datavorous/entrance-exam-dataset huggingface dataset:
In a photoelectric effect experiment, the threshold wavelength of light is 380 nm . If the wavelength of incident light is 260 nm , the maximum kinetic energy of emitted electrons will be
Given E (in eV ) =1237λ(in nm)

Options:
A 4.5 eV
B 3.0 eV
C 1.5 eV
D 15.1 eV

The correct answer is:
1.5 eV

Solution:
From Einstein’s photoelectric equation,
K.Emax=hcλ-hcλ0 (here, λ0= threshold wavelength)
=1237260-1237380
=1237380-260380×260
=1.5 eV

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2019 (10 Apr Shift 1)
In a photoelectric effect measurement, the stopping potential for a given metal is found to be V0 volt when radiation of wavelength λ0 is used. If radiation of wavelength 2λ0 is used with the same metal then the stopping potential (in volt) will be

Options:
A V02
B 2V0
C V0+hc2eλ0
D V0-hc2eλ0

The correct answer is:
V0-hc2eλ0

Solution:
e V 0 = hc λ 0 − W 0 and eV′= hc 2 λ 0 − W 0
Subtracting them we have
e( V 0 − V ′ )= hc λ 0 [ 1− 1 2 ]= hc 2 λ 0
Or V ′ = V 0 − hc 2e λ 0

Topic: Physics, Dual Nature of Matter, JEE Main
In a photoelectric effect measurement, the stopping potential for a given metal is found to be V0 volt when radiation of wavelength Î»0 is used. If radiation of wavelength 2Î»0 is used with the same metal then the stopping potential (in volt ) will be

Options:
A V02
B 2V0
C V0+hc2eλ0
D V0-hc2eλ0

The correct answer is:
V0-hc2eλ0

Solution:
From Einstein's photoelectric equation
eV0=hcÎ»0-W0
eVâ²=hc2Î»0-W0
Subtracting eV0-Vâ²= hcÎ»01-12=hc2Î»0
or Vâ²= V0-hc2eÎ»0

Topic: Physics, Dual Nature of Matter, JEE Main
In a photoelectric experiment, a graph of maximum kinetic energy $\left(\mathrm{KE}_{\max }\right)$ against the frequency of incident radiation $(v)$ is plotted. If $\mathrm{A}$ and $\mathrm{B}$ are the intercepts on the $\mathrm{X}$ and $\mathrm{Y}$ axis respectively the Planck's constant is given by

Options:
A A+B
B $\frac{\mathrm{B}}{\mathrm{A}}$
C $A \times B$
D $\frac{\mathrm{A}}{\mathrm{B}}$

The correct answer is:
$\frac{\mathrm{B}}{\mathrm{A}}$

Solution:
Kinetic energy is given by
$\text { (K.E.) })_{\max }=h v-h v_0$
$\text { Comparing }\mathrm{y}=\mathrm{mx}+\mathrm{c}$
We get $\mathrm{x}$-intercept when
$\text { (K.E. })_{\max }=0 \text {, i.e., } \mathrm{h} v-\mathrm{h} v_0=0$
Or $v=v_0=\mathrm{A}$
We get y-intercept when $v=0$
$\therefore(\text { K.E. })_{\max }=-\mathrm{h} v_0=\mathrm{B}$
$\therefore\left|\frac{\mathrm{B}}{\mathrm{A}}\right|=\frac{\mathrm{h} \mathrm{v}_0}{\mathrm{v}_0}=\mathrm{h}$

Topic: Physics, Dual Nature of Matter, JEE Main
In a photoelectric experiment, a monochromatic light is incident on the emitter plate \(E\), as shown in the figure. When switch \(S_1\) is closed and switch \(S_2\) is open, the photoelectrons strike the collector plate \(C\) with a maximum kinetic energy of \(1 \mathrm{eV}\). If switch \(S_1\) is open and switch \(S_2\) is closed and the frequency of the incident light is doubled the photoelectrons strike the collector plate with a maximum kinetic energy of \(20 \mathrm{eV}\). The threshold wavelength of the emitter plate is

