datavorous/entrance-exam-dataset huggingface dataset:
A body is at rest at $x=0$. At $t=0$, it starts moving in the positive $x$-direction with a constant acceleration. At the same instant another body passes through $x=0$ moving in the positive $x$ direction with a constant speed. The position of the first body is given by $\mathrm{x}_1(\mathrm{t})$ after time ' $\mathrm{t}$ ' and that of the second body by $x_2(t)$ after the same time interval. Which of the following graphs correctly describes $\left(x_1-x_2\right)$ as a function of time ' $t$ '?

Options:
A

B

C

D

The correct answer is:

Solution:
$$
\begin{aligned}
& x_1(t)=\frac{1}{2} a t^2 \\
& x_2(t)=v t \\
& x_1-x_2=\frac{1}{2} a t^2-v t
\end{aligned}
$$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2008
A body is dropped on ground from a height h1 and after hitting the ground, it rebounds to a height h2. If the ratio of velocities of the body just before and after hitting ground is 4, then percentage loss in kinetic energy of the body is x4. The value of x is _____.

Options:

The correct answer is:
375

Solution:
The given ratio of the velocities is

v1v2=4

The percentage loss in kinetic energy is

mv212-mv222mv212×100=x4⇒v21-v22v21=x400⇒1-v22v21=x400⇒1-116=x400⇒x=375

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (06 Apr Shift 2)
A body is executing S.H.M. At a displacement x its potential energy is 9 J and at a displacement y its potential energy is 16 J. The potential energy at displacement x+y is

Options:
A 25 J
B 5 J
C 49 J
D 7 J

The correct answer is:
49 J

Solution:
Consider an oscillating spring-mass system.

Potential energy at displacement x will be written as, 12kx2.

Now,

12kx2=9 ...Eq1

and

12ky2=16 ...Eq2

Then, potential energy at the displacement is given by,

=12kx+y2=12kx2+y2+2xy=12kx2+12ky2+212kx212ky2=9+16+2×3×4=49 J

Topic: Physics, Oscillations, JEE Main
A body is falling freely under the action of gravity alone in vaccum. Which of the following quantities remain constant during the fall?

Options:
A Kinetic energy

B Potential energy

C Total mechanical energy

D Total linear momentum

The correct answer is:
Total mechanical energy

Solution:
As the body is falling freely under gravity and no external force act on body in vaccum so law of conservation, the potential energy decreases and kinetic energy increases because total mechanical energy (PE + KE) of the body and earth system will be remain constant.

Topic: Physics, Work Power Energy
A body is heated to a temperature 40 Â°C and kept in a chamber maintained at 20 Â°C. If the temperature of the body decreases to 36 Â°C in 2 minutes, then the time after which the temperature will further decrease by 4 Â°C, is

Options:
A 2 minutes
B 2 minutes 33 seconds
C 2 minutes 55 seconds
D 3 minutes

The correct answer is:
2 minutes 33 seconds

Solution:
Rate of cooling,
dÎ¸dt=-kÎ¸-Î¸0
From 40 Â°C to 36 Â°C,

Î¸1=40+362=38 Â°C
dÎ¸dt1=42=2 Â°C min-1
From 36 Â°C to 32 Â°C,

Î¸2=36+322=34 Â°C
dÎ¸dt2=4t
dÎ¸dt1dÎ¸dt2=kÎ¸1âÎ¸0kÎ¸2âÎ¸0
â 424t=38-2034-20
â t=2min 33 s

Topic: Physics, Thermal Properties of Matter, NEET
A body is in simple harmonic motion with time period T=0.5 s and amplitude A=1 cm . Find the average velocity in the interval in which it moves from equilibrium position to half of its amplitude.

