \%$

B
$10 \%$

C
$5.2 \%$

D
$2.7 \%$

The correct answer is:
$2.7 \%$

Solution:
$\Delta T_{\mathrm{f}}=K_{\mathrm{f}} \times m \times i$
$$
\begin{aligned}
& i=\frac{\Delta T_{\mathrm{f}}}{K_{\mathrm{f}} \times m}=\frac{1.91}{1.86 \times 1}=1.02 \\
& \text { For } \mathrm{HI} \leftrightharpoons \mathrm{H}^{+}+\mathrm{I}^{-} \\
& \quad(1-\alpha) \quad \alpha \quad \alpha \\
& 1-\alpha+\alpha+\alpha=i=1.027 \\
& 1+\alpha=1.027 \\
& \alpha=0.027 \text { or } 2.7 \%
\end{aligned}
$$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2012 (26 May Online)
The freezing point of a diluted milk sample is found to be $-0.2^{\circ} \mathrm{C},$ while it should have been $-0.5^{\circ} \mathrm{C}$ for pure milk. How much water has been added to pure milk to make the diluted sample?

Options:
A 1 cup of water to 2 cups of pure milk
B 3 cups of water to 2 cups of pure milk
C 1 cup of water to 3 cups of pure milk
D 2 cups of water to 3 cups of pure milk

The correct answer is:
3 cups of water to 2 cups of pure milk

Solution:
Freezing point of diluted milk $=-0.2^{\circ} \mathrm{C}$

$$

\Delta \mathrm{T}_{\mathrm{f}}^{*}=0.2^{\circ} \mathrm{C}

$$

Free ring point ol pure milk $=-0.5^{\circ} \mathrm{C}$

$$

\Delta \mathrm{T}_{\mathrm{f}}=0.5^{\circ} \mathrm{C}

$$

$\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\Delta \mathrm{T}_{\mathrm{f}}^{\prime}}=\frac{\mathrm{K}_{\mathrm{f}} \times m}{\mathrm{~K}_{\mathrm{f}} \times m^{\prime}} ; m=\frac{\text { mole of solute }}{\text { mass of solvent }(\mathrm{kg})}$

Moles of solute are same in both samples.

$$

\therefore \frac{0.5}{0.2}=\frac{\mathrm{W}^{n}}{\mathrm{~W}}

$$

$\frac{\mathrm{W}^{\prime}}{\mathrm{W}}=\frac{5}{2} ; \mathrm{W}^{\prime}=\frac{5}{2} \mathrm{~W}$

2 cups of pure milk is mixed with 3 cups of water to make

5 cups of diluted milk.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2019 (11 Jan Shift 1)
The freezing point of a solution composed of $10.0$ gof $\mathrm{KCl}$ in $100 \mathrm{~g}$ of water is $4.5^{\circ} \mathrm{C}$. Calculate the van't Hoff factor, i for this solution.

Options:
A $2.50$
B $1.8$
C $1.2$
D $1.3$

The correct answer is:
$1.8$

Solution:
$$
\begin{array}{l}
\Delta \mathrm{T}_{f}=i K_{f} m \\
m=\frac{10 \times 1000}{74.55 \times 100}=1.34 \mathrm{~m} \\
{[\mathrm{Molecular} \mathrm{Mass} \text { of } \mathrm{KCl}=74.55]} \\
\mathrm{i}=\frac{\Delta \mathrm{T}_{f}}{\mathrm{~K}_{f} \mathrm{~m}}=\frac{4.5^{\circ} \mathrm{C}}{\left(1.86^{\circ} \mathrm{C} / \mathrm{m}\right)(1.34 \mathrm{~m})}=1.8
\end{array}
$$

Topic: Chemistry, Solutions, JEE Main
The freezing point of a solution prepared from $1.25 \mathrm{gm}$ of a non-electrolyte and $20 \mathrm{gm}$ of water is $271.9 \mathrm{~K}$. If molar depression constant is $1.86 \mathrm{~K} \mathrm{~mole}^{-1}$, then molar mass of the solute will be

