Volume 24 Managing Editor Mahabir Singh Editor Anil Ahlawat (BE, MBA)
No. 8
August 2016
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AIIMS Topper Interview
8
CONTENTS
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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Physics For you | AUGUST ‘16
7
Topper Interview
AIIMS 2016
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.
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.
Nikhil Bajiya
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-
fledged preparation from Class XI. I had joined a Kota based would like to follow? coaching institute for medical entrance exam preparation. I had Nikhil Bajiya: I plan to take admission in AIIMS Delhi. I want strictly followed the guidelines given by my faculty members. to be a neurosurgeon. Initially I concentrated on gaining stronghold “Systematic preparation You had followed a strict study schedule. over the concepts followed by taking regular How did you unwind yourself? Could strategy and focus practice tests and solving previous years you find time to pursue your hobbies? are key factors for my question papers and taking lessons of Nikhil Bajiya: Since I stayed in Kota, away success, says Nikhil improvement from the mistakes. Apart from from family, I did not find my scope to pursue Bajiya, AIR 2” coaching, I used to put around 6 to 7 hours my hobby, which is a sport. I love playing of self-study. badminton, table tennis and computer games. Occasionally I 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
10
Physics For you | AUGUST ‘16
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
NEET JEE
Class XI
ESSENTIALS
Unit
Kinematics
2
If all three coordinates change with time, motion is three dimensional (3 - D) or motion in space.
) Motion •
•
•
•
12
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. ¾ The coordinates (x, y, z) of the particle specify the position of the particle with respect to origin of that frame. ¾ If all the three coordinates of the particle remain unchanged as time passes it means the particle is at rest with respect to this frame. ¾ The reference frame is chosen according to problems. ¾ If frame is not mention, then ground is taken as 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. Physics For you | august ‘16
) Distance anD DisPlaceMent •
•
•
•
•
Total length of path C covered by the particle, in definite time interval B is called distance. The A 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|
•
Displacement in terms of position vector : Let a body is Y A(x1, y1, z1) displaced from r A(x1, y1, z1) to rB B(x2, y2, z2) then B(x2, y2, z2) rA its displacement is X O given by vector AB. From DOAB Z rA + AB = rB or AB = rB − rA Q rB = x2 i + y2 j + z2 k and rA = x1 i + y1 j + z1 k \ AB = (x2 − x1 )i + ( y2 − y1 )j + (z2 − z1 )k 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 ¾ It is defined for a time interval. Average speed of a trip Total distance travelled vav = Total time taken ¾ Let Ds be the distance travelled in the time interval t to t + Dt. The average speed in this time interval is Ds vav = Dt ¾ If a particle travels distances s1, s2, s3 etc. with speeds v1, v2, v3 etc. respectively, then total distance travelled s = s1 + s2 + s3 + ...... + sn s s s s Total time taken = 1 + 2 + 3 + .... + n v1 v2 v3 vn
s + s + s + ..... + sn Average speed of a trip = 1 2 3 sn s1 s2 v + v + ..... + v 1 2 n instantaneous speed ¾ The speed at a particular instant is defined as instantaneous speed (or speed). ¾ If Dt approaches zero, average speed becomes instantaneous speed. Ds ds v = lim = Dt →0 Dt dt i.e., instantaneous speed is the time derivative of distance.
•
average velocity ¾ Velocity is the rate of change of position vector or change in position per unit time. Y ¾ Suppose a particle A is at a point A at time t1 and B at r2 r1 time t2. Position B vectors of A and B O X are r1 and r2 . The Z displacement in this time interval is the vector AB = (r2 − r1 ). The average velocity in this time interval is, displacement vav = time interval r −r AB vav = = 2 1 t −t t −t 2 1 2 1 Here, AB = r2 − r1 = change in position vector. ¾ For small time interval between t and t + Dt, change in position vector is Dr then average Dr velocity in Dt time interval is, vav = Dt instantaneous velocity ¾ It is the velocity at a particular instant. If time interval approaches zero then average velocity become instantaneous velocity. Dr dr v = lim = dt Dt →0 Dt i.e., instantaneous velocity is the time derivative of position vector. ¾ Magnitude of instantaneous velocity is the instantaneous speed. ¾ When a particle moves with constant velocity, 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 ¾ Let velocity of a particle at t1 is v1 and at t2 it is v2 . The change in velocity in time interval (t2 – t1) is (v2 − v1 ). v2 − v1 a = . \ av t2 − t1 Physics For you | august ‘16
13
For small time interval between t and t + Dt, change in velocity is taken Dv then average Dv acceleration in the time interval Dt is, aav = . Dt instantaneous acceleration ¾ If Dt approaches zero, the average acceleration becomes the instantaneous acceleration (or Dv dv acceleration) a = lim = dt Dt →0 Dt i.e., instantaneous acceleration is the time derivative of velocity. ¾
•
•
•
) 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 1 2 ¾ s = ut + at ..... without v 2 2 2 ¾ v = u + 2as ..... without t 1 2 ¾ s = vt – at ..... without u 2
1 (u + v)t ..... without a 2 a th ¾ Distance travelled in n sec, sn = u + (2n – 1) 2 If the motion is not along a straight line ¾ v = u + at 1 s = ut + at 2 ¾ 2 v ⋅ v = u ⋅ u + 2a ⋅ s ¾ Equations of motion are valid only when acceleration remains constant during motion, otherwise we have to write the equations as under : ¾
s=
v = ∫ a(t ) dt and s = ∫ v(t ) dt
) eFFective use oF MatheMatical tools in solvinG ProBleMs oF oneDiMensional 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 displacementtime equation by integration.
) KineMatic GraPhs Displacement-time Graph for various types of Motion of a Body Description of motion For a stationary body, the displacement-time graph is a straight line AB parallel to the time axis.
Shape of graph Displacement
•
A
s = constant
When a body is moving with constant velocity, the displacement-time graph will be a straight line OA, inclined to time axis.
Displacement
O
Displacement O
14
Physics For you | august ‘16
B
The slope of straight line AB (representing instantaneous velocity) is zero.
Time A
Greater is the slope of straight line OA, greater will be the velocity.
s = vt
O
When a body is moving with a constant acceleration, the displacement-time graph is a curve which bend upwards.
Feature of graph
Time
s=
1 2 at 2 Time
The slope of displacement-time curve (i.e., instantaneous velocity) increases with time.
Displacement
When a body is moving with constant retardation, the displacement-time graph is a curve which bend downwards.
s =– O
Time B
Displacement
When a body is moving with infinite velocity, the displacement-time curve is a straight line AB parallel to displacement axis.
A
s=
A
Such a motion of the body is never possible. Time
The displacement of the body decreases with time with respect to the reference point, till it becomes zero.
s = –vt > 90°
O
B
Time
velocity-time Graph for various types of Motion of a Body Description of motion
When a body is moving with a constant velocity, the velocity-time graph is a straight line AB parallel to time axis.
Shape of graph
v = constant
Velocity
•
Displacement
O
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.
The slope of displacement-time curve (i.e., instantaneous velocity) decreases with time.
1 2 at 2
A O
When a body is moving with increasing acceleration, the velocity-time graph is a curve which bend upwards.
Velocity O
A O
A
Velocity
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.
Time
v
+ =u
at B
Time
v = kt 2
O
(i) Here OA represents the initial velocity of the body. (ii) The area enclosed by the velocitytime graph with time axis represents the distance travelled by the body. The slope of velocity-time graph (i.e. instantaneous acceleration increases with time.
Time
v = –at B
O
The slope of this graph (representing the instantaneous acceleration) is zero.
Greater is the slope of straight line OA, greater will be the instantaneous acceleration.
v = at
Velocity
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.
B
Time A
Velocity
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.
Feature of graph
The slope of this straight line with time axis, makes an angle q > 90°.
Time Physics For you | august ‘16
15
•
acceleration-time Graph for various types of Motion of a Body
When a body is moving with constant acceleration, the acceleration-time graph is a straight line AB parallel to time axis.
Shape of graph Acceleration
Description of motion
A
a = constant
When a body is moving with constant increasing acceleration, the acceleration-time graph is a straight line OA.
Acceleration
O
When a body is moving with constant decreasing acceleration, the acceleration-time graph is a straight line.
Acceleration
O
A
a=
Time
O
•
Freely falling body released from a height h above the ground
¾
Taking initial position as u = 0, t = 0 origin and direction of h motion (i.e., downward v y direction) positive y-axis. As body is just released/dropped, u = 0 acceleration along +y axis, a = g The body acquires velocity v(downward) after falling a distance h in time t, then equations of motion become v = gt h=
1 2 gt 2
v2 = 2gh
16
•
2h or t = g
•
v = 2 gh or h = v 2 / 2 g
Physics For you | august ‘16
)
The body is moving with negative acceleration and slope of straight line makes an angle q > 90° with time axis.
Body projected vertically upward y v=0 ¾ Take initial position as origin and direction of (v, t) motion (i.e., vertically h u 0, t = 0 upward) as positive y-axis. y = 0 v = 0 at maximum height, at t = T, a = – g (because directed downward) ¾ Equations of motion of the particle at any latter time t become v = u – gt 1 h = ut – gt2 2 v2 = u2 – 2gh ¾ At time t = T, u = gT 2hmax 1 hmax = gT 2 or, T = 2 g u2 2g After attaining maximum height, body comes back at the ground. During complete flight acceleration is constant, a = –g. Time taken during up flight v=0 y and down flight are equal. u Time for one side, T = g h 2u u and total flight time = 2T = g u2 = 2ghmax
v or t = g
(or
The body is moving with constant acceleration and slope of straight line OA, makes an angle q < 90° with time axis.
Time
•
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.
¾
kt
•
B
The area enclosed by accelerationtime graph for the given time gives the velocity of the body.
Time
a = –kt
) vertical Motion unDer Gravity
Feature of graph
•
or, hmax =
•
At each equal height from ground speed of body will be same whether going up or coming down.
¾
) relative Motion •
•
•
•
•
¾
v
vR
vm = v – vR
v
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
R
B
C vR
d
v v m
A
(For minimum time)
along resultant velocity, i.e., his displacement will not be minimum but time taken to cross the river will be minimum. d t min = v In time tmin swimmer travels distance BC along the river with speed of river vR. \ BC = tmin vR = distance travelled along river flow d = drift of man = v R . v
) 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. v1
u
vm
If the swimming is in the direction opposite to the flow of water or along the upstream, then
•
v
If the swimming is in the direction of flow of water or along the downstream, then vm = v + vR
Case (I) : For shortest path, resultant velocity vm = (v + v R ) is in the direction of displacement AB. B To reach at B, v sin q = vR vR v d vm ⇒ sin q = R v v cos = vm v Component of velocity v sin = v A R along AB = v cos q (For mimimum displacement) d d So time taken, T = = v cos q v2 − v2
Differentiate the equation w.r.t. time, S S drPS drPS ′ drS ′S dr P = + but v = dt dt dt dt rPS rPS So, v PS = v PS ′ + vS ′S rSS i.e., vactual = v relative + vreference or vrelative = vactual − vreference
¾
vR
vm = v + v R
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 rPS while position of frame S′ relative to frame S is rS ′S at a moment. According to vector law of addition rPS = rPS ′ + rS ′S .
swimming in the river ¾ A man can swim with velocity v i.e., the velocity of man w.r.t. still water. If water is also flowing with velocity v R then velocity of man relative to ground vm = v + v R
If the man swims across the river i.e., v and v R are not collinear then use the vector algebra
vR
θ
g
g
v2 g
v3 g
v4 g
Physics For you | august ‘16
17
In ascending motion 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. ¾
•
v θ θ
at = g sin θ
•
•
• •
g
at = 0
ar = g ar
ar = g cosθ
O
θ
A
=
18
¾
¾
θ va = g sin θ t gc os g θ
•
•
2g
,R=
2ux u y g
B
HA
HB
X
O
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°
Here H =
u2 u2 sin 2(45°) u2 sin 90° = ⇒ Rmax = g g g
u2 sin2 45° u2 = 2g 4g
Rmax = 4 ⇒ Rmax : H = 4 : 1 H For R = H \
•
u2 (2 sin q cos q) u2 sin2 q sin q = ⇒ 2 cos q = g 2g 2
⇒ tanq = 4 \ q = tan–1(4) = 76° θ
X
Initial velocity ui = u cos qi + u sin qj Final velocity, u f = u cos qi − u sin qj Total change in its velocity =| Du | = 2u sin q Total change in momentum = m | Du | = 2mu sin q If K0 is initial kinetic energy, then kinetic energy at highest point = K0cos2q Physics For you | august ‘16
u2y
T and H depend only upon initial vertical speed uy. If two projectiles thrown in different directions, have equal time of flight then their initial vertical speeds are same so that their maximum height attained is also same. If HA = HB then (uy)A = (uy)B and TA = TB
and Rmax =
v = u cosθ
H
g
,H =
H
⇒
u
•
¾
2uy
Y
that at maximum height v = v 2x + v 2y = u cos q 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. u
T=
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
Y
•
) 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 vector of a particle moving w.r.t. some fixed point is
called angular displacement. Dq = angular displacement ArcPQ Arc ; Dq = Angle = r Radius
•
Fixed point
Q ∆θ r
•
P
¾ Small angular displacement dq is a vector quantity, but large angular displacement is scalar quantity. ¾ Its direction is perpendicular to plane of rotation and given by right hand screw rule. ¾ It is dimensionless and its S.I. unit is radian 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. Angle traced Dq dq w= = lim = Time taken Dt →∞ Dt dt
•
dv d dw dr but a = = (w × r ) = ×r +w× dt dt dt dt or a = α × r + w × v or a = aT + aC ¾ ( aT = α × r is tangential acceleration and aC = w × v is centripetal acceleration) ¾ aT and aC are two mutually perpendicular components of net linear acceleration.