Options:
A 16âcm/s
B 6 cm/s
C 4 cm/s
D 12âcm/s

The correct answer is:
12 cm/s

Solution:
∴x=Asin⁡ωt
A2=Asin⁡ωt
ωt=π6
t=π6ω
=π6×2π
t=T12
∴Time taken to reach from x=0 to x=A2 isT12
∴Average velocity=DisplacementTime
=A/2T/12
=AT×6
=6×10.5
=12 cm/s

Topic: Physics, Oscillations, JEE Main, JEE Main 2014 (19 Apr Online)
A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time ' $t$ ' is proportional to

Options:
A
$t^{3 / 4}$

B
$t^{3 / 2}$

C
$t^{1 / 4}$

D
$t^{1 / 2}$

The correct answer is:
$t^{3 / 2}$

Solution:
Power $=$ F. V
$\mathrm{F}=\mathrm{m}\left(\frac{\mathrm{dV}}{\mathrm{dt}}\right) \Rightarrow \mathrm{m} \cdot \mathrm{v} \cdot \frac{\mathrm{dV}}{\mathrm{dt}}=$ constant $=\mathrm{C}$
$\Rightarrow \frac{\mathrm{dV}}{\mathrm{dt}}=\frac{\mathrm{C}}{\mathrm{m}}=\mathrm{k} \Rightarrow \mathrm{vdv}=\mathrm{kdt} \quad \Rightarrow \int \mathrm{vdv}=\int \mathrm{kdt} \Rightarrow \frac{\mathrm{V}^2}{2}=\mathrm{kt}+\mathrm{c}$
$\Rightarrow \mathrm{v} \propto(\mathrm{t})^{1 / 2}$ $\frac{\mathrm{ds}}{\mathrm{dt}}=\mathrm{c} \cdot \mathrm{t}^{1 / 2}$
$\Rightarrow \int \mathrm{ds}=\int\left(\mathrm{c} \cdot \mathrm{t}^{1 / 2}\right) \mathrm{dt} \Rightarrow \mathrm{S}=\mathrm{C} \cdot \frac{2}{3} \mathrm{t}^{3 / 2}$ $\Rightarrow \mathrm{S}=\frac{\mathrm{ct}^{3 / 2}}{3 / 2} \Rightarrow \mathrm{s} \propto \mathrm{t}^{3 / 2}$

Topic: Physics, Work Power Energy, JEE Main
A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time ' $t$ ' is proportional to

Options:
A
$t^{3 / 4}$

B
$t^{3 / 2}$

C
$t^{1 / 4}$

D
$t^{1 / 2}$

The correct answer is:
$t^{3 / 2}$

Solution:
Power $=$ F. V
$\mathrm{F}=\mathrm{m}\left(\frac{\mathrm{dV}}{\mathrm{dt}}\right) \Rightarrow \mathrm{m} \cdot \mathrm{v} \cdot \frac{\mathrm{dV}}{\mathrm{dt}}=$ constant $=\mathrm{C}$
$\Rightarrow \frac{\mathrm{dV}}{\mathrm{dt}}=\frac{\mathrm{C}}{\mathrm{m}}=\mathrm{k} \Rightarrow \mathrm{vdv}=\mathrm{kdt} \quad \Rightarrow \int \mathrm{vdv}=\int \mathrm{kdt} \Rightarrow \frac{\mathrm{V}^2}{2}=\mathrm{kt}+\mathrm{c}$
$\Rightarrow \mathrm{v} \propto(\mathrm{t})^{1 / 2}$ $\frac{\mathrm{ds}}{\mathrm{dt}}=\mathrm{c} \cdot \mathrm{t}^{1 / 2}$
$\Rightarrow \int \mathrm{ds}=\int\left(\mathrm{c} \cdot \mathrm{t}^{1 / 2}\right) \mathrm{dt} \Rightarrow \mathrm{S}=\mathrm{C} \cdot \frac{2}{3} \mathrm{t}^{3 / 2}$ $\Rightarrow \mathrm{S}=\frac{\mathrm{ct}^{3 / 2}}{3 / 2} \Rightarrow \mathrm{s} \propto \mathrm{t}^{3 / 2}$

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2003
A body is moving according to the equation $x=a t+b t^2-c t^3$ where $x=$ displacement and ${ }^{a, b}$ and $c$ are constants. The acceleration of the body is

