Physics Spectrum

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1. An equa

tion is given as p

a

V

b

V

+

æ

è

ç

ö

ø

÷ =

2

q

, where,

p = pres sure,

V = vol

ume and q =

ab so lute

tem per a ture. If a and b are con

stants, then

di men sions of a will be

(a) [ML T5 - 2] _{(b) [M L T ]}- 1 5 2 (c) [ML T-5 - 1] (d) [ML T]5

2. A man throws the balls with the same speed

ver ti cally up wards one af ter the other at an in ter val

of 2 s. What should be the speed of the throw, so that

more than two balls are in the sky at any time?

(take g = 9 8

. m/s)

(a) Any speed less than 19.6 m/s (b) Only with speed 19.6 m/s (c) More than 19.6 m/s (d) Atleast 9.8 m/s

3. A stone is at tached to one end of a string and ro tated

in a ver ti cal cir cle. If string breaks at the po si tion of

max i mum ten sion, then it will break at

(a) A (b) B

(c) C (d) D

4. A body of 1 kg ex plodes into three frag ments. The

ra tio of their masses is 1 1 3

: : . The frag ments of same

mass move per

pen

dic

u

lar to each other with the

speed 30 m/s, while the heavier part re mains in the

intial di rec tion. The speed of heavier part is

(a) 10

2 m/s (b) 10 2 m/s

(c) 20 2 m/s (d) 30 2 m/s

5. Equa tion of a trans verse wave trav el ling in a rope is

given by y

=

5 sin ( .

4 0

t

-

0 02

.

x

) where, y and x are

ex pressed in cm and time in sec onds. Cal cu late the

am pli tude, fre quency and ve loc ity of the wave.

(a) 8 cm, 0.8673 cy cle s- 1, 200 cms- 1

(b) 5 cm, 0.673 cy cle s- 1, 200cms-1 (c) 5.8 cm, 0.673 cy cle s- 1, 250cms-1 (d) None of the above

6. A ball of 0.5 kg mov ing with a speed of 12

m/s strikes a hard wall at an an gle of 30°

with the wall. It is re flected with the same

speed and at the same an gle. If the ball is

in con

tact with the wall for 0.25s, the

av er age force act ing on the wall is

(a) 48 N (b) 24 N

(c) 12 N (d) 96 N

7. A bomb of mass 30 kg at rest ex plodes into two

pieces of masses 18 kg and 12 kg. The ve loc ity of 18

kg mass is 6

ms

-1

. The ki netic en ergy of the other

mass is

(a) 256 J (b) 486 J

(c) 524 J (d) 324 J

8. A wheel of bi cy cle is roll ing with out slip ping on a

level road. The ve loc ity of the cen tre of mass is v

cm

,

then true state ment is

(a) the ve loc ity of point A is 2 vcm and ve loc ity of point B is zero (b) the ve loc ity of point A is zero and ve loc ity of point B is 2 vcm (c) the ve loc ity of point A is 2 vcm and ve loc ity of point B is - vcm (d) the ve loc i ties of both A and B are Vcm

A D C B 3 0 ° 3 0 ° Vcm cm B A

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9. Imag ine a new planet hav ing the same den sity as

that of Earth but it is 3 times big ger than the Earth in

size. If the ac cel er a tion due to grav ity on the sur face

of Earth is g and that on the sur face of the new

planet is g’, then,

(a) g¢ = 3g (b) g¢ = g 9 (c) g¢ = 9g (d) g¢ = 27g

10. The ra tio of the ra dii of gy ra tion of a cir cu lar disc

about a tan gen tial axis in the plane of the disc and of

a cir cu lar ring of the same ra dius about a tan gen tial

axis in the plane of the ring is

(a) 2 3: (b) 2 1:

(c) 5: 6 (d) 1: 2

11. The es cape ve loc ity of a body on the sur face of the

Earth is 11.2 km/s. If the Earth’s mass in creases to

twice its pres ent value and the ra dius of the Earth

be

comes half, then the es

cape ve

loc

ity would

become

(a) 44.8 km/s (b) 22.4 km/s

(c) 11.2 km/s (re main un changed) (d) 5.6 km/s

12. A tube of length L is filled com

pletely with an

in com press ible liq uid of mass M and closed at both

the ends. The tube is then ro tated in a hor i zon tal

plane about one of its end with a uni form an gu lar

ve loc ity w. The force ex erted by the liq uid at the

other end is

(a) MLw 2 2 (b) ML2 2 w (c) MLw2 _(d)ML2 2 2 w

13. The

wave length

cor re spond ing

to

max i mum

in

ten

sity of ra

di

a

tion emit

ted by a source at

tem per a ture 2000 K is l, then what is the

wave length cor re spond ing to max i mum in ten sity of

ra di a tion at tem per a ture 3000 K?

(a) 2 3l (b) 16 81l (c) 81 16l (d) 4 3l

14. Two sim ple pen du lums of lengths 0.5 m and 2.0 m,

re spec tively are given small lin ear dis place ment in

one di rec tion at the same time. They will again be in

the same phase when the pen

du

lum of shorter

length has com pleted os cil la tions.

(a) 5 (b) 1

(c) 2 (d) 3

15. An en gine has an ef fi ciency of

1

6 . When the tem

per-ature of sink is re duced by 62°C, then its ef fi ciency

will be dou bled. Tem per a ture of the source is

(a) 124°C (b) 37°C (c) 62°C (d) 99°C

16. Two sim

ple har

monic mo

tions given by,

x

=

a

sin (

w

t

+

d and y

)

=

a

æ

t

+

è

ç

ö

_ø

÷

sin w

d

p

2 act on a

par ti cle si mul ta neously. Then, the mo tion of

par ti cle will be

(a) cir cu lar anti-clock wise (b) cir cu lar clock wise (c) el lip ti cal anti-clock wise (d) el lip ti cal clock wise

17. A mass of 2.0 kg is put on a flat pan at tached

to a ver ti cal spring fixed on the ground as

shown in the fig ure. The masses of the spring

and the pan are neg

li

gi

ble. When pressed

slightly and re

leased the mass ex

e

cutes a

sim ple har monic mo tion. The spring con stant

is 200 N/m. What should be the min i mum

am

pli

tude of the mo

tion, so that the mass gets

de tached from the pan? (take g = 10

m s

/ )

2

(a) 8.0 cm (b) 10.0 cm

(c) Any value less than 12.0 cm(d) 4.0 cm

18. For a black body at tem per a ture 727° C, its ra di at ing

power is 60 W and tem per a ture of sur round ing is

227° C. If the tem

per

a

ture of the black body is

changed to 1227° C, then its ra di at ing power will be

(a) 120 W (b) 240 W

(c) 304 W (d) 320 W

19. A point per forms sim ple har monic os cil la tion of

pe riod T and the equa tion of mo tion is given by

x

=

a

sin (w

t

+

p / )

6 . Af ter the elapse of what frac tion

of the time pe riod the ve loc ity of the point will be

equal to half of its max i mum velocity?

