datavorous/entrance-exam-dataset huggingface dataset:
A plane polarized monochromatic EM wave is travelling a vacuum along $z$ direction such that at $\mathrm{t}=\mathrm{t}_1$ it is found that the electric field is zero at a spatial point $z_1$. The next zero that occurs in its neighbourhood is at $z_2$. The frequency of the electromagnetic wave is:

Options:
A
$\frac{3 \times 10^8}{\left|z_2-z_1\right|}$

B
$\frac{6 \times 10^8}{\left|\mathrm{z}_2-\mathrm{z}_1\right|}$

C
$\frac{1.5 \times 10^8}{\left|\mathrm{z}_2-\mathrm{z}_1\right|}$

D
$\frac{1}{t_1+\frac{\left|z_2-z_1\right|}{3 \times 10_8}}$

The correct answer is:
$\frac{3 \times 10^8}{\left|z_2-z_1\right|}$

Solution:
Using $E=E_0-e^i(k z-\omega t)$
Given, at $t=t_1, z=z_1, E=0$
the next zero that occurs in it's neighborhood is at $z_2$, the frequency of the electromagnetic wave at $t_2$
$$
\begin{aligned}
&e^{i\left(k z_1-\omega t_1\right)}=e^{i\left(k z_2-\omega t_2\right)} \\
&k z_1-\omega t_1=k z_2-\omega t_2 \\
&\left(t_2-t_1\right) \omega=k\left(z-z_1\right) \\
&\text { where } k=\frac{2 \pi}{\lambda}=2 \pi v \\
&\left(t_2-t_1\right)=\frac{2 \pi}{\lambda \times 2 \pi v}\left(z_2-z_1\right)
\end{aligned}
$$
$$
\begin{aligned}
&\quad=\frac{1}{x \times v}\left(z_2-z_1\right) \\
&\Rightarrow \lambda \times v=\frac{\left(z_2-z_1\right)}{\left(t_2-t_1\right)}=C \\
&\quad\left(t_2-t_1\right)=\frac{\left(z_2-z_1\right)}{C} \\
&\text { Frequency is } f \propto \frac{1}{t} \text { then } \frac{1}{\left(t_2-t_1\right)}=\frac{C}{\left(z_2-z_1\right)} \\
&\therefore \text { Frequency, } f=\frac{3 \times 10^8}{\left(z_2-z_1\right)}
\end{aligned}
$$

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
A plane progressive wave is given by $y=2 \cos 2 \pi(330 \mathrm{t}-x) \mathrm{m}$. The frequency of the wave is :

Options:
A $330 \mathrm{~Hz}$
B $660 \mathrm{~Hz}$
C $340 \mathrm{~Hz}$
D $165 \mathrm{~Hz}$

The correct answer is:
$330 \mathrm{~Hz}$

Solution:
$\begin{aligned} & y=2 \cos 2 \pi(330 t-x) m \\ & y=A \cos (\omega t-k x) \\ & \text { by comparing } \omega=2 \pi \times 330 \\ & 2 \pi f=2 \pi \times 330 \\ & f=330\end{aligned}$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2024 (08 Apr Shift 2)
A plane wave \(y=a \sin (\omega t-k x)\) propagates through a stretched string. The particle velocity versus \(x\) graph at \(t=0\) is

Options:
A

B

C

D

The correct answer is:

