1. No category

Test Booklet Code
C
Regd. Office : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075
Ph.: 011-47623456 Fax : 011-47623472
Time : 3 hrs.
JEE (MAIN)-2015
M.M. : 360
(Mathematics, Physics and Chemistry)
Important Instructions :
1.
The test is of 3 hours duration.
2.
The Test Booklet consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Mathematics, Physics and
Chemistry having 30 questions in each part of equal weightage. Each question is allotted 4 (four)
marks for each correct response.
4.
Candidates will be awarded marks as stated above in Instructions No. 3 for correct response of each
question. ¼ (one-fourth) marks will be deducted for indicating incorrect response of each question.
No deduction from the total score will be made if no response is indicated for an item in the answer
sheet.
5.
There is only one correct response for each question. Filling up more than one response in each
question will be treated as wrong response and marks for wrong response will be deducted accordingly
as per instruction 4 above.
6.
Use Blue/Black Ball Point Pen only for writing particulars/marking responses on Side-1 and Side-2
of the Answer Sheet. Use of pencil is strictly prohibited.
7.
No candidate is allowed to carry any textual material, printed or written, bits of papers, pager, mobile
phone, any electronic device, etc. except the Admit Card inside the examination room/hall.
8.
The CODE for this Booklet is
C. Make sure that the CODE printed on Side-2 of the Answer Sheet
and also tally the serial number of the Test Booklet and Answer Sheet are the same as that on this
booklet. In case of discrepancy, the candidate should immediately report the matter to the Invigilator
for replacement of both the Test Booklet and the Answer Sheet.
JEE (MAIN)-2015
PART–A : MATHEMATICS
1.
A complex number z is said to be unimodular if
|z| = 1. Suppose z1 and z2 are complex numbers
1.
z1  2 z2
such that
is unimodular and z 2 is not
2  z1 z2
unimodular. Then the point z1 lies on a
(1) Circle of radius
2
2
(1)
(4) Circle of radius 2
2.
(1) Meets the curve again in the fourth quadrant
(2) Does not meet the curve again
(3) Meets the curve again in the second quadrant
(4) Meets the curve again in the third quadrant
The sum of first 9 terms of the series
13 1 3  2 3 13  2 3  3 3


 ........ is
1
1 3
135
3.
z2 ,dekikadh
ugha gS] rks
(2) x-v{k
ds lekarj ,d js[kk ijA
(3) y-v{k
ds lekarj ,d js[kk ijA
okys o`Ùk ijA
oØ
x2 + 2xy – 3y2 = 0
(1)
oØ dks nksckjk prqFkZ prqFkk±'k esa feyrk gS
(2)
oØ dks nksckjk ugha feyrk
(3)
oØ dks nksckjk f}rh; prqFkk±'k esa feyrk gS
(4)
oØ dks nksckjk r`rh; prqFkk±'k esa feyrk gS
Js.kh
z1  2 z2
2  z1 z2
fcUnq z1
f=kT;k okys o`Ùk ijA
(4) 2 f=kT;k
The normal to the curve, x2 + 2xy – 3y2 = 0 at (1,1) :
|z| = 1
,dekikadh dgykrh gS ;fn
rFkk z2 ,slh lfEeJ la[;k,¡ gSa fd
,dekikadh gS rFkk
fLFkr gS
(3) Straight line parallel to y-axis
3.
z1
gSA ekuk
(2) Straight line parallel to x-axis
2.
z
,d lfEeJ la[;k
ds fcUnq (1,1) ij vfHkyEc
13 1 3  2 3 13  2 3  3 3


 ........
1
1 3
135
ds izFke
9 inksa
dk ;ksx gS
4.
(1) 192
(2) 71
(3) 96
(4) 142
Let f(x) be a polynomial of degree four having extreme
4.
f (x) ⎤
⎡
values at x = 1 and x = 2. If lim ⎢ 1 
⎥  3 , then
x 0 ⎣
x2 ⎦
f(2) is equal to
(1) 192
(2) 71
(3) 96
(4) 142
ekuk
f(x)
x = 2 ij
?kkr
4
dk ,d cgq i n gS ftlds
pje eku gSAa ;fn
f (x) ⎤
⎡
lim ⎢ 1 
⎥3
x 0 ⎣
x2 ⎦
x=1
rFkk
gS] rks
f(2)
cjkcj gS
5.
6.
(1) 4
(2) –8
(3) –4
(4) 0
The negation of ~ s  (~ r s) is equivalent to
(1) s  r
(2) s  ~ r
(3) s  (r  ~ s)
(4) s  (r  ~ s)
5.
⎡1 2 2 ⎤
⎢
⎥
If A = ⎢ 2 1 2 ⎥ is a matrix satisfying the
⎢⎣ a 2 b ⎥⎦
equation AAT = 9I, where I is 3 × 3 identity matrix,
then the ordered pair (a, b) is equal to
6.
(1) 4
(2) –8
(3) –4
(4) 0
~ s  (~ r s) dk
fu"ks/ lerqY; gS
(1) s  r
(2) s  ~ r
(3) s  (r  ~ s)
(4) s  (r  ~ s)
;fn
⎡1 2 2 ⎤
A = ⎢⎢ 2 1 2 ⎥⎥
⎢⎣ a 2 b ⎥⎦
,d ,slk vkO;wg gS tks vkO;wg
lehdj.k AAT = 9I, dks larq"V djrk gS] tgk¡ I, 3 × 3 dk
rRled vkO;wg gS] rks Øfer ;qXe (a, b) dk eku gS
(1) (–2, –1)
(2) (2, –1)
(1) (–2, –1)
(2) (2, –1)
(3) (–2, 1)
(4) (2, 1)
(3) (–2, 1)
(4) (2, 1)
2
JEE (MAIN)-2015
7.
The integral
1
4
⎛ x  1⎞
(1)  ⎜ 4 ⎟  c
⎝ x ⎠
4
1
4
8.
1
⎛ x4  1⎞ 4
(1)  ⎜ 4 ⎟  c
⎝ x ⎠
1
4
(3) ( x 4  1) 4  c
1
⎛ x4  1⎞ 4
(2) ⎜ 4 ⎟  c
⎝ x ⎠
1
1
8.
(4) ( x 4  1) 4  c
means between l and n, then G14  2G24  G34 equals.
;fn nks fofHk okLrfod la[;kvksa l rFkk n (l, n > 1) dk lekarj
ekè; (A.M.) m gS vkSj l rFkk n ds chp rhu xq.kksÙkj ekè;
(G.M.) G1, G2 rFkk G3 gSa] rks G14  2G24  G34 cjkcj gS
(1) 4 l2m2n2
(1) 4 l2m2n2
(2) 4 l2mn
(3) 4 lm2n
(4) 4 lmn2
If m is the A.M. of two distinct real numbers l and
n (l, n > 1) and G1, G2 and G3 are three geometric
(3) 4
9.
4
lekdy ∫ 2 4dx 3/4 cjkcj gS
x ( x  1)
1
4
⎛ x  1⎞
(2) ⎜ 4 ⎟  c
⎝ x ⎠
4
(4) ( x  1)  c
(3) ( x  1)  c
4
7.
dx
∫ x 2 (x 4  1)3/4 equals
(2) 4 l2mn
lm2n
(4) 4
lmn2
Let y(x) be the solution of the differential equation
 x log x 
9.
dy
 y  2 x log x , ( x  1).
dx
ekuk vody lehdj.k
 x log x 
Then y(e) is equal to
dy
 y  2 x log x , ( x  1)
dx
(1) 2e
(2) e
y(e) cjkcj
(3) 0
(4) 2
(1) 2e
(2) e
(3) 0
(4) 2
10. The number of integers greater than 6,000 that can
be formed, using the digits 3, 5, 6, 7 and 8, without
repetition, is
(1) 72
(3) 192
10.
vadksa
iz;ksx ls] fcuk nksgjk;s] cuus okys
6,000 ls cM+s iw.kk±dksa dh la[;k gS
(2) 216
(4) 120
(3) 192
(4) 120
(1) 780
(2) 901
(3) 861
(4) 820
12. Let  and  be the roots of equation
x2
– 6x – 2 = 0.
If an = n – n, for n  1, then the value of
12.
a10 – 2 a8
2 a9
f=kHkqt] ftlds 'kh"kZ (0, 0), (0, 41) rFkk (41, 0) gSa] ds
vkUrfjd Hkkx esa fLFkr mu fcUnqvksa dh la[;k ftuds nksuska
funsZ'kkad iw.kk±d gSa] gS
(1) 780
(2) 901
(3) 861
(4) 820
ekuk  rFkk  f}?kkr lehdj.k
;fn
is equal to
n1
x2 – 6x – 2 = 0
ds fy,] an = n – n gS] rks
eku gS
(1) –3
(2) 6
(1) –3
(2) 6
(3) –6
(4) 3
(3) –6
(4) 3
⎛ 2x ⎞
13. Let tan 1 y  tan 1 x  tan 1 ⎜
⎝ 1  x 2 ⎟⎠
where |x|
1
3
13.
ekuk
⎛ 2x ⎞
tan 1 y  tan 1 x  tan 1 ⎜
⎝ 1  x 2 ⎟⎠
tgk¡ |x|
. Then a value of y is
1
3
gS] rks
y dk
,d eku gS
(1)
3x  x 3
1  3x 2
(2)
3x  x 3
1  3x 2
(1)
3x  x 3
1  3x 2
(2)
3x  x 3
1  3x 2
(3)
3x  x 3
1  3x 2
(4)
3x  x 3
1  3x 2
(3)
3x  x 3
1  3x 2
(4)
3x  x 3
1  3x 2
3
gS] rks
3, 5, 6, 7 rFkk 8 ds
(1) 72
11.
y(x)
gS
(2) 216
11. The number of points, having both co-ordinates as
integers, that lie in the interior of the triangle with
vertices (0, 0), (0, 41) and (41, 0), is
dk gy
ds ewy gSaA
a10 – 2 a8
2 a9
dk
JEE (MAIN)-2015
14. The distance of the point (1, 0, 2) from the point of
14.
x2 y1 z2


