84 JEE MAIN CHAPTERWISE EXPLORER
At resonance,0 power factor cosf = 1 or e = –(20t – 50) = –[(20 × 3)– 50] = –10 volt
or e = – 10 volt.
\ P = Vr 2m s = (220 V) 2 = 242 W.
Z (200 W ) 16. (a) : In an a.c. generator, maximum emf = NABw.
10. (d) : For the given R, L circuit the potential difference across 17. (d) : Current I = ZE
AD = VBC as they are parallel. where E = V 2 + (V - V ) 2
I1 = E/R1. R L C
I2 = I0(1– e –t/t) where t = mean life or L/R.
t = t0 (given). A B Z = R2 + ( X - X ) 2
L C
E (across BC) = L dI 2 + R2I 2 12 V I 1 I 2 At resonance, X L = X C
dt R 1 L \ Z = R
I2 = I0(1– e –t/t ). Again at resonance, V L = V C
\ E = V R
But I 0 = E = 12 = 6 A. S
R2 2 R 2
C V R 100
´ 10-3 \ I = R = 1´103 = 0.1 A
2 W
t = t 0 = L = 400 H = 0.2 s D
R
\ I 0.1
\ I2 = 6(1 – e –t/0.2 ) VL = I Lw = Cw = (2 ´10-6 ) ´ ( 200)
Potential drop across L = E – R2I2 \ VL = 250 volt.
= 12 – 2 × 6(1 – e –t/0.2 ) = 12e –t/0.2 = 12e –5t V.
18. (d) : R and L cause phase difference to lie between 0 and
11. (a) : M = m 0n 1n 2pr 12 l. p/2 but never 0 and p/2 at extremities.
From f2 = pr 1 2 (m 0 ni)n2 l. R 1125 = 0.8.
Z
A = pr1 2 = 10 cm 2, l = 20 cm, N 1 = 300, N 2 = 400. 19. (c) : Power factor cosf = =
M = m0 N1N2 A = 4p ´ 10-7 ´ 300´ 400 ´10 ´ 10- 4
l 0.20
20. (a) : For maximum power, L w = 1
= 2.4p× 10 –4 H Cw
12. (a) : During the growth of current in LR circuit is given by \ C = 1 = 10 ´ ( 2 1
Lw2 p ´ 50) 2
æ - R t ö or I = E æ - R t ö = 55 ççèæ1 - e - 5 ´2 ö = 10 1 ( p )2 = 10 -6 F
I = I0 ççè1 - e L ÷÷ø R ççè1 - L ÷÷ø 10 ÷÷ ´104 ´
I = (1 – e– 1) . e
ø
or C = 1 mF.
21. (d) : During growth of charge in an inductance,
13. (c) : Given : E = E 0 sin wt I = I 0 (1 – e –Rt/L )
I = I0 sin çèæ wt - p2 ÷öø or I 0 = I0 (1 - e- Rt / L )
Since the phase difference (f) between voltage and current 2
is p2 . or e- Rt / L = 1 = 2 -1 or Rt = ln 2Þ t = L ln 2
2 LR
\ Power factor cos f = cos p = 0 t = 300 ´10 -3 ´ (0.693)
2 2
Power consumption = Er ms Ir ms cos f = 0. or t = 0.1 sec.
14. (b) : Maximum current I 0 = E = 100 = 1 A 22. (a) : The emf induced in the circuit is zero because the two
R 100 emf induced are equal and opposite when one U tube slides
The current decays for 1 millisecond = 1 × 10– 3 sec inside another tube.
During decay, I = I 0 e –tR/L 23. (b) : Induced emf = 1 Bwl 2 = 1 ´ (0.2 ´10-4 )(5)(1) 2
2 2
(-1´10-3 )´100
I = (1) e 100´10- 3 \ Induced emf = 10-4 = 100 ´ 10 -6 = 50 m V.
2 2
1
or I = e -1 = e A . 24. (c) : At resonance, w = 1
LC
15. (c) : f = 10t 2 – 50t + 250
when w is constant,
d f
\ dt = 20t - 50 \ 1 = 1 Þ 1 = 1 = 1
L1C1 L2C2 LC L2 (2C ) 2 L2C
-d f
Induced emf e = dt \ L2 = L/2.
Electromagnetic Induction and Alternating Current 85
25. (b) : Magnetic flux linked 30. (c) : Let Q denote maximum charge on capacitor.
Let q denote charge when energy is equally shared
= BAcos wt = Bpr2 cos wt
2 \ 1 æ 1 Q2 ö = 1 q 2 Þ Q 2 = 2 q 2
2 ççè 2 C ÷÷ø 2 C
\ Induced emf e = - d f = -21 Bpr 2 wsin wt
dt
\ q = Q / 2.
\ Power = e2 = B2p2r 4w2 sin 2 wt
R 4 R 31. (d) : L = - e = -8 ´ 0.05 = 0.1 H .
di / dt - 4
( Bpr 2 w) 2 sin 2 wt
= 4 R
32. (c) : Mutual inductance between two coils depends on the
Q <sin 2 wt> = 1/2 materials of the wires of the coils.
\ Mean power generated 33. (d) : Induced emf = vBL.
= ( Bpr2w)2 ´ 21 = ( Bpr 2w) 2 . 34. (b) : I 2 N2 = I 1 N1 for a transformer
4R 8 R
I1 N1
(b) : Induced current I = -n d f = -n dW where \ I 2 = N2 = 4 ´140 = 2 A .
R¢ dt R¢ dt 280
26.
f = W = flux × per unit turn of the coil 35. (b) : Power factor = R .
\ I = - 1 n(W2 - W1 ) = - n(W2 - W1 ) R2 + L2w 2
+ 4R) t 5 Rt
(R . 36. (d) : Three inductors are in parallel
27. (c) : Average value of A.C. for complete cycle is zero. 3 H
Hence A.C. can not be measured by D.C. ammeter.
3 H
28. (d) : In an LCR series a.c. circuit, the voltages across A D
components L and C are in opposite phase. The voltage
across LC combination will be zero. 3 H
29. (a) : The energy loss due to eddy currents is reduced by \ 1 = 1 + 1 + 1 = 3 = 1
using laminated core in a transformer. L(eq ) 3 3 3 3 1
\ L (eq) = 1 H.
86 JEE MAIN CHAPTERWISE EXPLORER
CHAPTER ELECTROMAGNETIC WAVES
15
1. The magnetic field in a travelling electromagnetic wave 5. The rms value of the electric field of the light coming from
has a peak value of 20 nT. The peak value of electric field the sun is 720 N/C. The average total energy density of the
strength is electromagnetic wave is
(a) 12 V/m (b) 3 V/m (a) 3.3 × 10– 3 J/m3 (b) 4.58 × 10 –6 J/m3
(c) 6 V/m (d) 9 V/m (2013) (c) 6.37 × 10 –9 J/m3 (d) 81.35 × 10 –12 J/m3 .
2. A radar has a power of 1 kW and is operating at a frequency (2006)
of 10 GHz. It is located on a mountain top of height 500 m. 6. An electromagnetic wave of frequency u = 3.0 MHz passes
from vacuum into a dielectric medium with permitivity
The maximum distance upto which it can detect object located e = 4.0. Then
(a) wavelength is doubled and the frequency remains
on the surface of the earth (Radius of earth = 6.4 × 106 m) is unchanged
(b) wavelength is doubled and frequency becomes half
(a) 16 km (b) 40 km (c) wavelength is halved and frequency remains unchanged
(d) wavelength and frequency both remain unchanged.
(c) 64 km (d) 80 km (2012) (2004)
3. An electromagnertic warv e in vacuum has the electric and 7. Consider telecommunication through optical fibres. Which
magnetic fields E and B , which are always perpendicurl ar to
each other. The direction of pro larization is given by X and of the following statements is not true?
that orf warve prropargatiorn by k . Thren r r r r
(a) Xr P rE and kr P Er ´ Br (b) Xr PBr and kr P Er ´ Br (a) Optical fibres can be of graded refractive index.
(c) X PE and k P B ´ E (d) X PB and k P B ´ E
(2012) (b) Optical fibres are subject to electromagnetic interference
4. This question has Statement1 and Statement2. Of the four from outside.
choices given after the statements, choose the one that best (c) Optical fibres have extremely low transmission loss.
describes the two statements. (d) Optical fibres may have homogeneous core with a suitable
Statement1 : Sky wave signals are used for long distance cladding. (2003)
radio communication. These signals are in general, less stable 8. Which of the following are not electromagnetic waves?
than ground wave signals. (a) cosmic rays (b) gamma rays
Statement2 : The state of ionosphere varies from hour to (c) brays (d) Xrays. (2002)
hour, day to day and season to season.
(a) Statement1 is true, statement2 is false. 9. Electromagnetic waves are transverse in nature is evident by
(b) Statement1 is true, Statement2 is true, Statement2 is (a) polarization (b) interference
the correct explanation of Statement1. (c) reflection (d) diffraction. (2002)
(c) Statement1 is true, Statement2 is true, Statement2 is 10. Infrared radiation is detected by
not the correct explanation of Statement1. (a) spectrometer (b) pyrometer
(d) Statement1 is false, Statement2 is true. (2011) (c) nanometer (d) photometer. (2002)
Answer Key
1. (c) 2. (d) 3. (a) 4. (b) 5. (b) 6. (c)
7. (b) 8. (c) 9. (a) 10. (b)
Electromagnetic Waves 87
1. (c) : In electromagnetic wave, the peak value of electric = 1 e0 E r2m s + 1 æ E r2m s ö
2 2 m0 ççè c2 ÷÷ø
field (E 0 ) and peak value of magnetic field (B 0 ) are
related by = 1 e0 Er2ms + 1 Er2m se0m0
2 2 m 0
E0 = B 0 c
E 0 = (20 × 10 –9 T) (3 × 108 m s –1 ) = 6 V/m 1 Er2ms 12 e0 Er2ms e 0 Er2m s
2
2. (d) : Maximum distance on earth = e0 + =
where object can be detected is d,
h = (8.85 × 10 –12 ) × (720) 2
then d
(h + R) 2 = d 2 + R 2 = 4.58 × 10 –6 Jm –3 .
d 2 = h 2 + 2Rh 90° R
R 6. (c) : During propogation of a wave from one medium to
Q h < < R another, frequency remains constant and wavelength changes
\ d = 2 Rh m= e = 4=2
d = 2 ´ 6.4 ´ 106´ 500 = 8 ´ 104 m = 80 km e 0
Since m µ 1
l
3. (a) :rTher direction of polarization is parallel to electric field. \ Wavelength is halved
\ X P E
Hence option (c) holds good.
Trhe drirection of wave propagation is parallel to
E ´ B. 7. (b) : Optical fibres are subject to electromagnetic intereference
from outside.
rr r
\ k P E ´ B 8. (c) : brays are not electromagnetic waves.
4. (b)
9. (a) : Polarization proves the transverse nature of
1 1 electromagnetic waves.
2 2 m 0
5. (b) : u = e0 Er2ms + Br2m s 10. (b) : Infrared radiation produces thermal effect and is detected
by pyrometer.
88 JEE MAIN CHAPTERWISE EXPLORER
CHAPTER OPTICS
16
1. Diameter of a planoconvex lens is 6 cm and thickness at the 6. An object 2.4 m in front of a lens forms a sharp image on a
centre is 3 mm. If speed of light in material of lens is
2 × 10 8 m/s, the focal length of the lens is film 12 cm behind the lens. A glass plate 1 cm thick, of
(a) 10 cm (b) 15 cm (c) 20 cm (d) 30 cm
(2013) refractive index 1.50 is interposed between lens and film
with its plane faces parallel to film. At what distance (from
lens) should object be shifted to be in sharp focus on film?
2. Two coherent point sources (a) 2.4 m (b) 3.2 m
S 1 and S2 are separated by a (c) 5.6 m (d) 7.2 m (2012)
small distance ‘d’ as shown.
7. Direction : The question has a paragraph followed by two
The fringes obtained on the statements, Statement1 and Statement2. Of the given four
screen will be alternatives after the statements, choose the one that describes
(a) concentric circles (b) points the statements.
(c) straight lines (d) semicircles (2013) A thin air film is formed by putting the convex surface of a
3. A beam of unpolarised light of intensity I 0 is passed through planeconvex lens over a plane glass plate. With
a polaroid A and then through another polaroid B which is
monochromatic light, this film gives an interference pattern
oriented so that its principal plane makes an angle of 45° due to light reflected from the top (convex) surface and the
relative to that of A. The intensity of the emergent light is bottom (glass plate) surface of the film.
(a) I0 /8 (b) I 0 (c) I0 /2 (d) I0 /4 Statement1 : When light reflects from the airglass plate
(2013)
interface, the reflected wave suffers a phase change of p.
Statement2 : The centre of the interference pattern is dark.
4. The graph between angle of deviation (d) and angle of (a) Statement1 is true, Statement2 is false.
incidence (i) for a triangular prism is represented by
(b) Statement1 is true, Statement2 is true, Statement2 is
the correct explanation of Statement1.
(c) Statement1 is true, Statement2 is true, Statement2 is
not the correct explanation of Statement1.