Solution:
Plane wave is given as
\(y=a \sin (\omega t-k x)\)
at \(t=0\),
\(\begin{aligned}
& y=a \sin (\omega \times 0-k x) \\
& \Rightarrow \quad=a \sin (-k x) \Rightarrow-a \sin k x \\
\end{aligned}\)
Particle velocity,
\(\begin{aligned}
v_{\mathrm{pa}} & =\frac{d y}{d t}=\frac{d}{d t}(-a \sin k x) \\
& =-a k \cos k x=-a k \cos \frac{2 \pi}{\lambda} x\left[\because k=\frac{2 \pi}{\lambda}\right] \\
& =-\frac{2 \pi a}{\lambda} \cos \frac{2 \pi x}{\lambda} \quad \ldots (i)
\end{aligned}\)
From Eq. (i),
When \(x=0, v_{\mathrm{pa}}=\frac{-2 \pi a}{\lambda} \cdot \cos 0=\frac{-2 \pi a}{\lambda}\)
When \(x=\frac{\lambda}{4}, v_{\mathrm{pa}}=\frac{-2 \pi a}{\lambda} \cdot \cos \frac{\pi}{2}=0\)
When \(x=\frac{\lambda}{2}, v_{\mathrm{pa}}=\frac{-2 \pi a}{\lambda} \cdot \cos \pi=\frac{2 \pi a}{\lambda}\)
When \(x=\frac{3 \lambda}{4}, v_{\mathrm{pa}}=\frac{-2 \pi a}{\lambda} \cos \left(\frac{-\pi}{2}\right)=0\)
When \(x=\lambda, v_{\mathrm{pa}}=\frac{-2 \pi a}{\lambda} \cdot \cos 2 \pi=\frac{-2 \pi a}{\lambda}\)
Hence, correct graph is

Topic: Physics, Waves and Sound, JEE Main
A plane which bisects the angle between the two given planes 2x-y+2z-4=0 and x+2y+2z-2=0, passes through the point

Options:
A 2, 4, 1
B (1,-4, 1)
C 1, 4,-1
D 2,-4, 1

The correct answer is:
2,-4, 1

Solution:
Equation of bisector of angle
2x-y+2z-43=±x+2y+2z-23
(+) gives x-3y=2
- gives 3x+y+4z=6
therefore 2,-4, 1 satisfies the above equation.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (12 Apr Shift 2)
A plane which passes through the point $(3,2,0)$ and the line $\frac{x-4}{1}=\frac{y-7}{5}=\frac{z-4}{4}$ is

Options:
A
x - y + z = 1

B
x + y + z = 5

C
x + 2y - z = 1

D
2x - y + z = 5

The correct answer is:
x - y + z = 1

Solution:

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2002
A plane $x$ passes through the point $(1,1,1)$. If $b, c, a$ are the direction ratios of a normal to the plane, where $a, b, c(a < b < c)$ are the factors of 2001 , then the equation of the plane is

Options:
A $29 x+31 y+3 z=63$
B $23 x+29 y-29 z=23$
C $23 x+29 y+3 z=55$
D $31 x+37 y+3 z=71$

The correct answer is:
$23 x+29 y+3 z=55$

Solution:
We have,
$$
\begin{aligned}
& 2001=3.23 .29 \\
& \text { so, } a=3, b=23, c=29
\end{aligned}
$$
Since, direction ratios of a normal are $b, c, a$ then the equation of plane is
$$
\begin{aligned}
b x+c y+a z+d & =0 \\
\Rightarrow \quad 23 x+29 y+3 z+d & =0
\end{aligned}
$$
It passes through $(1,1,1)$
$$
\begin{aligned}
& \therefore \quad 23+29+3+d=0 \\
& \Rightarrow \quad 55+d=0 \\
& \Rightarrow \quad d=-55 \\
&
\end{aligned}
$$
Thus, equation of the plane is
$$
\begin{aligned}
& 23 x+29 y+3 z-55=0 \\
& \Rightarrow \quad 23 x+29 y+3 z=55 \\
&
\end{aligned}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A planet has double the mass of the earth. Its average density is equal to that of the earth. An object weighing W on earth will weigh on that planet:

Options:
A 214W
B 213W
C 2W
D 223W

The correct answer is:
213W

Solution:
The formula for acceleration due to gravity is

g=GMR2

Let the density of earth and planet be ρ, and the radius of the planet be R'. Let R be the radius of the earth and g be the acceleration due to gravity of earth and g' be the acceleration due to gravity of planet.