intersection of the line
and the
3
4
12
plane x – y + z = 16, is
x2 y1 z2


3
4
12
js[kk
ds izfrPNsn fcUnq dh] fcUnq
rFkk lery
(1, 0, 2) ls
nwjh gS
(1) 13
(1) 13
(2) 2 14
(2) 2 14
(3) 8
(4) 3 21
(3) 8
(4) 3 21
15. {(x, y) : y2  2x
rFkk y  4x – 1} }kjk ifjHkkf"kr {ks=k dk
{ks=kiQy (oxZ bdkb;ks)a esa gS
15. The area (in sq. units) of the region described by
{(x, y) : y2  2x and y  4x – 1} is
9
(1)
32
7
(2)
32
(1)
9
32
(2)
7
32
5
(3)
64
15
(4)
64
(3)
5
64
(4)
15
64
16.
16. Let O be the vertex and Q be any point on the
parabola, x2 = 8y. If the point P divides the line
segment OQ internally in the ratio 1 : 3, then the
locus of P is
(1) x 2  2 y
(2) x 2  y
(3) y 2  x
(4) y 2  2 x
(1) 14.0
(2) 16.8
(3) 16.0
(4) 15.8
ekuk ijoy; x2 = 8y dk 'kh"kZ O rFkk ml ij dksbZ fcUnq
Q gSA ;fn fcUnq P] js[kk[kaM OQ dks 1 : 3 ds vkarfjd
vuqikr esa ck¡Vrk gS] rks P dk fcUnqiFk gS %
(1) x 2  2 y
(2) x 2  y
(3) y 2  x
(4) y 2  2 x
17. 16 iz{s k.kksa
okys vk¡dM+kas dk ekè; 16 gSA;fn ,d iz{s k.k ftldk
eku 16 gS] dks gVk dj] 3 u;s izs{k.k ftuds eku 3, 4 rFkk
5 gSa] vk¡dM+ksa esa feyk fn;s tkrs gSa] rks u;s vk¡dM+ksa dk
ekè; gS
17. The mean of the data set comprising of 16
observations is 16. If one of the observation valued
16 is deleted and three new observations valued 3,
4 and 5 are added to the data, then the mean of the
resultant data, is
18.
18. The area (in sq. units) of the quadrilateral formed by
the tangents at the end points of the latera recta to
(1) 14.0
(2) 16.8
(3) 16.0
(4) 15.8
nh?kZo`Ùk
x2 y2

1
9
5
ds ukfHkyEcksa ds fljksa ij [khaph xbZ
Li'kZ js[kkvksa }kjk fufeZr prqHkqZt dk {ks=kiQy (oxZ bdkb;ksa
es)a gSa
x2 y2

 1 , is
the ellipse
9
5
(1) 27
(1) 27
27
4
(3) 18
(2)
27
(2)
4
(3) 18
(4)
x – y + z = 16
(4)
27
2
19.
19. The equation of the plane containing the line 2x –
5y + z = 3; x + y + 4z = 5, and parallel to the plane,
x + 3y + 6z = 1, is
(1) 2 x  6 y  12 z  13
(2) 2 x  6 y  12 z  13
(3) x  3y  6 z  7
(4) x  3y  6 z  7
4
27
2
js[kk 2x – 5y + z = 3; x + y + 4z = 5 dks varfoZ"V djus
okys lery] tks lery x + 3y + 6z = 1 ds lekarj gSa]
dk lehdj.k gS
(1) 2 x  6 y  12 z  13
(2) 2 x  6 y  12 z  13
(3) x  3y  6 z  7
(4) x  3y  6 z  7
JEE (MAIN)-2015
20. The number of common tangents to the circles
x2 + y2 – 4x – 6y – 12 = 0 and
x2 + y2 + 6x + 18y + 26 = 0, is
(1) 4
(2) 1
(3) 2
(4) 3
20.
21. The set of all values of  for which the system of
linear equations
x2 + y2 – 4x – 6y – 12 = 0
rFkk
x2 + y2 + 6x + 18y + 26 = 0 dh mHk;fu"B Li'kZ js[kkvksa
dh la[;k gS
(1) 4
(2) 1
(3) 2
(4) 3
21.  ds
lHkh ekuksa dk leqPp;] ftuds fy, jSf[kd lehdj.k
fudk;
2x1 – 2x2 + x3 = x1
2x1 – 2x2 + x3 = x1
2x1 – 3x2 + 2x3 = x2
2x1 – 3x2 + 2x3 = x2
–x1 + 2x2 = x3
–x1 + 2x2 = x3
has a non-trivial solution
dk ,d vrqPN gy gS]
(1) Contains more than two elements
(2) Is an empty set
(3) Is a singleton
(4) Contains two elements
11
(2)
55 ⎛ 2 ⎞
⎜ ⎟
3 ⎝3⎠
22.
11
12
10
2
⎛1⎞
(3) 55 ⎛⎜ ⎞⎟
(4) 220 ⎜ ⎟
⎝3⎠
⎝3⎠
23. The sum of coefficients of integral powers of x in

