datavorous/entrance-exam-dataset huggingface dataset:
A beam of light has two wavelengths of $4972 Å$ and $6216 Å$ with a total intensity of $3.6 \times 10^{-3}$ $\mathrm{Wm}^{-2}$ equally distributed among the two wavelengths. The beam falls normally on an area of $1 \mathrm{~cm}^2$ of a clean metallic surface of work function $2.3 \mathrm{eV}$. Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photoelectrons liberated in $2 \mathrm{~s}$ is approximately:

Options:
A
$6 \times 10^{11}$

B
$9 \times 10^{11}$

C
$11 \times 10^{11}$

D
$15 \times 10^{11}$

The correct answer is:
$9 \times 10^{11}$

Solution:
Given, $\lambda_1=4972 Å$
and $\lambda_2=6216 Å$
and $\mathrm{I}=3.6 \times 10^{-3} \mathrm{Wm}^{-2}$
Intensity associated with each wavelength
$$
=\frac{3.6 \times 10^{-3}}{2}=1.8 \times 10^{-3} \mathrm{Wm}^{-2}
$$
Work function $\phi=\mathrm{hv}=\frac{\mathrm{hc}}{\lambda}$
$$
\begin{aligned}
&=\frac{\left(6.62 \times 10^{-34}\right)\left(3 \times 10^8\right)}{\lambda} \\
&=\frac{12.4 \times 10^3}{\lambda} \mathrm{ev}
\end{aligned}
$$
for different wavelengths
$$
\begin{array}{r}
\phi_1=\frac{12.4 \times 10^3}{\lambda_1}=\frac{12.4 \times 10^3}{4972}=2.493 \mathrm{eV} \\
=3.984 \times 10^{-19} \mathrm{~J} \\
\phi_2=\frac{12.4 \times 10^3}{\lambda_2}=\frac{12.4 \times 10^3}{6216}=1.994 \mathrm{eV} \\
=3.184 \times 10^{-19} \mathrm{~J}
\end{array}
$$
Work function for metallic surface $\phi=2.3$ $\mathrm{eV}$ (given)
$\phi_2 < \phi$
Therefore, $\phi_2$ will not contribute in this process.
Now, no. of electrons per $\mathrm{m}^2-\mathrm{s}=$ no. of photons per $\mathrm{m}^2-\mathrm{s}$
no. of electrons per $\mathrm{m}^2-\mathrm{s}$
$$
\begin{aligned}
&=\frac{1.8 \times 10^{-3}}{3.984 \times 10^{-19}} \times 10^{-4} \\
&\left(\because 1 \mathrm{~cm}^2=10^{-4} \mathrm{~m}^2\right)=0.45 \times 10^{12}
\end{aligned}
$$
So, the number of photo electrons liberated
in 2 sec.
$=0.45 \times 10^{12} \times 2$
$=9 \times 10^{11}$

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2014 (12 Apr Online)
A beam of light is converging towards a point $I$ on a screen. A plane glass plate whose thickness in the direction of the beam $=t$, refractive index $=\mu$, is introduced in the path of the beam. The convergence point is shifted by

Options:
A $t\left(1-\frac{1}{\mu}\right)$ away
B $t\left(1+\frac{1}{\mu}\right)$ away
C $t\left(1-\frac{1}{\mu}\right)$ nearer
D $t\left(1+\frac{1}{\mu}\right)$ nearer

The correct answer is:
$t\left(1-\frac{1}{\mu}\right)$ away

Solution:
The correct option is $\mathbf{1} t\left(1-\frac{1}{\mu}\right)$ away
Normal shift $\Delta x=\left(1-\frac{1}{\mu}\right) t$ and shift takes place in direction of ray.

