datavorous/entrance-exam-dataset huggingface dataset:
An object of mass 1 kg executes simple harmonic oscillations along the x-axis, with a frequency of 2Ï Hz. At the position x=1 m, the object has a kinetic energy of 24 J. The amplitude of the oscillation is

Options:
A 32 m
B 2 m
C 4Â m
D 8Â m

The correct answer is:
2 m

Solution:
K.E.=12mÏ2A2-x2=24 J
12Ã1Ã16A2-1=24
A=2 m

Topic: Physics, Oscillations, JEE Main
An object of mass $0.2 \mathrm{~kg}$ executes simple harmonic motion along $\mathrm{x}$ axis with frequency of $\left(\frac{25}{\pi}\right) \mathrm{Hz}$. At the position $x=0.04 \mathrm{~m}$ the object has kinetic energy $0.5 \mathrm{~J}$ and potential energy $0.4 \mathrm{~J}$. The amplitude of oscillation is _____$\mathrm{cm}$.

Options:

The correct answer is:
6

Solution:
$\begin{aligned} & \text { Total energy }=\text { K.E. }+ \text { P.E. } \\ & \text { at } x=0.04 \mathrm{~m}, \text { T.E. }=0.5+0.4=0.9 \mathrm{~J} \\ & \begin{aligned} \text { T.E } & =1 \mathrm{~m} \omega^2 \mathrm{~A}^2=0.9 \\ & =\frac{1}{2} \times 0.2\left(2 \pi \times \frac{25}{\pi}\right)^2 \times \mathrm{A}^2=0.9 \\ \Rightarrow \mathrm{A} & =0.06 \mathrm{~m} \\ \mathrm{~A} & =6 \mathrm{~cm}\end{aligned}\end{aligned}$

Topic: Physics, Oscillations, JEE Main, JEE Main 2024 (08 Apr Shift 2)
An object of mass $2 \mathrm{~kg}$ is attached to a spring with spring constant $8 \mathrm{Nm}^{-1}$. If the object is executing simple harmonic motion then the number of cycles it completes in $66 \mathrm{~s}$ is

Options:
A $21$
B $16$
C $28$
D $12$

The correct answer is:
$21$

Solution:
Time period of a spring mass system is
$T=2 \pi \sqrt{\frac{m}{k}}$
Here, $\quad m=2 \mathrm{~kg}$ and $k=8 \mathrm{~N} / \mathrm{m}$
So, time for 1 oscillation,
$T=2 \pi \sqrt{\frac{2}{8}}=2 \times \pi \times \frac{1}{2}$
or $\quad T=\pi \mathrm{s}$
Number of cycles in 66 seconds is
$n=\frac{t}{T}=\frac{66}{\pi}$
$\Rightarrow \quad n=\frac{66 \times 7}{22}=21$
Hence, object complete 21 oscillations in given time.

Topic: Physics, Oscillations, JEE Main
An object of mass $2 m$ is projected with a speed of $100 \mathrm{~ms}^{-1}$ at an angle $\theta=\sin ^{-1}\left(\frac{3}{5}\right)$ to the horizontal. At the highest point, the object breaks into two pieces of same mass $m$ and the first one comes to rest. The distance between the point of projection and the point of landing of the bigger piece (in metre) is
(Given, $g=10 \mathrm{~m} / \mathrm{s}^2$ )

Options:
A $3840$
B $1280$
C $1440$
D $960$

The correct answer is:
$1440$

Solution:
Horizontal range of the object fired,
$R=\frac{u^2 \sin 2 \theta}{g}$
At the highest point, when object is exploded into two equal masses, then
$2 m u \cos \theta=m(0)+m v$
or
$v=2 u \cos \theta$
It means, the horizontal velocity becomes double at the highest point, hence it will cover double the distance during the remaining flight.
$\therefore$ Total horizontal range of the other part
$\begin{aligned} & =\frac{R}{2}+R=\frac{3 R}{2} \\ & =\frac{3}{2} \frac{u^2 \sin 2 \theta}{g}=\frac{3}{2} \times \frac{(100)^2 \times 2 \sin \theta \cos \theta}{g} \\ & =\frac{3}{2} \times \frac{(100)^2 \times 2 \times \frac{3}{5} \times \frac{4}{5}}{10}=1440 \mathrm{~m}\end{aligned}$