(a) T 8 (b) T 6 (c) T 3 (d) T 12

20. A par al lel plate ca pac i tor with oil (di elec tric

con stant 2) be tween the plates has ca pac i tance C. If

oil is re moved, the ca pac i tance of ca pac i tor

becomes

(a) 2 C (b) 2C (c) C 2 (d) C 2

21. A hos

pi

tal uses an ul

tra

sonic scan

ner to lo

cate

tu mour in a tis sue. The op er at ing fre quency of the

scan ner is 4.2 MHz. The speed of sound in a tis sue

is 1.7 km/s. The wave length of sound in tis sue is

close to

(a) 4´10-4 m (b) 8´10-4 m (c) 4 10´ -3 m (d) 8´10-3 m

22. A charged wire is bent in the form of a semi-cir cu lar

arc of ra

dius a. If charge per unit length is

l coloumb/m, then the elec tric field at the cen tre O

(a) l p e 2 a2 0 (b) l p e 4 2 0a (c) l pe0 2 a (d) zero m m

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AIPMT RIDE 1

23. Two trains move to wards each other with the same

speed. The speed of sound is 340 m/s. If the height

of the tone of the whis tle of one of them heard on the

other changes

9

8 times, then the speed of each train

should be

(a) 20 m/s (b) 2 m/s (c) 200 m/s (d) 2000 m/s

24. Three con cen tric spher i cal shells have ra dii a b

, and

c (

a

<

b

< and have sur face charge den si ties s,

c

)

-

s

and s re spec tively. If V

A

,

V

B

and V

C

de

note the

po ten tials of the three shells, then for c

=

a

+ , we

b

have

(a) VC =VA ¹VB (b) VC =VB ¹VA (c) VC ¹VB ¹VA (d) VC =VB =VA

25. The driver of a car trav el ling with speed 30 ms

-1

to wards a hill sounds a horn of fre quency 600 Hz. If

the ve loc ity of sound in air is 330 ms

-1

, then the

fre quency of re flected sound as heard by driver is

(a) 550 Hz (b) 555.5 Hz

(c) 720 Hz (d) 500 Hz

26. There are three cop

per wires of length and

cross-sec tional area (L, A) [2

L A

, (

/ )], [( / ),

2 L

2 2

A

]

. In

which case is the re sis tance min i mum?

(a) It is the same in all three cases (b) Wire of cross-sec tional area 2 A (c) Wire of cross-sec tional area A (d) Wire of cross-sec tional area 1 2 A

27. Two bat

ter

ies, one of emf 18 V and in

ter

nal

re sis tance 2 W and the other emf of 12 V and in ter nal

re sis tance 1 W are con nected as shown in the fig ure.

The volt me ter V will re cord a re cord ing of

(a) 15 V (b) 30 V (c) 14 V (d) 18 V

28. Fuse wire is a wire of

(a) low re sis tance and low melt ing point (b) low re sis tance and high melt ing point (c) high re sis tance and high melt ing point (d) high re sis tance and low melt ing point

29. A 10 eV elec tron is cir cu lat ing in a plane at right

an

gles to a uni

form field of mag

netic in

duc

tion

10

-4

Wb/m

2

. The or bital ra dius of the elec tron is

(a) 12 cm (b) 16 cm

(c) 11 cm (d) 18 cm

30. A beam of elec

trons passes undeflected through

mu tu ally per pen dic u lar elec tric and mag netic

fields. If the elec tric field is switched off and the

same mag

netic field is main

tained, then the

elec tron move

(a) in an el lip ti cal or bit (b) in a cir cu lar or bit (c) along a par a bolic path (d) along a straight line

31. Two bar mag

nets hav

ing same ge

om

e

try with

mag netic mo ments M and 2M, are firstly placed in

such a way that their sim i lar poles are on the same

side, then its pe riod of os cil la tion is T

1

. Now, the

po lar ity of one of the mag nets is re versed the time

pe riod of os cil la tions be comes T

2

. Then,

(a) T1<T2 (b) T1>T2 (c) T1=T2 (d) T2= ¥

32. Two iden ti cal con duct ing wires AOB and COD are

placed at right an gles to each other. The wire AOB

car

ries an elec

tric cur

rent I

1

and COD car

ries a

cur rent I

2

. The mag netic field on a point ly ing at a

dis tance d from O, in a di rec tion per pen dic u lar to the

plane of the wire AOB and COD, will be given

(a) m p 0 1 2 1 2 2 I I d + æ è ç ö ø ÷ / (b) m p 0 1 2 2 2 1 2 2 d (I I ) / + (c) m p 0 1 2 2 d(I + I ) (d) m p 0 1 2 2 2 2 d (I + I )

33. For a given in ci dent ray as shown in fig ure, the

con di tion of to tal in ter nal re flec tion of the ray will

be sat is fied if the re frac tive in dex of block will be

(a) 3 1 2 + (b) 2 1 2 + (c) 3 2 (d) 7 6

34. A car is fit ted with a con vex side view mir ror of fo cal

length 20 cm. A sec ond car 2.8 m be hind the first car

is over

tak

ing the first car at a rel

a

tive speed of

15 m/s. The speed of the im age of the sec ond car as

seen in the mir ror of the first one is

(a) 1

15 m/s (b) 10 m/s

(c) 15 m/s (d) 1

10 m/s

35. A body is lo cated on a wall. Its im age of equal size is

to be ob tained on a par al lel wall with the help of a

con vex lens. The lens is placed at a dis tance d ahead

of sec ond wall, then the re quired fo cal length will be

(a) only d/4

(b) only d/2

(c) more than d/4 but less than d/2 (d) less then d/4 V 1W 12 V 2 W 18 V q 45° r incident ray

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36. Fo cal length of a con vex lens of re frac tive in dex 1.5

is 2 cm. Fo cal length of lens when im mersed in a

liq uid of re frac tive in dex of 1.25 will be

(a) 10 cm (b) 2.5 cm

(c) 5 cm (d) 7.5 cm

37. The pho

to

elec

tric work func

tion for a metal

sur face is 4.125 eV. The cut-off wave length for this

sur face is

(a) 4125 Å (b) 3000 Å (c) 6000 Å (d) 2062 5. Å

38. The work func tion for met als A B

, and C are 1.92 eV,

2 0

. eV and 5 eV re spec tively. Ac cord ing to Ein stein’s

equa tion, the met als which will emit photoelectrons

for a ra di a tion of wave length 4100 Å is/are

(a) None (b) A only (c) A and B only (d) All the three met als

39. The to tal en ergy of elec tron in the ground state of

hy dro gen atom is - 13 6

. eV. The ki netic en ergy of an

elec tron in the first ex cited state is

(a) 3.4 eV (b) 6.8 eV

(c) 13.6 eV (d) 1.7 eV

40. The bind ing en ergy of deutron is 2 2

. MeV and that of

2

4

_{He is 28 MeV. If two deu ter ons are fused to form}

one

24

He, then the en ergy re leased is

(a) 25 8. MeV (b) 23 6. MeV (c) 19.2 MeV (d) 30.2 MeV

41. The ra dius of ger ma nium (

Ge nu clide is mea sured

)

to be twice the ra dius of

₄9

Be. The num ber of nu cleon

in Ge are

(a) 73 (b) 74

(c) 75 (d) 72

42. Zener di ode is used for

(a) pro duc ing os cil la tions in an os cil la tor (b) am pli fi ca tion

(c) sta bili sa tion (d) rec ti fi ca tion

43. A tran sis tor-os cil la tor us ing a res o nant cir cuit with

an in duc tor L (of neg li gi ble re sis tance) and a

ca pac i tor C in se ries pro duce os cil la tions of

fre quency F. If L is dou bled and C is changed to 4C,

the fre quency will be

(a) f / 4 (b) 8f

(c) f / 2 2 (d) f / 2

44. The logic cir

cuit shown be

low has the in

put

wave forms A and B. Pick out the cor

rect out

put

wave forms.

45. For a given cir cuit of ideal p-n junc tion di ode, which

of the fol low ing is cor rect?

(a) In for ward bi as ing, the volt age across R is V (b) In re verse bi as ing, the volt age across R is V (c) In for ward bi as ing, the volt age across R is 2 V (d) In re verse bi as ing, the volt age across R is 2 V

Answers

1. (a) 2. (c) 3. (b) 4. (b) 5. (b) 6. (b) 7. (b) 8. (a) 9. (a) 10. (c)

11. (b) 12. (a) 13. (a) 14. (c) 15. (d) 16. (b) 17. (b) 18. (d) 19. (d) 20. (d)

21. (a) 22. (c) 23. (a) 24. (a) 25. (c) 26. (b) 27. (c) 28. (d) 29. (c) 30. (b)

31. (a) 32. (b) 33. (c) 34. (a) 35. (b) 36. (c) 37. (b) 38. (c) 39. (a) 40. (b)

41. (d) 42. (c) 43. (c) 44. (a) 45. (a)

Detailed solutions of these questions are available on http://www.arihantbooks.com/Physics%20Spectrum.pdf A B y Input A Input B (a) (b) (c) (d) Diode V R

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1.