Topic: Physics, Ray Optics, JEE Main
A beam of light is incident on a glass plate at an angle of $60^{\circ}$. The reflected ray is polarized. If angle of incidence is $45^{\circ}$ then angle of refraction is

Options:
A $\sin ^{-1}\left(\frac{1}{\sqrt{6}}\right)$
B $\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
C $\sin ^{-1}\left(\sqrt{\frac{3}{2}}\right)$
D $\cos ^{-1}\left(\sqrt{\frac{3}{2}}\right)$

The correct answer is:
$\sin ^{-1}\left(\frac{1}{\sqrt{6}}\right)$

Solution:
According to Brewster's law,
$$
\begin{array}{ll}
& \tan \theta_{\mathrm{B}}=\mathrm{n} \\
\therefore \quad & \tan 60^{\circ}=\mathrm{n}
\end{array}
$$
$\begin{array}{ll}\therefore \quad & n=\sqrt{3} \\ & \text { Now, } \frac{\sin \mathrm{i}}{\sin \mathrm{r}}=\mathrm{n} \\ \therefore \quad & \sin \mathrm{r}=\frac{\sin \mathrm{i}}{\mathrm{n}} \\ \therefore \quad & \sin \mathrm{r}=\frac{\sin 45^{\circ}}{\sqrt{3}} \\ \therefore \quad & \sin \mathrm{r}=\frac{1}{\sqrt{6}} \\ \therefore & \mathrm{r}=\sin ^{-1}\left(\frac{1}{\sqrt{6}}\right) . \ldots\left(\because \sin 45^{\circ}=\frac{1}{\sqrt{2}}\right)\end{array}$

Topic: Physics, Wave Optics, JEE Main
A beam of light travelling along X-axis is described by the electric field Ey=900sinωt-xc. The ratio of electric force to magnetic force on a charge q moving along Y-axis with a speed of 3×107 m s-1 will be :

[Given speed of light =3×108 m s-1 ]

Options:
A 1:1
B 1:10
C 10:1
D 1:2

The correct answer is:
10:1

Solution:
Given, Ey=900sinωt-ωxc

Here, E0=900

Electric field and electric force are related as FE=qE0

Magnetic force of a moving charge is given by FB=qvB0

Thus, the ratio FEFB=E0vB0

Using relation between magnetic and electric field, we have

FEFB=cv=3×1083×107=10:1

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2022 (27 Jul Shift 1)
A beam of monochromatic light is used to excite the electron in Li++ from the first orbit to the third orbit. The wavelength of monochromatic light is found to be x×10-10 m. The value of x is _____ .

[Given hc=1242 eV nm]

Options:

The correct answer is:
114

Solution:
Energy released in the transition will be,

ΔE=13.6×Z2112-132 =13.6×32×89 eV=108.8 eV

Now energy of photon can be written as,

hcλ=108.8 eV ⇒λ=12400108.8 Ao≈114 Ao

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2022 (27 Jun Shift 1)
A beam of monoenergetic electrons, which have been accelerated from rest by a potential $\mathrm{U}$, is used to form an interference pattern in a Young's Double Slit experiment. The electrons are now accelerated by potential $4 \mathrm{U}$. Then the fringe width

Options:
A remains the same
B is half the original fringe width
C is twice the original fringe width
D is one-fourth the original fringe width

The correct answer is:
is half the original fringe width

Solution:
$\mathrm{eV}=\frac{1}{2} \mathrm{mv}^{2} \mathrm{so}, \mathrm{eV}=\frac{p^{2}}{2 \mathrm{~m}}$
and
$$
\lambda=\frac{h}{p}, \Rightarrow V \propto \frac{1}{\lambda^{2}} \text { and } \beta=\frac{n \lambda D}{d}
$$
So as potential U becomes $4 \mathrm{U}$ then $\beta$ becomes half

Topic: Physics, Wave Optics, KVPY, KVPY 2015 (SB/SX)
A beam of parallel rays is brought to a focus by a plano-convex lens. A thin concave lens of the same focal length is joined to the first lens. The effect of this is

Options:
A the focal point shifts away from the lens by a small distance
B the focus remains undisturbed
C the focus shifts to infinity
D the focal point shifts towards the lens by a small distance