Options:
A \(5233.3 Å\)
B \(4133.3 Å\)
C \(4166.7 Å\)
D \(5336.7 Å\)

The correct answer is:
\(4133.3 Å\)

Solution:
Let threshold frequency of emitter plate \(=v_0\)
Energy of photon in first case is \(\mathrm{E}\).
When switch \(S_1\) is closed and switch \(S_2\) is open,
So, \(E=h v_0+(5+1) \mathrm{eV}\) ...(i)
For second case, when switch \(S_1\) is open and switch \(S_2\) is closed and frequency of incident light is doubled.
then, \(\quad 2 E=h v_0+(20-5) \mathrm{eV}\) ...(ii)
From Eqs. (i) and (ii), we get
\(\begin{aligned}
& \Rightarrow \quad 2\left(h v_0+6 \mathrm{eV}\right)=h v_0+15 \mathrm{eV} \\
& \Rightarrow \quad 2 h v_0+12 \mathrm{eV}=h v_0+15 \mathrm{eV} \\
& \Rightarrow \quad h v_0=3 \mathrm{eV} \\
& \Rightarrow \quad v_0=\frac{3 \times 1.6 \times 10^{-19}}{6.62 \times 10^{-34}}=7.25 \times 10^{14} \mathrm{~Hz} \\
& \lambda_0=\frac{c}{v_0}=\frac{3 \times 10^8}{7.25 \times 10^{14}} \\
& \lambda_0=41333 Å \\
\end{aligned}\)
Hence, the threshold wavelength of the emitter plate is \(4133.3 Å\).

Topic: Physics, Dual Nature of Matter, JEE Main
In a photoelectric experiment, a parallel beam of monochromatic light with the power of 200ââW is incident on a perfectly absorbing cathode of work function 6.25 eV. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of 500ââV is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force F=nÃ10-4 N due to the impact of the electrons. The value of n is __________. Mass of the electron me=9Ã10-31 kg and 1.0 eV=1.6Ã10-19 J.

Options:

The correct answer is:
24

Solution:
Power =NhÎ½
N= number of photons per second
Since KE=0,hÎ½=Ï
200=N6.25Ã1.6Ã10-19 Joule
N=2006.25Ã1.6Ã10-19
As photon is just above threshold frequency KEmax is zero and they are accelerated by potential difference of 500 V
KEf=qâV
P22m=qâVâP=2mqâV
Since efficiency is 100%, number of electrons emitted is equal to number of photons falling per second as electrons are completely absorbed, force exerted nmv
= 2006.25Ã1.6Ã10-19Ã29Ã10-31Ã1.6Ã10-19Ã500
= 3Ã200Ã10-25Ã16006.25Ã1.6Ã10-19=2Ã406.25Ã1.6Ã10-4Ã3=24 Ã 10 â4
n=24

Topic: Physics, Dual Nature of Matter, JEE Main
In a photoelectric experiment for $4000 Å$ incident radiation, the potential difference to stop the ejection is $2 \mathrm{~V}$. If the incident light is changed to $3000 Å$, then the potential required to stop the ejection of electrons will be

Options:
A $2$ $V$
B Less than $2 \mathrm{~V}$
C Zero
D Greater than $2 \mathrm{~V}$

The correct answer is:
Greater than $2 \mathrm{~V}$

Solution:
Greater than $2 \mathrm{~V}$
According to Einstein's photoelectric equation
$V_0=\frac{h c}{e}\left[\frac{1}{\lambda}-\frac{1}{\lambda_0}\right]$
Hence if $\lambda$ decreases $V_0$ increases

Topic: Physics, Dual Nature of Matter, JEE Main
In a photoelectric experiment, if both the intensity and frequency of the incident light are doubled, then the saturation photoelectric current

Options:
A Remains constant
B Is halved
C Is doubled
D Becomes four times

The correct answer is:
Is doubled

Solution:
The saturation photoelectric current is directly proportional to the intensity of incident radiation but it is independent of its frequency. Therefore, the saturation photoelectric current becomes doubled, when both the intensity and frequency of the incident light are doubled.

