datavorous/entrance-exam-dataset huggingface dataset:
A point object is placed as shown. The two pieces are of the same lens of focal length 10 cm. Find the distance (in cm) between the two images formed by two pieces of the lens.

Options:

The correct answer is:
6

Solution:
The distance of image formed by each piece can be calculated by

1f=1v-1u

110=1v-1-15

⇒v=30 cm

m=3015=2

Topic: Physics, Ray Optics, JEE Main
A point object is placed at a distance of $10 \mathrm{~cm}$ and its real image is formed at a distance of $20 \mathrm{~cm}$ from a concave mirror. If the object is moved by $0.1 \mathrm{~cm}$ towards the mirror, the image will shift by about

Options:
A $0.4 \mathrm{~cm}$ away from the mirror
B $0.4 \mathrm{~cm}$ towards the mirror
C $0.8 \mathrm{~cm}$ away from the mirror
D $0.8 \mathrm{~cm}$ towards the mirror

The correct answer is:
$0.4 \mathrm{~cm}$ away from the mirror

Solution:
Mirror formula
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u} \Rightarrow \frac{1}{f}=\frac{1}{-20}+\frac{1}{(-10)} \Rightarrow f=\frac{20}{3} \mathrm{~cm}$. If object moves towards the mirror by $0.1 \mathrm{~cm}$ then
$u=(10-0.1)=9.9 \mathrm{~cm}$. Hence again from mirror formula $\frac{1}{-20 / 3}=\frac{1}{v}+\frac{1}{-9.9} \Rightarrow v^{\prime}=20.4 \mathrm{~cm}$ i.e. image
shifts away from the mirror by $0.4 \mathrm{~cm}$.

Topic: Physics, Ray Optics, JEE Main
A point object is placed at a distance of 10 cm from a concave mirror and its real image is formed at a distance of 20 cm from the concave mirror. If the object is moved by 0.1 cm towards the mirror, calculate the shift of image in mm .

Options:

The correct answer is:
-4

Solution:
dv=-v2u2du=--202-102×0.1=-0.4 cm=-4 mm

Topic: Physics, Ray Optics, JEE Main
A point object is placed at a distance of 60 cm from a convex lens of focal length 30 cm. If a plane mirror were put perpendicular to the principal axis of the lens and at a distance of 40 cm from it, the final image would be formed at a distance of :

Options:
A 20 cm from the lens, it would be a real image.
B 30 cm from the lens, it would be a real image.
C 30 cm from the plane mirror, it would be a virtual image.
D 20 cm from the plane mirror, it would be a virtual image.

The correct answer is:
20 cm from the plane mirror, it would be a virtual image.

Solution:
first, for image formation from lens
u=-60 cm

f=+30 cm

⇒v=ufu+f=-60×30-60+30=60 cm

this real image formed by lens acts as virtual object for mirror

Real image from plane mirror is formed 20 cm in front of mirror, hence at 20 cm distance from lens. Now, for second refraction from lens, u=-20 cm

f=+30 cm

v=ufu+f=-20×30-20+30=-60 cm

So, final virtual image is 60 cm from lens, or 20 cm behind mirror

Topic: Physics, Ray Optics, JEE Main
A point object is placed at the center of a glass sphere of radius $6 \mathrm{~cm}$ and refractive index 1.5 . The distance of the virtual image from the surface of the sphere is

Options:
A 2 cm
B 4 cm
C 6 cm
D 12 cm

The correct answer is:
6 cm

Solution:
Using refraction formula $\frac{1 \mu_2-1}{R}=\frac{{ }_1 \mu_2}{v}-\frac{1}{u}$
in given case, medium (1) is glass and (2) is air
So $\frac{{ }_g \mu_a-1}{R}=\frac{{ }_g \mu_a}{v}-\frac{1}{u} \Rightarrow \frac{\frac{1}{1.5}-1}{-6}=\frac{1}{1.5 v}-\frac{1}{-6}$
$\begin{aligned} & \Rightarrow \frac{1-1.5}{-6}=\frac{1}{v}+\frac{1.5}{6} \Rightarrow \frac{0.5}{6}=\frac{1}{v}+\frac{1}{4} \\ & \Rightarrow \frac{1}{v}=\frac{1}{12}-\frac{1}{4}=-\frac{2}{12}=-\frac{1}{6} \Rightarrow v=6 \mathrm{~cm}\end{aligned}$