The correct answer is:
the focus shifts to infinity

Solution:
The combined focal length of plano-convex lens is
$$
\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}
$$
where, $f_{1}=\infty$ for the plane surface and $f_{2}=f$
$$
\begin{array}{ll}
\therefore \quad & \frac{1}{f}=\frac{1}{\infty}+\frac{1}{f} \\
\Rightarrow \quad f & =f
\end{array}
$$
Now, when concave lens of same focal length is joined to first lens, then combined focal length
$$
\begin{array}{ll}
\Rightarrow & \frac{1}{F}=\frac{1}{F_{1}}+\frac{1}{F_{2}} \\
\Rightarrow \quad \frac{1}{F} & =\frac{1}{f}-\frac{1}{f} \\
\Rightarrow \quad & \frac{1}{F}=0 \Rightarrow F=\infty
\end{array}
$$
Thus, the image can be focused on infinity or focus shifts to infinity.

Topic: Physics, Ray Optics, COMEDK, COMEDK 2019
A beam of parallel rays is brought to focus by a plano-convex lens. A thin concave lens of the same focal length is joined to the first lens. The effect of this is

Options:
A the focus shifts to infinity
B the focal point shifts towards the lens by a small distance
C the focal point shifts away from the lens by a small distance
D the focus remains undisturbed

The correct answer is:
the focus shifts to infinity

Solution:
The focal length of the combination is given by
$\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$
$$
\begin{aligned}
f_{1} &=\text { focal length of plano-convex lens }=f \\
f_{2} &=\text { focal length of concave lens }=-f \\
\therefore \quad \frac{1}{F} &=\frac{1}{f}+\frac{1}{-f}=0 \\
\Rightarrow \quad F &=\infty
\end{aligned}
$$
Hence, focus shifts to infinity.

Topic: Physics, Ray Optics, COMEDK, COMEDK 2013
A beam of parallel rays is brought to focus by a plano-convex lens. Then, a concave lens of the same focal length is joined to the first lens. The effect of this is

Options:
A The focus shifts to infinity
B The focal point shifts towards the lens by a small distance
C The focal point shifts away from the lens by a small distance
D The focus remains undisturbed

The correct answer is:
The focus shifts to infinity

Solution:
The focal length of plano-convex lens
F 1 =f
Now when concave lens of same focal length is joined to first lens , then combined focal length
1Fâ²=1F1+1F2
=1f-1f (âµF1=f, F2=-f)
=0
Fâ²= â
Thus, the image can be focused on infinity (â) or focus shifts to infinity.

Topic: Physics, Ray Optics, NEET
A beam of unpolarised light of intensity I0 is passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of 45° relative to that of A. The intensity of emergent light is :

Options:
A I04
B I0
C I02
D I08

The correct answer is:
I04

Solution:
Intensity of polarised light from polaroid A will be I02. Therefore,

Intensity of emergent light
=I02cos245°=I04

Topic: Physics, Wave Optics, JEE Main, JEE Main 2024 (30 Jan Shift 2)
A beam of unpolarised light of intensity I0 is passed through a polaroid A and then through another polaroid B which is oriented so that its principle plane makes an angle of 45 o relative to that of A. The intensity of the emergent light is:

Options:
A I04
B I08
C I0
D I02

The correct answer is:
I04

Solution:
Relation between intensities is
I R = I 0 2 cos 2 45 o = I 0 2 × 1 2 = I 0 4

Topic: Physics, Wave Optics, JEE Main, JEE Main 2013 (07 Apr)
A benzenoid organic compound A C 8 H 8 O gives B and white crystalline solid C with Cl2 and NaOH. On heating compound B gives a compound with unpleasant smell with CH3–CH2–NH2 and alcoholic KOH. Compound A is

Options:
A
B
C
D

The correct answer is:

Solution:
CHCl 3 + CH 3 − CH 2 − NH 2 +alc. KOH → Δ CH 5 CH 2 −NC+ H 2 O+KCl (unpleaseant smell)

Topic: Chemistry, Aldehydes and Ketones, JEE Main
A bi convex lens of focal length 10 cm is cut in two identical parts along a plane perpendicular to the principal axis. The power of each lens after cut is _____ D.