Its direction is same as that of angular displacement i.e., perpendicular to the plane of rotation and along the axis according to right hand screw rule. ¾ Its unit is radian/second. Relation between Linear and Angular Velocity Angle =
Arc Radius
r r
Q
s
•
Ds or Ds = r Dq Dq = r P dq ds Ds r Dq if Dt → 0 then = r or v = wr \ = dt dt Dt Dt In vector form v = w×r (direction of v is found according to right hand rule) Average Angular Velocity (wav)
•
•
Total angle of rotation q2 − q1 Dq = = Total time taken t2 − t1 Dt 2p = = 2pn T 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. Dq dq dq w = lim = ; w= dt dt Dt →0 Dt
•
wav =
It is an axial vector quantity. Its direction is along 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.) ¾
¾
•
Angular Acceleration Rate of change of angular velocity is called angular acceleration. dw Dw dw α = lim = ; α= dt dt Dt →0 Dt
) uniForM circular Motion •
• • •
When a body moves along a circular path with uniform speed, its motion is said to be uniform circular motion. ¾ Position vector (r ) is always perpendicular to the velocity vector (v ) i.e. r ⋅ v = 0 ¾ Velocity vector is always perpendicular to the 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. Physics For you | august ‘16
19
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 –2 (c) 2000 m s (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 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
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
v(m s–1)
(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) (c)
2000 2 m 1000 2 m
(b) 2000 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 v(n −1) 2v(n − 1) (a) (b) n n 2v(n + 1) v(n + 1) (c) (d) n n 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 20
Physics For you | August ‘16
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
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? p2 (a) m s −2 and direction along the radius towards 4 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 m1r1 m r (a) 1 (b) 1 (c) m r (d) 1 m2 r2 22
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 km h −1 (c) 8 3 km h–1 (d) 7 km h–1 13. A particle moves so that its position vector is given ^ ^ by r = cos wt x + sin wt 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 by R = 4 sin(2pt )i + 4 cos(2pt )j Where R is in meters, t is in seconds and i and 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. v2 (d) Magnitude of acceleration vector is , where R 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 j)m s 1 , ^ where, i is along the ground and j is along the vertical. Physics For you | August ‘16
21
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, 2 gh2 + 2 gh1 v − (−v1 ) v2 + v1 a= 2 = = t t t 2 × 10 × 20 + 2 × 10 × 5 20 + 10 = = = 1500 m s −2 0.02 0.02 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 1 1 = (3 × 2) − (1 × 2) + (1 × 2) = 4 m 2 2 4. (b) : Time taken by the body to strike the ground is 1 2h 2 × 1960 given by h = gt 2 or t = = = 20 s 2 9. 8 g 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 v −u v −0 v = = 5. (a) : Since, a = t n n Displacement in last 2 s, 1 1 sn − sn−2 = an2 − a(n − 2)2 2 2 v 2v(n − 1) = 2a(n − 1) = 2 × (n − 1) = n n 6. (b) : If v (in km h–1) is the constant speed of the trains then the distance between the successive trains x vx km =v× = 60 60 When a train moves in the same direction as that of vx / 60 20 the bus passenger, = (v − 30) 60 or vx = 20(v – 30) ...(i) When a train moves in a direction opposite to the bus passenger, vx / 60 10 or vx = 10(v + 30) ...(ii) = (v + 30) 60 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), 22
Physics For you | August ‘16
i.e., R = v02 sin 2q / g 2v sin q 2v sin(90° − q) 2v0 cos q = , As t1 = 0 and t2 = 0 g g g 4v 2 sin q cos q 2 v02 sin 2q 2R t1t2 = 0 = = g g g g2 8. (c) : Since, x = 3t2 – 6t \
vx =
At t = 1 s, vx = 6 × 1 – 6 = 0 Also, y = t2 – 2t \
vy =
At t = 1 s, vy = 2 × 1 – 2 = 0
dx = 6t − 6 dt
dy = 2t − 2 dt
2
2 9. (c) : a = r w = r (2pn / t )
=r×
4 p2n2 2
=
1 × 4 × p2 × (22)2 2
= p2 m s −2
(44) t 10. (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, 1 ...(i) 5 = g (2T )2 = 2 gT 2 2 Further, distance covered by the second drop in time T (from t = T to t = 2T), 1 ...(ii) y = gT 2 2 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 sin a = 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), 3 sin(30° + a) 12 × 3 or cot a = = 2 sin a 8×2 8 From (ii), vrain, road = = 4 7 km h −1 sin a 13. (a) : Given, r = cos wt x + sin wt y dr \ v= = −w sin wt x + w cos wt y dt dv a= = −w2 cos wt x − w2 sin wt y = −w2r dt 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, at2 + ar2 = 12 + 82 = 65 ≈ 8 m s −2 1 15. (a) : Using h = ut + at2 2 1 For first stone, y1 = 10t – gt2 2 1 For second stone, y2 = 40t – gt2 2 Relative position of the second stone with respect to 1 1 the first stone Dy = y2 – y1 = 40t – gt2 – 10t + gt2 2 2 = 30t Therefore, it will be a straight line. After 8 seconds, first stone reaches to the ground, i.e., y1 = –240 m 1 \ Dy = y2 – y1 = 40t – gt2 + 240 2 Therefore, it will be a parabolic curve till the second a=
stone reaches to the ground. 16. (c) : Speed v = 10 m s–1 We know, centripetal acceleration is given by, v2 a= r Q
→
| v | = constant
so, a ∝
1 or ar = constant r
This represents a rectangular hyperbola. 17. (a) 18. (d) : Time taken by the particle to reach the top most point is, u t= ... (i) g Time taken by the particle to reach the ground = nt Using, s = ut + 1 at 2 2 1 ⇒ − H = u(nt ) − g (nt )2 2 ⇒
2
u u 1 − H = u × n − gn2 [using (i)] g g 2
⇒ –2gH = 2nu2 – n2u2 ⇒ 2gH = nu2(n – 2) 19. (a) : The equation of trajectory is gx 2 y = x tan q − 2u2 cos2 q 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 g projection, = constant u2 9. 8 g ′ As per question, = 52 32 where g′ is acceleration due to gravity on the planet. 9.8 × 9 = 3.5 m s −2 g′ = 25 20. (c) : Given : u = i + 2 j As u = ux i + u y j \ ux = 1 and uy = 2 Also x = uxt 1 2 \ x = t and y = u y t − gt 2 .. . y = 2t − 1 × 10 × t 2 = 2t − 5t 2 2 Equation of trajectory is y = 2x – 5x2.
Physics For you | August ‘16
23
CLASS XI Series 2
CBSE Motion in a Plane Laws of Motion
Time Allowed : 3 hours Maximum Marks : 70
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.
section-A 1. Fifteen vectors, each of magnitude 5 units, are
2. 3.
4.
5.
represented by the sides of a closed polygon, all taken in same order. What will be their resultant? Can an object be accelerated without speeding up or slowing down? Explain. 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. 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°? 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 24
Physics For you | AUGUST ‘16
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 2 j + 4.0 k m where t is in seconds and the coefficients have the proper units for r to be in metres. (a) Find v and a of the particle? (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 y maximum height of the projectile. Calculate . 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
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
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.
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?
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. Physics For you | AUGUST ‘16
25
section-e
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., f F Mg − f = =g− a= M M M Net acceleration acting on the body of mass m, i.e., f a′ = g − 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
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.
Now R = A2 + B2 + 2 AB cos q \
or 28 = 9 x 2 + 25x 2 + 15x 2 = 7 x or x = 28/7 = 4 \ Forces are; A = 3 × 4 = 12 N and B = 5 × 4 = 20 N 7.
2.
2 2 −1 \ v = (3.0) + (−8) = 73 = 8.54 m s If q is the angle which makes with x-axis, then v y −8 tan q = = = −2.667 vx 3
26
Zero vector. The vector sum of all the vectors represented by the sides of a closed polygon taken in the same order is zero. 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. Physics For you | AUGUST ‘16
(a) Velocity, dr d v= = (3.0 t i − 2.0 t 2 j + 4.0 k ) dt dt = [3.0 i − 4.0 t j] m s −1 dv d a = = (3.0 i − 4.0 t j) Acceleration, dt dt = 0 − 4.0 j = − 4.0 j m s −2 (b) At time t = 2 s, v = 3.0 i − 4.0 × 2 j = 3.0i − 8.0 j
solutions 1.
28 = (3x )2 + (5x )2 + 2(3x )(5x )cos 60°
\ 8.
q = tan–1(–2.667) – 70° with x-axis
Here, range of projectile = 2 × maximum height
v 2 sin 2q v 2 sin2 q or 2v2 sin q cos q = v2 sin2q =2 g 2g or 2(v sin q)(v cos q) = (v sin q)(v sin q)
But v cos q = x and v sin q = y y \ 2yx = y2 or 2x = y or = 2 x
(given)
OR
m − m2 g m1 − m2 As a = 1 g \ = g 8 m1 + m2 m1 + m2 (m + m2 ) + (m1 − m2 ) 8 + 1 m1 − m2 1 = or 1 = or m1 + m2 8 (m1 + m2 ) − (m1 − m2 ) 8 − 1 m1 9 or = or m1 : m2 = 9 : 7 m2 7 9.
Suppose force FB is required in applying brakes to stop the truck in distance d. Then mv 2 v2 FB = ma = a = 2d 2d
The magnitude of the resultant velocity is 2 v = v R2 + vW = (35)2 + (12)2 = 37 m s −1 Let resultant velocity v makes an angle q with the vertical. AC vW Then, tan q = = = 0.343 OA v R \ q = tan −1(0.343) 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.
Suppose FT force is required in taking a turn of radius d. Then
mv 2 1 = 2FB or FB = FT d 2 Clearly, it is better to apply brakes than to take a circular turn. FT =
10. Let F be retarding force applied on the motor car.
So retarding acceleration, a =
F = constant m
Using kinematics equation, v2 = u2 + 2as
u2 2a Now speed of the car is doubled so distance covered by it will be s′ (say). u2 \ 0 = (2u)2 + 2(–a)s′ ⇒ s′ = 4 × = 4 s 2a 11. In figure, Velocity of rain, v R = OA = 35 m s −1, vertically downward Here v = 0, \ 0 = u2 + 2(–a)s ⇒ s =
Thus Ax , A y and Az are the three rectangular 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 = Ax + A y + Az = Ax i + A y j + Az k If a, b and g are the angles between A and X-, Yand Z-axes repectively, then Ax = A cos a, Ay = A cos b, Az = A cos g We note that OP 2 = OT 2 + TP 2 = OQ 2 + QT 2 + TP 2 or
A2 = Ax2 + A2y + Az2 or A = Ax2 + A2y + Az2
13. The scalar or dot product of
Velocity of wind, vW = OB = 12 m s −1 , east to west
two vector A and B is defined as the product of the magnitudes of A and B and cosine of the angle q between them. Thus Physics For you | AUGUST ‘16
27
A ⋅ B = A B cos q = AB cos q As A, B and cos q are all scalars, so the dot product of A and B is a scalar quantity. Both A and B have direction but their dot product has no direction. Geometrical interpretation of scalar product : As shown in figure (a), suppose two vectors A and B are represented by OP and OQ respectively and ∠POQ = q. A ⋅ B = AB cos q = A(B cos q) = A(OR ) = A × Magnitude of component of B in the direction of A
t2 t2 t J = ∫ Fav dt = Fav ∫ dt = Fav [t ]t2 = Fav (t2 − t1 ) t1
or
t1
1
J = Fav × ∆t , where ∆t = t2 − t1
Thus, the impulse of a force is equal to the product of the average force and the time interval for which it acts. Dimensions of impulse = [MLT–1] SI unit of impulse = kg m s–1 15. The horizontal range of a projectile projected at an
angle q with the horizontal velocity u is given by u2 sin 2q R= g Replacing q by (90° – q), we get
u2 sin 2(90° − q) u2 sin(180° − 2q) u2 sin 2q = = g g g i.e., R′ = R Hence for a given velocity of projection, a projectile has the same horizontal range of the angles of projection q and (90° – q). As shown in figure, the horizontal range is maximum of 45°. Clearly, R is same for q = 15° and 75° but less than Rm. Again R is same for q = 30° and 60°. R′ =
From figure (b), we have A ⋅ B = AB cos q = B( A cos q) = B(OS) = B × Magnitude of component of A in the direction of B. Thus the scalar product of two vectors is equal to the product of magnitude of one vector and the magnitude of component of other vector in the direction of first vector. 14. Impulse is defined as the product of the force and
the time for which it acts and is equal to the total change in momentum. Impulse = Force × time duration = Total change in momentum Impulse is a vector quantity denoted by J . Its direction is same as that of force. The impulse of a force is positive, negative or zero depending on the momentum of the body increases, decreases or remains unchanged. Suppose a force F acts for a small time dt. The impulse of the force is given by dJ = Fdt If we consider a finite interval of time from t1 to t2, then the impulse will be t2 J = ∫ dJ = ∫ Fdt t1
If Fav is the average force, then 28
Physics For you | AUGUST ‘16
Maximum height attained in first case, u2 sin2 q 2g Maximum height attained in second case, H1 =
H2 =
u2 sin2 (90° − q) u2 cos2 q = 2g 2g
Again, H (30°) u2 sin2 30° 2g = ⋅ 2 H (60°) 2g u sin2 60° =
(1 / 2)2
2
( 3 / 2)
=
1 3
16. Let the ball of mass m
moving along AO with initial speed u hits the bat PQ and is deflected by the batsman along OB (without change in the speed of the ball), such that ∠AOB = 45°.
Let ON be the normal to the bat such that ∠AON = ∠BON =
45° = 22.5° 2
The initial momentum of the ball can be resolved into two components (i) mu cos22.5° along NO and (ii) mu sin22.5° along PQ Also, the final momentum of the ball can be resolved into two components (i) mu cos22.5° along ON and (ii) mu sin22.5° along PQ The component of the momentum of the ball along PQ remains unchanged (both in magnitude and direction). However, the components of the momentum of the ball along ON are equal in magnitude but opposite in direction. Since the impulse imparted by the batsman to the ball is equal to the change in momentum of the ball along ON, Impulse = mucos22.5° – (–mucos22.5°) = 2 mucos22.5° Here, m = 0.15 kg ; u = 54 km h–1 = 15 m s–1 Therefore, impulse = 2 × 0.15 × 15 × cos 22.5° = 2 × 0.15 × 15 × 0.9239 = 4.16 kg m s–1
prevents skidding. If road is straight up then q is large then f is small. Due to this reason, mountain roads generally made winding upward rather than going straight up. 18. Let v be the velocity acquired by stone after falling
through a height of 50 m.
Clearly, v = 2 gh = 2 × 9.8 × 50 = 980 m s −1 Now v is the velocity with which the stone starts burying itself into the sand and finally comes to rest in it after travelling a distance of 1 m. From vf2 – vi2 = 2as, (as vf = 0, vi = v) 0 – v2 = 2as
v2 (neglecting negative sign) 2s 980 or a = (as s = 1 m) = 490 m s −2 2(1) Let F be the average resistance offered by the sand to the stone. Since F acts vertically upwards and mg acts vertically downwards, F – mg = ma or F = m(g + a) = (5)(9.8 + 490) = 2499 N Let t be the time taken by the stone to penetrate the sand. From vf = vi + at ; 0 = 31.30 + (−490)t 31.30 or t = = 0.06 s 490 19. Let us draw free-body diagrams for the weights A, B and C as shown in figure (a), (b) and (c). or
a=
17. (a) When porcelain objects are wrapped in paper
or straw, the time of impact between themselves is very much increased during jerk in transportation. F=
∆P ∆t
Hence, force on the porcelains is reduced during transportation, and saves them from breakage. (b) A small portion of the road is shown in the figure. Friction on the track, f = mN = mmgcosq As roads winding upwards, q is small, f is large, which
From figure (a), (b) and (c) T1 – mg = ma ...(i) T2 + mg – T1 = ma ...(ii) and mg – T2 = ma ...(iii) Adding equations (i), (ii) and (iii), g mg = 3ma or a = 3 From equation (i), T1 = mg + ma = mg + m(g/3) 4 = mg 3 Physics For you | AUGUST ‘16
29
From equation (iii), T2 = mg – ma = mg – m(g/3) 2 = mg 3 20. Here, m1 = 4 kg, m2 = 5 kg Force applied on block A = 12 N This force must atleast be equal to the kinetic friction applied on A by B. \ 12 = f k = mk R = mk m1 g or 12 = mk 4 g 12 3 = or mk = 4g g The block B is on a smooth surface. Hence to move A and B together, the force F that can be applied on B is equal to the frictional forces applied on A by B and applied on B by A. F = mk m1 g + mk m2 g = mk (m1 + m2)g 3 = (4 + 5) g = 27 N g As this force moves both the blocks together on a smooth table, so the acceleration produced is F 27 a= = = 3 m s −2 m1 + m2 4 + 5 OR
1 Here m = 4 kg, sin q = , g = 9.8 m s −2 3 (i) Various forces acting on the box are shown in figure. When the box just begins to slide, the forces are in equilibrium. f = mg sin q and R = mg cos q f mg sin q = = tan q R mg cos q 1 1 = = = 0.35 8 32 − 12
m=
(ii) When the block moves up the inclined plane, friction f acts down the plane. So minimum force needed to just move the box up the inclined plane F = mg sin q + f = mg sin q + mR [... R = mg cos q] = mg(sin q + m cos q) 1 1 8 = 4 × 9. 8 + ⋅ = 26.13 N 8 3 3 21. First law is contained in the second law : According
to Newton's second law of motion, the force acting on a body is given by F = ma
30
Physics For you | AUGUST ‘16
In the absence of any external force, F = 0 and ma = 0 As m ≠ 0, therefore, a = 0. Thus there is no acceleration when no force is applied. That is in the absence of any external force a body at rest will remain at rest and a body in uniform motion will continue to move uniformly along the same straight path. Hence first law of motion is contained in the second law. dp Third law is contained in the second law : Let B dt be the resulting change of momentum of B. Let FAB be the force (reaction) exerted by B on A. Let dpA be the resulting change of momentum of A. dt According to Newton's second law, dp dp FBA = B and FAB = A dt dt \ FBA + FAB = dpB + dpA = d ( p B + pA ) dt dt dt In the absence of external force, the rate of change d of momentum must be zero i.e., ( pB + pA ) must dt be zero. \ FBA + FAB = 0 or FBA = −FAB or Action = – Reaction Hence third law of motion is contained in the second. As both first and third laws of motion are contained in the second law, we can say that Newton's second law is the real law of motion. 22. In a closed cage, the inside air is bound with the cage.
(i) As the acceleration is zero, there is no change in the weight of the cage. (ii) In this cage, the reaction R is given by R – Mg = Ma or R = M(g + a) Thus the cage will appear heavier than before. (iii) In this case, the reaction R is given by Mg – R = Ma or R = M(g – a) Thus the cage will appear lighter than before.
23. (a)
Quest for knowledge (b) Concern for the child and scientific thinking. (c) The child experiences the velocity of the rain relative to himself. To protect himself from the rain, the child should hold umbrella in the direction of relative velocity of rain with respect to the child.