Topic: Physics, Nuclear Physics, NEET
In a photoelectric experiment, increasing the intensity of incident light:

Options:
A increases the number of photons incident and also increases the K.E. of the ejected electrons.
B increases the frequency of photons incident and increases the K.E. of the ejected electrons.
C increases the number of photons incident and the K.E. of the ejected electrons remains unchanged.
D increases the frequency of photons incident and the K.E. of the ejected electrons remains unchanged.

The correct answer is:
increases the number of photons incident and the K.E. of the ejected electrons remains unchanged.

Solution:
The relation between the energy of photons and the intensity of photons is, hv=IAn, where n is the number of photons, if we increase the number of photons then frequency does not change it means total energy and kinetic energy of photon remains unchanged but due to to this property the intensity of light is increased.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2021 (27 Aug Shift 1)
In a photoelectric experiment, the relation between the applied potential difference V and the photoelectric current I was found to be as shown in the graph below. If the work function of the cathode plate is 28.8 eV and h=6.6Ã10-34 J s, the frequency of incident radiation would be nearly (in s-1)

Options:
A 0.436×1018 Hz
B 0.436×1017 Hz
C 0.775×1016 Hz
D 0.775×1015 Hz

The correct answer is:
0.775×1016 Hz

Solution:
For the photoelectric effect,
hf=Ï+KEmax
From the graph KEmax=3.2âeV
â´hf=(28.8+3.2)âeV
f=32h eV=ââ32Ã1.6Ã10â196.6Ã10â34 Hz
=0.775Ã1016 Hz

Topic: Physics, Dual Nature of Matter, JEE Main
In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. Choose the closest value of change in the stopping potential from given options hce=1240 nmÂ·V

Options:
A 2.0 V
B 0.5 V
C 1.0 V
D 1.5 V

The correct answer is:
1.0 V

Solution:
vs1=1240300-ϕ

vs2=1240400-ϕ

vs1-vs2=1240300-1240400

=4.13-3.1
=1.03
=1

Topic: Physics, Dual Nature of Matter, JEE Main
In a photoelectric experiment, the wavelength of the light incident on a metal is changed from $300 \mathrm{nm}$ to $400 \mathrm{nm}$. The decrease in the stopping potential is close to :

$\left(\frac{\mathrm{hc}}{\mathrm{e}}=1240 \mathrm{nm}-\mathrm{V}\right)$

Options:
A $0.5 \mathrm{~V}$
B $1.5 \mathrm{~V}$
C $1.0 \mathrm{~V}$
D $2.0 \mathrm{~V}$

The correct answer is:
$1.0 \mathrm{~V}$

Solution:
Let $\phi=$ work function of the metal,

$\frac{\mathrm{hc}}{\lambda_{1}}=\phi+\mathrm{eV}_{1}$

$\frac{\mathrm{hc}}{\lambda_{2}}=\phi+\mathrm{eV}_{2} \quad \ldots \ldots$

(ii)

Sutracting (ii) from (i) we get

$\mathrm{hc}\left(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}\right)=\mathrm{e}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)$

$\Rightarrow V_{1}-V_{2}=\frac{h c}{e}\left(\frac{\lambda_{2}-\lambda_{1}}{\lambda_{1} \cdot \lambda_{2}}\right)\left[\begin{array}{l}\lambda_{1}=300 \mathrm{nm} \\ \lambda_{2}=400 \mathrm{nm} \\ \frac{\mathrm{hc}}{\mathrm{e}}=1240 \mathrm{nm}-\mathrm{V}\end{array}\right]$

$=(1240 \mathrm{nm}-\mathrm{v})\left(\frac{100 \mathrm{nm}}{300 \mathrm{nm} \times 400 \mathrm{nm}}\right)$

