datavorous/entrance-exam-dataset huggingface dataset:
An object is floating in a liquid, kept in a container. The container is placed in a lift. Choose the correct option.

Options:
A Buoyant force increases as the lift accelerates up
B Buoyant force decreases as the lift accelerates up
C Buoyant force remains constant as the lift accelerates
D The fraction of solid submerged into liquid decreases

The correct answer is:
Buoyant force increases as the lift accelerates up

Solution:
As pressure gradient changes dPdy=ρg+a,
Buoyant force also changes, as it is due to pressure difference. But as body also accelerates with lift, so the fraction of volume submerged does not change and the extra buoyant force will provide the resultant force 'ma' to the body.

Topic: Physics, Mechanical Properties of Fluids, NEET
An object is gradually moving away from the focal point of a concave mirror along the axis of the mirror. The graphical representation of the magnitude of linear magnification m versus distance of the object from the mirror x is correctly given by
(Graphs are drawn schematically and are not to scale)

Options:
A
B
C
D

The correct answer is:

Solution:
At focus, magnification is ∞ .

Topic: Physics, Ray Optics, JEE Main, JEE Main 2020 (08 Jan Shift 2)
An object is kept at a distance of 16 cm from a thin lens and the image formed is real. If the object is kept at a distance of 6 cm from the same lens, the image formed is virtual. If the sizes of the images formed are equal the focal length of the lens will be

Options:
A 21 cm
B 11 cm
C 15 cm
D 17 cm

The correct answer is:
11 cm

Solution:
As a convex lens alone can form a real images as well as a virtual images, therefore, the lens in the present question is a convex lens. Let f be the focal length of the lens and m be the magnification produced
In the first case, when image is real,
u=-16 cm, v=(m×16)cm
As 1v-1u=1f
∴116m+116=1f or 1+1m=16f ...(i)
In the second case, when image is virtual
u=-6 cm,v=(-6 m)cm
From 1v-1u=1f
1-6m+16=1f or 1-1m=6f, ...(ii)
Add Eq.(i) and Eq.(ii) we have
2=22f or f=222=11 cm

Topic: Physics, Ray Optics, JEE Main
An object is kept on a smooth inclined plane of inclination 1 in l. Find the horizontal acceleration that should be imparted to the inclined plane such that the object remains stationary relative to the inclined plane.

Options:
A gl2-1
B g(l2-1)
C gl2-1
D gl2+1

The correct answer is:
gl2-1

Solution:
The object moves a height of 1 m when it travels a distance lalong the incline.

Here, sinâ¡Î¸=1l
Let required acceleration of inclined plane be a for the object to remain stationary relative to incline, we have
sinâ¡Î¸=1l
tanâ¡Î¸=1l2-1
macosâ¡Î¸=mgsinâ¡Î¸
a=gtanâ¡Î¸ =gl2-1

Topic: Physics, Laws of Motion, NEET
An object is located at 2 km beneath the surface of the water. If the fractional compression ΔVV is 1.36%, the ratio of hydraulic stress to the corresponding hydraulic strain will be ______[Given: density of water is 1000 kg m-3 and g=9.8 m s-2.

Options:
A 1.96×107 N m-2
B 1.44×107 N m-2
C 2.26×109 N m-2
D 1.44×109 N m-2

The correct answer is:
1.44×109 N m-2

Solution:
P=hρg

β=pΔVV=2×103×103×9.81.36×10-2

=1.44×109 N m-2

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2021 (17 Mar Shift 2)
An object is located in a fixed position in front of a screen. Sharp image is obtained on the screen for two positions of a thin lens separated by 10 $\mathrm{cm}$. The size of the images in two situations are in the ratio $3: 3$. What is the distance between the screen and the object?