Options:

The correct answer is:
5

Solution:
The focal length of the bi-convex lens f can be calculated as follows:

1f=(μ-1)1R-1-R=2μ-1R ...1

When the lens is cut into two halves, the focal length f' of each part can be calculated as follows:

1f'=(μ-1)1R-1∞= μ-1R ...2

Hence, the ratio of the powers for both the lenses can be given by

P'P=1f'1f= ff'=μ-1R2μ-1R=12⇒P'= P2 ...3

Substitute the value of the known parameter into equation (3) to calculate the required power.

P'=10 D2= 5 D

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (13 Apr Shift 2)
A biased coin is tossed 10 times. The head is 2 times more likely to appear than the tail. The probability that 2nd tail and 4th tail occur at 4th and 10th tosses respectively is

Options:
A 1639
B 320310
C 32039
D 160310

The correct answer is:
32039

Solution:
PH=23,PT=13
2nd tail and 4th tail occur at 4th and 10th tosses respectively implies that there is one tail in first three tosses, 2nd tail at 4th toss, one tail at next five tosses and 4th tail at 10th toss.
So, the required probability = C 3 1 1 3 2 3 2 Ã 1 3 Ã C 5 1 1 3 1 2 3 4 Ã 1 3
=32039

Topic: Mathematics, Probability, JEE Main
A biased coin is tossed repeatedly until a tail appears for the first time. Heads is 3 times as likely to appear as tails. Let x be the number of tosses required. Assume that all the trials of tossing a biased coin are independent. Then, the conditional probability that xâ¥6, given that x>3, is equal to

Options:
A 916
B 116
C 12
D 35

The correct answer is:
916

Solution:
PH=34,PT=14
Px≥6x>3=Px≥6∩x>3Px>3=Px≥6Px≥4
Px≥6=34514+34614+34714+....=345141-34=345
Px≥4=34314+34414+34514+......=343141-34=343
Px≥6x>3=342=916

Topic: Mathematics, Probability, JEE Main
A biased coin is tossed repeatedly until a tail appears for the first time. The head is 3 times likely to appear as the tail. The probability that the number of tosses required will be more than 4, given that in first two toss no tail has occur, is

Options:
A 34
B 78
C 916
D 1132

The correct answer is:
916

Solution:
PH=34,PT=14
Let, x be the number of tosses required
Px≥5=34414+34514+34614+.…
=344141-34=344
Px≥3=34214+34314+34414+.…
=342141-34=342
Px≥5x≥3=Px≥5∩x≥3Px≥3=Px≥5Px≥3=344342=916

Topic: Mathematics, Probability, JEE Main
A biased coin that has a probability of getting heads as p ( 0<p<1 ) is tossed until a head appears for the first time. If the probability that the number of tosses required is even, is 25, then 9p is equal to

Options:

The correct answer is:
3

Solution:
p: probability of getting head
q: probability of getting tail

Such that, p+q=1
According to question we can get head in 2nd, 4th, 6th … tail.
Hence required probability =qp1+q2+q4+…=25
⇒ qp1-q2=25
⇒5pq=2−2 q 2
⇒5p( 1−p )=2−2 ( 1−p ) 2
⇒5p−5 p 2 =2−2( 1+ p 2 −2p )
⇒5p−5 p 2 =2−2−2 p 2 +4p
⇒3 p 2 =p
⇒p( 3p−1 )=0
⇒p=0,  p=13
∵p≠0
∴p= 1 3
⇒9p=3

Topic: Mathematics, Probability, JEE Main
A biased coin with probability \( p, 0 < p < 1 \) of heads is tossed until a head appears for the first time. If the probability that the number of tossed required is even is
\( \frac{2}{5} \), then \( p \) equal to

Options:
A \( \frac{1}{3} \)
B \( \frac{2}{3} \)
C \( \frac{2}{5} \)
D \( \frac{3}{5} \)

The correct answer is:
\( \frac{1}{3} \)