Topic: Physics, Motion In Two Dimensions, AP EAMCET, AP EAMCET 2007
An object of mass $2 m$ is projected with a speed of $100 \mathrm{~ms}^{-1}$ at an angle $\theta=\sin ^{-1}\left(\frac{3}{5}\right)$ to the horizontal. At the highest point, the object breaks into two pieces of same mass $m$ and the first one comes to rest. The distance between the point of projection and the point of landing of the bigger piece (in metre) is
(Given, $g=10 \mathrm{~m} / \mathrm{s}^2$ )

Options:
A $3840$
B $1280$
C $1440$
D $960$

The correct answer is:
$1440$

Solution:
Horizontal range of the object fired,
$R=\frac{u^2 \sin 2 \theta}{g}$
At the highest point, when object is exploded into two equal masses, then
$2 m u \cos \theta=m(0)+m v$
or
$v=2 u \cos \theta$
It means, the horizontal velocity becomes double at the highest point, hence it will cover double the distance during the remaining flight.
$\therefore$ Total horizontal range of the other part
$\begin{aligned} & =\frac{R}{2}+R=\frac{3 R}{2} \\ & =\frac{3}{2} \frac{u^2 \sin 2 \theta}{g}=\frac{3}{2} \times \frac{(100)^2 \times 2 \sin \theta \cos \theta}{g} \\ & =\frac{3}{2} \times \frac{(100)^2 \times 2 \times \frac{3}{5} \times \frac{4}{5}}{10}=1440 \mathrm{~m}\end{aligned}$

Topic: Physics, Motion In Two Dimensions, TS EAMCET, TS EAMCET 2007
An object of mass $3 \mathrm{~kg}$ is at rest. Now a force $F=6 t^2 \hat{i}+4 \hat{j}$ is applied on the object then the velocity of the object at $t=3 \mathrm{~s}$ is:

Options:
A $18 \hat{i}+3 \hat{j}$
B $18 \hat{i}+6 \hat{j}$
C $3 \hat{i}+6 \hat{j}$
D $18 \hat{i}+4 \hat{j}$

The correct answer is:
$18 \hat{i}+6 \hat{j}$

Solution:
From question $\vec{a}=\frac{\vec{F}}{m}=2 t^2 \hat{i}+\frac{4}{3} t \hat{j}$
$\therefore d \vec{V}=\left(2 t^2 \hat{i}+\frac{4}{3} t \hat{j}\right) d t$
Integrated both sides, We have,
$\vec{V}=2\left[\frac{t^3}{3}\right] \hat{i}+\frac{4}{3}\left[\frac{t^2}{2}\right] \hat{j}$
at $t=3 \mathrm{sec}, \vec{V}=\frac{2}{3}(3)^3 i+\frac{4}{6}(3)^2 j$
$=18 \hat{i}+6 \hat{j}$

Topic: Physics, Motion In One Dimension, JEE Main
An object of mass $3 \mathrm{~kg}$ is executing simple harmonic motion with an amplitude $\frac{2}{\pi} \mathrm{m}$. If the kinetic energy of the object when it crosses the mean position is $6 \mathrm{~J}$, the time period of oscillation of the object is

Options:
A $1 \mathrm{~s}$
B $2 \mathrm{~s}$
C $3 \mathrm{~s}$
D $4 \mathrm{~s}$

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

Solution:
Mass of object, $\mathrm{m}=3 \mathrm{~kg}$
Amplitude, $A=\frac{2}{\pi} \mathrm{m}$
Kinetic energy $=6 \mathrm{~J}$
Kinetic energy $=\frac{1}{2} m \omega^2 A^2=6$
$\begin{aligned}
& \frac{1}{2} \times 3 \times \frac{4 \pi^2}{\mathrm{~T}^2} \times \frac{4}{\pi^2}=6 \\
& \Rightarrow \mathrm{T}^2=4 ; \mathrm{T}=2 \mathrm{~s}
\end{aligned}$