(a) (b) (d) Statement1 is false, Statement2 is true. (2011)
8. Let the x–z plane be the boundary between two transparent
media. Medium 1 in z ³ 0 has a refractive index of 2 and
medium 2 with z < 0 has a refractive index of 3 . A ray
(c) (d) (2013) orf li ght in mediu m 1 given b y th e vector
A= 6 3 i$ + 83 $j -10 k$ is in cident on t he pl ane of
separation. The angle of refraction in medium 2 is
5. In Young’s double slit experiment, one of the slit is wider (a) 30° (b) 45°
than other, so that the amplitude of the light from one slit is (c) 60° (d) 75° (2011)
double of that from other slit. If I m be the maximum intensity, 9. A car is fitted with a convex sideview mirror of focal length
the resultant intensity I when they interfere at phase 20 cm. A second car 2.8 m behind the first car is overtaking
the first car at a relative speed of 15 m s –1. The speed of the
difference f is given by image of the second car as seen in the mirror of the first one is
I m ( ) (a) f ( ) (b) I m f
3 1 + 2cos 2 2 5 1 + 4cos 2 2 (a) 1 m s - 1 (b) 1 m s - 1
10 15
( ) (c)I m f Im
9 1 + 8cos 2 2 (d) 9 (4 + 5 cos f ) (2012) (c) 10 m s– 1 (d) 15 m s– 1 (2011)
Optics 89
Directions : Questions number 1012 are based on the following v (cm) v (cm)
paragraph.
An initially parallel cylindrical beam travels in a medium of (c) (d)
refractive index m(I) = m0 + m 2I , where m 0 and m 2 are positive
constants and I is the intens ity of the light beam. The intensity O u (cm) O u (cm)
of the beam is decreasing with increasing radius.
10. The initial shape of the wavefront of the beam is (2008)
(a) planar (b) convex 16. Two lenses of power –15 D and +5 D are in contact with each
(c) concave other. The focal length of the combination is
(d) convex near the axis and concave near the periphery (a) + 10 cm (b) –20 cm
11. The speed of light in the medium is (c) –10 cm (d) + 20 cm (2007)
(a) maximum on the axis of the beam
(b) minimum on the axis of the beam 17. In a Young’s double slit experiment the intensity at a point
(c) the same everywhere in the beam
(d) directly proportional to the intensity I where the path difference is l (l being the wavelength of
6
I
light used) is I. If I0 denotes the maximum intensity, I 0 is
12. As the beam enters the medium, it will equal to
(a) travel as a cylindrical beam 3 1 3 1
4 (b) 2 2 2
(b) diverge (c) converge (a) (c) (d)
(d) diverge near the axis and converge near the periphery (2007)
(2010) 18. The refractive index of glass is 1.520 for red light and 1.525
13. A mixture of light, consisting of wavelength 590 nm and an for blue light. Let D1 and D2 be angles of minimum deviation
for red and blue light respectively in a prism of this glass.
unknown wavelength, illuminates Young’s double slit and
gives rise to two overlapping interference patterns on the Then
screen. The central maximum of both lights coincide. Further, (a) D 1 > D2 (b) D 1 < D 2
it is observed that the third bright fringe of known light (c) D 1 = D 2
coincides with the 4t h bright fringe of the unknown light. (d) D1 can be less than or greater than depending upon the
From this data, the wavelength of the unknown light is angle of prism. (2006)
(a) 393.4 nm (b) 885.0 nm 19. A thin glass (refractive index 1.5) lens has optical power of
–5 D in air. Its optical power in a liquid medium with refractive
(c) 442.5 nm (d) 776.8 nm (2009) index 1.6 will be
(a) 25 D (b) –25 D (c) 1 D (d) –1 D (2005)
14. A transparent solid cylindrical rod has a refractive index of
2 . It is surrounded by air. A light ray is incident at the mid
3
point of one end of the rod as shown in the figure. 20. A fish looking up through the R
water sees the outside world
q contained in a circular q Cq C 12 cm
horizon. If the refractive (2005)
The incident angle q for which the light ray grazes along the index of water is 4/3 and the
wall of the rod is fish is 12 cm below the
( ) (a) 1 ( ) (b) 3 surface, the radius of this
2 2
sin -1 sin - 1 circle in cm is
( ) (c) sin - 1 2 ( ) (d) sin - 1 1 (2009) (a) 36 5 (b) 4 5
3 3
(c) 36 7 (d) 36 / 7
15. A student measures the focal length of a convex lens by 21. Two point white dots are 1 mm apart on a black paper. They
putting an object pin at a distance u from the lens and are viewed by eye of pupil diameter 3 mm. Approximately,
measuring the distance v of the image pin. The graph between what is the maximum distance at which these dots can be
u and v plotted by the student should look like resolved by the eye? [Take wavelength of light = 500 nm]
v (cm) v (cm) (a) 6 m (b) 3 m (c) 5 m (d) 1 m
(2005)
(a) (b) 22. When an unpolarized light of intensity I 0 is incident on a
polarizing sheet, the intensity of the light which does not get
O u (cm) O u (cm) transmitted is
90 JEE MAIN CHAPTERWISE EXPLORER
(a) zero (b) I 0 (2005) 28. The maximum number of possible interference maxima for
(c) 12 I 0 (d) 14 I 0 slitseparation equal to twice the wavelength in Young’s
doubleslit experiment is
23. If I0 is the intensity of the principal maximum in the single (a) infinite (b) five (c) three (d) zero. (2004)
slit diffraction pattern, then what will be its intensity when
29. To get three images of a single object, one should have two
the slit width is doubled? plane mirrors at an angle of
(a) I 0 (b) I 0/ 2 (c) 2I0 (d) 4I 0 (a) 60° (b) 90°
(2005)
(c) 120° (d) 30°. (2003)
24. A Young’s double slit experiment uses a monochromatic 30. The image formed by an objective of a compound microscope
is
source. The shape of the interference fringes formed on a
screen (a) virtual and diminished (b) real and diminished
(a) straight line (b) parabola (c) real and enlarged (d) virtual and enlarged.
(c) hyperbola (d) circle (2005) (2003)
25. A plano convex lens of refractive index 1.5 and radius of 31. To demonstrate the phenomenon of interference we require
curvature 30 cm is silvered at the curved surface. Now this
lens has been used to form the image of an object. At what two sources which emit radiation of
distance from this lens an object be placed in order to have
a real image of the size of the object? (a) nearly the same frequency
(a) 20 cm (b) 30 cm (c) 60 cm (d) 80 cm.
(2004) (b) the same frequency
(c) different wavelength
(d) the same frequency and having a definite phase
relationship. (2003)
26. A light ray is incident 32. An astronomical telescope has a large aperture to
perpendicular to one face of a (a) reduce spherical aberration
(b) have high resolution
90° prism and is totally internally (c) increase span of observation
reflected at the glassair
interface. If the angle of (d) have low dispersion. (2002)
45° 33. Which of the following is used in optical fibres?
reflection is 45°, we conclude 45° (a) total internal reflection (b) scattering
that the refractive index n
(c) diffraction (d) refraction. (2002)
n < 1
(a) 2 (b) n > 2 34. Wavelength of light used in an optical instrument are l 1 =
(d) n < 2 .
(c) n > 1 (2004) 4000 Å and l 2 = 5000 Å, then ratio of their respective
2
resolving powers (corresponding to l 1 and l 2) is
27. The angle of incidence at which reflected light in totally (a) 16 : 25 (b) 9 : 1
polarized for reflection from air to glass (refractive index n), (c) 4 : 5 (d) 5 : 4. (2002)
is 35. If two mirrors are kept at 60° to each other, then the number
(a) sin –1( n)
(c) tan –1( 1/n) (b) sin– 1( 1/n) of images formed by them is
(d) tan– 1 (n).
(a) 5 (b) 6
(2004) (c) 7 (d) 8. (2002)
Answer Key
1. (d) 2. (a) 3. (d) 4. (d) 5. (c) 6. (c)
7. (c) 8. (b) 9. (b) 10. (a) 11. (b) 12. (c)
13. (c) 14. (d) 15. (d) 16. (c) 17. (a) 18. (b)
19. (*) 20. (d) 21. (c) 22. (c) 23. (a) 24. (a)
25. (a) 26. (b) 27. (d) 28. (b) 29. (b) 30. (c)
31. (d) 32. (b) 33. (a) 34. (d) 35. (a)
Optics 91
1. (d): According to lens maker’s formula 5. (c) : Here, A2 = 2A 1
. . . Intensity µ (Amplitude) 2
1 1) êëé 1 1 ù
f = (m - R1 - R2 ûú I2 æ A2 ö 2 æ 2A A11 ö÷ø 2
I1 èç A1 ÷ø èç
\ = = = 4
As the lens is planoconvex
\ R 1 = R, R 2 = ¥ I2 = 4 I1
2
I1 + I2
1 (m - 1) ( ) Maximum intensity, Im =
f R
\ = ( ) ( ) 2 2
= I1 + 4I1 = 3 I1 = 9 I1
or f = R ...(i) I m
(m - 1) 9
or I1 = ...(i)
As speed of light in the medium of lens is 2 × 10 8 m/s Resultant intensity, I = I1 + I2 + 2 I1I2 cos f
\ m = c = 3 ´ 108 m/s = 3 ...(ii) = I1 + 4I1 + 2 I1(4I1 ) cos f
v 2 ´ 108 m/s 2
= 5I1 + 4I 1 cosf = I 1 + 4I 1 + 4I 1c osf
If r is the radius and t is the thickness of lens (at the centre), = I 1 + 4I 1 (1 + cosf)
the radius of curvature R of its curved surface in accordance f ( ) Q f
2 2
with figure will be given by = I1 + 8I1 cos2 1 + cos f = 2 cos 2
R2 = r 2 + (R – t) 2 ( ) = I1 1 + 8cos 2 f2
R2 = r2 + R 2 + t 2 – 2Rt
2Rt = r2 + t 2 Putting the value of I 1 from eqn. (i), we get
r 2 ( ) I = I m f
R = 2t (Q r >> t ) 9 1 + 8 cos 2 2
Here, r = 3 cm, t = 3 mm = 0.3 cm
(3 cm ) 2 6. (c) : 1 cm
´ 0.3 cm
\ R = 2 = 15 cm Film
On substituting the values of m and R from Eqs. (ii) and O
(iii) in (i), we get
f = 15 cm = 30 cm. 2.4 m 12 cm
(1.5 - 1)
2. (a) : When the screen is placed perpendicular to the line According to thin lens formula
joining the sources, the fringes will be concentric circles.
1 = 1 - 1
f v u
3. (d) : Intensity of light after passing polaroid A is Here, u = – 2.4 m = – 240 cm, v = 12 cm
I1 = I 0 \ 1 = 1 - 1 = 1 + 1
2 f 12 (- 240) 12 240
Now this light will pass through the second polaroid B 1 = 21 or f = 22410 cm
f 240
whose axis is inclined at an angle of 45° to the axis of
polaroid A. So in accordance with Malus law, the inten‑ When a glass plate is interposed between lens and film, so
sity of light emerging from polaroid B is shift produced by it will be
( )( ) I2 = I1 cos2 45° = I0 1 2 I 0 ( ) ( ) ( ) Shift 1 1 2 = 31 cm
2 2 4 m 1.5 3
= = t 1- = 1 1 - =1 1-
4. (d) : The graph between angle of To get image at film, lens should form image at distance
deviation(d) and angle of incidence (i)
1 335 cm
for a triangular prism is as shown in v¢ = 12 - 3 =
the adjacent figure. Again using lens formula
92 JEE MAIN CHAPTERWISE EXPLORER
\ 21 = 3 - 1 or 1 = 3 - 21 = 1 é3 - 21 ù If q c has to be the critical angle, qc = sin -1 1
240 35 u¢ u¢ 35 240 5 ë7 48 û m
1 1 é144 - 147 ù 1 3 But qc = 90° – f, q i = q.
u¢ 5 ëê 336 ûú u¢ 1680
= or = - sin qi 2 sin q m
sin f 3 cos qc m2 - 1
= m = Þ = m. 1
u¢ = – 560 cm = – 5.6 m qc
|u¢| = 5.6 m
But
7. (c) m2 -1 m2 - 1
m m
cosqc = \ sin q = m = m2 - 1.
8. (b) :
( ) \ q = sin-1 1
34 - 1 = sin -1 3
So that qc is making total internal reflection.
15. (d) : According to the new cartesian v is +ve
r system used in schools, 1 - 1 = 1 for
Here, A = 6 3 i^ + 8 3 j^ - 10 k^ v u f
a convex lens. f
f
cos i = 10 = 10 u has to be negative. u negative
20
(6 3 )2 + (8 3) 2 + (-10) 2 If v = ¥, u = f and if u = ¥, v = f.
A parallel beam (u = ¥) is focussed at f and if
( ) cosi 1 1 the object is at f, the rays are parallel. The point which
= 2 or i = cos-1 2 = 60 °
meets the curve at u = v gives 2f. Therefore v is +ve, u is
Using Snell’s law, m1 s ini = m2 sinr negative, both are symmetrical and this curve satisfies all
the conditions for a convex lens.
2 sin 60° = 3 sin r Þ r = 45° 16. (c) : Power of combination = P 1 + P2
= – 15 D + 5 D = – 10 D.
9. (b)
10. (a) : As the beam is initially parallel, the shape of wavefront Focal length of combination F = 1 = 1 D
is planar. P - 10
= – 0.1 m = – 10 cm.