Options:
A $a+2 b t$
B $2 b+6 c t$
C $2 b-6 c t$
D $3 b-6 c t^2$

The correct answer is:
$2 b-6 c t$

Solution:
$x=a t+b t^2-c t^3, a=\frac{d^2 x}{d t^2}=2 b-6 c t$

Topic: Physics, Motion In One Dimension, JEE Main
A body is moving from rest under constant acceleration and let S1 be the displacement in the first P-1 seconds and S2 be the displacement in the first P seconds. The displacement in P2-P+1th second will be

Options:
A S1+S2
B S1S2
C S1-S2
D None of these

The correct answer is:
S1+S2

Solution:
From the equation of motion
S=ut+12 at2
S1=0+12 aP-12
And S2=0+12 aP2
From Sn=u+a22n-1
SP2-P+1th=a22P2-P+1-1
=a22P2-2P+1=S1+S2

Topic: Physics, Motion In One Dimension, JEE Main
A body is moving in a low circular orbit about a planet of mass M and radius R. The radius of the orbit can be taken to be R itself. Then the ratio of the speed of this body in the orbit to the escape velocity from the planet is:

Options:
A 12
B 2
C 1
D 2

The correct answer is:
12

Solution:
The orbital velocity of the body in an orbit of the radius R is vo=GmR

The escape velocity from the surface is ve=2GmR

Ratio, v0ve=GmR2GmR=12

Topic: Physics, Gravitation, JEE Main, JEE Main 2020 (04 Sep Shift 2)
A body is moving is \( x-y \) plane as shown in a circular path of radius \( 2 \mathrm{~m} \). At a certain instant when the body is crossing the positive \( \mathrm{y} \)-axis its acceleration is \( (6 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}) \mathrm{m} \mathrm{s}^{-2} \). Then its angular acceleration and angular velocity at this instant will be-

Options:
A \( -3 \widehat{\mathrm{k}} \mathrm{rad} \mathrm{s}^{-2} \) and \( -2 \widehat{\mathrm{k}} \mathrm{rad} \mathrm{s}^{-1} \) respectively
B \( +3 \widehat{\mathrm{k}} \mathrm{rad} \mathrm{s}^{-2} \) and \( +2 \widehat{\mathrm{k}} \mathrm{rad} \mathrm{s}^{-1} \) respectively
C \( -4 \widehat{\mathrm{k}} \mathrm{rad} \mathrm{s}^{-2} \) and \( -\sqrt{3} \widehat{\mathrm{k}} \mathrm{rad} \mathrm{s}^{-1} \) respectively
D \( +4 \widehat{\mathrm{k}} \mathrm{rad} \mathrm{s}^{-2} \) and \( +\sqrt{3} \widehat{\mathrm{k}} \mathrm{rad} \mathrm{s}^{-1} \) respectively

The correct answer is:
\( -3 \widehat{\mathrm{k}} \mathrm{rad} \mathrm{s}^{-2} \) and \( -2 \widehat{\mathrm{k}} \mathrm{rad} \mathrm{s}^{-1} \) respectively

Solution:
Given that, the radius of the circular path is, r=2 m and the net acceleration of the particle is,

a→=6i^-8j^ m s-2 ⋯1

Compare the above equation with

a=ati^-arj^ m s-2 ⋯2

By comparing equation 1 & 2, we can write,

Radial acceleration as, ar=-8j^ m s-2

Tangential acceleration as, at=6i^ m s-2

Relation between the tangential acceleration and angular acceleration is,

at=6=rα ⇒ 2α=6

α=3 rad s-2 ⇒ α=-3k^ rad s-2

Relation between the radial acceleration and angular velocity is,

ar=rω2 ⇒ ω=arr ⇒ ω=82

ω=2 rad s-1 ⇒ ω=-2k^ rad s-1

Topic: Physics, Motion In Two Dimensions, JEE Main
A body is moving unidirectionally under the influence of a source of constant power. The square of its displacement in time t is proportional to