Options:
A 105.7
B 106.7
C 115.3
D 93.9

The correct answer is:
105.7

Solution:
Molar mass $=\frac{K_f \times 1000 \times W}{\Delta T_f \times W}=\frac{1.86 \times 105.7 \times 1.25}{20 \times 1.1}$

Topic: Chemistry, Solutions, JEE Main
The freezing point of benzene decreases by 0.45°C on adding 0.2 g of acetic acid to 20 g of benzene. If acetic acid associates to form a dimer in benzene, then what is the percentage association of acetic acid in benzene?

Kf for benzene=5.12 K kg mol-1

Options:
A 80.4%
B 74.6%
C 94.6%
D 64.6%

The correct answer is:
94.6%

Solution:
2CH3COOH⇌CH3COOH21-αα2
i=1-α2
ΔTf=i×Kf×m
0.45=i×5.12×0.2×100060×20
i=0.527=1-α2

⇒α=0.946 or 94.6%

Topic: Chemistry, Solutions, JEE Main, JEE Main 2017 (02 Apr)
The freezing point of equimolal aqueous solution will be highest for

Options:
A C6H5N+H3Cl- (aniline hydrochloride)
B Ca(NO3)2
C La(NO3)3
D C6H12O6 (Glucose)

The correct answer is:
C6H12O6 (Glucose)

Solution:
ΔTf∝m×i
m = morality, i = Vant Hoff factor
Except glucose, all other a electrolytes are undergoing dissociation so (i>1) . Glucose is non electrolyte, i=1 . so it has least value of ΔTf hence its freezing point is maximum.

Topic: Chemistry, Solutions, JEE Main
The frequencies at which the current amplitude in an LCR series circuit becomes 12 times its maximum value, are 212 rad s-1 and 232 rad s-1. The value of resistance in the circuit is R=5 Ω. The self inductance in the circuit is _____ mH.

Options:

The correct answer is:
250

Solution:
Bandwidth of RLC circuit is BW=f2-f1=R2πL

Or, BW=ω2-ω1=RL

BW=232-212=RL

Thus, self inductance in the circuit is L=520=250 mH

Topic: Physics, Alternating Current, JEE Main, JEE Main 2022 (28 Jul Shift 1)
The frequencies of alleles 'A' and 'a' in a population at Hardy–Weinberg equilibrium are 0.7 and 0.3, respectively. In a random sample of 250 individuals taken from the population, how many are expected to be heterozygous?

Options:
A 112
B 81
C 105
D 145

The correct answer is:
105

Solution:
The HardyâWeinberg equation is a mathematical equation that can be used to calculate the genetic variation of a population at equilibrium. The equation is an expression of the principle known as HardyâWeinberg equilibrium, which states that the amount of genetic variation in a population will remain constant from one generation to the next in the absence of disturbing factors. The HardyâWeinberg equation is expressed as:
p2 + 2pq + q2 =1
Where p is the frequency of the 'A' allele and q is the frequency of the 'a' allele in the population. In the equation, p2 represents the frequency of the homozygous genotype AA, q2 represents the frequency of the homozygous genotype aa, and 2pq represents the frequency of the heterozygous genotype Aa. In the given question heterozygote will be 2pq X 250, i.e., 2 X 0.7 X 0.3 X 250 = 105

Topic: Biology, Evolution, NEET
The frequency distribution of daily working expenditure of families in a locality is as follows:
If the mode of the distribution is $₹ 140$, then the value of $b$ is