11. (b) : Given m = m 0 + m 2 I 17. (a) : In Young’s double slit experiment intensity at a point is
As m = Speed of light in vacuum given by
Speed of light in medium
I0 cos 2 çæè fö
c c c I = 2 ÷ø
v m m0 + m 2 I
m = or v = =
As the intensity is maximum on the axis of the beam, therefore where f = phase difference, I 0 = maximum intensity
v is minimum on the axis of the beam. or I = cos 2 çæè fö ... (i)
I 0 2 ÷ø
12. (c) Phase difference f = 2 p ´ path difference
l
13. (c) : For interference, by Young’s double slits, the path
\ f = 2p ´ l or f = p ... (ii)
xd xd (2n + 1) l2 l 6 3
difference D = n l for bright fringes and D =
Substitute eqn. (ii) in eqn. (i), we get
for getting dark fringes. I = cos 2 çæè 6p ÷öø I = 3
I 0 I 0 4 .
The central fringes when x = 0, coincide for all wavelengths. or
The third fringe of l1 = 590 nm coincides with the fourth 18. (b) : Angle of minimum deviation D = A(m – 1)
bright fringe of unknown wavelength l.
\ xd = 3 × 590 nm = 4 × l nm D1 for red = mR -1
D D2 for blue mB - 1
\ l = 3 ´ 590 = 442.5 nm . Since mB > mR ,
4
D1 < 1
f qc m = ref. index of the rod \ D2
14. (d) : q \ D 1 < D2 .
Optics 93
19. (* ) : 1= (a m g - 1) èçæ 1 - 1 ö Hence intensity remains constant at I 0
fa R1 R2 ÷ø I = I 0 (1) = I0 .
1 = (l m g - 1) çèæ 1 - 1 ö 24. (a) : Straight line fringes are formed on screen.
fl R1 R2 ø÷
25. (a) : A planoconvex lens behaves like a concave mirror
\ f a = (l m g - 1) = (m g / ml ) - 1 when its curved surface is silvered.
fl (a m g - 1) (m g - 1) \ F of concave mirror so formed
= mg - ml = 1.5 -1.6 = R = 2 ´3 01 .5 = 10 cm
ml (mg -1) 1.6 (1.5 - 1) 2m
or Pl = - 0.1 = -1 To form an image of object size, the object should be placed
Pa 1.6 ´ 0.5 8 at (2F) of the concave mirror.
\ Distance of object from lens = 2 × F
Þ Pl = - Pa = - ( -5) = 85
8 8 = 2 × 10 = 20 cm.
or Optical power in liquid medium = 85 Dipotre. 26. (b) : Total internal reflection occurs in a denser medium
N.B. : This answer is not given in the four options provided when light is incident at surface of separation at angle
exceeding critical angle of the medium.
in the question. Given : i = 45° in the medium and total internal reflection
occurs at the glass air interface
20. (d) : For total internal reflection,
m = 1 Þsin qC = 1 = 3 \ n > 1 > 1 > 2 .
sin qC m 4 sin C sin 45°
\ tan qc = sin qc 27. (d) : According to Brewster's law of polarization, n = tan i p
1 - sin 2 q c where ip is angle of incidence
\ ip = tan –1 (n).
= 3/ 4 = 3 ´ 4 = 3 28. (b) : For interference maxima, dsinq = nl
9 4 7 7 \ 2lsinq = nl
1 - 16
or sin q = n
2
R 3 Þ R = 36 cm.
\ 12 = 7 7 This equation is satisfied if n = –2, –1, 0, 1, 2.
21. (c) : Resolution limit = 1.2d2 l sinq is never greater than (+1), less than (–1)
\ Maximum number of maxima can be five.
Again resolution limit = sin q = q = y 29. (b) : n = 360° - 1
D q0
\ y = 1.2d2 l \ 3 = 360° - 1 Þ 4q° = 360°Þ q° = 90°.
D q°
D = yd y
1.22 l
or q 30. (c) : The objective of compound microscope forms a real
(10-3 ) ´ 10-3 ) D and enlarged image.
(1.22) ´ 10-7 )
or D = ´ (3 = 63.01 » 5 m. 31. (d) : For interference phenomenon, two sources should emit
´ (5
radiation of the same frequency and having a definite phase
22. (c) : Intensity of polarized light = I 0/ 2 relationship.
\ Intensity of light not transmitted
32. (b) : Large aperture leads to high resolution of telescope.
I0 = I 0 .
= I 0 - 2 2 33. (a) : Total internal reflection is used in optical fibres.
23. (a) : For diffraction pattern 34. (d) : Resolving power is proportional to l– 1
I = I æ sin f 2 f denotes path difference \ R.P . for l1 = l2 = 5000 = 5 .
0 çè f R.P. for l2 l1 4000 4
ö
÷ where
ø
For principal maxima, f = 0. Hence æ sin f ø÷ö = 1 35. (a) : n = 360° -1 = 360° -1 = 5.
çè f q° 60 °
94 JEE MAIN CHAPTERWISE EXPLORER
CHAPTER DUAL NATURE OF MATTER
17 AND RADIATION
1. The anode voltage of a photocell is kept fixed. The wavelength (c) Statement1 is true, Statement2 is true, Statement2 is
l of the light falling on the cathode is gradually changed. The
plate current I of the photocell varies as follows not the correct explanation of Statement1.
I I (d) Statement1 is false, Statement2 is true. (2011)
4. If a source of power 4 kW produces 10 20 photons/second, the
radiation belongs to a part of the spectrum called
(a) grays (b) Xrays
(a) (b) (c) ultraviolet rays (d) microwaves (2010)
O O
l l 5. Statement1 : When ultraviolet light is incident on a photocell,
I I
its stopping potential is V 0 and the maximum kinetic energy
(c) (d) (2013) of the photoelectrons is K max . When the ultraviolet light is
replaced by Xrays, both V0 and Km ax increase.
Statement2 : Photoelectrons are emitted with speeds ranging
O l O l from zero to a maximum value because of the range of
frequencies present in the incident light.
2. This question has Statement 1 and Statement 2. Of the four (a) Statement1 is true, Statement2 is false.
choices given after the statements, choose the one that best (b) Statement1 is true, Statement2 is true; Statement2 is
describes the two statements. the correct explanation of Statement1.
Statement 1 : Davisson Germer experiment established the (c) Statement1 is true, Statement2 is true; Statement2 is
wave nature of electrons. not the correct explanation of Statement1.
Statement 2 : If electrons have wave nature, they can interfere (d) Statement1 is false, Statement2 is true. (2010)
and show diffraction. 6. The surface of a metal is illuminated with the light of
(a) Statement 1 is true, Statement 2 is false. 400 nm. The kinetic energy of the ejected photoelectrons was
(b) Statement 1 is true, Statement 2 is true, Statement 2 is found to be 1.68 eV. The work function of the metal is
the correct explanation for Statement 1. (hc = 1240 eV nm)
(c) Statement 1 is true, Statement 2 is true, Statement 2 is (a) 3.09 eV (b) 1.41 eV
not the correct explanation of Statement 1. (c) 1.51 eV (d) 1.68 eV (2009)
(d) Statement 1 is false, Statement 2 is true. (2012)
3. This question has Statement1 and Statement2. Of the four Directions : Questions 7, 8 and 9 are based on the following
choices given after the statements, choose the one that best paragraph.
describes the two statements : Wave property of electrons implies that they will show diffraction
Statement1 : A metallic surface is irradiated by a effects. Davisson and Germer demonstrated this by diffracting
monochromatic light of frequency u > u 0 (the threshold electrons from crystals. The law governing the diffraction from
frequency). The maximum kinetic energy and the stopping a crystal is obtained by requiring that electron waves reflected
potential are Km ax and V0 respectively. If the frequency incident from the planes of atoms in a crystal interfere constructively (see
on the surface is doubled, both the K max and V0 are also figure).
doubled.
Statement2 : The maximum kinetic energy and the stopping (2008)
potential of photoelectrons emitted from a surface are linearly
dependent on the frequency of incident light. incoming i outgoing
(a) Statement1 is true, statement2 is false. electrons electrons
(b) Statement1 is true, Statement2 is true, Statement2 is
the correct explanation of Statement1. d
crystal plane
Dual Nature of Matter and Radiation 95
7. Electrons accelerated by potential V are diffracted from a I I
crystal. If d = 1 Å and i = 30°, V should be about (a) (b)
(h = 6.6 × 10 –34 Js, m e = 9.1 × 10 –31 kg, e = 1.6 × 10 –19 C)
(a) 1000 V (b) 2000 V (c) 50 V (d) 500 V. O O
l l
8. If a strong diffraction peak is observed when electrons are
incident at an angle i from the normal to the crystal planes
with distance d between them (see figure), de Broglie I I
wavelength l dB of electrons can be calculated by the (c) (d) . (2006)
relationship (n is an integer)
O l O l
(a) d cosi = nl dB (b) d sini = nl dB 13. The threshold frequency for a metallic surface corresponds
(c) 2d cosi = nl dB (d) 2d sini = nl dB
9. In an experiment, electrons are made to pass through a narrow to an energy of 6.2 eV, and the stopping potential for a
slit of width d comparable to their de Broglie wavelength.
They are detected on a screen at a distance D from the slit. radiation incident on this surface 5 V. The incident radiation
lies in
(a) Xray region (b) ultraviolet region
(c) infrared region (d) visible region. (2006)
14. The time by a photoelectron to come out after the photon
d y =0 strikes is approximately
(a) 10– 1 s (b) 10– 4 s
D (c) 10– 10 s (d) 10– 16 s. (2006)
Which of the following graphs can be expected to represent 15. If the kinetic energy of a free electron doubles, its de Broglie
the number of electrons N detected as a function of the detector wavelength changes by the factor
position y (y = 0 corresponds to the middle of the slit)?
(a) 1/ 2 (b) 2 (c) 1/2 (d) 2. (2005)
y y
16. A photocell is illuminated by a small bright source placed 1 m
away. When the same source of light is placed (1/2) m away,
(a) N d (b) N d the number of electrons emitted by photocathode would
(a) decrease by a factor of 2
(b) increase by a factor of 2
y (c) decrease by a factor of 4
(d) increase by a factor of 4 (2005)
(c) (d) N d 17. A charged oil drop is suspended in a uniform field of
3 × 104 V/m so that it neither falls nor rises. The charge on the
drop will be (take the mass of the charge = 9.9 × 10– 15 kg and
10. If gE and gM are the accelerations due to gravity on the surfaces g = 10 m/s 2)
of the earth and the moon respectively and if Millikan’s oil
drop experiment could be performed on the two surfaces, one (a) 3.3 × 10 –18 C (b) 3.2 × 10 –18 C
will find the ratio
(c) 1.6 × 10 –18 C (d) 4.8 × 10 –18 C. (2004)
18. The work function of a substance is 4.0 eV. The longest
electronic charge on the moon wavelength of light that can cause photoelectron emission
electronic charge on the earth to be
from this substance is approximately
(a) g M /gE (b) 1 (a) 540 nm (b) 400 nm
(c) 0 (d) gE /gM
(2007) (c) 310 nm (d) 220 nm. (2004)
11. Photon of frequency u has a momentum associated with it. 19. According to Einstein’s photoelectric equation, the plot of
the kinetic energy of the emitted photo electrons from a
If c is the velocity of light, the momentum is metal vs the frequency, of the incident radiation gives a straight
line whose slope
(a) hu/c (b) u/c (a) depends on the nature of the metal used
(b) depends on the intensity of the radiation
(c) huc (d) hu/c 2 (2007) (c) depends both on the intensity of the radiation and the
metal used
12. The anode voltage of a photocell is kept fixed. The
wavelength l of the light falling on the cathode is gradually
changed. The plate current I of the photocell varies as follows
96 JEE MAIN CHAPTERWISE EXPLORER
(d) is the same for all metals and independent of the intensity v12 v22 2 h ( 1
m
of the radiation. (2004) (c) + = f1 + f 2 ) (d) v1 - v2 = êéë2m h ( f1 - f 2 )ûúù 2 .
20. Two identical photocathodes receive light of frequencies f 1 (2003)
and f 2 . If the velocities of the photoelectrons (of mass m) 21. Sodium and copper have work functions 2.3 eV and 4.5 eV
coming out are respectively v 1 and v 2 , then respectively. Then the ratio of the wavelengths is nearest to
v12 v22 2 h ( 1 (a) 1 : 2 (b) 4 : 1 (c) 2 : 1 (d) 1 : 4.
m
(a) - = f1 - f 2 ) (b) v1 + v2 = êëé2m h ( f1 + f 2 ) ùûú 2 (2002)
Answer Key
1. (a) 2. (b) 3. (d) 4. (b) 5. (a) 6. (b)
7. (c) 8. (c) 9. (a) 10. (b) 11. (a) 12. (c)
13. (b) 14. (c) 15. (a) 16. (d) 17. (a) 18. (c)
19. (d) 20. (a) 21. (c)
Dual Nature of Matter and Radiation 97
1. (a) m e = 9.1 × 10 –31 kg, e = 1.6 × 10 –19 C.
Bragg’s equation for Xrays, which is also used in electron
2. (b) : DavissonGermer experiment showed that electron diffraction gives nl = 2d sinq.
beams can undergo diffraction when passed through atomic
crystals. This shows the wave nature of electrons as waves \ l = 2 ´1(A° ) ´ sin 60 ° (assuming first order)
can exhibit interference and diffraction. 1
3. (d) : The maximum kinetic energy of the electron l = 3 A° , V = (12.27 ´10-10 )
Km ax = hu – hu0 3 ´ 10 -10
Here, u 0 is threshold frequency.