Topic: Physics, Ray Optics, JEE Main
A point object is placed on the principal axis of a concave mirror, at a distance of 15 cm from the pole. The radius of curvature of the mirror is 20 cm and the object is made to oscillate along the principal axis with an amplitude of 2 mm. The amplitude of its image will be

Options:
A 2 mm
B 4 mm
C 8 mm
D 16 mm

The correct answer is:
8 mm

Solution:
Using mirror formula,
1 - 1 0 = 1 v + 1 - 1 5 ⇒ v = - 3 0 cm
Axial magnification = V 2 u 2 = 3 0 1 5 2 = 4
Amplitude of image = 4 x 2 = 8 mm.

Topic: Physics, Ray Optics, JEE Main
A point object moves on a circular path such that distance covered by it is given by function S=t22+2t meter ( t in second). The ratio of the magnitude of acceleration at t=2 s and t=5 s. is 1:2 then the radius of the circle is

Options:
A 1 m
B 351Â m
C 51Â m
D 3 m

The correct answer is:
351 m

Solution:
at=1 m/s2 v=t+2
at t=2 s a2=12+(2+2)2R2
a5=12+(5+2)2R2
21+162R2=1+492R2
4+4162R2=1+492R2
3=492R2-4162R2âR=351 m

Topic: Physics, Laws of Motion, JEE Main
A point object $O$ is placed in front of a glass rod having spherical end of radius of curvature $30 \mathrm{~cm}$. The image would be formed at

Options:
A $30 \mathrm{~cm}$ left
B Infinity
C $1 \mathrm{~cm}$ to the right
D $18 \mathrm{~cm}$ to the left

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

Solution:
By using formula $\frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R}$
$\Rightarrow \frac{1.5}{v}-\frac{1}{(-15)}=\frac{(1.5-1)}{+30} \Rightarrow v=-30 \mathrm{~cm}$.

Topic: Physics, Ray Optics, JEE Main
A point object $O$ is placed on the axis of a cylindrical piece of glass of refractive index 1.6 as shown in the figure. One surface of the glass piece is convex with radius of curvature $3 \mathrm{~mm}$. The point appeared to be at $5 \mathrm{~mm}$ on the axis when viewed along the axis and from right side of convex surface. The distance of the point object from the convex surface is :

Options:
A $4 \mathrm{~mm}$
B $6 \mathrm{~mm}$
C $3 \mathrm{~mm}$
D $2.5 \mathrm{~mm}$

The correct answer is:
$4 \mathrm{~mm}$

Solution:
A point object $O$ is placed on the axis of a cylindrical glass as shown in the figure,
For a curved spherical surface,
$$
\frac{\mu_2}{v}+\frac{\mu_1}{u}=\frac{\mu_1-\mu_2}{R}
$$
where, $v=5 \mathrm{~mm}, R=-3 \mathrm{~mm}, \mu_1=1.6$ and $\mu_2=1$
Putting the given values in the above expression, we get
$$
\begin{gathered}
\frac{1}{5}+\frac{1.6}{u}=\frac{1.6-1}{-3} \\
\frac{1.6}{u}=-0.2-\frac{1}{5}=-\frac{2}{5} \\
\Rightarrow \quad u=-\frac{1.6 \times 5}{2}=-4 \mathrm{~mm}
\end{gathered}
$$
Hence, the correct option is (a).

