datavorous/entrance-exam-dataset huggingface dataset:
A cavity of the radius R2 is made inside a solid sphere of radius R. The centre of the cavity is located at a distance R2 from the centre of the sphere. The gravitational force on a particle of mass m at a distance R2 from the centre of the sphere on the line joining both the centres of sphere and cavity is (opposite to the centre of the cavity).
[Here g=GMR2, where M is the mass of the sphere]

Options:
A mg2
B 3 mg8
C mg16
D None of these

The correct answer is:
3 mg8

Solution:
Gravitation field at mass m due to a full solid sphere
E → 1 = - g 0 r → R = - g 0 r ^ 2

Gravitational field due to the cavity at mass m

M'=R23R3M ⇒M'=M8
E→2=--GM' r^R2,
= GM r ^ 8 R 2 = g 0 r ^ 8
E → net = - g 0 2 + g 0 8 r ^
= - 3 g 0 r ^ 8
⇒ F → net = - 3 mg 0 r ^ 8

Topic: Physics, Gravitation, JEE Main
A ceiling fan rotates about its own axis with some angular velocity. When the fan is switched off, the angular velocity becomes 14th of the original in time t and n revolutions are made in that time. The number of revolution made by the fan during the time interval between switch off and rest are (Angular retardation is uniform)

Options:
A 4n15
B 8n15
C 16n15
D 32n15

The correct answer is:
16n15

Solution:
Ï42=Ï2-2Î±n(2Ï)
â´ 2Î±n2Ï=Ï2-Ï216
â´ 2Ïn=1516Ï22Î±
0= Ï2-2Î±nâ²
â´ 2Ïnâ²=Ï22Î±
â´ nâ²=1615n

Topic: Physics, Laws of Motion, JEE Main
A cell can be balanced against $110 \mathrm{~cm}$ and $100 \mathrm{~cm}$ of potentiometer wire, respectively with and without being short circuited through a resistance of $10 \Omega$. Its internal resistance is

Options:
A Zero
B $1.0 \mathrm{ohm}$
C $0.5 \mathrm{ohm}$
D $2.0 \mathrm{ohm}$

The correct answer is:
$1.0 \mathrm{ohm}$

Solution:
Internal resistance $=\left(\frac{110}{100}-1\right) \times 10 \Omega$

Topic: Physics, Current Electricity, NEET, NEET 2008 (Mains)
A cell can be balanced against $110 \mathrm{~cm}$ and $100 \mathrm{~cm}$ of potentiometer wire, respectively with and without being short circuited through a resistance of $10 \Omega$ Its internal resistance is

Options:
A $1.0 \Omega$
B $0.5 \Omega$
C $2.0 \Omega$
D zero

The correct answer is:
$1.0 \Omega$

Solution:
Key Idea : This problem is based on the application of potentiometer in which we find the internal resistance of a cell.
In potentiometer experiment in which we find internal resistance of a cell, let $E$ be the emf of the cell and $V$ the terminal potential difference, then
$\frac{E}{V}=\frac{l_1}{l_2}$
where $l_1$ and $l_2$ are lengths of potentiometer wire with and without short circuited through a resistance.
Since, $\frac{E}{V}=\frac{R+r}{R}[\because E=I(R+r)$ and $V=I R]$
$\therefore \frac{R+r}{R}=\frac{l_1}{l_2}$
$\begin{aligned}
1+\frac{r}{R} & =\frac{110}{100} \\
\frac{r}{R} & =\frac{10}{100} \\
r & =\frac{1}{10} \times 10=1 \Omega
\end{aligned}$
$r=\frac{1}{10} \times 10=1 \Omega$

Topic: Physics, Current Electricity, JEE Main
A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of 10-6 M hydrogen ion. The EMF of the cell is 0.118 V at 298 K. The concentration of H+ ion at the positive electrode is 10-x ,The value of 'x' is

Options:

