DPP_EP_PC

Target

  PHYSICS

AIEEE - 2013 Course Name : ANOOP (EP)

Date : 02-07-2012

DPP No. : 13 to 14

Test Syllabus : PT-2 : (15- 07-12) : G.O upto Prism upto electric field due to disc and sheet This DPP is to be discussed in the week (02-07-12 to 07-07-12)

DPP No. : 13 To ta l M a r k s : 3 0

M a x. T ime : 30 m i n .

Single choice Objective (' –  – 1' negative marking) Q.1 to Q.10

( 3 m a rk s 3 m i n. )

[3 0 , 3 0]

1.

A ray of light falls on a plane mirror. When the mirror is turned, about an axis which is at right angle to the plane of the mirror through 20 º the angle between the incident ray and new r eflected ray is 45º. The angle between the incident ray and original reflected ray was therefore : (1) 65 º (2) 25º or 65º (3) 25 º  (4) 45 º

2.

An object moves in front of a fixed plane mirror. The velocity of the image of the object is (1) Equa Equall in the the ma magn gnit itud ude e and and in the the dire direct ctio ion n to to that that of the the obj objec ect. t. (2) (2) Equal Equal in the the mag magnit nitude ude and and oppo opposit site e in in direc directio tion n to to that that of the the object object.. (3) (3) Equal Equal in the the magnit magnitude ude and and the the di direc recti tion on will will be ei eithe therr same same or oppos opposite ite to that that of the the objec object. t. (4) (4) Equal Equal in in magnit magnitude ude and and makes makes any any angl angle e with with that that of the the objec objectt depend depending ing on on direct direction ion of of motion motion of of the object.

3.

Two plane mirrors are inclined to each other at 90 º. A ray of light is incident on one mirr or and the reflected light goes to the other mirro r. The ray will undergo a total deviatio n of : (1) 180 º (2) 90 º (3) 45 º (4) cannot be found because angle of incidenc e is not given.

4.

A ray of light is incident at an

!

of 30º on a plane mirror M1. Another plane mirror M 2 is inclined at angle # to

M1. What is the value of angle # so that light reflected from M 2 is parallel to M 1.

(1) 60 º 5.

 (3) 67.5 º

(4) none of these

A person AB of height 170 cm is standing infront of a plane mirror. His eyes are at height 164 cm. At what distance from P should a hole be made in the mirror so that he cannot see the top of his head.

( 1) 167 cm 6.

(2) 75 º

( 2) 1 61 c m

( 3) 1 63 c m

(4) none of these

Two plane mirrors are parallel to each other and spaced 20 cm apart. An object object is ke pt in between them at 15 cm from A. Out of the following at which point(s) image(s) is/are not formed in mirror A (distance (distance measured from mirror A): ( 1) 1 5 cm ( 2) 2 5 c m ( 3) 4 5 c m (4 ) 5 5 c m

7.

The figure shows two parallel plane mirrors placed 10 cm apart. An object O is placed 7 cm from mirror M 1. Then the distance of final image formed from mirror M 1 is :

(1) 7 cm 8.

(2) 13 cm

(3) 14 cm

(4)

$

A particle m oves in a plane fr om A to E along the shown path. It is gi ven that AB = BC = CD = DE = 10 metre. Then the magnitude of net displacement of particle is : D 108

°

E

108

°

C

108

°

B

A

(1) 10 m

9.

(2) 15 m

(3) 5 m

(4) 20 m

The dependence of variable y on variable x i s defined by the equation y =

x . Then the area occupied by this 2

curve and the x-axis in between x = 1 to x = 4 will be : (1) 10.

5 units 3

(2) 2 units

7  units 3

(3)

(4) 4 units

Three charges + 4q, -q and +4q are kept on a straight line at position (0, 0, 0), (a, 0, 0) and (2a, 0, 0) respectively. Considering that they are free to move along the x-axis only (1) all the charges are in stable equilibrium (2) all the charges are in unstable equilibrium (3) only the middle charge is in stable equilibrium (4) only middle charge is in unstable equilibrium

DPP No. : 14 Total Marks : 31

Max. Time : 32 min.

Single choice Objective (' – 1' negative marking) Q.1 to Q.9

(3 marks 3 min.)

[27, 27]

Subjective Questions (' – 1' negative marking) Q.10

(4 marks 5 min.)

[4, 5]

1.

The reflecting surface of a plane mirror is vertical. A particle is projected in a vertical plane which is also perpendicular to the mirror. The initial speed of the particle is 10 m/s and the angle of projection is 60 ° from the normal of the mirror. The point of projection is at a distance 5m from the mirror. The particle moves towards the mirror. Just before the particle touches the mirror, the velocity of approach of the particle and the image is ; (1) 10 m/s

(2) 5 m/s

(3)

10 3m /  s

(4)

5 3m /  s

2.