(a) Ac cord ing to prin ci ple of ho mo ge ne ity of di men sions, Dimensions of p = dimensions of a V2 p a V = 2 \ a=pV2₌_[ML-1_T-2_]_{[ ]}_L3 2₌_{[ML T}5 -2_]

2.

(c) Time taken by ball to reach max i mum height, v u= -gt

at maximum height, final speed is zero i.e. v = 0

So, u=gT

T u

g

= In 2 s, u =2´9 8 19 6. = . m/s

If man throws the ball with velocity of 19.6 m/s, then after 2 s it will reach the maximum height. When he throws 2nd ball, 1st is at top. When he throws third ball, 1st will come to ground and 2nd will be at the top. There fore only 2 balls are in air. If he wants to keep more than 2 balls in air he should throw the ball with a speed greater than 19.6 m/s.

3.

(b) When string makes an an gle q with the ver ti cal in a ver ti cal cir cle, then

T mg mv l - cos q= 2 or T mg mv l = cos q+ 2

Tension is maximum when cos q = + 1

i.e, q = 0

Thus, q is zero at lowest point B. At this point tension is maximum. So, string will break at point B.

4.

(b) Let v be the ve loc ity and q the di rec tion of the third peice as shown.

Equating the momentum of the system along OA and OB to zero, we get

m´30-3m´vcos q=0 ... (i) m´30-3m´vsin q=0 ... (ii) From Eqs (i) and (ii), we get

3mvcosq=3mvsinq

or cosq=sinq

q =45°

Thus, ÐAOC= ÐBOC =180° -45° =135° Putting the value of q in Eq. (i), we get

30m=3mvcos45° =3 2

mv v = 10 2 m/s

So, speed of heavier part of a fragment is 10 2 m/s.

5.

(b) Given, y=5sin ( .4 0t-0 02. x) com par ing this with the stan dard equa tion of wave mo tion, y=A é ft- x

ë ê ù û ú sin 2p 2p l where

A, f and l are am pli tude, fre quency and wave length re spec tively.

Thus, amplitude, A = 5 cm, 2pf =4 frequency, f = 4 = -2 0 673 1 p . cycle s Again, 2p 0 02 l = . or wavelength, l= 2p = p 0 02. 100 cm Velocity of the transverse wave, v= ´ lf

= 4 ´ = -2 2 0 02 200 1 p p . cms

6.

(b) The vec tor OA rep re sents the mo men tum of the ob ject be fore the col li sion and the vec tor OB that af ter the col li sion. The vec tor AB rep re sents the change in mo men tum of the ob ject DP

As the magnitudes of OA and OB are equal, the components of OA and OB along the wall are equal and in the same direction, while those perpendicular to the wall are equal and opposite. Thus, the change in momentum is due to only the change in direction of the perpendicular components. Hence, DP=OBsin30° - -( OAsin30°)

=mvsin30° - -( mvsin30°) =2mv sin30°

Its time rate will appear in the form of average force acting on the wall. \ F t´ =2mvsin30° F mv t =2 sin30° Given, m = 0 5. kg, v = 12 m/s, t= 0 25. s q =30° Hence, Net average froce on well

F =2 ´0.5 ´12 sin 30°

0.25 = 24 N

7.

(b) Ap ply ing con ser va tion of lin ear mo men tum m u1 1=m u2 2 Here, m1=18 kg, m2=12 kg, u1=6 ms u-1 2=? \ 18´6 12= u₂ Þ u2 1 18 6 12 9 = ´ = ms -Thus, kinetic energy of 12 kg mass

K2 m u2 22 1 2 = =1´ ´ = ´ = 2 12 9 6 81 486 2 _J A D C B mg mg sin q mg cos q O q q C 3m q m B A m O 3 0 ° 3 0 ° 60° 60° B A O

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8.

(a) Ve loc ity of point A is

vA=vcm+ wR =vcm+vcm (Q vcm= w)R = 2vcm

Velocity of point B is,

v_B=v_cm-Rw =v_cm-v_cm=0

Thus, the velocity of point A is 2 vcm and velocity of point B is zero.

9.

(a) The acceleration due to gravity on the new planet can be

found using the relation

g GM R = 2 ...(i) But M=4´R 3

3_{r, where r is density of the planet}

Thus, Eq. (i) becomes

\ g G R R = ´4 3 3 2 p r =G´4´ R 3 p r Þ gµR \ g g R R ¢₌ ¢ Þ g g R R ¢ =3 =3 Þ g¢ = 3g

So, acceleration due to gravity g ¢ on the new planet is 3g.

10.

(c) Moment of inertia of a disc and circular ring about a tangential

axis in their planes are respectively

I_d=5M R_d 4 2 Ir= M Rr 3 2 2 But, I MK= 2_Þ_K I M = \ K K I I M M d r d r r d = ´ or I I M R M R M M d r d r r d = ´ = 2 ( / ) ( / ) 5 4 3 2 5 6 2 \ Id:Ir= 5: 6

11.

(b) Escape velocity on the Earth’s surface is given by ¢ = v GM R es e e 2

where, G is gravitational constant, Me and Re are the mass and

radius of Earth respectively.

\ ¢ = ¢´ ¢ v v M M R R es es e e e e but, M¢ = 2e Me and R R e e ¢ = 2 ves= 11.2 km/s \ v v M M R R es es e e e e ¢ = 2 ´ = = / 2 4 2 \ v¢es=2ves=2´11.2=22.4 km/s

12.

(a) Let the length of a small element of tube be dx. Mass of this element

dm M

Ldx

=

where, M is mass of filled liquid and L is length of tube. Force on this element

dF dM= ´ wn 2 dF M L xdx F =

_ò

ò

w2 0 2 0 or F M L L = é ë ê ê ù û ú ú w2 2 2 F=MLw 2 2

13.

(a) Wien’s displacement law is given by

lmT = constant or l1 1T =l2 2T l2 l1 1 2 = æ è ç ö ø ÷ T T Here, T1=2000K,T2=3000K, l1=l Then, wavelength corresponding to maximum intensity of radiation, l₂ l 2000 l

3000 2

3

= ´ = .

14.

(c) For the pendulum to be again in the same phase, there should be difference of one complete oscillation. If smaller pendulum complete n oscillations the larger pendulum will complete (n - 1) oscillations.

So, time period of n oscillations of first

= time period of (n - 1 oscillation of second) i.e. nT1=(n-1)T2 n l g n l g 2_p 1 ₌₍ _-1 2₎ _p 2 or n l1=(n-1) l2 or n n l l -1= = 2 0 0 5 2 1 . . or n n -1=2 n=2n-2 \ n = 2

15.

(d) Efficiency of engine is given by h = -1 2 1 T T \ T T 2 1 1 1 1 6 5 6 = -h= - = ...(i) In other case, T T 2 1 62 1 1 2 6 2 3 -= -h= - = ...(ii)

Using Eq. (i),

T2 62 T1 T2 2 3 2 3 6 3 - = = ´ or 1 5T =2 62 T2=310K =310 273 C =- ° 37 C° Here, T1 T2 6 5 6 5 310 = = ´ =372K=372-273 =99 C° Thus, initial temperature of the source is

T1=99°C dx F F+dF r w L

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16.

