datavorous/entrance-exam-dataset huggingface dataset:
A particle is projected from ground at some angle with horizontal (Assuming point of projection to be origin and the horizontal and vertical directions to be the x and y-axis ) the particle passes through the points (3 m, 4 m) and (4 m, 3 m) during its motion then the range of the particle would be: g=10ââ m/s 2

Options:
A 36 m
B 377 m
C 4 m
D 16 m

The correct answer is:
377 m

Solution:
Using equation of trajectory:
y = x tan α 1 - x R ; For points (3, 4)
⇒ 4=3tanα1-3R ...(i)
For point (4, 3)
⇒ 3=4tanα1-4R ...(ii)
dividing (i) and (ii)
⇒ 43=341-3R1-4R⇒161-4R=91-3R
⇒ 7=64R-27R⇒R=377 m

Topic: Physics, Motion In Two Dimensions, NEET
A particle is projected from the ground with an initial speed of vâ¡ at an angle Î¸ with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is

Options:
A vâ¡ 2 1 + 2 cos 2 θ
B vâ¡ 2 1 + cos 2 θ
C vâ¡ 2 1 + 3 cos 2 θ
D vâ¡ cos θ

The correct answer is:
vâ¡ 2 1 + 3 cos 2 θ

Solution:
Average velocity = displacement time

vâ¡ av = H 2 + R 2 4 T / 2 ...(i)
Here, H = maximum height = vâ¡ 2 sin 2 θ 2 g
R = range = vâ¡ 2 sin 2 θ g
and T = time of flight = 2 vâ¡ sin θ g
Substituting in Eq. (1) we get
vâ¡ av = vâ¡ 2 1 + 3 cos 2 θ

Topic: Physics, Motion In Two Dimensions, NEET
A particle is projected from the mid-point of the line joining two fixed particles each of mass m. If the separation between the fixed particles is l , the minimum velocity of projection of the particle so as to escape is equal to

Options:
A Gml
B Gm2l
C 2Gml
D 22Gml

The correct answer is:
22Gml

Solution:
The gravitational potential at the mid-point P=V=V1+V2
=-Gml2-Gml2=-4Gml
The gravitational potential energy U=-4Gmm0l, m0 = mass of particle
When it is projected with a speed v, it just escapes to infinity, and the potential and kinetic energy will become zero.
â âK.E.+âP.E.=0
â 0-12m0v2+0--4Gmm0l=0
âv=22Gml

Topic: Physics, Gravitation, NEET
A particle is projected from the surface of the earth with a velocity of 5 ms-1 at an angle θ with the horizontal. Another particle is projected on the surface of some other planet with a velocity of 3 ms-1 at the same angle and it follows a trajectory which is identical to the trajectory of the projectile which is fired on earth. The value of the acceleration due to gravity on that planet is [given that the acceleration due to gravity on the surface of earth, g=9.8 ms-2]

Options:
A 16.3 ms-2
B 1.8 ms-2
C 3.5 ms-2
D 5.9 ms-2

The correct answer is:
3.5 ms-2

Solution:
y=x tan Î¸-gx22u2cos2Î¸
For equal trajectories for same angle of projection gu2= constant
â9.852=g32
g'=9.8Ã925=3.528 m/s2=3.5 m/s2

Topic: Physics, Motion In Two Dimensions, NEET
A particle is projected in air at an angle β to a surface which itself is inclined at an angle α to the horizontal Find an expression of range on the plane surface

Options:
A $L=\frac{2{{v}_{o}}^{2}sin\beta cos\left( \alpha +\beta \right)}{gco{{s}^{2}}\beta }$
B $L=\frac{2{{v}_{o}}^{2}sin\beta cos2\beta }{gco{{s}^{2}}\alpha }$
C $L=\frac{2{{v}_{o}}^{2}sin\beta cos\left( \alpha +\beta \right)}{gco{{s}^{2}}\alpha }$
D $L=\frac{2{{v}_{o}}^{2}sin\beta cos2\alpha }{gco{{s}^{2}}\beta }$

The correct answer is:
$L=\frac{2{{v}_{o}}^{2}sin\beta cos\left( \alpha +\beta \right)}{gco{{s}^{2}}\alpha }$