Options:
A
$124.5 \mathrm{~cm}$

B
$144.5 \mathrm{~cm}$

C
$65.0 \mathrm{~cm}$

D
$99.0 \mathrm{~cm}$

The correct answer is:
$99.0 \mathrm{~cm}$

Solution:
Given: Separation of lens for two of its position, $d=10 \mathrm{~cm}$ Ratio of size of the images in two positions
$$
\frac{I_1}{I_2}=\frac{3}{2}
$$
Distance of object from the screen, $D=$ ? Applying formula,
$$
\begin{aligned}
&\frac{I_1}{I_2}=\frac{(D+d)^2}{(D-d)^2} \\
&\Rightarrow \quad \frac{3}{2}=\frac{(D+10)^2}{(D-10)^2} \\
&\Rightarrow \quad \frac{3}{2}=\frac{D^2+100+20 D}{D^2+100-20 D} \\
&\Rightarrow 3 D^2+300-60 D=2 D^2+200+40 D \\
&\Rightarrow D^2-100 D+100=0
\end{aligned}
$$
On solving, we get $D=99 \mathrm{~cm}$
Hence the distance between the screen and the object is $99 \mathrm{~cm}$.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2014 (11 Apr Online)
An object is moving towards a mirror with a velocity of 10 m s-1 as shown in the figure. If the collision between the mirror and the object is perfectly elastic, then the velocity of the image after collision with mirror in vector form is

Options:
A -10 j^
B -10cosâ¡15j^+10sinâ¡15j^
C -10 i^
D -10cosâ¡30j^-10sinâ¡30i^

The correct answer is:
-10cos⁡30j^-10sin⁡30i^

Solution:
Vâim=Vobjcosâ¡2Î¸-j^+Vobjsinâ¡2Î¸-i^

Vâim=-10cosâ¡30j^-10sinâ¡30i^

Topic: Physics, Ray Optics, JEE Main
An object is placed $40 \mathrm{~cm}$ from a concave mirror of focal length $20 \mathrm{~cm}$. The image formed is

Options:
A Real, inverted and same in size
B Real, inverted and smaller
C Virtual, erect and larger
D Virtual, erect and smaller

The correct answer is:
Real, inverted and same in size

Solution:
Real, inverted and same in size because object is at the centre of curvature of the mirror.

Topic: Physics, Ray Optics, JEE Main
An object is placed at $20 \mathrm{~cm}$ from a convex mirror of focal length $10 \mathrm{~cm}$. The image formed by the mirror is

Options:
A Real and at $20 \mathrm{~cm}$ from the mirror
B Virtual and at $20 \mathrm{~cm}$ from the mirror
C Virtual and at $20 / 3 \mathrm{~cm}$ from the mirror
D Real and at $20 / 3 \mathrm{~cm}$ from the mirror

The correct answer is:
Virtual and at $20 / 3 \mathrm{~cm}$ from the mirror

Solution:
$u=-20 \mathrm{~cm}, f=+10 \mathrm{~cm}$ also $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
$\Rightarrow \frac{1}{+10}=\frac{1}{v}+\frac{1}{(-20)} \Rightarrow v=\frac{20}{3} \mathrm{~cm} ;$ virtual image.

Topic: Physics, Ray Optics, JEE Main
An object is placed at a distance of 12 cm in front of a plane mirror. The virtual and erect image is formed by the mirror. Now the mirror is moved by 4 cm towards the stationary object. The distance by which the position of image would be shifted, will be

Options:
A 4 cm towards mirror
B 8 cm towards mirror
C 8 cm away from mirror
D 2 cm towards mirror

The correct answer is:
8 cm towards mirror

Solution:
When an object is placed in front of a plane mirror, object distance is equal to the image distance. In the first case, the image distance will be 12 cm from the mirror. In the next case, as the mirror is moved towards the object, the object distance becomes 8 cm. Hence, image distance will also be 8 cm.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (10 Apr Shift 1)
An object is placed at a distance of 15 cm from a convex lens of focal length 10 cm. On the other side of the lens, a convex mirror is placed at its focus such that the image by the combination coincides with the object itself. What is the focal length (in cm) of the convex mirror?