(b) Two simple harmonic motions can be written as, x=asin (wt+d) ... (i) and

y=a æ t+ + è ç ö_ø÷ sin w d p 2 or y=acos (wt+d) ... (ii)

Squaring and adding Eqs. (i) and (ii), we get

x2+y2=a2[sin (2 wt+d) cos (+ 2 wt+d)] or x2₊y2₌a2 _{(Q sin}2_q₊_cos2_q₌₁₎ This is the equation of a circle

At wt + = 0 , x = 0, yd =a At wt + =d p 2, x=a y, =0 At wt+ =d p,x=0,y= -a At wt+ =d 3p x= -a y= 2 , , 0 At wt+ =d 2p,x=0,y=a

Thus, it is obvious that motion of particle is traversed in clockwise direction.

17.

(b) Let the minimum amplitude of SHM is a restoring force on spring

F=ka

Restoring force is balanced by weight Mg of block. For mass to execute simple harmonic motion of amplitude a,

\ Ka=Mg or a Mg K = where, M = 2 kg, K = 200 N/m g = 10 m/s2 a =2´10= 200 10 100m = ´ 10 100 100 cm a = 10 cm

Hence, minimum amplitude of the motion should be 10 cm, that the mass gets detached from the pan.

18.

(d) Boltzmann corrected Stefan’s law and stated that the amount of radiations emitted by the body, not only depends upon the temperature of the body but also on the temperature of the surrounding. The power radiated by the body is given by,

P=s (T4-T04) ... (i) where, T0 is the absolute temperature of the surrounding.

\ P P T T T T 2 1 24 04 14 04 = -æ è ç ö ø ÷ ...(ii) Here, P1=60 W, T1=727° =C 1000K T₀=227°C=500K, T2=1227°C=1500K Substituting in Eq. (ii)

Radiated power by a block body P2 4 4 4 4 1500 500 1000 500 60 = -- ´ ( ) ( ) ( ) ( ) = ´ -æ è ç ö ø ÷ ´ ( ) ( ) 500 500 3 1 2 1 60 4 4 4 4 =80´ = 15 60 320 W

19.

(d) When a object performs SHM of period T and equation of motion is given by x=a æ t+ è ç ö_ø÷ sin w p 6 v dx dt a t = = æ + è ç ö_ø÷ wcos w p 6 a a t w w w p 2 = +6 æ è ç ö ø ÷ cos ; 1 2= +6 æ è ç ö ø ÷ cos wt p wt +p=p 6 3 Þ w p t = 6 2 6 p p T t = , t T = 12

Time period of the velocity of the point will be equal to half it maximum velocity is T/12

20.

(d) The capacitance of a parallel plate capacitor with dielectric (oil) between its plates is

C K A

d

= e0 _...(i)

where, e0= electric permittivity of free space

K = dielectric constant

A = area of each plate of capacitor d = distance between two plates

When dielectric (oil) is removed, so capacitance,

C A

d

0= 0 e

...(ii) Comparing Eqs. (i) and (ii)

C=KC₀ Net capacitance becomes C C

K C

0 2

= = (K = 2)

21.

(a) Wavelength of a wave is the length of one wave. It is equal to the distance travelled by the wave during the time, any one particle of the medium completes one vibration about the mean position. Wavelength of a wave is l

n =v where v = velocity of wave (sound)

n = frequency of wave (sound) Here, v =1.7 10´ 3_{m/s, n =}_4.2_´₁₀6_Hz Wavelength of sound in tissue is

\ l = ´ ´ = ´ -1.7 4.2 10 10 4 10 3 6 4_m

22.

(c) Considering symmetric elements each of length dl at A and B. We note that electric field perpendicular to PO are cancelled and those along PO are added. The electric field due to an element of length dl(=adq) along PO,

dE dq a = 1 4pe0 2 q cos (Qdl=adq) = 1 4_pe 2 l q 0 dl a cos = 1 4_pe 2 l( q) q 0 ad a cos

Net electric field at O

E dE a d a =

_ò

=

_ò

- 2 1 4 0 2 0 2 2 2 pe l q q p/ p p/ cos / = 2 1 4 ₀ 0 2 . [sin ] pe l qp/ a =2 = 1 4 ₀ 1 2 . . . pe l l pe₀ a a

23.

(a) According to Doppler’s effect, whenever there is a relative motion between a source of sound and listner, the apparent frequency of sound heard by the listner is different from the actual frequency of sound emitted by the source. Apparent frequency of sound wave heard by the listner is

¢ = -- ´ v v v v v v l s

where, V is actual frequency of sound emitted by the source, V_s is the velocity of source and V_l is velocity of listner. According to aY x a –a O wt+d –a

(8)

problem ¢ =v 9v

8 and source and observer are moving in opposite direction with same speed (say V), then apparent frequency ¢ = ´ + + æ è ç ö ø ÷ v v v v v v l s \ 9 8 340 340 v v v v = ´ + -\ 17v =340 or v =340= 17 20m/s

24.

(a) Here, potential at point A

V a a b b c c A= - + 2 1 4 4 1 4 4 1 4 4 0 2 2 pe s p pe s p pe s p 0 0 . = s - + = e s e 0 0 (a b c) (2a) (Qc=a+b) Potential at B, V a a b b c c B= - + 1 4 4 1 4 4 1 4 4 2 2 2 pe s p pe s p pe s p 0 0 0 . . Potential at c = æ - + è ç ö ø ÷ = s e s e 0 0 a c b c a 2 2 ( ) (Qc=a+b) and V a c b c c c C= - + 1 4 1 4 4 1 4 4 0 2 2 2 pe s4p pe s p pe s p 0 0 . = æ - + è ç ö ø ÷ = s e s e 0 0 a c b c c a 2 2 2 ( ) (Qc=a+b) Hence, VA=VC¹VB

25.

(c) Whenever there is a relative motion between a source of sound and the observer (listner), the frequency of sound heard by the observer is different from the actual frequency of sound emitted by source. Case I n n S· ¾ ¾¾ ¾ ¾30 ms- ® ¢ hill 1 ... | 0 Case II n¢¢ · ¾¾ ¾¾_- ®¬¾¾ ¢n S 30 ms 1 For Case I, n V V n ¢ = - 30 ... (i)

For Case II, n¢¢ =V+ ¢

V n

30

... (ii) From Eqs. (ii) and (i), we get,

¢¢ = + - = ´ n V V n 30 30 360 300 600 = 720Hz

26.

(b) The relation between length and area is

R L

A

=r ...(i)

r being specific resistance is the proportionality constant and depends on nature of material.

(i) length =L

2, area = 2 A

Putting in Eq. (i)

R L A L A =r( / )2 =r 2 4

(ii) Length = L, area = A Putting in Eq. (i)

R L

A

=r (iii) Length = 2 L, area = A / 2

Putting in Eq. (i), R L

A L A =r 2 = r 2 4 /

As it is understood from above, resistance is minimum only in option (b).

27.

(c) It is clear that the two cells oppose each other hence, the effective emf inclosed in circuit is 18 12- = V and net resistance is 6 1 2+ = W. 3

The current in circuit will be in direction of arrow shown in figure.

I = effective emf

total resistance = = 6 3 2 A

The potential difference across V will be same as the terminal voltage of either cell. Since current is drawn from the cell of 18 V, hence, V1=E1-i r1 =18-(2´2)=18-4 14= V

Similarly, current enters in the cell of 12 V, hence V2=E2+i r2 =12+2´1 = 14 V

28.

(d) The electric fuse is a device which is used to limit the current in an electric circuit. Thus, the use of fuse safeguards the circuit and the appliances connected in the circuit from being damaged. It is always connected with the live (or phase) wire. The fuse wire is a short piece of wire made of a material of high resistance and low melting point so, that it may easily melt due to over heating when excessive current passes through it.

29.