Options:
A t3
B t2
C t
D t

The correct answer is:
t3

Solution:
By work-energy theorem,

W=P×t=12mv2

⇒v2=2Ptm

v=2Ptm1/2

or dsdt=2Ptm1/2

∫ds=∫2Ptm1/2dt

s=2Pm1/2·23t3/2

i.e. s∝t3/2

s2∝t3

Topic: Physics, Work Power Energy, JEE Main
A body is moving up an inclined plane of angle $\theta$ with an initial kinetic energy $E$. The coefficient of friction between the plane and the body is $u$. The work done against friction before the body comes to rest is

Options:
A $\frac{\mu \cos \theta}{E \cos \theta+\sin \theta}$
B $E$
C $\frac{\mu E \cos \theta}{\mu \cos \theta-\sin \theta}$
D $\frac{\mu E \cos \theta}{\mu \cos \theta+\sin \theta}$
Increase in KE, $\begin{aligned} \frac{p_2^2-p_1^2}{2 m} & =\frac{(10)^2-(8)^2}{2 \times 4} \\ & =\frac{100-64}{8}=\frac{36}{8} \\ & =4.5 \mathrm{~J}\end{aligned}$

The correct answer is:
$E$

Solution:
Acceleration of a body up a rough inclined plane
$a=g(\mu \cos \theta+\sin \theta)$
From equation of motion
$\begin{aligned}
v^2 & =u^2-2 a s \\
0 & =u^2-2 a s \\
u^2 & =2 a s
\end{aligned}$
Initial $\mathrm{KE}, E=\frac{1}{2} m u^2$
$\begin{aligned}
E & =\frac{1}{2} m(2 a s) \\
s & =\frac{E}{m a}
\end{aligned}$
$\begin{aligned} & \text { Work done } W=F \cdot s \\ & =(m g \sin \theta+\mu R) \cdot \frac{E}{m a} \\ & =(m g \sin \theta+\mu m g \cos \theta) \cdot \frac{E}{m(\mu \cos \theta+\sin \theta)} \\ & =g(\sin \theta+\mu \cos \theta) \cdot \frac{E}{g(\mu \cos \theta+\sin \theta)}=E\end{aligned}$

Topic: Physics, Work Power Energy, JEE Main
A body is moving with constant speed, in a circle of radius 10 m. The body completes one revolution in 4 s. At the end of 3rd second, the displacement of body (in m) from its starting point is:

Options:
A 30
B 15 π
C 5 π
D 102

The correct answer is:
102

Solution:
Give: r=10 m, T=4 s.

Displacement is defined to be the change in position of an object. As we can see from the above figure, displacement will be d=r2+r2

⇒d=102 m

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2023 (31 Jan Shift 2)
A body is moving with velocity $30 \mathrm{~m} / \mathrm{s}$ towards east. After $10 \mathrm{~s}$ its velocity becomes $40 \mathrm{~m} / \mathrm{s}$ towards north. The average acceleration of the body is

Options:
A $7 \mathrm{~m} / \mathrm{s}^2$
B $\sqrt{7} \mathrm{~m} / \mathrm{s}^2$
C $5 \mathrm{~m} / \mathrm{s}^2$
D $1 \mathrm{~m} / \mathrm{s}^2$

The correct answer is:
$5 \mathrm{~m} / \mathrm{s}^2$

Solution:
$$
\begin{aligned}
& \text { Average acceleration }=\frac{\text { Change in velocity }}{\text { Total time }} \\
& \qquad \begin{aligned}
a & =\frac{\left|\overrightarrow{\mathbf{v}}_f-\overrightarrow{\mathbf{v}}_i\right|}{\Delta t} \\
& =\frac{\sqrt{30^2+40^2}}{10}=\frac{\sqrt{900+1600}}{10} \\
& =5 \mathrm{~ms}^{-2}
\end{aligned}
\end{aligned}
$$

Topic: Physics, Motion In One Dimension, JEE Main
A body is performing simple harmonic with an amplitude of 10 cm. The velocity of the body was tripled by air Jet when it is at 5 cm from its mean position. The new amplitude of vibration is x cm. The value of x is _____ .

