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Volume 24 No. 8 August 2016
CONTENTS
AIIMS Topper Interview 8 Class 11
NEET | JEE Essentials 12
Ace Your Way CBSE 24
MPP-2 34
Brain Map 46 Class 12
NEET | JEE Essentials 38
Brain Map 47
Ace Your Way CBSE 52
Core Concept 61
Exam Prep 67
MPP-2 76
Competition Edge
Physics Musing Problem Set 37 80
You Ask We Answer 82
Live Physics 83
Physics Musing Solution Set 36 84
Crossword 85
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Editor : Anil Ahlawat
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Many congratulations for your achievement in cracking the AIIMS MBBS entrance exam! Were you expecting to be among the top rankers?
Nikhil Bajiya: Many thanks for your wishes! I was confident about my performance and had expected to be among the top rankers but never expected to be the AIR 2. Also, to be frank, I never speculated about the rank. I had only focused on my performance and after the exam, I did not think about the results.
Why did you decide to pursue medicine?
Nikhil Bajiya: I was inspired to be a doctor from my paternal uncle since childhood. I was motivated by his practice as a pediatrician and the respect he earned. I made sure that I work hard to reach my goal.
Which other medicine entrance exam did you appear for?
Nikhil Bajiya: Apart from AIIMS MBBS 2016, I had appeared for the NEET 1 exam.
This year, there has been lot of confusion regarding AIPMT and NEET. Did the ongoing NEET row affect your preparation?
Nikhil Bajiya: No I was not much bothered about the AIPMT and NEET exam row. I had aimed for AIPMT/ NEET and AIIMS and I was very confident about cracking the exams and get admission to a good college since my preparation was moving in right direction. At the end of the day, all that matters is how you perform, results reflect automatically. I ensured that I do not take any undue stress during the preparation phase.
Tell us something about your family members and their contributions in your success. Nikhil Bajiya: My family consists of my parents and elder sister. Their contributions behind my success are enormous and I can’t express them in words. Since I had joined coaching in Kota, my father, who is the Deputy Registrar with the Cooperative Societies Department took transfer in the city just to be with me. My mother is a government school teacher and my elder sister is studying medicine in Bikaner. All of them have been immensely supportive throughout.
What was your preparation strategy for the exam?
Nikhil Bajiya: I had started full-Nikhil Bajiya, AIR 2 at AIIMS MBBS 2016 exam dreamt of being a doctor from his very childhood and with determination and perseverance, he achieved the top rank in the highly competitive medical entrance exam. With dedicated preparation for 2 years, he hardly let anything distract him from his aim. The aspiring neurosurgeon is aiming to take admission in AIIMS Delhi. Hailing from Jhunjhunu, Rajasthan, Nikhil went to G.B Modi Vidya Mandir and moved to Kota, the hub of coaching institutes for competitive entrance exams. He believes that self-study is equally important along with the guidance of coaching mentors and followed a systematic preparation plan. Nikhil would be the second doctor in his family after his sister who is presently pursuing medicine. He loves to unwind by playing badminton, table tennis or computer games.
AIIMS 2016
Topper Interview
fledged preparation from Class XI. I had joined a Kota based coaching institute for medical entrance exam preparation. I had strictly followed the guidelines given by my faculty members. Initially I concentrated on gaining stronghold
over the concepts followed by taking regular practice tests and solving previous years question papers and taking lessons of improvement from the mistakes. Apart from coaching, I used to put around 6 to 7 hours of self-study.
What were your strong and weak areas among Physics, Chemistry and Biology?
Nikhil Bajiya: Physics have always been my strong point and it is my favourite subject. However, that does not mean I am weak in any of the two other subjects. I am equally good in Chemistry and Biology too and there is no such weak subject or areas for me.
Which is your dream medical college where you aspire to take admission? Is there any specialization you
would like to follow?
Nikhil Bajiya: I plan to take admission in AIIMS Delhi. I want to be a neurosurgeon.
You had followed a strict study schedule. How did you unwind yourself? Could you find time to pursue your hobbies? Nikhil Bajiya: Since I stayed in Kota, away from family, I did not find my scope to pursue my hobby, which is a sport. I love playing badminton, table tennis and computer games. Occasionally I played these games whenever I found some time. Otherwise I used to take short naps within the day to catch up on my sleep.
What is your message for the medicine aspirants who would be appearing for the exam next year?
Nikhil Bajiya: I would suggest them to never over burden themselves. Follow a systematic preparation schedule with ample practice. Courtesy : careers360.com
“
Systematic preparationstrategy and focus are key factors for my success, says Nikhil Bajiya, AIR 2
”
Motion
)
Motion is a combined property of the object under
•
study and the observer. There is no meaning of rest or motion without the viewer.
If a body changes its position with time, it is said to
•
be moving else it is at rest. Motion is always relative to the observer.
To locate the position of a particle we need a
•
reference frame. A commonly used reference frame is cartesian coordinate system or x-y-z coordinate system.
The coordinates (
¾ x, y, z) of the particle specify
the position of the particle with respect to origin of that frame.
If all the three coordinates of the particle remain
¾
unchanged as time passes it means the particle is at rest with respect to this frame.
The reference frame is chosen according to
¾
problems.
If frame is not mention, then ground is taken as
¾
reference frame.
If only one coordinate changes with time, motion
•
is one dimensional motion (1- D) or rectilinear motion.
If only two coordinates change with time, motion is two dimensional (2 - D) or motion in a plane.
If all three coordinates change with time, motion is three dimensional (3 - D) or motion in space.
Distance anD DisPlaceMent
)
Total length of path
•
covered by the particle, in definite time interval is called distance. The
length of path ACB is called the distance travelled by the body.
Overall, body is displaced from
• A to B. A vector
for A to B i.e., AB is its displacement vector or displacement that is the minimum distance. Displacement is directed from initial position to final position.
For a moving body, distance can not have zero or
•
negative values but displacement may be +ve, –ve or zero.