the binomial expansion of 1  2 x

50
(1)
esa nks ls vf/d vo;o gSAa
(2)
,d fjDr leqPp; gSA
(3)
,d ,dy leqPp; gSA
(4)
22. If 12 identical balls are to be placed in 3 identical
boxes, then the probability that one the boxes
contains exactly 3 balls is
⎛1⎞
(1) 22 ⎜ ⎟
⎝3⎠
o`Ùkksa
23.
is
esa nks vo;o gSa
;fn 12 ,d tSlh xsans] 3 ,d tSls cDlksa esa j[kh tkrh gSa]
rks buesa ls ,d cDls esa Bhd 3 xsna as gksus dh izkf;drk gS
⎛1⎞
(1) 22 ⎜ ⎟
⎝3⎠
11
2
(3) 55 ⎛⎜ ⎞⎟
⎝3⎠
10
1  2 x 
(2)
50
55 ⎛ 2 ⎞
⎜ ⎟
3 ⎝3⎠
11
⎛1⎞
(4) 220 ⎜ ⎟
⎝3⎠
ds f}in izlkj esa
x
12
dh iw.kk±dh; ?kkrksa ds
xq.kkadksa dk ;ksx gS
(1)
1 50
(2  1)
2
(2)
1 50
(3  1)
2
(1)
1 50
(2  1)
2
(2)
1 50
(3  1)
2
(3)
1 50
(3 )
2
(4)
1 50
(3  1)
2
(3)
1 50
(3 )
2
(4)
1 50
(3  1)
2
4
24. The integral
to
∫
2 log x
log x 2
2
 log(36 – 12 x  x 2 )
(1) 6
(2) 2
(3) 4
(4) 1
4
dx is equal
24.
25.
25. If the function.
(1) 4
(2) 2
16
(3)
5
10
(4)
3
dx cjkcj gS
lekdy ∫
log x 2  log(36 – 12 x  x 2 )
2
(1) 6
(2) 2
(3) 4
(4) 1
;fn iQyu
⎪⎧ k x  1
g( x )  ⎨
⎪⎩ mx  2
⎪⎧ k x  1 , 0  x  3
g( x )  ⎨
⎪⎩ mx  2 , 3  x  5
is differentiable, the value of k + m is
vodyuh; gS] rks
(1) 4
(3)
5
log x 2
16
5
, 0x3
, 3x5
k + m dk
eku gS
(2) 2
(4)
10
3
JEE (MAIN)-2015
26.
26. Locus of the image of the point (2, 3) in the line
(2x – 3y + 4) + k(x – 2y + 3) = 0, k  R, is a
(1) Circle of radius
3
(2, 3)
3
(1)
(2) Straight line parallel to x-axis
(3) Straight line parallel to y-axis
(4) Circle of radius
ds js[kk (2x – 3y + 4) + k(x – 2y + 3) = 0,
k  R esa izfrfcac dk fcUnqiFk ,d
fcUnq
(2) x-v{k
ds lekarj js[kk gSAa
(3) y-v{k
ds lekarj js[kk gSAa
2
2
(4)
27.
lim
 1  cos 2 x  3  cos x 
x tan 4x
x 0
is equal to
27.
1
2
(2) 4
(3) 3
(4) 2
(1)
(1) 2 : 3
3 :
(3)
(2)
2
lim
28.
3 : 1
(4) 1 :
(2) 219
(3) 256
(4) 275
 1  cos 2 x  3  cos x 
x tan 4x
3
(2) 4
(3) 3
(4) 2
rhu lajs[k fcUnqvksa A, B rFkk C, ,d ,slh js[kk ij fLFkr gSa
tks ,d ehukj ds ikn dh fn'kk esa ys tkrh gS] ls ,d
ehukj ds f'k[kj ds mÂ;u dks.k Øe'k% 30º, 45º rFkk 60º
gS]a rks AB : BC dk vuqikr gS
30. Let a , b and c be three non-zero vectors such that
no two of them are collinear and
30.
1 ( a  b )  c  |b ||c | a . If  is the angle between
3
vectors b and c , then a value of sin  is
(2)
3 :
(3)
29.
3 : 1
(4) 1 :
2
3
ekuk A rFkk B nks leqPp; gSa ftuesa Øe'k% pkj rFkk nks
vo;o gS]a rks leqPp; A × B ds mu mileqPp;ksa dh la[;k]
ftuesa izR;sd esa de ls de rhu vo;o gS]a gS
(1) 510
(2) 219
(3) 256
(4) 275
ekuk
a, b
rFkk c rhu 'kwU;srj ,sls lfn'k gSa fd muesa ls
dksbZ nks laj[s k ugha gSa rFkk
lfn'kksa
eku gS
b
rFkk
(1)
2 3
3
(1)
2 3
3
(2)
2 2
3
(2)
2 2
3
(3)
 2
3
(3)
 2
3
(4)
2
3
(4)
2
3
6
cjkcj gS
1
2
(1) 2 : 3
29. Let A and B be two sets containing four and two
elements respectively. Then the number of subsets of
the set A × B, each having at least three elements is
(1) 510
f=kT;k dk o`Ùk gSA
x 0
(1)
28. If the angles of elevation of the top of a tower from
three collinear points A, B and C, on a line leading
to the foot of the tower, are 30º, 45º and 60º
respectively, then the ratio, AB : BC, is
f=kT;k dk o`Ùk gSA
c
1 ( a  b )  c  |b ||c | a
3
ds chp dk dks.k
 gS]
rks
gSA ;fn
sin  dk
,d
JEE (MAIN)-2015
PART–B : PHYSICS
31. In the circuit shown, the current in the 1  resistor
is
6V
31.
n'kkZ;s x;s ifjiFk esa
6V
2
P
1
3
1 izfrjksèkd
1
3
9V
3
Q
Q
(1) 0.13 A, P ls Q dh
(1) 0.13 A, from P to Q (2) 1.3 A, from P to Q
(3) 0 A
(4) 0.13 A, from Q to P
32. Distance of the centre of mass of a solid uniform
cone from its vertex is z0. If the radius of its base is
R and its height is h then z0 is equal to
(1)
3h 2
8R
(2)
h2
4R
(3)
3h
4
(4)
5h
8
32.
33. Match List-I (Fundamental Experiment) with List-II
(its conclusion) and select the correct option from
the choices given below the list:
List -I
(A) Franck-Hertz
experiment
(B) Photo-electric
experiment
(C) Davison-Germer
experiment
2
P
9V
3
(3) 0 A
ls izokfgr èkkjk gksxh%
33.
List-II
Particle nature
of light
(ii) Discrete energy
levels of atom
(iii) Wave nature of
electron
(iv) Structure of
atom
vksj (2)
1.3 A, P ls Q dh
(4) 0.13 A, Q ls P dh
(1)
3h 2
8R
(2)
h2
4R
(3)
3h
4
(4)
5h
8
lwph&I (ewy iz;ksx) dk lwph&II (mlds ifj.kke) ds lkFk
lqesyu (eSp) dhft;s vkSj fuEukafdr fodYiksa esa ls lgh
fodYi dk p;u dhft;s %
(A)
lwph -I
izsaQd gV~Zl iz;ksx
(i)
(B)
izdk'k fo|qr iz;ksx
(ii)
(C)
Msohlu teZj iz;ksx
(iii)
lwph-II
izdk'k dh df.kdk
izÑfr
v.kq ds fofoDr ÅtkZ
Lrj
bysDVªkWu dh rjax
izÑfr
ijek.kq dh lajpuk
(1) (A) - (iv)
(B) - (iii)
(C) - (ii)
(2) (A) - (i)
(B) - (iv)
(C) - (iii)
(1) (A) - (iv)
(B) - (iii)
(C) - (ii)
(3) (A) - (ii)
(B) - (iv)
(C) - (iii)
(2) (A) - (i)
(B) - (iv)
(C) - (iii)
(4) (A) - (ii)
(B) - (i)
(C) - (iii)
(3) (A) - (ii)
(B) - (iv)
(C) - (iii)
(4) (A) - (ii)
(B) - (i)
(C) - (iii)
(iv)
L
g . Measured value of L is 20.0 cm known
to 1 mm accuracy and time for 100 oscillations of
the pendulum is found to be 90 s using a wrist
watch of 1 s resolution. The accuracy in the
determination of g is
T  2
34.
vksj
fdlh ,dleku Bksl 'kadq ds nzO;eku dsUæ dh mlds 'kh"kZ ls
nwjh z0 gSA ;fn 'kadq ds vkèkkj dh f=kT;k R rFkk 'kadq dh Å¡pkbZ
h gks rks z0 dk eku fuEukafdr esa ls fdlds cjkcj gksxk\
(i)
34. The period of oscillation of a simple pendulum is
vksj
fdlh ljy yksyd dk vkorZ]
T  2
L
g
gSA
L
dk
ekfir eku 20.0 cm gS] ftldh ;FkkFkZrk 1 mm gSA bl yksyd
ds 100 nksyuksa dk le; 90 s gS] ftls 1 s foHksnu dh
?kM+h ls ukik x;k gSA rks] g ds fuèkkZj.k esa ;FkkFkZrk gksxh %
(1) 5%
(2) 2%
(1) 5%
(2) 2%
(3) 3%
(4) 1%
(3) 3%
(4) 1%
7
JEE (MAIN)-2015
35. A red LED emits light at 0.1 watt uniformly around
it. The amplitude of the electric field of the light at
a distance of 1 m from the diode is
35.
,d yky jax dk ,y-bZ-Mh- (izdk'k mRltZd Mk;ksM) 0.1 okV
ij] ,dleku izdk'k mRlftZr djrk gSA Mk;ksM ls 1 m nwjh ij]
bl izdk'k ds fo|qr {ks=k dk vk;ke gksxk %
(1) 7.75 V/m
(2) 1.73 V/m
(1) 7.75 V/m
(2) 1.73 V/m
(3) 2.45 V/m
(4) 5.48 V/m
(3) 2.45 V/m
(4) 5.48 V/m
36. In the given circuit, charge Q2 on the 2 F capacitor
changes as C is varied from 1 F to 3 F. Q2 as a
function of C is given properly by : (Figures are
drawn schematically and are not to scale)
36.
fn;s x;s ifjiFk esa] C ds eku ds 1F ls 3F ifjofrZr gksus
ls] 2F laèkkfj=k ij vkos'k Q2 esa ifjorZu gksrk gSA 'C' ds iQyu
ds :i esa Q2 dks dkSu lk vkys[k lgh n'kkZrk gS\ (vkys[k
dsoy O;oLFkk vkjs[k gSa vkSj Ldsy ds vuqlkj ugha gSaA)
1 F
1 F
C
C
2 F
2 F
E
E
Charge
Charge
Q2
Q2
(1)
1 F
3 F
1 F
3 F
(1)
C
1 F
(3)
C
1 F
3 F
C
37.
gL
(2) sin   cos 
0
gL
(3) 2 sin   cos 
0
(4) 2
I
L
(4)
3 F
C
1 F
gL
tan 
0

gL
(2) sin   cos 
0
I
gL
(3) 2 sin   cos 
0
gL
tan 
0
(4) 2
8
C
3 F
C
nks irys yEcs rkjksa esa izR;sd ls I èkkjk izokfgr gks jgh gSA
bUgsa L yEckbZ ds fo|qrjksèkh èkkxksa ls yVdk;k x;k gSA bu
èkkxksa esa izR;sd ds }kjk ÅèokZèkj fn'kk ls  dks.k cukus dh
fLFkfr esa] ;s nksuksa rkj lkE;koLFkk esa jgrs gSaA ;fn bu rkjksa
dh izfr bdkbZ yEckbZ nzO;eku  gS rFkk g xq#Roh; Roj.k
gS rks] I dk eku gksxk %
(1)