The stopping potential is V = 50.18 Volt.
eV 0 = K max = hu – hu 0
Therefore, if u is doubled K max and V 0 is not doubled. 8. (c) : Bragg’s relation nl = 2d sinq for having an intensity
maximum for diffraction pattern.
4. (b) : Here,
i
Power of a source, P = 4 kW = 4 × 10 3 W q
Number of photons emitted per second, N = 10 20 But as the angle of incidence is given,
nl = 2d cosi is the formula for finding a peak.
Energy of photon, E = hu = hc 9. (a) : The electron diffraction pattern from a single slit will
l be as shown below.
. . . E = P d
N
q
hc = P D
l N
or l = Nhc = 1020 ´ 6.63 ´ 10-34 ´ 3 ´ 108
P 4 ´ 103
= 4.972 × 10 –9 m = 49.72 Å
It lies in the Xray region.
5. (a) : According to Einstein’s photoelectric equation
Km ax = hu – f 0
where, x
u = frequency of incident light d sinq = 2lp .
The line of maximum intensity for the zeroth order will exceed
f 0 = work function of the metal d very much.
Since K max = eV 0
V0 = hu - f0 10. (b) : Since electronic charge (1.6 × 10 –19 C) universal constant.
e e It does not depend on g.
\ Electronic charge on the moon = electronic
As u Xrays > uU ltraviolet charge on the earth
Th er efore, b ot h K max an d V 0 in cr ea s e wh en ultraviolet light
is replaced by Xrays. electronic charge on the moon
or electronic charge on the earth = 1.
Statement2 is false.
6. (b) : The wavelength of light illuminating the photoelectric 11. (a) : Energy of a photon E= hu ... (i)
surface = 400 nm. Also E = pc ... (ii)
where p is the momentum of a photon
i.e., hu = 1240 eV nm = 3.1 eV . From (i) and (ii), we get
400 nm
Max. kinetic energy of the electrons = 1.68 eV. hu = pc or p = hc u .
hu = Wf + kinetic energy 12. (c) : The graph (c) depicts the variation of l with I.
\ Wf, the work function = hu – kinetic energy
= 3.1 – 1.68 eV = 1.42 eV.
7. (c) : For electron diffraction, d = 1 Å, i = 30° 13. (b) : For photoelectron emission,
i.e., grazing angle q = 60°, h = 6.6 × 10 –34 J s. (Incident energy E) = (K.E.) max + (Work function f)
98 JEE MAIN CHAPTERWISE EXPLORER
or E = K m + f 18. (c) : Let l m = Longest wavelength of light
or E = 5 + 6.2 = 11.2 eV
\ hc = f (work function)
= 11.2 × (1.6 × 10 –19 ) J lm
\ hc = 11.2 ´1.6 ´ 10 -19 \ lm = hc = (6.63´10-34 ) ´ (3´108 )
l f 4.0 ´1.6 ´ 10- 19
(6.63´10-34 ) ´ (3´108 ) or l m = 310 nm.
or l = 11.2 ´1.6 ´ 10-19 m
19. (d) : According to Einstein's equation,
or l = 1110 × 10 –10 m = 1110 Å. Kinetic energy = hf – f where kinetic energy and f (frequency)
are variables, compare it with equation, y = mx + c
The incident radiation lies in ultra violet region.
14. (c) : Emission of photoelectron starts from the surface Kinetic energy
after incidence of photons in about 10 –10 sec.
15. (a) : de Broglie wavelength l = h / p = h / (2mK )
\ l = h where K = kinetic energy of q f
2 mK f
particle
\ l2 = K1 = K1 = 1 . Work function
l 1 K2 2 K1 2
\ slope of line = h
16. (d) : P of source = P h is Planck's constant.
I = 4p(distance)2 4 pd 2 Hence the slope is same for all metals and independent of
the intensity of radiation.
Here, we assume light to spread uniformly in all directions. Option (d) represents the answer.
Number of photoelectrons emitted from a surface depend 20. (a) : For photoelectric effect, according to Einstein's equation,
Kinetic energy of emitted electron = hf – (work function f)
on intensity of light I falling on it. Thus the number of \ 12 mv12 = hf1 - f
12 mv22 = hf2 - f
electrons emitted n depends directly on I. P remains constant \ 12 m(v12 - v22 ) = h ( f1 - f2 )
\ v12 - v22 = 2m h ( f1 - f 2 ).
as the source is the same.
21. (c) : Work function = hc/l
\ I2 = n2 Þ P2 æ d1 ö2 = n2 W Na = 4.5 = 2 .
I1 n1 P1 èç d2 ÷ n1 WCu 2.3 1
ø
\ n2 = çæè PP ø÷ö çæè 1 ø÷ö 2 = 14 .
n1 1/2
17. (a) : For equilibrium of charged oil drop,
qE = mg
\ q = mg = (9.9 ´10-15 ) ´ 10 = 3.3 ´10-18 C.
E ( 3 ´ 104 )
Atoms and Nuclei 99
CHAPTER ATOMS AND NUCLEI
18
1. In a hydrogen like atom electron makes transition from an A - Z - 4 A - Z - 8
(a) Z - 2 (b) Z - 4
energy level with quantum number n to another with quantum
A - Z - 4 A - Z - 12
number (n – 1). If n > > 1, the frequency of radiation emitted (c) Z - 8 (d) Z - 4
is proportional to (2010)
1 (b) 1 1 1
(a) n 3 n (c) n 2 (d) n 3/2
Directions : Questions number 89 are based on the following
(2013) paragraph.
2. Hydrogen atom is excited from ground state to another state A nucleus of mass M + Dm is at rest and decays into two daughter
with principal quantum number equal to 4. Then the number
of spectral lines in the emission spectra will be nuclei of equal mass M each. Speed of light is c.
2
(a) 3 (b) 5 (c) 6 (d) 2
(2012) 8. The speed of daughter nuclei is
3. Assume that a neutron breaks into a proton and an electron. c Dm c M Dm
M + D m + D m
The energy released during this process is (a) (b)
(Mass of neutron = 1.6725 × 10– 27 kg
Mass of proton = 1.6725 × 10– 27 kg (c) c 2 D m (d) c D m (2010)
M M
Mass of electron = 9 × 10– 31 kg)
(a) 7.10 MeV (b) 6.30 MeV 9. The binding energy per nucleon for the parent nucleus is E 1
(c) 5.4 MeV (d) 0.73 MeV (2012) and that for the daughter nuclei is E 2 . Then
4. A diatomic molecule is made of two masses m 1 and m2 which (a) E1 = 2E2 (b) E2 = 2E 1
are separated by a distance r. If we calculate its rotational
energy by applying Bohr’s rule of angular momentum (c) E1 > E2 (d) E2 > E 1 (2010)
quantization, its energy will be given by (n is an integer)
10. The transition from the state n = 4 to n = 3 in a hydrogen like
atom results in ultraviolet radiation.Infrared radiation will be
(a) n2h 2 (b) 2n 2h 2 obtained in the transition from
2(m1 + m2 ) r 2 (m1 + m2 ) r2
(a) 2 ® 1 (b) 3 ® 2
(d) (m1 + m2 ) 2 n2h 2
(c) (m1 + m2 ) n2h 2 2 m12m22 r 2 (2012) (c) 4 ® 2 (d) 5 ® 4 (2009)
2 m1m2 r 2
5. Energy required for the electron excitation in Li+ + from the 11. Eb B C D E
first to the third Bohr orbit is A F
(a) 12.1 eV (b) 36.3 eV
(c) 108.8 eV (d) 122.4 eV (2011) M
6. The half life of a radioactive substance is 20 minutes. The The above is a plot of binding energy per nucleon Eb , against
the nuclear mass M; A, B, C, D, E, F correspond to different
approximate time interval (t2 – t 1) between the time t 2 when nuclei. Consider four reactions:
2 of it has decayed and time t 1 when 13 of it had decayed is (i) A + B ® C + e (ii) C ® A + B + e
3
(a) 7 min (b) 14 min
(c) 20 min (d) 28 min (2011) (iii) D + E ® F + e (iv) F ® D + E + e
7. A radioactive nucleus (initial mass number A and atomic where e is the energy released? In which reactions is e positive?
number Z) emits 3aparticles and 2 positrons. The ratio of
number of neutrons to that of protons in the final nucleus (a) (i) and (iv) (b) (i) and (iii)
will be
(c) (ii) and (iv) (d) (ii) and (iii) (2009)
100 JEE MAIN CHAPTERWISE EXPLORER
Directions : Question 12 contains statement1 and statement2. 19. The ‘rad’ is the correct unit used to report the measurement of
Of the four choices given, choose the one that best describes the
two statements. (a) the rate of decay of radioactive source
(a) Statement1 is true, statement2 is false.
(b) Statement1 is false, statement2 is true. (b) the ability of a beam of gamma ray photons to produce
(c) Statement1 is true, statement2 is true; statement2 is a
ions in a target
correct explanation for statement1.
(d) Statement1 is true, statement2 is true; statement2 is not a (c) the energy delivered by radiation to target
correct explanation for statement1. (d) the biological effect of radiation. (2006)
20. When 3 Li7 nuclei are bombarded by protons, and the resultant
nuclei are 4B e8 , the emitted particles will be
(a) neutrons (b) alpha particles
12. Statement1 : Energy is released when heavy nuclei undergo (c) beta particles (d) gamma photons. (2006)
fission or light nuclei undergo fusion. 21. The energy spectrum of bparticles [number N(E) as a function
of benergy E] emitted from a radioactive source is
Statement2 : For heavy nuclei, binding energy per nucleon
increases with increasing Z while for light nuclei it decreases
with increasing Z. (2008) N(E ) N(E )
13. Suppose an electron is attracted towards the origin by a force (a) (b)
k/r where k is a constant and r is the distance of the electron E 0 E E
E 0
from the origin. By applying Bohr model to this system, the
radius of the nt h orbital of the electron is found to be r n and (c) N(E ) N(E ) . (2006)
the kinetic energy of the electron to be T n. Then which of
the following is true? (d)
1 , µ n 2 1 rn µ n 2 E 0 E E 0 E
n n 2 ,
(a) Tn µ rn (b) Tn µ 22. An alpha nucleus of energy 1 mv 2 bombards a heavy nuclear
2
(c) Tn independent of n, rn µ n
target of charge Ze. Then the distance of closest approach
1 ,
(d) Tn µ n rn µ n (2008) for the alpha nucleus with be proportional to
(a) 1/Ze (b) v 2 (c) 1/m (d) 1/v4 . (2006)
14. Which of the following transitions in hydrogen atoms emit 23. A nuclear transformation is denoted by X(n, a) 37 Li . Which
of the following is the nucleus of element X?
photons of highest frequency ?
(a) n = 1 to n = 2 (b) n = 2 to n = 6
(c) n = 6 to n = 2 (d) n = 2 to n = 1 (2007) (a) 95 B (b) 141 Be (c) 162 C (d) 105 B (2005)
15. The halflife period of a radioactive element X is same as the 24. The diagram shows the energy levels for an electron in a
mean life time of another radioactive element Y. Initially they certain atom. Which transition shown represents the emission
have the same number of atoms. Then of a photon with the most energy?
(a) X and Y decay at same rate always n = 4
n = 3
(b) X will decay faster than Y
(c) Y will decay faster than X
(d) X and Y have same decay rate initially (2007) n = 2
16. In gamma ray emission from a nucleus
(a) only the proton number changes
(b) both the neutron number and the proton number change
(c) there is no change in the proton number and the neutron n = 1
III IV
number I II
(c) III (d) IV
(d) only the neutron number changes (2007) (a) I (b) II (2005)
17. If M O is the mass of an oxygen isotope 8O 17, MP and MN are 25. Starting with a sample of pure 66 Cu, 7/8 of it decays into
the masses of a proton and a neutron respectively, the nuclear
Zn in 15 minutes. The corresponding halflife is
binding energy of the isotope is
(b) 7 12 minutes
(a) (MO – 17 M N) c2 (b) (M O – 8 MP ) c 2 (a) 5 minutes (d) 14 minutes (2005)
(c) (MO – 8 M P – 9 MN ) c 2 (c) 10 minutes
(d) MO c2
(2007) 26. The intensity of gamma radiation from a given source is I.
18. If the binding energy per nucleon in 37 Li and 42 He nuclei are On passing through 36 mm of lead, it is reduced to I/8. The
5.60 MeV and 7.06 MeV respectively, then in the reaction :
thickness of lead which will reduce the intensity to I/2 will be
p + 7 Li ® 2 42 He , energy of proton must be (a) 18 mm (b) 12 mm
3
(a) 39.2 MeV (b) 28.24 MeV (c) 6 mm (d) 9 mm (2005)
(c) 17.28 MeV (d) 1.46 MeV. (2006)
Atoms and Nuclei 101
27. If radius of the 1237 Al nucleus is estimated to be 3.6 fermi [Boltzmann's constant k = 1.38 × 10 –23 J/K]
(a) 10 7 K (b) 105 K (c) 103 K (d) 109 K. (2003)
then the radius of 12552 Al nucleus be nearly
(a) 4 fermi (b) 5 fermi 35. Which of the following cannot be emitted by radioactive
(c) 6 fermi (d) 8 fermi (2005) substances during their decay?