Topic: Physics, Ray Optics, JEE Main
A point object O is placed on the principal axis of a convex lens of focal length f=20 cm at a distance of 40 cm to the left of it. The diameter of the lens is 10 cm. An eye is placed 60 cm to the right of the lens and a distance h below the principal axis. The maximum value of h to see the image is. (assume paraxial approximation to be valid here)

Options:
A 0
B 2.5 cm
C 5 cm
D 10 cm

The correct answer is:
2.5 cm

Solution:
1 v = 1 fâ¡ + 1 u
= 1 2 0 - 1 4 0 = 1 4 0
v = 40 cm
To locate field of view ray passing through upper end should be used. At location of eye, its height should not be below 20 tan Î¸ from axis to see the image i.e., 2 0 5 4 0 = 2.5 cm

Topic: Physics, Ray Optics, JEE Main
A point on the curve $\frac{x^2}{A^2}-\frac{y^2}{B^2}=1$ is

Options:
A $(A \cos \theta, B \sin \theta)$
B $(A \sec \theta, B \tan \theta)$
C $\left(A \cos ^2 \theta, B \sin ^2 \theta\right)$
D None of these

The correct answer is:
$(A \sec \theta, B \tan \theta)$

Solution:
\(\begin{aligned}
& \text {If }(x, y) \equiv(A \sec \theta, B \tan \theta) \\
& \text {So, } \frac{(A \sec \theta)^2}{A_2}-\frac{(B \tan \theta)^2}{B^2}=1 \\
& \Rightarrow \sec ^2 \theta-\tan ^2 \theta=1 \text {(identity) }
\end{aligned}\)

Topic: Mathematics, Hyperbola, JEE Main
A point on the ellipse, $4 x^2+9 y^2=36$, where the normal is parallel to the line, $4 x-2 y-5=0$, is :

Options:
A
$\left(\frac{9}{5}, \frac{8}{5}\right)$

B
$\left(\frac{8}{5},-\frac{9}{5}\right)$

C
$\left(-\frac{9}{5}, \frac{8}{5}\right)$

D
$\left(\frac{8}{5}, \frac{9}{5}\right)$

The correct answer is:
$\left(-\frac{9}{5}, \frac{8}{5}\right)$

Solution:
Given ellipse is $4 x^2+9 y^2=36$
$$
\Rightarrow \frac{x^2}{9}+\frac{y^2}{4}=1
$$
Normal at the point is parallel to the line
$$
4 x-2 y-5=0
$$
Slope of normal $=2$
Slope of tangent $=\frac{-1}{2}$
Point of contact to ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
and line is $\left(\frac{a^2 m}{\sqrt{a^2 m^2+b^2}}, \frac{b}{\sqrt{a^2 m^2+b^2}}\right)$
Now, $a^2=9, b^2=4$
$$
\therefore \quad \text { Point }=\left(\frac{-9}{5}, \frac{8}{5}\right)
$$

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2013 (25 Apr Online)
A point on the parabola $y^2=18 x$ at which the ordinate increases at twice the rate of the abscissa is

Options:
A
$(2,4)$

B
$(2,-4)$

C
$\left(\frac{-9}{8}, \frac{9}{2}\right)$

D
$\left(\frac{9}{8}, \frac{9}{2}\right)$

The correct answer is:
$\left(\frac{9}{8}, \frac{9}{2}\right)$

Solution:
Any point be $\left(\frac{9}{2} t^2, 9 t\right) ;$ differentiating $y^2=18 x$
$$
\begin{aligned}
& \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{9}{\mathrm{y}}=\frac{1}{\mathrm{t}}=2 \text { (given) } \Rightarrow \mathrm{t}=\frac{1}{2} . \\
& \Rightarrow \text { Point is }\left(\frac{9}{8}, \frac{9}{2}\right)
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2004
A point on the plane that passes through the points $(1,-1,6),(0,0,7)$ and perpendicular to the plane $x-2 y+z=6$ is