Solution:
Correct Option is : (C)
$L=\frac{2{{v}_{o}}^{2}sin\beta cos\left( \alpha +\beta \right)}{gco{{s}^{2}}\alpha }$

Topic: Chemistry, Motion In Two Dimensions, NEET
A particle is projected in the X-Y plane. 2 s after projection, the velocity of the particle makes an angle 45Â° with the X-axis. 4 s after projection, it moves horizontally. Find the velocity of projection (use g = 10 m s-2)

Options:
A 205 m s-1
B 105 m s-1
C 55 m s-1
D 25 m s-1

The correct answer is:
205 m s-1

Solution:
vy=uy+ayt
at the highest point of motion, vy=0
0=uy-gt
uy=4g=40 m s-1
at t=2 s
vy=uy+ayt
vy=4g-gt=20 m s-1

given that vy=ux=20 m s-1
the velocity of projection u=ux2+uy2

⇒u=402+202
u=2000 ⇒205m s-1

Topic: Physics, Motion In One Dimension, NEET
A particle is projected vertically upwards with a speed of 16 m s-1. After some time, when it again passes through the point of projection, its speed is found to be 8 m s-1. It is found that the work done by air resistance is the same during the upward and downward motions. Then the maximum height attained by the particle is

Options:
A 8 m
B 4.8 m
C 17.6 m
D 12.8 m

The correct answer is:
8 m

Solution:
From work energy theorem for upward motion,
12m 162=mgh+W
For downward motion,
12m 82=mgh â W
h=8 m

Topic: Physics, Work Power Energy, NEET
A particle is projected vertically upwards with a speed of 16 m s-1, after some time, when it again passes through the point of projection, its speed is found to be 8 m s-1. It is found that the work done by air resistance is the same during upward and downward motion. Then the maximum height attained by the particle is â

Options:
A 8 m
B 4.8 m
C 17.6 m
D 12.8 m

The correct answer is:
8 m

Solution:
From work energy theorem
for upward motion,
12m 162=mgh+W
for downward motion
12m 82=mgh â W
h=8 m

Topic: Physics, Work Power Energy, NEET
A particle is projected vertically upwards with a velocity u and from a point O. When it returns to the point of projection, which of the following is incorrect ?

Options:
A average velocity is zero
B displacement is zero
C average speed is u
D average speed is u2

The correct answer is:
average speed is u

Solution:
Total displacement is 0 as the particle returns to the original position, thus the average velocity is zero.
Total distance travelled is 2s and total time taken is 2t
0 2 = u 2 - 2 gs
â´ s = u 2 2 g
Also 0 = u - g t
â t = u / g
â´ 2 t = 2 u g

speed = u 2 g / 2 u g = u 2 â.

Hence only option 3 has incorrect option and hence the option to be selected is 3.

Topic: Physics, Motion In Two Dimensions, NEET
A particle is projected vertically with speed V from the surface of the earth. Maximum height attained by the particle, in terms of the radius of earth R,V and g is (V< escape velocity, g is the acceleration due to gravity on the surface of the earth)

Options:
A 3RV22gR-2V2
B 2RV23gR-V2
C RV22gR-V2
D RV2gR-V2

The correct answer is:
RV22gR-V2

Solution:
12mV2-GMmR=-GMmR+h
12V2-GMR2⋅R=-GmR2R2(R+h)
12V2-gR=-gR2(R+h)
R+h=-2gR2V2-2gR
h=-2gR2V2-2gR-R
h=-RV2V2-2gR
h=RV22gR-V2

Topic: Physics, Gravitation, NEET
A particle is projected with a speed 102 m s-1 and at an angle 45Â° with the horizontal. The rate of change of speed with respect to time at t=1 s is g=10 m s-2

Options:
A 102 m s-2
B 10 m s-2
C zero
D 5 m s-2

The correct answer is:
zero

Solution:
At t=1 s, the particle is at maximum height.
At maximum height velocity is horizontal and acceleration is vertical. Hence, the angle between the velocity vector and acceleration vector is 90Â°. Therefore at this position tangential acceleration, i.e., the rate of change of speed with respect to time is zero.