A luminous point object is moving along the principal axis of a concave mirror of focal length 12 cm towards it. When its distance from the mirror is 20 cm its velocity is 4 cm/s. The velocity of the image in cm/s at that instant is (1) 6, towards the mirror (2) 6, away from the mirror (3) 9, away from the mirror (4) 9, towards the mirror.

3.

The distance of an object from the pole of a concave m irror is equal to its radius of curvature. The image must be: (1) real (2) inverted (3) same sized (4) erect

4.

The distance of an object from a spherical mirror is equal to the focal length of the mirror. Then the image: (1) must be at infinity (2) may be at infinity (3) may be at the focus (4) none

5.

The incorrect statement for a concave mirror prod ucing a virtual image of an object. (1) The linear magnification is always greater than one, Except at the pole (2) The linear magnification is always less than one. (3) The magnification tends to one as the object moves nearer to the pole of the mirror. (4) The distance of the object from the pole of the mirror is less than the focal length of mirror.

6.

The image (of a real object) formed by a concave mirror is twice the size of the object. The focal length of the mirror is 20 cm. The distance of the object from the mirror is (are) (1) 10 cm (2) 30 cm (3) 25 cm (4) 15 cm

7.

A semicircular wire is uniformly charged with linear charge density dependent on the angle # from y-direction as % = %0 |sin# |, where %0 is a constant. The electric field intensity at the centre of the arc is

(1)

%0

( & j )

(2)

ˆ

2 ('0 R

%0

( & j )

(3)

ˆ

4 ('0 R

%0

( & j )

(4)

ˆ

6 ('0 R

%0

The resultant electric field at centre of a ring due to ring is zero. Which of the following is (1) The total charge of the ring may be zero, although every part of the ring has charge. (2) The charge on the ring must be uniformly distributed. (3) The charge on the ring may be distributed nonuniformly. (4) Total charge on the ring may be positive.

9.

The vector joining the points A (1, 1,  –1) and B (2,  –3, 4) & pointing from A to B is (1)  – i  + 4 j  – 5 k ˆ

ˆ

ˆ

(3) i  – 4  j  + 5 k

(2) i  + 4  j + 5 k ˆ

ˆ

ˆ

ˆ

ˆ

incorrect  :

(4)  – i  – 4 j  – 5 k .

ˆ

ˆ

ˆ

2 2 ('0 R

8.

( & j )

ˆ

ˆ

10.

The forces, each numerically equal to 5 N, are acting as shown in the Figure. Find the angle between forces?

 1.

(1)

2.

(4)

3.

(2)

8.

(3)

9.

(2)

10.

(3)

1.

(3)

2.

(4)

3.

(4)

 8.

(1)

9.

(4)

10.

 1.

(2)

2.

(1)

3.

8.

(2)

9.

(4)

4.

(2)

5.

(2)

6.

(2)

7.

(1)

4.

(1)

5.

(4)

6.

(1)

7.

(2)

(3)

5.

(3)

6.

(2)

7.

(2)

2 i ) 3 j ) 4k ˆ

ˆ

ˆ

29

(3)

4.

  PHYSICS

Target AIEEE - 2013 Course Name : ANOOP (EP)

Date : 02-07-2012

DPP No. : 13 to 14

Test Syllabus : PT-2 : (15- 07-12) : G.O upto Prism upto electric field due to disc and sheet This DPP is to be discussed in the week (02-07-12 to 07-07-12)

DPP No. : 13 Total Marks : 30

Max. Time : 30 min.

Single choice Objective (' – 1' negative marking) Q.1 to Q.10

1.

(3 marks 3 min.)

[30, 30]

,d izdk'k dh fdj.k lery niZ.k ij fxjrh gSA tc niZ.k dks blds ry ds yEcor~ v{k ds lkis{k 20 º ls ?kqek;k tkrk gS rks vkifrr fdj.k o u;h ijkofrZr fdj.k ds chp dks.k 45 º gSA vkifrr fdj.k o ewy ijkofrZr fdj.k ds chp dks.k Fkk (1) 65 º

(2) 25 º or 65º

(3) 25 º

 (4) 45 º

2.

,d fcEc ,d fLFkj lery niZ.k ds lkeus xfr djrk gSA fcEc ds izfrfcEc d k osx (1) fcEc ds osx ds ifjek.k vk Sj fn'kk ds leku gSA (2) fcEc ds osx ds ifjek.k ds leku gS vkSj fn'kk ds foijhr g SA (3) fcEc ds osx ds ifjek.k ds leku gS vkSj fn'kk ;k rks mlds ¼fcEc ds½ leku g ksxh ;k mlds ¼fcEc ds½ foijhr gksxhA (4) fcEc ds osx ds ifjek.k ds leku gS vkSj bldh fn'kk fcEc ds lkFk ,slk dksbZ Hkh dks.k cukrh gS tks fcEc dh xfr dh fn'kk ij fuHkZj djrh gSA

3.

nks lery niZ.k ,d nwljs ls 90º ds dks.k ij j[ks gSa A ,d izdk'k fdj.k ,d niZ.k ij vkifrr gksrh gS rFkk ijkofrZr fdj.k nwljs niZ.k ij tkrh gSA fdj.k dk dqy fopyu gksxk & (1) 180 º (3) 45 º

4.