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

The correct answer is:
$(1,1,2)$

Solution:
Equation of plane passes through the points $(1,-1,6),(0,0,7)$ and perpendicular to the plane $x-2 y+z=6$ is
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y+1 & z-6 \\
0-1 & 0+1 & 7-6 \\
1 & -2 & 1
\end{array}\right|=0 \\
\Rightarrow \quad & \left|\begin{array}{ccc}
x-1 & y+1 & z-6 \\
-1 & 1 & 1 \\
1 & -2 & 1
\end{array}\right|=0 \\
\Rightarrow & (x-1)(1+2)-(y+1)(-1-1) \\
& +(z-6)(2-1)=0 \\
\Rightarrow \quad 3 x-3+2 y+2+z-6 & =0 \\
\Rightarrow \quad 3 x+2 y+z-7 & =0
\end{aligned}
$$
This plane is passes through $(1,1,2)$.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A point on the straight line, 3x+5y=15 which is equidistant from the coordinate axes will lie only in:

Options:
A 1st and 2nd quadrants
B 1st,2nd and 4th quadrants
C 1st quadrant
D 4th quadrant

The correct answer is:
1st and 2nd quadrants

Solution:
Let a,a be a point on the line 3x+5y=15

⇒3a+5a=15

⇒a=158

∴ Point is 158 ,158 which lies in 1st quadrant

Again, Let -a,a be a point on the line

3x+5y=15

⇒-3a+5a=15

⇒a=152

∴ Point is -152 ,152 which lies in 2nd quadrant

Hence, required point lies in1st and 2nd quadrants.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2019 (08 Apr Shift 1)
A point P consider at the contact point of a wheel on the ground which roll on the ground without slipping, then the value of the displacement of point $\mathrm{P}$ when wheel completes half of the rotation (if radius of wheel is $1 \mathrm{~m}$ )is:

Options:
A $2 \mathrm{~m}$
B $\sqrt{\pi^2+4} \mathrm{~m}$
C $\pi \mathrm{m}$
D $\sqrt{\pi^2+2} \mathrm{~m}$

The correct answer is:
$\sqrt{\pi^2+4} \mathrm{~m}$

Solution:
Displacement $P Q$
$\begin{aligned}
& =\sqrt{(P M)^2+(Q M)^2} \\
& =\sqrt{(\pi R)^2+(2 R)^2} \\
& =\sqrt{\pi^2+4}\{\because R=1 \mathrm{~m}\}
\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
A point P lies on the axis of a ring of mass M and radius a at a distance a from its centre O. A small particle of mass m starts from P and reaches O under gravitational attraction only. Its speed when it reaches O is

Options:
A 2GMa
B GMa
C 2GMa2-1
D 2GMa1-12

The correct answer is:
2GMa1-12

Solution:
The gravitational potential at P, VP=-GMAP
VP=-GMa2+a2=-GM2a and gravitational potential at O, VO=-GMa
Let v be the velocity of the particle when it reaches O.
∴ K.E. at O=12mv2
Gravitational P.E. = Gravitational potential × mass
∴ P.E. at P=-GMm2a and P.E. at O=-GMma
By the principle of conservation of mechanical energy
12mv2-GMma=-GMm2a
∴v2=2GMma-GMm2a=2GMa1-12
∴v=2GMa1-12

Topic: Physics, Gravitation, JEE Main
A point P moves in a counter-clockwise direction on a circular path as shown in the figure. The movement of P is such that it sweeps out a length s=t3+5 where s is in metre and t is in seconds. The radius of the path is 27 m. The acceleration of P when t=3 s is ____ m/s2. Takeâ13=3.6

Options:

The correct answer is:
32.40

Solution:
As s=t3+5
dsdt=3t2=v
â´âat=dvdt=6t
aâ=ac2+at2
At t=3s
aâ=v2R2+at2=9t4R2+6t2
=272+324=1053=913
â´aâ=32.4 m/s2