Topic: Physics, Motion In Two Dimensions, NEET
A particle is projected with a speed v at 45°with the horizontal. The magnitude of angular momentum of the projectille about the point of projection when the particle is at its maximum height h is

Options:
A Zero
B mvh22
C mv2h2
D mvh2

The correct answer is:
mvh2

Solution:
When a particle is projected with a speed v at 45° with the horizontal then velocity of the projectile at maximum height.
v′=vcosâ¡45°=v2
Angular momentum of the projectile about the point of projection
=mv′h
=mv2h=mvh2

Topic: Physics, Rotational Motion, NEET
A particle is projected with a velocity u making an angle Î¸ with the horizontal. The instantaneous power of the gravitational force

Options:
A Varies linearly with time.
B Is constant throughout
C Is negative for complete path
D None of these

The correct answer is:
Varies linearly with time.

Solution:
At any time, v→=u cosθi^+u sinθ-gtj^
P=F→.v→=–mgj^.u cosθi^+u sinθ-gtj^
=mg2t-mgu sinθ
Hence, power varies linearly with time.

Topic: Physics, Work Power Energy, NEET
A particle is projected with a velocity v, so that its range on a horizontal plane is twice the greatest height attained. If g is the acceleration due to gravity, then its range is -

Options:
A 4v25g
B 4g5v2
C 4v25g2
D 4v5g2

The correct answer is:
4v25g

Solution:
∵R=2H
∴2v2sinθcosθg=2v2sin2 θ2g⇒tanθ=2
⇒R=2v2sinθ(cosθ)g=2v225 15g=4v25g

Topic: Physics, Motion In Two Dimensions, NEET
A particle is projected with speed 20 m s-1 at an angle 30Â° with horizontal. After how much time the angle between velocity and acceleration will be 90Â°

Options:
A 1 s
B 2 s
C 1.5 s
D never

The correct answer is:
1 s

Solution:
Acceleration g is vertically downward
To make velocity perpendicular to acceleration, velocity must be horizontal. This is possible at maximum height.
For this
vy=0
uy=usinâ¡Î¸
uy=20sin30Â°=10 m s-2
g=-10 ms-2
We know that
vy=uy+gt
By putting all value, we get t=1 s

Topic: Physics, Motion In Two Dimensions, NEET
A particle is projected with velocity 20 m s-1 at angle 60Â° with horizontal. The radius of curvature of trajectory, at the instant when velocity of projectile becomes perpendicular to velocity of projection, is
(g = 10 m s-1)

Options:
A 603Â Â m
B 803Â Â m
C 403Â Â m
D 8033 m

The correct answer is:
8033 m

Solution:
Vcos30Â°=ucos60Â°â V=203 m s-1
ac=gcos30Â°=10Ã32
â´ r=V2ac=8033 m

Topic: Physics, Motion In Two Dimensions, NEET
A particle is projected with velocity kve in the vertically upward direction from the ground into space (ve is escape velocity and k<1). The maximum height from the centre of the earth to which it can reach is

Options:
A Rk2+1
B k2R1-k2
C R1-k2
D k2Rk2+1

The correct answer is:
R1-k2

Solution:
Total energy of surface = Total energy at h height
⇒12mkve2+-GMmR=0+-GMmR+h
⇒k212mve2+-GMmR=-GMmR+h
⇒k212m2GMR-GMmR=-GMmR+h
⇒1R-k2R=1R+h
⇒1-k2R=1R+h
⇒R+h=R1-k2

Topic: Physics, Gravitation, NEET
A particle is projected with velocity kve in vertically upward direction from the ground into the space ( ve is escape velocity & k<1 ) then the maximum height from the centre of earth to which it can go, will be

Options:
A Rk2+1
B Rk2–1
C R1–k2
D Rk+1

The correct answer is:
R1–k2

Solution:
Total energy of surface = Total energy at h height
⇒ 12mkve2+–GMmR=U+–GMmR+h
⇒ h=Rk21–k2
From the centre
r=R+h=R1–k2

Topic: Physics, Gravitation, NEET
A particle is projected with velocity v0 along x-axis. The deceleration on the particle is proportional to the square of the distance from the origin i.e. a=-x2. The distance at which particle stops is -