3 F
Q2
1 F
(g = gravitational acceleration)
gL
tan 
0
1 F
vkos'k
(3)
37. Two long current carrying thin wires, both with
current I, are held by insulating threads of length L
and are in equilibrium as shown in the figure, with
threads making an angle  with the vertical. If wires
have mass  per unit length then the value of I is
(1)
C
Q2
(4)
3 F
3 F
vkos'k
Q2
Q2
1 F
(2)
Charge
Charge
Q2
Q2
(2)
C
vkos'k
vkos'k
gL
tan 
0
I
L
I
JEE (MAIN)-2015
38. A particle of mass m moving in the x direction with
speed 2v is hit by another particle of mass 2m
moving in the y direction wth speed v. If the
collision is perfectly inelastic, the percentage loss in
the energy during the collision is close to
39.
38. x–fn'kk esa 2v pky ls
ls] y–fn'kk esa v osx
pyrs gq, m æO;eku ds ,d d.k
ls pyrk gqvk 2m æO;eku dk ,d
d.k] Vdjkrk gSA ;fn ;g la?kV~V (VDdj) iw.kZr% vizR;kLFk
gS rks] VDdj ds nkSjku ÅtkZ dk {k; (gkfu) gksxh%
(1) 62%
(2) 44%
(1) 62%
(2) 44%
(3) 50%
(4) 56%
(3) 50%
(4) 56%
F
F
A B
A B
39.
;gk¡ vkjs[k esa nks CykWd (xqVds) A vkSj B n'kkZ;s x;s gSa
ftuds Hkkj Øe'k% 20 N rFkk 100 N gSaA bUgsa] ,d cy F
}kjk fdlh nhokj ij nck;k tk jgk gSA ;fn ?k"kZ.k xq.kkad
dk eku] A rFkk B ds chp 0.1 rFkk B vkSj nhokj ds chp
0.15 gS rks] nhokj }kjk CykWd B ij yxk cy gksxk %
Given in the figure are two blocks A and B of
weight 20 N and 100 N, respectively. These are
being pressed against a wall by a force F as shown.
If the coefficient of friction between the blocks is 0.1
and between block B and the wall is 0.15, the
frictional force applied by the wall on block B is
(1) 150 N
(2) 100 N
(3) 80 N
(4) 120 N
40.
40. Consider an ideal gas confined in an isolated closed
chamber. As the gas undergoes an adiabatic
expansion, the average time of collision between
molecules increases as Vq, where V is the volume of
the gas. The value of q is
(1) 150 N
(2) 100 N
(3) 80 N
(4) 120 N
,d vkn'kZ xSl fdlh cUn (laor` )] fo;qDr (foyfxr) d{k
esa lhfer (j[kh) gSA bl xSl esa #n~èkks"e izlkj gksus ij]
blds v.kqvksa ds chp VDdj dk vkSlr dky (le;) V q
ds vuqlkj c<+ tkrk gS] tgk¡ V xSl dk vk;ru gSA rks q
dk eku gksxk %
⎛
CP ⎞
⎜ 
⎟
Cv ⎠
⎝
⎛
CP ⎞
⎜ 
⎟
Cv ⎠
⎝
1
(1)
2
3  5
(2)
6
(1)
1
2
(2)
3  5
6
3  5
(3)
6
1
(4)
2
(3)
3  5
6
(4)
1
2
41. 10 cm
rFkk 5 cm Hkqtkvksa ds ,d vk;rkdkj ywi (ik'k)
ls ,d fo|qr èkkjk] I = 12 A, izokfgr gks jgh gSA bl ik'k
dks vkjs[k esa n'kkZ;s x;s vuqlkj fofHkUu vfHkfoU;klksa
(fLFkfr;ks)a esa j[kk x;k gSA
41. A rectangular loop of sides 10 cm and 5 cm carrying
a current I of 12 A is placed in different orientations
as shown in the figures below:
z
z
I
B
I
(a)
y
I
x
I
y
I
x
B
z
B
I
I
x
I
(a)
z
(b)
I
B
I
I
I
(b)
y
x
9
I
I
I
I
y
JEE (MAIN)-2015
z
z
I
I
I
(c)
x
I
B
I
x
z
(d)
I
I
I
x
y
I
(1) (b) and (c), respectively
(2) (a) and (b), respectively
42.
(3) (a) and (c), respectively
(4) (b) and (d), respectively
42. Consider a spherical shell of radius R at
temperature T. The black body radiation inside it
can be considered as an ideal gas of photons with
U
4
internal energy per unit volume u   T and
V
1 U
pressure P  ⎛⎜ ⎞⎟ . If the shell now undergoes an
3⎝V ⎠
adiabatic expansion the relation between T and R is
1
R3
y
I
z
I
I
I
x
If there is a uniform magnetic field of 0.3 T in the
positive z direction, in which orientations the loop
would be in (i) stable equilibrium and (ii) unstable
equilibrium?
(1) T 
B
B
B
(d)
I
(c)
y
I
y
I
;fn ogk¡ 0.3 T rhozrk dk dksbZ ,dleku pqEcdh; {ks=k]
èkukRed z fn'kk esa fo|eku gS rks] n'kkZ;s x;s fdl vfHkfoU;kl
es]a ;g ik'k (ywi) (i) LFkk;h laryq u rFkk (ii) vLFkk;h laryq u
es]a gksxk\
(1) Øe'k% (b) rFkk (c) esa
(2) Øe'k% (a) rFkk (b) esa
(3) Øe'k% (a) rFkk (c) esa (4) Øe'k% (b) rFkk (d) esa
fdlh xksyh; dks'k ('kSy) dh f=kT;k R gS vkSj bldk rki
T gSA blds Hkhrj Ñf".kdk fofdj.kksa dks iQksVkWuksa dh ,d
,slh vkn'kZ xSl ekuk tk ldrk gS ftldh izfr bdkbZ
vk;ru vkUrfjd ÅtkZ ]
1⎛U ⎞
p  ⎜ ⎟ gSA ;fn
3⎝V ⎠
rFkk R ds chp lacèa k
(1) T 
1
R3
(3) T  e–3R
(2) T  e–R
43.
1
(3) T 
(4) T 
R
43. As an electron makes a transition from an excited
state to the ground state of a hydrogen-like atom/ion
e–3R
(1) Kinetic energy and total energy decrease but
potential energy increases
(2) Its kinetic energy increases but potential energy
and total energy decrease
(3) Kinetic energy, potential energy and total energy
decrease
(4) Kinetic energy decreases, potential energy
increases but total energy remains same
44.
44. On a hot summer night, the refractive index of air is
smallest near the ground and increases with height
form the ground. When a light beam is directed
horizontally, the Huygen's principle leads us to
conclude that as it travels, the light beam
10
u
U
 T4
V
rFkk nkc]
bl dks'k esa #n~èkks"e izlkj gks rks]
T
gksxk %
(2) T  e–R
(4) T 
1
R
tc dksbZ bysDVªkuW ] gkbMªkt
s u tSls ijek.kq@vk;u dh mÙksftr
voLFkk ls U;wure ÅtkZ voLFkk esa laØe.k djrk gS rks
mldh%
(1) xfrt ÅtkZ o dqy ÅtkZ de gks tkrh gSa fdUrq]
fLFkfrt ÅtkZ c<+ tkrh gSA
(2) xfrt ÅtkZ esa o`f¼ rFkk fLFkfrt ÅtkZ rFkk dqy ÅtkZ
esa deh gksrh gSA
(3) xfrt ÅtkZ] fLFkfrt ÅtkZ rFkk dqy ÅtkZ esa deh
gks tkrh gSA
(4) xfrt ÅtkZ de gksrh gS] fLFkfrt ÅtkZ c<+rh gS vkSj
dqy ÅtkZ ogh jgrh gSA
xzh"e ½rq dh xeZ jkf=k esa] Hkw&ry ds fudV] ok;q dk
viorZukad U;wure gksrk gS vkSj Hkw&ry ls Å¡pkbZ ds lkFk
c<+rk tkrk gSA ;fn] dksbZ izdk'k&fdj.k&iqat {kSfrt fn'kk
esa tk jgk gks rks] gkbxsUl ds fl¼kUr ls ;g ifj.kke izkIr
gksrk gS fd] pyrs gq, izdk'k&fdj.k iaqt %
JEE (MAIN)-2015
(1) Bends upwards
(1)
(2) Becomes narrower
(2)
(3) Goes horizontally without any deflection
(3)
(4) Bends downwards
(4)
45. From a solid sphere of mass M and radius R, a
R
spherical portion of radius
is removed, as
2
shown in the figure. Taking gravitational potential
V = 0 at r = , the potential at the centre of the
cavity thus formed is
45.
(3)
2GM
R
(2)
GM
R
(4)
,d Bksl xksys dk nzO;eku
rFkk f=kT;k
GM
2R
(1)
(2)
2GM
3R
2GM
R
GM
2R
(3)
GM
R
(4)
2GM
3R
46. Monochromatic light is incident on a glass prism of
angle A. If the refractive index of the material of the
prism is , a ray, incident at an angle , on the face
AB would get transmitted through the face AC of the
prism provided.
46.
R
gSA blls
R
2
dk¡p ds fdlh fizTe dk dks.k 'A' gSA bl ij ,do.khZ izdk'k
vkifrr gksrk gSA ;fn] fizTe ds inkFkZ dk viorZukad  gS
rks] fizTe ds AB iQyd ij]  dks.k vkifrr izdk'k dh
fdj.k] fizTe ds iQyd AC ls ikjxr gksxh ;fn %
A
A

B
M
f=kT;k dk ,d xksyh; Hkkx] vkjs[k esa n'kkZ;s x;s vuqlkj
dkV fy;k tkrk gSA r =  (vuUr) ij xq#Roh; foHko
ds eku V dks 'kwU; (V = 0) ekurs gq,] bl izdkj cus
dksVj (dSfoVh) ds dsUnz ij] xq#Roh; foHko dk eku gksxk%
(G = xq#Roh; fLFkjk¡d gS)
(G = gravitational constant)
(1)
Åij dh vksj >qd tk;sxkA
ladfq pr (ladh.kZ) gks tk;sxkA
fcuk fo{ksfir gq,] {kSfrt fn'kk esa pyrk jgsxkA
uhps dh vksj >qd tk;sxkA