So, the masses are

M=43πR3ρ ...(i)

M'=43πR'3ρ ...(ii)

So, g'g=GM'R'2GMR2=M'MRR'2 ...(iii)

From equation (i) and (ii) we get R'=213R

Substituting the value in equation (iii),

g'g=212-23=213g

Weight is given by W'=mg'=213W

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (06 Apr Shift 1)
A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is $11 \mathrm{kms}^{-1}$, the escape velocity from the surface of the planet would be

Options:
A
$1.1 \mathrm{kms}^{-1}$

B
$11 \mathrm{kms}^{-1}$

C
$110 \mathrm{kms}^{-1}$

D
$0.11 \mathrm{kms}^{-1}$

The correct answer is:
$110 \mathrm{kms}^{-1}$

Solution:
$$
v_{\mathrm{esc}}=\sqrt{\frac{2 \mathrm{GM}}{R}}=\sqrt{\frac{2 \mathrm{G} \times 10 \mathrm{M}}{\mathrm{R} / 10}}=10 \times 11=110 \mathrm{~km} / \mathrm{s}
$$

Topic: Physics, Gravitation, JEE Main, JEE Main 2008
A planet is revolving around the sun in an elliptical orbit having eccentricity e=Ï4. If the time period of revolution is 16 months then find the time taken by the planet in going from A to B in months.

Options:

The correct answer is:
6

Solution:
t=Tπabπab4+12abe
t=T14+e2π
t=1814+18
t = 6 months

Topic: Physics, Gravitation, JEE Main
A planet moves around the sun in an elliptical orbit with the sun at one of its foci. The physical quantity associated with the motion of the planet that remains constant with time is

Options:
A velocity
B centripetal force
C linear momentum
D angular momentum

The correct answer is:
angular momentum

Solution:
A planet revolves around the sun, is an elliptical orbit under the effect of gravitational pull on the planet.

So, torque, $C=r \times F=r F \sin 180^{\circ} n=0$
$$
\text{As } \mathrm{C}=\frac{\mathrm{dL}}{d t} ; \mathrm{soL}=\mathrm{a} \text { constant }
$$
$\Rightarrow$ Angular momentum is constant.

Topic: Physics, Rotational Motion, JEE Main
A planet moves around the sun in an elliptical orbit with the sun at one of its foci. The physical quantity associated with the motion of the planet that remains constant with time is

Options:
A velocity
B centripetal force
C linear momentum
D angular momentum

The correct answer is:
angular momentum

Solution:
A planet revolves around the sun, is an elliptical orbit under the effect of gravitational pull on the planet.

So, torque, $C=r \times F=r F \sin 180^{\circ} n=0$
$$
\text{As } \mathrm{C}=\frac{\mathrm{dL}}{d t} ; \mathrm{soL}=\mathrm{a} \text { constant }
$$
$\Rightarrow$ Angular momentum is constant.

Topic: Physics, Rotational Motion, WBJEE, WBJEE 2013
A planet moves around the sun (mass = Ms) in an elliptical orbit such that its minimum and maximum distance from the sun are r and R respectively. The period of revolution of this planet around the sun is

Options:
A T=πr+R32GMs
B T=πr+R33GMs
C T=πr+R3GMs
D T=π2r+R3GMs

The correct answer is:
T=πr+R32GMs

Solution:
The length of the semi-major axis of the elliptical orbit of the planet is

a=r+R2

If we assume that there is a hypothetical planet which moves around the sun in a circular orbit of radius r0=r+R2, then the time period of this hypothetical planet and our given planet will be same. The time period of a planet around the sun in circular orbit is

T=2πr30GMs=2πr+R23GMs
T=πr+R32GMs

Topic: Physics, Gravitation, JEE Main
A planet moving around sun sweeps area $\mathrm{A}_{1}$ in 2 days, $\mathrm{A}_{2}$ in 3 days and $\mathrm{A}_{3}$ in 6 days. Then, the relation between $\mathrm{A}_{1}, \mathrm{~A}_{2}$ and $\mathrm{A}_{3}$ is

Options:
A $6 \mathrm{~A}_{1}=3 \mathrm{~A}_{2}=2 \mathrm{~A}_{3}$
B $3 A_{1}=2 A_{2}=A_{3}$
C $2 \mathrm{~A}_{1}=3 \mathrm{~A}_{2}=6 \mathrm{~A}_{3}$
D $3 A_{1}=2 A_{2}=6 A_{3}$

The correct answer is:
$3 A_{1}=2 A_{2}=A_{3}$

Solution:
By Kepler's second law of motion,
$\frac{A_{1}}{t_{1}}=\frac{A_{2}}{t_{2}}=\frac{A_{3}}{t_{3}}$
$\frac{A_{1}}{2}=\frac{A_{2}}{3}=\frac{A_{3}}{6}$
$\frac{3 A_{1}}{6}=\frac{2 A_{2}}{6}=\frac{A_{3}}{6}$
$3 A_{1}=2 A_{2}=A_{3}$