The correct answer is:
4

Solution:
Cell reaction would be like, assume the concentration of hydrogen ion in positive electrode is 'x'
Pt H2 1 atm H+10-6M H+x M Pt H2 1 atm , if Ecell= 0.118 V.
At cathode: 2H+xM+2e-âH2(g)
At anode: H2gâ2H+10-6M+2e-
ECell=ECellÎ¿-0.5912logâ¡H+anode2H+cathode2
ECell=ECellÎ¿-0.5912logâ¡H+anode H+cathode 2
ECell=ECellÎ¿-0.05912logâ¡10-6x2
0.118=0-0.05912logâ¡10-6x2
0.118Ã2=-0.0591logâ¡10-6x2
0.118Ã20.0591=-logâ¡10-6x2
-logâ¡10-6x2=0.118Ã20.0591
-2logâ¡10-6x =4
logâ¡10-6x =-2
10-6x =10-2
x=10-4

Topic: Chemistry, Electrochemistry, JEE Main
A cell divides when its karyocytoplasmic index or karyoplasmic index

Options:
A Decreases
B Increased
C Restored
D Remain same

The correct answer is:
Decreases

Solution:
Karyoplasmic index is the ratio of the volume of the nucleus to the volume of the cytoplasm. For each cell fixed volume of nucleus can control the fixed volume of cytoplasm, when the volume of the cytoplasm increases more than the volume of the nucleus the karyoplasmic index decreases. Thus the cell starts dividing.

Topic: Biology, Cell Cycle and Cell Division, NEET
A cell E1 of emf 6 V and internal resistance 2Ω is connected with another cell E2 of emf 4 V and internal resistance 8 Ω (as shown in the figure). The potential difference across points X and Y is:

Options:
A 10.0 V
B 5.6 V
C 2.0 V
D 3.6 V

The correct answer is:
5.6 V

Solution:
I=6-410=15 A

Vx+4+8×15-Vy=0

Vx-Vy=-5.6⇒Vx-Vy=5.6 V

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (24 Feb Shift 1)
A cell from the following which converts electrical energy into chemical energy

Options:
A Dry cell
B Electrochemical cell
C Electrolytic cell
D None of these

The correct answer is:
Electrolytic cell

Solution:
In the electrolytic cell electrical energy change into chemical energy.

Topic: Chemistry, Electrochemistry, JEE Main
A cell has 32 chromosomes. It undergoes mitotic division. What will be the chromosome number (n) during metaphase? What would be the DNA content $(\mathrm{C})$ during anaphase?

Options:

None

Solution:
Mitosis helps in the growth of organism and its development. It also plays a vital role in a sexually reproducing organisms. The mitotic cell division occurs in somatic cells of an organism. The chromosome number in the daughter cells remains same as that of the parent (dividing) cell, so even at metaphase or anaphase, the chromosome number does not change. The DNA content gets doubled at the synthetic phase of interphase and gets divided at anaphase but the chromosome number remains same.

Topic: Biology, Cell Cycle and Cell Division
A cell having an $\operatorname{emf} \varepsilon$ and internal resistance $r$ is connected across a variable external resistance $R$. As the resistance $R$ is increased, the plot of potential difference $V$ across $R$ is given by

Options:
A

B

C

D

The correct answer is:

Solution:
Here, $E=I(R+r) \Rightarrow E=I R+I r$ and
$$
\begin{aligned}
& E=V+I r \\
& E=V+\frac{E r}{R+r} \\
& V=E-\frac{E}{R+r} \times r \\
& y=c-\frac{1}{x}
\end{aligned}
$$
This equation represents option (3).

Topic: Physics, Current Electricity, NEET, NEET 2012 (Mains)
A cell is connected between the points A and C of a circular conductor ABCD with O as centre and angle AOC=60Â°. If B1 and B2 are the magnitudes of the magnetic fields at O due to the currents in ABC and ADC respectively, then ratio B1B2 is

Options:
A 1
B 2
C 5
D 6

The correct answer is:
1

Solution:
currents in ABC and ADC will not be the same because of the resistance and B also depends on length .
i 1 = 1 6 Ã1= 1 6 A i 2 = 5 6 Ã1= 5 6 A B ABC = Î¼ 0 i 1 4Ïr = Î¼ 0 4Ïr ( 1 6 )( 15 9 Ï ) B ADC = Î¼ 0 i 2 4Ïr = Î¼ 0 4Ï ( 5 6 )( Ï 3 )
equal and opposite magnetic field due to arc ABC & ADC , so net magnetic field at O is zero.