C
B
C
⎡
⎤
⎛
1
1 ⎛ 1 ⎞ ⎞
(1)   cos ⎢ sin ⎜ A  sin ⎜  ⎟ ⎟ ⎥
⎢⎣
⎝ ⎠ ⎠ ⎥⎦
⎝
⎡
⎤
⎛
1
1 ⎛ 1 ⎞ ⎞
(1)   cos ⎢ sin ⎜ A  sin ⎜  ⎟ ⎟ ⎥
⎢⎣
⎝ ⎠ ⎠ ⎥⎦
⎝
⎡
⎤
⎛
1
1 ⎛ 1 ⎞ ⎞
(2)   sin ⎢ sin ⎜ A  sin ⎜  ⎟ ⎟ ⎥
⎝ ⎠ ⎠ ⎦⎥
⎝
⎣⎢
⎡
⎤
⎛
1
1 ⎛ 1 ⎞ ⎞
(2)   sin ⎢ sin ⎜ A  sin ⎜  ⎟ ⎟ ⎥
⎝ ⎠ ⎠ ⎦⎥
⎝
⎣⎢
⎡
⎤
⎛
1
1 ⎛ 1 ⎞ ⎞
(3)   sin ⎢ sin ⎜ A  sin ⎜  ⎟ ⎟ ⎥
⎝ ⎠ ⎠ ⎦⎥
⎝
⎣⎢
⎡
⎤
⎛
1
1 ⎛ 1 ⎞ ⎞
(3)   sin ⎢ sin ⎜ A  sin ⎜  ⎟ ⎟ ⎥
⎢⎣
⎝ ⎠ ⎠ ⎥⎦
⎝
⎡
⎤
⎛
1
1 ⎛ 1 ⎞ ⎞
(4)   cos ⎢ sin ⎜ A  sin ⎜  ⎟ ⎟ ⎥
⎝ ⎠ ⎠ ⎦⎥
⎝
⎣⎢
⎡
⎤
⎛
1
1 ⎛ 1 ⎞ ⎞
(4)   cos ⎢ sin ⎜ A  sin ⎜  ⎟ ⎟ ⎥
⎢⎣
⎝ ⎠ ⎠ ⎥⎦
⎝
47.
47. Two stones are thrown up simultaneously from the
edge of a cliff 240 m high with initial speed of
10 m/s and 40 m/s respectively. Which of the
following graph best represents the time variation of
relative position of the second stone with respect to
the first? (Assume stones do not rebound after
hitting the ground and neglect air resistance, take g
= 10 m/s2)
11
fdlh 240 m Å¡ph pksVh ds ,d fdukjs ls] nks iRFkjksa dks
,dlkFk Åij dh vksj isaQdk x;k gS] budh izkjafHkd pky
Øe'k% 10 m/s rFkk 40 m/s gS] rks] fuEukafdr esa ls dkSulk
xzkiQ (vkys[k) igys iRFkj ds lkis{k nwljs iRFkj dh fLFkfr
ds le; fopj.k (ifjorZu) dks lokZfèkd lgh n'kkZrk gS\
(eku yhft, fd] iRFkj tehu ls Vdjkus ds i'pkr Åij
dh vksj ugha mNyrs gSa rFkk ok;q dk izfrjksèk ux.; gS]
fn;k gS g = 10 m/s2)
JEE (MAIN)-2015
(The figures are schematic and not drawn to scale)
240
(;gk¡
xzkiQ dsoy O;oLFkk vkjs[k gSa vkSj Ldsy ds vuqlkj
ugha gSa)
(y2 – y1) m
240
(y2 – y1) m
(1)
8
240
(1)
t (s)
12
8
12
t 8
(y2 – y1) m
12
(y2 – y1) m
240
t (s)
(y2 – y1) m
(2)
t 8
(y2 – y1) m
(2)
t (s)
12
240
t (s)
240
(3)
12
240
(3)
t (s)
t (s)
12
(y2 – y1) m
240
(y2 – y1) m
(4)
(4)
t (s)
8
12
48. For a simple pendulum, a graph is plotted between
its kinetic energy (KE) and potential energy (PE)
against its displacement d. Which one of the
following represents these correctly?
8
48.
(Graphs are schematic and not drawn to scale)
E
E
KE
PE
(1)
KE
E
KE
E
KE
PE
(2)
d
PE
(2)
d
E
(3)
E
PE
KE
(3)
d
12
t (s)
fdlh ljy yksyd ds fy;s] mlds foLFkkiu d rFkk mldh
xfrt ÅtkZ ds chp vkSj foLFkkiu d rFkk mldh fLFkfrt
ÅtkZ ds chp xzkiQ [khaps x;s gSaA fuEukafdr esa ls dkSu lk
xzkiQ (vkys[k) lgh gS\
(;gk¡ xzkiQ dsoy O;oLFkk vkjs[k gSa vkSj Ldsy ds vuqlkj
ugha g)aS
PE
(1)
12
PE
KE
d
JEE (MAIN)-2015
E
E
KE
d
KE
d
(4)
(4)
PE
PE
49.
49. A train is moving on a straight track with speed
20 ms–1. It is blowing its whistle at the frequency of
1000 Hz. The percentage change in the frequency
heard by a person standing near the track as the
train passes him is (speed of sound = 320 ms–1)
close to
(1) 24%
(2) 6%
(3) 12%
(4) 18%
,d Vªsu (jsyxkM+h) lhèkh iVfj;ksa ij 20 ms–1 dh pky ls
xfr dj jgh gSA bldh lhVh dh èofu dh vko`fÙk 1000
Hz gSA ;fn èofu dh ok;q esa pky 320 ms–1 gks rks] iVfj;ksa
ds fudV [kM+s O;fDr ds ikl ls Vªsu ds xqtjus ij] ml
O;fDr }kjk lquh xbZ lhVh dh èofu dh vko`fÙk esa izfr'kr
ifjorZu gksxk yxHkx :
(1) 24%
(3) 12%
(2) 6%
(4) 18%
50. LCR
(,y-lh-vkj) ifjiFk fdlh voeafnr yksyd ds rqY;
gksrk gSA fdlh LCR ifjiFk esa laèkkfj=k dks Q0 rd vkosf'kr
fd;k x;k gS] vkSj fiQj bls vkjs[k esa n'kkZ;s x;s vuqlkj L
o R ls tksMk+ x;k gSA
50. An LCR circuit is equivalent to a damped
pendulum. In an LCR circuit the capacitor is
charged to Q0 and then connected to the L and R as
shown below :
R
L
R
L
C
C
If a student plots graphs of the square of maximum
2
charge  QMax
 on the capacitor with time (t) for two
different values L1 and L2 (L1 > L2) of L then which
of the following represents this graph correctly?
(Plots are schematic and not drawn to scale)
2
QMax
;fn ,d fo|kFkhZ L ds] nks fofHkUu ekuksa] L1 rFkk L2 (L1 >
L2) ds fy;s] le; t rFkk laèkkfj=k ij vfèkdre vkos'k ds
2
oxZ QMax
ds chp nks xzkiQ cukrk gS rks fuEukafdr esa ls
dkSu lk xzkiQ lgh gS\ (IykWV dsoy O;oLFkk IykWV gSa rFkk
Ldsy ds vuqlkj ugha gSa)
2
QMax
Q0 (For both L1 and L2)
2
QMax
2
(2)
(1)
L1
QMax
L2
t
t
Q0 (L1 vkSj L2 nksuksa ds fy,)
(1)
(2)
L2
t
2
t
2
QMax
(3)
L1
QMax
L2
L1
(4)
t
L1
L2
2
2
QMax
(3)
t
51. A solid body of constant heat capacity 1 J/°C is
being heated by keeping it in contact with reservoirs
in two ways :
51.
(i) Sequentially keeping in contact with 2
reservoirs such that each reservoir supplies
same amount of heat.
(ii) Sequentially keeping in contact with 8
reservoirs such that each reservoir supplies
same amount of heat.
13
QMax
L2
L1
(4)
t
L1
L2
t
,d Bksl fiaM (oLrq) dh fLFkj Å"ek /kfjrk 1 J/°C gSA bldks
Å"edksa (Å"ek HkaMkjks)a ds lEidZ esa j[kdj fuEu nks izdkj
ls xeZ fd;k tkrk gS]
(i) vuqØfed :i ls 2 Å"edksa ds lEidZ esa bl izdkj
j[kdj fd izR;sd Å"ed leku ek=kk esa Å"ek nsrk gS]
(ii) vuqØfed :i ls 8 Å"edksa ds lEidZ esa bl izdkj
j[kdj fd izR;sd Å"ed leku ek=kk esa Å"ek nsrk gS]
JEE (MAIN)-2015
In both the cases body is brought from initial
temperature 100°C to final temperature 200°C.
Entropy change of the body in the two cases
respectively is
nksuksa fLFkfr;ksa esa fiaM dk izkjafHkd rki 100°C rFkk vfUre
rki 200°C gSA rks] bu nks fLFkfr;ksa esa fiaM dh ,UVªkWih esa
ifjorZu gksxk] Øe'k%
(1) 2ln 2, 8ln 2
(2) ln 2, 4ln 2
(1) 2ln 2, 8ln 2
(2) ln 2, 4ln 2
(3) ln 2, ln 2
(4) ln 2, 2ln 2
(3) ln 2, ln 2
(4) ln 2, 2ln 2
52.
52. An inductor (L = 0.03 H) and a resistor (R = 0.15
k) are connected in series to a battery of 15 V EMF
in a circuit shown below. The key K1 has been kept
closed for a long time. Then at t = 0, K1 is opened
and key K2 is closed simultaneously. At t = 1 ms,
the current in the circuit will be ( e 5  150)
0.03 H
n'kkZ;s x;s ifjiFk esa] ,d izsjd (L = 0.03 H) rFkk ,d
izfrjksèkd (R = 0.15 k) fdlh 15 V fo|qr okgd cy
(bZ-,e-,iQ) dh cSVjh ls tqM+s gSaA dqath K1 dks cgqr le;
rd cUn j[kk x;k gSA blds i'pkr~ le; t = 0 ij] K1
dks [kksy dj lkFk gh lkFk] K2 dks cUn fd;k tkrk gSA
le; t = 1 ms ij] ifjiFk esa fo|qr èkkjk gksxh % ( e 5  150)
0.03H
0.15 k
K2
K2
15 V
K1
(1) 0.67 mA
(2) 100 mA
(3) 67 mA
(4) 6.7 mA
0.15 k
15V
K1
(1) 0.67 mA
(2) 100 mA
(3) 67 mA
(4) 6.7 mA
53. R
f=kT;k ds fdlh ,dleku vkosf'kr Bksl xksys ds i`"B
dk foHko V0 gS (ds lkis{k ekik x;k)A bl xksys ds fy;s]
53. A uniformly charged solid sphere of radius R has
potential V0 (measured with respect to ) on its
surface. For this sphere the equipotential surfaces
3V0 5V0 3V0
V
,
,
rFkk 0 foHkoksa okys lefoHkoh
2
4
4
4
f=kT;k;sa] Øe'k% R1, R2, R3 rFkk R4 gSaA rks]
V
3V0 5V0 3V0
,
,
and 0 have radius
2
4
4
4
R1, R2, R3 and R4 respectively. Then
with potentials
(1) 2R < R4
(2) R1 = 0
rFkk R2 > (R4 – R3)
(2) R1 = 0 and R2 > (R4 – R3)
(3) R1  0
rFkk (R2 – R1) > (R4 – R3)
(3) R1  0 and (R2 – R1) > (R4 – R3)
(4) R1 = 0
rFkk R2 < (R4 – R3)
(1) 2R < R4
(4) R1 = 0 and R2 < (R4 – R3)
54.
54. A long cylindrical shell carries positive surface
charge  in the upper half and negative surface
charge – in the lower half. The electric field lines
around the cylinder will look like figure given in
fdlh yEcs csyukdkj dks'k ds Åijh Hkkx esa èkukRed i`"B
vkos'k  rFkk fupys Hkkx esa ½.kkRed i`"B vkos'k – gSaA
bl csyu (flfyUMj) ds pkjksa vksj fo|qr {ks=k&js[kk;sa] ;gk¡
n'kkZ;s x;s vkjs[kksa esa ls fdl vkjs[k ds leku gksxa hA
(;g vkjs[k dsoy O;oLFkk vkjs[k gS vkSj Ldsy ds vuqlkj
ugha gS)
(figures are schematic and not drawn to scale)
(1)
(1)
++++
+
+
––
––
––
––
(2)
14
++++
+
+
––
––
––
––
(2)
i`"Bksa dh
JEE (MAIN)-2015
(3)
(3)
++ ++
+
+
–– ––
–– ––
(4)
++ ++
+
+
–– ––
–– ––
(4)
55. Assuming human pupil to have a radius of 0.25 cm