Options:
A 3v02
B v02313
C 2v02313
D 3v02213

The correct answer is:
3v02213

Solution:
Given initial velocity = v0
final velocity = 0
a = - x 2
now we know that a = dv dt and v = dx dt
thus a = dv dt = dv dt Â· dx dx = vdv dx
vdv dx = - x 2
vdv = - x 2 dx
integrating we get
â« v 0 0 vdv = â« 0 x - x 2 dx
v 2 2 v 0 0 = - x 3 3 0 x
- v 0 2 2 = - x 3 3 â x = 3 v 0 2 2 1 3

Topic: Physics, Mathematics in Physics, NEET
A particle is pulled along a curved surface very slowly. The coefficient of kinetic friction between the particle and surface is 0.4. The heat generated during the pulling of the particle from the lowest point A to the topmost point B equals (The mass of the particle is 5 g)

Options:
A 100 J
B 0.1 J
C 50 J
D 23 J

The correct answer is:
0.1 J

Solution:
Heat generated = work done by friction =Î¼mgL
=0.4 Ã5 Ã10-3Ã10Ã5
=10-1 joule
=0.1 J

Topic: Physics, Laws of Motion, NEET
A particle is released from height S from the surface of the Earth. At a certain height its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively___

Options:
A S4,3gS2
B S4,3gS2
C S4,3gS2
D S2,3gS2

The correct answer is:
S4,3gS2

Solution:
∵ At Every pt.

KE+PE= constant

so,

(KE)A+(PE)A=(KE)B+(PE)B

0+mgS=3(PE)B+(PE)B

mgS=4mgh

h=S4

Now,

KE at p t . B-

∵ At p t . B-

KE=3PE

12mv2=3mgs4

v=32gS

Topic: Physics, Work Power Energy, NEET, NEET 2021
A particle is released from rest from a tower of height 3h. The ratio of time taken to fall equal heights h i.e., t 1: t 2: t 3 is

Options:
A 3 : 2 : 1
B 3 : 2 : 1
C 9 : 4 : 1
D 1 : 2 - 1 : 3 - 2

The correct answer is:
1 : 2 - 1 : 3 - 2

Solution:
h⁡ = 1 2 g t 1 2
2 h⁡ = 1 2 g t 1 + t 2 2
and 3 h⁡ = 1 2 g t 1 + t 2 + t 3 2
i.e., t 1 : t 1 + t 2 : t 1 + t 2 + t 3 = 1 : 2 : 3
or t 1 : t 2 : t 3 = 1 : 2 - 1 : 3 - 2

Topic: Physics, Motion In One Dimension, NEET
A particle is subjected to two simple harmonic motions along x and and y-directions according to the equations x = 4 sin 1 0 0 Ï t and y = 3 sin 1 0 0 Ï t . Choose the correct statements

Options:
A Motion of particle will be on a ellipse
B Motion of the particle will be on a straight line
C Particle will execute SHM of amplitude 5
D Particle will not execute SHM

The correct answer is:
Motion of the particle will be on a straight line

Solution:
dxdt=400π cos t100πt
dydt=300π cos100πt
both are the same path
so Motion of the particle will be on a straight line

Topic: Physics, Oscillations, NEET
A particle is subjected to two simple harmonic motions along x and y directions according to and x=3 sin 100 Ït; y=4 sin 100 Ït

Options:
A Motion of particle will be an ellipse traversing it in clockwise direction.
B Motion of particle will be on a straight line with slope 4/3.
C Motion will be a simple harmonic motion along x-axis with amplitude 5.
D Phase difference between two motions is π / 2 .

The correct answer is:
Motion of particle will be on a straight line with slope 4/3.

Solution:
x=3sin( 100 Ït ) ... (i)
y=4sin( 100âÏt ) ...(ii)
dividing (ii) by (i), we get
y= 4 3 x
This represents a straight line with slope 4 3

Topic: Physics, Oscillations, NEET
A particle is thrown over a triangle from one end of a horizontal base and after grazing the vertex falls on the other end of the base. If 30Â° and 60Â° be the base angles and Î¸ the angle of projection then tanÎ¸ is

Options:
A 23
B 43
C 13
D 3

The correct answer is:
43

Solution:
tan⁡θ=tan⁡α+tan⁡β
tan⁡θ=tan⁡30o+tan⁡60o
tan⁡θ=13+3
tan⁡θ=43

Topic: Physics, Motion In Two Dimensions, NEET
A particle is vibrating in simple harmonic motion with an amplitude of 4 cm. The displacement of the particle when kinetic energy becomes half of total energy is