24. An acceleration which is directed along the radius
towards the centre of the circular path is called centripetal acceleration. Consider a particle moving on a circular path of radius r and centre O, with a uniform speed v as shown in figure (a). Suppose at time t the particle is at P and at time t + Dt, the particle is at Q. Let v1 and v2 be the velocity vectors at P and Q.
Applying triangle law of vector addition in DABC, AB + BC = AC \ BC = AC − AB = v 2 − v1 Thus the change in velocity in time Dt is given by BC = Dv Also OP = OQ = r, radius of the circle. i.e., AB = AC = v v1 = v2 = v And ∠POQ = Dq, ∠BAC = Dq Thus the two triangles POQ and BAC are similar. PQ BC Hence, = OP AB v or Ds = Dv or Dv = Ds r r v Dividing both sides by Dt, we get Dv v Ds = Dt r Dt Taking the limit Dt → 0 on both sides, we get Dv v Ds lim = lim r Dt →0 Dt Dt →0 Dt Dv dv But lim = = a, is the instantaneous acceleration dt Dt →0 Dt Ds ds and lim = = v , is the instantaneous velocity. Dt →0 Dt dt v v2 \ a = ⋅ v or a = = w2r [... v = wr] r r This gives the magnitude of the acceleration of the particle in uniform circular motion. Direction of acceleration : As Dt tends to zero, the
angle Dq also approaches zero. In this limit, as AB = AC, so ∠ABC = ∠ACB = 90°. Thus the change in velocity Dv and hence the acceleration a is perpendicular to the velocity vector v1. But v1 is directed along tangent at point P, so acceleration a acts along the radius towards the centre of the circle. OR
Any object launched in an arbitrary direction in space with an initial velocity and then allowed to move under the influence of gravity alone is called a projectile. Suppose a body is projected with initial velocity u, making an angle q with the horizontal. The velocity u has two rectangular components (i) The horizontal component u cos q, which remains constant throughout the motion. (ii) The vertical component u sin q, which changes with time under the effect of gravity. Y vy
A v Path of projectile v u P(x, y) x u sin y Maximum height = hm vx = u cos X O u cos B R v vy = u sin x
Equation of trajectory of a projectile : Suppose the body reaches the point P(x, y) after time t. \ The horizontal distance covered by the body in time t, x = Horizontal velocity × time = u cos q t x or t = u cos q For vertical motion : u = u sin q, a = – g, so the vertical distance covered in time t is given by 1 s = ut + at 2 2 x 1 x2 − g. or y = u sin q ⋅ u cos q 2 u2 cos2 q g x2 or y = x tan q − 2 2 2u cos q Thus y is a quadratic function of x. Hence the trajectory of a projectile is a parabola. Time of flight : It is the time taken by the projectile from the instant it is projected till it reaches a point Physics For you | AUGUST ‘16
31
in the horizontal plane of its projection. The body reaches the point B after the time of flight Tf. \ Net vertical displacement covered during the time of flight = 0 1 As s = ut + at 2 2 1 \ 0 = u sinqT f − gT f2 2 2 u sin q or T f = g Maximum height of a projectile : It is the maximum vertical distance attained by the projectile above the horizontal plane of projection. At the highest point A, vertical component of velocity = 0 As v2 – u2 = 2as u2 sin2 q \ 02 – (u sin q)2 = 2 (–g)hm or hm = 2g Horizontal range : It is the horizontal distance travelled by the projectile during its time of flight. So, Horizontal range = Horizontal velocity × time of flight or
R = u cos q ×
or
R=
2u sin q u2 = ⋅ 2 sin q cos q g g
u2 sin 2q g
[... 2 sin q cos q = sin 2 q]
25. The process of raising the outer edge of a curved
road above its inner edge through certain angle is called banking of the road. Consider a car of weight mg going along a curved path of radius r with speed v on a road banked at an angle q. The forces acting on the vehicle are 1. Weight mg acting vertically downwards 2. Normal reaction R of the road acting at an angle q with the vertical. 3. Force of friction f acting downwards along the inclined plane.
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Physics For you | AUGUST ‘16
Equating the forces along horizontal and vertical direction respectively, we get mv 2 ...(i) r mg + f sin q = R cos q, where f = mR or R cos q – f sin q = mg ...(ii) Dividing equation (i) by equation (ii), we get R sin q + f cos q =
R sin q + f cos q v 2 = R cos q − f sin q rg Dividing numerator and denominator of LHS by R cos q, we get f 2 tan q + 2 f R = v or tan q + m = v m = 1 − m tan q rg R f 1 − tan q rg R m + tan q m + tan q 2 or v = rg or v = rg . 1 − m tan q ...(iii) 1 − m tan q For m = 0 in equation (iii) v = (rg tan q)1/2 OR
(a) Consider a body of weight mg placed on an inclined plane. Suppose the angle of inclination q be greater than the angle of repose. Let a be the acceleration with which the body slides down the inclined plane. The weight mg has two rectangular components : (i) mg cos q perpendicular to the inclined plane. It balances the normal reaction R. Thus R = mg cos q (ii) mg sin q down the inclined plane. If f k is the kinetic friction, then the net force acting down the plane is F = mg sin q – f k But f k = mk R = mk mg cos q \ ma = mg sin q – mk mg cos q Hence a = g(sin q – mk cos q) (b) As shown in figure (a), suppose a force F is applied to pull a lawn roller of weight W. The force F has two rectangular components (i) Horizontal component F cos q helps to move the roller foraward. (ii) Vertical component F sin q acts in the upward direction. If R is the normal reaction, then R + F sin q = W or R = W – F sin q Force of kinetic friction, f k = mk R = mk(W – F sin q) ...(i)
F
20 × 5 × 10–3 × v = 10 × 10–3 × 9.8 10 × 9.8 v= = 0.98 m s–1 = 98 cm s–1 100
R
F cos
f k
F sin
F
W (b)
As shown in figure (b), if a force F is applied to push a roller of weight W, then the normal reaction is R′ = W + F sin q Force of kinetic friction, ...(ii) f k′ = mk R′ = mk (W + F sin q) Comparing (i) and (ii) we find that f k′ > f k i.e., the force of friction is more in case of push than in case of pull. So it is easier to pull a lawn roller than to push it. 26. (a) In the absence of any external force, the vector
sum of the linear momentum of a number of particles in an isolated system remains constant and remains unchanged by their mutual interaction. Consider an isolated system of n particles. Suppose the n particles have masses m1, m2, m3, ..., mn and are moving with velocities v1 , v2 , v3 ,..., vn , respectively. Then total linear momentum of the system is p = m1v1 + m2v2 + m3v3 + .... + mnvn = p1 + p2 + p3 + ... + pn If F is the external force acting on the system, then dp according to Newton's second law, F = dt For an isolated system, F = 0 or dp = 0 dt As the derivative of a constant is zero, so p = constant or p1 + p2 + p3 + .... + pn = constant Thus in the absence of any external force, the total linear momentum of the system is constant. (b) Mass of each marble piece, m = 5 g = 5 × 10–3 kg Number of marbles thrown per second = 10 Let velocity of impact of each marble = v Change in momentum of each marble = mv – (– mv) = 2 mv Change in momentum per second = 2 mv × 10 \ Force exerted by marbles on the disc = 20 mv But the disc can be kept floating if this force balances the weight of the disc. \ 20 mv = Mg
OR
(a) Laws of limiting friction : (i) The limiting friction depends on the nature of the surfaces in contact and their state of polish. (ii) The limiting friction acts tangential to the two surfaces in contact and in a direction opposite to the direction of motion of the body. (iii) The value of limiting friction is independent of the area of the surface in contact so long as the normal reaction remains the same. (iv) The limiting friction (fs)max is directly proportional to the normal reaction R between the two surfaces. i.e., ( f s )max ∝ R or ( f s )max = m s R ( f s )max Limiting friction = or m s = R Normal reaction The proportionality constant ms is called coefficient of static friction. It is defined as the ratio of limiting friction to the normal reaction. As f k < ( f s )max or mk R < m s R \ mk < m s Thus the coefficient of kinetic friction is less than the coefficient of static friction. (b) W is the weight of the body, R is the normal reaction, (fs)max is the limiting friction, F is the applied force and OC is the resultant of (fs)max and R. C
B
R A (fs)max
O
Applied force F
W
\
tan q =
BC OA ( f s )max = = OB OB R
( f s )max = m s = Coefficient of static friction R \ tan q = ms Thus the coefficient of static friction is equal to the tangent of the angle of friction. But
Physics For you | AUGUST ‘16
33
Class XI
T
his specially designed column enables students to self analyse their extent of understanding of specified chapters. Give yourself four marks for correct answer and deduct one mark for wrong answer. Self check table given at the end will help you to check your readiness.
Kinematics Total Marks : 120
Time Taken : 60 min
NEET / AiiMs / PMTs Only One Option Correct Type
1. Points P,Q and R are in a vertical line such that PQ = QR. A ball at P is allowed to fall freely. The ratio of times of descent through PQ and QR is (a) 1 : 1 (b) 1 : 2 (c) 1 : ( 2 − 1)
(d) 1 : ( 2 + 1)
2. Two particles A and B are moving in x-y plane. Particle A moves along a line with equation y = x while B moves along x-axis such that their x-coordinates are always equal. If B moves with a uniform speed 3 m s–1, the speed of A is 3 (a) 3 2 m s −1 (b) m s −1 2 1 (c) (d) 3 m s–1 m s −1 3 3. Two cars are moving in the same direction with a speed of 30 km h–1. They are separated from each other by 5 km. Third car moving in the opposite direction meets the two cars after an interval of 4 minutes. The speed of the third car is (a) 30 km h–1 (b) 35 km h–1 –1 (c) 40 km h (d) 45 km h–1 4. A particle is projected from the ground with an initial speed of u at an angle of projection q with horizontal. The average velocity of the particle between its time of projection and time it reaches highest point of trajectory is u u (a) (b) 1 + 2 cos2 q 1 + 2 sin2 q 2 2 34
Physics For you | AUGUST ‘16
(c) u 1 + 3 cos2 q (d) ucosq 2 5. A man running at a speed of 5 km h–1 finds that the rain falls vertically. When he stops running, he finds that the rain is falling at an angle of 60° with the horizontal. The velocity of rain with respect to running man is (a)
5 3
km h −1
(b)
5 3 km h −1 2
4 3 (d) 5 3 km h −1 km h −1 5 6. A certain vector in the x-y plane has an x component of 4 m and a y component of 10 m. It is then rotated in the x-y plane so that its x component is doubled. Then its new y component is (a) 20 m (b) 7.2 m (c) 5.0 m (d) 4.5 m (c)
7. A police party is moving in a jeep at a constant speed v. They saw a thief at a distance x on a motorcycle which is at rest. At the same moment the thief saw the police and he started at constant acceleration a. Which of the following relations is true, if the police is able to catch the thief? (a) v2 < ax (b) v2 < 2ax (c) v2 ≥ 2ax (d) v2 = ax 8. A car starts from rest and accelerates at uniform rate of 6 m s–2 for some time, then moves with constant speed for some time and retards at the same uniform rate and comes to rest. Total time for the journey is 24 s and average speed for journey is 20 m s–1.
How long does the car move with constant speed? (a) 4 s (b) 8 s (c) 12 s (d) 16 s 9. A particle P is moving in a circle of radius R with uniform speed v. C is the centre of the circle and AB is a diameter. The angular velocity of P about A and C are in the ratio (a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) 4 : 1 10. The distance travelled by a particle in time t is given by x = kt3, where k = 10 m s–3. The average speed of the particle from t = 1 s to t = 5 s is (a) 280 m s–1 (b) 300 m s–1 (c) 310 m s–1 (d) 320 m s–1 11. The maximum height attained by a projectile is increased by 1% by increasing its speed of projection without changing the angle of projection. Then the percentage increase in the horizontal range will be (a) 2% (b) 1% (c) 0.5% (d) 0.2% 12. Given below are four curves describing variation of velocity with time of a particle. Which one of these describe the motion of a particle initially in positive direction with constant negative acceleration?
(a) W
(b) X
(c) Y
13. Assertion : Two balls of different masses are thrown vertically upward with same speed. They will pass through their point of projection in the downward direction with the same speed. Reason : The maximum height and downward velocity attained at the point of projection are independent of the mass of the ball. 14. Assertion : A body can have acceleration even if its velocity is zero at a given instant of time. Reason : The body is momentarily at rest when it reverses it direction of motion. 15. Assertion : A particle in x-y plane is related by x = a sin wt and y = a(1 – cos wt), where a and w are constants, then the particle will have parabolic motion. Reason : A particle under the influence of two perpendicular velocities has parabolic motion. JEE MAiN / JEE AdvANcEd / PETs Only One Option Correct Type
16. A particle moves rectilinearly. Its displacement x at time t is given by x2 = at2 + b, where a and b are constants. Its acceleration at time t is proportional to 1 1 (a) 1 (b) − 3 x x2 x t 1 (c) (d) − 2 x2 x 17. Bamboo strips are hinged to form three rhombi as shown. Point A0 is fixed to a rigid support. The lengths of the sides of the rhombi are in the ratio 3 : 2 : 1. Point A3 is pulled with a speed v. Let vA1 and vA2 be the speeds with which points A1 and A2 move. Then, the ratio vA1 : vA2 is
(d) Z
Assertion & Reason Type
Directions : In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as : (a) If both assertion and reason are true and reason is the correct explanation of assertion. (b) If both assertion and reason are true but reason is not the correct explanation of assertion. (c) If assertion is true but reason is false. (d) If both assertion and reason are false.
(a) 2 : 3
(b) 3 : 5
(c) 3 : 2
(d) 5 : 3
18. A point P moves in counter-clockwise direction on a circular path as shown in the figure. The movement of P is such that it sweeps out a length s = t3 + 5, where s is in metres and t is in seconds. The radius of the path of 20 m. The acceleration of P when t = 2 s is nearly Physics For you | AUGUST ‘16
35
(b) 13 m s–2 (d) 7.2 m s–2
19. Two NCC cadets simultaneously aim their guns at a target put on a tower. The first cadet fired the bullet with a speed of 100 m s–1 at an angle of 30° with the horizontal. The second cadet, ahead of the first cadet by a distance of 50 m, fired the bullet with a speed of 80 m s–1 at an angle of q with the horizontal. Both the bullets hit the target simultaneously. Which of the following is correct? 5 5 (a) sin q = (b) cos q = 8 6 5 3 (c) tan q = (d) tan q = 3 5 More than One Options Correct Type
20. The distances covered by a freely falling body in its first, second, third, ... nth seconds of its motion (a) forms an arithmetic progression (b) form a series corresponding to the squares of the first n natural numbers (c) do not form any well defined series (d) forms a series corresponding to the difference of the squares of the successive natural numbers.
23. The magnitude of the vector product of two vectors A and B may be (a) greater than AB (b) equal to AB (c) less than AB (d) equal to zero Integer Answer Type
24. A racing car is travelling along a track at a constant speed of 40 m s–1. A TV cameraman is recording the event from a distance of 30 m directly away from the track as shown in figure. In order to keep the car under view, what should be angular speed of the camera (in rad s–1) when q is 30°? 40 m s–1
30 m
(a) 14 m s–2 (c) 12 m s–2
(b) the displacement of the bolt during the free fall relative to the elevator shaft is 0.75 m. (c) the distance covered by the bolt during the free fall relative to the elevator shaft is 1.38 m. (d) the distance covered by the bolt during the free fall relative to the elevator shaft is 2.52 m.
25. A particle moving in a straight line covers half the distance with speed 3 m s–1. The other half of the distance is covered in two equal intervals with speeds 4.5 m s–1 and 7.5 m s–1 respectively. The average speed of the particle (in m s–1) during this motion is
21. Velocity of a particle moving in a curvilinear path varies with time as v = (2t i + t 2 j) m s −1 . Here, t is in second. At t = 1 s, (a) acceleration of particle is 8 m s–2.
(b) tangential acceleration of particle is (c) radial acceleration of particle is
2 5
6 5
m s −2 .
m s −2.
5 5 m. 2 22. An elevator ascends with an upward acceleration of 2.0 m s–2. At the instant its upward speed is 2.5 m s–1, a loose bolt is dropped from the ceiling of the elevator 3.0 m from the floor. If g = 10 m s–2, then (a) the time of the flight of the bolt from the ceiling to floor of the elevator is 0.71 s. (d) radius of curvature to the path is
36
Physics For you | AUGUST ‘16
0124-6601200 for further assistance.