Topic: Physics, Gravitation, JEE Main
A planet moving around the sun sweeps area A1 in 2 days, A2 in 4 days and A3 in 9 days. Then, relation between them

Options:
A A1=A2=A3
B 9A1=3A2=2A3
C 18A1=9A2=4A3
D 3A1=4A2=6A3

The correct answer is:
18A1=9A2=4A3

Solution:
From Kepler's law of area, the aerial velocity remains constant, for a planet moves around the sun.
We have, A12=A24=A39
⇒18A1=9A2=4A3

Topic: Physics, Gravitation, JEE Main
A planet of mass m is in an elliptical orbit about the sun m<<Msun with an orbital period T. If A be the area of orbit, then its angular momentum would be

Options:
A 2mAT
B mAT

C mA2T
D 2mAT

The correct answer is:
2mAT

Solution:
using Kepler's IInd law ; areal velocity = constant =L2m
⇒dAdt = L2m
⇒ L=2mdAdt
L=2mAT

Topic: Physics, Gravitation, JEE Main
A planet of the sun is in a circular orbit around the sun, midway between the sun and the earth. Then

Options:
A the period of the planet is nearly 229 days
B the period of the planet is nearly 129 days
C the speed of the planet is greater than the escape velocity of the earth
D the acceleration of the planet is two times the acceleration of the earth

The correct answer is:
the period of the planet is nearly 129 days

Solution:
T2â r3
T1T2=rr/23/2
365 daysT=22; T=36522 days â129 days

Topic: Physics, Gravitation, JEE Main
A planet revolves about the sun in an elliptical orbit of semi-major axis 2Ã1012 m. The areal velocity of the planet when it is nearest to the sun is 4.4Ã1016 m2 s-1. If the least distance between the planet and the sun is 1.8Ã1012 m, then the minimum speed of the planet is

Options:
A vmin=40 km s-1
B vmin=30 km s-1
C vmin=10 km s-1
D vmin=20 km s-1

The correct answer is:
vmin=40 km s-1

Solution:
The areal velocity of a planet around the sun is

dAdt=L2m, where L is the angular momentum of the planet around the sun and m is its mass.

4.4×1016=mvminrmax2m

2×4.4×1016=vmin4×1012-1.8×1012

vmin=40 km s-1

Topic: Physics, Gravitation, JEE Main
A planet revolves about the sun in elliptical orbit of semi-major axis 2Ã 10 12 m. The areal velocity of the planet when it is nearest to the sun is 4.4Ã1016 m2/s.The least distance between planet and the sun is 1.8Ã1012 m. The minimum speed of the planet in km/s is 10K. Value of K is

Options:
A 4
B 3
C 1
D 2

The correct answer is:
4

Solution:
Area covered by line joining planet and sun in time dt is ds=12x2dθ; Areal velocity =dsdt=12x2dθ/dt=12x2ω, where x = distance between planet and sun ω = angular speed of planet about the sun. From kepler's second low, areal velocity of planet is constant. At farthest position
A = ds dt = 1 2 2R - r 2 ω = 1 2 2R - r V B
A=122R - rVB
or V B = 2A 2R - r
(least speed). (Using values)
V B = 40 km/s
Thus, K = 4

Topic: Physics, Gravitation, JEE Main
A planet revolving in elliptical orbit has :

A. a constant velocity of revolution.

B. has the least velocity when it is nearest to the sun.

C. its areal velocity is directly proportional to its velocity.

D. areal velocity is inversely proportional to its velocity.

E. to follow a trajectory such that the areal velocity is constant.

Choose the correct answer from the options given below:

Options:
A A only
B C only
C E only
D D only

The correct answer is:
E only

Solution:
As per Kepler’s second law “The areal velocity of a planet revolving around the sun in elliptical orbit remains constant which implies the angular momentum of a planet remains constant”. Hence (option E) is the correct option.

Topic: Physics, Gravitation, JEE Main, JEE Main 2021 (26 Feb Shift 1)
A planet takes 200 days to complete one revolution around the Sun. If the distance of the planet from Sun is reduced to one fourth of the original distance, how many days will it take to complete one revolution?