Options:
A 10Â cm
B 2Â cm
C 2Â cm
D 22 cm

The correct answer is:
22 cm

Solution:
The total energy E of a particle vibrating in SHM is given by
E=12mÏ2a2 .....(1)
The kinetic energy K is given by K=12mÏ2a2-y2 where y= displacement of the particle but
K=E2=1212mÏ2a2
â´ 1212mÏ2a2=12mÏ2a2-y2
or a22=a2-y2 or y2=a22
â´ y=a2
Hence the kinetic energy is half of the total energy when displacement of the particle is a2.
Given that a=4 cm
â´ y=42=22 cm

Topic: Physics, Oscillations, NEET
A particle located at x = 0 at time tâ¡ = 0 starts moving along the positive x-direction with the velocity v that varies as Î½ = Î± x . The displacement of the particle varies with time as

Options:
A tâ¡ 3
B t

C tâ¡ 1 / 2
D tâ¡ 2

The correct answer is:
t⁡ 2

Solution:
ν = α x
dâ¡ x dt = α x
dâ¡ x x α dt
∫ dâ¡ x x = α ∫ dt
2 x 1 / 2 = α t
x = ( α 2 ) 2 tâ¡ 2

Topic: Physics, Motion In One Dimension, NEET
A particle moves a distance $\mathrm{x}$ in time $\mathrm{t}$ according to equation $x=(t+5)^{-1}$. The acceleration of particle is proportional to

Options:
A (velocity) ${ }^{3 / 2}$
B $(\text { distance })^2$
C (distance $)^{-2}$
D (velocity) ${ }^{2 / 3}$

The correct answer is:
(velocity) ${ }^{3 / 2}$

Solution:
Given, distance $x=(t+5)^{-1}$
Differentiating Eq. (i) w.r.t. $t$, we get
$$
\frac{\mathrm{dx}}{\mathrm{dt}}=(\mathrm{v})=\frac{-1}{(\mathrm{t}+5)^2}
$$
Again, differentiating Eq. (i) w.r.t. t, we get
$$
\frac{\mathrm{d}^2 \mathrm{x}}{\mathrm{dt}^2}=(\mathrm{a})=\frac{2}{(\mathrm{t}+5)^3}
$$
Comparing Eqs. (ii) and (iii), we get
(a) $\propto(v)^{3 / 2}$

Topic: Physics, Motion In One Dimension, NEET, NEET 2010 (Screening)
A particle moves according to the law, =rcosÏt2. The distance covered by it during the time interval between t= 0 to t=3s is

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

The correct answer is:
3r

Solution:
When t=0, x=rcos⁡π×02=r;
When t=3s, x=rcos⁡π×32=0
Here ω=π2 or 2πT=π2 or T=4s
∴ In 3 sec, the particle goes from one extreme to other extreme and then back to mean position. So the distance travelled =2r+r=3r

Topic: Physics, Oscillations, NEET
A particle moves along a circle $\left(\frac{20}{\pi}\right) \mathrm{m}$ with constant tangential acceleration. If the velocity of the particle is $80 \mathrm{~m} / \mathrm{s}$ at the end of the second revolution after after motion has begun, the tangential acceleration is:

Options:
A $40 \mathrm{~m} / \mathrm{s}^2$
B $640 \pi \mathrm{m} / \mathrm{s}^2$
C $160 \pi \mathrm{m} / \mathrm{s}^2$
D $40 \pi \mathrm{m} / \mathrm{s}^2$

The correct answer is:
$40 \mathrm{~m} / \mathrm{s}^2$

Solution:
From question
$$
\begin{aligned}
& r=\frac{20}{\pi} m, v=80 \mathrm{~m} / \mathrm{s} \\
& \therefore \theta=2 \text { revolutions }=2 \times 2 \pi=4 \pi \mathrm{rad} \\
& \therefore \omega^2=\omega_0{ }^2+2 \alpha \theta \\
& \therefore \omega^2=2 \alpha \theta \ldots(\mathrm{i}) \text { since } \omega_0=0 \\
& \text { Since } \alpha=\frac{a}{r} \text { and } \omega=\frac{\mathrm{V}}{r}
\end{aligned}
$$
Putting these values in eq. (i)
$$
\begin{aligned}
& \frac{V^2}{r^2}= \frac{2 a \theta}{r} \\
& \Rightarrow a= \frac{V^2}{2 r \theta}=\frac{80 \times 80}{2 \times\left(\frac{20}{\pi}\right) \times 4 \pi} \\
&=40 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2003
A particle moves along a circle of radius r with constant tangential acceleration. If the velocity of the particle is v at the end of second revolution after the revolution has started then the tangential acceleration is