26. A particle is projected at an angle of 60° above the horizontal with a speed of 10 m s–1. After some time, the direction of its velocity makes an angle of 30° above the horizontal. The speed of the particle 2n m s −1 . What is the value of n? at this instant is 3 Comprehension Type
A balloon of mass 100 kg with a basket of mass 25 kg attached to it is descending with a uniform downward velocity 8 m s–1. A person of mass 50 kg is also standing in the basket. He throws a packet of mass 2 kg with an initial velocity 6 m s–1 relative to him and perpendicular to the motion of the descending balloon. The person observes that the packet hits the ground 4 s after being thrown. Assume that the balloon continues to move with the same uniform velocity vertically downward and g = 10 m s–2. 27. Height of the balloon when the packet hits the ground is nearly (a) 92 m (b) 72 m (c) 80 m (d) 112 m 28. Velocity of the packet 1.6 s after being thrown, relative to the person in the basket, has a magnitude nearly equal to (a) 17 m s–1 (b) 12 m s–1 (c) 21 m s–1 (d) 10 m s–1 Matrix Match Type
(B) A − C = B
(Q)
(C) B − A = C
(R)
(D) A + B + C = 0
(S)
(a) (b) (c) (d)
A P R S Q
B R Q R P
C Q S P R
D S P Q S
30. Four balls of same masses are projected with the equal speeds at angles 1x5°, 30°, 60° and 75° with the horizontal direction from the ground. Match the quantities of Column I with the angle of projection given in Column II. Column I Column II
29. Given below in Column I are the relations between vectors A, B and C and in Column II are the orientations of A, B and C in the x-y plane. Match the relations in Column I to correct orientations in Column II. Column I Column II (P) (A) A + B = C
(A) Horizontal range is (P) maximum for angle (B) The vertical height is (Q) maximum for angle (C) The time of flight is (R) minimum for angle (D) The horizontal range is (S) minimum for angle A B C D (a) P, R P S Q, R (b) Q, R S P P, S (c) R, S Q R P, Q (d) P, S R Q R, S
15° 30° 60° 75°
Keys are published in this issue. Search now! J
Check your score! If your score is > 90%
ExcEllENT work !
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No. of questions attempted
……
90-75%
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No. of questions correct
……
74-60%
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Physics For you | AUGUST ‘16
37
F Electric current and current Density •
•
Electric current is the rate of flow of electric charge through a cross section of a conductor dq I= dt
If A is not normal to I but makes an angle q with the normal to current, then I = JA cos q J=
• • • •
•
•
38
For steady current, I =
q t
DQ For average current, Iav = Dt Electric current is a scalar quantity because it does not obey the law of vectors. Its SI unit is ampere (A) or coulomb per second (C s–1) One ampere of current means the flow of 6.25 × 1018 electrons per second through any cross section of conductor. Current density at a point in a conductor is the ratio of the current at that point in the conductor to the area of cross section of the conductor of that point, dI J= n or, I = ∫ J ⋅ dA dA Physics For you | AUGUST ‘16
•
I A cos q
Current density is a vector quantity and its SI unit is A m–2. If a steady current flows in a metallic conductor of non uniform cross section. f
f
Along the wire, current is same, I1 = I2 Current density depends inversely on area, so J1 > J2, as A1 < A2
F Thermal Velocity and Drift Velocity •
• •
Thermal velocities of free electrons are randomly distributed in all possible direction of a metallic conductor. Average thermal velocity is zero but average thermal speed is non zero. Drift velocity is defined as the velocity with which
the free electrons get drifted towards the positive terminal under the effect of the applied electric field. eE f vd = a t ⇒ vd = − t m f Order of drift velocity is 10–4 m s–1. f t is the relaxation time which is average time elapsed between two successive collisions. It decreases with increase in temperature.
•
The first two coloured rings from the left end indicate the first two significant figures of the resistance in ohms. The third colour ring indicates the decimal multiplier and the last colour ring stands for the tolerance in percent.
•
The colour code of a resistor is as shown in the table.
F relation between current Density, conductivity
Colour Number Multiplier Black 0 100 Brown 1 101 Red 2 102 Orange 3 103 Yellow 4 104 Green 5 105 Blue 6 106 Violet 7 107 Gray 8 108 White 9 109 Gold 10–1 Silver 10–2 No colour
and Electric Field
•
Let the number of free electrons per unit volume in a conductor = n Total number of electrons in dx distance = n(Adx) Total charge dQ = n(Adx)e Cross sectional area = A
• •
dQ dx = nAe ⇒ I = neAvd dt dt I Current density J = = nevd ⇒ J = ne eE t A m ne 2 t J = E ⇒ J = sE m Current I =
ne 2 t m In vector form J = sE (Another form of Ohm's law) s depends only on the material of the conductor and its temperature. Conductivity s =
•
I 1V rl = ⇒ V = I = RI A r l A At a given temperature, current is directly proportional to the applied potential difference. The substances which obey ohm's law are called ohmic.
Ohm's law, J = sE ⇒ f
f
F colour code of resistors •
A colour code is used to indicate the resistance value and its percentage accuracy.
Tolerance
5% 10% 20%
F resistance and its combinations •
•
When a potential difference is applied across a conductor, free electrons get accelerated and collide with positive ions and their motion is thus opposed. This opposition offered by the ions is called resistance of the conductor. Resistance is the property of a conductor by virtue of which it opposes the flow of current in it.
rl rl 2 rV = = A V A2 Resistance of the conductor depends on the temperature and varies as Rt = R 0 (1 + a Dt), where Rt = Resistance at t °C, R0 = resistance at 0 °C, Dt = change in temperature, a = temperature coefficient of resistance. For metal : a is positive and for semiconductors and insulators : a is negative. Resistance of the conductor decreases linearly with decrease in temperature and becomes zero at R=
•
• •
Physics For you | AUGUST ‘16
39
a specific temperature. This temperature is called critical temperature, at this temperature conductor becomes a superconductor.
f
– Resistance between square faces distance between faces a RAB = r =r area of square b2
• •
•
40
If a wire is stretched to n times of its original length, its new resistance will be n2 times. If a wire is stretched such that its radius is reduced 1 to of its original values, then resistance nth will increase n 4 times similarly resistance will decrease n4 time if radius is increased n times by contraction. Resistance of different shaped conductors.
Physics For you | AUGUST ‘16
– Resistance between rectangular faces b r (does not depend on b) RXY = r = a ⋅b a • A frame of cube is made with wires each of equal resistance r then f Resistance between two nearer 7 corners R12 = r 12 3 f Resistance across face diagonal R13 = r 4 5 f Resistance across main diagonal R17 = r 6 EMF ( e ) and Terminal Voltage ( V ) F • The potential difference across the terminals of a cell when it is not producing any current is called emf of the cell. • The energy given by the cell in the flow of unit charge in the whole circuit (including the cell) is called the emf of the cell. • emf depends on f nature of electrolyte f metal of electrodes • emf does not depend on f area of plates f distance between the electrodes f quantity of eletrolyte f size of cell • When current drawn through the cell or current is supplied to cell, the potential difference across its terminals = terminal voltage. f When current I is drawn from the cell, then terminal voltage is less than its emf. V = e – Ir
At the time of charging a cell, current is supplied to the cell, the terminal voltage is greater than emf e. V = e + Ir Efficiency of cell is defined by Total potential difference η= × 100% emf
f
•
•
f
f
f
Internal resistance of a cell
f
f
f
F Kirchhoff's Laws •
First law (Kirchhoff 's current law) f In an electric circuit, the algebraic sum of the current meeting at any junction in the circuit is zero.
f
•
– V4
+ V2 –
+
From the figure, I1 – I2 – I3 – I4 + I5 = 0 I1 + I5 = I2 + I3 + I4
+ V3 – f
This is based on law of conservation of charge.
Second law (Kirchhoff 's voltage law) f The algebraic sum of all the potential differences along a closed loop is zero. SIR + SEMF = 0
The closed loop can be traversed in any direction. While traversing a loop if potential increases, put a positive sign in expression and if potential decreases put a negative sign. + V1 –
∑I = 0 f
Resistance offered by the electrolyte of the cell when the electric current flows through it is known as internal resistance. Distance between two electrodes increases ⇒ r increases Area dipped in electrolyte increases ⇒ r decreases Concentration of electrolyte increases ⇒ r decreases Temperature increases ⇒ r decreases
f
From the figure, –V1 – V2 + V3 – V4 = 0. Boxes may contain resistor or battery or any other element (linear or nonlinear). This law is based on conservation of energy.
F Wheatstone Bridge •
It consists of four resistances R1, R2, R3 and R4 which are connected to form a quadrilateral ABCD as shown in figure. Physics For you | AUGUST ‘16
41
f
The experiment should be repeated by interchanging the positions of X and R to minimize an error due to contact resistance.
F Potentiometer •
• •
•
In a balanced condition even though current flows in the rest of the circuit, galvanometer will not show any deflection (i.e. Ig = 0). R R Also, 1 = 3 R2 R4 The measurement of resistance by Wheatstone's bridge is not affected by the internal resistance of the cell.
F Meter-Bridge •
•
•
Meterbridge is the modification of Wheatstone's network used to determine unknown resistance. It is an instrument which works on the principle of Wheatstone's network. So it is also called Wheatstone's meterbridge. The length of wire used is one metre, so it is called meterbridge.
From the balancing condition, X Resistance of wire AD of length lx = R Resistance of wire DC of length lr lx X = R 100 − lx
•
42
Minimization of errors in measurement of X f The wire used must be uniform i.e., of same cross section. f The value of R should be so chosen that the null point is obtained as near (close) to the centre of wire as possible. Physics For you | AUGUST ‘16
Primary circuit contains constant source of voltage, rheostat of resistance box and resistance of the potentiometer of wire. f Secondary circuit contains unknown circuit with galvanometer. Let r = Resistance per unit length of potentiometer wire Application of Potentiometer f Comparision of emf of two cells (i) plug only in (1 –2) Jockey is at position J balancing length AJ = l1 e1 = kl1 (ii) Plug only in (2 –3) Jockey is at position J′ balancing length AJ′ = l2 e2 = kl2 e1 l1 = \ e2 l2 f
•
•
Potentiometer is an ideal voltmeter because it draws no current from the circuit at the instant of measurement. Working : It is based on the fact that the fall of potential across any portion of the wire is directly proportional to the length of that portion provided the wire is of uniform area of cross-section and a constant current is flowing through it. i.e., V ∝ l (If I and A are constants.) or V = kl where k is known as potential gradient i.e., fall of potential per unit length of the given wire. Circuit of potentiometer
f
Internal resistance of given primary cell e −V e = V + Ir ⇒ r = I e −V or r = R V
Key K open e = kl1 (AJ = l1) Key K closed total potential difference V = kl2 (AJ′ = l2) l1 − l2 r = R l2
• •
1 hp = 746 watt. If one heater boils a certain mass of water in time t1 and another heater boils the same mass of water in time t2, then both the heaters are connected in series, the same water will boil in time (t1 + t2); when both the heaters are connected in parallel the same water will boil in time t =
•
• f
f
f f
The emf of battery (driver cell or auxiliary cell) must be greater than emf 's to be compared i.e., e > e1, e > e2 and for the combination method e > (e1 + e2)... The positive terminal of e 1 or e 2 or of the combination must be connected to that end of potentiometer wire where positive terminal of the battery (driving cell) is connected. The potentiometer wire must be uniform. The resistance of potentiometer wire should be high.
F Joule's Law of heating •
According to Joule’s heating effect of current, the amount of heat produced H in a conductor of resistance R, carrying current I for time t is H = I2Rt (in joule) or H =
I 2 Rt (in calorie) J
t1 × t2 . t1 + t2
If P1, P2, P3 .... are the powers of electric appliances in series with source of rated voltage V, the effective power consumed is 1 1 1 1 = + + + ...... PS P1 P2 P3
If P1, P2, P3 are the powers of electric appliances in parallel with a source of rated voltage V, the effective power consumed is PP = P1 + P2 + P3 + .....
F charging and Discharging of capacitors •
Charging a capacitor Charge on the capacitor at any time t,
f
f
f f f f f
q = CV(1– e– t/RC) Current in the circuit at any time t, V I = e(−t /RC ) R Maximum charge on capacitor, q0 = CV. Maximum current in the circuit, I0 = V/R. RC is known as time constant (tC) of circuit. VC + VR = V ...(i) Charging means q increases. So if q increase then VC increases so from equation (i) V R decreases and hence I decreases.
where J is Joule’s mechanical equivalent of heat (= 4.2 J cal–1).
F Electric Power •
It is defined as the rate at which work is done by the source of emf in maintaining the current in the electric circuit. Electric power P = P = VI = I 2 R =
• • •
electric work done time taken
V2 . R
The SI unit of power is watt (W). The practical unit of power is kilowatt (kW) and horse power (hp). 1 kilowatt = 1000 watt. Physics For you | AUGUST ‘16
43
•
Discharging a capacitor f Charge on the capacitor at any time t, q = q0 e–t/RC ;
q0 = CV = initial charge on the capacitor at t = 0 f
Current during discharging I=
dq = − I0e −t /RC dt
I0 = f
f
V = initial current at t = 0 R
VC + VR = 0 |VC| = |VR| ...(i) Discharging means q decreases so VC decreases. Hence from eqn (i) |VR| decreases and correspondingly I decreases.
In discharing all the parameters (q, VC, VR, I) decay exponentially. In both charging and discharging current decreases exponentially i.e. capacitor blocks D.C. In time t = RC = tC (time constant) any parameter changes by 63%. If time constant (tC) of a circuit is very less, then initial changes in any parameter (q, VC, VR, I) is so rapid.
f
• • •
1. An electron moves in a circle of radius 10 cm with a constant speed of 4.0 × 106 m s–1. The electric current at a point on the circle will be (a) 2.0 × 10–12 A (b) 1.0 × 10–12 A (c) 3.2 × 10–12 A (d) 6.4 × 10–12 A 2. A metal rod of length 10 cm and a rectangular 1 cross-section of 1 cm × cm is connected to a 2 battery across opposite faces. The resistance will be (a) maximum when the battery is connected 1 across 1 cm × cm faces. 2 (b) maximum when the battery is connected across 10 cm × 1 cm faces. (c) maximum when the battery is connected 1 across 10 cm × cm faces. 2 (d) same irrespective of the three faces. 3. Two batteries of emf 4 V and 8 V with internal resistance 1 W and 2 W are connected in a circuit with resistance of 9 W as shown in figure. The current and potential difference between the points P and Q are 44
Physics For you | AUGUST ‘16
P
1
8V
4V
2 r2
r1
Q
9 R
(a)
1 A and 3 V 3
(b) 1 A and 4 V
(b)
1 A and 9 V 9
(d)
6
1 A and 12 V 12
4. n batteries are connected to form a circuit as shown in the figure. The resistances denote the internal resistances of the batteries which are related to the emf ’s as ri = kei where k is a constant with proper SI units. The solid dots represent the terminals of the batteries. Then 1
r1 2
r2 3
r3
n
rn
n . k (b) the potential difference between the terminals (a) the current through the circuit is of the ith battery is zero.
