UPKAR'S

Tips in Science(Unique for PMT, PET and 10 + 2 students for

General Awareness)

● It puts important topics of your syllabus at yourfinger-tips.

● It crystal clears the intelligently selected problemswith clear-cut hints and explanations.

● It provides indepth knowledge which enables youto solve problems quickly and precisely.

● Only a trial will convince you that it can do to youthe un-do-able.

Edited byThe Board of Competition Science Vision

UPKAR PRAKASHAN, AGRA–2

© Publishers

PublishersUPKAR PRAKASHAN2/11A, Swadeshi Bima Nagar, AGRA–282 002Phone : 2530966, 2531101, 2602653, 2602930; Fax : (0562) 2531940e-mail : upkar1@sancharnet.in

Branch Office4840/24, Govind Lane, Ansari Road,Daryaganj, New Delhi–110 002Phone : 23251844, 23251866

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Price : Rs. 55·00(Rs. Fifty Five Only)

Code No. 499

Printed at : Upkar Prakashan (Printing Unit) Bye-pass, AGRA

Preface

Under a carefully thought out plan, this book has been prepared

to cater to the needs of Pre-Medical, Pre-Engineering, and 10 + 2

students. It covers the salient topics in the syllabi of Physics,

Chemistry, Botany and Zoology for these examinations. The book is

unique in all respects and creates its own standard.

The plan of the book has been to carefully select problems, mostly

asked in related examinations, covering all important aspects of the

subjects, give their to-the-point short answers and provide clear-cut

hints, explanations and even solutions to enable readers to grasp and

understand them without leaving any confusion.

The book offers a never-before opportunity to the students to

acquire indepth knowledge of the often-asked problems so as to be

able to answer them quickly and precisely.

Only a survey trial will convince you about the utility of the book

and you will realise that it can be unbelievably good to you.

Our word of advice to you, our young aspirants is that you should

have full faith in your abilities and should never quit of you want to

be a winner because winners never quit and quitters never win.

—Authors

Contents

Section – 1

Physics 3—70Section – 2

Chemistry 71—106Section – 3

Zoology 107—140Section – 4

Botany 141—171

⎯⎯⎯⎯⎯⎯⎯

Tips in Science

S & T | 5

Hence, the tension becomes

T′ = mg + mv

2

L

Clearly T′ is maximum as cos θ decreases when θ increases. ]

� The angular momentum of a body about the axis of rotation isgiven by J = Iω. What is its rate of change with time equal to ?

➠ Torque C

[ Hints—Torque C = I α

where α is angular acceleration.

∴ C = I Δω

Δt

Also J = I ω

∴ΔJ

Δt= I

Δω

Δt = C ]

� A body falls on earth from infinity, what will be its velocity onreaching the earth ?

➠ 11·2 km/sec

[ Hints—A body projected up with the escape velocity ve will go toinfinity, therefore, the velocity of the body falling on earth from infinity willbe ve . Now escape velocity on the earth is

ve = √⎯⎯⎯⎯2g Re

= √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯2 × 9·8 × 6·4 × 106

= 11·2 × 103 m/s = 11·2 km/s ]

� What is impulse-momentum theorem ?

➠ ∫0

t

→→→→F dt =

→→→→p 2 –

→→→→p 1

[ Hints—From Newton’s second law of motion

→F =

d→p

dt

d→p =

→F d t

Integrating ∫p1

p2 d

→p = ∫0

t

→F d t

8 | S & T

or,NN0

=1e

Thus, λ can also be defined as the reciprocal of the time when N/N0falls to 1/e. ]

� If t is the time of n half lives of a radioactive substance whosehalf-life is T, starting from N0 atoms in the beginning what will bethe number of atoms N left after n half-lives ?

➠ N = N0 ( )12

t /T

[ Hints—We know that N = N0 ( )12

n

If t is the time corresponding to n half lives, thent = n T

or, n =tT

∴ N = N0 ( )12

t /T ]

� A certain element has a half life of 30 days. What is its averagelife ?

➠ 43·28 day

[ Hints—Average life τ is given by

τ =1λ where λ is decay constant

Also half-life T =0·6931

λ = 0·6931 τ

τ =T

0·6931 = 30

0·6931

= 43·28 day ]

� Name five elements which lie on peaks in binding energy curve.➠ 2He4, 4Be8, 6C12, 8O16 and 10Ne20

[ Hints—Between mass number 4 and 20, the binding energy curveshows cyclic recurrence of peaks at 2He4, 4Be8, 6C12, 8O16 and 10Ne20.This shows that the binding energy per nucleon of these nuclides isgreater than those of their immediate neighbours. So, these are morestable as compared to their neighbours. Note that mass number of thesenuclides are multiples of 4 and they contain equal number of protons andneutrons. Each of these nuclei can be formed by adding an α-particle tothe preceding nucleus. ]

S & T | 9

� Momentum p can be associated with a particle and λ is asso-ciated only with a wave. Who established a relation between thetwo ?