Options:
A v28πr
B v26πr
C v24πr
D v210πr

The correct answer is:
v28πr

Solution:
Using v2-u2=2as and u=u1 and s=4πr
∴ 2as=v2⇒v22s=a=v22×4πr=v28πr

Topic: Physics, Laws of Motion, NEET
A particle moves along a straight line $\mathrm{OX}$. At a time $t$ (in seconds) the distance $x$ (in metres) of the particle from $\mathrm{O}$ is given by $x=40+12 t-t^3$. How long would the particle travel before coming to rest ?

Options:
A $16 \mathrm{~m}$
B $24 \mathrm{~m}$
C $40 \mathrm{~m}$
D $56 \mathrm{~m}$

The correct answer is:
$16 \mathrm{~m}$

Solution:
The given equation is
$x=40+127-t^3$
Differentiating we get
$u=\frac{d x}{d t}=12-3 t^2$
When particle comes to rest means
$\begin{array}{l}
& v = 0 \\
& 12-3 t^2 =0 \\
& t =2 \mathrm{sec} .
\end{array}$
The distance travelled before coming to rest
$\begin{aligned}
\int_0^5 d s & ={ }_0^2 v d t \\
5 & =\int_0^2\left(12-3 t^2\right) d t \\
& =\left[12 t-\frac{3 t^3}{3}\right]_0^2 \\
& =12 \times 2-(8-0) \\
& =16 \mathrm{~m} .
\end{aligned}$

Topic: Physics, Motion In One Dimension, NEET, NEET 2006
A particle moves along a straight line OX. At a time t (in seconds) the distance x (in metres) of the particle from O is given by x=40+12tât3. How long would the particle travel before coming to rest?

Options:
A 16 m

B 24 m

C 40 m

D 56 m

The correct answer is:
16 m

Solution:
x=40+12t–t3
∴ Velocity, v = d x d t = 1 2 - 3 t 2
When particle come to rest, d x d t = v = 0
∴ 12–3t2=0 ⇒ 3t2=12 ⇒ t=2 s.
Distance travelled by the particle before coming to rest,
xi=40 t=0
xf=40+24-8 for t=2 s
xf-xi=16 m

Topic: Physics, Motion In One Dimension, NEET
A particle moves along a straight line path. Its speed decreases continuously and after some time it comes to rest. The motion is with constant acceleration whose direction with respect to the direction of velocity is

Options:
A positive throughout motion
B negative throughout motion
C first positive then negative
D first negative then positive

The correct answer is:
negative throughout motion

Solution:
As the speed decreases continuously therefore the motion is retarded and acceleration and velocity must be in opposite direction.

Topic: Physics, Motion In One Dimension, NEET
A particle moves along the x-axis from x=0 to x=5 m under the influence of a force given by F= 7 â 2x + 3x2. The work done in the process is-

Options:
A 70 J
B 270 J
C 35 J
D 135 J

The correct answer is:
135 J

Solution:
W=â«F.dx=â«05(7-2x+3x2) dx
=7x-2x22+3x3305
=35-25+125=135 J

Topic: Physics, Work Power Energy, NEET
A particle moves along x-axis as x=4t-2+at-22. Which of the following is true?

Options:
A The initial velocity of the particle is 4 m s-1
B The acceleration of particle is 2a
C The particle is at origin at t=0
D None of the above

The correct answer is:
The acceleration of particle is 2a

Solution:
Given, x=4 t-2+at-22
v=dxdt=4+2at-2
At t=0, v=41-a
Acceleration a=d2xdt2=2a

Topic: Physics, Motion In One Dimension, NEET
A particle moves from a point - 2i^+5j^ to 4j^+3k^ when a force of 4i^+3j^ N is applied. How much work has been done by the force?