(2) 75 º

 (3) 67.5 º

(4)

(2) 161 cm

(3) 163 cm

(4)

M2, M1

164 cm

Å¡pkbZ ij gSA

(2) 25 cm

(3) 45 cm

P ls

fdl

buesa ls dksbZ ugh

nks lery niZ.k 20 cm nwjh ij ,d nwljs ds lkekukUrj j[ks tkrs gSA ,d oLrq dks mu nksuks ds chp esa ij j[krs gSA fuEu esa ls fdu fcUnqvksa ij niZ.k A esa izfrfcEc ugha curk (1) 15 cm

ls " dks.k cukrk

buesa ls dksbZ ugh

,d O;fDr ftldh Å¡pkbZ 170 cm gS ,d lery niZ.k ds lkeus [kM+k gSA mldh vk¡[ks nwjh ij Nsn cuk;k tk;s ftlls og viuk flj dk Åijh fljk ugh ns[k ldrk A

(1) 167 cm 6.

ugha dj ldrs  D;ks afd vkiru dks.k ugha fn;k x;k gS A

,d izdk'k fdj.k lery niZ.k M1 ij 30º ds dks.k ij vkifrr gksrh gSA ,d nwljk lery niZ.k gSA dks.k " dk eku D;k gks rkfd izdk'k M2 ls ijkorZu ds i'pkr~ M1 ds lekukUrj gksA

(1) 60 º 5.

(2) 90 º (4) Kkr

(4) 55 cm

A ls 15 cm

nwjh

7.

nks lekUrj lery niZ.k 10 cm dh nwj  h ij j[ks gq, gSa ,d oLrq cus vfUre izfrfcEc dh nwjh gksxh :

(1) 7 cm 8.

(2) 13 cm

O niZ.k M1 ls 7 cm dh

(3) 14 cm

fp=kkuqlkj ,d d.k fdlh lery esa iFk A ls rc d.k ds dqy foLFkkiu dk ifjek.k gksxkA

E ds

(4)

nwjh ij j[kk gqvk gSA rks niZ.k

M1 ls

AB = BC = CD = DE = 10

ehVj]

#

vuqfn'k xfr djrk gSA fn;k x;k gS D 108

°

E

108

°

C

108

°

B

A

(1) 10 m

9.

pj

x ij]

(2) 15 m

pj y dh fuHkZjrk lehdj.k

(3) 5 m

y=

x 2

}kjk ifjHkkf"kr dh tkrh gS rc

(4) 20 m

x = 1 vkSj x = 4 ds

chp x v{k vkSj oØ }kjk

f?kjk gqvk {ks=kQy gksxkA (1) 10.

5 3

units

(2) 2 units

7  units 3

(3)

(4) 4 units

+ 4q, -q o +4q ds

rhu vkos'k ,d lh/kh js[kk ij Øe'k% (0, 0, 0), (a, 0, 0) vkSj (2a, 0, 0) fLFkfr;ksa ij j[ks gSA ;g ekurs  gq, fd ;s vkos'k dsoy x-v{k ds vuqfn'k xfr ds fy;s LorU=k gS  rc (1) lHkh vkos'k LFkk;h larqyu esa gS A (2) lHkh vkos'k vLFkk;h larqyu esa gSA (3) dsoy e/; vkos'k LFkk;h lUrqyu esa gSA (4) dsoy e/; vkos'k vLFkk;h lUrqyu esa gSA DPP No. : 14

Total Marks : 31

Max. Time : 32 min.

Single choice Objective (' – 1' negative marking) Q.1 to Q.9

(3 marks 3 min.)

[27, 27]

Subjective Questions (' – 1' negative marking) Q.10

(4 marks 5 min.)

[4, 5]

1.

lery niZ.k dh ijkorZd lrg Å/okZ/kj gSA ,d d.k Å/okZ/kj ry esa iz{ksfir fd;k tkrk gS  rFkk ;g Å/okZ /kj ry niZ.k ds yEcor~ gSA d.k dh izkjfEHkd pky 10 m/s rFkk niZ.k ds vfHkyEc ds lkFk iz{ksi.k dks.k 60º gSA iz{ksi.k fcUnq niZ.k ls 5m dh nwjh ij gSA d.k niZ.k dh rjQ xfr djrk gSA d.k ds niZ.k dks Li'kZ djus ds Bhd igys  d.k rFkk izfrfcEc dk vyxko osx gksxk ; (1) 10 m/s

(2) 5 m/s

(3)

10 3m /  s

(4)

5 3m /  s

2.