(c) If charged particle is moving perpendicular to the direction of

B, it experiences a maximum force which acts perpendicular to

the direction of B as well as V. Hence, this force will provide the required centripetal force and the charged particle will describe a circular path in the magnetic field of radius r,

mv r qB 2 = Now, KE of electron = 10 eV 1 2 10 2 mV = eV \ 1 2 9 1 10 10 10 31 2 19 ´( . ´ - ) v = ´1.6´ v2 19 31 2 10 10 10 = ´ ´ ´ ´ -1.6 9.1 v2₌_3.52_´₁₀12 v =1.88 10´ 6_m Now, radius of circular path,

r mv qB = = ´ ´ ´ ´ ´ -- -9 1 10 10 10 10 31 6 19 4 . 1.88 1.6 = 11 cm a b c V 1W 12 V 2 W 18 V

(9)

30.

(b) If both electric and magnetic fields are present and perpendicular to each other and the particle is moving perpendicular to both of them with Fe=Fm. In this situation E ¹ 0

and B ¹ 0. But if electric field becomes zero, then only force due to magnetic field exists. Under this force the charge moves along a circle.

31.

(a) The time period of bar magnet

T I

MH

= 2p where, M = magnetic moment of magnet

I = moment of inertia

H = horizontal component of magnetic field

When the same poles of magnets are placed on same side, then net magnetic moment

M1=M+2M=3M \ T I M H I MH 1 1 2 2 3 = p = p ...(i)

When opposite poles of magnets are placed on same side, then net magnetic moment.

M2=2M-M=M \ T I M H I MH 2 2 2 2 = p = p ...(ii)

From Eqs. (i) and (ii),

T₁<T₂

32.

(b) The magnetic field induction at a point P, at a distance d from

O in a direction perpendicular to the plane ABCD due to currents

through AOB and COD are perpendicular to each other is

B= B12+B22 = æ è ç ö_ø÷ +æ_èç ö_ø÷ é ë ê ê ù û ú ú m p m p 0 1 2 0 2 2 1 2 4 2 4 2 I d I d / = m + 2p 0 d(I I ) / 12 22 1 2

33.

(c) For total internal reflection to take place, angle of incidence > critical angle

i.e. q > C sinq >sinC sin C =1 m From figure, q =90° -r So, sin (90-r)> 1 m i.e. m > 1 cos r ...(i)

From Snell’s law,

sin sin 45° = r m sin r = 1 2m \ cosr= 1-sin r= 1- 1 2 2 2 m Thus, Eq. (i) becomes,

m m2 > -1 1 1 2 \ m m2 2 1 1 1 2 = or m2 1 2 1 - = m = 3 2 Refractive index of a block.

34.

(a) According to mirror formula, we get

1 1 1 u+v= Þ -f - = 1 1 0 2 2 u du dt v dv dt du dt v u du dt =- æ è ç ö_ø÷ 2 2 But, v u f u f = - \ dv dt f u t du dt = -æ è ç ö ø ÷ 2 = - -æ è ç ö ø ÷ ´ 0.2 2.8 0.2 2 15 =1 15 ms - 1

35.

(b) The lens formula can be written as

1 1 1

f=v-u ...(i)

Given, v d=

For equal size image

v = u =d By sign convention, u= -d \ 1 1 1 f=d +d or f d = 2

36.

(c) Focal length of convex lens

1 ₁ 1 1 1 2 fa R R g = - æ -è ç ö ø ÷ (m ) ...(i)

Focal length of convex lens when immersed in liquid of refractive index ml 1 1 1 1 1 2 f_l= l g- R -R æ è ç ö ø ÷ (m ) ...(i) 1 ₁ 1 1 1 2 fl R R g l =æ -è ç ö ø ÷æ -è ç ö ø ÷ m m ...(ii)

From Eqs. (i) and (ii) 1 1 1 1 f f a l g g l = -æ è ç ö ø ÷ (m ) m m or f f l a g g l = -æ è ç ö ø ÷ (m ) m m 1 1 f f l a = -æ è ç ö_ø÷ = = ( ) / / 1.5 1.5 1.25 1 1 1 2 1 5 5 2

Focal length of a conven lens when immersed in a liquid, \ f_l=5f_a= ´ =

2 5

2 2 5 cm

37.

(b) The maximum wavelength above which no photoelectron can emit from metal surface is called cutt-off wavelength and is given by q _D O d C B A I2 I1 P

(10)

Work function = hc Cutt -off wave length or Cut-off wavelength = hc Work function \ l0 0 =hc W ...(i) Given, h =_{6 6 10}_. ´ -34_J-s C =3 10´ 8 m/s W0= .4 125eV =4 125. ´1.6´10-19J Substituting the given values in Eq. (ii),

l0 34 8 19 10 3 10 10 = ´ ´ ´ ´ ´ -6.6 4.125 1.6 Å = ´ -3 10 7 m

Cut off vave length to be used l0=3000Å

38.

(c) Work function for wavelength of 4100 Å is W0=hc l = ´ ´ ´ ´ -6 -62 10 3 10 4100 10 34 8 10 . =_{4.8 10}´ -19_{J =} ´ ´ = -4 8 10 10 3 19 19 . 1.6 eV eV Now, WA= 1.92 eV W_B= 2 0. eV WC= 5eV Since, WA<W W_B<W

Hence, A and B will emit photoelectrons

39.

(a) The energy of hydrogen atom when the electron revolves in

nth orbit is E n =-13 6= 2 . eV In the ground state, n = 1

E =-13 6= -12 13 6 . _{. eV} for n = 2, E =-13 6= -22 3 4 . . eV

So, kinetic energy of electron in the first excited state (i.e. for n = 2) is

KE = -E= - -( 3 4. )=3 4 eV.

40.

(b) The reaction can be written as

1H H 2+1 2¾® He energy2 4+

The energy released in the reaction is difference of binding energies of daughter and parent nuclei.

Hence, energy released,

= binding energy of ₂He4 _-2_{´ binding energy of} 1H2 =28 2- ´2.2=23 6. MeV

41.

(d) Let radius of ₄9_{Be nucleus be r. Then radius of germanium (Ge)} nucleus will be 2 r.

Radius of a nucleus is given by

R=R A0 1 3/ R R A A 1 2 1 2 1 3 =æ è ç ö ø ÷ / Þ r r A 2 9 2 1 3 =æ è ç ö ø ÷ / (Q A1=9) or 1 2 9 3 2 æ è ç ö ø ÷ = A Hence, A₂₌₉_´_{( )}₂3₌₇₂

Thus in germanium (Ge) nucleus number of nucleons is 72.

42.

(c) Zener diode is a silicon crystal diode having an reverse current

characteristic which is particularly suitable for voltage regulatory purposes. Due to this characteristic, it is used as voltage stabiliser in many applications in electronics appliances.

43.

(c) In a series L-C circuit, frequency of LC oscillations is given by

f LC = 1 2p or f LC µ 1 f f L C L C 1 2 2 2 1 1 = Given, L1= , CL 1= , LC 2=2L, C2=4C, f1=f \ f f L C LC 2 2 4 ₈ = ´ = f2= /f 2 2

44.

(a) Truth Table

A B A¢ B y = ¢ + ¢¢ A B 1 1 0 0 1 0 0 1 1 0 0 1 1 0 0 1 0 0 1 0 Output waveform

45.

(a) In forward biasing the diode conducts. For ideal junction diode, the forward resistance is zero. Therefore, entire applied voltage occurs across resistance R i.e. there occurs no voltage drop.

While in reverse biasing, the diode does not conduct. So, it has infinite resistance, thus, voltage across R is zero in reverse biasing. A B y A B 1 0 0 0 0 0 1

(11)

Paper 1

One or More than One Option Correct Type

1.

A heavy particle is projected with a velocity at an angle

with the horizontal into a uniform gravitational field. The slope of the trajectory of the particle varies not according to which of the following curves?

2.