➠ de Broglie

[ Hints—Combining Einstein’s mass-energy equivalence (E = mc 2)

and Planck’s quantum theory (E = h υ) we get

E = h υ = mc 2

But c = υλ for waves, thus for a quantum of light (photon)

E = h cλ = mc

2

λ =h

mc = hp

This relation is for a photon, it occurred to de Broglie that what istrue for a photon, may also be true for a material particle of mass m withvelocity v (not c).

∴ λ =h

mv ]

� What is the frequency of a photon of energy 3·3 × 10–20 J ?➠ 0·5 ×××× 1014 Hz

[ Hints— υ =Eh =

3·3 × 10–20

6·6 × 10–34 = 0·5 × 1014 Hz ]

� With increase in temperature, the electrical conductivity of intrin-sic semi-conductor

➠ Increases

[ Hints—Resistance of an intrinsic semiconductor decreases withincrease in temperature. Hence, its electrical conductivity increases. ]

� There is no atmosphere on the moon because➠ Escape velocity of gas molecules is less than their

root mean square velocity

[ Hints—The escape velocity at the moon is only 2·38 km/s,while the rms velocity of gas molecules at the temperature of the moon isgreater than this. Therefore the gas molecules can not stay at moon. ]

� In boiling water reactor, the boiling water is used as➠ Coolant

� In nuclear reactors controlling rods are made of➠ Cadmium

10 | S & T

[ Hints—Cadmium is a good absorber of neutrons. It effectively con-trols the rate of fission. ]

� One can not see through fog because➠ Light is scattered by the droplets

� Name two phenomena exhibiting dual nature of light➠ Diffraction and photoelectric effect

[ Hints—Diffraction exhibits wave nature of light while photo-electriceffect exhibits its particle (quantum) nature. ]

� Kinetic energy, with any reference is always➠ Positive

[ Hints—K. E. = 12 mv

2 where v is velocity and its square is always

positive. ]

� At what angle of incidence will the light reflected from a glass (μ =1·5) plate be completely plane polarised ?

➠ 57°°°° (nearly)

[ Hints— μ = tan ipThis relation is known as Brewster’s law and gives the value of the

polarising angle ip. ]

� Two lenses of power + 12 D and – 2D are combined together.Their equivalent focal length will be

➠ + 10 cm

[ Hints— P = 12 – 2 = + 10

∴ Equivalent focal length F =1P =

110 m = + 10 cm ]

� Write the statement ‘Y equals A and B negated OR A negatedAND B’ as Boolean expression

➠ Y = A · –B +

–A · B

[ Hints—It is Boolean expression for XOR gate. XOR gate gives theoutput for dissimilar values of inputs. It is clear from the truth table.]

A B Y

0 0 00 1 11 0 11 1 0

S & T | 11

� In what manner does the escape velocity of a particle dependupon its mass ?

➠ It is independent of mass (or it depends on m°°°°)

[ Hints—ve = √⎯⎯⎯⎯2g Re

Thus escape velocity does not depend on the mass of the particle. ]

� The angle of a prism is 6° and its refractive index for green light is1·5. If a green ray passes through the prism, its deviation will be

➠ 3°°°°

[ Hints—For a thin prism δ = (μ – 1) A

= (1·5 – 1) 6 = 3° ]

� Which quantity increases in a step down transformer?➠ Current

[ Hints—In a step down transformer voltage decreases. Since, thepower (V × I) remains constant, current increases. ]

� In an L–R circuit, the time constant is the time in which currentgrows from zero value to

➠ 0·632 I0

[ Hints—In L–R circuit growth of current is represented byI = I0 [1 – e

– (R/L) t ]

If t =LR = time constant, then

I = I0 ( )1 – 1e

= I0 × 0·632

Thus, time constant is equal to time in which the current growth to 0·632times the maximum value. ]

� In p-type semiconductors the majority and the minority chargecarriers are

➠ Holes and electrons respectively

� The vertical component of earth’s magnetic field is zero whereangle of dip is

➠ 0°°°°

[ Hints—If angle of dip is φ, then

tan φ =VH

12 | S & T

V is zero where tan φ = 0

or, φ = 0

This happens at magnetic equator. ]

� A person can not see objects clearly beyond 2·0 m. The power ofthe lens to correct his vision will be

➠ – 0·5 D

[ Hints— u = ∞

v = – 2·0f = ?