If motion is in straight line without change in
•
direction, then
distance = |displacement|
If motion is not in straight line, then distance > |displacement|
Magnitude of displacement may be equal or less
•
than distance but never greater than distance. distance ≥ |displacement| A B C
Unit
2
NEET JEE
ESSENTIALS
Kinematics
Class
XI
Displacement in terms of position vector : • Let a body is displaced from A(x1, y1, z1) to B(x2, y2, z2) then its displacement is given by vector AB . From DOAB rA+ AB r = or AB r rB = −B A Q rB =x i y j z k2+ 2+ 2 and rA =x i y j z k1+ 1+ 1 \ AB x x i y =( 2− 1) (+ 2−y j z1) (+ 2−z k1) or AB =Dxi+Dy j+Dzk
sPeeD anD velocity
)
Speed is related to distance and it is a scalar while
•
velocity is related to displacement and it is a vector. For a moving body speed can not have zero or
•
negative values but velocity can have.
average speed
•
It is defined for a time interval. Average speed
¾
of a trip
vav=Total distance travelled Total time taken Let
¾ Ds be the distance travelled in the time
interval t to t + Dt. The average speed in this time interval is
v s
t
av= DD
If a particle travels distances
¾ s1, s2, s3 etc. with
speeds v1, v2, v3 etc. respectively, then total distance travelled s = s1 + s2 + s3 + ... + sn
Total time taken = +s + + +
v s v s v s vnn 1 1 2 2 3 3 ....
Average speed of a trip = + + + +
+ + + s s s s s v s v s v n n n 1 2 3 1 1 2 2 ... ... instantaneous speed •
The speed at a particular instant is defined as
¾
instantaneous speed (or speed). If
¾ Dt approaches zero, average speed becomes
instantaneous speed. v s t ds dt t = = → lim D D D 0
i.e., instantaneous speed is the time derivative
of distance. O X Z Y A x y z( , , )1 1 1 B x y z( , , )2 2 2 rA r B r average velocity •
Velocity is the rate of change of position vector
¾
or change in position per unit time. Suppose a particle ¾ is at a point A at time t1 and B at time t2. Position vectors of A and B are r1 and r2. The
displacement in this
time interval is the vector AB r r =( 2−1). The average velocity in this time interval is,
vav =displacementtime interval v AB t t r r t t av = − = −− 2 1 2 1 2 1
Here, AB r r = −2 1 = change in position vector. For small time interval between
¾ t and t + Dt,
change in position vector is Dr then average velocity in Dt time interval is, v r
t
av= DD
instantaneous velocity
•
It is the velocity at a particular instant. If
¾
time interval approaches zero then average velocity become instantaneous velocity.
v r t dr dt t = = → lim D D D 0
i.e., instantaneous velocity is the time derivative
of position vector.
Magnitude of instantaneous velocity is the
¾
instantaneous speed.
When a particle moves with constant velocity,
¾
its average velocity, its instantaneous velocity and its speed all are equal.
acceleration
)
The acceleration is rate of change of velocity or
•
change in velocity per unit time interval.
Velocity is a vector quantity hence a change in its
•
magnitude or in direction or in both, will give the acceleration (or non uniform motion).
average acceleration
•
Let velocity of a particle at
¾ t1 is v1 and at
t2 it is v2. The change in velocity in time
interval (t2 – t1) is (v2−v1). \ a v v t t av= 2−− 1 2 1. O X Z Y r1 r 2 A B
For small time interval between
¾ t and t + Dt,
change in velocity is taken Dv then average acceleration in the time interval Dt is, a v
t
av = DD .
instantaneous acceleration
•
If
¾ Dt approaches zero, the average acceleration
becomes the instantaneous acceleration (or acceleration) a v t dv dt t = = → lim D D D 0
i.e., instantaneous acceleration is the time
derivative of velocity.
equations oF Motion
)
If
• u = initial velocity of the body,
a = uniform acceleration of the body, s = displacement in time t.
v = velocity of the body after a time t, then the following equations hold good, in order to describe the motion of the body.
If the motion is along a straight line,
• ¾ v = u + at ... without s ¾ s = ut + 1 2 at2 ... without v v ¾ 2 = u2 + 2as ... without t ¾ s = vt – 1 2 at2 ... without u s ¾ = 1 2 (u + v)t ... without a D
¾ istance travelled in nth sec, sn = u + a
2(2n – 1) If the motion is not along a straight line
• v u at= + ¾ s ut= + 1at 2 2 ¾ v v u u a s⋅ = ⋅ + ⋅2 ¾
Equations of motion are valid only when
•
acceleration remains constant during motion, otherwise we have to write the equations as under : v = a t dt
∫
( ) and s = v t dt∫
( )eFFective use oF MatheMatical
)
tools in solvinG ProBleMs oF
one-DiMensional Motion
If displacement-time equation is given, we can
•
get velocity-time equation with the help of differentiation. Again, we can get acceleration-time equation with the help of differentiation.
If acceleration-time equation is given, we can
•
get velocity-time equation by integration. From velocity-time equation, we can get displacement-time equation by integration.
KineMatic GraPhs
)
Displacement-time Graph for various types of Motion of a Body
•
Description of motion Shape of graph Feature of graph
For a stationary body, the displacement-time graph is a straight line AB parallel to the time
axis. A B O Time D isp lac em en t
s = constant The slope of straight line AB (representing instantaneous velocity) is zero.
When a body is moving with constant velocity, the displacement-time graph will be a straight
line OA, inclined to time axis. s vt=
O Time D isp lac em en t A
Greater is the slope of straight line OA, greater will be the velocity.
When a body is moving with a constant acceleration, the displacement-time graph is a curve which bend upwards.
O Time D isp lac em en t s = at1 2 2
The slope of displacement-time curve (i.e., instantaneous velocity) increases with time.
When a body is moving with constant retardation, the displacement-time graph is a curve which bend downwards.
O Time D is p la ce m en t s =– at1 2 2
The slope of displacement-time curve (i.e., instantaneous velocity) decreases with time.