The half lives of two radioactive samples A and B are 15 min and 30 min, respectively. If A and B have same number of particles, then ratio of residual particles in the samples A and B,

(a) after 60 min is 1/8 (b) after 60 min is 1/4 (c) after 80 min is 4 4_{.( )}1 3/ _{(d) after 80 min is 88}

3.

Two boats A and B are moving in a river. Boat A moves normal to the river currents as observed by an observer moving with river currents. Boat B moves normal to the river as observed by the observer on the ground. Let u and v be the velocities of boat with respect to water and boat with respect to the ground respectively. Then, to stationary observer on the ground

(a) boat B reaches exactly opposite point on the bank (b) boat A reaches a point ub

v æ è

ç ö

ø

÷ distance downstream from the starting point

(c) boat A travels a longer distance

(d) to a ground observer, boat A moves faster than boat B.

4.

A particle of mass m is suspended by the mean of a

string of length l and describes a circle of radius, where a is the angle the string makes with the vertical axis. For this situation,

(a) tension developed in the string is mg / cos a. (b) speed of mass m is l g sin

cos 2

a a (c) the time period is 2 p l a

g cos

(d) the centripetal acceleration is given by a= sina where, T T = tension in the string

5.

Consider the situation given below.

A pulley is smooth, massless and string is inextensible light mass. The frictional coefficient between block x and y is m while friction between block x and contact surface of wedge is zero.

For this situation, choose the correct alternatives. (a) The acceleration of the system if

m³(m -m )tanq m

y x

y

2 , where my>mx

(b) The friction between x and y is zero when mX=mY (c) Block will move up if mx>my

(d) Tension in the string is mg (sinq- cosm q) if mx=my=m.

6.

A block of mass m is kept on a smooth surface and

connected to the spring of stiffness K which in turn connected to a fixed wall as shown in the figure below.

A bullet of mass m₀ is moving with the horizontal velocity v, strikes the block gets embedded to it and their common velocity is V. The spring is compressed by an amount x then, (a) V K m m x m = [ ( + 0)]/ × 1 2 0 (b) V K m m x m = [2 ( 0)]/ 0 + 1 2× (c) v K m m x = + æ è ç ö ø ÷ × 0 1 2/ (d) v K m m x = 3 0 1 2 + æ è ç ö ø ÷ × /

7.

The diagram shows a cyclic process ABCA for a sample of 5 mole ideal gas. The temperatures of the gas at A

and B are 250K and 450 K

respectively. An amount of heat 1000 J is withdrawn from the gas in the process. Take the gas constant, R = 8.3 J K mol-1 -1 for this cyclic process. t S lo p e (a) _x S lo p e (b) x S lo p e (d) (c) _t S lo p e y x Fixed wedge q m m0 K v C A B P re ss u re (p ) Volume(V)

(12)

JEE ADVANCED RIDE 1

(a) Work done during process AB is 8300 J (b) Work done during the process BC is - 9300 J (c) Work done during process CA is zero (d) Work done during process is + 9300 J

8.

Some amount of an ideal gas is enclosed in a vertical cylindrical container support a piston of mass M. The piston is of area of cross-section A and is free to move.

At equilibrium of the piston the volume of the gas is V0, while the pressure is p0. Now, piston is slightly displaced from its mean position x and then released. Assuming the system is completely isolated.

(a) The restoring force of the piston is A p x V 2 0 0 g in magnitude

(b) The restoring force of the piston is A p

V x 3 0 0 g in magnitude

(c) The frequency of oscillations is A p MV

2 0 0 g

(d) The time period would be A p M V 3 0 0 3 g

9.

The temperature of a body falls from 50°C to 40°C in 10 min when its surrounding temperature is constant at 20°C. Then,

(a) the time taken by the body to cool upto temperature 30°C is given by t = 10 ln (0.50)

ln(0.66)

(b) the time taken by a body is given by t = 10 ln (0.20) ln (0.33) (c) the constant used to determine the above meintioned time

is given by K = ln (0.5) 10

(d) the constant is given by K = ln (0.66) 10

10.

A proton enters a uniform magnetic field in direction region of 0.2T with a velocity 2´105 m/s. The velocity makes an angle of 45° with the magnetic field induction. Then,

(a) radius of path followed by proton is 7.34 ´₁₀-2_m (b) radius of path followed by proton is 7 34. ´10-5 m (c) radius of path followed by proton is 3.26 ´10-7 s (d) the time period to complete revolution of the proton on the

circular track is 3.26 ms

Integer Type

11.

An inductance 20 mH, a capacitor 100 mF, a resistor 50 W and an AC source of 12V, 50Hz are forms an series

AC circuit. The average power dissipated in the circuit expressed as the nearest integer is

12.

An inductor having inductance L and a resistance R = 5 W are connected in series with a battery of emf 10V. The maximum rate at which the energy is stored in the magnetic field would be

13.

Long cylindrical iron core of area of cross-section 4 cm2 is inserted into a long solenoid having 1000 turn per metre and carrying a current of 4 A. If the magnetic field induction within the core of solenoid is set to be 1.7 T, then neglecting the end effects, the order of magnitude of pole strength while expressed in unit (Am-1) is

14.

A moving coil galvanometer having a coil of resistance 15 W needs 30 mA current for showing full deflection on the scale. In order to pass a maximum current of 3A through the galvanometer. If R is the resistance R should be added as a shunt then what is the order of magnitude of R ?

15.

Two parallel wires A and B are separated at a distance of 6 cm. The current in the wires A and B are 5A and 2A respectively. The distance in cm from the wire B where, the magnetic field induction is zero, would be?

16.

In a capacitive circuit, the electric field between plates

of a parallel plate capacitor is drops from 2_´10-6_{F to} its one third in 4 4 10. ´ -6s while the plates are connected through a thin wire. After disconnecting the supply, the resistance of the connecting wire would be...(in ohm).

17.

A block of mass 20 kg is kept on a horizontal surface. The coefficient of friction exerting between block and surface is 0.6. A horizontal force is applied on the block which varies with the graph shown below.

If the speed of the block after 10 s is 8v then find the speed of a block, v is (take, =10msg -2)

18.

Two soap bubbles x and y are in a closed chamber with air pressure 8N / m2. The radii of the bubbles are 2cm and 4 cm respectively. The surface tension of soap water is 0.004 N/m. If NA and NB are the number of moles of air in the bubbles x and y, respectively, then N

N x y

will be... (neglecting the effect of gravity)

19.

A uni form string hav ing mass 0.1 kg and length 2.45 m hangs from a ceiling. If the speed of trans verse wave pro duced in string at a point x m(x < 2 45 m. ) dis tant from its lower end is v. Then cal cu late the time taken by trans verse wave travel the full length of the string?

20.

The band en ergy gap in ger ma nium is DE = 0 68. eV. The numbers of hole-electron pair for the material is directly proportional to e

E kt -D

2 _{. If the percentage} increment in number of charge carriers in pure germanium as the temperature increased from 300K to 320 K is given by 63 5. x, then the value of x would be

F(N)

200

5 10 t(s)

(13)

JEE ADVANCED RIDE 1

Paper 2

1.

A boy releases an arrow with the help of bow horizontally from the top of a 20 m height of a tower. What should be the minimum initial horizontal velocity of arrow, so as to hit the ground at a distance of 30 m from the bottom of the tower (take g= 10 m/s2) (a) 15 m/s (b) 20 m/s (c) 30 m/s (d) 7.5 m/s

2.

There is an equilateral triangle PQR of side length a. At

time t = 0, three particles A B, and C are the corners P Q, and R respectively. Now particles are moving along the sides of the triangle in anti-clockwise with a constant speed v. The time after which the particles meet one another is given by (a) 3 2 a v (b) 2 3 a v (c) 3 2 a v (d) 2 a v

3.

The change in volume of an sphere while the radius is increased from 20.0 cm to 20.1 cm is

[assume that the rate does not appreciably change during the given change in radius]

(a) 1600 p cm3 (b) 16. ´104 p cm3 (c) 16 p cm3 _{(d) 160 p cm}3

4.