$=1,03 \mathrm{~V} \approx 1 \mathrm{~V}$

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2019 (11 Jan Shift 2)
In a photoelectric experiment three different lights are incident on a metal of work function $1.5 \mathrm{eV}$. Light $A$ is $200 \mathrm{~nm}$ wavelength with intensity $1.8 \mathrm{~W} / \mathrm{m}^2$, light $B$ is $400 \mathrm{~nm}$ with $1 \mathrm{~W} / \mathrm{m}^2$ and light $C$ is $600 \mathrm{~nm}$ with $0.5 \mathrm{~W} / \mathrm{m}^2$. The current versus voltage is measured. Which graphs correspond to which light?

Options:
A I III II
B I II IV
C I III IV
D III II IV

The correct answer is:
III II IV

Solution:
Given that, work-function, $\phi_0=1.5 \mathrm{eV}$
Let $\lambda_0=$ threshold wavelength.
Then, $\phi_0=\frac{h c}{\lambda_0}$
$$
\begin{aligned}
\Rightarrow \quad 1.5 \mathrm{eV} & =\frac{1242}{\lambda_0}(\mathrm{eV}-\mathrm{nm}) \\
\lambda_0 & =828 \mathrm{~nm}
\end{aligned}
$$
Now, photocurrent is directly proportional to intensity of incident light.
$V_1, V_2, V_3$ and $V_4$ are cut off voltage, which is required to stop the electron having maximum kinetic energy.
Given that, $\lambda_A=200 \mathrm{~nm}$, its corresponding cut off voltage is $V_2$.
$\lambda_B=400 \mathrm{~nm}$, its corresponding cut off voltage is $V_3$.
$\lambda_C=600 \mathrm{~nm}$, its corresponding cut off voltage is $V_4$.
and corresponding photocurrent,
$$
\begin{aligned}
I_A & =1.8 \mathrm{~W} / \mathrm{m}^2 \rightarrow \mathrm{III} \\
I_B & =1 \mathrm{~W} / \mathrm{m}^2 \rightarrow \mathrm{II} \\
I_C & =0.5 \mathrm{~W} / \mathrm{m}^2 \rightarrow \mathrm{IV}
\end{aligned}
$$
Hence, consequently option (d) is correct.

Topic: Physics, Dual Nature of Matter, TS EAMCET, TS EAMCET 2021 (06 Aug Shift 1)
In a photoelectric experiment, ultraviolet light of wavelength 280 nm is used with lithium cathode having work function ϕ=2.5 eV. If the wavelength of incident light is switched to 400 nm, find out the change in the stopping potential.

h=6.63×10-34 J s, c=3×108 m s-1

Options:
A 1.1 V
B 0.6 V
C 1.3 V
D 1.9 V

The correct answer is:
1.3 V

Solution:
From the photoelectric equation,

eV0=hcλ−ϕ

Using this equation for the given two cases,

eV1=hc280 nm−ϕ

eV2=hc400 nm−ϕ

On subtracting,

eV1−V2=1240120280×400 =1.3 eV

⇒V1−V2=1.3 V

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2021 (26 Aug Shift 1)
In a photoelectric experiment, with light of wavelength $\lambda,$ the fastest electron has speed $v$. If the exciting wavelength is changed to $3 \lambda / 4$, the speed of the fastest emitted electron will become

Options:
A $v \sqrt{\frac{3}{4}}$
B $v \sqrt{\frac{4}{3}}$
C less than $v \sqrt{\frac{4}{3}}$
D greater than $v \sqrt{\frac{4}{3}}$

The correct answer is:
greater than $v \sqrt{\frac{4}{3}}$

Solution:
$\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{hc}}{\lambda}-\phi$
$$
\begin{array}{l}
\frac{1}{2} \mathrm{~m}^{\prime}=\frac{\mathrm{hc}}{(3 \lambda / 4)}-\phi=\frac{4 \mathrm{hc}}{3 \lambda}-\phi \\
\text { Clearly, } \mathrm{v}^{\prime}>\sqrt{\frac{4}{3} \mathrm{v}}
\end{array}
$$

Topic: Physics, Dual Nature of Matter, JEE Main
In a physical balance working on the principle of moments, when 5 mg weight is placed on the left pan, the beam becomes horizontal. Both the empty pans of the balance are of equal mass. Which of the following statements is correct?