Topic: Physics, Oscillations, JEE Main
An object of mass $3 \mathrm{~kg}$ is tied by a string of negligible mass to a ceiling and held such that the string is taut. The object is released suddenly such that the string remains taut. It's acceleration when released is (acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

Options:
A $3.5 \mathrm{~ms}^{-2}$
B $4.9 \mathrm{~ms}^{-2}$
C $7.5 \mathrm{~ms}^{-2}$
D $6.9 \mathrm{~ms}^{-2}$

The correct answer is:
$4.9 \mathrm{~ms}^{-2}$

Solution:
When object is released, net force acting on it is $\mathrm{mg}$ $\sin 30^{\circ}$ as shown, because $\mathrm{T}=\mathrm{mg} \cos 30^{\circ}$
So, $\mathrm{a}=\frac{\mathrm{mg} \sin 30^{\circ}}{\mathrm{m}}=\mathrm{g} \sin 30^{\circ}=4.9 \mathrm{~m} / \mathrm{s}^2$

Topic: Physics, Motion In Two Dimensions, AP EAMCET, AP EAMCET 2022 (05 Jul Shift 2)
An object of mass 5 kg is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of 10 N throughout the motion. The ratio of time of ascent to the time of descent will be equal to : [Use g=10 m s-2].

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

The correct answer is:
2:3

Solution:
Net force while going upward will be mg+R=50+10=60 N

⇒aup=605=12 m s-2

Similarly, Net force while going downward will be mg-R=40 N

⇒ad=405=8 m s-2

Since

h=12at2⇒auptup2adtd2=1⇒tuptd=812=23

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2022 (24 Jun Shift 2)
An object of mass 8 kg starts to move from rest under the action of a variable force F=3x N where x is the distance (in m) covered by the object. If initially, the position of the object is x=2 m, then find its speed (in m s-1) when it crosses x=10 m.

Options:

The correct answer is:
6

Solution:
Increment in kinetic energy = work done

12mv2-u2=∫x1x2F.dx=∫2103x dx

12mv2=32x2210=32100-4

12×8×v2=32×96

v=6 m s-1

Topic: Physics, Work Power Energy, JEE Main
An object of mass 10 kg is connected to the lower end of a massless string of length 4 m hanging from the ceiling. If a force F is applied horizontally at the mid-point of the string, the top half of the string makes an angle of 45° with the vertical, then the magnitude of F is

Options:
A 75 N
B 90 N
C 100 N
D 70 N

The correct answer is:
100 N

Solution:
T2=100
T2=F
F=100 N

Topic: Physics, Laws of Motion, JEE Main
An object of mass $500 \mathrm{~g}$, initially at rest acted upon by a variable force whose $\mathrm{X}$ component varies with $\mathrm{X}$ in the manner shown. The velocities of the object a point $X=8 \mathrm{~m}$ and $X=12 \mathrm{~m}$, would be the respective values of (nearly)

Options:
A $18 \mathrm{~m} / \mathrm{s}$ and $24.4 \mathrm{~m} / \mathrm{s}$
B $23 \mathrm{~m} / \mathrm{s}$ and $24.4 \mathrm{~m} / \mathrm{s}$
C $23 \mathrm{~m} / \mathrm{s}$ and $20.6 \mathrm{~m} / \mathrm{s}$
D $18 \mathrm{~m} / \mathrm{s}$ and $20.6 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$23 \mathrm{~m} / \mathrm{s}$ and $20.6 \mathrm{~m} / \mathrm{s}$