A bob of mass m is attached to light inextensible string of length L is kept in the horizontal position. What will be the angle made by spring with the vertical axis so, as to the weight of the bob balances the tension produced in the string? (a) cos- æ è ç ö ø ÷ 1 2 3 (b) cos - æ è ç ö ø ÷ 1 1 6 (c) cos - æ è ç ö ø ÷ 1 1 3 (d) cos - æ è ç ö ø ÷ 1 2 3

5.

A skate board negotiating the circular surface of 3.5m

as shown in the figure. At the angle a = 30° the speed of centre of mass is 4m/s. If the mass of the person along with the skate board is 60 kg while, the

centre of mass is 0.60m from the surface. The normal reaction exerting between the surface and the skate board wheel is (take, g = 10 m/s2) (a) 512 N (b) 691 N (c) 864 N (d) 794 N

6.

An inclined plane having angle q with the horizontal. A body starts from rest to slide down the incline. If the height of incline is h then the time taken by the body to reach the bottom of the incline is

(a) sin q 2 h g æ è ç ö ø ÷ (b) 1 2 sin q h g (c) 2 h g cos q (d) 2 h g cos q

7.

As starting from rest a body slides down to an inclined plane having angle of inclination 30° is thrice the time taken by it to slide down the same distance in absence of friction. In this situation, the coefficient of friction between the body and the inclined surface would be (a) 7 9 (b) 5 6 (c) 8 9 (d) 3 8

8.

A log of weight W con nected to a rope of length L which is pulled by a force in such a way that the velocity of the log is constant. The distance between the free end of

the rope and ground is H. If the thickness of the log is assumed to be negligible, the frictional coefficient between log and ground is

(a) FL H FH WL 2 2 -- (b) WL FH F L H -2 2 (c) F L H WL FH 2 2 -- (d) WL FH FL H F -2 2

9.

There is a mass m which is oscillating inside the spherical shell along its diameter whose radius is R. The kinetic energy of oscillating mass at any time is K, the force applied by the mass on the shell at the same instant is (a) 7 9 R K (b) 3K R (c) 2 K R (d) 2 3 R K

10.

Figure shows three blocks of masses m m₁, ₂ and m₃ are

connected with one another. All the surfaces are frictionless. The string and pulleys are light in mass. The acceleration of mass m₁ is

(a) m g m m m 1 1 2 3 1 1 1 + + æ è ç ö ø ÷ (b) 4 1 2 3 2 3 m g m m m m + æ è ç ö ø ÷ (c) g m m m 1 4 1 1 1 2 3 + æ + è ç ö ø ÷ (d) m g m m 1 2 3 4æ 1 + 1 è ç ö ø ÷

Passage 1

A heavy particle whirling by means of a incompressible light string of length L. The particle is given to speed v₀ at its minimum height. The string becomes slack at some angle, and the particle proceeds on a parabola as shown in the figure.

11.

Determine the angle a if particle passes through point of suspension.

(a) tan- 1( ₃) _{(b) tan}- 1₍ ₂₎ (c) tan- 1( 5) (d) tan- æ è ç ö ø ÷ 1 3 2

12.

Determine the value of v0 if particle passes through point of suspension. (a) (g l + 2 3) 1 2 _{(b) [}_{g l 2}₍ ₃_)] 1 2 + (c) [ 3 2] 1 2 g l + (d) [ 2 3] 1 2 g l + m2 m3 m1 B A V a O X Y Parabolic path v P(Point to Slack) v0 Circular path

(14)

Passage 2

A wooden plank of length 2 m and of uniform area of cross section is hinged at one end to the bottom

of a tank as shown in figure.

The tank is filled with water upto a height of 1m. The specific gravity of the plank is 0.6. Excluding the case for which q is zero.

13.

Determine the buoyant force applied by the liquid. (a) 133. cos dg l q (b) 133. sin d lg q (c) 166. cos d lg q (d) 14. cos d l g q

14.

Find the an gle q that the plank makes with ver ti cal in the equi lib rium po si tion.

(a) cos ( .-10 64) (b) sin ( .-10 85) (c) cos ( . )-1₀₇₅ _{(d) tan ( . )}-1₀₇₀

Pas sage 3

A parallel plate capacitor is placed horizontally in such a way, that its linear plate is dipped into a liquid having dielectric constant K and density d. The area of plates is A. Now, the plates are connected to a battery that supplies a charge +Q to the upper plate of capacitor.

15.

The force on up per plate of the ca pac i tor will be? (a) (K )Q A K 2 2 0 2 1 2 -e (b) (K )Q KA - 1 2 2 0 e (c) 2 1 0 2 2 A K Q e ( - ) (d) 2 1 2 0 2 K A K Q e ( - )

16.

The rise in liq uid level h, is the space be tween plates of

ca pac i tor, is (a) 2 1 0 2 2 2 2 e A d g K Q ( - ) (b) 2 1 0 2 2 A d g K Q e ( - ) (c) (K )Q A K dg 2 2 2 2 0 1 2 -e (d) (K )Q AK dg - 1 2 2 2 0 e

17.

Match the Col umn Type If a 100 W sodium lamp is

radiating light of wavelength 5890 Å uniformly in all directions then match the following columns (all the magnitudes are in their SI units)

Column I Column II

A. Rate of emission of photon from the lamp (i.e.,

photon per second) 1. 4 9 10

7 . ´ B. The distance from lamp where average flux is 1

photon/cm2-s

2. 5 9 10_{. ´} 14

C. Flux of photon at a distance 2 m from the lamp 3. 2´104 D. Average density (photon/cm2_{) of photons at a}

distance 2 m from the lamp.

4. 3 10_´ 20

A B C D A B C D

(a) 4 1 2 3 (b) 4 1 3 2

(c) 3 1 4 2 (d) 1 2 3 4

18.

Match the following col umn re gard ing phys i cal Situations

A. Transition between two atomic energy levels

1. Photoelectric effect B. Moseley’s law 2. Characteristics of X-ray C. Change of photon energy into

kinetic energy of electron.

3. Hydrogen spectrum D. Electron emission. 4. b-decay

A B C D

(a) (1, 3) 2 1 (1, 4) (b)(1, 3) 3 1 (1, 4)

(c) 3 1 3 (1, 4)

(d)(1, 3) (1, 4) 1 (1, 3)

19.

An amount of 1 mole of a monoatomic ideal gas is taken along in a cy clic pro cess.

B A D B® ® ® and B A C B® ® ® as shown below in pV-diagram.

Relate to the above diagram match the column

Column I Column II A. D B® 1. 160 p V0 0ln2 B. D C® 2. 36p V0 0 C. A C® 3. 24p V0 0 D. A D® 4. 31p V0 0 A B C D A B C D (a) 4 3 2 1 (b) 4 3 1 2 (c) 3 1 2 4 (d) 1 3 2 4

20.

Match the fol low ing col umns with ref er ence to ex pres sions for the phys i cal quan ti ties.

A. Half- life of radioactive sample 1. N n

0( / )1 2 B. Atoms remaining after n half-lives 2. A e t

0 - l C. Decay constant, K 3. 0 693 1 2 0 . log / t N N D. Radioactivity 4. 0 693. / l A B C D A B C D (a) 4 1 3 2 (b) 4 2 3 1 (c) 1 4 3 2 (d) 3 4 2 1 a F P R mg Q C I A B C D Pressures(p) Volume(V) V0 VC VD p =32pA 0 p =pB 0

Detailed solutions of these questions are available on http://www.arihantbooks.com/Physics%20Spectrum.pdf

Answers

Paper 1

1. (b,c,d) 2. (b,c) 3. (a,b,c,d) 4. (a,b,c) 5. (a,b) 6. (a,c) 7. (a,b,c) 8. (a,c) 9. (a,d) 10. (a,d)

11. (3) 12. (5) 13. (2) 14. (1) 15. (4) 16. (2) 17. (3) 18. (6) 19. (1) 20. (2)

Paper 2

1. (a) 2. (b) 3. (d) 4. (c) 5. (d) 6. (b) 7. (c) 8. (c) 9. (b) 10. (c)

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Paper 1

1.