When a body is moving with infinite velocity, the displacement-time curve is a straight line
AB parallel to displacement axis.
O Time D isp lac em en t A B s =
Such a motion of the body is never possible.
When a body returns back towards the original point of reference while moving with uniform negative velocity, the displacement-time graph is an oblique straight line AB, making an angle
q > 90° with the time axis. O Time
D is p la ce m en t B A > 90°
s= –vt The displacement of the body decreases with time with respect to the reference point, till it becomes zero.
velocity-time Graph for various types of Motion of a Body
•
Description of motion Shape of graph Feature of graph
When a body is moving with a constant velocity, the velocity-time graph is a straight line AB parallel to time axis.
O Time
Ve
loci
ty A v = constant B The slope of this graph (representing the instantaneous acceleration) is zero.
When a body is moving with a constant acceleration and its initial velocity is zero, the velocity-time graph is an oblique straight line, passing through the origin.
O Time Ve loci ty A v at=
Greater is the slope of straight line
OA, greater will be the instantaneous
acceleration. When a body is moving with a constant
acceleration and its initial velocity is not zero, the velocity-time graph is an oblique straight line AB not passing through the origin.
O Time Ve loci ty A B v u at = +
(i) Here OA represents the initial velocity of the body.
(ii) The area enclosed by the velocity-time graph with velocity-time axis represents the distance travelled by the body. When a body is moving with increasing
acceleration, the velocity-time graph is a curve which bend upwards.
O Time
Ve
loci
ty v kt= 2 The slope of velocity-time graph
(i.e. instantaneous acceleration increases with time.
When a body is moving with a constant retardation and its initial velocity is not zero, the velocity-time graph is an oblique straight line not passing through the origin.
O Time V eloci ty A B
v= –at The slope of this straight line with time axis, makes an angle q > 90°.
acceleration-time Graph for various types of Motion of a Body
•
Description of motion Shape of graph Feature of graph
When a body is moving with constant acceleration, the acceleration-time graph is a
straight line AB parallel to time axis. A B
O Time
Accelera
tio
n
a = constant The area enclosed by acceleration-time graph for the given time gives the velocity of the body.
When a body is moving with constant increasing acceleration, the acceleration-time graph is a straight line OA.
A
O Time
a=kt
Accelera
tio
n The body is moving with constant
acceleration and slope of straight line OA, makes an angle q < 90° with time axis.
When a body is moving with constant decreasing acceleration, the acceleration-time graph is a straight line.
O Time a= –kt A ccelera tio
n The body is moving with negative
acceleration and slope of straight line makes an angle q > 90° with time axis.
vertical Motion unDer Gravity
)
If air resistance is neglected and a body is freely
•
moving along vertical line near the earth’s surface then an acceleration acts downward which is 9.8 m s–2 or 980 cm s–2.
Freely falling body released from a height
• h
above the ground
Taking initial position as
¾
origin and direction of motion (i.e., downward direction) positive y-axis.
As body is just released/dropped, u = 0 acceleration along +y axis, a = g The body acquires velocity
¾ v(downward) after
falling a distance h in time t, then equations of motion become v = gt or t v g = h gt t h g = = 1 2 2 2 or v2 = 2gh or
(
v= 2gh or h v= 2/2g)
y v u= 0, = 0t hBody projected vertically upward
•
Take initial position as
¾
origin and direction of motion (i.e., vertically upward) as positive y-axis.
v = 0 at maximum height, at t = T, a = – g (because directed downward)
Equations of motion of the particle at any latter
¾ time t become v = u – gt h = ut – 1 2 gt 2 v2 = u2 – 2gh At time ¾ t = T, u = gT h gT T h g
max=12 2 or, = 2 max
u2 = 2ghmax or, hmax = u
g
2
2 After attaining maximum height, body
• comes back
at the ground. During complete flight acceleration is constant, a = –g.
Time taken during up flight
•
and down flight are equal. Time for one side,
• T = u
g
and total flight time = 2T = 2u
g y = 0 u0, = 0t y v = 0 h( , )v t y h u v = 0
At each equal height from ground speed of body
•
will be same whether going up or coming down.
relative Motion
)
There is no meaning of motion without reference or
•
observer. If reference is not mentioned then we take the ground as a reference of motion. Velocity or displacement of the particle with respect to ground is called actual velocity or actual displacement of the body.
If we describe the motion of a particle with respect
•
to an object which is also moving w.r.t. ground then velocity of particle w.r.t. ground is its actual velocity (vact) and velocity of particle w.r.t. moving object is its relative velocity (vrel) and the velocity of moving object (w.r.t. ground) is the reference velocity (vref). In the figure let
• S is ground frame and S′ is frame
of moving object. Position of particle P relative to frame S is rPSwhile position of frame S′ relative to frame S is rS S′ at a moment. According to vector law of addition rPS=rPS′+rS S′ .
Differentiate the equation w.r.t. time,
• dr dt dr dt dr dt PS PS S S = ′ + ′ but v dr dt = So, vPS=vPS′+vS S′
i.e., vactual=vrelative+vreference
or vrelative=vactual−vreference
swimming in the river
•
A man can swim with velocity
¾ v i.e., the
velocity of man w.r.t. still water. If water is also flowing with velocity vR then velocity of man relative to ground vm= +v v R
If the swimming is in the direction of flow of
¾
water or along the downstream, then
vm = v + vR v
vR
I
¾ f the swimming is in the direction opposite to
the flow of water or along the upstream, then
vm = v – vR v vR S S rPS rPS rS S P
If the man swims across the river
¾ i.e., v and vR
are not collinear then use the vector algebra vm= +v vR v vR vm
Case (I) : For shortest path, resultant velocity
vm= +(v vR) is in the direction of displacement
AB. To reach at B, v sin q = vR ⇒ sinq =vvR Component of velocity along AB = v cos q So time taken, T d v d v vR = = − cosq 2 2
Case (II) : For minimum time, man should start swimming perpendicular to water current. Due to effect of river velocity, man will reach at point C
along resultant velocity, i.e., his displacement will not be minimum but time taken to cross the river will be minimum.
t d
v
min=
In time tmin swimmer travels distance BC along the river with speed of river vR.