Topic: Physics, Magnetic Effects of Current, JEE Main
A cell is equally permeable to sucrose solution and $\mathrm{NaCl}$ solution. First the cell is put in 0.6 $\mathrm{M}$ sucrose solution there is no change in size but when put in $0.6 \mathrm{M} \mathrm{NaCl}$ solution the size will

Options:
A increase
B decrease
C remain same
D can't be said.

The correct answer is:
decrease

Solution:
$0.6 \mathrm{M} \mathrm{NaCl}$ solution is hypertonic to $0.6 \mathrm{M}$ sucrose solution because there are more solutes per unit volume in $\mathrm{NaCl}$ solution as compared to sucrose solution. Hence, when the cell is placed in $0.6 \mathrm{M} \mathrm{NaCl}$ solution after being placed in $0.6 \mathrm{M}$ sucrose solution water will move out of it and its size will decrease on account of exosmosis.

Topic: Biology, Transport in Plants, AIIMS, AIIMS 2019 (26 May)
A cell of constant e.m.f. first connected to a resistance $R_1$ and then connected to a resistance $R_2$. If power delivered in both cases is then the internal resistance of the cell is

Options:
A $\sqrt{R_1 R_2}$
B $\sqrt{\frac{R_1}{R_2}}$
C $\frac{R_1-R_2}{2}$
D $\frac{R_1+R_2}{2}$

The correct answer is:
$\sqrt{R_1 R_2}$

Solution:
Power dissipated $=I^2 R=\left(\frac{E}{R+r}\right)^2 R$
$\begin{aligned} & \therefore\left(\frac{E}{R+r}\right)^2 R_1=\left(\frac{E}{R_2+r}\right)^2 R_2 \\ & \Rightarrow R_1\left(R_2^2+r_2+2 R_2 r\right)=R_2\left(R_1^2+r^2+2 R_1 r\right) \\ & \Rightarrow R_2^2 R_1+R_1 r^2+2 R_2 r=R_1^2 R_2+R_2 r^2+2 R_1 R_2 r \\ & \Rightarrow\left(R_1-R_2\right) r^2=\left(R_1-R_2\right) r^2=\left(R_1-R_2\right) R_1 R_2 \\ & \Rightarrow r=\sqrt{R_1 R_2}\end{aligned}$

Topic: Physics, Current Electricity, JEE Main
A cell of e.m.f. $1.5 \mathrm{~V}$ having a finite internal resistance is connected to a load resistance of $2 \Omega$. For maximum power transfer the internal resistance of the cell should be

Options:
A $4 \mathrm{ohm}$
B $0.5 \mathrm{ohm}$
C $2 \mathrm{ohm}$
D None of these

The correct answer is:
$2 \mathrm{ohm}$

Solution:
For maximum power, external resistance $=$ internal resistance

Topic: Physics, Current Electricity, JEE Main
A cell of emf 90 V is connected across series combination of two resistors each of 100 Ω resistance. A voltmeter of resistance 400 Ω is used to measure the potential difference across each resistor. The reading of the voltmeter will be:

Options:
A 40 V
B 45 V
C 80 V
D 90 V

The correct answer is:
40 V

Solution:
As we know, voltmeter is connected in parallel to the element being measured.

The equivalent resistance of the above circuit is Req=400×100400+100+100=80+100=180 Ω

Current through the circuit is i=90180=12 A

Reading of the voltmeter is VR=i400×100400+100=12×80=40 V

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (24 Jan Shift 2)
A cell of emf E and internal resistance r is connected in series with an external resistance nr. Then what will be the ratio of the terminal potential difference to emf, if n=9.

Options:

The correct answer is:
0.9

Solution:
i=En+1r

VT=iR=inr= En+1r× nr
VTE=nn+1=99+1=0.9

Topic: Physics, Current Electricity, JEE Main
A cell of emf E and internal resistance r supplies currents for the same time t through external resistance R1=100 Î© and R2=40 Î© separately. If the heat developed in both cases is the same, then the internal resistance of the cell is given by

Options:
A 28.6 Ω
B 70 Ω
C 63.3 Ω
D 140Â Î©

The correct answer is:
63.3 Ω

Solution:
Current drawn from the cell in resistance R1 will be I=E/(R1+r)
Therefore, the heat produced in R1 i.e.
H1=E2R1tR1+r2
Heat produced in R2 , H2=E2R2tR2+r2
As per question H1=H2
or E2R1tR1+r2= E2R2tR2+r2
On solving we get;
r=R1R2
=100Ã40=63.25 Î©