28. An aparticle of energy 5 MeV is scattered through 180° by (a) protons (b) neutrinos
a fixed uranium nucleus. The distance of the closest approach (c) helium nuclei (d) electrons. (2003)
is of the order of 36. A nucleus with Z = 92 emits the following in a sequence:
a, a, b – , b – , a, a, a, a, b – , b –, a, b +, b +, a. The Z of the
(a) 1 Å (b) 10– 10 cm resulting nucleus is
(a) 76 (b) 78 (c) 82 (d) 74. (2003)
(c) 10– 12 cm (d) 10– 15 cm. (2004)
( ) 29. The binding energy per nucleon of deuteron 12 H and helium 37. A radioactive sample at any instant has its disintegration rate
( ) nucleus 42 He is 1.1 MeV and 7 MeV respectively. If two
5000 disintegrations per minute. After 5 minutes, the rate is
deuteron nuclei react to form a single helium nucleus, then
the energy released is 1250 disintegrations per minute. Then, the decay constant
(a) 13.9 MeV (b) 26.9 MeV (per minute) is
(c) 23.6 MeV (d) 19.2 MeV. (2004) (a) 0.4 ln 2 (b) 0.2 ln 2
30. A nucleus disintegrates into two nuclear parts which have (c) 0.1 ln 2 (d) 0.8 ln 2. (2003)
their velocities in the ratio 2 : 1. The ratio of their nuclear 38. When U 238 nucleus originally at rest, decays by emitting an
alpha particle having a speed u, the recoil speed of the residual
sizes will be nucleus is
(a) 2 1/3 : 1 (b) 1 : 31 /2 (c) 31 /2 : 1 (d) 1 : 21 /3. (2004)
31. If the binding energy of the electron in a hydrogen atom is (a) 4u (b) - 243u 4 (c) 4u (d) - 243u 8 . (2003)
238 234
13.6 eV, the energy required to remove the electron from the
first excited state of Li+ + is 39. Which of the following radiations has the least wavelength?
(a) 30.6 eV (b) 13.6 eV (a) grays (b) brays
(c) 3.4 eV (d) 122.4 eV. (2003) (c) arays (d) Xrays. (2003)
32. The wavelengths involved in the spectrum of deuterium (12 D) 40. If N0 is the original mass of the substance of halflife period
are slightly different from that of hydrogen spectrum, because t 1/2 = 5 years, then the amount of substance left after 15 years
is
(a) size of the two nuclei are different
(b) nuclear forces are different in the two cases (a) N 0 /8 (b) N0 /16 (c) N 0 /2 (d) N 0/ 4. (2002)
41. If 13.6 eV energy is required to ionize the hydrogen atom,
(c) masses of the two nuclei are different
(d) attraction between the electron and the nucleus is different then the energy required to remove an electron from n = 2
in the two cases. (2003) is
33. Which of the following atoms has the lowest ionization (a) 10.2 eV (b) 0 eV
potential? (c) 3.4 eV (d) 6.8 eV. (2002)
42. At a specific instant emission of radioactive compound is
(a) 147 N (b) 13535 Cs (c) 1480 Ar (d) 186 O. (2003) deflected in a magnetic field. The compound can emit
34. In the nuclear fusion reaction, (i) electrons (ii) protons
2 H +13 H ® 42 He + n given that the repulsive potential (iii) He2 + (iv) neutrons
1
The emission at the instant can be
energy between the two nuclei is ~ 7.7 × 10 –14 J, the
(a) i, ii, iii (b) i, ii, iii, iv
temperature at which the gases must be heated to initiate the
(c) iv (d) ii, iii. (2002)
reaction is nearly
Answer Key
1. (a) 2. (c) 3. (*) 4. (c) 5. (c) 6. (c)
7. (c) 8. (c) 9. (d) 10. (d) 11. (a) 12. (a)
13. (c) 14. (d) 15. (b) 16. (c) 17. (c) 18. (c)
19. (d) 20. (d) 21. (d) 22. (c) 23. (d) 24. (c)
25. (a) 26. (b) 27. (c) 28. (c) 29. (c) 30. (d)
31. (a) 32. (c) 33. (b) 34. (d) 35. (a) 36. (b)
37. (a) 38. (b) 39. (a) 40. (a) 41. (c) 42. (a)
102 JEE MAIN CHAPTERWISE EXPLORER
1. (a) : In a hydrogen like atom, when an electron makes an On rearranging, we get
transition from an energy level with n to n – 1, the frequency m2r
m1 + m2
of emitted radiation is r1 =
u= RcZ 2 é (n 1 - 1 ù m1r
êë - 1) 2 n 2 úû m1 + m2
Similarly, r2 =
= RcZ 2 é n2 - (n - 1) 2 ù RcZ 2 (2n - 1) Therefore, the moment of inertia can be written as
ëê (n2 )(n - 1)2 úû = n2 (n - 1)
As n > > 1 m1 æçè m2r øö÷ 2 m2 æçè m1m+1 r m 2 ÷öø 2
m1 + m2
RcZ 2 2n 2 RcZ 2 I = +
n4 n3
\ u= =
or u µ 1 = m1m2 r 2 ...(i)
n3 m1 + m2
2. (c) : Number of spectral lines in the emission spectra, According to Bohr’s quantisation condition
n(n - 1) L = nh
2 2p
N =
Here, n = 4 or L2 = n2h 2 ...(ii)
4 p 2
\ N = 4(4 - 1) = 6 Rotational energy, E = L2
2 2 I
3. (*) : Mass defect, Dm = m p + m e – m n E = n2h 2 (Using (ii))
= (1.6725 × 10 –27 + 9 × 10 –31 – 1.6725 × 10 –27 ) kg 8p 2 I
= 9 × 10 –31 kg = n2h2 (m1 + m2 ) (Using (i))
8p 2 (m1m2 ) r 2
Energy released = Dmc 2
= 9 × 10 –31 × (3 × 10 8 )2 J
9 ´ 10-31 ´ 9 ´ 101 6 = n2h2 (m1 + m2 ) ( ) Q h = h
= 1.6 ´ 10-13 MeV 2 m1m2 r 2 2 p
= 0.51 MeV In the question instead of h, h should be given.
* None of the given option is correct.
4. (c) : A diatomic molecule consists of two atoms of masses 5. (c) : Using, E n = - 13.n62 Z 2 eV
m 1 and m 2 at a distance r apart. Let r1 and r 2 be the distances Here, Z = 3 (For Li+ +)
of the atoms from the centre of mass.
\ E1 = - 13.(61)( 23 ) 2 eV
m1 m 2 E 1 = – 122.4 eV
COM r2
r 1 -13.6 ´ (3) 2
r (3) 2
The moment of inertia of this molecule about an axis passing and E3 = = -13.6 eV
through its centre of mass and perpendicular to a line joining
the atoms is DE = E3 – E1 = – 13.6 + 122.4 = 108.8 eV
I = m1r12 + m2r2 2 6. (c) : Number of undecayed atoms after time t2 ,
As m1r1 = m2r2
or r1 = m2 r2 N 0 = N 0 e - lt 2 ...(i)
Q r1 + r2 = r m1 3 ...(ii)
Number of undecayed atoms after time t 1,
m2 2 N 0 = N0 e- lt1
m1 3
\ r1 = (r - r1 )
Dividing (ii) by (i), we get
Atoms and Nuclei 103
2 = el(t2 -t1 ) or ln 2 = l(t 2 – t 1) Transition 4 ® 3 is in Paschen series. This is not in the
ultraviolet region but this is in infrared region.
or (t2 - t1 ) = ln 2 Transition 5 ® 4 will also be in infrared region (Brackett).
l
As per question, t 1/2 = half life time = 20 min 11. (a) : When two nucleons combine to form a third one, and
energy is released, one has fusion reaction. If a single nucleus
\ t 2 – t 1 = 20 min éëêQ t1/2 = ln 2 ù splits into two, one has fission. The possibility of fusion is
l ûú more for light elements and fission takes place for heavy
elements.
7. (c) : When a radioactive nucleus emits an alpha particle, its Out of the choices given for fusion, only A and B are light
elements and D and E are heavy elements. Therefore
mass number decreases by 4 while the atomic number A + B ® C + e is correct. In the possibility of fission is only
for F and not C. Therefore
decreases by 2. F ® D + E + e is the correct choice.
When a radioactive nucleus, emits a b + particle (or positron
(e+ ) ) its mass number remains unchanged while the atomic
number decreases by 1.
\ A X ¾3¾a ® A- -126Y ¾2¾ e +¾® A - 12 W 12. (a) : Statement1 states that energy is released when heavy
Z Z Z -8 nuclei undergo fission and light nuclei undergo fusion is
correct. Statement2 is wrong.
In the final nucleus, The binding energy per nucleon, B/A, starts at a small value,
rises to a maximum at 62N i, then decreases to 7.5 MeV for
Number of protons, Np = Z – 8 the heavy nuclei. The answer is (a).
Number of neutrons, N n = A – 12 – (Z – 8)
= A – Z – 4
\ N n = A - Z - 4 . 13. (c) : Supposing that the force of attraction in Bohr atom
N p Z - 8
does not follow inverse square law but inversely proportional
( ) 8. M M ù to r,
(c) : Mass defect, DM = êéë(M + Dm) - 2 + 2 ûú
1 e 2 would have been = mv 2
= [M + Dm – M] = Dm 4 pe 0 r r
Energy released, Q = DMc 2 = Dmc2 ...(i) \ mv 2 = e 2 = k Þ 1 mv 2 = 12 k.
4pe 0 2
According to law of conservation of momentum, we get
This is independent of n.
(M + Dm) ´ 0 = M ´ v1 - M ´ v2 or v1 = v2 From mvrn = 2n ph ,
2 2 as mv is independent of r, r n µ n.
( ) ( ) Also,Q = 1 M v12 + 1 M v22 - 12 (M +Dm) ´ (0) 2
2 2 2 2
= M v12 (Q v1 = v2 ) ...(ii) 14. (d) : n = 2 –13.6 eV
2 2 2
hu 2 ® 1
Equating equations (i) and (ii), we get n = 1 – 13.6 eV
( ) M v12 = Dm c2 hu2®1 = -13.6 æ 1 - 1 ö eV
2 èç 22 12 ÷ø
v1 2 = 2 Dmc 2 = +13.6 ´ 3 eV = 10.2 eV .
M 4
2 Dm Emission is n = 2 ® n = 1 i.e., higher n to lower n.
M
v1 = c Transition from lower to higher levels are absorption lines.
9. (d) : After decay, the daughter nuclei will be more stable, -13.6 æ 1 - 1 ÷öø = +13.6 ´ 2
hence binding energy per nucleon of daughter nuclei is more çè 62 22 9
than that of their parent nucleus.
Hence, E 2 > E 1. This is <E n = 2 ® E n = 1.
15. (b) : T 1/2, half life of X = tY , mean life of Y
ln 2 = 1 Þ lX = lY ln 2
lX lY
10. (d) : n = 5 Pfund
n = 4 Brackett l X > l Y
Balmer n = 3 Paschen
Lyman n = 2 Balmer \ AX = A0 e-lX t ; AY = A0 e-lY t
n = 1 Lyman
X will decay fas ter than Y.
16. (c) : gray emission takes place due to deexcitation of the
nucleus. Therefore during gray emission, there is no change
in the proton and neutron number.
104 JEE MAIN CHAPTERWISE EXPLORER
17. (c) : Binding energy = [ZMP + (A – Z) MN – M]c2 26. (b) : Q I = I 0 e –kx Þ I = e - kx
= [8M P + (17 – 8) MN – MO ]c2 I 0
= (8M P + 9MN – M O )c 2
\ æ I ö = -kx
[But the option given is negative of this]. ln èç I0 ø÷
18. (c) : Binding energy of In first case
37 Li = 7 ´ 5.60 = 39.2 Me V
ln æ 1 ö = -k ´ 36
Binding energy of 42 He = 4 ´ 7.06 = 28.24MeV è 8 ø
\ Energy of proton = Energy of éë2(42 He) -37 Liùû
ln (2 –3) = – k × 36
= 2 × 28.24 – 39.2
= 17.28 MeV. or 3ln2 = k × 36 ........(i)
19. (d) : The 'rad' the biological effect of radiation. In second case, ln èæ 1 ö = - k ´ x
2 ø
20. (d) : 3 Li7 + 1H1 ® 4 Be 8 + Z X A
Z for the unknown X nucleus = (3 + 1) – 4 = 0 or ln(2 –1 ) = – kx
A for the unknown X nucleus = (7 + 1) – 8 = 0
Hence particle emitted has zero Z and zero A or ln2 = kx ......... (ii)
It is a gamma photon.
From (i) and (ii)
21. (d) : Graph (d) represents the variation.
3 × (kx) = k × 36
22. (c) : For closest approach, kinetic energy is converted into
potential energy or x = 12 mm.
27. (c) : R is proportional to A 1/3 where A is mass number
3.6 = R 0 (27) 1/3 = 3R0 , for 1237 Al .
Again R = R0 (125) 1/3, for 15225 Al
\ R = (33× 6 ) ´ 5 = 6 fermi .