\ BC = tmin vR = distance travelled along river flow
= drift of man = d
vvR.
Projectile Motion
)
A particle thrown in the space which moves under
•
the effect of gravity only is called a projectile. The motion of this projectile is referred as projectile motion.
Projectile motion = Horizontal motion + Vertical
•
motion
Angle between velocity vector and acceleration
•
vector during the flight. g g g g v1 u θ g v2 v3 v4 vm v vR vcos = vm B A vsin = vR
(For mimimum displacement) d
B
A
(For minimum time) vm v vR d C
In ascending motion
¾ q is an obtuse angle. At the
top of the flight q = 90°, and during descending motion, q is an acute angle. During the flight q decreases continuously with increase in time and it always lies between 0° and 180°.
In projectile motion, magnitude as well as direction
•
of velocity continuously changes so there must be presence of tangential (at) as well as radial
component (ar) acceleration. cosθ a =r g cosθ = 0 v at θ θ g = g ar g θ at= gsinθ sinθ = g at = g ar v vx In ascending motion, • at is against v, so v decreases
and in descending motion at is in the direction of v, so v increases.
Since deceleration in
• y-component of velocity
in ascending motion is equal to acceleration in
y-component of velocity during descending motion,
so time of ascending motion is equal to the time of descending motion and magnitude of y-component of velocity at same height in ascending as well as in descending motion is same but opposite in directions.
At maximum height,
• vy = 0 and vx = ux = ucosq so that at maximum height v= vx2+v2y =ucosq When particle again returns to ground at point
•
B, its y coordinate is zero and its magnitude of
velocity is u at angle q with ground. Angular change (or difference) between initial velocity and final velocity = 2q. Y u O A H v u= cosθ X θ u θ B
Initial velocity u ui = cosqi u+ sinqj
Final velocity, uf =ucosqi u− sinqj
Total change in its velocity =|Du|=2usinq Total change in momentum =m u|D 2|= musinq If
• K0 is initial kinetic energy, then kinetic energy at highest point = K0cos2q
T=2ug Hy =u2g Ry =2u ux yg
2
, ,
•
T
¾ and H depend only upon initial vertical
speed uy.
If two projectiles thrown in different directions,
¾
have equal time of flight then their initial vertical speeds are same so that their maximum height attained is also same.
If ¾ HA = HB then (uy)A = (uy)B and TA = TB Y H HA O HB X
When two projectiles are thrown with equal speeds
•
at angle q and (90° – q), then their ranges are equal but maximum heights attained are different and time of flights are also different.
For maximum range,
• q = 45°
and Rmax =u2sin (g2 45° =) u2sing90° ⇒Rmax =ug2
Here H u= 2sin2g245° =u4g2 \ RmaxH = ⇒4 Rmax:H=4 1: For
• R = H
⇒ u2( sin cos )2 gq q =u2sin2g2q⇒2cosq=sin2q ⇒ tanq = 4 \ q = tan–1(4) = 76°
circular Motion
)
When a particle moves in a plane such that its
•
distance, from fixed (or moving) point remains constant then its motion is called as circular motion with respect to that fixed (or moving) point. That point is called centre and the distance is called radius of circular path.
Angular Displacement :
• Angle traced by position
called angular displacement. Dq = angular displacement point r∆θ P Q Fixed
Angle=RadiusArc ;Dq ArcPQ= r Small angular displacement
¾ dq is a vector
quantity, but large angular displacement is scalar quantity.
Its direction is perpendicular to plane of
¾
rotation and given by right hand screw rule. It is dimensionless and its S.I. unit is radian
¾
while other units are degree or revolution. 2p radian = 360° = 1 revolution
Angular Velocity :
• It is defined as the rate of change of angular displacement of moving particle.
w= = q= q
→∞
Angle traced Time taken Dlim
D D
t t
d dt
Its direction is same as that of angular
¾
displacement i.e., perpendicular to the plane of rotation and along the axis according to right hand screw rule.
Its unit is radian/second.
¾
Relation between Linear and Angular Velocity
•
Angle=RadiusArc
s r P r Q Dq=Drs or Ds r= Dq \ DDst=rDDtqifDt→0thendsdt r= ddtqorv=wr In vector form v= ×w r
(direction of v is found according to right hand rule)
Average Angular Velocity (
• wav) w q q q p av t t t T = = −− = = Total angle of rotation
Total time taken 22 11 2
D D == 2pn
where q1 and q2 are angular positions of the particle at instants t1 and t2 respectively.
Instantaneous Angular Velocity
•
The angular velocity at some particular instant of time is called instantaneous angular velocity.
w= q= q w q= → lim ; D D D t t d dt ddt 0 Angular Acceleration •
Rate of change of angular velocity is called angular acceleration. α= w= w α= w → lim D D D t t d dt ddt 0 ;
It is an axial vector quantity. Its direction is along
¾
the axis according to right hand screw rule.
Relation between Angular and Linear
•
Acceleration
v= ×w r( v is a tangential vector, w is a axial vector and r is a radial vector.)
but a dv=dt=dtd (w × =r) ddt rw × + × w drdt or a= × + ×α r w v
or a a= T+aC
(
¾ aT= ×α ris tangential acceleration and
aC= ×w vis centripetal acceleration)
aT andaC
¾ are two mutually perpendicular
components of net linear acceleration.
uniForM circular Motion
)
When a body moves along a circular path with
•
uniform speed, its motion is said to be uniform circular motion.
Position vector
¾ ( )r is always perpendicular to
the velocity vector ( ) . .v i e r v ⋅ = 0
Velocity vector is always perpendicular to the
¾
acceleration. v a⋅ = 0 For circular motion, force
•
towards centre (centripetal force) must act so that direction of v keeps on changing.