Topic: Physics, Current Electricity, JEE Main
A cell of emf E and internal resistance r supplies currents for the same time t through external resistances R1=100 Î© and R2=40 Î© separately. If the heat developed in both cases is the same, then the internal resistance of the cell is

Options:
A 28.6 Ω
B 80 Ω
C 63.3 Ω
D 140 Ω

The correct answer is:
63.3 Ω

Solution:
Current drawn from the cell in resistance R1 will be I=ER1+r.
Therefore, heat produced in R1 will be H1=E2R1tR1+r2.
Heat produced in R2 will be H2=E2R2tR2+r2.
As per question, H1=H2
E2R1tR1+r2=E2R2tR2+r2
On solving we get,
r=R1R2
=100×40=63.25 Ω

Topic: Physics, Current Electricity, JEE Main
A cell of emf E having an internal resistance 'r' is connected to an external resistance R. The potential difference 'V' across the resistance R varies with R as shown by the curve:

Options:
A A
B B
C C
D D

The correct answer is:
B

Solution:
V = E - ir = - Er R + r = E R + r - r R + r
V = ER R + r
⇒ V = 0 at R = 0
V = E at R = ∞

Topic: Physics, Current Electricity, JEE Main
A cell of emf $E$ is connected to a resistance $R_{1}$ for time $t$ and the amount of heat generated in it is $H$. If the resistance $R_{1}$ is replaced by another resistance $R_{2}$ and is connected to the cell at the same time $t,$ the amount of heat generated in $R_{2}$ is $4 \mathrm{H}$. Then internal resistance of the cell is

Options:
A $\frac{2 R_{1}+R_{2}}{2}$
B $\sqrt{R_{1} R_{2}} \frac{2 \sqrt{R_{2}}-\sqrt{R_{1}}}{\sqrt{R_{2}}-2 \sqrt{R_{1}}}$
C $\sqrt{R_{1} R_{2}} \frac{\sqrt{R_{2}}-2 \sqrt{R_{1}}}{2 \sqrt{R_{2}}-\sqrt{R_{1}}}$
D $\sqrt{R_{1} R_{2}} \frac{\sqrt{R_{2}}-\sqrt{R_{1}}}{\sqrt{R_{2}}+\sqrt{R_{1}}}$

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

Solution:
We have, $H=I^{2} R$ According to given condition, $I_{1}^{2} R_{1}=H$
and
$$
I_{2}^{2} R_{2}=4 H
$$
$$
\Rightarrow \quad \frac{E^{2}}{\left(R_{1}+r\right)^{2}} R_{1}=H \text { and } \frac{E^{2}}{\left(R_{2}+r\right)^{2}} R_{2}=4 H
$$
$\therefore \quad \frac{R_{2}}{\left(R_{2}+r\right)^{2}}=\frac{4 R_{1}}{\left(R_{1}+r\right)^{2}}$
$\Rightarrow \quad \sqrt{R_{2}}\left(R_{1}+r\right)=2 \sqrt{R_{1}}\left(R_{2}+r\right)$
On solving. $r=\frac{\sqrt{R_{1} R_{2}}\left[\sqrt{R_{1}}-2 \sqrt{R_{2}}\right]}{2 \sqrt{R_{1}}-\sqrt{R_{2}}}$

Topic: Physics, Current Electricity, JEE Main
A cell of e.m.f. $E$ is connected with an external resistance $R$, then p.d. across cell is $V$. The internal resistance of cell will be

Options:
A $\frac{(E-V) R}{E}$
B $\frac{(E-V) R}{V}$
C $\frac{(V-E) R}{V}$
D $\frac{(V-E) R}{E}$

The correct answer is:
$\frac{(E-V) R}{V}$

Solution:
Let the current in the circuit $=i=\frac{V}{R}$
Across the cell,
$E=V+i r \Rightarrow r=\frac{E-V}{i}=\frac{E-V}{V I R}=\left(\frac{E-V}{V}\right) R$

Topic: Physics, Current Electricity, JEE Main
A cell of internal resistance r drives current through an external resistance R. The power delivered by the cell to the external resistance will be maximum when:

Options:
A R=2r
B R=r
C R=1000 r
D R=0.001 r

The correct answer is:
R=r

Solution:
Current i=Er+R
Power generated in R

P=i2R
P=E2R(r+R)2

for maximum power dPdR=0
E2(r+R)2×1-R×2(r+R)(r+R)4=0
⇒when, r=R then, dPdR=0

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (08 Apr Shift 2)
A cell, shunted by a 8 Ω resistance, is balanced across a potentiometer wire of length 3 m. The balancing length is 2 m when the cell is shunted by 4 Ω resistance. The value of internal resistance of the cell will be _____ Ω.