Options:

The correct answer is:
700

Solution:
In SHM, velocity of the particle at any displacement, x is given by, v=A2-x2.

Therefore,

⇒v3v=A12-25A22-25 ⇒A22-25=3×75

⇒A22=700 ⇒A2=700 cm

Topic: Physics, Oscillations, JEE Main, JEE Main 2022 (29 Jun Shift 1)
A body is projected at an angle of 60Â° with the horizontal. If its kinetic energy at its maximum height is 10 J, then the height at which its potential energy and kinetic energy have equal values is (consider potential energy at the point of projection to be zero)

Options:
A half of the maximum height
B two third of the maximum height
C one sixth of the maximum height
D insufficient data to solve the problem

The correct answer is:
two third of the maximum height

Solution:
KE at maximum height
1 2 mu 2 cos 2 6 0 â = 1 0 J
1 2 mu 2 = 4 0 J
Mg H max = 3 0 J ...(i)
When PE = KE
â mgh = 2 0 = ...(ii)
from (i) and (ii)
h H max = 2 3

Topic: Physics, Work Power Energy, JEE Main
A body is projected at an angle of 60° with the horizontal such that the vertical component of its initial velocity is 40 m s-1. The magnitude of velocity of the projectile at one quarter of its time of flight is nearly (Acceleration due to gravity =10 m s-2)

Options:
A 3.54 m s-1
B 35.40 m s-1
C 30.54 m s-1
D 34.5 m s-1

The correct answer is:
30.54 m s-1

Solution:
Suppose initial velocity is v and angle of projectionθ is given 60°.

Given,

Vertical component of initial velocity v sin θ=40 m s-1

⇒ v=46.19 m s-1

Time of flight T=2 v sin θg=2×4010=8 s

Given, time t=T4=2 s

Suppose vertical component of velocity after time t is v1.

Applying first eqaution of motion-

v1=v sin 60°-gt

⇒ v1=40-20=20 m s-1

Horizontal component remains uncharged.

∴ vH=v cos 60°=23.09 m s-1

Magnitude of velocity after time t vt=v12+vH2

⇒ vt=202+23.092=30.54 m s-1

Topic: Physics, Motion In Two Dimensions, AP EAMCET, AP EAMCET 2019 (21 Apr Shift 2)
A body is projected at an angle θ to the horizontal with kinetic energy Ek. The potential energy at the highest point of the trajectory is

Options:
A Ek
B Ekcos2θ
C Eksin2θ
D Ektan2θ

The correct answer is:
Eksin2θ

Solution:
Kinetic energy at highest point = 1 2 m v cos θ 2 = E k cos 2 θ
Total energy remains constant ⇒ PE = T.E. - KE k
= E k - E k cos 2 θ = E k sin 2 θ

Topic: Physics, Motion In Two Dimensions, NEET
A body is projected at $t=0$ with a velocity $10 \mathrm{~ms}^{-1}$ at an angle of $60^{\circ}$ with the horizontal. The radius of curvature of its trajectory at $t=1$ s is $R$. Neglecting air resistance and taking acceleration due to gravity $\mathrm{g}=10 \mathrm{~ms}^{-2}$, the value of $R$ is:

Options:
A $10.3 \mathrm{~m}$
B $2.8 \mathrm{~m}$
C $2.5 \mathrm{~m}$
D $5.1 \mathrm{~m}$

The correct answer is:
$2.8 \mathrm{~m}$

Solution:
Horizontal component of velocity $\mathrm{v}_{x}=10 \cos 60^{\circ}=5 \mathrm{~m} / \mathrm{s}$

vertical component of velocity $v_{y}=10 \cos 30^{\circ}=5 \sqrt{3} \mathrm{~m} / \mathrm{s}$

After $\mathrm{t}=1 \mathrm{sec}$

Horizontal component of velocity $\mathrm{v}_{x}=5 \mathrm{~m} / \mathrm{s}$ Vertical component of velocity $v_{y}=|(5 \sqrt{3}-10)| \mathrm{m} / \mathrm{s}=10-5 \sqrt{3}$