The work done by centripetal force is always zero.
•
Kinetic energy = constant
•
Since
• | |v = constant, so tangential acceleration at = 0
In projectile motion, both the magnitude and the
•
direction of acceleration (g) remain constant, while in uniform circular motion the magnitude remains constant but the direction continuously changes.
1. A ball falls from 20 m height on the floor and rebounds to 5 m. Time of contact is 0.02 s. Find the acceleration during impact.
(a) 1200 m s–2 (b) 1000 m s–2 (c) 2000 m s–2 (d) 1500 m s–2
2. The displacement-time (x-t) graph of a body is shown in figure. The body is accelerated along the path
(a) OA only (b) BC only (c) CD only (d) OA and CD only
3. The velocity versus v(m s )–1 time graph of a body
moving in a straight line is shown in figure. The displacement of the body in 5 s is
(a) 3 m (b) 5 m (c) 4 m (d) 2 m
4. An aeroplane is flying in a horizontal direction with a velocity of 360 km h–1 and at a height of 1960 m. When it is vertically above the point A on the ground, a body is dropped from it. The body strikes the ground at a point B. The distance AB is (a) 2000 2 m (b) 2000 m
(c) 1000 2 m (d) 1000 m
5. A body initially at rest is moving with uniform acceleration a. Its velocity after n seconds is v. The displacement of the body in last 2 s is
(a) 2v n(n−1) (b) v n(n−1) (c) v n( n+1 ) (d) 2v n(n+1)
6. Two cities C1 and C2 are connected on the opposite ends of a long straight parallel track. The cities are connected by a train service as well as a bus service. The trains leave with constant speed v for either city at regular frequency of one train every x minute. The buses ply on a parallel road at a constant speed of 30 km h–1. A bus passenger going from city C1 to city
C2 observes a train going past him every 20 minutes while a train goes in the opposite direction every
10 minutes. What are the values of x and v ? (a) x = 15 min, v = 90 km h–1
(b) x = 13 min 20 s, v = 90 km h–1 (c) x = 15 min, v = 75 km h–1 (d) x = 13 min 20 s, v = 70 km h–1
7. A particle is projected with a certain velocity at two different angles of projection with respect to the horizontal plane so as to have the same range R on the horizontal plane. If t1 and t2 are the times taken for the two paths, then which one of the following relations is correct?
(a) t1t2 = 2R/g (b) t1t2 = R/g (c) t1t2 = R/2g (d) t1t2 = 4R/g
8. The position of a particle moving in the x-y plane at any time t is given by; x = (3t2 – 6t) metres;
y = (t2–2t) metres. Select the correct statement. (a) Acceleration is zero at t = 0
(b) Velocity is zero at t = 0 (c) Velocity is zero at t = 1 s
(d) Velocity and acceleration of the particle are never zero.
9. A stone tied to the end of a 1 m long string, is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in 44 s, what is the magnitude and direction of acceleration of the stone?
(a) p2 2
4 m s− and direction along the radius towards the centre.
(b) p2 m s–2 and direction along the radius away from the centre.
(c) p2 m s–2 and direction along the radius towards the centre.
(d) p2 m s–2 and direction along the tangent to the circle.
10. Water drops fall at regular intervals from a tap which is 5.0 m above the ground. The third drop is leaving the tap at the instant the first drop reaches the ground. How far above the ground is the second drop at that instant?
(a) 1.25 m (b) 2.50 m (c) 3.75 m (d) 5.00 m
11. Two cars having masses m1 and m2 move in circles of radii r1 and r2 respectively. If they complete the
circle in equal time, the ratio of their angular speed w1/w2 is (a) mm1 2 (b) r r12 (c) m r m r2 21 1 (d) 1
12. A man running on a horizontal road at 8 km h–1 finds the rain falling vertically. He increases his speed to 12 km h–1 and finds that the drops make angle 30° with the vertical. What is the speed of the rain ? (a) 8 km h–1 (b) 4 7 1
km h− (c) 8 3 km h–1 (d) 7 km h–1
13. A particle moves so that its position vector is given by r=coswt x^+sinwt y^, where w is a constant. Which of the following is true?
(a) Velocity is perpendicular to r and acceleration is directed towards the origin.
(b) Velocity is perpendicular to r and acceleration is directed away from the origin.
(c) Velocity and acceleration both are perpendicular to r.
(d) Velocity and acceleration both are parallel to r. [NEET Phase - 1 2016] 14. A uniform circular disc of radius 50 cm at rest is free
to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0 rad s–2. Its net acceleration in m s–2 at the end of 2.0 s is approximately
(a) 6.0 (b) 3.0 (c) 8.0 (d) 7.0
[NEET Phase - 1 2016] 15. Two stones are thrown up simultaneously from
the edge of a cliff 240 m high with initial speed of 10 m s–1 and 40 m s–1 respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first stone?
(Assume stones do not rebound after hitting the ground and neglect air resistance, take g = 10 m s–2)
(a) (b)
(c) (d)
[JEE Main 2015]
16. If a body is moving in a circular path maintains constant speed of 10 m s–1, then which of the following correctly describes relation between acceleration and radius ?
(a) (b)
(c) (d) a
[JEE Main Online 2015] 17. The position vector of a particle R as a function of
time is given byR=4sin(2pt i)+4cos(2pt j)Where
R is in meters, t is in seconds and iand j denote unit vectors along x and y-directions, respectively. Which one of the following statements is wrong for the motion of the particle?
(a) Magnitude of the velocity of particle is 8 m s–1. (b) Path of the particle is a circle of radius 4 m. (c) Acceleration vector is along −R.
(d) Magnitude of acceleration vector is vR2,where
v is the velocity of particle.