n2 . k (d) the potential difference between the terminals e of the ith battery is . k 5. Two non-ideal batteries are connected in parallel. Consider the following statements : (A) The equivalent emf is smaller than either of the two emfs. (B) The equivalent internal resistance is smaller than either of the two internal resistances. Which one is correct regarding the given statements? (a) Both A and B are correct. (b) A is correct but B is wrong. (c) B is correct but A is wrong. (d) Both A and B are wrong. (c) the current through the circuit is
6. A metal wire of diameter 2 mm and length 100 m has a resistance of 0.5475 W at 20 °C and 0.805 W at 150 °C. Then (a) the temperature coefficient of resistance is 3.6 × 10–3 °C–1 (b) resistance at 0°C is 0.5107 W (c) resistivities at 20°C is 1.72 × 10–8 W m (d) All of above 7. The current in a conductor varies with time t as I = 2t + 3t2, where I is in ampere and t in seconds. Electric charge flowing through a section of the conductor during t =2 s to t = 3 s is (a) 10 C
(b) 24 C
(c) 33 C
9. A network of resistors of resistances R1 and R2 extends off to infinity to the right as shown in the figure. R1 C R2 B
R1 D
R1
R2 R1
R1
R2 R1
(a)
R1 + R12 + 2R1R2
(c)
R2 + R22 + 2R1R2 (d)
(b) R1 − R12 + R1R2 R2 − R22 + R1R2
10. The equivalent resistance (RAB) between the point A and B is C
2
4 3
A
B 12
6 D
(a) 6 W (c) 4.5 W
(b) 7.5 W (d) 8 W
11. A fully charged capacitor C discharges through a resistor R. After how many time constants does the stored energy drop to half its initial value? (a) 0.20 (b) 0.02 (c) 0.35 (d) 0.16 12. Refer to the circuit shown in the figure. 3 A 2 9V
8 2 D 2
(d) 44 C
8. I-V characteristics of a copper wire of length L and area of cross-section A is shown in figure. The slope of the curve becomes (a) more if the experiment is performed at higher I temperature (b) more if a wire of steel of V same dimension is used (c) more if the length of the wire is increased (d) less if the length of the wire is increased
A
The total resistance of the network between points A and B is equal to
The (a) (b) (c) (d)
B 2 8
4
C 2
current through the 3 W resistor is 0.50 A 3 W resistor is 0.25 A 4 W resistor is 0.50 A 4 W resistor is 0.25 A
13. In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be (a) 14 A (b) 8 A (c) 10 A (d) 12 A [JEE Main 2014] 14. Two metal wires of identical dimensions are connected in series. If s1 and s2 are the conductivities of the metal wires respectively, the effective conductivity of the combination is Physics For you | AUGUST ‘16
45
(a)
s1 + s2 s1s2
(b)
(c)
2s1s2 s1 + s2
(d)
s1s2 s1 + s2 s1 + s2 2s1s2 [AIPMT 2015]
15. A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. E0 and a resistance r1. An unknown e.m.f. E is balanced at a length l of the potentiometer wire. The e.m.f. E will be given by (a)
E0 l L
(b)
LE0r (r + r1 )l
(c)
LE0r lr1
(d)
E0 r l . (r + r1 ) L [AIPMT 2015]
16. When 5 V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 × 10–4 m s–1. If the electron density in the wire is 8 × 1028 m–3, the resistivity of the material is close to (a) 1.6 × 10–6 W m (b) 1.6 × 10–5 W m (c) 1.6 × 10–8 W m (d) 1.6 × 10–7 W m [JEE Main 2015] 17. In the circuit shown, the resistance r is a variable resistance. If for r = f R, the heat generation in r is maximum then the value of f is (a) (c)
1 2 1 4
(b) 1 3
(d) 4 [JEE Main Online 2016]
18. The charge flowing through a resistance R varies with time t as Q = at – bt2, where a and b are positive constants. The total heat produced in R is (a) (c)
a3 R 2b a3 R 6b
3 (b) a R b
(d)
a3 R 3b
[NEET Phase-1 2016]
19. Two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. 48
Physics For you | AUGUST ‘16
The fall of potential per km is 8 V and the average resistance per km is 0.5 W. The power loss in the wire is (a) 19.2 W (b) 19.2 kW (c) 19.2 J (d) 12.2 kW [AIPMT 2014] 20. The resistances in the two arms of the meter 5 R bridge are 5 W and R W respectively. When the G resistance R is shunted with B an equal resistance, the new A l 100 – l1 1 balance point is at 1.6l1. The resistance R is (a) 10 W (b) 15 W (c) 20 W (d) 25 W [AIPMT 2014] SolutionS 1. (b) : Consider a point A on the circle. The electron crosses this point once in every revolution. In one revolution, the electron travels 2p × (10 cm) distance. Hence, the number of revolutions made by the electron in one second is 4.0 × 106 2 = × 107. −2 p 20p × 10 The charge crossing the point A per second is 2 3. 2 × 107 × 1.6 × 10−19 C = × 10−12 C. p p Thus, the electric current at this point is 3. 2 × 10−12 A ≈ 1.0 × 10−12 A p l , resistance will be maximum A when l is maximum and A is least. This holds good
2. (a) : Since R = r
for option (a) only. 3. (a) : I =
8−4 4 1 = = A 1 + 2 + 9 12 3
1 VP − VQ = 4 − × 3 = 3 V 3 4. (b) : Suppose the current is I in the indicated direction. Applying Kirchoff ’s loop law, e1 – Ir1 + e2 – Ir2 + e3 – Ir3 + ... + en – Irn = 0 or,
I=
e1 + e2 + e3 + ... + en r1 + r2 + r3 + ...rn
A
e + e + e + ... + en 1 = 1 2 3 = k(e1 + e2 + e3 + ...en ) k The potential difference between the terminals of the ith battery is 1 ei – Iri = ei − (kei ) = 0 k 5. (c) 6. (d) : Here r = 1 mm = 10–3 m, l = 100 m, t1 = 20 °C, R1 = 0.5475 W, t2 = 150 °C, R2 = 0.805 W Temperature coefficient of resistance is R − R1 0.805 − 0.5475 = = 3.6 × 10− 3°C −1 . a= 2 R1 (t 2 − t1 ) 0.5475(150 − 20)
Resistance at 0°C, R1 0.5475 0.5475 = = = 0.5107 W R0 = −3 1 + at1 1 + 3.6 × 10 × 20 1.072 Resistivity at 0°C, R A R × pr 2 0.5107 × 3.14 × (10−3 )2 = r0 = 0 = 0 100 l l
= 1.60 ×10–8 W m Resistivity at 20° C, r20 = r0 (1 + a t) = 1.60 × 10–8 (1 + 3.6 × 10–3 × 20) = 1.60 × 10–8 × 1.072 = 1.72 × 10–8 W m
7. (b) : Given I = 2t + 3t2 dq As I = \ dq = Idt = (2t + 3t2)dt dt Charge passed from t = 2 s to t = 3 s is 3
q = ∫ dq = ∫ (2t + 3t 2 )dt = [t 2 ]32 + [t 3 ]32 2
= (9 – 4) + (27 – 8) = 24 C DI 1 = DV R If the experiment is performed at a higher temperature, the resistance R of copper increases and hence slope decreases, so the option (a) is wrong. In options (b) and (c), the resistances increase and so their slope become less. In option (d), the resistance R increases and so slope decreases. Hence only option (d) is correct.
8. (d) : Slope of I – V graph =
9. (a) : Since the network is infinite, the resistance of the network to the right of points C and D is also equal to RT (net resistance). We can redraw the circuit as show in figure.
RT
B
Clearly, RT = 2R1 +
A R1 C R2 B
R2 RT R2 + RT
or
RT2 − 2R1RT − 2R1R2 = 0
or
RT = R1 ± R12 + 2R1R2
RT
R1 D
[ R2 || RT ]
Since RT > 0, RT = R1 + R12 + 2R1R2 2W 6W = 4 W 12 W Hence, the Wheatstone bridge is balanced. RACB = 2 + 4 = 6 W RADB = 6 + 12 = 18 W 6 × 18 RAB = = 4. 5 W 6 + 18 11. (c) : In discharging the capacitor the charge on the capacitor varies as q = q0e–t/t The energy stored in the capacitor is q 2 q02 −2t /t U= = e = U 0 e −2t /t 2C 2C in which U0 is the initial stored energy. The time at 1 which U = U 0 is found from 2 1 U 0 = U 0e −2t /t or 2 = e 2t /t 2 ln 2 ln 2 = 2t/t or t =t = 0.35t 2 8×8 12. (d) : RBC (right hand side) = W 8+8 = 4 W (as 2 W + 4 W + 2 W = 8 W) RAD (right hand side) is again 4 W. Equivalent resistance of the circuit, R=3W+4W+2W=9W V 9 Current drawn from battery I = = = 1 A R 9 At A, I is equally divided (I/2) between 8 W resistance and the remaining circuit of 8 W. At B, (I/2) is equally divided (I/4) between the 8 W resistor and the remaining circuit of resistance 8 W. Thus, current through 4 W resistor is I/4, i.e., 0.25 A. 10. (c) : As
13. (d) : Power of 15 bulbs of 40 W = 15 × 40 = 600 W Power of 5 bulbs of 100 W = 5 × 100 = 500 W Power of 5 fans of 80 W = 5 × 80 = 400 W Physics For you | AUGUST ‘16
49
Power of 1 heater of 1 kW = 1000 W \ Total power, P = 600 + 500 + 400 + 1000 = 2500 W When these combination of bulbs, fans and heater are connected to 220 V mains, current in the main fuse of building is given by P 2500 I= = = 11.36 A ≈12 A V 220 14. (c) : As both metal wires are of identical dimensions, so their length and area of cross-section will be same. Let them be l and A respectively. Then The resistance of the first wire is l ... (i) R1 = s1 A and that of the second wire is l ... (ii) R2 = s2 A
The current through the potentiometer wire is E0 I= (r + r1 ) and the potential difference across the wire is Er V = Ir = 0 (r + r1 ) The potential gradient along the potentiometer wire is E0 r V k= = L (r + r1 )L As the unknown e.m.f. E is balanced against length l of the potentiometer wire, Er l \ E = kl = 0 (r + r1 ) L 16. (b) : V = IR As I = neAvd and R = \
As they are connected in series, so their effective resistance is Rs = R1 + R2 =
l s1 A
+
l s2 A
... (iii)
If seff is the effective conductivity of the combination, then 2l ... (iv) Rs = seff A Equating eqns. (iii) and (iv), we get 1 l 1 = + seff A A s1 s2 2l
s + s1 2 2s s or s eff = 1 2 = 2 s eff s1s 2 s1 + s 2
15. (d) :
E0
r1
rl A
or r =
= 0.156 × 10–4 W m 1.6 × 10–5 W m 17. (a) : Let the source voltage be V. Equivalent resistance of the circuit when r = fR Req = R +
\
fR ( 2 f + 1)R r×R =R+ = f +1 ( f + 1) r+R
Current in the circuit, I = V = V ( f + 1) Req
I2 =
I V = f + 1 R( 2 f + 1)
Now, heat generated per unit time in r H = I 22 r =
V2 f R( 2 f + 1)2 df
L
⇒
50
G
Physics For you | AUGUST ‘16
R( 2 f + 1)
Current in the resistance r(= fR)
For maximum H , dH = 0
l E
V ne vd l
Here, V = 5 V, n = 8 × 1028 m–3, vd = 2.5 × 10 –4 m s–1, l = 0.1 m, e = 1.6 × 10–19 C 5 \ r= 28 −19 8 × 10 × 1.6 × 10 × 2.5 × 10 −4 × 0.1
(using (i) and (ii))
1 l 1 = + A s1 s2
V = neAvd ×
rl A
or
4f 1 =0 − 2 3 ( 2 f + 1) ( 2 f + 1) 1 2f + 1 – 4f = 0 ⇒ f = 2
V2 R
18. (c) : Given, Q = at – bt2 dQ \ I= = a – 2bt dt At t = 0, Q = 0 ⇒ I = 0 Also, I = 0 at t = a/2b \ Total heat produced in resistance R, H=
a /2b
∫
2
I Rdt = R
0
= R
a /2b
∫
20. (b) : In the first case, At balance point l1 5 = ...(i) R 100 − l1 () 5
2
(a − 2bt ) dt
0
a /2b
∫ 0
2
R
G
2 2
(a + 4b t − 4abt )dt
A
a /2b
l1
3 2 = R a2t + 4b2 t − 4ab t 3 2 0
In the second case, At balance point
a 4b2 a3 4ab a2 = R a 2 × + × − × 2b 3 8b3 2 4b2
1.6l1 5 = (R / 2) 100 − 1.6l1
a3 R 1 1 1 a3 R + − = b 2 6 2 6b 19. (b) : Here, Distance between two cities = 150 km Resistance of the wire, R = (0.5 W km–1)(150 km) = 75 W Voltage drop across the wire, V = (8 V km–1)(150 km) = 1200 V Power loss in the wire is 2 V 2 (1200 V) P= = = 19200 W = 19.2 kW 75 W R =
Mathematics Today 2015
()
R R
5
...(ii)
Divide eqn. (i) by eqn. (ii), we get 1 100 − 1.6l1 = 2 1.6(100 − l1 )
B
100 – l1
G A
1.6l1
100 –1.6l1
B
160 – 1.6l1 = 200 – 3.2l1 40 1.6l1 = 40 or l1 = = 25 cm 1. 6 Substituting this value in eqn. (i), we get 5 25 = R 75
or
R=
375 W = 15 W 25
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51
CLASS XII Series 3
CBSE
Moving Charges and Magnetism Magnetism and Matter
Time Allowed : 3 hours Maximum Marks : 70
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.
Previous Years Analysis 2016 2015 2014 Delhi AI Delhi AI Delhi AI VSA SA-I SA-II VBQ LA
1 –
1 –
– –
– –
2 1
1 –
2 _
1 _ _
1 _
1 _
– _
1
1
1 _ _
–
1
section-A
section-B
1. Give two points to compare the magnetic properties of steel and soft iron.
6. Do magnetic forces obey Newton’s third law. Verify ^ for two current elements dl1 = dl i located at the ^ origin and dl2 = dl j located at (0, R, 0). Both carry current I. 7. A long wire is bent into a circular coil of one turn and then into a circular coil of smaller radius having n turns. If the same current passes in both the cases, find the ratio of the magnetic fields produced at the centres in the two cases.
2. Two identical charged particles moving with the same speed enter a region of uniform magnetic field. If one of them enters normal to the field direction and the other enters in a direction at 30° with the field, what would be the ratio of their angular frequencies? 3. What is the function of radial magnetic field in a moving coil galvanometer? 4. A solenoid coil of 300 turns m–1 is carrying a current of 5 A. The length of the solenoid is 0.5 m and has a radius of 1 cm. Find the magnitude of the magnetic field inside the solenoid. 5. In a hydrogen atom, the electron moves in an orbit of radius 0.5 Å, making 1016 rps. Calculate the magnetic moment associated with the relative motion of electron. 52
Physics For you | AUGUST ‘16
8. A magnetising field of 1500 A m–1 produces a magnetic flux of 2.4 × 10–5 Wb in a bar of iron of cross section 0.5 cm2. Calculate permeability and susceptibility of the iron bar used. OR Out of the two magnetic materials, A has relative permeability slightly greater than unity while B has less than unity. Identify the nature of the materials A and B. Will their susceptibilities be positive or negative?
9. Using the relation for potential energy of a current carrying planar loop, in a uniform magnetic field, obtain the expression for the work done in moving the planar loop from its unstable equilibrium position to stable equilibrium position. 10. A beam of alpha particles and of protons having the same velocity enters a uniform magnetic field B at right angles to the direction of field. The particles describe circular paths. What is the ratio of the radii of circular paths described by alpha particles and protons. section-c 11. Define current sensitivity and voltage sensitivity of a galvanometer. Increase in the current sensitivity may not necessarily increase the voltage sensitivity of a galvanometer. Justify. 12. The following figure shows the variation of intensity of magnetisation versus the applied magnetic field intensity, H, for two magnetic materials A and B. (a) Identify the materials A and B. (b) For the material A, plot the variation of intensity of magnetisation versus temperature. 13. A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45° with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east. (a) Determine the horizontal component of the Earth's magnetic field at the location. (b) The current in the coil is reversed , and the coil is rotated about its vertical axis by an angle of 90° in the anticlockwise sense looking from the above. Predict the direction of the needle. Take the magnetic declination at the places to be zero. 14. The following figure shows a part of an electric circuit. PQRS is a rectangular loop of uniform conducting wire. The lengths PQ and SR are quite long but PS = QR = 2.0 cm. If a current of 10 A is recorded by the ammeter A, find the magnetic force per unit length between the wires PQ and SR.
Is the magnetic force repulsive or attractive ?
15. An electron moves around the nucleus in a hydrogen atom of radius 0.51 Å, with a velocity of 2.0 × 105 m s–1. Calculate (a) the equivalent current due to orbital motion of electrons, (b) the magnetic field produced at the centre of the nucleus, and (c) the magnetic moment associated with the electrons. 16. A galvanometer having 30 divisions has a current sensitivity of 20 mA per division. It has a resistance of 20 W. How will you convert it into an ammeter measuring upto 1 ampere? How will you convert this ammeter into voltmeter reading upto 1 V? 17. Figure shows a long straight wire of a circular crosssection with radius a carrying steady current I. The current I is uniformly distributed across this cross-section. Calculate the magnetic field in the region r1 < a and r2 > a and show its variation with distance from the axis of wire.
I r1
P r2
Q
a I
18. Equal currents are flowing through two infinitely long parallel wires. Will there be a magnetic field at a point exactly halfway between the wires when the currents in them are (a) in the same direction, and (b) in opposite direction? 19. A combination of two long straight wires AB and DE and a semicircular arc BCD of radius R is shown in the figure. If a current I flows as shown in the figure, then find the net magnetic field at point O.