Centripetal, acceleration $\mathrm{a}_{\mathrm{n}}=\frac{\mathrm{v}^{2}}{\mathrm{R}}$

$\Rightarrow \mathrm{R}=\frac{\mathrm{v}_{\mathrm{x}}^{2}+\mathrm{v}_{\mathrm{y}}^{2}}{\mathrm{a}_{\mathrm{n}}}=\frac{25+100+75-100 \sqrt{3}}{10 \cos \theta}$

From figure (using (i)) $\tan \theta=\frac{10-5 \sqrt{3}}{5}=2-\sqrt{3} \Rightarrow \theta=15^{\circ}$

$\mathrm{R}=\frac{100(2-\sqrt{3})}{10 \cos 15}=2.8 \mathrm{~m}$

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2019 (11 Jan Shift 1)
A body is projected from the Earth's surface with a velocity equal to half the escape velocity from that point at an angle \( 45^{\circ} \) with horizontal. The maximum height attained by the body above the Earth's surface is (nearly)

Options:
A \( 0.125 \mathrm{R} \)
B \( 0.19 \mathrm{R} \)
C \( \frac{\sqrt{3}}{\frac{\pi}{3}} \)
D \( 0.9 \mathrm{R} \)

The correct answer is:
\( 0.19 \mathrm{R} \)

Solution:
Let h be the maximum height attained and v be the velocity at maximum height. From conservation of angular momentum

mVe2cos 45×R=mvr (where r=R+h

Ve2×12R=vr⇒v=Ve22Rr=1222GMR×Rr⇒v=GMR4r2

From conservation of energy

-GMmR+12mVo24=-GMmr+12mv2⇒-GMR+12×14×2GMR=-GMr+12×GMR4r2

-34R=-8r+R8r2⇒-6r2=-8rR+R2⇒6r2-8rR+R2=0

r=8R±64R2-4×6×R212⇒r=8R±210R12⇒R+h=4R+10R6⇒h=0.19R

Topic: Physics, Gravitation, JEE Main
A body is projected from the ground at an angle of 45° with the horizontal. Its velocity after 2 s is 20 m s-1. The maximum height reached by the body during its motion is _____ m. (use g=10 m s-2)

Options:

The correct answer is:
20

Solution:
After 2 s the velocity is 20 m s-1

⇒vx2+vy2=v2=400

⇒u2cos245°+usin45°-gt2=u22+u2-202=400

⇒u22+u22+400-402u=400

⇒u2=402u⇒u=402 m s-1

Now the maximum height of the projectile is given by,

H=u2sin2θ2g⇒H=160020×2×2=20 m

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2022 (24 Jun Shift 2)
A body is projected from the ground with some speed at some angle with the horizontal. Taking the horizontal and vertical direction to be x and y axes respectively and the point of projection as origin, calculate the minimum speed (in m s-1) of projection so that it can pass through a point whose x and y coordinates are 30 m and 40 m respectively? Take g=10 m s-2

Options:

The correct answer is:
30

Solution:
In projectile motion, the equation of the trajectory is
y=xtan⁡θ-gx22u2cos2⁡θ
=xtan⁡θ-gx22u21+tan2⁡θ
Substituting the given values, we have
40=30tan⁡θ-10×3022u21+tan2⁡θ
or 900tan2⁡θ-6u2tan⁡θ+900+8u2=0
The value of tan⁡θ will be real if
6u22≥4×900×900+8u2
or u4≥100900+8u2
or u4-800u2≥90000
or u2-4002-160000≥90000
or u2-4002≥250000
u2-400≥500⇒u2≥900
⇒ u≥30 m s-1

Topic: Physics, Motion In Two Dimensions, JEE Main
A body is projected horizontal on the top of the Mount Everest with speed much greater than \( \sqrt{2 g r} \), where \( r \) is the distance of the projection point from the centre of the earth and \( g \) is the acceleration due to gravity there. The trajectory of the body will be a

Options:
A straight line
B parabola
C hyperbola
D ellipse

The correct answer is:
hyperbola

Solution:
Escape velocity if given by ve=2gr

In astrodynamics or celestial mechanics, a hyperbolic trajectory is the trajectory of any object around a central body with more than enough speed to escape the central object's gravitational pull. The name derives from the fact that according to Newtonian theory such an orbit has the shape of a hyperbola

According to question projection speed is much greater than 2gr. So, body escape from earth along hyperbola.