Options:
A Every object that is weighed using this balance appears lighter than its actual weight
B Left arm is shorter than the right arm
C Both the arms are of same length
D Left arm is longer than the right arm

The correct answer is:
Left arm is shorter than the right arm

Solution:
Let m 1 and m 2 be the masses of pan
It is given that m1=m2
Since balance is achieved, so net torque about O should be zero.
5 mg+ m 1 g r 1 = m 2 g r 2
⇒5mg+m1gr1=m1gr2
⇒r1r2=m1gm1g+5mg
⇒r1r2<1

So, left arm is shorter than the right arm.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2017 (08 Apr Online)
In a plane electromagnetic wave, the directions of electric field and magnetic field are represented by k^ and 2i^-2j^, respectively. What is the unit vector along direction of propagation of the wave.

Options:
A 12(i^+j^)
B 12(j^+k)
C 15(i^+2j^)
D 15(2i^+j^)

The correct answer is:
12(i^+j^)

Solution:
E→×B→‖C→
E^×B^=12i^j^k^0011-10=i^+j^2⇒c^=i^+j^2

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2020 (02 Sep Shift 2)
In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5×1010 Hz and an amplitude of 50 V m–1. The total average energy density of the electromagnetic field of the wave is : [Use ε0=8.85×10–12 C2 N-1 m-2]

Options:
A 1.106×10–8 J m–3
B 4.425×10–8 J m–3
C 2.212×10–8 J m–3
D 2.212×10–10 J m–3

The correct answer is:
1.106×10–8 J m–3

Solution:
Total average energy density of the electromagnetic field is given by, UE=12ε0E02.

⇒UE=12×8.85×10-12×502

=1.106×10–8 J m–3

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2024 (31 Jan Shift 1)
In a plant, the colour of grains is controlled by 3 genes located on separate chromosomes. Gene 'C' and 'R' independently do not form any colour, but when together, they impart greenish-brown colour to the stem, while the grains remain colourless. In presence of an additional allele 'A', the stem as well as grains become violet. In a trihybrid cross, what phenotypic ratio is expected in grains?

Options:
A 48 coloured : 16 colourless
B 36 coloured : 28 colourless
C 27 coloured : 37 colourless
D 40 coloured : 24 colourless

The correct answer is:
27 coloured : 37 colourless

Solution:
According to the data given, coloured seed can be produced in the plant only when the genotype has at least one dominant C gene and one dominant R gene and one dominant A gene, that is C_R_A_
A trihybrid cross is CcRrAa x CcRrAa . The three genes are on separate chromosomes. Thus, they are independently assorted. If we consider the three genes separately: Cc x Cc will produce 1 CC, 2 Cc and 1 cc. Thus, the proportion of individuals having CC or Cc is 34 .
Similarly, Rr x Rr will produce 1 RR, 2 Rr and 1 rr. Thus, the proportion of individuals having RR or Rr is 34 .
Similarly, Aa x Aa will produce 1 AA, 2 Aa and 1 aa. Thus, the proportion of individuals having AA or Aa is 34 .
Thus, the proportion of plants with genotype C_R_A_ in trihybrid cross will be 34Ã 34 Ã 34 = 2764 . Thus, 27 out of 64 progenies will have coloured grains. The remaining 37 out of 64 will have colourless grains.

Topic: Biology, Principles of Inheritance and Variation, NEET
In a p-n junction diode not connected to any circuit

Options:
A the potential is the same everywhere
B the p-type side is at a higher potential than the n-type side
C there is an electric field at the junction directed from the n-side to the p-side
D there is an electric field at the junction directed from the p-type side to the n-type side

The correct answer is:
there is an electric field at the junction directed from the n-side to the p-side

Solution:
At the junction, a potential barrier/depletion layer is formed, with n-side at higher potential and p-side at lower potential. Therefore, there is an electric field at the junction directed from the n-side to the p-side.