Options:

The correct answer is:
10

Solution:
â
The retrace of the path after reflection from mirror rays after refraction from lens must incident towards the centre of curvature of the mirror
Image by lense 1v-1u=1f⇒110+2fm-1-15=110
1 1 0 + 2 f m = 1 1 0 - 1 1 5 ⇒ 1 1 0 + 2 f m = 3 - 2 3 0
1 0 + 2 f m = 3 0 ⇒ f m = 1 0 cm

Topic: Physics, Ray Optics, JEE Main
An object is placed at a distance of 40 cm from a concave mirror of focal length 15 cm. If the object is displaced through a distance of 20 cm towards the mirror, the displacement of the image will be

Options:
A 30 cm towards the mirror
B 36 cm away from the mirror
C 30 cm away from the mirror
D 36 cm towards the mirror

The correct answer is:
36 cm away from the mirror

Solution:
By mirror formula,v1=ufu-f=-40-15-40+15=600-25=-24 cm
When object is displaced by 20 cm towards mirror v2=ufu-f=-20 -15-20+15=-60 cm
Displacement of image =v2-v1=-36 cm
= 36 cm away from the mirror

Topic: Physics, Ray Optics, JEE Main
An object is placed at a distance of 40 cm from a convex lens of focal length 20 cm. On the far side of the lens, a concave mirror of focal length 10 cm is placed such that the distance of the object from the concave mirror is 100 cm. Then the final image which is formed after refraction from the lens, reflection from the mirror and again refraction from the lens, will be

Options:
A at a distance of 40 cm from the lens and of the same size of the object
B at a distance of 20 cm from the lens and of the same size of the object
C at a distance of 20 cm from the lens and of half the size of the object
D at a distance of 40 cm from the lens and of half the size of the object

The correct answer is:
at a distance of 40 cm from the lens and of the same size of the object

Solution:
Image is same size, inverted, real and coinciding with object.

Topic: Physics, Ray Optics, JEE Main
An object is placed at a distance of $f / 2$ from a convex lens. The image will be

Options:
A At one of the foci, virtual and double its size
B At $3 f / 2$, real and inverted
C At $2 f$, virtual and erect
D None of these

The correct answer is:
At one of the foci, virtual and double its size

Solution:
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ (Given $u=\frac{-f}{2}$ )
$\begin{aligned} & \Rightarrow \frac{1}{f}=\frac{1}{v}+\left(\frac{1}{f / 2}\right) \Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{2}{f} \\ & \Rightarrow \frac{1}{v}=\frac{-1}{f} \text { and } m=\frac{v}{u}=\frac{f}{f / 2}=2\end{aligned}$
So virtual at the focus and of double size

Topic: Physics, Ray Optics, JEE Main
An object is placed at the focus of concave lens having focal length f. What is the magnification and distance of the image from the optical centre of the lens?

Options:
A 14,f4
B 12,f2
C Very high, ∞
D 1,∞

The correct answer is:
12,f2

Solution:
1 F=1v-1u
Put u=-f and F=-f
1-f=1v-1-f; -1f=1f-1v;-2f=1v⇒v=-f2
Magnification =v/u=-f2(-f)=12

Topic: Physics, Ray Optics, JEE Main, JEE Main 2021 (31 Aug Shift 1)
An object is placed beyond the centre of curvature C of the given concave mirror. If the distance of the object is d1 from C and the distance of the image formed is d2 from C, the radius of curvature of this mirror is:

Options:
A 2 d1 d2 d1+d2
B 2 d1 d2 d1-d2
C d1 d2 d1+d2
D d1 d2 d1-d2

The correct answer is:
2 d1 d2 d1-d2

Solution:
xy=f2

f+d1f-d2=f2

f=d1d2d1-d2

R=2f=2d1d2d1-d2

Topic: Physics, Ray Optics, JEE Main, JEE Main 2021 (27 Aug Shift 1)
An object is placed first at infinity and then at $20 \mathrm{~cm}$ from the object side focal plane of the convex lens. The two images thus formed are $5 \mathrm{~cm}$ apart. The focal length of the lens is

Options:
A 5 cm
B 10 cm
C 15 cm
D 20 cm

The correct answer is:
10 cm

Solution:
For first case : $\frac{1}{f}=\frac{1}{v}-\frac{1}{\infty} \Rightarrow f=v$
For second case $\frac{1}{f}=\frac{1}{(f+5)}-\frac{1}{-(f+20)} \Rightarrow f=10 \mathrm{~cm}$
Alternative sol. $-f^2=x_1 x_2 \Rightarrow f=10 \mathrm{~cm}$.