Options:
A
34

B
31

C
26

D
36

The correct answer is:
36

Solution:
$$
\text { Frequency distribution is given as }
$$
Clearly, modal class is $100-150$, as the maximum frequency occurs in this class.
Given, Mode $=140$
We have
$$
\text { Mode }=\ell+\frac{f_0-f_{-1}}{2 f_0-f_{-1}-f_1} \times i
$$
where
$$
\begin{aligned}
& \ell=100, f_0=37, f_{-1}=33, f_1=\mathrm{b} \\
& i=50
\end{aligned}
$$
Thus, we get
$$
140=100+\left[\frac{37-33}{2(37)-33-b}\right] \times 50
$$
$$
\begin{aligned}
& =100+\left[\frac{4}{74-33-b}\right] \times 50 \\
& =100+\frac{200}{41-b} \\
\Rightarrow & 5740=4300+40 b \Rightarrow b=36
\end{aligned}
$$

Topic: Mathematics, Statistics, JEE Main, JEE Main 2012 (07 May Online)
The frequency distribution of the age of students in a class of 40 students is given below.
$\begin{array}{|l|c|c|c|c|l|l|}
\hline \text{Age} & 15 & 16 & 17 & 18 & 19 & 20 \\
\hline \text{No of Students} & 5 & 8 & 5 & 12 & x & y \\
\hline
\end{array}$
If the mean deviation about the median is 1.25, then $4 x+5 y$ is equal to :

Options:
A 46
B 43
C 44
D 47

The correct answer is:
44

Solution:
$\begin{aligned} & \mathrm{x}+\mathrm{y}=10 \ldots \ldots \ldots .(1) \\ & \text {Median }=18=\mathrm{M} \\ & \text {M.D. }=\frac{\sum \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right|}{\sum \mathrm{f}_{\mathrm{i}}} \\ & 1.25=\frac{36+\mathrm{x}+2 \mathrm{y}}{40} \\ & \mathrm{x}+2 \mathrm{y}=14 \ldots \ldots \ldots .(2) \\ & \text { by (1) & (2) } \\ & \mathrm{x}=6, \mathrm{y}=4 \\ & \Rightarrow 4 \mathrm{x}+5 \mathrm{y}=24+20=44\end{aligned}$
$\begin{array}{|l|l|l|l|}
\hline \operatorname{Age}\left(\mathrm{x}_{\mathrm{i}}\right) & \mathrm{f} & \left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right| & \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right| \\
\hline 15 & 5 & 3 & 15 \\
\hline 16 & 8 & 2 & 16 \\
\hline 17 & 5 & 1 & 5 \\
\hline 18 & 12 & 0 & 0 \\
\hline 19 & \mathrm{x} & 1 & \mathrm{x} \\
\hline 20 & \mathrm{y} & 2 & 2 \mathrm{y} \\
\hline
\end{array}$

Topic: Mathematics, Statistics, JEE Main, JEE Main 2024 (09 Apr Shift 1)
The frequency for a series limit of Balmer and Paschen series respectively are f1 and f3. If the frequency of the first line of Balmer series is f2 then the relation between f1, f2 and f3 is

Options:
A f1-f2=f3
B f1+f3=f2
C f1+f2=f3
D f2-f3=2f1

The correct answer is:
f1-f2=f3

Solution:
v=fÎ»,1Î»=fv
1Î»=R1P2-1n2
f=Rc1P2-1n2
f2=Rc122-132=Rc14-19
f1=Rc122=Rc4
f3=Rc132=Rc9
â´ f1-f3=Rc14-19
â´ f1-f3=f2
â´ f1-f2=f3

Topic: Physics, Atomic Physics, JEE Main
The frequency of a car horn encountered a change from 400 Hz to 500 Hz. When the car approaches a vertical wall. If the speed of sound is 330 m s-1. Then the speed of car is ___ km h-1.

Options:

The correct answer is:
132

Solution:
Wall as an observer
Frequency received by wall,

f1=f0CC-V
Again wall as a source
Frequency received by observer on car,
f2=f1C+VC
f2=f0C+VC-V
500=400C+VC-V
54=C+VC-V
C=9 V

V=C9=3309 m s-1
V=3309×185=132 km h-1

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2021 (20 Jul Shift 1)
The frequency of a light wave in a material is $2 \times 10^{14} \mathrm{~Hz}$ and wavelength is $5000 Å$. The refractive index of material will be.