Solution:
Let q=1-p. Since, head appears first time in an even throw 2 or 4 or 6
∴ 25=qp+q3p+q5p+...
∴ 25=qp1-q2
⇒ 25=1-pp1-1-p2
⇒ 25=1-p2-p
⇒ 4-2P=5-5P ⇒ p=13

Topic: Mathematics, Probability, JEE Main
A biconvex lens has radii of curvature 20 cm each. If the refractive index of the material of the lens is 1.5, the power of the lens is:

Options:
A +20 D
B +5 D
C infinity
D +2 D

The correct answer is:
+5 D

Solution:
Power of the lens is,

P=1f=μ-11R1-1R2⇒P=1.5-110.2-1-0.2=0.5×10⇒P=+5 D

Topic: Physics, Ray Optics, JEE Main
A biconvex lens of focal length f1=10 cm is placed 40 cm in front of a concave mirror of focal length f2=7.50 cm, as shown in the figure. An object, 2 cm high, is placed 20 cm to the left of the lens.

The image formed by the lens, as rays travel to the right, is

Options:
A Real and erect
B Virtual and erect
C Real and inverted
D Virtual and inverted

The correct answer is:
Real and inverted

Solution:
From lens equation,
1v-1-20=110 â v=+20 cm
Magnification, m1=vu= +20-20 =-1
Image is real, inverted, and same size as object.

Topic: Physics, Ray Optics, JEE Main
A bicycle tyre is filled with air having pressure of 270 kPa at 27°C. The approximate pressure of the air in the tyre when the temperature increases to 36°C is

Options:
A 270 kPa
B 262 kPa
C 278 kPa
D 360 kPa

The correct answer is:
278 kPa

Solution:
Given here, P1=270 kPa, T1=27+273=300 K, T2=36+273=309 K

Ideal gas equation is given by PV=nRT.

Considering isochoric process, i.e., constant volume so P∝T

Thus, P1T1=P2T2

⇒P2=P1T1×T2=270×309300

=278 kPa

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2023 (29 Jan Shift 1)
A big circular coil of 1000 turns and average radius 10 m is rotating about its horizontal diameter at 2 rad s-1. If the vertical component of earth's magnetic field at that place is 2×10-5 T and electrical resistance of the coil is 12.56 Ω, then the maximum induced current in the coil will be:

Options:
A 1.5 A
B 1 A
C 2 A
D 0.25 A

The correct answer is:
1 A

Solution:
The magnetic flux is given by,

ϕ=BAcosωt. Therefore, induced EMF,

ε=-dϕdt=-BAωsinωt

For maximum current sinωt=1

⇒imax=BAωR=BNπr2ωR=2×10-5×1000×3.14×100×212.56=1 A

Topic: Physics, Electromagnetic Induction, JEE Main
A big drop is formed by coalescing 1000 small droplets of water. The ratio of surface energy of 1000 droplets to that of energy of big drop is $\frac{10}{x}$. The value of $x$ is __________

Options:

The correct answer is:
1

Solution:
$\begin{aligned} 1000 \frac{4}{3} \pi r^3 & =\frac{4}{3} \pi R^3 \\ 10 r & =R \\ R & =10 r\end{aligned}$
$\begin{aligned} \frac{\text { S.E. of } 1000 \text { drops }}{\text { S.E. of Big drop }} & =\frac{1000\left(4 \pi \mathrm{r}^2\right) \mathrm{T}}{4 \pi \mathrm{R}^2 \mathrm{~T}} \\ & =\frac{1000 \times \mathrm{r}^2}{(10 \mathrm{r})^2}=10=\frac{10}{\mathrm{x}} \\ \therefore \quad x & =1\end{aligned}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2024 (06 Apr Shift 1)
A big drop is formed by coalescing 1000 small droplets of water. The surface energy will become :

Options:
A 100 times
B 10 times
C 1100th
D 110th

The correct answer is:
110th

Solution:
Lets say radius of small droplets is r and that of big drop is R. Total volume of the drops will remains unchanged. Therefore,

43πR3=100043πr3

⇒R=10r

Now, initial surface energy is, Ui=10004πr2 S and the final surface energy is,

Uf=4πR2 S

=1004πr2S

Clearly, Uf=110Ui.