Topic: Physics, Ray Optics, JEE Main
An object is placed on the principal axis of a concave mirror at a distance of 1.5f (f is the focal length). The image will be at,

Options:
A -3 f
B 1.5 f
C -1.5 f
D 3 f

The correct answer is:
-3 f

Solution:
Given, u=-1.5f, now use the mirror formula for the concave mirror, in terms of distance between object and pole, the distance between image and pole and focal length,

1u+1v=1f⇒1-1.5f+1v=1-f

⇒1v=-1f+11.5f⇒1v=-1.5+11.5f=-0.51.5f

⇒v=-3f.

Topic: Physics, Ray Optics, JEE Main
An object is placed on the principal axis of convex lens of focal length 10 cm as shown. A plane mirror is placed on the other side of lens at a distance of 20 cm. The image produced by the plane mirror is 5 cm inside the mirror. The distance of the object from the lens is ____ cm.

Options:

The correct answer is:
30

Solution:
Given here: Focal length of convex lens, f=10 cm

Now, as seen in figure, image from lens is formed at point I1, whose location is 5 cm from mirror. Thus, I1 is at 15 cm from lens.

Using lens equation, we have

1v-1u=1f

Putting the value,

115-1-u=110

⇒1u=110-115⇒u=10×1510-15

Thus, object is placed at 30 cm from lens.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (25 Jan Shift 2)
An object is projected from the earth's surface with escape velocity at 30° with horizontal. What is the angle made by the velocity with horizontal when the object reaches a height 2R from the earth's surface? R is the radius of the earth. Horizontal can be considered as a line parallel to the tangent at the earth's surface just below the object.

Options:
A 30°
B 45°
C 60°
D 15°

The correct answer is:
60°

Solution:
-GMmR + 12mx2GMR=-GMm3R+12mv12
⇒ v1=2GM3R
By conservation of angular momentum:
m×2GMRcos30°×R=m×v1cosθ×3R
⇒cosθ=12 ⇒ θ=60°

Topic: Physics, Gravitation, JEE Main
An object is projected with a speed 10 m s-1 at an angle of 30Â° with the horizontal. It breaks into n equal fragments during its motion. One fragment strikes the ground at a distance of 3 m from the point of projection. The centre of mass of the remaining fragments strikes the ground at a distance of 73 m from the point of projection. If all fragments strike the ground at the same time, Find the value of n .

Options:
A 2
B 3
C 4
D 525

The correct answer is:
3

Solution:
Position of overall center of mass of the system,
Rcm=102sinâ¡60g=53 m
If m is the mass of one fragment then,
53=m3+m(n-1)73mn
n=3

Topic: Physics, Rotational Motion, NEET
An object is projected with a velocity of $20 \mathrm{~m} / \mathrm{s}$ making an angle of $45^{\circ}$ with horizontal. The equation for the trajectory is $h=A x-B x^2$ where $h$ is height, $x$ is horizontal distance $A$ and $B$ are constants. The ratio $A: B$ is $\left(g=10 \mathrm{~ms}^{-2}\right)$

Options:
A $1: 5$
B $5: 1$
C $1: 40$
D $40: 1$

The correct answer is:
$40: 1$

Solution:
$u=20 \mathrm{~m} / \mathrm{s}, \theta=45^{\circ}$
Equation of trajectory $h=A x-B x^2$ Standard equation
$h=x \tan \theta-\frac{g x^2}{2 u^2 \cos ^2 \theta}$
On comparing the two equations
$\begin{aligned} A & =\tan \theta, B=\frac{g}{2 u^2 \cos ^2 \theta} \\ \frac{A}{B} & =\frac{\tan \theta}{\frac{g}{2} u^2 \cos ^2 \theta}=\frac{2 u^2 \cos ^2 \theta \tan \theta}{g} \\ & =\frac{2 \times(20)^2 \cos ^2 45^{\circ} \tan 45^{\circ}}{10} \\ & =\frac{2 \times 400 \times\left(\frac{1}{\sqrt{2}}\right)^2 \times 1}{10} \\ & =2 \times 40 \times \frac{1}{2}=40\end{aligned}$
$A: B=40: 1$