Options:
A 25
B 50
C 100
D 20

The correct answer is:
25

Solution:
The Law of Periods: The square of the period of any planet is proportional to the cube of the semimajor axis of its orbit.

T2∝r3

Therefore, we can write

T12r13=T22r23

⇒(200)2r3=T22r43

⇒200×2004×4×4=T22

⇒T2=25 days

Topic: Physics, Gravitation, JEE Main, JEE Main 2024 (29 Jan Shift 2)
A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches 30°, the box starts to slip and slides 4.0 m down the plank in 4.0 s. The coefficients of static and kinetic friction between the box and the plank will be, respectively:

Options:
A 0.6 and 0.5
B 0.5 and 0.6
C 0.4 and 0.3
D 0.6 and 0.6

The correct answer is:
0.6 and 0.5

Solution:
μs=tan⁡30o=13
⇒μs=0.57=0.6
S=ut+12at2
4=12a42⇒a=12=0.5
a=g sinθ-μkgcosθ
⇒μK=0.93≈0.5

Topic: Physics, Laws of Motion, JEE Main
A plano - convex lens (focal length f2 , refractive index μ2, radius of curvature R) fits exactly into a plano - concave lens (focal length f1, refractive index μ1, radius of curvature R). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be:

Options:
A Rμ2 - μ1
B f1-f2
C f1+f2
D 2f1f2f1+f2

The correct answer is:
Rμ2 - μ1

Solution:
1f1=Î¼1-11-R-1Ï 1f2=Î¼2-11R-1Ï
When joined together
1feq=1f1+1f2=Î¼2-1-R+Î¼2-1R=Î¼2-Î¼1R
âfeq=RÎ¼2-Î¼1

Topic: Physics, Ray Optics, JEE Main, JEE Main 2019 (12 Jan Shift 2)
A plano-convex lens Î¼ =32 has a radius of curvature R=15 cm and is placed at a distance z from a concave lens of focal length 20 cm as shown. At what distance x0 should a point object be placed from the plano-convex lens, so that position of the final image is independent of z?

Options:
A 20Â cm
B 30 cm
C 40Â cm
D 60Â cm

The correct answer is:
30 cm

Solution:
The position of the final image will be independent of z, only if the rays of light are incident on the concave lens as a beam parallel to the principal axis. This will be possible only when the object is kept at focus of the plano-convex lens.
â´1x0=1f1=(Î¼-1)1R1-1R2=32-1115-1â
âx0=30 cm

Topic: Physics, Ray Optics, JEE Main
A plano-convex lens Î¼ = 1.5 has maximum thickness of 1 mm. If the diameter of its aperture is 4 cm, then what is the value of its focal length in the air (in cm)?

Options:

The correct answer is:
40

Solution:
R 2 = 2 2 + R - 0.1 2 ⇒ R 2 = 4 + R 2 + 0.01 - 0.2 R
0.2R=4.01 ⇒R=4.010.20
=40120⇒R=20cm
from lens formula
1 f = a μ g - 1 ( 1 R 1 - 1 R 2 ) ⇒ 1 f = 1 . 5 - 1 ( 1 2 0 - 1 ∞ )
1 f = 1 4 0 ⇒ f = 4 0 cm

Topic: Physics, Ray Optics, JEE Main
A plano-convex lens becomes an optical system of 28 cm focal length when its plane surface is silvered and illuminated from left to right as shown in fig-A If the same lens is instead silvered on the curved surface and illuminated from another side as in fig-B, it acts as an optical system of focal length 10 cm. The refractive index of the material of the lens is:

Options:
A 1.55
B 1.50
C 1.75
D 1.51

The correct answer is:
1.55

Solution:
Case-1
1 f 1 = (μ−1) R ,f=-28 ⇒P=2p1+p2
⇒128=2μ-1R
Case-2
1f1=(μ−1)R, f2=R2 , f =−10 cm
⇒P=2P1+P2⇒110=2μ-12+2R⇒110=228+2R
2R=110-228=18280⇒R=2809 cm⇒128=2μ-12809
μ-1=59⇒μ=1+59=149=1.55.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2018 (15 Apr)
A planoconvex lens becomes an optical system of $28 \mathrm{~cm}$ focal length when its plane surface is silvered and illuminated from left to right as shown in Fig-A. If the same lens is instead silvered on the curved surface and illuminated from other side as in Fig. B, it acts like an optical system of focal length $10 \mathrm{~cm}$. The refractive index of the material of lens is