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2024 (01 Feb Shift 2)
A big drop is formed by coalescing 1000 small identical drops of water. If E1 be the total surface energy of 1000 small drops of water and E2 be the surface energy of single big drop of water, the E1 : E2 is x: 1, where x=________.

Options:

The correct answer is:
10

Solution:
Let radius of the small drop be r and the radius of the big drop be R. Then, total volume will remain constant, therefore

1000×ρ43πr3=43πR3ρ

⇒R=10r

Initial total surface energy, E1=1000×4πr2×S and

E2=4π(10r)2 S

⇒E1E2=101Hence, x=10.

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2024 (30 Jan Shift 2)
A big explosion on the moon cannot be heard on the earth because

Options:
A The explosion produces high frequency sound waves which are inaudible
B Sound waves required a material medium for propagation
C Sound waves are absorbed in the moon's atmosphere
D Sound waves are absorbed in the earth's

The correct answer is:
Sound waves required a material medium for propagation

Solution:
Sound waves require a material medium for propagation. There is vacuum between Earth and Moon. In vacuum, there are no molecules (material medium) through which sound waves produced by the mechanical vibrations of the explosion can propagate. Therefore, big explosion on the Moon cannot be heard on the Earth.

Topic: Physics, Waves and Sound, JEE Main
A big water drop is divided into 8 equal droplets. $\Delta \mathrm{P}_{\mathrm{S}}$ and $\Delta \mathrm{P}_{\mathrm{B}}$ be the excess pressure inside a smaller and bigger drop respectively. The relation between $\Delta \mathrm{P}_{\mathrm{S}}$ and $\Delta \mathrm{P}_{\mathrm{B}}$ is

Options:
A $\Delta \mathrm{P}_{\mathrm{B}}=\Delta \mathrm{P}_{\mathrm{S}}$
B $\Delta \mathrm{P}_{\mathrm{B}}=\frac{1}{2} \Delta \mathrm{P}_{\mathrm{S}}$
C $\Delta \mathrm{P}_{\mathrm{B}}=\frac{1}{4} \Delta \mathrm{P}_{\mathrm{S}}$
D $\Delta \mathrm{P}_{\mathrm{B}}=2 \Delta \mathrm{P}_{\mathrm{S}}$

The correct answer is:
$\Delta \mathrm{P}_{\mathrm{B}}=\frac{1}{2} \Delta \mathrm{P}_{\mathrm{S}}$

Solution:
Volume of 8 smaller drop $=$ Volume of the bigger drop
$$
\begin{aligned}
& \therefore 8 \times \frac{4}{3} \pi r^3=\frac{4 \pi}{3} R^3 \\
& \therefore 2 r=R \text { or } r=\frac{R}{2}
\end{aligned}
$$
Excess pressure $\Delta \mathrm{P}_{\mathrm{s}}=\frac{2 \mathrm{~T}}{\mathrm{r}}$ and $\Delta \mathrm{P}_{\mathrm{B}}=\frac{2 \mathrm{~T}}{\mathrm{R}}$
$$
\therefore \frac{\Delta \mathrm{P}_{\mathrm{B}}}{\Delta \mathrm{P}_{\mathrm{S}}}=\frac{\mathrm{r}}{\mathrm{R}}=\frac{1}{2}
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A bimetallic strip is made of aluminium and steel $\left(\alpha_{\mathrm{Al}}>\alpha_{\text {steel }}\right)$. On heating, the strip will

Options:
A remain straight
B get twisted
C bend with aluminium on concave side
D bend with steel on concave side

The correct answer is:
get twisted

Solution:
As we know that,
$$
\begin{array}{rlrl}
\Delta l & =l_0 \alpha \Delta T \\
\Rightarrow & \alpha & =\frac{\Delta l}{l_0 \Delta T}
\end{array}
$$
where, $\Delta l$ is change in length, $l_0$ is initial length,
$\Delta T$ is change in temperature and $\alpha$ is coefficient of linear expansion.
Now,
$$
\begin{array}{ll}
\because & \alpha_{\mathrm{Al}}>\alpha_{\text {stecl }} \\
\therefore & \Delta l_{\text {Al }}>\Delta l_{\text {stecl }}
\end{array}
$$
That means aluminium expands more than steel.