Topic: Physics, Motion In Two Dimensions, JEE Main
An object is taken to a height above the surface of earth at a distance 54R from the centre of the earth. Where radius of earth, R=6400 km. The percentage decrease in the weight of the object will be

Options:
A 36%
B 50%
C 64%
D 25%

The correct answer is:
36%

Solution:
Acceleration due to gravity at height h is given by, g'=g1+hR2

For distance from the centre r=54R height h=R4,

⇒g'=g1+142=1625g

Change in weight

∆W=mg-mg'⇒∆W=mg1-1625

⇒∆W=925mg

Percentage change =925×100%

=36%

Topic: Physics, Gravitation, JEE Main, JEE Main 2022 (25 Jul Shift 2)
An object is thrown vertically upwards. At its maximum height, which of the following quantity becomes zero ?

Options:
A Momentum
B Potential energy
C Acceleration
D Force

The correct answer is:
Momentum

Solution:
When a ball is thrown vertically upwards, at the maximum height the velocity is zero.

As momentum is p=mv, so momentum is also zero at maximum height.

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2022 (26 Jun Shift 1)
An object moves along the circle with normal acceleration proportional to tα, where t is the time and is α a positive constant. The power developed by all the forces acting on the object will have time dependence proportional to

Options:
A tα-1
B tα/2
C t1+α2
D t2α

The correct answer is:
tα-1

Solution:
Given,

Normal acceleration =tα

We know, acceleration =an=v2R=tα

∴ ν2=tαR

K.E=12 mv2=m2tαR

⇒P=ddtK.E=ddtm2tαR

⇒P=mR2αtα-1

∴ P∝tα-1

Topic: Physics, Motion In Two Dimensions, JEE Main
An object moves at a constant speed along a circular path in a horizontal plane with centre at the origin. When the object is at x=+2 m, its velocity is -4j^ m s-1 . The object’s velocity (v) and acceleration (a) at x=–2 m will be

Options:
A v=4i^ m s-1, a=8j^ m s-2
B v=4j^ m s-1, a=8i^ m s-2
C v=-4j^ m s-1, a=8i^ m s-2
D v=-4i^ m s-1, a=-8j^ m s-2

The correct answer is:
v=4j^ m s-1, a=8i^ m s-2

Solution:
As the speed is constant, only centripetal acceleration will act(towards the centre of the circular path), ac=v2r=422=8 m s-2 towards the centre which at the given time will be towards right, hence a→c=8i^ m s-2. Velocity of the particle at this instant will in upward direction and hence v→=4j^ m s-1.

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2023 (29 Jan Shift 2)
An object moves in a straight line with deceleration whose magnitude varies with velocity as $3 v^{2 / 3}$. If at an initial point, the velocity is $8 \mathrm{~m} / \mathrm{s}$, then the distance travelled by the object before it stops is

Options:
A $2 \mathrm{~m}$
B $4 \mathrm{~m}$
C $6 \mathrm{~m}$
D $8 \mathrm{~m}$

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

Solution:
Given that, $a=$ deceleration $=-3 v^{2 / 3} \mathrm{~m} / \mathrm{s}^2$
At initial point $(t=0)$ velocity $=8 \mathrm{~m} / \mathrm{s}$
Distance travelled by the object before it stop is We know that,
$$
\begin{aligned}
a & =\frac{d v}{d t} \Rightarrow a=\frac{d v}{d s} \cdot \frac{d s}{d t} \\
a & =v d v / d s \\
a d s & =v d v \\
-3 v^{2 / 3} d s & =v d s \\
d \quad \quad d s & =-\frac{1}{3} v^{1 / 3} d v
\end{aligned}
$$
Integrate both sides, we get
$$
\begin{aligned}
& \int d s=\frac{-1}{3} \int v^{1 / 3} d v \\
& \Rightarrow \quad s=\frac{-1}{4} v^{4 / 3}+C \\
& \Rightarrow \text { At } \quad t=0, u=8 \mathrm{~m} / \mathrm{s}, s=0 \\
& C=\frac{1}{4}(8)^{4 / 3}=\frac{16}{4}=4 \\
&
\end{aligned}
$$
So,
Therefore, from Eq. (i)
$$
s=\frac{-1}{4} v^{4 / 3}+4
$$
Body stops, when $v=0$.
So,
$$
s=4 \mathrm{~m}
$$