20. Two parallel coaxial circular coils of equal radius R and equal number of turns N carry equal current I in the same direction and are separated by a distance 2R as shown in the figure. Find the magnitude and direction of the net magnetic field produced at the mid-point O of the line joining their centres. Physics For you | AUGUST ‘16
53
about this, he decided to take help of his family, friends and neighbours and arranged for the cost. He convinced his uncle to undergo this test so as to enable the doctor to diagnose the disease. He got the test done and the resulting information greatly helped the doctor to give him proper treatment.
OR An electron beam passes through a magnetic field of 2.0 mT and an electric field of 3.4 × 104 V m–1 both acting simultaneously in mutually perpendicular directions. If the path of electrons remains undeflected, calculate the speed of the electrons. If the electric field is switched off, what will be the radius of the electron path? Given that me = 9.1 × 10–31 kg and e = 1.60 × 10–19 C. 21. The figure shows the path of an electron that passes through two regions containing uniform magnetic fields of magnitudes B1 and B2. The electron describes a semicircular path in each region. (a) Which magnetic field is stronger? (b) What are the directions of fields B1 and B2 ? (c) Is the time spent by the electron in region B1 greater than, less than or the same as the time spent in region B2 ? B1
e� B2
22. Two identical short magnetic dipoles P and Q of magnetic moment m each, are placed at a separation of d with their axes mutually perpendicular to each other as shown in figure. Find the magnetic field at the point O midway between the dipoles. P
N S
N
O
Q
S
d
section-D
23. Sumit's uncle was advised by his doctor to undergo an MRI scan test of his chest and gave him an estimate of the cost. Not knowing much about the significance of this test and finding it to be too expensive he first hesitated. When Sumit learnt 54
Physics For you | AUGUST ‘16
Based on the above paragraph, answer the following: (a) What according to you, are the values displayed by Sumit and his family, friends and neighbours? (b) Assuming that the MRI scan of his uncle's chest was done by using a magnetic field of 1.5 T, find the maximum and minimum values of force that this magnetic field could exert on a proton (charge = 1.6 × 10–19 C) moving with a speed of 104 m s–1. State the condition under which the force can be minimum. section-e
24. (a) Using Biot-Savart’s law, derive the expression for the magnetic field in the vector form at a point on the axis of a circular current loop. (b) What does a toroid consist of ? Find out the expression for the magnetic field inside a toroid for N turns of the coil having the average radius r and carrying a current I. Show that the magnetic field in the open space inside and exterior to the toroid is zero. OR (a) Using Ampere’s circuital law, obtain an expression for the magnetic field along the axis of a current carrying solenoid of length l and having N number of turns. (b) A toroid has a core (non-ferromagnetic) of inner radius 19 cm and outer radius 21 cm, around which 3000 turns of a wire are wound. If the current in the wire is 10 A, then find the magnetic field (i) outside the toroid, (ii) inside the core of the toroid, and (iii) in the empty space surrounded by the toroid. 25. (a) Deduce an expression for the frequency of revolution of a charged particle in a magnetic field and show that it is independent of velocity or energy of the particle. (b) Draw a schematic sketch of a cyclotron. Explain, giving the essential details of its construction, how it is used to accelerate the charged particles.
OR Distinguish the magnetic properties of dia-, para- and ferro-magnetic substances in terms of (i) susceptibility, (ii) magnetic permeability and (iii) coercivity. Give one example of each of these materials. Draw the field lines due to an external magnetic field near a (i) diamagnetic, (ii) paramagnetic substance. 26. (a) Draw a labelled diagram of a moving coil galvanometer. Describe briefly its principle and working. (b) Which one of the two, an ammeter or a milliammeter, has a higher resistance and why? OR (a) A small compass needle of magnetic moment m is free to turn about an axis perpendicular to the direction of uniform magnetic field B. The moment of inertia of the needle about the axis is I. The needle is slightly disturbed from its stable position and then released. Prove that it executes simple harmonic motion. Hence deduce the expression for its time period. (b) A compass needle whose magnetic moment is 60 A m2 pointing geographical north at a certain place where the horizontal component of earth's magnetic field is 40 mWb m–2 experiences a torque of 1.2 × 10–3 N m. What is the declination of the place? solutions solutions
1. (i) Steel has high retentivity and soft iron has low retentivity. (ii) Hysteresis loss is more in case of steel than in case of soft iron. qB . It is independent of m angle q. For same q, B and m, w remains the same, i.e., the ratio of the angular frequencies of the two charged particles is w1 : w2 = 1 : 1.
2. Angular frequency, w =
3. Radial magnetic field provides constant magnetic field and hence the torque on the coil becomes constant in all positions of the coil for the given current. This provides a linear current scale. 4. Here, n = 300 turns m–1, I = 5 A,
m0 = 4p × 10–7 T m A–1 So, B = m0nI = 4p × 10–7 × 300 × 5 = 1.9 × 10–3 T
5. Here, r = 0.5 Å = 0.5 × 10–10 m, e = 1.6 × 10–19C e As m = IA = × pr 2 = eυpr 2 T –19 = 1.6 × 10 × 1016 × 3.14 × (0.5 × 10–10)2 A m2 = 1.256 × 10–23 A m2 6. Given situation is shown in the figure. The magnetic field due to
current element dl2 does not affect current element dl1 . Hence, force on dl1 due to dl2 is zero. The magnetic field due to current element dl1 affects current element dl2 . Hence force on dl2 due to dl1 is non-zero. Hence, the magnetic forces do not obey Newton’s third law. 7. Let l be the length of the wire. When the wire is first bent in the form of one turn circular coil, l l = 2 pr1 , or r1 = , N =1 2p m NI m × 1 × I m 0 pI = \ B1 = 0 = 0 l 2r1 2 × (l / 2 p) When the wire is bent in the form of n-turns coil, l l = n × 2 pr2 or r2 = ,N =n 2np m NI m 0nI m pn 2 I \ B2 = 0 = = 0 l 2r2 2(l / 2np) B1 1 So, = or, B1 : B2 = 1 : n2 B2 n2 8. Here, H = 1500 A m–1, f = 2.4 × 10–5 Wb, A = 0.5 cm2 = 0.5 × 10–4 m2 Magnetic induction, f 2.4 × 10−5 B= = = 0.48 Wb m −2 − 4 A 0.5 × 10 B 0.48 Permeability, m = = = 3.2 × 10−4 T m A −1 H 1500 As m = m0 (1 + cm) m −1 \ Susceptibility, cm = m0 3.2 × 10−4 = −1 4 × 3.14 × 10−7 = 254.77 – 1 = 253.77 Physics For you | AUGUST ‘16
55
OR Relative permeability, mr = 1 + cm \ cm = mr – 1 As the relative permeability of A is slightly greater than 1, so its susceptibility c is small and positive. Hence A is a paramagnetic substance. As the relative permeability of B is slightly less than 1, so its susceptibility c is small and negative. Hence B is a diamagnetic substance. 9. Potential energy of current loop when its dipole moment m makes angle q with the field B is given by U = –mB cos q In unstable equilibrium position, q = 180° \ U1 = –mB cos 180° = +mB In stable equilibrium position, q = 0° \ U2 = –mB cos0° = –mB Required work, W = U2 – U1 = –mB – mB = –2mB = –2IAB [Q m = IA] 10. We know that radius of circular path described by a charged particle moving with a speed v at right angle to uniform magnetic field B is given by ma q p r mv r= ⇒ a = qB rp m p qa But we know that, ma = 4mp and qa = 2 qp ra 4m p q p 2 = = or ra = 2rp ⇒ rp m p 2q p 1 11. Current sensitivity of a galvanometer is defined as the deflection produced in the galvanometer when a unit current flows through it. Voltage sensitivity of a galvanometer is defined as the deflection produced in the galvanometer when a unit voltage is applied across the two terminals of the galvanometer. Let q be the deflection produced in the galvanometer on applying voltage V, then q nBA Current sensitivity, I s = = I k q nBA Voltage sensitivity, Vs = = V kR Thus, the current sensitivity can be increased by increasing, n, B, A and by decreasing k. If n is increased, it will increase the resistance of conductor. The voltage sensitivity can be increased by increasing n, B, A and by decreasing k and R. 56
Physics For you | AUGUST ‘16
Therefore, the increase in current sensitivity of galvanometer may not necessarily increase the voltage sensitivity of the galvanometer. If n → 2n then R → 2R. Hence, I′s = 2Is and V′s = Vs . 12. (a) Slope of I-H graph gives susceptibility I cm = of the material. H For material A, the slope is positive and smaller, it is likely to be paramagnetic. For material B, the slope is positive and larger, it is likely to be ferromagnetic. (b) For the paramagnetic material A, B I =C 0 T B0 = applied magnetic field \ I T = const This represents a rectangular hyperbola. 13. (a) As per question, number of turns in the coil, N = 30, radius of the coil, R = 12 cm = 0.12 m and current flowing in the coil, I = 0.35 A \ Magnetic field at the centre of the coil, m NI B= 0 2R 4 p × 10−7 × 30 × 0.35 ⇒ B= = 5.5 × 10−5 T 2 × 0.12 As the coil is in a vertical plane making an angle of 45° with the magnetic meridian, hence magnetic field B is in horizontal plane making an angle 45° east of south. B can be resolved into two components (i) B cos 45° due south, and (ii) B sin 45° due east. As the compass needle points west to east, it is clear that B cos 45° due south is just balanced by BH due north and net field in horizontal plane is B sin 45° due east. \ BH = B cos 45° 1 = 5.5 × 10−5 × = 3.9 ×10–5 T 2 (b) When direction of current in the coil is reversed and the coil is rotated about its vertical
axis by an angle of 90° in the anticlockwise sense, the direction of field B will be in horizontal plane making an angle of 45° west of south. Thus, it can be easily shown that compass needle points east to west. 14. Here, current flowing through each wire of PQ and SR, 10 I= = 5 A , d = 2.0 cm = 2.0 × 10–2 m 2 \ Force per unit length between the wires PQ and SR, 2
−7
m 2I 10 × 2 × 5 × 5 F= 0⋅ = 4p d (2 × 10−2 ) = 2.5 × 10–4 N m–1 By right hand thumb rule the force between the wires PQ and SR is attractive in nature. –11
15. Radius of electron orbit, R = 0.51 Å = 5.1 × 10 m; velocity of electron, v = 2.0 × 105 m s–1; mass of electron, m = 9.1 × 10–31 kg and charge on electron, e = 1.6 × 10–19 C (a) Equivalent current due to orbital motion of electron, I=
1.6 × 10−19 × 2.0 × 105 ev = = 1 × 10−4 A 2 pR 2 × 3.14 × 5.1 × 10−11
(b) Magnetic field at the centre of the nucleus, m I 4 p × 10−7 × 1 × 10−4 B= 0 = 2R 2 × 5.1 × 10−11
= 1.23 T (c) Magnetic moment associated with the electron, | m | = IA = IpR2 = 1 × 10–4 × 3.14 × (5.1 × 10–11)2 = 8.2 × 10–25 A m2 16. Here n = 30, Rg = 20 W Current sensitivity, k = 20 mA per division \ Current required for full-scale deflection is Ig = nk = 30 × 20 = 600 mA = 6 × 10–4 A = 0.0006 A
The resistance of the ammeter formed is R g Rs 20 × 0.012 R′g = = = 0.012 W R g + Rs 20 + 0.012
Current for full scale deflection, I ′g = 1 A Voltage range, V = 1 V \ Required series resistance, V 1 R= − R′g = − 0.012 = 0.988 W I ′g 1
17. The current I is uniformly distributed in the wire, I current per unit area can be calculated as . pa2 (a) To calculate magnetic field at P, let us consider a loop of radius r1, and apply Ampere’s law. B . dl = m0Inet
∫
I B × 2pr1 = m0 pr 2 2 1 pa m0 Ir1 B= 2pa2
B ∝ r1 (b) To calculate magnetic field at Q, let us consider a loop of radius r2, and apply Ampere’s law. B . dl ∫ = m0Inet B × 2pr2 = m0I m I B= 0 2p r2 1 B∝ r2
B
Br
B 1/ r r
A graph can be plotted showing variation of magnetic field B with distance r from the axis of wire. 18. Let there be two infinitely long parallel wires carrying a current I each and separation between then is d.
(i) For conversion into ammeter, I = 1 A Ig 0.0006 × 20 \ RS = × Rg = = 0.012 W I − Ig 1 − 0.0006 i.e., a shunt of 0.012 W should be connected across the galvanometer. (ii) For conversion of resulting ammeter into voltmeter.
(a) If both wires carry current in same direction as shown in figure (a), then the magnetic field at mid-point due to both wires have same magnitude, Physics For you | AUGUST ‘16
57
m0 I m I = 0 d pd 2p 2 But, BP and BQ are in opposite directions. Hence, net magnetic field at the mid-point between the two wires, m I m I B = BP + BQ = 0 ⊗ + 0 = 0 pd pd (b) If two wires carry current in mutually opposite directions as shown in figure (b), then the net magnetic field at the mid-point between the two wires, m I m I 2m I B = BP + BQ = 0 ⊗ + 0 ⊗ = 0 ⊗, pd pd pd which is finite and nonzero. 19. Let B1 , B2 and B3 be the magnetic fields at point O due to current carrying straight wire AB, semicircular wire BCD and straight wire DE respectively. m I Then, B1 = 0 4 pR [Since point O lies near one end of long wire at a normal distance R] m I m I B2 = 0 and B3 = 0 4R 4 pR \ Net magnetic field at O is given by m I m I m I B = B1 + B2 + B3 = 0 + 0 + 0 4 pR 4 R 4 p R m I ⇒ | B | = 0 [2 + p] 4 pR and B is directed perpendicular to plane of paper directed out of it. 20. We know that the magnetic field at a point on the axial line of a circular current carrying coil is given by BP = BQ =
B=
m 0 NIR 2
2(R 2 + x 2 )3/ 2
Here, x = R. As both coils are identical and carry current in the same direction, hence B1 = B2 =
m 0 NIR 2
m NIR 2 m NI = 0 = 0 2 2 3/ 2 3 4 2R 2(R + R ) 4 2R
Directions of B1 and B2 are same along the axial line towards the right side. m NI Hence, B = B1 + B2 = B1 + B1 = 2B1 = 0 2 2R along the direction A1OA2. 58
Physics For you | AUGUST ‘16
OR Here, me = 9.1 × 10 kg, e = 1.6 × 10–19 C, B = 2.0 mT = 2.0 × 10–3 T and E = 3.4 × 104 V m–1 As electron beam goes undeflected, –31
E 3.4 × 104 = = 1.7 × 107 m s–1 B 2.0 × 10−3 If the electric field is switched off, then under the magnetic field, electron beam will describe a circular path of radius, mv 9.1 × 10−31 × 1.7 × 107 r= e = eB 1.6 × 10−19 × 2.0 × 10−3 Hence, v =
= 4.8 × 10–2 m = 4.8 cm 21. (a) We know that radius of circular path of a charge mv in a magnetic field is given by r = qB In the given figure, radius of semicircular path in region of B1 is less than that in region B2 , hence it is obvious that B1 > B2. (b) Applying Fleming's left-hand rule, we find that B1 is directed into the plane of paper but B2 is directed out of the plane of paper. (c) Time of revolution of a charged particle in a 2pm magnetic field, T = qB Hence, time to complete a semicircular path, T pm t= = 2 qB As mass and charge of electron are constant, t B Hence, 1 = 2 t2 B1 As B1 > B2, hence we conclude that t1 < t2. So, the time spent by the electron in region B1 is less than that in region B2 . d 22. The point O lies at a distance, r = from the centre 2 of either magnet (short). For magnet P, the point O lies on axial line but for magnet Q, the point O lies on the equatorial line. m 2m m 0 2m m 16m Hence, | BP | = 0 = = 0 3 3 4 p d3 4p r 4p d 2 m0 m m0 m m 8m and | BQ | = = = 0 4 p r 3 4 p d 3 4 p d 3 2
Directions of BP and BQ are shown in figure.