Topic: Physics, Semiconductors, JEE Main
In a $P N$-junction diode

Options:
A The current in the reverse biased condition is generally very small
B The current in the reverse biased condition is small but the forward biased current is independent of the bias voltage
C The reverse biased current is strongly dependent on the applied bias voltage
D The forward biased current is very small in comparison to reverse biased current

The correct answer is:
The current in the reverse biased condition is generally very small

Solution:
In forward biased $P N$-junction, external voltage decreases the potential barrier, so current is maximum. While in reversed biased PN-junction, extemal voltage increases the potential barrier, so the current is very small.

Topic: Physics, Semiconductors, JEE Main
In a PN junction photo cell, the value of photo-electromotive force produced by monochromatic light is proportional to

Options:
A The voltage applied at the PN junction
B The barrier voltage at the PN junction
C The intensity of the light falling on the cell
D The frequency of the light falling on the cell

The correct answer is:
The intensity of the light falling on the cell

Solution:
When a light (wavelength sufficient to break the covalent bond) falls on the junction, new hole electron pairs are created. No. of produced electron hole pair deponed upon no. of photons. So photo emf or current proportional to intensity of light.

Topic: Physics, Semiconductors, JEE Main
In a poisson distribution with unit mean, \(\sum_{x=0}^{\infty}|x-\bar{x}| P(X=x)=(\bar{x}\) is the mean of the distribution)

Options:
A e
B \(\frac{1}{\mathrm{e}}\)
C \(\frac{2}{\mathrm{e}}\)
D \(\frac{2}{3 e}\)

The correct answer is:
\(\frac{2}{\mathrm{e}}\)

Solution:
For the unit mean, \(\bar{x}=1\)
\(\begin{aligned}
& \text {so } \sum_{x=0}^{\infty}|x-\bar{x}| P(X=x)=\sum_{x=0}^{\infty}|x-1| \frac{e^{-1}}{x !} \\
& =\frac{1}{e}\left[\frac{1}{0 !}+\sum_{x=1}^{\infty} \frac{x-1}{x !}\right]=\frac{1}{e}\left[1+\sum_{x=1}^{\infty} \frac{1}{(x-1) !}-\sum_{x=1}^{\infty} \frac{1}{x !}\right]
\end{aligned}\)
\(=\frac{1}{e}[1+e-(e-1)]=\frac{1}{e}(1+e-e+1)=\frac{2}{e}\)
Hence, option (3) is correct.

Topic: Mathematics, Statistics, JEE Main
In a polar molecule, the ionic charge is $4.8 \times 10^{-10}$ e.s.u. If the inter ionic distance is one $Å$ unit, then the dipole moment is

Options:
A 41.8 debye
B 4.18 debye
C 4.8 debye
D 0.48 debye

The correct answer is:
4.8 debye

Solution:
Given ionic charge $=4.8 \times 10^{-10}$ essul. and ionic distance $=1 A^{\circ}=10^{-8} \mathrm{~cm}$ we know that dipole moment $=$ ionic charge $\times$ ionic distance
$=4.8 \times 10^{-10} \times 10^{-8}$
$=4.8 \times 10^{-8}$ e.s.u per $\mathrm{cm}$
$=4.8$ debye.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
In a polygon, no three diagonals are concurrent. If the total number of points of intersection of diagonals interior to the polygon is \( 70 \), then the number of diagonals of the polygon is

Options:
A \( 20 \)
B \( 28 \)
C \( 8 \)
D \( 7 \)

The correct answer is:
\( 20 \)

Solution:
A selection of 4 vertices of the polygon gives an intersection of two diagonals.

Let the number of sides be n.

⇒C4n=70

⇒nn-1n-2n-3=24×70

=8×7×6×5

∴ n=8.

∴ The number of diagonals =C28-8=20.