Solution:
The area under the force displacement curve give the amount of work done.
From work-energy theorem,
$$
\begin{aligned}
& \mathrm{W}=\Delta \mathrm{KE} \\
& \therefore \text { At } \mathrm{x}=8 \mathrm{~m} \text {, } \\
& \mathrm{W}=\text { Area } \mathrm{ABDO}+\text { Area CEFD } \\
& =20 \times 5+10 \times 3=130 \mathrm{~J} \\
& \text { Using Eq. (i) } \\
& \Rightarrow \quad 130=\frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \times \frac{500}{1000} \mathrm{v}^2 \\
& \Rightarrow \quad \mathrm{v}=2 \sqrt{130}=22.8 \mathrm{~ms}^{-1} \approx 23 \mathrm{~ms}^{-1} \\
& \text { At } \quad \mathrm{x}=12 \mathrm{~m} \\
& \mathrm{~W}=\text { Area } \mathrm{ABDO}+\text { Area CEFD + Area FGHIJ } \\
& + \text { Area KLMJ } \\
& \mathrm{W}=20 \times 5+10 \times 3+(-20 \times 2)+\left(\frac{1}{2} \times-5 \times 2\right) \\
& +10 \times 2 \quad[\because \text { Area FGHIJ }=\text { Area } \\
& \mathrm{FGIJ}+\text { Area } \mathrm{GHI}] \\
& =100+30-40-5+20=105 \mathrm{~J} \\
& \text { Using Eq. (i) } \\
& \therefore \quad 105=\frac{1}{2} \times \frac{1}{2} \times v^2 \\
& \Rightarrow \quad \mathrm{v}=2 \sqrt{105} \simeq 20.6 \mathrm{~ms}^{-1} \\
&
\end{aligned}
$$

Topic: Physics, Work Power Energy, JEE Main
An object of mass m1 collides with another object of mass m2, which is at rest. After the collision the objects move with equal speeds in opposite direction. The ratio of the masses m2:m1 is :

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

The correct answer is:
3:1

Solution:
m1v1=-m1v+m2v

v1=-v+m2 m1v

v1+vv=m2 m1

e=2vv1=1

v=v12

v1+v1/2v1/2=m2 m1

3=m2 m1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2021 (18 Mar Shift 2)
An object of mass m initially at rest on a smooth horizontal plane starts moving under the action of force F=2 N. In the process of its linear motion, the angle θ (as shown in figure) between the direction of force and horizontal varies as θ=kx, where k is a constant and x is the distance covered by the object from its initial position. The expression of kinetic energy of the object will be E=nksin θ. The value of n is ______.

Options:

The correct answer is:
2

Solution:
As seen in the free body diagram of the object, normal force N, weight mg and force in parallel and perpendicular directions are acting.

Using Newton's law of motion

Along horizontal direction, Fcosθ=ma

⇒2cos(kx)=mvdvdx

Integrating the above relation, we have

∫0vvdv=2∫0xcos(kx)dx

⇒mv22=2ksinkx

Thus, kinetic energy of the object will be K.E.=2ksinθ

Thus, the value of n=2.

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (25 Jan Shift 1)
An object of mass $M$ is at the origin and is at rest. Suddenly an explosion happens in the object and object is divided three equal parts.
The first part goes with the velocity $\left(v_0 \mathrm{i}+\mathrm{v}_0 \mathrm{j}\right)$.
The second part goes with the velocity $\mathrm{v}_0 \mathrm{j}$
What is the velocity vector for the third part?

Options:
A $\left(v_0 i+v_0 j\right)$
B $\left(-v_0 i-2 v_0 j\right)$
C $\left(\mathrm{v}_0 \mathrm{i}-\mathrm{v}_0 \mathrm{j}\right)$
D None of these

The correct answer is:
$\left(-v_0 i-2 v_0 j\right)$

Solution:
Applying law of conservation of linear momentum
$\mathbf{0}=\frac{M}{3}\left(v_0 \mathbf{i}+v_0 \mathbf{j}\right)+\frac{M}{3}\left(v_0 \mathbf{j}\right)+\frac{M}{3} \mathbf{v}$
Or
$\mathbf{v}=-v_0 \mathbf{i}-2 v_0 \mathbf{j}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
An object of mass m is being moved with a constant velocity under the action of an applied force of 2 N along a frictionless surface with following surface profile.

The correct applied force vs distance graph will be :

Options:
A
B
C
D

The correct answer is:

Solution:
When the block will be moving up the incline, the external force will be up the incline and will be balancing the down the incline component of gravity.

When the block will be moving down the incline, the external force will be again up the incline and will be balancing the down the incline component of gravity. But in this case the external force will be on the back side of the block.