and a comfortable viewing distance of 25 cm, the
minimum separation between two objects that
human eye can resolve at 500 nm wavelength is
55.
(1) 300 m
;fn ekuo us=k dh iqryh dh f=kT;k 0.25 cm, vkSj Li"V
lqfoèkk tud ns[kus dh nwjh 25 cm gks rks] 500 nm rjaxnSè;Z
ds izdk'k esa] nks oLrqvksa ds chp fdruh U;wure nwjh rd
ekuo us=k mu nksuksa ds chp foHksnu dj ldsxk\
(2) 1 m
(1) 300 m
(2) 1 m
(3) 30 m
(3) 30 m
(4) 100 m
(4) 100 m
56.
56. A signal of 5 kHz frequency is amplitude modulated
on a carrier wave of frequency 2 MHz. The
frequencies of the resultant signal is/are
ds fdlh ladrs (flXuy) dk 2 MHz vko`fÙk
dh okgd rjax ij vk;ke ekWMy
q u fd;k x;k gSA rks] ifj.kkeh
flXuy (ladrs ) dh vko`fÙk gksxh %
5 kHz vko`fÙk
(1) 2000 kHz
(1) 2000 kHz and 1995 kHz
rFkk 1995 kHz
(2) 2 MHz dsoy
(2) 2 MHz only
(3) 2005 kHz and 1995 kHz
(3) 2005 kHz
(4) 2005 kHz, 2000 kHz and 1995 kHz
(4) 2005 kHz, 2000 kHz
57. Two coaxial solenoids of different radii carry
current I in the same direction. Let F1 be the
57.
magnetic force on the inner solenoid due to the
outer one and F2 be the magnetic force on the outer
rFkk 1995 kHz
nks lek{kh ifjukfydkvksa esa] izR;sd ls I èkkjk ,d gh fn'kk
esa izokfgr gks jgh gSA ;fn] ckgjh ifjukfydk ds dkj.k]
Hkhrjh ifjukfydk ij pqEcdh; cy
F1
rFkk Hkhrjh ifjukfydk
ds dkj.k] ckgjh ifjukfydk ij pqEcdh; cy
solenoid due to the inner one. Then
(1) F1
(1) F1 is radially outwards and F2 = 0
ckgj dh vksj o vjh; gS rFkk
F2
gks rks %
F2 = 0 gSA
(2) F1 = F2 = 0
(2) F1 = F2 = 0
(3) F1 is radially inwards and F2
rFkk 1995 kHz
(3) F1
is radially
Hkhrj dh vksj o vjh; (f=kT;) gS rFkk
dh vksj o vjh; gSA
outwards
(4) F1
(4) F1 is radially inwards and F2 = 0
15
Hkhrj dh vksj o vjh; gS rFkk
F2 = 0 gSA
F2
ckgj
JEE (MAIN)-2015
58. A pendulum made of a uniform wire of crosssectional area A has time period T. When an
additional mass M is added to its bob, the time
period changes to TM. If the Young's modulus of the
material of the wire is Y then
58.
1
is equal to
Y
fdlh ,dleku rkj dh vuqizLFk dkV dk {ks=kiQy 'A' gSA
blls cuk;s x;s ,d yksyd dk vkorZdky T gSA bl yksyd
ds xksyd ls ,d vfrfjDr M nzO;eku tksM+ nsus ls yksyd
dk vkorZdky ifjofrZr gksdj TM gks tkrk gSA ;fn bl rkj
ds inkFkZ dk ;ax xq.kkad 'Y' gks rks
(g = gravitational acceleration)
(g = xq#Roh;
⎡ ⎛ T ⎞2 ⎤ A
(1) ⎢1  ⎜ T ⎟ ⎥ Mg
⎢⎣ ⎝ M ⎠ ⎥⎦
dk eku gksxk%
Roj.k)
⎡ ⎛ T ⎞2 ⎤ A
(1) ⎢1  ⎜ T ⎟ ⎥ Mg
⎢⎣ ⎝ M ⎠ ⎥⎦
⎡⎛ TM ⎞ 2
⎤ A
(2) ⎢⎜ T ⎟  1⎥ Mg
⎠
⎥⎦
⎣⎢⎝
⎡⎛ TM ⎞ 2
⎤ A
(2) ⎢⎜ T ⎟  1⎥ Mg
⎠
⎣⎢⎝
⎦⎥
⎡⎛ TM ⎞ 2
⎤ Mg
(3) ⎢⎜ T ⎟  1⎥ A
⎠
⎣⎢⎝
⎦⎥
⎡⎛ TM ⎞ 2
⎤ Mg
⎢
(3) ⎜ T ⎟  1⎥ A
⎠
⎥⎦
⎣⎢⎝
⎡ ⎛ TM ⎞ 2 ⎤ A
(4) ⎢1  ⎜ T ⎟ ⎥ Mg
⎠ ⎦⎥
⎣⎢ ⎝
⎡ ⎛ TM ⎞ 2 ⎤ A
(4) ⎢1  ⎜ T ⎟ ⎥ Mg
⎠ ⎥⎦
⎢⎣ ⎝
59.
59. From a solid sphere of mass M and radius R a cube
of maximum possible volume is cut. Moment of
inertia of cube about an axis passing through its
center and perpendicular to one of its faces is
fdlh Bksl xksys dk nzO;eku M rFkk bldh f=kT;k R gSA blesa
ls vfèkdre laHko vk;ru dk ,d D;wc (?ku) dkV fy;k
tkrk gSA bl D;wc dk tM+Ro vk?kw.kZ fdruk gksxk] ;fn]
bldh ?kw.kZu&v{k] blds dsUnz ls gksdj xqtjrh gS rFkk blds
fdlh ,d iQyd ds yEcor~ gS\
4 MR 2
3 3
(1)
4 MR 2
3 3
(2)
MR 2
32 2 
(2)
MR 2
32 2 
(3)
MR 2
16 2 
(3)
MR 2
16 2 
4 MR 2
(4)
9 3
(4)
4 MR 2
9 3
(1)
1
Y
60. 0.1 m yacs
fdlh rkj ds fljksa ds chp 5 V foHkokarj vkjksfir
djus ls bysDVªkWuksa dh viokg pky 2.5 × 10–4 ms–1 gksrh
gSA ;fn bl rkj esa bysDVªkWu ?kuRo 8 × 1028 m–3 gks rks]
bl ds inkFkZ dh izfrjksèkdrk gksxh] yxHkx%
60. 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 ms–1. If the electron density in the wire is
8 × 1028 m–3, the resistivity of the material is close to
(1) 1.6 × 10–5 m
(1) 1.6 × 10–5 m
(2) 1.6 × 10–8 m
(2) 1.6 × 10–8 m
(3) 1.6 × 10–7 m
(3) 1.6 × 10–7 m
(4) 1.6 × 10–6 m
(4) 1.6 × 10–6 m
16
JEE (MAIN)-2015
PART–C : CHEMISTRY
61. The vapour pressure of acetone at 20°C is 185 torr.
When 1.2 g of a non-volatile substance was
dissolved in 100 g of acetone at 20°C, its vapour
pressure was 183 torr. The molar mass (g mol–1) of
the substance is
61. 20°C
ij ,sflVksu dh ok"i nkc 185 torr gSA tc 20°C
ij] 1.2 g vok"i'khy inkFkZ dks 100 g ,sflVksu esa ?kksyk
x;k] rc ok"i nkc 183 torr gks x;kA bl inkFkZ dk eksyj
nzO;eku (g mol–1 es)a gS :
(1) 488
(2) 32
(1) 488
(2) 32
(3) 64
(4) 128
(3) 64
(4) 128
62.
62. 3 g of activated charcoal was added to 50 mL of
acetic acid solution (0.06N) in a flask. After an hour
it was filtered and the strength of the filtrate was
found to be 0.042 N. The amount of acetic acid
adsorbed (per gram of charcoal) is
(1) 54 mg
(2) 18 mg
(3) 36 mg
(4) 42 mg
63. Which of the following is the energy of a possible
excited state of hydrogen?
(1) +6.8 eV
(2) +13.6 eV
(3) –6.8 eV
(4) –3.4 eV
63.
64. Which among the following is the most reactive?
(1) ICl
(2) Cl2
(3) Br2
(4) I2
64.
65. Which polymer is used in the manufacture of paints
and lacquers?
(1) Poly vinyl chloride
(2) Bakelite
(3) Glyptal
(4) Polypropene
65.
66. The molecular formula of a commercial resin used
for exchanging ions in water softening is
C8H 7SO 3Na (mol. wt. 206). What would be the
maximum uptake of Ca2+ ions by the resin when
expressed in mole per gram resin?
66.
,d ÝykLd esa 0.06N ,flfVd vEy ds 50 mL foy;u
esa 3 g lfØ;r~ dk"B dks;yk feyk;k x;kA ,d ?kaVs ds
i'pkr~ mls Nkuk x;k vkSj fuL;an dh izcyrk 0.042 N
ikbZ xbZA vf/'kksf"kr ,flfVd vEy dh ek=kk (dk"B&dks;yk
ds izfr xzke ij) gS:
(1) 54 mg
(2) 18 mg
(3) 36 mg
(4) 42 mg
fuEufyf[kr esa ls gkbZMªkstu dh laHko mÙksftr voLFkk dh
mQtkZ dkSulh gS\
(1) +6.8 eV
(2) +13.6 eV
(3) –6.8 eV
(4) –3.4 eV
fuEufyf[kr esa ls dkSu lokZf/d vfHkfØ;k'khy gS\
(1) ICl
(2) Cl2
(3) Br2
(4) I2
fdl cgqyd dk mi;ksx izyis vkSj izyk{k cukus esa gksrk gS\
(1)
ikWfy okbfuy DyksjkbM
(2)
csdsykbV
(3)
fXyIVky
(4)
ikWfyizksihu
,d okf.kT; jsft+u dk vkf.od lw=k C8H7SO3Na gS
(vkf.od Hkkj = 206) bl js f t+ u dh Ca2+ vk;u dh
vfèkdre varxzgZ .k {kerk (eksy izfr xzke jsft+u) D;k gS\
1
(1)
412
1
(2)
103
(1)
1
412
(2)
1
103
1
(3)
206
2
(4)
309
(3)
1
206
(4)
2
309
67.
67. In Carius method of estimation of halogens, 250 mg
of an organic compound gave 141 mg of AgBr. The
percentage of bromine in the compound is
gSykstu ds vkdyu dh dSfjvl fof/ esa 250 mg dkcZfud
;kSfxd 141 mg AgBr nsrk gSA ;kSfxd esa czksehu dh
izfr'krrk gSS :
Ag = 108; Br = 80)
(At. mass Ag = 108; Br = 80)
(ijekf.od
(1) 60
(2) 24
(1) 60
(2) 24
(3) 36
(4) 48
(3) 36
(4) 48
17
nzO;eku
JEE (MAIN)-2015
68. Assertion : Nitrogen and Oxygen are the main
components in the atmosphere but these
do not react to form oxides of nitrogen.
Reason :
68.
The reaction between nitrogen and
oxygen requires high temperature.
(1) Both the assertion and reason are incorrect
(2) Both assertion and reason are correct, and the
reason is the correct explanation for the
assertion
(3) Both assertion and reason are correct, but the
reason is not the correct explanation for the
assertion
(4) The assertion is incorrect, but the reason is correct
69. The following reaction is performed at 298 K.
69.
2NO (g)
2NO(g) + O2(g) 2
2NO (g)
2NO(g) + O2(g) 2
The standard free energy of formation of NO(g) is
86.6 kJ/mol at 298 K. What is the standard free
energy of formation of NO2(g) at 298 K?
298 K ij NO(g) ds la H kou dh ekud
86.6 kJ/mol gSA 298 K ij NO2(g) dh
mQtkZ D;k gS\(Kp = 1.6 × 1012)
(Kp = 1.6 × 1012)
12
(1) 0.5 ⎡⎣ 2  86, 600  R  298  ln 1.6  10 ⎤⎦
(3) 86600 + R(298) ln(1.6 × 1012)
(4) 86600 