1v –

1u =

1f

– 1

2·0 =1f

⇒ f = – 2·0 m

P =1f = –

120 = – 0·5 D

i.e., a concave lens of focal length 2·0 m or power 0·5 D. ]

� A coolie carrying a suitcase on his head moves from one place ona horizontal ground to another place and finally comes to rest.How much work does he do?

➠ Zero work

[ Hints—Here the force (weight of the suitcase) and displacementare perpendicular to each other. Hence workdone

= Fd cos θ = Fd cos 90° = 0 ]

� If →P ·

→Q = PQ, then the angle between

→P and

→Q is

➠ 0°°°°

[ Hints—Let the angle between →P and

→Q be θ, then

→P .

→Q = PQ cos θ

If θ = 0,cos θ = 1

then→P .

→Q = PQ ]

� If the maximum range of a projectile is R, then what is thegreatest height attained by it in terms of R ?

➠ R2

S & T | 13

[ Hints—Maximum range of a projectile is

R =u

2

gThe maximum height attained by a projectile (for vertical projection) is

h =u

2

2g = R2 ]

� What is the impedance of LCR circuit at resonance equal to ?➠ Resistance of the circuit

[ Hints—Impedance Z = √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯R2 + (XL – XC)2

At resonance XL = XC

or ωL =1

ωC

∴ Z = R ]

� At electrical resonance what is the phase difference betweenvoltage and current in the circuit ?

➠ Phase difference is zero

[ Hints—The phase difference between the current and the voltagein LCR circuit is given by

tan φ =XL – XC

RAt resonance XL = XC

∴ tan φ = 0

⇒ φ = 0

i.e., the current and the voltage are in the same phase or phase diffe-rence between them is zero. ]

� When is the current in an LCR circuit maximum ?➠ At resonance

[ Hints— I =V

√⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯R2 + (XL – XC)2

At resonance XL = XC

Thus I =VR

This is the maximum value of current since impedance is minimum(equal to R) ]

� Write expression for resonance frequency.

➠ f = 12ππππ

√⎯ 1LC

14 | S & T

[ Hints—At resonance XL = XC

or, ωL =1

ωC

ω2 =1

LC

(2πf)2 =1

LC

⇒ f =12π √⎯ 1

LC ]

� A lens acts as convex lens in air and when immersed in water itacts as a concave lens. The refractive index of the material of thelens is

➠ Greater than μair but less than μwater

[ Hints—A lens when immersed in a medium whose refractive indexis greater than that of the material of the lens, it changes its nature, i.e.,convex lens acts as a concave lens and vice-versa.

In the present case the lens changes nature when it is immersed inwater. It means that the refractive index of its material must be smallerthan μ water. ]

� What is the unit of inductive reactance XL ?➠ Ohm

[ Hints— XL = ωL = 2πT · L = Hs – 1

but, 1 H =1 V × 1 s

1 A

∴ 1 Hs – 1 =1 V1 A = 1Ω ]

� What is the opposition offered by a purely inductive circuit to theflow of D.C. through it ?

➠ Zero[ Hints— XL = ωL = 2πf L

For D.C., f = 0

∴ XL = 0 ]

� What is the capacitive reactance to the flow of D.C. through thecircuit containing a capacitor ?

➠ Infinite

[ Hints— XC =1

ωC =

12πf C

16 | S & T

� What will be the magnitude of change in velocity in the above pro-blem (11) in half a round ?

➠ 20 m/s

[ Hints— Δ→V =

→V2 –

→V1

= 10 – (– 10) = 20 m/s ]

� The acceleration due to gravity decreases as we go above orbelow the surface of the earth. What will be its value at a distanceRe above and below earth’s surface ?

➠ g4

, Zero

[ Hints—Value of g at a height h above earth’s surface

g ′ = ( )1 + h Re

2

When h = Re , g ′ = 14 g

Value of g at a depth h below earth’s surface

g " = g ( )1 – hRe

When h = Re , g " = 0 ]

� What is the value of escape velocity from earth’s surface ?➠ 11·2 km/sec

[ Hints— ve = √⎯⎯⎯⎯2g Re = √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯2 × 9·8 × 6·4 × 106

= 11·2 × 103 m/s = 11·2 km/s ]

� What is the distance of the geo-stationary orbit above earth’s sur-face ?

➠ 36,000 km (nearly)

[ Hints—The time period of a satellite in geo-stationary state is 24hours so that it may appear stationary relative to earth. Now

T =2π (Re + h)3/2

Re √⎯ g

Substituting T = 24 × 3600 sec,

Re = 6·37 × 106 m

and g = 9·8 ms– 2

On solvingh ≈ 36,000 km ]

Tips in Science Physics, Chemistry,Zoology and Botany

Publisher : Upkar Prakashan Author : Editorial Board ofCompetition Science

Type the URL : http://www.kopykitab.com/product/5122

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