Topic: Physics, Thermal Properties of Matter, AP EAMCET, AP EAMCET 2021 (23 Aug Shift 1)
A bimetallic strip is made of aluminium and steel $\left(\alpha_{\mathrm{Al}}\right\rangle$ $\left.\alpha_{\text {steel }}\right)$. On heating, the strip will

Options:
A
remain straight

B
get twisted

C
will bend with aluminium on concave side

D
will bend with steel on concave side

The correct answer is:
will bend with steel on concave side

Solution:
If strips of aluminium and steel are fixed together on metallic strip and both are heated then $\left(\alpha_{\mathrm{Al}}>\alpha_{\text {steel }}\right)$ aluminium will exapnd more because the metallic strip with higher coefficient of linear expansion $\left(\alpha_{\mathrm{Al}}\right)$ will expand more Thus it should have larger radius of curvature. Hence, aluminium will be on convex side.

Topic: Physics, Thermal Properties of Matter
A binary liquid solution is prepared by mixing $n$-heptane and ethanol. Which one of the following statements is correct regarding the behaviour of the solution?

Options:
A
The solution formed is an ideal solution

B
The solution is non-ideal, showing +ve deviation from Raoult's law.

C
The solution is non-ideal, showing -ve deviation from Raoult's law.

D
n-heptane shows +ve deviation while ethanol shows -ve deviation from Raoult's law.

The correct answer is:
The solution is non-ideal, showing +ve deviation from Raoult's law.

Solution:
The interactions between $\mathrm{n}$-heptane and ethanol are weaker than that in pure components.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2009
A binary star consists of two stars A (mass= 22 Ms) and B (mass= 11 Ms), where Ms is the mass of the sun. They are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star system to the angular momentum of star B about the centre of mass is

Options:

The correct answer is:
1.5

Solution:
L Total L B = I A + I B ω I B · ω

(as ω will be same in both cases)
LTotalLB=IAIB+1=mArA2mBrB2+1

as mArA=mBrB

LTotalLB=mBmA+1=0.5+1=1.5

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A binary star consists of two stars $A$ (mass $2.2 M_S$ ) and $B$ (mass $11 M_s$ ), where $M_S$ is the mass of the sun. They are separated by distance $d$ and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of $\operatorname{star} B$ about the centre of mass is

Options:

The correct answer is:
6

Solution:
$\frac{L_{\text {Total }}}{L_B}=\frac{\left(I_A+I_B\right) \omega}{I_B \cdot \omega}$ (as $\omega$ will be same in both cases)
$$
\begin{aligned}
& =\frac{I_A}{I_B}+1=\frac{m_A r_A^2}{m_B r_B^2}+1 \\
& \left.=\frac{r_A}{r_B}+1 \quad \quad \text { (as } m_A r_A=m_B r_B\right) \\
& =\frac{11}{2.2}+1 \quad \quad\left(\text { as } r \propto \frac{1}{m}\right) \\
& =6
\end{aligned}
$$
$\therefore$ The correct answer is 6 .

Topic: Physics, Gravitation, JEE Main
A biodegradable polyamide can be made from:

Options:
A Glycine and isoprene
B Hexamethylene diamine and adipic acid
C Glycine and aminocaproic acid
D Styrene and caproic acid

The correct answer is:
Glycine and aminocaproic acid

Solution:
A biodegradable polyamide nylon-2-Nylon-6
made from glycine and amino caproic acid

Topic: Chemistry, Polymers, JEE Main, JEE Main 2021 (25 Jul Shift 2)