Options:

The correct answer is:
8

Solution:
Potential difference across test battery is, ∵ V=ERR+r∝l; RR+r∝l

Therefore, we can write, 88+r44+r=32

⇒44+r=38+r ⇒16+4r=24+3r ⇒r=8 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (27 Jun Shift 1)
A central atom in a molecule has two lone pairs of electrons and forms three single bonds. The shape of this molecule is

Options:
A see-saw
B planar triangular
C T-shaped
D trigonal pyramidal

The correct answer is:
T-shaped

Solution:
sp3d hybridised

T-shaped

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2021 (17 Mar Shift 1)
A certain 12 -hour digital clock displays the hour and the minute of a day. Due to a defect in the clock whenever the digit 1 is supposed to be displayed it displays 7 . What fraction of the day will the clock show the correct time?

Options:
A $\frac{1}{2}$
B $\frac{5}{8}$
C $\frac{3}{4}$
D $\frac{5}{6}$

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

Solution:
The clock will show the incorrect time (between $1-2,10-11,11-12,12-1$ day and night both)
$\therefore$ in correct time $8 \times 60=480$ (each minute it will display 1 )
Remaining 20 hours it will show the incorrect time $16 \times 15=240$
Total incorrect time $=240+480=720$
correct time $=1-$ incorrect time
$=1-\frac{720}{24 \times 60}=1 / 2$

Topic: Mathematics, Basic of Mathematics, KVPY, KVPY 2011 (SA)
A certain amount of gas is taken through a cyclic process (A B C D A) that has two isobars, one isochore and one isothermal. The cycle can be represented on a $\mathrm{P}-\mathrm{V}$ indicator diagram as :

Options:
A

B

C

D

The correct answer is:

Solution:
P-V indicator diagram for isobaric
$\mathrm{P}-\mathrm{V}$ indicator diagram for isochoric process
$\mathrm{P}-\mathrm{V}$ indicator diagram for isothermal process

Topic: Physics, Thermodynamics, JEE Main
A certain amount of gas is taken through a cyclic process (A B C D A) that has two isobars, one isochore and one isothermal. The cycle can be represented on a $\mathrm{P}-\mathrm{V}$ indicator diagram as :

Options:
A

B

C

D

The correct answer is:

Solution:
P-V indicator diagram for isobaric
$\mathrm{P}-\mathrm{V}$ indicator diagram for isochoric process
$\mathrm{P}-\mathrm{V}$ indicator diagram for isothermal process

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2013 (22 Apr Online)
A certain buffer solution contains equal concentration of $X^{-}$and $H X$. The ${ }^{K_b}$ for $X^{-}$is $10^{-10}$. The $p H$ of the buffer is

Options:
A 4
B 7
C 10
D 14

The correct answer is:
4

Solution:
$\mathrm{X}^{-}+\mathrm{H}_2 \mathrm{O}=\mathrm{OH}^{-}+\mathrm{HX}$
$\begin{aligned} & K_b=\frac{\left[O H^{-}\right][H X]}{\left[X^{-}\right]} \\ & H X=H^{+}+X^{-} \\ & K_a=\frac{\left[H^{+}\right]\left[X^{-}\right]}{[H X]} \\ & \therefore K_a \times K_b=\left[H^{+}\right]\left[O H^{-}\right]=K_w=10^{-14}\end{aligned}$
Hence $K_\alpha=10^{-4}$
Now as $\left[X^{-}\right]=[H X], p H=p K_a=4$.

Topic: Chemistry, Ionic Equilibrium, JEE Main
A certain buffer solution contains equal concentration of X- and HX . The Ka for HX is 10-8. The pH of the buffer is

Options:

The correct answer is:
8.00

Solution:
The Henderson equation is
pH=pKa+log⁡ConjugatebaseAcid
=-log⁡Ka+log⁡X-HX
Hence, [HX]=X-
=-log⁡(10-8)+log⁡1
or pH = 8.0.