Options:
A 1.50
B 3.00
C 1.33
D 1.40

The correct answer is:
3.00

Solution:
Refractive index $\mu=\frac{C}{V}=\frac{C}{V \lambda}$
$=\frac{3 \times 10^8}{2 \times 10^{10} \times 5000 \times 10^{-10}}=3$

Topic: Physics, Ray Optics, JEE Main
The frequency of a sonometer wire is f, but when the weight producing the tensions are completely immersed in water the frequency becomes f2 and on immersing the weights in a certain liquid the frequency becomes f3. The specific gravity of the liquid is

Options:
A 43
B 169
C 1512
D 3227

The correct answer is:
3227

Solution:
f = 1 2 â W Î¼ ...(1)
Now weight is immeresed in water.
f 2 = 1 2 â W - B Î¼ ...(2)
From (1) and (2) 1 2 Ã 1 2 â W Î¼ = 1 2 â W - B Î¼ â B = 3 4 W
V Ï w g = 3 4 V Ï g â Ï w = 3 4 Ï ...(3)
Now weight is immeresed in liquid.
f 3 = 1 2 â W - B â² Î¼ ...(4)
From (1)and(4) B â² = 8 9 W â V Ï â g = 8 9 V Ï g â Ï â = 8 9 Ï ...(5)
From (3)and(5) ÏâÏw=3227

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The frequency of a sound wave is $n$ and its velocity is $v$. If the frequency is increased to $4 n$, the velocity of the wave will be

Options:
A $v$
B 2v
C 4v
D $v / 4$

The correct answer is:
$v$

Solution:
Velocity of sound is independent of frequency. Therefore it is same ( $v$ ) for frequency $n$ and $4 n$.

Topic: Physics, Waves and Sound, JEE Main
The frequency of AC mains in India is

Options:
A $110 \mathrm{Cs}^{-1}$
B $50 \mathrm{Cs}^{-1}$
C $60 \mathrm{Cs}^{-1}$
D $120 \mathrm{Cs}^{-1}$

The correct answer is:
$50 \mathrm{Cs}^{-1}$

Solution:
Frequency of AC mains in India is $50 \mathrm{C} / \mathrm{s}$.

Topic: Physics, Alternating Current, MHT CET, MHT CET 2012
The frequency of an electromagnetic radiation is $2 \times 10^6 \mathrm{~Hz}$. What is its wavelength in metref (Velocity of light $=3 \times 10^8 \mathrm{~ms}^{-1}$ )

Options:
A $6.0 \times 10^{14}$
B $1.5 \times 10^4$
C $1.5 \times 10^2$
D $0.66 \times 10^{-2}$

The correct answer is:
$1.5 \times 10^2$

Solution:
When $c=v \times \lambda$ than $\lambda=\frac{c}{v}=\frac{3 \times 10^8}{2 \times 10^6}=1.5 \times 10^2 \mathrm{~m}$

Topic: Chemistry, Structure of Atom, JEE Main
The frequency of an electromagnetic radiation is $2 \times 10^6 \mathrm{~Hz}$. What is its wavelength in metres (Velocity of light $=3 \times 10^8 \mathrm{~ms}^{-1}$ )

Options:
A $6.0 \times 10^{14}$
B $1.5 \times 10^4$
C $1.5 \times 10^2$
D $0.66 \times 10^{-2}$

The correct answer is:
$1.5 \times 10^2$

Solution:
When $c=v \times \lambda$ than $\lambda=\frac{c}{v}=\frac{3 \times 10^8}{2 \times 10^6}=1.5 \times 10^2 \mathrm{~m}$

Topic: Chemistry, Structure of Atom, JEE Main
The frequency of oscillation of current in the inductor is-