[AIPMT 2015] 18. From a tower of height H, a particle is thrown
vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is
(a) gH = (n – 2)u2 (b) 2gH = n2u2 (c) gH = (n – 2)2u2 (d) 2gH = nu2(n – 2)
[JEE Main 2014] 19. A projectile is fired from the surface of the earth
with a velocity of 5 m s–1 and angle q with the horizontal. Another projectile fired from another planet with a velocity of 3 m s–1 at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet (in m s–2) is (Given g = 9.8 m s–2) (a) 3.5 (b) 5.9 (c) 16.3 (d) 110.8
[AIPMT 2014] 20. A projectile is given an initial velocity of (i+ 2 m s ,j) _1
If g = 10 m s–2, the equation of its trajectory is (a) 4y = 2x – 25x2 (b) y = x – 5x2 (c) y = 2x – 5x2 (d) 4y = 2x – 5x2
[JEE Main 2013]
SolutionS
1. (d) : The acceleration during impact,
a=v2− −t( v1)=v2+t v1= 2gh2+t 2gh1
= 2 10 20× × +0 02. 2 10 5× × =20 100 02.+ =1500m s−2 2. (a) : The slope of x-t graph gives velocity and the
slope of the graph is increasing in part OA only.
3. (c) : Displacement = area of bigger triangle – area of smaller triangle + area of rectangle
= 12(3 2 1× −) 2(1 2× + ×) (1 2)=4m
4. (b) : Time taken by the body to strike the ground is given by h=21gt2 or t= 2gh = 2 1960×9 8. =20s Velocity of the aeroplane in horizontal direction,
vx0 = 360 km h–1 = 100 m s–1 Distance AB = vx0 × t = (100 × 20) m = 2000 m 5. (a) : Since, a v u= − = − =t vn0 vn Displacement in last 2 s, sn−sn−2=12an2−12a n( −2)2 =2a n( − = ×1 2) n nv( − =1 2) v n(n−1)
6. (b) : If v (in km h–1) is the constant speed of the trains then the distance between the successive trains
= ×v x60= vx60 km
When a train moves in the same direction as that of the bus passenger, vx(v−/3060)=2060
or vx = 20(v – 30) ...(i) When a train moves in a direction opposite to the bus passenger,
vx v /
( +3060)=1060 or vx = 10(v + 30) ...(ii) From eqns. (i) and (ii),
20(v – 30) = 10(v + 30) or v = 90 km h–1 From eqn. (i),
90x = 20(90 – 30) = 1200 or x = 13 min 20 s
7. (a) : R is same for angles of projection q and (90° – q),
i.e., R v= 02sin /2q g
Ast1=2v0gsinqandt2=2v0sin(g90° −q)=2v0cosg q,
t t v g g v g gR 1 2 0 2 2 0 2 4 2 2 2
= sin cosq q= sin q= 8. (c) : Since, x = 3t2 – 6t \ vx =dxdt = −6 6t At t = 1 s, vx = 6 × 1 – 6 = 0 Also, y = t2 – 2t \ vy =dydt = −2 2t At t = 1 s, vy = 2 × 1 – 2 = 0 9. (c) : a r= w2=r
(
2 /pn t)
2 = × = × × × = − r n t 4 1 4 22 44 2 2 2 2 2 2 2 2 p p ( ) p ( ) m s10. (c) : Let T be the time interval between the drops (1, 2, 3) falling from the tap as shown in the figure.
Since distance covered by the first drop in time 2T is 5 m,
5 12 2= g T( )2=2gT2 ...(i) Further, distance covered by the second drop in time T (from t = T to t = 2T),
y= 12gT2 ...(ii)
From eqns. (i) and (ii), y = 1.25 m
Distance of the second drop from the ground = 5 – y = 5 – 1.25 = 3.75 m
11. (d)
12. ( b) : We have, vrain road, =vrain man, +vman road, ...(i) The two situations given in the problem may be represented by the given figures.
vrain, road is same in magnitude and direction in both the figures.
Taking horizontal components in equation (i) for figure (a)
vrain road, sina = 8 ...(ii)
Now consider figure (b). Draw a line OA ^ vrain,man as shown.
Taking components in equation (i) along the line OA.
vrain,road sin(30° + a) = 12 cos30° ...(iii)
From (ii) and (iii), sin( )
sin30° +a a =128 2×× 3 or cota = 32 From (ii), vrain road, =sin8a=4 7km h−1 13. (a) : Given, r=coswt x+sinwt y
\ v dr dt t x t y = = −wsinw +wcosw a dv dt t x t y r = = −w2cosw −w2sinw = −w2 Since position vector ( )r is directed away from the origin, so, acceleration (−w2r is directed towards ) the origin. Also,
r v⋅ = –w sin wt cos wt + w sin wt cos wt = 0 ⇒ r v^
14. (c) : Given, r = 50 cm = 0.5 m, a = 2.0 rad s–2, w0 = 0 At the end of 2 s,
Tangential acceleration, at = ra = 0.5 × 2 = 1 m s–2
Radial acceleration, ar = w2r = (w0 + at)2r = (0 + 2 × 2)2 × 0.5 = 8 m s–2
\ Net acceleration,
a = at2+ar2 = 1 82+ 2 = 65 8≈ m s−2 15. (a) : Using h = ut + 1
2at2 For first stone, y1 = 10t – 12gt2 For second stone, y2 = 40t – 12gt2
Relative position of the second stone with respect to the first stone Dy = y2 – y1 = 40t – 1
2gt2 – 10t + 12gt2 = 30t
Therefore, it will be a straight line.
After 8 seconds, first stone reaches to the ground,
i.e., y1 = –240 m
\ Dy = y2 – y1 = 40t – 12gt2 + 240
Therefore, it will be a parabolic curve till the second
stone reaches to the ground.
16. (c) : Speed v = 10 m s–1
We know, centripetal acceleration is given by,
a v r = 2 Q | |→v = constant so, a ∝1 r or ar = constant
This represents a rectangular hyperbola.