Topic: Chemistry, Ionic Equilibrium, JEE Main
A certain charge Q is divided into two parts q and (Q-q). How should the charges Q and q be divided so that q and (Q-q) placed at a certain distance apart experience maximum electrostatic repulsion?

Options:
A Q=q2
B Q=2q
C Q=4q
D Q=3q

The correct answer is:
Q=2q

Solution:
Fq=kq(Q-q)L2=kL2qQ-q2
dFdq=0 when force is maximum
dFdq=kL2[Q-2q]=0
⇒Q-2q=0⇒Q=2q

Topic: Physics, Electrostatics, JEE Main, JEE Main 2021 (20 Jul Shift 1)
A certain elastic conducting material is stretched into a circular loop. It is placed with its plane perpendicular to a uniform magnetic field B=0.8 T. When released the radius of the loop starts shrinking at a constant rate of 2 cm s–1. The induced emf in the loop at an instant when the radius of the loop is 10 cm will be ______ mV.

Options:

The correct answer is:
10

Solution:
According to the Faraday's law of electromagnetic induction, the emf induced in a conducting loop is given by, e=-dϕdt, here, t is time and ϕ=BA is magnetic flux.

Here, e=dϕdt=dBAdt=Bdπr2dt

=Bπ2rdrdt=2×π×0.1×0.8×2×10-2

≈10 mV (rounded off)

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2023 (29 Jan Shift 1)
A certain element crystallises in a bcc lattice of unit cell edge length 27 Ao. If the same element under the same conditions crystallises in the fcc lattice, the edge length of the unit cell in Ao will be __________. (Round off to the Nearest Integer).

[Assume each lattice point has a single atom]

[Assume 3=1.73, 2=1.41]

Options:

The correct answer is:
33

Solution:
For BCC 3a=4r

so r=34×27

for FCC a=22r

=2×2×34×27

=32×27

=33

Topic: Chemistry, Solid State, JEE Main, JEE Main 2021 (16 Mar Shift 1)
A certain far-sighted person cannot see objects closer to the eye than \( 100 \mathrm{~cm} \). The power of lens which will enable him to read at a distance of \( 25 \mathrm{~cm} \) will be

Options:
A \( 3 \) dioptre
B \( 1 \) dioptre
C \( 4 \) dioptre
D \( 2 \) dioptre

The correct answer is:
\( 3 \) dioptre

Solution:
For the required lens,

Image distance v=-100 cm

Object distance u=-25 cm

Using the lens equation:

1v-1u=1f

1-100-1-25=1f

⇒f=1003=33.33 cm

Power of the lens:1f=10.33=3 D

Topic: Physics, Ray Optics, JEE Main
A certain gas takes three times as long to effuse out as helium. Its molecular mass will be

Options:
A $27 \mathrm{u}$
B $36 \mathrm{u}$
C $64 \mathrm{u}$
D $9 \mathrm{u}$

The correct answer is:
$36 \mathrm{u}$

Solution:
From Graham's diffusion law,
$$
\begin{aligned}
\frac{r_1}{r_2} & =\sqrt{\frac{M_2}{M_1}} \\
\text { or } \quad \frac{V_1}{t_1} \times \frac{t_2}{V_2} & =\sqrt{\frac{M_2}{M_1}}
\end{aligned}
$$
$\because$ Volume is same,
$\therefore \quad \frac{3}{1}=\sqrt{\frac{M_2}{4}}$
or $\quad 9=\frac{M_2}{4}$
or $\quad M_2=9 \times 4=36 \mathrm{u}$

Topic: Chemistry, States of Matter, NEET, NEET 2012 (Mains)
A certain ideal gas undergoes a polytropic process \( P V^{n}= \) constant such that the molar specific heat during the process is negative. If the ratio of the specific heats of the gas be \( \gamma \), then the range of values of \( n \) will be

Options:
A \( 0 < n < \gamma \)
B \( 1 < n < \gamma \)
C \( n=\gamma \)
D \( n>\gamma \)

The correct answer is:
\( 1 < n < \gamma \)

Solution:
Since PVn= constant and also PV=RT, taking 1 mol of the gas for simplicity,
dU=CvdT
where Cv→ molar specific heat at constant volume.
Now the molar specific heat capacity in a polytropic process PVn= constant is given by
C=Rγ-1-Rn-1=n-γRn-1γ-1 ...(i)
From this equation, we see that C will be negative when n<γ and

n > 1, simultaneously, i.e., 1<n<γ. Since γ for all ideal gases is greater than 1, if n>γ or n < 1, then Cv will be positive.