Topic: Physics, Laws of Motion, JEE Main
A point $\mathrm{P}$ moves in counter-clockwise direction on a circular path as shown in the figure. The movement of ' $\mathrm{P}$ ' is such that it sweeps out a length $s=t^3+5$, where $s$ is in metres and $t$ is in seconds. The radius of the path is $20 \mathrm{~m}$. The acceleration of ' $\mathrm{P}$ ' when $t=2 \mathrm{~s}$ is nearly

Options:
A
$13 \mathrm{~m} / \mathrm{s}^2$

B
$12 \mathrm{~m} / \mathrm{s}^2$

C
$7.2 \mathrm{~m} / \mathrm{s}^2$

D
$14 \mathrm{~m} / \mathrm{s}^2$

The correct answer is:
$14 \mathrm{~m} / \mathrm{s}^2$

Solution:
$$
S=t^3+5
$$
$\therefore \quad$ speed, $\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}=3 \mathrm{t}^2$
and rate of change of speed $=\frac{\mathrm{dv}}{\mathrm{dt}}=6 \mathrm{t}$
$\therefore \quad$ tangential acceleration at $\mathrm{t}=2 \mathrm{~s}, \mathrm{a}_{\mathrm{t}}=6 \times 2=12 \mathrm{~m} / \mathrm{s}^2$
at $t=2 \mathrm{~s}, \mathrm{v}=3(2)^2=12 \mathrm{~m} / \mathrm{s}$
$\therefore \quad$ centripetal acceleration, $\quad a_c=\frac{v^2}{R}=\frac{144}{20} \mathrm{~m} / \mathrm{s}^2$
$\therefore \quad$ net acceleration $=\sqrt{\mathrm{a}_t^2+\mathrm{a}_{\mathrm{i}}^2}$ $\approx 14 \mathrm{~m} / \mathrm{s}^2$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2010
A point P moves on the line 2x-3y+4=0. If Q1, 4 and R3, -2 are fixed points, then the locus of the centroid of ΔPQR is a line:

Options:
A with slope 23
B with slope 32
C parallel to y-axis
D parallel to x-axis

The correct answer is:
with slope 23

Solution:
Let, coordinate of centroid is (h, k) and coordinate of point P is (α, β)

We know that the centroid of a triangle with vertices x1, y1, x2, y2 and x3, y3 is x1+x2+x33, y1+y2+y33.

Hence, in ∆PQR, we have α+1+33=h and β+4-23=k
⇒α=3h-4, β=3k-2

Since (α, β) lies on the line 2x-3y+4=0

Hence, 23h-4-33k-2+4=0

⇒6h-9k+2=0

For finding the locus replace h, k by x, y

Hence, the locus of centroid is 6x-9y+2=0 which has slope =-6-9=23 since the slope of a line ax+by+c=0 is -ab.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2019 (10 Jan Shift 1)
A point $P$ moves such that its distances from $(1,2)$ and $(-2,3)$ are equal. Then the locus of $P$ is

Options:
A straight line
B Parabola
C ellipse
D hyperbola

The correct answer is:
straight line

Solution:
Let moving point be $\mathrm{P}(\mathrm{x}, \mathrm{y})$
$\sqrt{(\mathrm{y}-2)^{2}+(\mathrm{x}-1)^{2}}=\sqrt{(\mathrm{y}-3)^{2}+(\mathrm{x}+2)^{2}}$
$\Rightarrow(\mathrm{y}+2)^{2}+(\mathrm{x}-1)^{2}=(\mathrm{y}-3)^{2}+(\mathrm{x}+2)^{2}$
$\Rightarrow y^{2}+4+4 y+x^{2}+1-2 x=y^{2}+9-6 y+x^{2}+4 x+4$
$\Rightarrow 10 \mathrm{y}-6 \mathrm{x}-8=0$
$\therefore$ Locus of $\mathrm{P}$ is a straight line.