Options:
A 1 3 LC
B 1 6 Ï LC
C 1 LC
D 1 2 π LC

The correct answer is:
1 6 π LC

Solution:
Here L eq = L + 2 L = 3 L (in series)
C eq = C +2 C = 3 C (in parallel)
Ï = 1 L eq C eq
= 1 3 LC
frequency f= Ï 2 Ï = 1 6 Ï LC

Topic: Physics, Alternating Current, JEE Main
The frequency of the first overtone of a closed pipe of length $l_{1}$ is equal to that of the first. overtone of an open pipe of length $l_{2}$. The ratio of their lengths $\left(l_{1}: l_{2}\right)$ is

Options:
A $2: 3$
B $4: 5$
C $3: 5$
D $3: 4$

The correct answer is:
$3: 4$

Solution:
Here, $\quad v_{1}=\frac{(2 n-1)}{4 l_{1}} v$
$$
v_{2}=\frac{n}{2 l_{2}} v
$$
Hence, $y_{1}=v_{2}$
$$
\frac{l_{1}}{l_{2}}=\frac{3}{4}
$$

Topic: Physics, Waves and Sound, JEE Main
The frequency of the second overtone of the open pipe is equal to the frequency of the first overtone of the closed pipe. The ratio of the lengths of the open pipe and the closed pipe is

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

The correct answer is:
$2: 1$

Solution:
The frequency of first overtone for closed pipe,
$f_{c}=\frac{3 \mathrm{v}}{4 I_{c}}$
Given, i.e.,
$\begin{aligned}
f_{c} &=f_{0} \\
\frac{3 v}{4 I_{c}} &=\frac{3 v}{2 I_{o}} \Rightarrow \frac{I_{o}}{I_{c}}=\frac{2}{1}
\end{aligned}$

Topic: Physics, Waves and Sound, JEE Main
The frequency of vibration f of a mass m suspended from a spring of spring constant k is given by a relation of the type f=c mx ky, where c is a dimensionless constant. The values of x and y are

Options:
A x=12,Â y=12
B x=Â -12,Â y=Â -12
C x=12,Â y=Â -12
D x= -12, y=12

The correct answer is:
x= -12, y=12

Solution:
f=c mx ky;
Spring constant k= force/length.
M0 L0 T-1=MxMT-2y=Mx+y T-2y
⇒ x+y=0, -2y= -1 or y=12
Therefore, x= -12

Topic: Physics, Units and Dimensions, JEE Main
The frequency of vibration of string is given by

$\mathrm{v}=\frac{\mathrm{p}}{2 l}\left[\frac{\mathrm{F}}{\mathrm{m}}\right]^{1 / 2}$

Here $\mathrm{p}$ is number of segments in the string and $l$ is the length. The dimensional formula for $\mathrm{m}$ will be

Options:
A $\left[\mathrm{M}^{0} \mathrm{LT}^{-1}\right]$
B $\left[\mathrm{ML}^{0} \mathrm{~T}^{-1}\right]$
C $\left[\mathrm{ML}^{-1} \mathrm{~T}^{0}\right]$
D $\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]$

The correct answer is:
$\left[\mathrm{ML}^{-1} \mathrm{~T}^{0}\right]$

Solution:
$\mathrm{v}=\frac{\mathrm{p}}{2 \ell}\left[\frac{\mathrm{F}}{\mathrm{m}}\right]^{1 / 2}$

$v^{2}=\frac{p}{4 \ell^{2}} \frac{F}{m} \Rightarrow m=\frac{p^{2} F}{4 \ell^{2} v^{2}}$

Now, dimensional formula of R.H.S.

$=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^{2}\left(\frac{1}{\mathrm{~T}}\right)^{2}}$

[p will have no dimension as it is an integer] $=\mathrm{ML}^{-1} \mathrm{~T}^{0}$

So, dimensions of $\mathrm{m}$ will be $\mathrm{ML}^{-1} \mathrm{~T}^{0}$.

Topic: Physics, Units and Dimensions, JEE Main