Topic: Mathematics, Permutation Combination, JEE Main
In a population of 1000 individuals 360 belong to genotype AA, 480 to Aa, and the remaining 160 to aa, Based on this data, the frequency of allele A in the population is:

Options:
A 0.4
B 0.5
C 0.6
D 0.7

The correct answer is:
0.6

Solution:
According to Hardy–Weinberg principle:
Frequency of dominant allele (A) = p
Frequency of recessive allele (a) = q
Frequency of AA individuals = p × p = p2
Frequency of aa individuals = q × q = q2
Frequency of Aa individuals = 2pq
Total population = 1000
Frequency of AA individuals = p × p = p2 = 360/1000 = 0.36
Frequency of dominant allele (A) = p = 0.6

Topic: Biology, Evolution, NEET, NEET 2014
In a population of 1000 individuals, 360 belong to genotype AA, 480 to $\mathrm{Aa}$ and the remaining 160 to aa. Based on this data, the frequency of allele $A$ in the population is

Options:
A $0.4$
B $0.5$
C $0.6$
D $0.7$

The correct answer is:
$0.6$

Solution:
(c) : According to Hardy Weinberg principle
$\mathrm{p}^2+2 \mathrm{pq}+\mathrm{q}^2=1 ;(\mathrm{p}+\mathrm{q})^2=1$
(AA) $\mathrm{p}^2=360$ out of 1000 individuals or $\mathrm{p}^2=0.36$
Therefore, $\mathrm{p}=0.6$.

Topic: Biology, Organisms and Populations, AIIMS, AIIMS 2018 (26 May)
In a potentiometer, a uniform potential gradient of 0.8âVâm -1 is maintained across its 10 m wire. The potential difference across two points on the wire located at 65âcm and 2.45 m is

Options:
A 1.44 V
B 0.144 V
C 1.96 V
D 0.196Â V

The correct answer is:
1.44 V

Solution:
Here,
Length of the wire=10 m
Potential gradient of the wire, k=0.8  V m-1
l2=2.45,  l2=0.65
⇒Δl=l2−l1=1.8 m
ΔV=k·(Δl)
=1.8×0.8

∴ The potential difference between A and B is
ΔV=VB-VA=1.96  V-0.52  V=1.44  V

Topic: Physics, Current Electricity, NEET
In a potentiometer arrangement, a cell gives a balancing point at 75 cm length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emf's of two cells respectively is 3:2, the difference in the balancing length of the potentiometer wire in above two cases will be _____ cm.

Options:

The correct answer is:
25

Solution:
If the potential difference between the potentiometer is V and the potential of the battery is ε.

Then, εV=lL

Therefore, ε1ε2=l1l2⇒32=75x⇒x=50 cm

Therefore, the difference in the balancing length is 75-50=25 cm

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (24 Jun Shift 1)
In a potentiometer arrangement, a cell of emf 1.20 V gives a balance point at 36 cm length of wire. This cell is now replaced by another cell of emf 1.80 V. The difference in balancing length of potentiometer wire in above conditions will be _____ cm.

Options:

The correct answer is:
18

Solution:
In a potentiometer, null point or balance point is defined as the point where the deflection of the galvanometer is zero. If emf of the unknown battery is ε then,

ε=(Potential Gradient)×(Balancing length l). Therefore,

ε1ε2=l1l2

⇒1.201.80=36l2

⇒l2=18×3=54 cm

Hence, difference in balancing length is l2-l1=54-36=18 cm

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (25 Jul Shift 2)
In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5 Ω ,a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of the cell.

Options:
A 2.5 Î©
B 1 Î©
C 1.5 Î©
D 2 Î©

The correct answer is:
1.5 Î©

Solution:
Without shunting conditions,
Es=52×x .....(i), when, balanced, where, x= potential gradient of wire, with shunting condition
On balancing, Es-Er+Rr=40×x ....(ii).
On solving:
(i)(ii)⇒11-rr+R=5240
∴r=1.5 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2018 (08 Apr)