into it. Hence, current enclosed is zero. \ B=0 OR (a) Refer to point 3.2 (2) page no. 171 (MTG Excel in Physics)
Obviously, the magnitude of resultant magnetic field B will be | B | = BP2 + BQ2 2 5 m 0m m 8m (2)2 + (1)2 = = 0 3 4p d pd 3 The magnetic field B subtends an angle b from the axes joining two magnets P and Q, where m0 8 m BQ 1 4 p d3 tan b = = = = 0.50 m 0 16 m 2 BP 4 p d3 –1 ⇒ b = tan (0.50) = 26.6° 23. (a) (i) Presence of mind ; High degree of general awareness; Ability to take prompt decisions ; Concern for his uncle ; (ii) Empathy ; Helping and caring nature. (b) Maximum force = qvB = 1.6 × 10–19 × 104 ×1.5 = 2.4 × 10–15 N Force is maximum when v is perpendicular to B . Minimum force = 0 Force is minimum when v is parallel to B . 24. (a) Refer to point 3 (vi) page no. 170 (MTG Excel in Physics) (b) A long solenoid shaped in the form of closed ring is called a toroid. According to Amperes's circuital law ∫ B . dl = m0 I
\ B·2pr = m0NI B = m0nI
N = number of turns per unit length 2pr Inside : The current enclosed by the loop, in the open space inside the toroid, is zero. Hence B = 0. where n =
Exterior : The current coming out of the plane of the paper is cancelled exactly by the current going 60
Physics For you | AUGUST ‘16
(b) Mean radius r =
19 + 21
= 20 cm 2 3000 Number of turns per unit length = 2p(20 × 10−2 ) = 2388.5 turns m–1 (i) Outside the toroid, B = 0 (ii) Inside the toroid, B = m0nI = 4p × 10–7 × 2388.5 × 10 B = 0.03 T (iii) In the empty space surrounded by toroid, B = 0 25. (a) Refer to point 3.3 (2) page no. 172(MTG Excel in Physics) (b) Refer to point 3.3 (5) page no. 173 (MTG Excel in Physics) OR Refer to point 3.8 (8) page no. 180, 181 (MTG Excel in Physics) 26. (a) Refer to point 3.4 (2) page no. 175 (MTG Excel in Physics) I × Rg (b) Shunt resistance, Rs = g I − Ig Clearly, the shunt needed to convert galvanometer into a milliammeter has a larger value than that required to convert into an ammeter. As the shunt resistance is connected in parallel with the galvanometer, so the milliammeter will have a higher resistance than the ammeter. OR (a) Refer to point 3.6 (7) page no. 179 (MTG Excel in Physics) (b) In stable equilibrium, a compass needle points along magnetic north and experiences no torque. When it is turned through declination a, it points along geographic north and experiences torque, t = mB sin a \ sin a =
or a = 30°
t 1.2 × 10−3 1 = = − 6 mB 60 × 40 × 10 2
C RE C NCEPT
Electrostatic Potential Energy and Grounding of Conductors
When we keep two like charges close to each other, due to repulsion they move away from each other and gain kinetic energy. How do we explain this? One way is we say that the repulsive forces pushes them away from each other due to which they accelerate and gain energy. Another way is energy method. We know that energy can neither be created, nor be destroyed in an isolated system. So considering the two charges together as a system, initially from a zero KE (kinetic energy) they attain non-zero KE, which means there has been a gain of KE, therefore there has to be loss of some other form of energy, which is the potential energy here. It is a stored form of energy which depends upon the relative positions of the charges. Mathematically, PE (Potential energy) of any given system of charges is defined as the minimum amount of work that is required to be done to build the given system, beginning from a situation from where they can be assumed to be placed for away from each other. Let us first calculate the work required to be done in moving one charge q in the vicinity of another charge Q from a position A to position B in the path shown in the figure. A r1 Q
r
Fext r2
s ds co Fe dr = q ds B
ds is a small instantaneous displacement along the path. Fe and Fext are the electrostatic and external forces respectively.
Fext is almost equal and opposite to Fe since we are to move the charge q slowly (quasi statically) so our applied force should be such that it appears to be at almost equilibrium at all instant but over a prolonged period, the charge would appear to have been shifted. dr = ds cosq is the displacement along the radial vector, i.e., displacement in the direction of force. \ Work done by external agent, Wext = ∫ dwext = ∫ (−Fext )(dr ) r2
kqQ 1 dr ; where k = 2 r 4 πε 0 r1
= − ∫ Fe dr = − ∫ r
2 1 = −kqQ − r r1
\ Wext =
kqQ kqQ − r2 r1
This work clearly is only initial and final positions dependent but independent of path. Hence we conclude that the external agent was performing work against the conservative electrostatic forces. Therefore we can associate PE with it, which by definition is, Wext = DPE kqQ kqQ − r2 r1 \ PE at a separation r becomes, kqQ + U ∞ where U ∞ indicates the potential U(r) = r energy at infinite separation, that is conventionally taken to be zero until and unless specified. So, if U ∞ = 0 then \ U2 – U1 =
U (r ) =
kqQ r
1 \ U ∝ , so can we conclude that by increasing r
Contributed By: Bishwajit Barnwal, Aakash Institute, Kolkata
Physics For you | AUGUST ‘16
61
separation, PE decreases? No, not if the charges are kqQ unlike. If qQ < 0, then U = < 0, so increasing r, r gives a less negative PE, so obviously PE increases. Forget this trap of –ve and +ve. Let me give you a Golden rule! Whenever you release any system and allow them to move spontaneously, it always moves towards lower PE configuration. So, if q and Q are both +ve or both –ve, due to repulsion they would spontaneously move away which mean if r increases, PE decreases. If q and Q are unlike, they attract and move closer so if r decrease, PE decreases, hence if r increases, PE increases. Similarly, if we use the definition of PE as work done by external agent to build the system assuming they were initially placed far away. r1 = ∞ (far away) r2 = r \ Wext = kqQ − kqQ = kqQ = PE at a separation r r ∞ r
Let us see few examples of it before moving to next section. E1. Consider two point charges of given charges q1 and q2 and masses m1 and m2 which are released from an initial separation r0. Find their relative velocity when the separation between them becomes large. Sol.:
\ Usystem = W1 + W2 + W3
kq q kq q kq q kq q kq q kq q =0+ 1 2 + 1 3 + 2 3 = 1 2 + 2 3 + 3 1 r12 r23 r31 r23 r12 r13
A close introspection shows us that we need to form as many two charge pairs that we can and then for each kqi q j . two charge pair we write rij q a q For example consider four identical charges placed at the vertices of a square. We have a a 4 edge pairs and 2 diagonal pairs. q q a kq 2 kq 2 \ Usystem = 4 +2 r 2r 62
Physics For you | AUGUST ‘16
q2
r0
q1
v2 q2
Notice that on combined mass system, (there are no external forces, so linear momentum has to be conserved which means, m1v1 = m2v2 = p (say) \ Using energy conservation (DE = 0), KEi + PEi = KEf + PEf ⇒
kq1q2 1 p2 p2 1 = m1v12 + m2v22 = + r0 2 2 2m1 2m2
⇒
kq1q2 p2 1 1 p2 = + = 2 m1 m2 2µ r0
where µ =
PE of a System of three Point charges
Assume, that the charges are far away from each other. When we bring q1 from infinity to its position, no work would be done since there are no charges in vicinity which would exert force on the charge. Now keeping q1 q1 fixed q2 is brought till separation r12, so work would be required to r12 r31 be done against the force of q1. Now keeping both q1 and q2 fixed, q3 is r23 q3 moved for which work would again q2 be done against forces of q1 and q2.
q1
v1
m1m2 = reduced mass of the system m1 + m2
kq q ⇒ p = 2µ 1 2 = m1v1 = m2v2 r0 \
vrel = v1 + v2 =
p p p 2 (kq1q2 ) + = = m1 m2 µ µ r0
Alternatively, in COM (centre of mass) frame KEsystem =
1 2 µvrel 2
KEi + PEi = KEf + PEf ⇒
kq1q2 1 2 = µvrel r0 2
⇒ vrel =
2 (kq1q2 ) µ r0
E2. The charges q1 and q2 are fixed, while Q is moved in a circular path from A to B as shown in the figure. Find the work required to be done to do so?
y A 4a q2
3a
q1
4a
B
x
Sol.: We observe that the seperation of Q from q1 does not change, so no work would be required to be done for it, so consider it to be not existing for us. But for q2, ri = 5a and rf = 7a 1 1 \ Wext = DPE = kq2Q − 7a 5a −kq2Q(2) −2kq2Q = 35a 35a Now, let us define potential difference between two points. Potential difference between two points A and B is defined as minimum amount of work required to be done per unit test charge in moving the test charge from point A to point B. From work energy theorem (WET), Welectric field + Wext agent = DKE = 0 ⇒ Wext agent = – Welectric field rB Felec ⋅ dr ∫ W W r ⇒ VB – VA = ext = − elec . = − A q0 q0 q0
z
rB
VB – VA = − ∫ (Ex i + E y j + Ez k ) ⋅ (dxi + dy j + dzk ) rA
yB zB xB = − ∫ Ex dx + ∫ E y dy + ∫ Ez dz x A yA zA
=
rB
⇒ VB – VA = −
∫ q E ⋅ dr
rA
0
q0 \ VB − VA = − ∫ E ⋅ dr rB
rA
Therefore the line integral of E over any path, taken with a negative sign gives us the potential difference between the two points. Few note worthy points
Why would anyone wish to calculate potential difference between two points? Since if VB – VA is known, multiplying it with a charge q gives the work required to be done in moving the charge form A to B, and hence it also gives the change in potential energy. z What does negative in the expression of VB – VA on RHS indicate? Suppose we choose the points A and B in the direction of a field as below. E Clearly E ⋅ dr > 0 dr A B \ dV = VB – VA = – E ⋅ dr < 0 \ VB < VA \ Electric field points in the direction of decreasing potential. z
What if E is given in vector form, E = Ex i + E y j + Ez k ?
where (xA, yA, zA) and (xB, yB, zB) are the (x, y, z) coordinates of the respective points A and B. Specifically, as a special case, if electric field is uniform, that is, constant in magnitude as well as direction, Ex, Ey and Ez will be constant in which case they can be taken out of integration symbol, as below. yB zB xB VB − VA = − Ex ∫ dx + E y ∫ dy + Ez ∫ dz x A yA zA VB − VA = − Ex Dx + E y Dy + Ez Dz Instead of memorizing the formula, let us try to decode the expression. Ex Dx indicates component of E along x axis multiplied with distance between the two points along x axis and this gives us –ve of the potential difference between the two points. Let us find a logical way to do this. Electric field along any direction multiplied with the separation between any two points in this direction gives the magnitude of potential difference between the two points. Whether it is positive or negative can be decided by the direction of electric field. |VA – VB| = (E cosq) l E No t e t h e d i re c t i o n o f B E cosq – from A towards B, A E cos hence B has to beat a lower l potential \ VA – VB = (E cosq)l
What do we mean by potential at a point? The definition says that it refers to the minimum work done by external agent per unit test charge in bringing the test charge from infinity to the point under consideration. W U (r ) − U (∞) \ V(r) – V(∞) = ext = q0 q0 z
For example, potential due to a point charge Q at a distance r. Physics For you | AUGUST ‘16
63
R +x
2
P
kdq R2 + x 2
∫ dq =
kQ R2 + x 2
VP
kQ Potential at centre R
x Potential due to a uniformly charged thin spherical shell
We know from shell theorem that the field of a uniformly charged sphere matches with a point charge for external points so the results of potential will match too, i.e., for external points, we can assume the entire charge on sphere to be concentrated at the centre. But for internal points E = 0 , hence potential will become constant and equal to that of the surface since no further work is required to be done in moving the test charge from surface to internal point. +
+
+
+
+
+
+
R
kQ , rR R
+
+
V (r)=
+
Physics For you | AUGUST ‘16
k 2
x
R2 + x 2
\ VP = ∫ dVP = ∫ VP =
R
+
64
same distance from point P.
R2 x 2
+
The locus of all such point in space which are at identical potential is said to be equipotatial surface. There cannot be any non-zero component of electric field along equipotential surface else it would create a potential difference which is a contradiction. Hence E will always be perpendicular to equipotential surfaces. Saying a surface is equipotential is different from saying that the entire region of a space is equipotential. In the 1st case, E may or may not be present but if it is present, it has to be perpendicular to the surface but in 2nd case E = 0 everywhere in the region else in whichever direction E ≠ 0 along this direction potential difference would be created which is a contradiction.
made up of large number of elemental point charges dq. Each dq element is at
+
qn
Potential due to a charged ring at a point on its axis Q Consider the ring to be
+
kq kq kq V = 1 + 2 + ....... + n r1 r2 rn
Some standard results
+
Where V(∞) indicates potential due to +Q at infinity. This value is generally taken to be zero until and unless specified. One thing, we should remember here is that the potential at any point is dependent upon the choice of zero potential but potential difference between any two points is independent of this choice. So if V(∞) = 0, then kQ , which is the potential for a point charge Q VP = r at a point P at a distance r. Unlike electric field which is a vector quantity, electric potential is a scalar quantity, hence the positioning of the charge is immaterial as long as the separation remains unchanged. r1 P Potential at any point q1 due to multiple point r2 charges is a scalar sum of potential due rn to individual point q2 charges. \ At point P,
+
Q
+
r
This also means that the entire conductor under electrostatic condition is equipotential. Since if E ≠ 0 inside the bulk, they will drift free electrons inside the bulk and if E is not perpendicular to surface, again drifting of free electrons along the surface would happen, both of which are contradictions of electrostatic condition.
Vs
+
kQ \ VP = + V (∞) r
P
V kQ R
+
kq0Q kq0Q − ∞ = kQ VP − V (∞) = r q0 r
+
At P,
V
1 r
kQ , rR r
r
Potential due to a uniformly charged solid sphere
For external points we can again imagine the entire charge to be concentrated at the centre. + But for internal points let us see + + + + + how do we find it + + + + R + + + + P is an internal point at a distance + + + + + + 1+ r + + r from the centre (r < R) we have + + + + P divided the entire sphere into two 2+ + + parts namely 1 and 2 . For 1 , the point is external point for which potential. 3Q 4 πr 3 k 3 kq kQ VP1 = 1 = 4 πR 3 = 3 r 2 r r R For 2 , we can imagine P to be an internal point for large number of thin concentric shells. Charge dq in the elemental shell of radius x and Q 3Q 2 2 thickness dx is dq = 4 πx dx = 3 x dx 4 R 3 πR 3 \ VP 2 = ∫ dV2
(
kdq x R k 3Q = ∫ ⋅ 3 x 2dx x R r =∫
=
)
2 r
x dx
P
3kQ 2 2 (R − r ) 2R 3
kQ 3 kQ 2 2 \ VP = VP1 + VP 2 = 3 r 2 + (R − r ) R 2 R3 kQ VP = 3 [3R2 − r 2 ] (As r < R) 2R Potential at centre, for r = 0, VC =
3 kQ 2 R
and potential at surface, VS =
kQ R
3 \ VC = VS , this result is applicable only if potential 2 V at infinity is taken as zero. VC But since the difference in parabolic potential is not dependent VS hyperbolic on this choice kQ R r VC − VS = 2R 66
Physics For you | AUGUST ‘16
In general,
GroundinG concePt Grounding any conductor gives two informations z The potential of the conductor becomes zero. z Earth is considered to be an infinite source as well as sink of electrons, hence as many electrons can flow in either direction as required by the conductor to make its potential equal to zero. MisconcePtion Grounding a conductor means its charge becomes zero. Well this is true, had the conductor been isolated but not if there were other charged conductors in its vicinity. Let me explain. Case-I
Isolated grounded conductor (initially charged with Q) : The entire charge flows to earth, + +Q since if the conductor has any + + qflown = Q non zero charge, it will create + + + a field which will create a + potential difference between infinity and conductor which is again a contradiction since at infinity also the potential is zero. Case-II
+Q Concentric shell arrangement : Outer shell with charge Q and R q inner shell initially uncharged qflown = q and later grounded. r Let us assume a charge q has flown from ground to the inner shell. r kQ kq + = 0 ⇒ q = −Q . \ Vinnershell = 0 ⇒ R R r This clearly shows that charge on grounded conductor is non-zero. Answer KEY MPP-2 CLASS XI
1. 6. 11. 16. 21. 26.
(c) (b) (b) (a) (b,c,d) (5)
2. 7. 12. 17. 22. 27.
(a) (c) (c) (b) (a,b,c) (c)
3. 8. 13. 18. 23. 28.
(d) (d) (a) (a) (b,c,d) (a)
4. 9. 14. 19. 24. 29.
(c) (b) (a) (a) (1) (c)
5. 10. 15. 20. 25. 30.