Options:
A 8  J
B 11  J
C 5ââJ
D 2  J

The correct answer is:
5  J

Solution:
S→=r→2-r→1=4j^+3k^-- 2i^+5j^
=2i^-j^+3k^
F→=4i^+3j^
W= F→. s→=8-3=5 J

Topic: Physics, Work Power Energy, NEET
A particle moves from a point - 2i^+5j^ to 4j^+3k^ when a force of 4i^+3j^ N is applied. How much work has been done by the force?

Options:
A 8  J
B 11  J
C 5ââJ
D 2  J

The correct answer is:
5  J

Solution:
S→=r→2-r→1=4j^+3k^-- 2i^+5j^
=2i^-j^+3k^
F→=4i^+3j^
W= F→. s→=8-3=5 J

Topic: Physics, Work Power Energy, NEET, NEET 2016 (Phase 2)
A particle moves from position r1=3i^+2j^-6k^ m to position r2=14i^+13j^+9k^ m under the action of a force 4i^+j^-3k^ N, then the work done is

Options:
A 46 J
B 56Â J
C 86Â J
D 10 J

The correct answer is:
10 J

Solution:
Given, r1=3i^+2j^-6k^ m
r2=14i^+13j^+9k^ m
F=4i^+j^-3k^ N
Displacement, s=r2-r1
=14i^+13j^+9k^-3i^+2j^-6k^
or, s=11i^+11j^+15k^
∴ Work done, W=F.s
=4i^+j^-3k^⋅11i^+11j^+15k^
=44+11-45
or W=10 J

Topic: Physics, Work Power Energy, NEET
A particle moves in a circle of radius 0.5 m at a speed that uniformly increases. Find the angular acceleration of particle if its speed changes from 2 ms to 4 ms in 4 seconds?

Options:
A 1 rad / s 2
B 2ââ rad / s 2
C 4   rad / s 2
D 0   rad / s 2

The correct answer is:
1 rad / s 2

Solution:
The tangential acceleration of the particle is
at=dvdt=4.0-2.04.0
=0.5ââ m/s 2
The angular acceleration is
Î±=atr=0.50.5= 1   rad/s 2

Topic: Physics, Laws of Motion, NEET
A particle moves in a circle of radius $5 \mathrm{~cm}$ with constant speed and time period $0.2 \pi \mathrm{s}$. The acceleration of the particle is

Options:
A $25 \mathrm{~m} / \mathrm{s}^2$
B $36 \mathrm{~m} / \mathrm{s}^2$
C $5 \mathrm{~m} / \mathrm{s}^2$
D $15 \mathrm{~m} / \mathrm{s}^2$

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

Solution:
Given, $r=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$ and $T=0.2 \pi \mathrm{s}$
We know that acceleration
$$
\begin{aligned}
a & =r \omega^2 \\
& =\frac{4 \pi^2}{T^2} r \\
& =\frac{4 \times \pi^2 \times 5 \times 10^{-2}}{(0.2 \pi)^2}=5 \mathrm{~ms}^{-2}
\end{aligned}
$$

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2011 (Screening)
A particle moves in a circle of radius 0.25 m at angular speed of two revolutions per second. The acceleration of the particle in m s-2 is

Options:
A Ï2
B 8Â Ï2
C 4 Ï2
D 2Â Ï2

The correct answer is:
4 Ï2

Solution:
r=0.25 m
Ï=2 rev./s
Ï=2Ã2 Ï rad/s
We know that ac=Ï2r
â ac=2Ã2Ï2Ã0.25
ac=4π2 m s-2

Topic: Physics, Laws of Motion, NEET
A particle moves in a circle of radius 30 cm. Its liner speed is given by v=2t, where t is in second and v in m s-1. Find out its radial and tangential acceleration at t=3s, respectively,

Options:
A 220 m s-2,50 m s-2
B 100 m s-2,5 m s-2
C 120 m s-2,2 m s-2
D 110 m s-2,10 m s-2

The correct answer is:
120 m s-2,2 m s-2

Solution:
Given that, the radius of the circle, r=30 cm=0.3 m
linear speed v=2t
Now, radial acceleration aR=v2r=2t20.3
at t=3 s
aR=2×320.3
360.3=120 m s-2
or aR=120 m s-2
and tangential acceleration aT=dvdt=ddt2t=2 m s-2

Topic: Physics, Laws of Motion, NEET