Options:
A
1.50

B
1.55

C
1.75

D
1.51

The correct answer is:
1.55

Solution:
Case-1
$\frac{1}{f_1}=\left(\frac{\mu-1}{R}\right) \mathrm{f}=-28$
$\mathrm{P}=2 \mathrm{P}_1+\mathrm{P}_2 \Rightarrow \frac{1}{28}=2\left(\frac{\mu-1}{R}\right)$
$\left(\because\right.$ Power, $\left.P=\frac{1}{f} \& f_{\text {plane mirror }}=\infty\right)$
Case-2
$\frac{1}{f_1}=\left(\frac{\mu-1}{R}\right) \quad f_2=-\frac{R}{2} \quad f=-10 \mathrm{~cm}$
$\mathrm{P}=2 \mathrm{P}_1+\mathrm{P}_2 \Rightarrow \frac{1}{10}=2\left(\frac{\mu-1}{2}\right)+\frac{2}{R}$
or, $\quad \frac{1}{10}=\frac{1}{28}+\frac{2}{R} \Rightarrow \frac{2}{R}=\frac{1}{10}-\frac{2}{28}=\frac{18}{280}$
or, $\quad \mathrm{R}=\frac{280}{9} \mathrm{~cm}$
or, $\quad \frac{1}{28}=2\left(\frac{\mu-1}{280}\right) 9 \quad \Rightarrow \quad \mu-1=\frac{5}{9}$
$\therefore \quad \mu=1+\frac{5}{9}=\frac{14}{9}=1.55$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
A plano-convex lens fits exactly into a plano-concave lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices $\mu_1$ and $\mu_2$ and $R$ is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is

Options:
A $\frac{R}{2\left(\mu_1+\mu_2\right)}$
B $\frac{R}{2\left(\mu_1-\mu_2\right)}$
C $\frac{R}{\left(\mu_1-\mu_2\right)}$
D $\frac{2 R}{\left(\mu_2-\mu_1\right)}$

The correct answer is:
$\frac{R}{\left(\mu_1-\mu_2\right)}$

Solution:
Focal length of the combination
$$
\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}
$$
We have $f_1=\frac{R}{\left(\mu_1-1\right)}$ and $f_2=\frac{R}{\left(\mu_2-1\right)}$ or $\quad \frac{1}{f_1}=\frac{R}{\left(\mu_1-1\right)}$ or $\frac{1}{f_2}=-\frac{R}{\left(\mu_2-1\right)}$
Putting these values in Eq. (i), we get
$$
\begin{aligned}
\frac{1}{f} & =\frac{\left(\mu_1-1\right)}{R}-\frac{\left(\mu_2-1\right)}{R} \\
& =\frac{\left[\mu_1-1-\mu_2+1\right]}{R}=\frac{\mu_1-\mu_2}{R}
\end{aligned}
$$

Topic: Physics, Ray Optics, JEE Main
A plano-convex lens has a radius of curvature of the curved surface equal to 30 cm and a refractive index equal to 1.5, then what is the focal length (in cm ) of the lens?

Options:

The correct answer is:
60.00

Solution:
1f=(μ-1)1R1-1R2R1=∝R2=-30cm1f=(1.5-1)1∞-1-301f=0.530f=60cm

Topic: Physics, Ray Optics, JEE Main
A plano-convex lens has diameter 5 cm and its thickness at the centre is 0.25 cm. If the speed of light inside the lens is 2×108 ms-1, then what is the focal length (in cm) of the lens?