28. (c) : Kinetic energy is converted into potential energy at
1 mv 2 = 1 q1q2 = 1 (Ze)(2e ) closest approach
2 4pe0 r0 4 pe 0 r0
\ \ K.E. = P.E.
r0 = 4Ze2 = Ze 2 çæè m1 ÷øö \ 5 MeV = 1 q1q2
4 pe0 mv 2 pe 0 v 2 4 pe 0 r
or
or r 0 is proportional to çæè m1 ÷öø . or 5 ´10 6 ´ e = (9 ´109 ) ´ (92e)(2e )
r
or r = 9 ´109 ´ 92 ´ 2 ´ e
5 ´ 106
23. (d) : The nuclear transformation is given by
A X + 1 n ® 4He + 7 Li = 9 ´109 ´ 92 ´ 2 ´ (1.6 ´ 10 -19 )
Z 0 2 3 5 ´106
According to conservation of mass number \ r = 5.3 × 10 –14 m = 5.3 × 10 –12 cm.
A + 1 = 4 + 7
or A = 10 29. (c) : Total binding energy for (each deuteron)
= 2 × 1.1 = 2.2 MeV
According to conservation of charge number
Total binding energy for helium = 4 × 7 = 28 MeV
Z + 0 ® 2 + 3 \ Energy released = 28 –(2 × 2.2)
or Z = 5 = 28 – 4.4 = 23.6 MeV.
So the nucleus of the element be 105 B . 30. (d) : Momentum is conserved during disintegration
24. (c) : I is showing absorption photon. \ m 1v 1 = m2 v 2 .........(i)
From rest of three, III having maximum energy from
For an atom, R = R 0 A1 /3
æ 1 1 ö \ R1 = æ A1 ö1/ 3
DE µ ççè n12 - n2 2 ÷÷ø . R2 çè A2 ÷ø
N çæè 12 ÷öøt / T ö1/ 3 ö1 / 3
N 0 ÷ø ÷ø
25. (a) : = = æ m1 = æ m2 v 2 , from (i)
çè m2 çè m2v1
÷öø15 / T ö÷ø3 = çèæ 21 ÷øö15/ T
1 çèæ 1 çæè 1
\ 8 = 2 Þ 2 R1 æ 1 ö1/ 3 1 .
R2 è 2 ø 21/ 3
\ 1T5 = 3 Þ T = 5 min. \ = =
Atoms and Nuclei 105
- Z 2 E 0 -(3)2 ´13.6 Disintegration rate, initially = 5000
n2 (2 ) 2
31. (a) : Energy E 2 = = = -30.6 eV \ N 0l = 5000 ........... (i)
Disintegration rate, finally = 1250 ........... (ii)
\ energy required = 30.6 eV.
\ Nl = 1250
32. (c) : Masses of 1 H 1 and 1 D 2 are different. Hence the
corresponding wavelengths are different. \ N l = 1250 = 1
N 0 l 5000 4
33. (b) : 15353 Cs has the lowest ionization potential. Of the four or N = 1 Þ N0 e -5 l = 1 Þ e -5l = (4) -1
atoms given, Cs has the largest size. Electrons in the outer N0 4 N0 4
most orbit are at large distance from nucleus in a largesize
atom. Hence the ionization potential is the least. \ 5l = ln4 = 2ln2
\ l = 2 ln 2 = 0.4ln 2.
5
34. (d) : At temperature T, molecules of a gas acquire a kinetic
energy = 3 kT where k = Boltzmann's constant 38. (b) : Linear momentum is conserved
2 aparticle = 24H e
U 238 ® X 234 + He 4
\ To initiate the fusion reaction
3 kT = 7.7 ´10-14 J \ (238 × 0) = (238 × v) + 4u
2
or v = - 4u .
\ T = 7.7 ´10-14 ´ 2 = 3.7 ´109 K. 234
3´1.38 ´ 10 -23
39. (a) : Gamma rays have the least wavelength.
35. (a) : Protons are not emitted during radioactive decay.
36. (b) : The nucleus emits 8a particles i.e., 8( 2H e 4 ) 40. (a) : N = èçæ 1 ÷öøt / T = èæç 1 øö÷15/ 5 = æçè 1 ÷øö3 = 1
N 0 2 2 2 8
\ Decrease in Z = 8 × 2 = 16 ........(i)
\ N = N 0/ 8.
Four b – particles are emitted i.e., 4( –1b0 )
\ Increase in Z = 4 × 1 = 4 ........(ii) 41. (c) : En = 13.6 Þ E 2 = 13.6 = 3.4 eV
n2 ( 2 ) 2
2 positrons are emitted i.e., 2( 1b 0 )
\ Decrease in Z = 2 × 1 = 2 ......(iii) 42. (a) : Neutrons are electrically neutral. They are not deflected
\ Z of resultant nucleus = 92 – 16 + 4 – 2 = 78. by magnetic field.
37. (a) : Let decay constant per minute = l Hence (a) represents the answer.
106 JEE MAIN CHAPTERWISE EXPLORER
CHAPTER ELECTRONIC DEVICES
19
1. The IV characteristic of an LED is 4. The combination of gates shown below yields
(a) (b)
A
X
B
(a) NAND gate (b) OR gate (2010)
(c) NOT gate (d) XOR gate
5. The logic circuit shown below has the input waveforms ‘A’
and ‘B’ as shown. Pick out the correct output waveform.
(c) (d) A
Y
B
(2013)
Input A
2. A diode detector is used to detect an amplitude modulated
wave of 60% modulation by using a condenser of capacity
250 pico farad in parallel with a load resistance 100 kilo Input B
ohm. Find the maximum modulated frequency which could Output is
(a)
be detected by it.
(a) 5.31 kHz (b) 10.62 MHz
(c) 10.62 kHz (d) 5.31 MHz (2013)
3. Truth table for system of four NAND gates as shown in
figure is
A
Y (b)
B
A B Y A B Y (c)
0 0 0 0 0 1
0 1 0 0 1 1 (d) (2009)
(a) 1 0 1 (b) 1 0 0 6. A pn junction (D) shown in the figure can act as a rectifier.
1 1 1 1 1 0 D
R
A B Y A B Y (2012)
0 0 1 0 0 0 V
0 1 0 0 1 1
An alternating current source (V) is connected in the circuit.
(c) 1 0 0 (d) 1 0 1 The current(I) in the resistor(R) can be shown by
1 1 1 1 1 0
Electronic Devices 107
I + 5V 10 V
(a) (b)
(a)
t
I (c) (d) (2007)
–10 V – 5V
(b) t 11. The circuit has two oppositely connect ideal diodes in parallel.
What is the current following in the circuit?
I
4 W
(c) t D 1 D 2
3 W 2 W
I 12V
(d) (2009) (a) 1.33 A (b) 1.71 A
(c) 2.00 A (d) 2.31 A.
t
7. In the circuit below, A and B represent two inputs and C (2006)
represents the output. The circuit represents
12. In the following, which one of the diodes is reverse biased?
A + 5V + 10V
C R
B (a) (b) R
+ 5V
(a) OR gate (b) NOR gate (2008) – 12V
(c) AND gate (d) NAND gate.
(c)
8. A working transistor with its three legs marked P, Q and R R R
(d) . (2006)
is tested using a multimeter. No conduction is found between – 5V – 10V
P and Q. By connecting the common (negative) terminal of
the multimeter to R and the other (positive) terminal to P 13. If the lattice constant of this semiconductor is decreased,
then which of the following is correct?
or Q, some resistance is seen on the multimeter. Which of
the following is true for the transistor? Conduction band width E c
(a) It is an npn transistor with R as collector.
(b) It is an npn transistor with R as base. Band gap E g
(c) It is a pnp transistor with R as collector. Valence band width E v
(d) It is a pnp transistor with R as emitter. (2008)
9. Carbon, silicon and germanium have four valence electrons (a) all E c, E g , Ev decrease
(b) all Ec , Eg , Ev increase
each. At room temperature which one of the following (c) E c, and E v increase, but E g decreases
(d) Ec , and Ev decrease, but E g increases.
statements is most appropriate ? (2006)
(a) The number of free electrons for conduction is significant
only in Si and Ge but small in C. 14. In common base mode of a transistor, the collector current
is 5.488 mA for an emitter current of 5.60 mA. The value
(b) The number of free conduction electrons is significant in of the base current amplification factor (b) will be
(a) 48 (b) 49 (c) 50 (d) 51. (2006)
C but small in Si and Ge.
(c) The number of free conduction electrons is negligibly
small in all the three. 15. In the ratio of the concentration of electrons that of holes
in a semiconductor is 7/5 and the ratio of currents is 7/4 then
(d) The number of free electrons for conduction is significant what is the ratio of their drift velocities?
(a) 4/7 (b) 5/8 (c) 4/5 (d) 5/4. (2006)
in all the three. (2007)
10. If in a p n junction diode, a square input signal of 10 V is 16. A solid which is transparent to visible light and whose
applied as shown conductivity increases with temperature is formed by
(a) metallic binding (b) ionic binding
5 V (c) covalent binding (d) van der Waals binding
(2006)
R L
–5 V
Then the output signal across R L will be
108 JEE MAIN CHAPTERWISE EXPLORER
17. In a full wave rectifier circuit operating from 50 Hz mains 25. In the middle of the depletion layer of a reversebiased p
frequency, the fundamental frequency in the ripple would be n junction, the
(a) 100 Hz (b) 70.7 Hz (a) electric field is zero
(c) 50 Hz (d) 25 Hz (2005) (b) potential is maximum
18. In a common base amplifier, the phase difference between (c) electric field is maximum
the input signal voltage and output voltage is (d) potential is zero. (2003)
(a) 0 (b) p/2 26. The difference in the variation of resistance with temperature
in a metal and a semiconductor arises essentially due to the
(c) p/4 (d) p (2005) difference in the
(a) crystal structure
19. The electrical conductivity of a semiconductor increases when (b) variation of the number of charge carriers with
temperature
electromagnetic radiation of wavelength shorter than 2480 (c) type of bonding
(d) variation of scattering mechanism with temperature.
nm is incident on it. The band gap in (eV) for the (2003)
semiconductor is
(a) 0.5 eV (b) 0.7 eV
(c) 1.1 eV (d) 2.5 eV (2005)
20. When pn junction diode is forward biased, then 27. A strip of copper and another germanium are cooled from
(a) the depletion region is reduced and barrier height is
increased room temperature to 80 K. The resistance of
(b) the depletion region is widened and barrier height is
reduced. (a) each of these decreases
(c) both the depletion region and barrier height are reduced
(d) both the depletion region and barrier height are increased (b) copper strip increases and that of germanium decreases
(2004)
(c) copper strip decreases and that of germanium increases
(d) each of these increases. (2003)
28. Formation of covalent bonds in compounds exhibits
21. The manifestation of band structure in solids is due to (a) wave nature of electron
(a) Heisenberg’s uncertainty principle (b) particle nature of electron
(b) Pauli’s exclusion principle (c) both wave and particle nature of electron
(c) Bohr’s correspondence principle (d) none of these. (2002)
(d) Boltzmann’s law (2004) 29. The part of a transistor which is most heavily doped to produce
22. A piece of copper and another of germanium are cooled from large number of majority carriers is
room temperature to 77 K, the resistance of
(a) each of them increases (a) emitter
(b) each of them decreases
(c) copper decreases and germanium increases (b) base
(d) copper increases and germanium decreases. (2004)
(c) collector
(d) can be any of the above three. (2002)
30. The energy band gap is maximum in
23. For a transistor amplifier in common emitter configuration (a) metals (b) superconductors
for load impedance of 1 kW (h fe = 50 and h oe = 25) the (c) insulators (d) semiconductors. (2002)
current gain is
31. By increasing the temperature, the specific resistance of a
(a) –5.2 (b) –15.7
conductor and a semiconductor
(c) – 24.8 (d) – 48.78. (2004) (a) increases for both (b) decreases for both
24. When npn transistor is used as an amplifier (c) increases, decreases (d) decreases, increases.
(a) electrons move from base to collector
(b) holes move from emitter to base (2002)
(c) electrons move from collector to base
(d) holes move from base to emitter. 32. At absolute zero, Si acts as
(a) nonmetal (b) metal
(2004) (c) insulator (d) none of these. (2002)
Answer Key
1. (b) 2. (c) 3. (d) 4. (b) 5. (a) 6. (c)
7. (a) 8. (a) 9. (a) 10. (a) 11. (c) 12. (a)
13. (d) 14. (b) 15. (d) 16. (c) 17. (a) 18. (a)
19. (a) 20. (c) 21. (b) 22. (c) 23. (d) 24. (a)
25. (a) 26. (b) 27. (c) 28. (a) 29. (a) 30. (c)
31. (c) 32. (c)
Electronic Devices 109
1. (b) : The IV characteristics of a LED is similar to that of a This is same as the Boolean expression of OR gate.
Si junction diode. But the threshold voltages are much higher Alternative method
and slightly different for each colour. The truth table of the given circuit is as shown in the table
Hence, the option (b) represents the correct graph.
A B A B A × B X = A × B
2. (c) : The maximum frequency which can be detected is 0 0 1 1 1
0 1 1 0 0 0
u = 1 1 0 0 1 0 1
2p ma t 1 1 0 0 0 1
This is same as that of OR gate. 1
where, t = CR
Here, C = 250 pico farad = 250 × 10– 12 farad
R = 100 kilo ohm = 100 × 103 ohm A
m a = 0.6 5. (a) : Y
\ 1 B
250 ´10-12
u = 2p ´ 0.6 ´ ´ 100 ´ 103 By de Morgan’s theorem, (A + B) = A × B.