17. (a)
18. (d) : Time taken by the particle to reach the top most point is,
t ug= ... (i) Time taken by the particle to reach
the ground = nt Using, s ut= + 12at2
⇒ − =H u nt( )−12g nt( )2
⇒ − = × H u n ug −12gn ug2 2 [using (i)] ⇒ –2gH = 2nu2 – n2u2 ⇒ 2gH = nu2(n – 2)
19. (a) : The equation of trajectory is
y x gx u = tan − cos q q 2 2 2 2
where q is the angle of projection and u is the velocity with which projectile is projected.
For equal trajectories and for same angles of projection, g
u2 = constant
As per question, 9 8 5. = ′g2 32
where g′ is acceleration due to gravity on the planet. ′ = × = − g 9 8 9.25 3 5. m s 2 20. (c) : Given : u i= + 2j As u u i u j= x+ y \ ux = 1 and uy = 2 Also x = uxt \ x = t and y u t= y − 12gt2 ... y= − × × = −2 12 10t t2 2 5t t2 Equation of trajectory is y = 2x – 5x2.
section-A
1. Fifteen vectors, each of magnitude 5 units, are represented by the sides of a closed polygon, all taken in same order. What will be their resultant?
2. Can an object be accelerated without speeding up or slowing down? Explain.
3. A projectile is projected at an angle of 15° to the horizontal with speed v. If another projectile is projected with the same speed, then at what angle with the horizontal it must be projected so as to have the same range.
4. A wooden box is lying on an inclined plane. What is the coefficient of static friction if the box starts sliding when the angle of inclination is just 30°?
5. Two bodies of masses M and m are allowed to fall from the same height. If air resistance for each be the same, then will both the bodies reach the Earth simultaneously?
section-B
6. Two forces whose magnitude are in the ratio 3 : 5 give
a resultant of 28 N. If the angle of their inclination is 60°, find the magnitude of each force.
7. The position of a particle is given by
r=3 0. ti−2 0. t j2+4 0. k m
where t is in seconds and the coefficients have the proper units for r to be in metres.
(a) Find vand of the particle?a
(b) What is the magnitude and direction of velocity of the particle at t = 2.0 s?
8. A projectile is thrown with an initial velocity of
xi y j+ . The range of the projectile is twice the maximum height of the projectile. Calculate y
x.
OR
A light string passing over a smooth pulley connects two blocks of masses m1 and m2 (vertically). If the acceleration of the system is g/8, find the ratio of the two masses.
9. The driver of a truck travelling with a velocity v suddenly notices a brick wall in front of him at a
GENERAL INSTRUCTIONS
(i) All questions are compulsory.
(ii) Q. no. 1 to 5 are very short answer questions and carry 1 mark each. (iii) Q. no. 6 to 10 are short answer questions and carry 2 marks each. (iv) Q. no. 11 to 22 are also short answer questions and carry 3 marks each. (v) Q. no. 23 is a value based question and carries 4 marks.
(vi) Q. no. 24 to 26 are long answer questions and carry 5 marks each. (vii) Use log tables if necessary, use of calculators is not allowed.
Motion in a Plane
Laws of Motion
CBSE
CLASS XI
Series 2
Time Allowed : 3 hours Maximum Marks : 70
distance d. Is it better for him to apply brakes or to make a circular turn without applying brakes in order to just avoid crashing into a wall? Explain.
10. A retarding force is applied to stop a motor car. If the speed of the motor car is doubled, how much more distance will it cover before stopping under the same retarding force?
section-c
11. On a certain day, rain was falling vertically with a speed of 35 m s–1. A wind started blowing after sometime with a speed of 12 m s–1 in east to west direction. In which direction should a boy waiting at a bus stop hold his umbrella?
12. Explain how a vector can be resolved into its rectangular components in three dimensions.
13. Define scalar product of two vectors. Give its geometrical interpretation.
14. What do you mean by impulse of a force? Show that impulse is equal to the product of average force and the time interval for which the force acts. Give the units and dimensions of impulse.
15. Show that there are two angles of projection for a projectile to have the same horizontal range. What will be the maximum heights attained in the two cases? Compare the two heights for q = 30° and 60°.
16. A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km h–1. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)
17. Explain why
(a) Porcelain objects are wrapped in paper or straw before packing for transportation?
(b) Mountain roads are generally made winding upwards rather than going straight up?
18. A stone of mass 5 kg falls from the top of a cliff 50 m high and buries itself 1 m in sand. Find the average resistance offered by the sand and the time it takes to penetrate.
19. Three equal weights A, B and C of mass m each are hanging on a string over a fixed pulley as shown in figure.
What are the tensions in the strings connecting weights A to B and B to C?
20. A block A of mass 4 kg is placed on another block B of mass 5 kg, and the block B rests on a smooth horizontal table. For sliding the block A on B, a horizontal force of 12 N is required to be applied on it. How much maximum horizontal force can be applied on B so that both A and B move together? Also find out the acceleration produced by this force.
OR
A box of mass 4 kg rests upon an inclined plane. The inclination of the plane to the horizontal is gradually increased. It is found that when the slope of the plane is 1 in 3, the box starts sliding down the plane. Given g = 9.8 m s–2
(i) Find the coefficient of friction between the box and the plane.
(ii) What force applied to the box parallel to the plane will just make it move up the plane?
21. Show that Newton's second law of motion is the real law of motion.
22. A bird is sitting on the floor of a closed glass cage and the cage is in the hand of a girl. Will the girl experience any change in the weight of the cage when the bird (i) starts flying in the cage with a constant velocity (ii) flies upwards with acceleration (iii) flies downwards with acceleration?
section-D
23. Rahul was going to a nearby shopping complex along with his father who was a Physics teacher. Suddenly it started raining heavily. Rahul's father advised him to hold umbrella slightly inclined with the vertical in the direction of motion. Out of his curiosity, Rahul asked his father why should I hold umbrella slightly inclined even when the rain drops are falling vertically downwards. His father explained him the reason behind it satisfactorily. (a) What are the values being displayed by Rahul? (b) What are the values displayed by Rahul's
father?
(c) Give reason for holding the umbrella slightly inclined with the vertical.