Topic: Physics, Gravitation, JEE Main
A body is projected up along a rough inclined plane of inclination $45^{\circ}$. The coefficient of friction is $0.5$. Then the retardation of the block is

Options:
A $\frac{g}{2 \sqrt{2}}$
B $\frac{g}{2}$
C $\frac{3 g}{2 \sqrt{2}}$
D $\frac{g}{\sqrt{2}}$

The correct answer is:
$\frac{3 g}{2 \sqrt{2}}$

Solution:
Retardation
\begin{aligned}
&=g(\sin \theta+\mu \cos \theta) \\
&=g\left(\frac{1}{\sqrt{2}}+0.5 \frac{1}{\sqrt{2}}\right) \\
&=\frac{g}{\sqrt{2}}(1+0.5) \\
&=\frac{1.5 g}{\sqrt{2}}=\frac{3 g}{2 \sqrt{2}}
\end{aligned}

Topic: Physics, Laws of Motion, JEE Main
A body is projected up with a velocity equal to 3/4th of the escape velocity from the surface of the earth. The height it reaches is, (Radius of earth =R)

Options:
A 10R9
B 97R
C 98R
D 10R3

The correct answer is:
97R

Solution:
Escape velocity for the earth is
ve=2GMR
Given the velocity projection of the body
=v=34ve=34 2GMR
Total energy on the earth
=Total energy at maximum height h
12mv2+-GMmR=0+-GMmR+h
12m . 916. 2GMR-GMmR=-GMmR+h
916-1=- RR+h or -RR+h=-716
7R+7h=16 R
7h=9Râh=97R

Topic: Physics, Gravitation, JEE Main
A body is projected upwards with a velocity of 4Ã11.2 km s-1 from the surface of the earth. The velocity of the body when it escapes from the gravitational pull of earth will be

Options:
A 11.2 km s-1
B 2Ã11.2 km s-1
C 3Ã11.2 km s-1
D 15Ã11.2 km s-1

The correct answer is:
15Ã11.2 km s-1

Solution:
The minimum velocity of projection to achieve escape velocity can be calculated as,
Initial KE=12mv2
=12Ãm4Ã11.22=16Ã12mve2
As 12mve2 energy is used up in coming out from the gravitational pull of the earth, so
Final KE should be 15Ã12mve2
Hence, 12mv'2=15Ã12mve2
â´ vâ²2=15ve2
or vâ²=15ve
=15Ã11.2 km s-1

Topic: Physics, Gravitation, JEE Main
A body is projected vertically upwards at time $t=0$ and it is seen at a height $H$ at time $t_1$ and $t_2$ second during its flight. The maximum height attained is ( $g$ is acceleration due to gravity)

Options:
A $\frac{g\left(t_2-t_1\right)^2}{8}$
B $\frac{g\left(t_1+t_2\right)^2}{4}$
C $\frac{g\left(t_1+t_2\right)^2}{8}$
D $\frac{g\left(t_2-t_1\right)^2}{4}$