= 10.61 × 103 Hz = 10.61 kHz. Verify
A × B
A 0 0 1 0 Y AB AB A+B A + B
0 1 1100 0 1 1
1 1 0011 1 0 0
3. (d) : 0 1 1 1 011 0 1 0 0
0 1001 1 0 0
B 0
This is the same as AND Gate of A and B.
A 0 0 1 1 Y 6. (c) : (a) is original wave (b) is a fullwave rectified (c) is the
0 1 1 Y correct choice. The negative waves are cut off when the
1 1 0 Y diode is connected in reverse bias (d) is not the diagram for
1 1 0 0 alternating current.
B 1 1
0 7. (a) : It is OR gate. When either of them conducts, the gate
A 1 1 0 conducts.
1
1 1 1 1 8. (a) : It is npn transistor with R as collector. If it is connected
0 1 to base, it will be in forward bias.
B 0 0 1
A 1 1 9. (a) : C, Si and Ge have the same lattice structure and their
1 valence electrons are 4. For C, these electrons are in the second
1 1 1 orbit, for Si it is third and germanium it is the fourth orbit.
0 0 In solid state, higher the orbit, greater the possibility of
1 0 overlapping of energy bands. Ionization energies are also less
B 1 1 therefore Ge has more conductivity compared to Si. Both are
semiconductors.Carbon is an insulator.
4. (b) : A A
10. (a) : The current will flow through R L when diode is forward
B X biased.
B
11. (c) : Since diode D1 is reverse biased, therefore it will act
The Boolean expression of the given circuit is like an open circuit.
Effective resistance of the circuit is R = 4 + 2 = 6 W.
X = A×B Current in the circuit is I = E/R = 12/6 = 2 A.
= A + B (Using De Morgan theorem)
= A + B (Using Boolean identity) 12. (a) : Figure (a) represent a reverse biased diode.
13. (d) : Ec and E v decrease but Eg increases if the lattice constant
of the semiconductor is decreased.
110 JEE MAIN CHAPTERWISE EXPLORER
14. (b) : b = Ic = I c = 5.488 = 5.488 = 49. When cooled, the resistance of copper decreases and that of
Ib Ie - Ic 5.60 - 5.488 0.112 germanium increases.
I 23. (d) : In common emitter configuration, current gain is
nAe
15. (d) : Drift velocity v d = -(h fe ) = -50
+ (hoe )(RL ) 1 + (25 ´10-6 ) ´ (1´ 103 )
(vd ) electron æ Ie öæ nh ö A i = 1
(vd ) hole çè Ih ÷ø çè ne ÷ø
= = 7 ´ 5 = 54 . = - 50 = -50
4 7 1+ 0.025 1.025
= – 48.78.
16. (c) : Covalent binding.
17. (a) : Frequency of full wave rectifier 24. (a) : Electrons of ntype emitter move from emitter to base
= 2 × input frequency = 2 × 50 = 100 Hz. and then base to collector when npn transistor is used as
an amplifier.
18. (a) : In a common base amplifier, the phase difference between 25. (a) : Electric field is zero in the middle of the depletion
the input signal and output voltage is zero. layer of a reverse baised pn junction.
19. (a) : Band gap = Energy of photon of l = 2480 nm 26. (b) : Variation of number of charge carriers with temperature
is responsible for variation of resistance in a metal and a
\ Energy = hc J = lhce (eV) semiconductor.
l
(6.63 ´ 10-34 ) ´ (3 ´ 108 ) 27. (c) : Copper is conductor and germanium is semiconductor.
\ Band gap = (2480 ´10-9 ) ´ (1.6 ´ 10-19 ) eV When cooled, the resistance of copper strip decreases and
that of germanium increases.
= 0.5 eV.
20. (c) : When pn junction diode is forward biased, both the 28. (a) : Wave nature of electron and covalent bonds are correlated.
depletion region and barrier height are reduced.
29. (a) : The emitter is most heavily doped.
21. (b) : Pauli's exclusion principle explains band structure of 30. (c) : The energy band gap is maximum in insulators.
solids.
31. (c) : For conductor, r increases as temperature rises.
22. (c) : Copper is a conductor. For semiconductor, r decreases as temperature rises.
Germanium is a semiconductor.
32. (c) : Semiconductors, like Si, Ge, act as insulators at low
temperature.
Experimental Skills 111
CHAPTER EXPERIMENTAL SKILLS
20
1. A spectrometer gives the following reading when used to distance v, from the lens, is plotted using the same scale for
measure the angle of a prism. the two axes. A straight line passing through the origin and
Main scale reading : 58.5 degree making an angle of 45° with the xaxis meets the experimental
Vernier scale reading : 09 divisions curve at P. The coordinates of P will be
Given that 1 division on main scale corresponds to 0.5 degree. (a) (2f, 2f) (b) (f/2, f/2)
Total divisions on the vernier scale is 30 and match with 29 (c) (f, f) (d) (4f, 4f) (2009)
divisions of the main scale. The angle of the prism from the 5. An experiment is performed to find the refractive index of
glass using a travelling microscope. In this experiment
above data distances are measured by
(a) a screw gauge provided on the microscope
(a) 58.77 degree (b) 58.65 degree (b) a vernier scale provided on the microscope
(c) a standard laboratory scale
(c) 59 degree (d) 58.59 degree (2012) (d) a meter scale provided on the microscope. (2008)
2. A screw gauge gives the following reading when used to
measure the diameter of a wire.
Main scale reading : 0 mm
Circular scale reading : 52 divisions 6. While measuring the speed of sound by performing a
Given that 1 mm on main scale corresponds to resonance column experiment, a student gets the first
100 divisions of the circular scale. resonance condition at a column length of 18 cm during
The diameter of wire from the above data is : winter. Repeating the same experiment during summer, she
(a) 0.52 cm (b) 0.052 cm measures the column length to be x cm for the second
(c) 0.026 cm (d) 0.005 cm (2011) resonance. Then
3. In an experiment the angles are required to be measured (a) 36 > x > 18 (b) 18 > x
using an instrument. 29 divisions of the main scale exactly (c) x > 54 (d) 54 > x > 36. (2008)
coincide with the 30 divisions of the vernier scale. If the 7. Two full turns of the circular scale of a screw gauge cover
smallest division of the main scale is halfadegree (= 0.5°), a distance of 1 mm on its main scale. The total number of
then the least count of the instrument is divisions on the circular scale is 50. Further, it is found that
(a) one minute (b) half minute the screw gauge has a zero error of – 0.03 mm. While
(c) one degree (d) half degree (2009) measuring the diameter of a thin wire, a student notes the
4. In an optics experiment, with the position of the object fixed, main scale reading of 3 mm and the number of circular scale
a student varies the position of a convex lens and for each
position, the screen is adjusted to get a clear image of the divisions in line with the main scale as 35. The diameter of
object. A graph between the object distance u and the image
the wire is
(a) 3.38 mm (b) 3.32 mm
(c) 3.73 mm (d) 3.67 mm. (2008)
Answer Key
1. (b) 2. (b) 3. (a) 4. (a) 5. (b) 6. (c)
7. (a)
112 JEE MAIN CHAPTERWISE EXPLORER
1. (b) : 30 VSD = 29 MSD If u = radius of curvature, 2f, v = 2f
1 VSD = 29 MSD i.e., 1 + 1 = 1f .
30 2f 2 f
Least count = 1 MSD – 1 VSD v and u are have the same value when the object is at the
( ) = 29 1 centre of curvature. The solution is (a).
30 30
1- MSD = ´ 0.5° According the real and virtual system, u is +ve and v is also
Reading = Main scale reading + Vernier scale +ve as both are real. If u = v, u = 2f = radius of curvature.
reading × least count \ 1 + 1 = 1 Þ 1 + 1 = 1f .
v u f 2f 2 f
0.5 °
= 58.5° + 9 ´ 30 = 58.5° + 0.15° = 58.65° . The ans wer is the same (a).
(The figure given is according to New Cartesian system).
2. (b) : Least count of screw gauge 5. (b) : A travelling microscope moves horizontally on a main
= Pitch scale provided with a vernier scale, provided with the
divisions on
Number of circular scale microscope.
= 1 mm = 0.01 mm 6. (c) : v1 = g RT assuming M is the average molar mass of
100 M
Diameter of wire = Main scale reading the air (i.e., nitrogen) and g is also for nitrogen.
+ circular scale reading × Least count n n
= 0 + 52 × 0.01 = 0.52 mm = 0.052 cm
3. (a) : Least count = value of 1 main scale division L 1
The number of divisions L 2
on the vernier scale
as shown below. 1st resonance 2nd resonance
Here n vernier scale divisions = (n – 1) M.S.D.
g RT1 g RT2
v1 = M ; v 2 = M where T 1 and T2 stand for winter
n - 1
\ 1 V .S.D . = n M.S.D
and summer temperatures.
L.C. = 1 M.S.D. - 1 V.S.D v1 l4 = 18
n
= 1 M.S.D- (n n-1) M.S.D. L1 = = cm. At temperature T 1
Þ L .C. = 0.5° - 3290 ´ 0.5 °
At T 2, summer, v2 > v 1 .
0.5 1 1 1 L2 = v2 = 3l > 3 ´ 18.
30 30 2 60 n 4
Þ L.C. = = ´ = ° = 1 min .
\ L 2 > 54 cm.
4. (a) : According to New Cartesian v 7. (a) : Least count of the screw gauge
coordinate system used in our 12t h = 0.55m0 m = 0.01 mm
Main scale reading = 3 mm.
classes, for a convex lens, as u is P
Vernier scale reading = 35
negative, the lens equation is u 45° \ Observed reading = 3 + 0.35 = 3.35
zero error = –0.03
1 - 1 = 1f . \ actual diameter of the wire = 3.35 – (–0.03)
v u
= 3.38 mm.
One has to take that u is negative again for calculation, it
effectively comes to 1 + 1 = 1f .
v u
CHEMISTRY
Some Basic Concepts in Chemistry 1
CHAPTER SOME BASIC CONCEPTS
IN CHEMISTRY
1
1. The molarity of a solution obtained by mixing 750 mL of (c) remain unchanged
(d) be a function of the molecular mass of the substance.
0.5 M HCl with 250 mL of 2 M HCl will be
(2005)
(a) 0.975 M (b) 0.875 M
(c) 1.00 M (d) 1.75 M (2013) 5. What volume of hydrogen gas, at 273 K and 1 atm. pressure
2. In the reaction, will be consumed in obtaining 21.6 g of elemental boron (atomic
2Al( s) + 6HCl( aq) ® 2Al3 + (aq) + 6Cl– (aq) + 3H2 (g)
(a) 11.2 L H2 (g) at STP is produced for every mole HCl (aq) mass = 10.8) from the reduction of boron trichloride by
consumed
hydrogen?
(b) 6 L HCl( aq) is consumed for every 3 L H2 (g) produced
(c) 33.6 L H2 (g) is produced regardless of temperature and (a) 89.6 L (b) 67.2 L
(c) 44.8 L (d) 22.4 L
(2003)
pressure for every mole Al that reacts 6. With increase of temperature, which of these changes?
(d) 67.2 L H2 (g) at STP is produced for every mole Al that (a) Molality
reacts. (b) Weight fraction of solute
(2007) (c) Fraction of solute present in water
3. How many moles of magnesium phosphate, Mg3 ( PO 4) 2 will (d) Mole fraction
contain 0.25 mole of oxygen atoms?
(2002)
(a) 0.02 (b) 3.125 × 10– 2 7. Number of atoms in 558.5 gram Fe (at. wt. of
(c) 1.25 × 10 –2 (d) 2.5 × 10– 2 (2006) Fe = 55.85 g mol– 1) is
4. If we consider that 1/6, in place of 1/12, mass of carbon atom (a) twice that in 60 g carbon
is taken to be the relative atomic mass unit, the mass of one (b) 6.023 × 102 2
mole of a substance will (c) half that in 8 g He
(a) decrease twice (d) 558.5 × 6.023 × 102 3
(b) increase two fold
(2002)
Answer Key
1. (b) 2. (a) 3. (b) 4. (a) 5. (b) 6. (c)
7. (a)
2 JEE MAIN CHAPTERWISE EXPLORER
1. (b) : M mixVm ix = M1 V1 + M 2V 2 Now, atomic mass of an element
M mix = M1V1 + M2 V 2 =1 amMua (sHs eorfe oonne tahteo msc aolfe tohfe 1el e omf eCn-t 1 2)
Vmix 6
Mmix = 0.5 ´ 750 + 2 ´ 250 Mass of one atom of the element
1000 2 amu (Here on the scale of 1 of C-12)
=
Mmix = 0.875 M 12
2. (a) : 2Al( s) + 6HCl (aq) 2Al3 (+aq ) + 6Cl –(a q) + 3H 2(g) \ Numerically the mass of a substance will become half
of the normal scale.