3. To have the same range (with same speed), the sum of the two angles of projection should be 90°. Since one of the angles is 15°, the other should be 90° – 15° = 75°.
4. Here, angle of friction, q = 30° Coefficient of static friction, ms = tan q = tan 30° = 0.5774
5. Net force acting on the body of mass M, i.e.,
F = Mg – f, f being the force of air resistance.
Net acceleration acting on the body of mass M, i.e.,
a F M Mg f M g f M = = − = −
Net acceleration acting on the body of mass m, i.e., ′ = −
a g f m
If M > m, then a > a′. Hence the body of larger mass will reach the Earth earlier.
6. Let A and B be the two forces
Then A = 3x; B = 5x; R = 28 N and q = 60° Thus, A/B = 3/5
Now R= A2+B2+2ABcosq
\ 28= ( ) ( )3x 2+ 5x 2+2 3 5( )( )cosx x 60° or 28= 9x2+25x2+15x2 =7x or x = 28/7 = 4
\ Forces are; A = 3 × 4 = 12 N and B = 5 × 4 = 20 N 7. (a) Velocity, v dr dt d dt t i t j k = = ( .3 0 −2 0. 2 +4 0. ) =[ .3 0i−4 0. t j m s] −1 Acceleration, a dv dt d dt i t j = = ( .3 0 4 0− . ) = −0 4 0. j =−4 0. m sj −2 (b) At time t = 2 s, v=3 0. i−4 0 2 3 0 8 0. × =j . i− . j \ v = ( . ) ( )3 02+ −82 = 73 8 54= . m s−1 If q is the angle which makes with x-axis, then
tanq =v = − = − . v y x 8 3 2 667
\ q = tan–1(–2.667) – 70° with x-axis 8. Here, range of projectile = 2 × maximum height
v g v g 2 2 2 2 2 2
sin q= sin q or 2v2 sin q cos q = v2 sin2q or 2(v sin q)(v cos q) = (v sin q)(v sin q)
section-e
24. Define centripetal acceleration. Derive an expression for the centripetal acceleration of a body moving with uniform speed v along a circular path of radius
r. Explain how it acts along the radius towards the
centre of the circular path.
OR
What is a projectile? Derive the expression for the trajectory, time of flight, maximum height and horizontal range for a projectile thrown upwards, making an angle q with the horizontal direction.
25. What is meant by banking of roads? What is the need for banking a road? Obtain an expression for the maximum speed with which a vehicle can safely negotiate a curved road banked at an angle q. The coefficient of friction between the wheels and the road is m.
OR
(a) Obtain an expression for the acceleration of a body sliding down a rough inclined plane.
(b) Explain why it is easier to pull a lawn roller than to push it?
26. (a) State the law of conservation of linear momentum and derive it from Newton's second law of motion.
(b) A disc of mass 10 g is kept floating horizontally by throwing 10 marbles per second against it from below. If the mass of each marble is 5 g, calculate the velocity with which the marbles are striking the disc. Assume that the marbles strike the disc normally and rebound downward with the same speed.
OR
(a) State the laws of limiting friction. Is kinetic friction less than or greater than the coefficient of static friction?
(b) Show that the tangent of the angle of friction is equal to the coefficient of static friction.
solutions
1. Zero vector. The vector sum of all the vectors represented by the sides of a closed polygon taken in the same order is zero.
2. Yes. For example, when a body moves along a circular path with constant speed, it possesses centripetal acceleration due to continuous change in its direction of motion.
But v cos q = x and v sin q = y (given) \ 2yx = y2 or 2x = y or y x = 2 OR As a m m m m g g m m m m g = − + \ = − + 1 2 1 2 1 2 1 2 8 or m mm m1 2 1 2 1 8 − + = or ( ) ( ) ( ) ( ) m m m m m m11 22 m m11 22 8 1 8 1 + + − + − − = +− or m m12 m m1 2 9 7 9 7 = or : = :
9. Suppose force FB is required in applying brakes to stop
the truck in distance d. Then
F ma mv
d a vd
B = = =
2 2
2 2
Suppose FT force is required in taking a turn of
radius d. Then F mv d F F F T = = B B= T 2 2 1 2 or
Clearly, it is better to apply brakes than to take a circular turn.
10. Let F be retarding force applied on the motor car. So retarding acceleration, a F
m
= = constant Using kinematics equation, v2 = u2 + 2as Here v = 0, \ 0 = u2 + 2(–a)s ⇒ s = u
a
2
2
Now speed of the car is doubled so distance covered by it will be s′ (say). \ 0 = (2u)2 + 2(–a)s′ ⇒ s′ = 4×u22 =4 a s 11. In figure, Velocity of rain, vR=OA=35 m s , vertically downward−1 Velocity of wind, vW =OB=12 m s , east to west−1
The magnitude of the resultant velocity is
v= vR2+vW2 = ( ) ( )352+122 =37 m s−1
Let resultant velocity v makes an angle q with the vertical. Then, tanq =AC = = . OA v vWR 0 343 \ q =tan ( .−10 343 19) °
Hence the boy should hold umbrella bending it towards east making an angle of about 19° with the vertical.
12. Suppose vector A is represented by OP , as shown in figure. Taking O as origin, construct a rectangular parallelopiped with its three edges along the three rectangular axes i.e., X-, Y- and Z-axis.
Thus A A x, y and are the three rectangular Az
components of A.
Applying triangle law of vectors, OP OT TP = + Applying parallelogram law of vectors,
OT OR OQ = +
\ OP OR OQ TP = + + But TP OS =
Hence OP OR OQ OS = + +
or A A = x+Ay+Az =A i A j A kx+ y+ z
If a, b and g are the angles between A and X-, Y-
and Z-axes repectively, then
Ax = A cos a, Ay = A cos b, Az = A cos g
We note that
OP2=OT2+TP2=OQ QT2+ 2+TP2
or A2=Ax2+Ay2+Az2 or A= Ax2+Ay2+Az2
13. The scalar or dot product of two vector A and is B
defined as the product of the magnitudes of A and