R  298 

(4) 86600 
70.
70. Which one of the following alkaline earth metal
sulphates has its hydration enthalpy greater than
its lattice enthalpy?
(1) SrSO4
(2) CaSO4
(3) BeSO4
(4) BaSO4
71.
71. The number of geometric isomers that can exist for
square planar [Pt(Cl)(py)(NH 3 )(NH 2 OH)] + is
(py = pyridine)
(1) 6
(2) 2
(3) 3
(4) 4
eqDr mQtkZ
ekud eqDr
12
(1) 0.5 ⎡⎣ 2  86, 600  R  298  ln 1.6  10 ⎤⎦
(2) R(298) ln(1.6 × 1012) – 86600
(3) 86600 + R(298) ln(1.6 × 1012)
(2) R(298) ln(1.6 × 1012) – 86600
ln 1.6  1012
vfHkdFku : ukbVªkstu vkSj vkWDlhtu okrkoj.k ds eq[;
?kVd gSa ijUrq ;g fØ;k djds ukbVªkstu ds
vkWDlkbM ugha cukrsA
rdZ :
ukbVªkt
s u vkSj vkWDlhtu ds chp vfHkfØ;k ds
fy, mPp rki dh vko';drk gS
(1) vfHkdFku o rdZ nksuksa xyr gSAa
(2) vfHkdFku vkSj rdZ nksuksa lgh gSa vkSj rdZ vfHkdFku
dk lgh Li"Vhdj.k gSA
(3) vfHkdFku vkSj rdZ nksuksa lgh gSa ijUrq rdZ vfHkdFku
dk lgh Li"Vhdj.k ugha gSA
(4) vfHkdFku xyr gS ijUrq rdZ lgh gSA
fuEufyf[kr vfHkfØ;k dks 298 K ij fd;k x;kA

ln 1.6  10 12
R  298 
fuEufyf[kr esa ls dkSu ls {kkjh; e`nk /krq lYiQsV dh
ty;kstu ,sUFkkYih mlds tkyd ,sUFkkYih ls vf/d gS\
(1) SrSO4
(2) CaSO4
(3) BeSO4
(4) BaSO4
oxZ leryh;
[Pt (Cl) (py) (NH 3 )(NH 2 OH)] +
(py = pyridine)
72.
72. The synthesis of alkyl fluorides is best
accomplished by
ds T;kferh; leko;fo;ksa dh la[;k gS :
(1) 6
(2) 2
(3) 3
(4) 4
vYdkby ÝyksjkbM ds la'ys"k.k ds fy, lcls csgrjhu fofèk gS
(1)
LokVZl vfHkfØ;k
(1) Swarts reaction
(2)
eqDr ewyd Ýyksfjus'ku
(2) Free radical fluorination
(3)
lSUMek;j vfHkfØ;k
(4)
fiQadyLVkbu vfHkfØ;k
(3) Sandmeyer's reaction
(4) Finkelstein reaction
73.
73. The intermolecular interaction that is dependent on
the inverse cube of distance between the molecules
is
og varjk&v.kqd vU;ksU; fØ;k tks v.kqvksa ds chp dh nwjh
ds izfrykse ?ku ij fuHkZj gS] gS :
(1)
gkbZMªkstu ca/d
(2) Ion-ion interaction
(2)
vk;u&vk;u vU;ksU;
(3) Ion-dipole interaction
(3)
vk;u&f}/zoq vU;ksU;
(4) London force
(4)
yaMu cy
(1) Hydrogen bond
18