Topic: Physics, Motion In One Dimension, JEE Main
An object moves with speed v1, v2 and v3 along a line segment AB, BC and CD respectively as shown in figure. Where AB = BC and AD = 3 AB, then average speed of the object will be :

Options:
A v1+v2+v33
B v1v2v33v1v2+v2v3+v3v1
C 3v1v2v3v1v2+v2v3+v3v1
D v1+v2+v33v1v2v3

The correct answer is:
3v1v2v3v1v2+v2v3+v3v1

Solution:
Speed of an object is defined as the distance travelled by the object per unit time.

Let's assume that, AB= d. So, BC= d, AD= 3d. Also, assume that the time required by the object to travel AB, BC, CD are t1, t2, t3 respectively.

Now,

AD= 3ABAB+BC+CD= 3ABAB+AB+CD= 3ABCD= AB

Hence,

t1= ABv1= dv1t2= dv2t3= dv3

The formula to calculate the average speed va of the object is given by

va= ADt1+t2+t3........................(1)

Substitute the expressions for the parameters into equation (1) and simplify to obtain the required average speed of the object.

va= 3ddv1+dv2+dv3= 3v1v2v3v1v2+v2v3+v3v1

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2023 (01 Feb Shift 1)
An object, moving with a speed of $6.25 \mathrm{~m} / \mathrm{s}$, is decelerated at a rate given by :
$$
\frac{\mathrm{dv}}{\mathrm{dt}}=-2.5 \sqrt{\mathrm{v}}
$$
where $v$ is the instantaneous speed. The time taken by the object, to come to rest, would be:

Options:
A
$2 \mathrm{~s}$

B
$4 \mathrm{~s}$

C
$8 \mathrm{~s}$

D
$1 \mathrm{~s}$

The correct answer is:
$2 \mathrm{~s}$

Solution:
$$
\frac{d v}{d t}=-2.5 \sqrt{v}
$$
Integrating the above equation.
$$
\Rightarrow 2 \sqrt{v}=-2.5 \mathrm{t}+\mathrm{C}
$$
at $\mathrm{t}=0, \mathrm{v}=6.25 \Rightarrow \mathrm{C}=5$
at $v=0 \Rightarrow t=\frac{5}{2.5}=2 \mathrm{~s}$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2011
An object 'O' is placed at a distance of 100 cm in front of a concave mirror of radius of curvature 200 cm as shown in the figure. The object starts moving towards the mirror at a speed 2 cm s-1. The position of the image from the mirror after 10 s will be at _____ cm.

Options:

The correct answer is:
400

Solution:
In 10 s, the object will travel distance =2×10=20 cm

After 10 sec, the object will be at distance u=20-100=-80 m.

Focal length of the concave mirror is f=R2=2002=-100 cm

Using mirror formula,

1v+1u=1f

⇒1v=-1100+180⇒v=800020

⇒v=400 cm

Position of image from the mirror is at 400 cm.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2022 (28 Jul Shift 2)
An object of $1 \mathrm{~kg}$ mass has a momentum of $10 \mathrm{~kg} \mathrm{~m} / \mathrm{sec}$ then the kinetic energy of the object will be

Options:
A $100 \mathrm{~J}$
B $50 \mathrm{~J}$
C $1000 \mathrm{~J}$
D $200 \mathrm{~J}$

The correct answer is:
$50 \mathrm{~J}$

Solution:
$E=\frac{P^2}{2 m}=\frac{(10)^2}{2 \times 1}=50 \mathrm{~J}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
An object of density $2000 \mathrm{~kg}-\mathrm{m}^{-3}$ is hung from a thin light wire. The fundamental frequency of the transverse waves in the wire is $200 \mathrm{~Hz}$. If the object is immersed in water such that half of its volume is submerged, then the fundamental frequency of the transverse waves in the wire is