(b,c,d) If the particle is projected with velocity u at an angle q, then equation of its trajectory will be

y x gx u = tan -cos q q 2 2 2 2 We know slope is given by, m dy

dx = Therefore, slope, m gx u =tan -cos q q 2 2

It implies that the graph between slope and x will be straight line having negative slope and a non-zero positive intercept on y-axis. But x is directly proportional to the time t, therefore the shape of graph between slope and time will be same as that of the graph between slope and x. Hence, only option (a) is correct i.e. options (b), (c) and (d) are incorrect.

2.

(b,c) As we know that,

half-life of radioactive sample, N No n = æ è ç ö ø ÷ 1 2 for the time 60 min, and for sample A

NA No nA = æ è ç ö ø ÷ 1 2

where, n_A=total time= = half life 60 15 4 Þ NA=Noæ No è ç ö ø ÷ = 1 2 16 4 For sample B, N_B N_o nB = æ è ç1ö_ø÷ 2 where, nB= total time half life = = 60 30 2 Þ N_B₌N_oæ No è ç ö_ø÷ = 2 1 2 4

Now the ratio of residual praticles in the samples A and B

N N N N A B o o = = 16 4 1 4

Similarly for the time 80 min and for sample A,

N_A N_o nA = æ è ç1ö_ø÷ 2 where, nA= = 80 15 16 3 Þ NA=No æ è ç ö ø ÷ 1 2 16 3/ for sample B, N_B N_o N n o B = æ è ç1ö_ø÷ = æ_èç ö_ø÷ 2 1 2 8 9/ [as, n_B=80 30]

The ratio of residual sample i.e. N

N N N A B = æ è ç ö ø ÷ æ è ç ö_ø÷ 0 16 3 0 8 3 1 2 1 2 / / = 4 4_{.( )}1 3/

3.

(a,b,c,d) The speed of boat A with respect to ground

vA= v2+u2

The speed of boat B with respect to ground v_B= v2-u2 Þ vA>vB

To the ground observer, since the boat moves normal to the river currents therefore, it reaches the point exactly opposite point to the starting one.

The velocity of boat A with respect to ground is more that of boat

B. Therefore A travels longer distance with respect to ground

frame. The downward distance travelled by A

x ut u b v ub v = = æ è ç ö ø ÷ =

4.

(a, b, c) Consider the diagram

For equilibrium,

Tcos a =mg Þ T= mg cos a

Now the centripetal force acting along horizontal direction, we get mv r T 2 = sin a Þ v T r m = sin a = l g sin cos 2_a a (\ =r l sin a)

For the time period, T r v l g =2 = 2 2 p p a a a sin sin / cos l = 2p l a g cos

5.

(a, b) Consider the free body diagram of blocks x and y

where, f is the frictional force exerting between x and y is required to keep the blocks stationary and T is tension in the string. Then, for the equilibrium of the blocks

m g_y sin q =T+f …(i)

and m gx sin q =T-f …(ii)

Solving Eqs (i) and (ii), we get,

F₌(my-mx) sin qg

2 while f£f_max, then

( ) sin cos m m g m g y x y -£ q m q 2 or, m³(m -m ) tanq m y x y 2 a T O T cos a 2 mwr T sina mg y T _f mg S in q y mg c os q y mg x T f mg S in q x mg c osq x mg

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The blocks will remain stationary for this condition otherwise not. When, m_x=m_y, f = 0 and T=m g_x sin q and the system is in equilibrium. If m_y<m_x, the blocks will move only when m£(m_x-m_y)tan /q 2m_x.

6.

(a, c) Applying law of conservation of momentum for bullet-mass system, we get m v0 = (m+m V0) Þ V m v m m = + 0 0 ( ) …(i)

Now from energy conservation principle, we know 1 2 1 2 0 2 2 (m+m )V = Kx Þ V K m m x = + é ë ê ù û ú ( 0 ) …(ii) From Eqs (i) and (ii) we get.

K m m x m v m m + æ è ç ö ø ÷ = + 0 0 0 ( )

So, the velocity of spring which is compressed by an distance x,

we get v K m m x

m

=[ ( + )]/

0 1 2

0

7.

(a, b, c) As we know that the change in internal energy during a cyclic process is zero. This implies, that the heat supplied to the gas is equal to work done by the gas.

Þ WAB+WBC+WCA= - 1000 J …(i)

During process AB, the work done,

W_AB=p_A(V_B-V_A) =nR T(B-TA)

= 5 8.3 (450 250)´ ´ -= 5 8.3 200 ´ ´ = 8300 J

Now, the work done during process CA is zero, because the volume is constant

Therefore, 8300+WBC= -1000 J Þ WBC= - 9300 J

8.

(a, c) Let the piston is displaced through a distance x above its mean or equilibrium position. So, increase in volume of the gas DV=Ax

As the process is adiabatic Þ pVg= constant on differentiating it, we get. lnp+glnV=constant Þ Dp D p V V +g =0 Dp pD V V =g Now, we can write

Dp p D

V V

=g 0 0

The resultant force acting on the piston F =A p= -A p V V D g 0D 0 = -A p V x 2 0 0 g = -Kx where, K A p V = 2 0 0 g

Thus, it is clear that the motion of the piston is simple harmonic with angular frequency

w = K M = A p MV 2 0 0 g Now frequency, n = w p 2 = 1 2 2 0 0 p g A p MV

The time period, T =1 n T = 2 0 2 0 p g MV A p

9.

(a, d) According to Newton‘s law of cooling d dt K q_{= -} ₍_{q q}_- ₎ 0 …(i) Þ dq K dt q q- 0= -where, K is a constant on integrating, d K c c q q q- = ° °

ò

0 50 40 (10 min) Þ ln 40 C 20 C 50 C 20 C 10 ° - ° ° - ° é ë ê ù û ú = K Þ ln2 3 10/ = K K =ln( / )2 3 10 …(ii)

Let t be the time to cool the body upto temperature 30°C from 50°C. Then dq q q- = ° °

ò

40 30 C C Kt [from Eq (i)] Þ ln30 C 20 C 40 C 20 C ° - ° ° - ° = Kt Þ ln 1 2 æ è ç ö_ø÷ = Kt \ ln 1 2 ln (2 / 3) 10 æ è ç ö ø ÷ = ´ t [from Eq (ii)] Þ 10ln (1 / 2) ln(2 / 3) = t

10.

(a, d) When proton enters the magnetic field induction region with some inclination with it. The velocity has two components i.e. the component of velocity parallel to induction is

v||=2´105cos45° = 2´105m/s

and the component of velocity perpendicular and the magnetic induction, v_^=2´105_cos45° = 2´105

As, the magnetic, force f=q (v B´ ) is perpendicular to the magnetic induction

So, v|| is remains constant

Now, for radius r of the path followed by proton

qv B m v r ^ = ^ 2 Þ r mv qB = ^₌ ´ ´ ´ ´ ´ -( . ) . . 1 67 10 2 10 1 6 10 0 2 27 5 19 = ´ ´ = ´ = ´ -- -2 35 10 3 2 10 0 734 10 7 34 10 22 19 3 2 . . . . m

Now, time taken by proton to complete one revolution T = ^ 2pr v = ´ ´ ´ ´ -2 3 14 7 34 10 2 10 2 5 . . =46 09´ -1 4-1 10 7 . . =32 69 10. ´ -7=3 27 10. ´ -6s» 3.27ms C A B P re ss u re (p ) Volume(V)