The correct answer is:
$\frac{g\left(t_2-t_1\right)^2}{8}$

Solution:
Let time taken by the body to fal from point $C$ to $B$ is $t^{\prime}$.
Then $t_1+2 t^{\prime}=t_2$
$t^{\prime}=\left(\frac{t_2-t_1}{2}\right) \ldots(\mathrm{i})$
Total time taken to reach point $C$
$\begin{aligned}
T & =t_1+t^{\prime} \\
& =t_1+\frac{t_2-t_1}{2} \\
& =\frac{2 t_1+t_2-t_1}{2} \\
& =\left(\frac{t_1+t_2}{2}\right)
\end{aligned}$
Maximum height attained
$\begin{aligned}
H_{\max } & =\frac{1}{2} g(T)^2 \\
& =\frac{1}{2} g\left(\frac{t_1+t_2}{2}\right)^2 \\
& =\frac{1}{2} g \cdot \frac{\left(t_1+t_2\right)^2}{4} \\
\Rightarrow \quad H_{\max } & =\frac{1}{8} g \cdot\left(t_1+t_2\right)^2 \mathrm{~m}
\end{aligned}$

Topic: Physics, Motion In Two Dimensions, JEE Main
A body is projected vertically upwards from the surface of earth with a velocity equal to one third of escape velocity. The maximum height attained by the body will be
(Take radius of earth =6400 km and g=10 ms-2)

Options:
A 800 km
B 1600 km
C 2133 km
D 4800 km

The correct answer is:
800 km

Solution:
Total mechanical energy of the body at the surface of earth

Mi=-GMmR+12m132GMR2=-GMmR+GMm9R=-8GMm9R

Total mechanical energy of the body at the highest point

Mf=-GMmR+h

Using conservation of mechanical energy,

Mi=Mf⇒-GMmR+h=-8GMm9R

⇒R+h=9R8

⇒h=R8=64008=800 km

Topic: Physics, Gravitation, JEE Main, JEE Main 2022 (26 Jul Shift 2)
A body is projected with a speed $u$ at an angle $\theta$ with the horizontal. The radius of curvature of the trajectory, when it makes an angle $\left(\frac{\theta}{2}\right)$ with the horizontal is ( $g$-acceleration due to gravity)

Options:
A $\frac{u^2 \cos ^2 \theta \sec ^3\left(\frac{\theta}{2}\right)}{\sqrt{3} g}$
B $\frac{u^2 \cos ^2 \theta \sec ^3\left(\frac{\theta}{2}\right)}{2 g}$
C $\frac{2 u^2 \cos ^3 \theta \sec ^2\left(\frac{\theta}{2}\right)}{g}$
D $\frac{u^2 \cos ^2 \theta \sec ^3\left(\frac{\theta}{2}\right)}{g}$

The correct answer is:
$\frac{u^2 \cos ^2 \theta \sec ^3\left(\frac{\theta}{2}\right)}{g}$

Solution:
Let velocity of projectile is $v$ at an angle $\frac{\theta}{2}$ with horizontal
$$
\begin{aligned}
\therefore & v \cos \frac{\theta}{2}=u \cos \theta \\
\text { or } \quad v & =\frac{u \cos \theta}{\cos \frac{\theta}{2}}
\end{aligned}
$$
As horizontal component remains same. Also, centripetal force is provided by the component of weight.
So,
$$
\frac{m v^2}{r}=m g \cos \frac{\theta}{2}
$$
Hence, radius of curvature of path,
$$
\begin{aligned}
\Rightarrow \quad r & =\frac{v^2}{g \cos \frac{\theta}{2}} \\
& =\frac{\left(\cos \frac{\theta}{2}\right)^2}{g \cos \frac{\theta}{2}} \Rightarrow r=\frac{u^2 \cos ^2 \theta \cdot \sec ^3\left(\frac{\theta}{2}\right)}{g}
\end{aligned}
$$

Topic: Physics, Motion In Two Dimensions, AP EAMCET, AP EAMCET 2018 (22 Apr Shift 1)
A body is released from a height equal to the radius R of the earth. The velocity of the body when it strikes the surface of the earth will be: (Given g= acceleration due to gravity on the earth.)

Options:
A 2gR
B gR
C 4gR
D gR2

The correct answer is:
gR

Solution:
Using conservation of mechanical energy,

Ui+Ki=Uf+Kf

⇒-GMm2R+0=-GMmR+12mv2

⇒12mv2=GMm2R

⇒v=GMR=gR

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (15 Apr Shift 1)