Options:
A 200 Hz
B 173.2 Hz
C 100 Hz
D 141.4 Hz

The correct answer is:
173.2 Hz

Solution:
Fundamental frequency, initially
$$
n_1=\frac{1}{2 l} \sqrt{\frac{T}{\mu}}=\frac{1}{2 l} \sqrt{\frac{V g(2000)}{\mu}}
$$
Fundamental frequency, finally
$$
n_2=\frac{1}{2 l} \sqrt{\frac{V g\left(2000-\frac{1000}{2}\right)}{\mu}}
$$
(here we applied loss of weight due to upthrust)
$$
\begin{aligned}
& =\frac{1}{2 l} \sqrt{\frac{V g \times 1500}{\mu}} \\
\Rightarrow \quad \frac{n_1}{n_2} & =\sqrt{\frac{2000}{1500}}=\sqrt{\frac{4}{3}} \\
\Rightarrow \quad n_2 & =\frac{n_1 \times \sqrt{3}}{2}=100 \sqrt{3} \mathrm{~Hz}=173.2 \mathrm{~Hz}
\end{aligned}
$$

Topic: Physics, Waves and Sound, JEE Main
An object of height \( 3 \mathrm{~cm} \) kept in front of a concave mirror, gives a real image of size \( 6 \mathrm{~cm} \). The magnification of the
mirror is

Options:
A \( -2 \)
B \( 2 \)
C \( -0.5 \)
D \( 0.5 \)

The correct answer is:
\( -2 \)

Solution:
Given,

The height of the object, ho=3 cm

The height of the image, hi=-6 cm (Real images are always inverted and hence the negative sign)

The magnification, m=hiho-63=-2

Topic: Physics, Ray Optics, JEE Main
An object of length 10 cm is placed at right angles to the principal axis of a mirror of radius of curvature 60 cm such that its image is virtual, erect and has a length 6 cm. What kind of mirror it is and also determine the position of the object?

Options:
A – 20 cm

B 20 cm

C – 30 cm

D 30 cm

The correct answer is:
– 20 cm

Solution:
Since the image is virtual, erect and of a smaller
size, the given mirror is 'convex' (concave mirror does not form an image with the said description).
Given R = + 60 cm
â´ fâ¡ = Râ¡ 2 = 3 0 cm
Transverse magnification,
mâ¡ = Iâ¡ Oâ¡ = 6 1 0 = + 3 5
Further mâ¡ = - vâ¡ uâ¡ = 3 5
â´ vâ¡ = - 3 uâ¡ 5
Using 1 vâ¡ + 1 uâ¡ = 1 fâ¡
- 5 3 uâ¡ + 1 uâ¡ = 1 3 0
- 5 + 3 3 uâ¡ = 1 3 0
â´ u = - 20 cm
Thus the object is at a distance 20 cm ( from the pole) in front of the mirror.

Topic: Physics, Ray Optics, JEE Main
An object of $m \mathrm{~kg}$ with speed of $v \mathrm{~m} / \mathrm{s}$ strikes a wall at an angle $\theta$ and rebounds at the same speed and same angle. The magnitude of the change in momentum of the object will be

Options:
A $2 m v \cos \theta$
B $2 m v \sin \theta$
C $0$
D $2 m v$

The correct answer is:
$2 m v \cos \theta$

Solution:
$\vec{P}_1=m v \sin \hat{\theta} \hat{i}-m v \cos \theta \hat{j}$
and $\vec{P}_2=m v \sin \theta \hat{i}+m v \cos \theta \hat{j}$
So change in momentum
$\overrightarrow{\Delta P}=\vec{P}_2-\vec{P}_1=2 m v \cos \theta \hat{j},|\Delta \vec{P}|=2 m v \cos \theta$

Topic: Physics, Mathematics in Physics, JEE Main
An object of mass 0.5 kg is executing simple harmonic motion. It amplitude is 5 cm and time period (T) is 0.2 s. What will be the potential energy of the object at an instant t=T4 s starting from mean position. Assume that the initial phase of the oscillation is zero.

Options:
A 0.62 J
B 6.2×10-3 J
C 1.2×103 J
D 6.2×103 J

The correct answer is:
0.62 J

Solution:
T=2πmk

0.2=2π0.5k

k=50π2

≈500

x=Asin(ωt+ϕ)

=5 cmsinωT4+0

=5 cmsinπ2

=5 cm

PE=12kx2

=12(500)51002

=0.6255

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (27 Jul Shift 2)