Topic: Mathematics, Straight Lines, JEE Main
A point P moves such that the chord of contact of P with respect to the circle x2+y2=4 passes through the point 1,1. The coordinates of P when it is nearest to the origin are

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

The correct answer is:
2,2

Solution:
Let, the point P be a,b then
Equation of the chord of contact is xa+yb=4
Since, it passes through 1,1 so,
a+b=4âx+y=4 is the locus of the point P.
Hence, the nearest point on x+y=4 is 2,2.

Topic: Mathematics, Circle, JEE Main
A point \(P\) moves such that the sum of twice its distance from the origin and its distance from the \(y\)-axis is a constant equal to 3 . P describes

Options:
A A circle with its centre at \((-1,0)\) and radius \(2 \sqrt{3}\)
B An ellipse centred at \((-1,0)\) and of eccentricity \(\frac{1}{2}\)
C A hyperbola centred at \((1,0)\) and of eccentricity 2
D None of these

The correct answer is:
An ellipse centred at \((-1,0)\) and of eccentricity \(\frac{1}{2}\)

Solution:
\(\begin{aligned}
& 2 \sqrt{\mathrm{x}^2+\mathrm{y}^2}+\mathrm{x}=3 \\
& \Rightarrow \frac{(\mathrm{x}+1)^2}{4}+\frac{\mathrm{y}^2}{3}=1 .
\end{aligned}\)
This represents an ellipse centred at \((-1,0)\) and
\(\text { of eccentricity }=\sqrt{\frac{4-3}{4}}=\frac{1}{2} \text {. }\)

Topic: Mathematics, Ellipse, JEE Main
A point $P$ on a line is at a distance of 4 units from the origin $(0,0)$. If the line makes $60^{\circ}$ with the negative direction of the $X$-axis, then $P$ is

Options:
A $(2,2 \sqrt{3})$
B $(2 \sqrt{3}, 2)$
C $(1, \sqrt{3})$
D $(2 \sqrt{3}, 1)$

The correct answer is:
$(2 \sqrt{3}, 2)$

Solution:
$O P$ is perpendicular to given line.
$\therefore$ Slope of $O P=\frac{1}{\sqrt{3}}=\frac{y}{x} \Rightarrow x=\sqrt{3} y$
Also $x^2+y^2=16$
$\therefore 3 y^2+y^2=16 \Rightarrow y= \pm 2$
$\therefore$ Points can be $(2 \sqrt{3}, 2)$ or $(-2 \sqrt{3},-2)$.

Topic: Mathematics, Straight Lines, JEE Main
A point P on the parabola y2=4x, the foot of the perpendicular from it upon the directrix and the focus are the vertices of an equilateral triangle. If the area of the equilateral triangle is Î² sq. units, then the value of Î²2 is

Options:

The correct answer is:
48

Solution:
From the diagram,

PM=1+t2
PS=t2-12+4t2=t2+1
MS=4+4t2=21+t2
â21+t2=1+t2
â´PM=1+t2=4ât2=3
â´ Area of âPMS=Î²=3442=43 sq. units

Hence, Î²2=48

Topic: Mathematics, Parabola, JEE Main
A point particle is held on the axis of a ring of mass $m$ and radius $r$ at a distance $r$ from its centre C. When released, it reaches $\mathrm{C}$ under the gravitational attraction of the ring. Its speed at $C$ will be

Options:
A $\sqrt{\frac{2 \mathrm{G} m}{r}(\sqrt{2}-1)}$
B $\sqrt{\frac{\mathrm{G} m}{r}}$
C $\sqrt{\frac{2 \mathrm{G} m}{r}\left(1-\frac{1}{\sqrt{2}}\right)}$
D $\sqrt{\frac{2 \mathrm{G} m}{r}}$

The correct answer is:
$\sqrt{\frac{2 \mathrm{G} m}{r}\left(1-\frac{1}{\sqrt{2}}\right)}$

Solution:
Let ' $\mathrm{M}$ ' be the mass of the particle

Now, $E_{\text {initial }}=E_{\text {final }}$

i.e. $\frac{\mathrm{GMm}}{\sqrt{2} r}+0=\frac{\mathrm{GM} m}{r}+\frac{1}{2} \mathrm{MV}^{2}$

or, $\frac{1}{2} \mathrm{MV}^{2}=\frac{\mathrm{GM} m}{r}\left[1-\frac{1}{\sqrt{2}}\right]$