F=mgsinθ=2 N

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2021 (01 Sep Shift 2)
An object of mass m is suspended at the end of a massless wire of length L and area of cross-section, A. Young modulus of the material of the wire is Y. If the mass is pulled down slightly its frequency of oscillation along the vertical direction is :

Options:
A f=12πmLYA
B f=12πYAmL
C f=12πmAVL
D f=12πYLmA

The correct answer is:
f=12πYAmL

Solution:
f=12πkm=12πYAmL

f=12πYAmL

Topic: Physics, Oscillations, JEE Main, JEE Main 2020 (06 Sep Shift 1)
An object of mass m, weighs m1 in a liquid of density d1 and it weighs m2 in a liquid of density d2. The density d of the object is

Options:
A d=m2d2-m1d1m2-m1
B d=m1d1-m2d2m2-m1
C d=m2d1-m1d2m1-m2
D d=m1d2-m2d1m1-m2

The correct answer is:
d=m1d2-m2d1m1-m2

Solution:
Vd-d1g=m1g
Vd-d2g=m2g
d-d1d-d2=m1m2
∴ d=m1d2-m2d1m1-m2

Topic: Physics, Mechanical Properties of Fluids, JEE Main
An object P is kept in front of two plane mirrors as shown in the figure. The number of images formed is k. Find out the value of k.

Options:

The correct answer is:
5

Solution:
The above diagram represents the angular positions of possible images that can be formed in the given case. Since, images further than 180Â° will not be visible, we obtain the maximum number of visible images as 5.

Topic: Physics, Ray Optics, JEE Main
An object placed on the ground is in stable equilibrium. If the object is given a slight push, then initially the position of centre of gravity

Options:
A moves nearer to ground
B rises higher above the ground
C remains as such
D may remain at same level

The correct answer is:
rises higher above the ground

Solution:
In stable equilibrium, the centre of gravity of object, lies at minimum height from ground. As the object is given a slight push, its centre of gravity rises because it comes in unstable equilibrium.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
An object takes \( n \) times as much time to slide down a \( 45^{\circ} \) rough inclined plane as it takes to slide down a perfectly smooth \( 45^{\circ} \) inclined plane. The coefficient of kinetic friction between the rough plane and the object is

Options:
A \( n^{2}-1 \)
B \( 1-\frac{1}{n^{2}} \)
C \( n^{2}+1 \)
D \( 1+\frac{1}{n^{2}} \)

The correct answer is:
\( 1-\frac{1}{n^{2}} \)

Solution:
s=ut+12at2,a=gsinθ,u=0

Then, s=0+12at2

or t=2sa=2sgsinθ [for smooth plane]

For the rough plane. the effective value of acceleration along the incline is
a'=gsinθ-μkgcosθ

and t'=2sa=2sgsinθ-μkgcosθ

Now, t't=n=gsinθgsinθ-μkgcosθ

=sin45°sin45°-μkcos45°

=11-μk

⇒ 1n2=1-μk⇒μk=1-1n2

Topic: Physics, Laws of Motion, JEE Main
An object viewed from a near point distance of 25 cm, using a microscopic lens with magnification 6, gives an unresolved image. A resolved image is observed at infinite distance with a total magnification double the earlier using an eyepiece along with the given lens and a tube of length 0.6 m, if the focal length of the eyepiece is equal to ____ cm.

Options:

The correct answer is:
25

Solution:
For simple microscope,
m=1+Df0
6=1+Df0
5=25f0
f0=5 cm

For compound microscope,
m=ℓ·Df0·fe
12=60×255·fe
fe=25 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2021 (20 Jul Shift 1)
An objective is kept at a distance of \( 16 \mathrm{~cm} \) from a thin lens and the image formed is real. If the object is kept at a distance of \( 6 \mathrm{~cm} \) from the same lens the image formed is virtual. If the size of the images formed are equal, then find the focal length of the lens?