Topic: Physics, Thermodynamics, JEE Main
A certain metal sulphide, $M S_2$, is used extensively as a high temperature lubricant. If $\mathrm{MS}_2$ is $40.06 \%$ by mass sulphur, metal $M$ has atomic mass.

Options:
A
$160 \mathrm{u}$

B
$64 \mathrm{u}$

C
$40 \mathrm{u}$

D
$96 \mathrm{u}$

The correct answer is:

$96 \mathrm{u}$

Solution:
$M S_2=M+32 \times 2=M+64$
$\%$ of sulphur $=\left(\frac{64}{M+64}\right) \times 100=40.06$
$M+64=\frac{6400}{40.06}$
$M+64=160$
$M=160-64=96$ u

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
A certain metal when irradiated by light r=3.2Ã1016 Hz emtextts photoelectrons with twice kinetic energy as did photoelectrons when the same metal is irradiated by light r=2.0Ã1016 Hz. Then v0 of metal is

Options:
A 1.2Ã1014Hz
B 8Ã1015 Hz
C 1.2Ã1016Hz
D 4Ã1012Â Hz

The correct answer is:
8Ã1015 Hz

Solution:
KE1=hv1-hv0
KE2=hv2-hv0
As, KE1=2ÃKE2
â´ hv1-hv0=2hv2-hv0
Or, hv0=2hv2-hv1
Or , v0=2v1-v1
=2Ã2Ã1016-3.2Ã1016
=0.8Ã1016Hz=8Ã1015 Hz

Topic: Chemistry, Structure of Atom, JEE Main
A certain metallic surface is illuminated with monochromatic light of wavelength λ. The stopping potential for photo-electric current for this light is 3V0. If the same surface is illuminated with light of wavelength 2λ, the stopping potential is V0. The threshold wavelength for this surface for the photoelectric effect is

Options:
A 6λ
B 4λ
C λ4
D λ6

The correct answer is:
4λ

Solution:
Using Einstein's photoelectric equation

eVs=E-Ï
Vs=hcÎ»e-hcÎ»0e
3V0=hcÎ»e-hcÎ»0e ...(i)
V0=hc2Î»e-hcÎ»0e ...(ii)
by solving (i) and (ii)
0=-hc2Î»e+2hcÎ»0e
Î»0=4Î»

Topic: Physics, Dual Nature of Matter, JEE Main
A certain nuclide has a half-life period of 30 min. If a sample containing 600 atoms is allowed to decay for 90 min, how many atoms will remain ?

Options:
A 200 atoms
B 450 atoms
C 75 atoms
D 150 atoms

The correct answer is:
75 atoms

Solution:
$\begin{aligned} n &=\frac{\text { total time }}{\text { half- life period }}=\frac{90}{30}=3 \\ N &=N_{0} \times\left(\frac{1}{2}\right)^{n} \\ &=600 \times\left(\frac{1}{2}\right)^{3} \\ &=\frac{600}{8} \\ &=75 \text { atoms } \end{aligned}$

Topic: Chemistry, Chemical Kinetics, JEE Main
A certain nuclide has a half-life period of 30 min. If a sample containing 600 atoms is allowed to decay for 90 min, how many atoms will remain ?

Options:
A 200 atoms
B 450 atoms
C 75 atoms
D 150 atoms

The correct answer is:
75 atoms

Solution:
$\begin{aligned} n &=\frac{\text { total time }}{\text { half- life period }}=\frac{90}{30}=3 \\ N &=N_{0} \times\left(\frac{1}{2}\right)^{n} \\ &=600 \times\left(\frac{1}{2}\right)^{3} \\ &=\frac{600}{8} \\ &=75 \text { atoms } \end{aligned}$

Topic: Chemistry, Chemical Kinetics, MHT CET, MHT CET 2007