(d) (c) (d) (a,d) (4) (b)
chapterwise McQs for practice
Useful for All National and State Level Medical/Engg. Entrance Exams Wave oPtics
1. In Young’s double slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths l1 =12000 Å and l2 = 10000 Å. At what minimum distance from the common central bright fringe on the screen 2 m from the slit, will a bright fringe from one interference pattern coincide with a bright fringe from the other? (a) 3 mm (b) 8 mm (c) 6 mm (d) 4 mm 2. Air has refractive index 1.0003. The thickness of air column, which will have one more wavelength of yellow light (6000 Å) than in the same thickness of vacuum is (a) 2 mm (b) 2 cm (c) 2 m (d) 2 km 3. The percentage of incident unpolarised light passing through the two Nicol prisms oriented with their principal planes making an angle q is 25. Then q is (a) 60° (b) 30° (c) 45° (d) 90° 4. White light is used to illuminate the two slits in a Young’s double slit experiment. The separation between slits is b and the screen is at a distance d(>> b) from the slits. At a point on the screen directly in front of one of the slits, certain wavelengths are missing. Some of these missing wavelengths are
(a) l =
b2 b2 , d 3d
(b) l =
b2 3b2 , 2d 2d
2b2 3b2 (d) l = 3d 4d 5. An unpolarised beam of light is incident on a set of four polarising plates such that each plate makes an angle of p/3 with preceding sheet. The light transmitted through the combination is (a) 1/128 (b) 1/256 (c) 1/64 (d) 1/32 (c) l =
6. In a double slit pattern (l = 6000 Å), the 1st order and 10th order maxima fall at 12.50 mm and 14.75 mm from a particular reference point. If l is changed to 5500 Å, other arrangements remaining the same, position of 10th order maxima will be (a) 12.25 mm (b) 14.55 mm (c) 15.50 mm (d) 16.55 mm 7. The time period of rotation of the sun is 25 days and its radius is 7 × 108 m. The Doppler shift for the light of wavelength 6000 Å emitted from the surface of the sun will be (a) 0.04 Å (b) 0.40 Å (c) 4.00 Å (d) 40.0 Å 8. A rocket is going towards moon with a speed v. The astronaut in the rocket sends signals of frequency u towards moon and receives them back on reflection from the moon. What will be the frequency of signal received by astronaut? (Take v 4 I 2
I min = I ( n − 1) > 0
\ Intensities of both, maxima and minima increase. 15. (a) : Intensity of polarised light from tourmaline I crystal A = 0 2 As principal plane of crystal B is parallel to that of A, therefore, intensity of final emergent light is I = I0/2 cos2 0° = I0/2 16. (b) : Momentum of incident light per second E 60 p1 = = = 2 × 10−7 c 3 × 108 Momentum of reflected light per second 60 E 60 60 p2 = × = × = 1.2 × 10−7 8 100 c 100 3 × 10 \ Force on the surface = Change in momentum per second = p2 – (–p1) = p2 + p1 = (1.2 + 2) × 10–7 = 3.2 × 10–7 N 17. (c) : We know that Einstein equation, 2(E − W0 ) 1 E = W0 + mv 2 or v= 2 m and a charged particle placed in uniform magnetic field experience a force, 2m(E − W0 ) mv 2 mv mv 2 = ⇒ evB = ⇒r = F= r eB eB r p2 18. (a) : We know that kinetic energy, Ek = ; 2m so p = 2mEk De-Broglie wavelength, h h 6.6 × 10−34 l= = = p 2mEk 2 × 9.1 × 10−31 × 100 × 1.6 × 10−19 = 1.22 × 10–10 m = 1.2 Å
19. (d) : Here, E = E0 sin [1.57 × 107 m–1 (ct – x)] Comparing it with the equation of harmonic wave in electric field;
ANSWER KEY
MPP-2 CLASS XII 1.
(a)
2.
(a)
3.
(d)
4.
(b)
5.
6.
(d)
7.
(a)
8.
(d)
9.
(c)
10. (c)
(a)
11. (c)
12. (b)
13. (b)
14. (a)
15. (b)
16. (b)
17. (b)
18. (a)
19. (b)
20. (a,c)
21. (a,b,d) 22. (a,b,c) 23. (a,c,d) 24. (7)
25. (6)
26. (6)
30. (a)
27. (b)
28. (a)
29. (b)
Physics for you | August ‘16
73
2p (vt − x ), we have l 2p 2p m = 1.57 × 107 or l = l 1.57 × 107
E = E0 sin
hc Maximum kinetic energy (in eV) = − f0 el =
(6.63 × 10−34 ) × (3 × 108 ) × 1.57 × 107
1.6 × 10−19 × (2 × 3.141) = 3.1 – 2.39 = 0.71 eV hc 20. (a) : Energy of photon = l Power of lamp = P
− 2.39
P hc / l Number of photoelectrons emitted per sec
Number of photons emitted per sec, n = \
\ =
0. 5 1 Pl = n= × 100 200 hc 1 Pl Photoelectric current = × ×e 200 hc
1 (10−3 ) × (4560 × 10−10 ) × (1.6 × 10−19 ) × 200 (6.62 × 10−34 ) × (3 × 108 )
= 1.836 × 10–6 A 21. (c) : We know that energy of photon, l hc 1 E E= or E ∝ \ 2 = 1 l l E1 l2 l1 10000 or E2 = E1 × = 1.243 × = 2.486 eV l2 5000 1 2 Now hv − f0 = mvmax = eVs 2 or f0 = E2 – eVs = 2.486 – 1.36 = 1.126 ≈ 1.12 eV 1 22. (b) : Here, y = cm = 0.5 × 10−2 m; 2 l = 3 cm = 3 × 10–2 m; v = 3 × 107 m s–1; d = 1 cm = 10–2 m; V = 550 V Time to cross the tube, l 3 × 10−2 t= = = 10−9 s v 3 × 107 1 1 Ee y = at 2 = t 2 2 2 m
or = 74
e 2y 2y 2 yd = = = 2 2 m Et Vt 2 (V / d )t 2 × 0.5 × 10
−2
× 10
−9 2
550 × (10 )
−2
= 1.8 × 1011 C kg −1
Physics for you | August ‘16
23. (b) : Here, l1 = 100 nm = 1000 Å and l2 = 200 nm = 2000 Å Now hc 12375 eVÅ = 12.375 eV E1 = = 1000Å l1 12375 and E2 = = 6.1875 eV 2000 As eVs = E1 – f0 \ 7.7 eV = 12.375 – f0 and eV′s = 6.1875 – f0 Subracting (ii) from (i), we get (7.7 –V′s)e = 6.1875 eV 7.7 – V′s = 6.1875 or V′s = 7.7 – 6.1875 = 1.5 V
...(i) ...(ii)
24. (d) : From graph, u = 1015 Hz, Kmax = 3 eV = 3 × 1.6 × 10–19 J
As Kmax = hu – hu0 or hu0 = hu – Kmax \
K 3 × 1.6 × 10−19 u0 = u − max = 1015 − h 6.63 × 10−34 = 2.7 × 1014 Hz
25. (a) : Using Einstein’s photoelectric equation, 1240 hc − 4. 7 eV0 = − W = l 200 = 6.2 – 4.7 = 1.5 eV \ Stopping potential, V0 = 1.5 V 1 q 1 ne But V0 = ⋅ = ⋅ 4 pε0 r 4 pε0 r \
1. 5 =
or
n=
9 × 109 × n × 1.6 × 10−19 10−2
1.5 × 10−2
−10
= 1.04 × 108
9 × 1.6 × 10 On comparing with A × 10Z, we get Z = 8 h
26. (d) : As, l =
i.e., l ∝
1
2mK K Given li = 1 nm, lf = 0.5 nm
\ or
Kf li = lf Ki 1 × 10−9
−9
=
Kf Ki
or
Kf Ki
= (2)2 = 4
or
0.5 × 10 Kf = 4Ki
\
Energy to be added = Kf – Ki = 4 Ki – Ki = 3Ki
27. (b) : l = K=
h
2
h 2mK =
or
l2 =
h2 2mK
(6.6 × 10−34 )2
2ml2 2 × 9.1 × 10−31 × (10 × 10−12 )2 6.6 × 6.6 × 10−15 = k eV 15 k eV 2 × 9.1 × 1.6 × 10−16
J
28. (d) : According to Einstein’s photoelectric equation, when the exciting wavelength is l, hc 1 hu = = hu0 + mv 2 l 2 3l When the exciting wavelength is , 4 hc 1 = hu0 + mv ′2 3l / 4 2 1 4 4 1 hc or mv ′2 = − hu0 = (hu0 + mv 2 ) − hu0 2 3 l 3 2 1 1 2 or mv ′2 = hu0 + mv 2 2 3 3 1 2 2 4 2 ⇒ mv ′ > mv or v ′2 > v 2 2 3 3 or v′ > v(4/3)1/2
29. (a) : Energy emitted per sec by S1, hc P1 = n1 l1 Energy emitted per sec by S2,
P2 n2 l1 1.02 × 1015 5000 = ⋅ = ⋅ = 1. 0 5100 P1 n1 l2 1015 30. (a) : Energies of incident photons of different wavelengths will be hc 1240 E(550 nm) = = = 2.25 eV l 550 1240 E(450 nm) = = 2.75 eV 450 1240 E(350 nm) = = 3.54 eV 350 In case of plate p, all radiations will cause photoelectric emission. In case of plate q, wavelengths 450 nm and 350 nm will cause photoelectric emission. In case of plate r, only wavelength 350 nm will cause photoelectric emission. Hence saturation current will be maximum for plate p, intermediate for q and minimum for r. \
IIT dreams: 823 SC students in JEE (Advanced) general list, 22 in top 2,000 This year’s Joint Entrance Exam (Advanced) exams for entry into the IITs saw 823 SC students make it to the Common Rank List in which candidates across categories are given ranks. In other words, these students got a general rank without using the 50 per cent relaxation in marks they were entitled to. According to data provided to The Indian Express by IIT Guwahati, the organising IIT for this year’s JEE, 22 of these 823 students ranked between 1 and 2,000 in the Common Rank List. To be included in the CRL, students irrespective of category need to score a minimum 10 per cent in each subject and an aggregate 20 per cent. The 823 qualifiers are a marked increase from the 515 SC students who made the CRL last year. They account for 11.21 per cent of the total 7,339 SC students who had cleared JEE (Advanced), while the proportion of SC students who had got general ranks last year had been 20 per cent of the total 2,570 who had cleared the exam. “Until last year, SC students needed to score at least 3.5 per cent in each subject and an aggregate of 12.25 per cent to qualify for IITs. This year, the minimum aggregate percentage that SC students need to qualify has been brought down to 10,” said 2015 JEE chairperson Professor P V Balaji of IIT Bombay. “This explains why over 7,000 SC students qualified this year compared to 2,570 in 2015.” The IITs have 10,500 seats. Of these, 7.5 per cent are reserved for STs, 15 per cent for SCs, 27 per cent for OBCs and the remaining 50.5 per cent for general category students. “While these 823 students are eligible for the open seats [general category seats], it is likely that a majority of them will take admission under the SC category. This is because they have better chances of getting their preferred institute and course in the reserved seats. Unless someone from among these 823 has a rank in the top 20 [CRL], these students are unlikely to get into the college of their choice,” said Prof K V Krishna, chairman, JEE 2016, of IIT Guwahati. “Also, not all the 7,000-plus SC students who have qualified will make it to the IITs – remember, there are only 10,500 seats in all. But they can get admission to the other engineering colleges in the country through their JEE (Main) ranks,” Krishna said. Among ST students, 9,537 students appeared this year and 2,392 of them (25 per cent) qualified. Last year, 1,745 of 7,736 ST students (22.5 per cent) had cleared the exam. The cutoff for STs was the same as for SCs in both 2015 and 2016. This year, 212 ST students have found a place in the common merit list as opposed to 108 last year. Of the 212, two had ranks between 1 and 2,000 in the CRL The JEE (Advanced) 2016 results were declared on June 12. Courtesy : The Indian Express Physics for you | August ‘16
75
Class XII
T
his specially designed column enables students to self analyse their extent of understanding of specified chapters. Give yourself four marks for correct answer and deduct one mark for wrong answer. Self check table given at the end will help you to check your readiness.
Electrostatic Potential and Capacitance Total Marks : 120
Time Taken : 60 min NEET / AIIMS / PMTs
Only One Option Correct Type
1. Two identical positive charges are placed on the y-axis at y = – a and y = + a. The variation of electric potential V along x-axis is shown by graph (a)
(b)
(c)
(d)
molecule at a point 52.0 nm away along the axis of the dipole? Assume V = 0 at infinity. (a) 10.3 mV (b) 16.3 mV (c) 20.3 mV (d) 26.3 mV 5. Figure shows the variation of electric field intensity E versus distance x. What is the potential difference between the points at x = 2 m and at x = 6 m from O ?
(a) 30 V 2. In a typical lightning flash, the potential difference between discharge points is about 1.0 × 109 V and the quantity of charge transferred is about 30 C. If all the energy released could be used to accelerate a 1200 kg automobile from rest, what would be the final speed of the automobile? (a) 7100 m s–1 (b) 3600 m s–1 –1 (c) 9500 m s (d) 6800 m s–1 3. Two capacitors each having capacitance C and breakdown voltage V are joined in series. The capacitance and the breakdown voltage of the combination will be (a) 2C and 2V (b) C/2 and V/2 (c) 2C and V/2 (d) C/2 and 2V 4. The ammonia molecule NH3 has a permanent electric dipole moment equal to 1.47 D, where D is the debye unit with a value of 3.34 × 10–30 C m. What will be the electric potential due to an ammonia 76
Physics For you | AUGUST ‘16
(b) 60 V
(c) 40 V
(d) 80 V
6. A capacitor of 4 mF is connected as shown in the circuit. The internal resistance of the battery is 0.5 W. The amount of charge on the capacitor plates will be
(a) 0
(b) 4 mC
(c) 16 mC
(d) 8 mC
7. Three charges q, 2q and 8q are to be placed on a 9 cm long straight line. Where the charges should be placed so that the potential energy of this system is minimum. (a) charge q between charges 2q and 8q and 3 cm from charge 2q. (b) charge q between charges 2q and 8q and 5 cm from the charge 2q.
(c) charge 2q between charges q and 8q and 5 cm from the charge q. (d) charge 2q between charges q and 8q and 7 cm from the charge q. 8. Which of the following statements is incorrect regarding equipotential surfaces? (a) Equipotential surfaces are closer in regions of large electric fields compared to regions of lower electric fields. (b) Equipotential surfaces will be more crowded near sharp edges of a conductor. (c) Equipotential surfaces will be more crowded near regions of large charge densities. (d) Equipotential surfaces will always be equally spaced. 9. An electric dipole of length 2 cm is placed with its axis making an angle of 30° to a uniform electric field 105 N C–1. If it experiences a torque of 10 3 N m, then the potential energy of the dipole is (a) – 10 J (b) – 20 J (c) – 30 J (d) – 40 J 10. The arrangement consists of four identical plates. P Each plate has the area A and the plate separation is d. The equivalent capacitance Q between P and Q is 3 e0 A e A 3e0 A 2 e0 A (a) 0 (b) (c) (d) d 2 d 2d 3 d 11. Three charges –q, +q and +q are situated in x-y plane at points (0, –a) (0, 0) and (0, a) respectively. The potential at a point distant r (r > a) in a direction making an angle q from y-axis will be kq kq (a) (b) 2 (2a cos q) r r kq kq 2a cos q 1 + (d) a r r 12. A charge Q has been divided on two concentric conducting spheres of radii R1 and R2 (R1 > R2) such that the surface charge densities on both the spheres are equal. The potential at their common centre is 1 Q(R1 + R2 ) 1 Q(R12 + R22 ) (a) (b) 4 πe0 (R12 + R22 ) 4 πe0 (R1 + R2 ) (c)
1 QR1R2 (b) 4 πe0 (R1 + R2 )
1 Q(R1 + R2 ) (d) R1R2 4 πe0
Assertion & Reason Type
Directions : In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as : (a) If both assertion and reason are true and reason is the correct explanation of assertion. (b) If both assertion and reason are true but reason is not the correct explanation of assertion. (c) If assertion is true but reason is false. (d) If both assertion and reason are false. 13. Assertion : When two conductors charged to different potentials are connected to each other, the negative charge always flows from lower potential to higher potential. Reason : In the process of charging there is always a flow of electrons only. 14. Assertion : The potential difference between two concentric spherical shells depends only on the charge of inner shell. Reason : The electric field in the region in between two shells depends on the charge of inner shell and electric field is the negative of potential gradient. 15. Assertion : If the distance between parallel pates of a capacitor is halved and dielectric constant is made three times, then the capacitance becomes six times. Reason : Capacitance of the capacitor does not depend upon the nature of the material of the plates. JEE MAIN / JEE AdvANcEd / PETs
Only One Option Correct Type
16. A capacitor is made of two circular plates of radius R each, separated by a distance d