Options:
A \( 11 \mathrm{~cm} \)
B \( -11 \mathrm{~cm} \)
C \( \sqrt{11} \mathrm{~cm} \)
D \( \sqrt{-11} \mathrm{~cm} \)

The correct answer is:
\( 11 \mathrm{~cm} \)

Solution:
m = h 2 h 1 = v u = f f + u
for real image m is negative - m = f f - 1 6 ---- 1
for virtual image m is positivem=ff-6----2
from (1) & (2)
- f f - 6 = f f - 1 6 ⇒ 6 f - f 2 = - 16 f + f 2 ⇒ 2 f 2 = 2 2 f
f = 11 cm

Topic: Physics, Ray Optics, JEE Main
An observed event $B$ can occur after one of the three events $A_{1}, A_{2}, A_{3} .$ If
$P\left(A_{1}\right)=P\left(A_{2}\right)=0.4, P\left(A_{3}\right)=0.2$ and $P\left(B / A_{1}\right)=0.25, P\left(B / A_{2}\right)$
$=0.4, P\left(B / A_{3}\right)=0.125$, what is the probability of $A_{1}$ after observing $B$ ?

Options:
A $\frac{1}{3}$
B $\frac{6}{19}$
C $\frac{20}{57}$
D $\frac{2}{5}$

The correct answer is:
$\frac{20}{57}$

Solution:
By Baye's theorem Required probability $=P\left(A_{1} / B\right)$ $=\frac{P\left(A_{1}\right) P\left(B / A_{1}\right)}{P\left(A_{1}\right) P\left(B / A_{1}\right)+P\left(A_{2}\right) P\left(B / A_{2}\right)+P\left(A_{3}\right) P\left(B / A_{3}\right)}$
$=\frac{0.4 \times 0.25}{0.4 \times 0.25+0.4 \times 0.4+0.2 \times 0.125}$
$=\frac{0.1}{0.1+0.16+0.025}=\frac{0.1}{0.285}=\frac{20}{57}$

Topic: Mathematics, Probability, JEE Main
An observer and a vehicle, both start moving together from rest (towards the right) with acceleration 5 m s-2 and 2 m s-2, respectively. There is a 2 kg block on the floor of the vehicle and coefficient of friction is Î¼ = 0. 3 between their surface. Then the work done by the frictional force on the 2 kg block observed by the running observer, during the first 2 seconds of the motion

Options:
A 24 J

B -24 J
C 16 J

D -16 J

The correct answer is:
-24 J

Solution:
flimiting=μ mg=0.3×2×10=6 N
Force needed for acc. of 2 m s-2
F=ma=4 N
So friction force static, f=4 N
Displacement w.r.t. observer in t = 2 sec. is
S=urelt+12arelt2=0+12-3×4=-6 m
W=f→.S→=4-6=-24 J

Topic: Physics, Work Power Energy, JEE Main
An observer can see through a pinhole the top end of a thin rod of height h, placed as shown in the figure. The beaker height is 3h and its radius h. When the beaker is filled with a liquid up to a height 2h, he can see the lower end of the rod. Then the refractive index of the liquid is

Options:
A 5 2
B 5 2
C 3 2
D 3 2

The correct answer is:
5 2

Solution:
PQ=QR=2h ∴ ∠i=450
∴ ST=RT, h=KM=MN
So, KS=h2+2h2
=h5
∴ sin r=hh5
= 1 5

∴ μ= sini sinr= sin 45°1/5=52

Topic: Physics, Ray Optics, JEE Main
An observer can see through a small hole on the side of a jar (radius 15 cm ) at a point at height of 15 cm from the bottom (see figure). The hole is at a height of 45 cm. When the jar is filled with a liquid up to a height of 30 cm the same observer can see the edge at the bottom of the jar. If the refractive index of the liquid isN100, where N is an integer, the value of N is

Options:

The correct answer is:
158

Solution:
From Shell's Law

μ×sini=1×sinr

μ×15152+302=1×sin45°

μ=52=158×10-2

Topic: Physics, Ray Optics, JEE Main, JEE Main 2020 (03 Sep Shift 1)
An observer finds that the angular elevation of a tower is Î¸. On advancing 3m towards the tower, the elevation is 45o and on advancing 2m further more towards the tower, the elevation is 90o-Î¸. The height of the tower is (assume the height of observer is negligible and observer lies on the same level as the foot of the tower)

Options:
A 2m
B 4m
C 6m
D 8m

The correct answer is:
6m

Solution:
Let, PQ be the tower of height h meters
∠PAQ=θ,∠PBQ=45o and ∠PCQ=90o-θ
AB=3m and BC=2m
⇒3=AB=PA-PB=hcot⁡A-h
and 2=BC=PB-CP=h-htan⁡A
∴h+3h=cot⁡A,h-2h=tan⁡A
∴h+3h·h-2h=1⇒h2+h-6=h2⇒h=6m