$\Rightarrow \frac{1}{2} \mathrm{~V}^{2}=\frac{\mathrm{G} m}{r}\left[1-\frac{1}{\sqrt{2}}\right]$ or,

$\mathrm{V}=\sqrt{\frac{2 \mathrm{Gm}}{r}\left(1-\frac{1}{\sqrt{2}}\right)}$

Topic: Physics, Gravitation, JEE Main
A point particle is held on the axis of a ring of mass $m$ and radius $r$ at a distance $r$ from its centre $C$. When released, it reaches $C$ under the gravitational attraction of the ring. Its speed at $C$ will be

Options:
A
$\sqrt{\frac{2 G m}{r}(\sqrt{2}-1}$

B
$\sqrt{\frac{G m}{r}}$

C
$\sqrt{\frac{2 G m}{r}\left(1-\frac{1}{\sqrt{2}}\right)}$

D
$\sqrt{\frac{2 G m}{r}}$

The correct answer is:
$\sqrt{\frac{2 G m}{r}\left(1-\frac{1}{\sqrt{2}}\right)}$

Solution:
Let ' $M$ ' be the mass of the particle
Now, $\mathrm{E}_{\text {initial }}=\mathrm{E}_{\text {final }}$
i.e. $\frac{\mathrm{GMm}}{\sqrt{2} r}+0=\frac{\mathrm{GM} m}{r}+\frac{1}{2} M \mathrm{~V}^2$
$$
\begin{aligned}
& \text { or, } \frac{1}{2} M V^2=\frac{G M m}{r}\left[1-\frac{1}{\sqrt{2}}\right] \\
& \Rightarrow \frac{1}{2} V^2=\frac{G m}{r}\left[1-\frac{1}{\sqrt{2}}\right] \\
& \text { or, } V=\sqrt{\frac{2 G m}{r}\left(1-\frac{1}{\sqrt{2}}\right)}
\end{aligned}
$$

Topic: Physics, Gravitation, JEE Main, JEE Main 2012 (26 May Online)
A point particle of mass $0.5 \mathrm{~kg}$ is moving along the $\mathrm{x}$-axis under a force described by the potential energy $\mathrm{V}$ shown below. It is projected towards the right from the origin with a speed $\mathrm{v}$. What is the minimum value of $v$ for which the particle will escape infinitely fasr away from the origin?

Options:
A $2 \sqrt{2} \mathrm{~ms}^{-1}$
B $2 \mathrm{~ms}^{-1}$
C $4 \mathrm{~ms}^{-1}$
D The particle will never escape

The correct answer is:
$2 \mathrm{~ms}^{-1}$

Solution:
$\frac{1}{2} m v^{2}+0=0+1$
$v^{2}=4$ or $v=2 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Motion In One Dimension, JEE Main
A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals μ . The particle is released, from rest, from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR.
The values of the coefficient of friction μ and the distance x=QR , are respectively close to:

Options:
A 0.29 and 3.5 m
B 0.29 and 6.5 m
C 0.2 and 6.5 m
D 0.2 and 3.5 m

The correct answer is:
0.29 and 3.5 m

Solution:
tan⁡30o=hℓ
ℓ=h3=23 m
WPQ=-μmgℓ or WQR=-μmgx
μmgℓ=μmgx

x=ℓ≈3.5m
x=23 m

now according to work energy theorem

Wall=∆K
mgh-μmgℓ-μmgx=0
h-μℓ-μx=0
2=μℓ+x⇒μ=2ℓ+x=243=123≈0.29

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2016 (03 Apr)
A point performs simple harmonic oscillation of period $\mathrm{T}$ and the equation of motion is given by $x=a \sin (w t+\pi / 6)$. After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity?