JEE (MAIN)-2015
74. In the context of the Hall-Heroult process for the
extraction of Al, which of the following statement is
false?
74.
gkWy&gsjkWYV izØe ls ,syqfefu;e ds fu"d"kZ.k ds lanHkZ esa
dkSu lk dFku xyr gS\
(1) Na3AlF6 fo|qr
(1) Na3AlF6 serves as the electrolyte
(2)
(2) CO and CO2 are produced in this process
bl izØe esa
vi?kV~; dk dke djrk gSA
CO rFkk CO2
dk mRiknu gksrk gSA
(3) CaF2 dks Al2O3
esa feykus ij feJ.k dk xyukad de
gksrk gS vkSj mlesa pkydrk vkrh gSA
(3) Al2O3 is mixed with CaF2 which lowers the
melting point of the mixture and brings
conductivity
(4)
(4) Al 3+ is reduced at the cathode to form Al
75.
75. Which of the following compounds will exhibit
geometrical isomerism?
dSFkksM ij
Al3+
vipf;r gks dj Al cukrk gSA
fuEufyf[kr esa ls dkSu lk ;kSfxd T;kferh; leko;ork
n'kkZrk gS\
(1) 1, 1 - MkbZiQsfuy - 1 izksisu
(1) 1, 1 - Diphenyl - 1 propane
(2) 1 - iQsfuy - 2 - C;wVhu
(2) 1 - Phenyl - 2 - butene
(3) 3 - iQsfuy - 1 - C;wVhu
(3) 3 - Phenyl - 1 - butene
(4) 2 - iQsfuy - 1 - C;wVhu
(4) 2 - Phenyl - 1 - butene
76. N3–, O2– rFkk F– dh
76. The ionic radii (in Å) of N 3– , O 2– and F – are
respectively
(1) 1.71, 1.36 and 1.40
(2) 1.36, 1.40 and 1.71
vk;fud f=kT;k;sa
(1) 1.71, 1.36
rFkk 1.40
(2) 1.36, 1.40
rFkk 1.71
(3) 1.36, 1.71
rFkk 1.40
(4) 1.71, 1.40
rFkk 1.36
(Å
esa) Øe'k% gSa
:
(3) 1.36, 1.71 and 1.40
(4) 1.71, 1.40 and 1.36
77. H2O2
77. From the following statement regarding H 2O 2,
choose the incorrect statement
ds lanHkZ esa] fuEufyf[kr dFkuksa esa ls xyr dFku pqfu,
(1)
bls /wy ls nwj j[kuk pkfg,
(1) It has to be kept away from dust
(2)
;g dsoy vkWDlhdkjd gS
(2) It can act only as an oxidizing agent
(3)
izdk'k esa bldk vi?kVu gksrk gS
(3) It decomposes on exposure to light
(4)
bls IykfLVd ;k eksevVs dkap cksryksa esa va/js s esa laxfz gr
fd;k tkrk gS
(4) It has to be stored in plastic or wax lined glass
bottles in dark
78.
78. Higher order (>3) reactions are rare due to
mPp dksfV vfHkfØ;k (>3) nqyHZ k gS D;ksfa d
:
(1)
Vdjko ls lfØ; Lih'kht+ dk {k; gksrk gSA
(2)
izfrfØ;k esa lHkh iztkfr;ksa ds ,d lkFk VDdj dh
laHkkouk de gksrh gSA
(3)
vf/d v.kqvksa ds 'kkfey gkssus ls ,aVªkih vkSj lafØ;.k
mQtkZ esa o`f¼ gksrh gSA
(4)
ykspnkj Vdjko ds dkj.k vfHkdkjdksa dh fn'kk esa lkE;
dk LFkkukarj.k gksrk gSA
(1) Loss of active species on collision
(2) Low probability of simultaneous collision of all
the reacting species
(3) Increase in entropy and activation energy as
more molecules are involved
(4) Shifting of equilibrium towards reactants due to
elastic collisions
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JEE (MAIN)-2015
79. Match the catalysts to the correct processes :
Catalyst
79.
fn;s x, mRisjz dksa dks lgh izØe ds lkFk lqefs yr djsa
Process
mRizsjd
izØe
a.
TiCl3
(i) Wacker process
a.
TiCl3
(i)
okWdj izØe
b.
PdCl2
(ii) Ziegler-Natta
polymerization
b.
PbCl2
(ii)
RlhXyj&uêðk
cgqydhdj.k
c.
CuCl2
(iii) Contact process
c.
CuCl2
(iii)
laLi'kZ izØe
(iv) Deacon's process
d. V2O5
(iv)
Mhdu izØe
d. V2O5
(1) a(iii), b(i), c(ii), d(iv) (2) a(iii), b(ii), c(iv), d(i)
(1) a(iii), b(i), c(ii), d(iv) (2) a(iii), b(ii), c(iv), d(i)
(3) a(ii), b(i), c(iv), d(iii) (4) a(ii), b(iii), c(iv), d(i)
(3) a(ii), b(i), c(iv), d(iii) (4) a(ii), b(iii), c(iv), d(i)
80. Which one has the highest boiling point?
80.
fuEufyf[kr esa ls lokZf/d DoFkukad fdldk gS\
(1) Xe
(2) He
(1) Xe
(3) Ne
(4) Kr
(2) He
81. In the reaction
(3) Ne
NH2
(4) Kr
NaNO2/HCl
0-5°C
D
CuCN/KCN

81.
E + N2 ,
fn, x, vfHkfØ;k esa mRikn
E
gS
NH2
CH3
NaNO2/HCl
0-5°C
the product E is
D
CuCN/KCN

E + N2 ,
CH3
CH3
CH3
(1)
(1)
COOH
COOH
(2)
(2)
CH3
(3) H3C
CH3
CH3
(3) H3C
CN
CH3
CN
(4)
(4)
CH3
CH3
82. Which of the following compounds is not colored
yellow?
82.
fn, x, ;kSfxdksa esa dkSuls ;kSfxd dk jax ihyk ugha gS\
(1) BaCrO4
(1) BaCrO4
(2) Zn2[Fe(CN)6]
(2) Zn2[Fe(CN)6]
(3) K3[Co(NO2)6]
(3) K3[Co(NO2)6]
(4) (NH4)3[As (Mo3O10)4]
(4) (NH4)3[As (Mo3O10)4]
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JEE (MAIN)-2015
83.
83. Sodium metal crystallizes in a body centred cubic
lattice with a unit cell edge of 4.29 Å. The radius of
sodium atom is approximately
(1) 0.93 Å
(2) 1.86 Å
(3) 3.22 Å
(4) 5.72 Å
(2) 1.86 Å
(3) 3.22 Å
(4) 5.72 Å
ij vfHkfØ;k 2A B + C dh ekud fxCt+
mQtkZ 2494.2 J gSA fn, x, le; esa vfHkfØ;k feJ.k dk
reaction 2A B + C is 2494.2 J. At a given time,
1
1
,  B   2 vkSj  C  
gS A vfHkfØ;k
2
2
gS : [R = 8.314 J/K/mol, e = 2.718]
la ? kVu A  
is
1
1
,  B   2 and  C   . The reaction proceeds
2
2
in the : [R = 8.314 J/K/mol, e = 2.718]
(1) Reverse direction because Q < KC
(2) Forward direction because Q > KC
(3) Reverse direction because Q > KC
(4) Forward direction because Q < KC
85. Which compound would give 5-keto-2-methyl
hexanal upon ozonolysis?
CH3
CH3
A  
(1)
(1) 0.93 Å
84. 300 K
84. The standard Gibbs energy change at 300 K for the
the composition of the reaction mixture
lksfM;e /krq ,d var%dsfUnzr ?kuh; tkyd esa fØLVfyr gksrk
gS ftlds dksj dh yackbZ 4.29Å gSA lksfM;e ijek.kq dh f=kT;k
yxHkx gS :
85.
vxzflr gksrh
(1) foijhr fn'kk eas D;ksfa d Q < KC
(2) vxz fn'kk esa D;ksfa d Q > KC
(3) foijhr fn'kk esa D;ksfa d Q > KC
(4) vxz fn'kk esa D;ksfa d Q < KC
vkst+ksuksfyfll djus ij dkSulk ;kSfxd 5-dhVks-2-esfFky
gSDlkuSy nsrk gS\
CH3
H3C
(2)
CH3
(1)
CH3
(3)
CH3
CH3
CH3
H3C
(2)
CH3
CH3
(4)
(3)
CH3
CH3
CH3
86. Which of the following compounds is not an
antacid?
86.
(1) Ranitidine
(2) Aluminium Hydroxide
(3) Cimetidine
(4) Phenelzine
87. In the following sequence of reactions :
KMnO
SOCl
87.
H /Pd
4
2
2
Toluene 
 A 
 B 
 C,
BaSO
(3) C6H5CH3
(4) C6H5CH2OH
fuEufyf[kr esa ls dkSulk ;kSfxd izfrvEy ugha gS\
(1) jSfufVMhu
(2) ,syfq efu;e gkbMªkDlkbM
(3) flesfVMhu
(4) fiQufYtu
fn, x, vfHkfØ;k vuqØe esa mRikn C gS :
KMnO
SOCl
H /Pd
4
the product C is
(2) C6H5COOH
CH3
4
2
2
Toluene 
 A 
B 
 C,
BaSO
4
(1) C6H5CHO
(4)
88.
88. Which of the vitamins given below is water soluble?
(1) Vitamin K
(1) C6H5CHO
(2) C6H5COOH
(3) C6H5CH3
(4) C6H5CH2OH
fuEufyf[kr foVkfeuksa esa ty esa foys; gksus okyk gS
(1)
foVkfeu K
(2)
foVkfeu C
(3)
foVkfeu D
(4)
foVkfeu E
(2) Vitamin C
(3) Vitamin D
(4) Vitamin E
21
:
JEE (MAIN)-2015
89. The color of KMnO4 is due to
89. KMnO4 ds
(1)  - * transition
jax dk dkj.k gS
(1)  - * laØe.k
(2) M  L charge transfer transition
(2) M  L vkos'k
(3) d - d transition
LFkkukarj.k laØe.k
(3) d - d laØe.k
(4) L  M charge transfer transition
(4) L  M vkos'k
(1) 127 g
LFkkukarj.k laØe.k
CuSO4 ds ,d foy;u esa] nks iQSjkMs fo|qr izokfgr dh
xbZA dSFkksM ij fu{ksfir rkacs dk nzO;eku gS :
(Cu dk ijekf.od nzO;eku = 63.5 amu)
(2) 0 g
(1) 127 g
90. Two faraday of electricity is passed through a
solution of CuSO4. The mass of copper deposited at
the cathode is (at. mass of Cu = 63.5 amu)
90.
(2) 0 g
(3) 63.5 g
(3) 63.5 g
(4) 2 g
(4) 2 g
22