Options:

The correct answer is:
25.25

Solution:
R. I. of lens,

μ=cv=3×1082×108=32

From ΔACO, radius of curvature of lens is,

R2=R12+(R-0.25)2

∴ R2=(2.5)2+(R-0.25)2

∴ R2=6.25+R2+0.0625-0.5R

∴ 0.5R=6.3125

∴ R=12.625cm

F=Rμ-1=12.62532-1=25.25 cm

Topic: Physics, Ray Optics, JEE Main
A plant in which sporophytic generation is represented by zygote:

Options:
A Dryopteris
B Chlamydomonas
C Selaginella
D Pinus

The correct answer is:
Chlamydomonas

Solution:
The plant body of Chlamydomonas is gametophyte (haploid). It reproduces asexually by zoospore formation and sexually by gametes. Gametes are isogametes which fuse to produce diploid zygote which is the only structure representing sporophytic generation.
In Pinus main plant body is sporophyte, same is the case with Selaginella and Dryopteris.

Topic: Biology, Plant Kingdom, NEET
A plant in your garden avoids photorespiratory losses, has improved water use efficiency, shows high rates of photosynthesis at high temperatures and has improved efficiency of nitrogen utilization. In which of the following physiological groups would you assign this plant?

Options:
A C3
B C4
C CAM
D Nitrogen fixer

The correct answer is:
C4

Solution:
C4 plants have a high rate of photosynthesis at a higher temperature. These plants efficiently uses PEPCo enzyme to assimilate the atmospheric carbon dioxide and convert it into a four-carbon compound.

In C4 plants, RuBisCO capacity limits net CO2 assimilation rate below 20°C in chillingâtolerant species, but the control over net CO2 assimilation rate at elevated temperature remains high. Therefore, the ability of C4 plants acclimatization remains high.

Topic: Biology, Photosynthesis in Higher Plants, NEET, NEET 2016 (Phase 1)
A plant requires magnesium for:

Options:
A protein synthesis
B chlorophyll synthesis
C cell wall development
D holding cells together.

The correct answer is:
chlorophyll synthesis

Solution:
A plants require magnesium for chlorophyllsynthesis All four sings bind with metalatom magnesium $\left(\mathrm{Mg}^{++}\right)$, Which remain present at the centre of chlorophyll molecule. Nitrogen is the main component for the synthesis of protein. Cell wall is made up cellulose, hemicellulose and pectin components.
Related Theory
Magnesium is highly mobile in the plant and deficiency symptoms first appear on the lower leaves. Symptoms are more severe on the lower leaves because magnesium is moved to the new growth. Deficiency symptoms consist of interveinal chlorosis (leaf veins stay green while the regions between them turn yellow).

Topic: Biology, Mineral Nutrition, NEET, NEET 2007
A plant species shows several morphological variations in response to altitudinal gradient. When grown under similar conditions of growth, the morphological variations disappear and all the variants have common morphology. What are these variants called?

Options:

None

Solution:
These morphological variants are called biotypes. It includes group of genetically similar plants showing similarity when grown in same environmental and geographical regions. The same environment provides them the similar abiotic factors like soil, $\mathrm{pH}$, temperature, etc.
When growth in two different geographical regions, they are exposed to different abiotic characters which affects their growth, and development bringing changes in their external morphological features but, their genetic constitution remain same.

Topic: Biology, The Living World
A plate composed of welded sheets of aluminum and iron is connected to an electrical circuit as shown in the figure. What will happen if a fairly strong current be passed through the circuit?
Given that relation between linear coefficients of expansion is Î±Fe<Î±Ae.

Options:
A strip bends upward
B strip bends downward
C strip remains in its initial condition
D none of the above

The correct answer is:
strip bends upward

Solution:
When current is passing through the strip, the temperature of strip increases. So, thermal expansion takes place in the strip. Since, the value of α for iron is lesser than that of aluminium. So, strip bends upward.

Topic: Physics, Thermal Properties of Matter, NEET
A player caught a cricket ball of mass $150 \mathrm{~g}$ moving at a rate of $20 \mathrm{~m} / \mathrm{s}$. If the catching process is completed in $0.1 \mathrm{~s}$, the force of the blow exerted by the ball on the hand of the player is equal to

Options:
A
$300 \mathrm{~N}$

B
$150 \mathrm{~N}$

C
$3 \mathrm{~N}$

D
$30 \mathrm{~N}$

The correct answer is:
$30 \mathrm{~N}$

Solution:
$(m v-0) \Rightarrow 0.15 \times 20$
$\mathrm{F}=\frac{3}{0.1}=30 \mathrm{~N}$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2006