Options:
A $\frac{\mathrm{T}}{12}$
B $\frac{T}{8}$
C $\frac{T}{6}$
D $\frac{T}{3}$

The correct answer is:
$\frac{\mathrm{T}}{12}$

Solution:
$$
\begin{aligned}
& v=\omega \cos \left(\omega t+\frac{\pi}{6}\right) \\
& \Rightarrow \quad \frac{\omega a}{2}=\omega \cos \left(\frac{2 \pi}{T} t+\frac{\pi}{6}\right) \\
& \Rightarrow \quad \frac{\pi}{3}=\frac{2 \pi}{T} t+\frac{\pi}{6} \\
& \Rightarrow t=\frac{T}{12}
\end{aligned}
$$

Topic: Physics, Oscillations, JEE Main
A point performs simple harmonic oscillation of period $\mathrm{T}$ and the equation of motion is given by $x=a \sin (w t+\pi / 6)$. After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity?

Options:
A $\frac{\mathrm{T}}{12}$
B $\frac{T}{8}$
C $\frac{T}{6}$
D $\frac{T}{3}$

The correct answer is:
$\frac{\mathrm{T}}{12}$

Solution:
$$
\begin{aligned}
& v=\omega \cos \left(\omega t+\frac{\pi}{6}\right) \\
& \Rightarrow \quad \frac{\omega a}{2}=\omega \cos \left(\frac{2 \pi}{T} t+\frac{\pi}{6}\right) \\
& \Rightarrow \quad \frac{\pi}{3}=\frac{2 \pi}{T} t+\frac{\pi}{6} \\
& \Rightarrow t=\frac{T}{12}
\end{aligned}
$$

Topic: Physics, Oscillations, NEET, NEET 2008 (Mains)
A point α,β,γ satisfies the equation of the plane 3x+4y+7z=3. The value of β, such that p→=αi^+βj^+γk^ satisfies the relation j^×j^×p→=0→, is equal to

Options:

The correct answer is:
0.75

Solution:
3α+4β+7γ=3 …. 1
j^×j^×p→=j^⋅p→j^-j^⋅j^p→
=-αi^-γk^=0⇒α=γ=0
From equation 1, we get,
4β=3⇒β=0.75

Topic: Mathematics, Vector Algebra, JEE Main
A point source emits sound equally in all directions in a non-absorbing medium. Two points P and Q are at a distance of 2 m and 3 m respectively from the source. The ratio of the intensities of the waves at P and Q is

Options:
A 9:4
B 2:3
C 3:2
D 4:9

The correct answer is:
9:4

Solution:
The average power per unit area that is incident perpendicular to the direction of propagation is called the intensity, i.e.,
I=P4πr2
I∝Ir2
I2I2=r2r12
Here, r1=2 m, r2=3 m
∴I1I2 =322=94

Topic: Physics, Waves and Sound, JEE Main
A point source is emitting sound waves of intensity 16×10-8 W m-2 at the origin. The difference in intensity (magnitude only) at two points located at a distances of 2 m and 4 m from the origin respectively will be ________×10-8 W m-2.

Options:

The correct answer is:
3

Solution:
Intensity for a point source varies as I∝1r2.

Intensity at a distance of 2 m, I2=16×10-8×122=4×10-8 W m-2

Similarly, intensity at a distance of 4 m, I4=16×10-8×142=1×10-8 W m-2

Required difference, I2-I4=3×10-8 W m-2

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2024 (30 Jan Shift 2)
A point source of 100 W emits light with 5% efficiency. At a distance of 5 m from the source, the intensity produced by the electric field component is:

Options:
A 12πWm2
B 140πWm2
C 110πWm2
D 120Wm2

The correct answer is:
140πWm2

Solution:
Total power emitted =100×5100=5 W

Now intensity due to electric field will be half of the total intensity. Therefore,

IE=12×powerarea=12×54π×52

=140πWm2

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2023 (30 Jan Shift 2)