,d pedhyk fcUnq fcEc]  12 cm Qksdl nwjh ds vory niZ.k ds çdk'kh; v{k ds vuqfn'k niZ .k dh rjQ xfr'khy gS A tc ;g niZ.k ls 20 cm nwjh ij gS rks bldk osx  4 cm/s gksrk gSA bl le; çfrfcEc dk cm/s es a osx gksxk \ (1) 6, niZ.k dh rjQ  (2) 6, niZ.k ls nwj (3) 9, niZ.k ls nwj (4) 9, niZ.k dh rjQ 

3.

,d vory niZ.k ds /kqzo ls ,d oLrw dh nwjh bldh oØrk f=kT;k ds cjkcj g SA izfrfcEc gksxk& (1) okLrfod (2) mYVk (3) leku vkdkj dk (4) lh/kk

4.

fdlh oLrq dh ,d xksyh; niZ.k ls nw jh niZ.k dh Qksdl nwjh ds cjkcj gSA rc izfrfcEc (1) vuUr ij gksuk pkfg;s (2) vuUr ij gks ldrk gS (3) Qksdl ij gks ldrk gS (4) dksbZ ugh

5.

fdlh oLrq dk vkHkklh izfrfcEc c ukus okys ,d vory niZ.k ds fy;s xyr dFku gSA (1) /kzqo ds vykok ] jsf[kd vko /kZu lnSo ,d ls cM+k gksrk gSA (2) jsf[kd vko/kZu lnSo ,d ls NksVk gksrk gSA (3) oLrq ds niZ.k ds /kzqo ds fudV tkus ij vko/kZu ,d dh vksj vxzlj gksrk gSA (4) niZ.k ds /kzqo ls oLrq dh nwjh niZ.k dh Qksdl nwjh ls de gk srh gSA

6.

,d okLrfod fcEc dk ç frfcEc ,d vory niZ.k }kjk fcEc ds vkdkj dk nks xquk curk gSA niZ.k dh Qksdl nwjh gSA niZ.k ls fcEc dh nwjh@nwfj;ka gS@gSa & (1) 10 cm

7.

(2) 30 cm

(3) 25 cm

,d v)ZoÙkkdkj rkj js[kh; vkos'k ?kuRo tks y fn'kk ls dks.k " ij gSA pki ds dsUnz ij fo|qr {ks=k dh rhozrk gSA

(1)

$0

( % j )

(2)

ˆ

2 '&0 R

$0

( % j )

(3)

ˆ

4 '&0 R

(4) 15 cm

$ = $ 0 |sin" |,

$0

( % j )

ds vuqlkj fuHkZj djrk gSA tgk¡

(4)

ˆ

6 '&0 R

$0

2 2 '&0 R

8.

oy; ds dkj.k oy; ds dsUnz ij ifj.kkeh fo|qr {ks=k 'kwU; gSA fuEu esa ls  dkSulk vlR; gSA oy; dk dqy vkos'k 'kwU; gks ldrk gS ;|fi oy; dk izR;sd Hkkx vkosf'kr gSA (1) oy; ij vkos'k ,d leku :i ls gh forfjr gksA (2) oy; ij vkos'k vleku :i ls forfjr gks ldrk gSA (3) oy; ij dqy vkos'k /kukRed gks ldrk gSA (4)

9.

lfn'k

A (1, 1,  –1) rFkk

(1)  – i  + 4 j  – 5 k ˆ

ˆ

ˆ

lfn'k  B (2,  –3, 4) dks feykus okyk rFkk ˆ

5 N gS

ˆ

ˆ

ˆ

10.

izR;sd cy dk ifjek.k

 1.

(1)

2.

(4)

3.

(2)

8.

(3)

9.

(2)

10.

(3)

1.

(3)

2.

(4)

3.

(4)

A ls B dh

ˆ

$0

,d fu;rkad

( % j ) ˆ

rjQ funsZf'kr lfn'k gS

(3) i  – 4  j  + 5 k

(2) i  + 4  j + 5 k

(4)  – i  – 4 j  – 5 k . ˆ

ˆ

ˆ

ˆ

rFkk og fp=kkuqlkj dk;Zjr gS rks cyksa ds chp dk dks.k crkvks ?

 8.

(1)

9.

(4)

10.

 1.

(2)

2.

(1)

3.

8.

(2)

9.

(4)

4.

(2)

5.

(2)

6.

(2)

7.

(1)

4.

(1)

5.

(4)

6.

(1)

7.

(2)

(3)

5.

(3)

6.

(2)

7.

(2)

2 i ( 3 j ( 4k ˆ

ˆ

ˆ

29

(3)

4.

20 cm