Topic: Mathematics, Heights and Distances, JEE Main
An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer? (speed of light =3×108 m s-1)

Options:
A 15.3 GHz
B 10.1 GHz
C 12.1 GHz
D 17.3 GHz

The correct answer is:
17.3 GHz

Solution:
cf′=cf 1-β1+β , β=vcβ=12
⇒1f′=1f1-121+12= 131f
⇒f′= 3f
⇒f′=1.7310
⇒f′=17.3 GHz

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2017 (02 Apr)
An observer looks at a distant tree of height 10 m with a telescope of magnifying power of 20. To the observer, the tree appears as

Options:
A 20 times taller.
B 20 times nearer.
C 10 times taller.
D 10 times nearer.

The correct answer is:
20 times nearer.

Solution:
Angular magnification of the telescope is 20 so now the tree will subtend a larger angle at the observer's eye and hence appears 20 times nearer

Topic: Physics, Ray Optics, JEE Main, JEE Main 2016 (03 Apr)
An observer looks at a tree of height $15 \mathrm{~m}$ with a telescope of magnifying power 10 . To him, the tree appears

Options:
A 10 times taller
B 15 times taller
C 10 times nearer
D 15 times nearer

The correct answer is:
10 times nearer

Solution:
Telescope is used to see the distant objects. More magnifying power means more nearer image.

Topic: Physics, Ray Optics, JEE Main
An observer moves towards a stationary source of sound with a speed \(\frac{1}{5}^{\text {th }}\) of the speed of sound. The wavelength and frequency of the waves emitted by the source are \(\lambda\) and \(f\) respectively. The apparent frequency and wavelength heard by the observer are respectively,

Options:
A \(1.2 f, \lambda\)
B \(f, 1.2 \lambda\)
C \(0.8 f, 0.8 \lambda\)
D \(1.2 f, 1.2 \lambda\)

The correct answer is:
\(1.2 f, \lambda\)

Solution:
When an observer moves towards on a stationary source of sound, then apparent frequency heard by the observer increases. The apparent frequency heard in this situation.
\(f^{\prime}=\left(\frac{v+v_0}{v-v_s}\right) f,\) (As source is stationary)
Hence, \(v_s=0\)
\(\begin{aligned}
& \therefore \quad f^{\prime}=\left(\frac{v+v_0}{v}\right) f \\
& \text{Given, } v_0=\frac{v}{5} \\
& f^{\prime}=\left(\frac{v+\frac{v}{5}}{v}\right) f=\frac{6}{5} f=1.2 f \\
\end{aligned}\)
Motion of the observer does not affect the wavelength reaching the observer, hence wavelength remains \(\lambda\).
Hence, the apparent frequency and wavelength heard by the observes are respectively, \(1.2 f\) and \(\lambda\).

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2019 (20 Apr Shift 1)
An observer moves towards a stationary source of sound, with a velocity one fifth of the velocity of sound. What is the percentage increase in the apparent frequency?

Options:
A
zero

B
$0.5 \%$

C
$5 \%$

D
$20 \%$

The correct answer is:
$20 \%$

Solution:
$$
f=\frac{v+v / 5}{v} f=\frac{6 f}{5}
$$
$\%$ increase in frequency $=20 \%$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2005
An observer standing on a railway crossing receives frequency of 2.2 kHz and 1.8 kHz when the train approaches and recedes from the observer. Find the velocity (in m s-1) of the train. (The speed of the sound in air is 300 m s-1.)

Options:

The correct answer is:
30

Solution:
f 1 = f 0 ( V Â± V 0 ) ( V â V S ) = f 0 v V â V S
When approaches:2.2Ã103=f0300300-Vs ...(i)
When recedes:1.8Ã103=f0300300+Vs ...(ii)
Dividing (i) by (ii)
2 . 2 1 . 8 = 3 0 0 + V S 3 0 0 - V S â 4 0 . 4 = 6 0 0 2 V S
â VS = 30 m s-1

Topic: Physics, Waves and Sound, JEE Main