6 moles of HCl produced H2 at STP = 3 × 22.4 L
\ 1 mole of HCl will produce H2 at STP 5. (b) : 2BCl 3 + 3H 2 ® 6HCl + 2B
3 ´ 22.4 or BCl 3 + 23 H 2 ® 3HCl + B
= = 11.2 L
6
3. (b) : 1 mole of Mg 3 (PO 4 ) 2 10.8 g boron requires hydrogen = 3 ´ 22.4 L
Þ 3 moles of Mg atom + 2 moles of P atom 2
21.6 g boron will require hydrogen
+ 8 moles of O atom = 3 ´ 1202..84 ´ 21.6 = 67.2 L
8 moles of oxygen atoms are present in = 1 mole of Mg3 ( PO 4) 2 2
1´ 0.25 6. (c) : Volume increases with rise in temperature.
0.25 mole of oxygen atoms are present in = 8
= 3.125 × 10 –2 moles of Mg 3( PO 4 ) 2 7. (a) : Fe (no. of moles) = 5555.88.55 = 10 moles
4. (a) : 1 atomic mass unit on the scale of 1/6 of C (no. of moles) = 60/12 = 5 moles.
C12 = 2 amu on the scale of 1/12 of C12. (atomic weight of carbon = 12)
States of Matter 3
CHAPTER STATES OF MATTER
2
1. Experimentally it was found that a metal oxide has formula 8. If 10 –4 dm 3 of water is introduced into a 1.0 dm 3 flask at
M0 .98 O. Metal M, is present as M 2+ and M 3+ in its oxide. Fraction 300 K, how many moles of water are in the vapour phase
of the metal which exists as M3 + would be when equilibrium is established?
(a) 5.08% (b) 7.01% (Given : Vapour pressure of H 2 O at 300 K is 3170 Pa;
R = 8.314 J K –1 mol –1 )
(c) 4.08% (d) 6.05% (2013) (a) 1.27 × 10 –3 mol (b) 5.56 × 10 –3 mol
(c) 1.53 × 10 –2 mol (d) 4.46 × 10 –2 mol (2010)
2. For gaseous state, if most probable speed is denoted by C*,
average speed by C and mean square speed by C, then for 9. Percentages of free space in cubic close packed structure and
a large number of molecules the ratios of these speed are in body centred packed structure are respectively
(a) C* : C : C = 1 : 1.225 : 1.128 (a) 48% and 26% (b) 30% and 26%
(b) C* : C : C = 1.225 : 1.128 : 1
(c) C* : C : C = 1.128 : 1 : 1.225 (c) 26% and 32% (d) 32% and 48% (2010)
(d) C* : C : C = 1 : 1.128 : 1.225
(2013) 10. Copper crystallizes in fcc with a unit cell length of 361 pm.
What is the radius of copper atom?
3. Lithium forms body centred cubic structure. The length of (a) 108 pm (b) 127 pm
the side of its unit cell is 351 pm. Atomic radius of the lithium (c) 157 pm (d) 181 pm (2009)
will be 11. In a compound, atoms of element Y form ccp lattice and those
(a) 300 pm (b) 240 pm of element X occupy 2/3 rd of tetrahedral voids. The formula
(c) 152 pm (d) 75 pm (2012) of the compound will be
4. The compressibility factor for a real gas at high pressure is (a) X 3Y 4 (b) X 4 Y 3 (2008)
(c) X 2Y 3 (d) X 2 Y
(a) 1 (b) 1 + Pb/RT
(c) 1 – Pb/RT (d) 1 + RT/Pb (2012) 12. Equal masses of methane and oxygen are mixed in an empty
5. In a face centred cubic lattice, atom A occupies the corner container at 25°C. The fraction of the total pressure exerted by
positions and atom B occupies the face centre positions. If
oxygen is
(a) 1/2 (b) 2/3
one atom of B is missing from one of the face centred points, (c) 1 ´ 273 (d) 1/3 (2007)
the formula of the compound is 3 298
(a) A 2 B (b) AB 2 (2011) 13. Total volume of atoms present in a facecentred cubic unit cell
(c) A2 B 3 (d) A 2 B 5 of a metal is (r is atomic radius)
6. ‘a’ and ‘b’ are van der Waals’ constants for gases. Chlorine (a) 20 pr 3 (b) 24 p r3
is more easily liquefied than ethane because 3 3
(a) a and b for Cl 2 > a and b for C 2 H 6 (c) 12 pr 3 (d) 16 pr 3 (2006)
(b) a and b for Cl 2 < a and b for C 2 H 6 3 3
(c) a for Cl 2 < a for C 2 H6 but b for Cl 2 > b for C 2H 6 14. Which one of the following statements is not true about the
(d) a for Cl 2 > a for C 2 H6 but b for Cl 2 < b for C 2H 6 effect of an increase in temperature on the distribution of
(2011) molecular speeds in a gas?
7. The edge length of a face centred cubic cell of an ionic substance (a) The most probable speed increases.
is 508 pm. If the radius of the cation is 110 pm, the radius (b) The fraction of the molecules with the most probable speed
increases.
of the anion is (c) The distribution becomes broader.
(a) 144 pm (b) 288 pm (d) The area under the distribution curve remains the same as
(c) 398 pm (d) 618 pm (2010) under the lower temperature.
(2005)
4 JEE MAIN CHAPTERWISE EXPLORER
15. An ionic compound has a unit cell consisting of A ions at the 20. According to the kinetic theory of gases, in an ideal gas, between
two successive collisions a gas molecule travels
corners of a cube and B ions on the centres of the faces of the (a) in a circular path
(b) in a wavy path
cube. The empirical formula for this compound would be (c) in a straight line path
(d) with an accelerated velocity.
(a) AB (b) A 2B (2005) (2003)
(c) AB3 (d) A3 B
16. What type of crystal defect is indicated in the diagram below?
Na+ Cl– Na+ Cl– Na+ Cl–
Cl– Cl– Na+ Na+ 21. How many unit cells are present in a cubeshaped ideal crystal
of NaCl of mass 1.00 g?
Na+ Cl– Cl– Na + Cl– [Atomic masses : Na = 23, Cl = 35.5]
(a) 2.57 × 102 1 (b) 5.14 × 10 21
Cl– Na+ Cl– Na+ Na+ (c) 1.28 × 10 21 (d) 1.71 × 10 21
(2003)
(a) Frenkel defect
(b) Schottky defect
(c) Interstitial defect
(d) Frenkel and Schottky defects (2004) 22. Na and Mg crystallize in bcc and fcc type crystals respectively,
17. In van der Waals equation of state of the gas law, the constant then the number of atoms of Na and Mg present in the unit cell
b is a measure of of their respective crystal is
(a) intermolecular repulsions (a) 4 and 2 (b) 9 and 14
(b) intermolecular attraction (c) 14 and 9 (d) 2 and 4
(c) volume occupied by the molecules (2002)
(d) intermolecular collisions per unit volume. (2004) 23. For an ideal gas, number of moles per litre in terms of its pressure
18. As the temperature is raised from 20°C to 40°C, the average P, gas constant R and temperature T is
kinetic energy of neon atoms changes by a factor of which of (a) PT/R (b) PRT
the following? (c) P/RT (d) RT/P
(a) 1/2 (b) 313/ 293 (2002)
(c) 313/293 (d) 2
(2004) 24. Kinetic theory of gases proves
(a) only Boyle's law
19. A pressure cooker reduces cooking time for food because (b) only Charles' law (2002)
(a) heat is more evenly distributed in the cooking space (c) only Avogadro's law
(b) boiling point of water involved in cooking is increased (d) all of these.
(c) the higher pressure inside the cooker crushes the food
material 25. Value of gas constant R is (2002)
(d) cooking involves chemical changes helped by a rise in (a) 0.082 L atm (b) 0.987 cal mol– 1 K– 1
temperature. (c) 8.3 J mol– 1 K– 1 (d) 83 erg mol– 1 K– 1
(2003)
Answer Key
1. (c) 2. (d) 3. (c) 4. (b) 5. (d) 6. (d)
7. (a) 8. (a) 11. (b) 12. (d)
13. (d) 14. (b) 9. (c) 10. (b) 17. (c) 18. (c)
19. (b) 20. (c) 23. (c) 24. (d)
25. (c) 15. (c) 16. (b)
21. (a) 22. (d)
States of Matter 5
1. (c) : Let the fraction of metal which exists as M3 + be x. 9. (c) : The packing efficiency in a ccp structure = 74%
\ Percentage free space = 100 – 74 = 26%
Then the fraction of metal as M 2+ = (0.98 – x) Packing efficiency in a body centred structure = 68%
Percentage free space = 100 – 68 = 32%
\ 3x + 2(0.98 – x) = 2
x + 1.96 = 2
x = 0.04
\ % of M3 + = 0.04 ´ 100 = 4.08% 10. (b) : Since Cu crystallizes in fcc lattice,
0.98 \ radius of Cu atom,
2. (d) : C* : C : C = 2RT = 8RT = 3 RT r = a (a = edge length)
M pM M 2 2
= 2 : 8 = 3 Given, a = 361 pm
3.14
\ r = 361 » 127 pm
\ C* : C : C = 1 : 1.128 : 1.225 2 2
3. (c) : a = 351 pm
For bcc unit cell, a 3 = 4r 11. (b) : Number of Y atoms per unit cell in ccp lattice (N) = 4
Number of tetrahedral voids = 2N = 2 × 4 = 8
r = a 3 = 351 ´ 3 = 152 pm Number of tetrahedral voids occupied by X = 2/3 rd of the
44 tetrahedral void = 2/3 × 8 = 16/3
Hence the formula of the compound will be
4. (b) : For real gases, æ P + a ÷öø (V - b) = RT X1 6/3 Y 4 = X4 Y 3
èç V 2
At high pressure, P >> a/V 2 12. (d) : Let the mass of methane and oxygen be m g.
Thus neglecting a/V2 gives
P(V – b) = RT or PV = RT + Pb Mole fraction of oxygen, x O 2
m
or PV = Z = RT + Pb Þ Z = 1 + Pb/RT
RT RT 32 m ´ 32 1
= m + m = 32 3m = 3
5. (d) : A B
8 ´ 81 5 ´ 21 32 16
Formula of the compound is A2 B5 . Let the total pressure be P.
\ Partial pressure of O 2, p O 2 = P × x O 2
6. (d) : a (dm 3 atm mol –2 ) b(dm 3 mol –1 )
= P × 1 = 1 P
Cl 2 6.49 0.0562 3 3
C 2 H 6 5.49 0.0638 13. (d) : In case of a facecentred cubic structure, since four
From the above values, a for Cl 2 > a for ethane (C 2 H6 ) atoms are present in a unit cell, hence volume
b for ethane (C 2 H6 ) > b for Cl 2 . V = 4 èæç 4 pr 3 ÷øö = 16 pr 3
3 3
7. (a) : In fcc lattice,
Given, a = 508 pm 14. (b) : Most probable velocity is defined as the speed possessed
by maximum number of molecules of a gas at a given
rc = 110 pm temperature. According to Maxwell's distribution curves, as
temperature increases, most probable velocity increases and
\ 110 + r a = 508 Þ r a = 144 pm fraction of molecule possessing most probable velocity
2 decreases.
8. (a) : The volume occupied by water molecules in vapour
phase is (1 – 10 –4) dm 3 , i.e., approximately 1 dm 3 . 15. (c) : Number of A ions per unit cell = 1 ´ 8 = 1
8
PV = nRT
Number of B ions per unit cell = 1 ´ 6 = 3
3170 × 1 × 10 –3 = n H 2O × 8.314 × 300 2
nH2 O = 3170 ´ 10 -3 = 1.27 ´ 10-3 mol Empirical formula = AB3
8.314 ´ 300
6 JEE MAIN CHAPTERWISE EXPLORER
16. (b) : When an atom or ion is missing from its normal lattice site, 24. (d) : Explanation of the Gas Laws on the basis of Kinetic
a lattice vacancy is created. This defect is known as Schottky
defect. Molecular Model
Here equal number of Na + and Cl– ions are missing from their
regular lattice position in the crystal. So it is Schottky defect. One of the postulates of kinetic theory of gases is
Average K.E. µ T
or, 1 mnCr2m s µ T or, 1 mnCr2m s = kT
2 2
17. (c) : van der Waals constant for volume correction b is the
measure of the effective volume occupied by the gas molecules. Now, PV = 1 mnCr2ms = 2 ´ 12 mnCr 2m s = 2 kT
3 3 3
18. (c) : Kb = 3/2 RT (i) Boyle’s Law :
K40 T4 0 l Constant temperature means that the average kinetic energy
K20 T20
= = 273 + 40 = 313 of the gas molecules remains constant.
273 + 20 293
l This means that the rms velocity of the molecules, Cr ms
19. (b) : According to Gay Lussac's law, at constant pressure of a remains unchanged.
given mass of a gas is directly proportional to the absolute
temperature of the gas. Hence, on increasing pressure, the l If the rms velocity remains unchanged, but the volume
temperature is also increased. Thus in pressure cooker due to
increase in pressure the boiling point of water involved in increases, this means that there will be fewer collisions
cooking is also increased.
with the container walls over a given time.
l Therefore, the pressure will decrease
20. (c) : According to the kinetic theory of gases, gas molecules are i.e. P µ 1
always in rapid random motion colliding with each other and V
with the wall of the container and between two successive
collisions a gas molecule travels in a straight line path. or PV = constant.
(ii) Charles’ Law :
21. (a) : Mass (m) = density × volume = 1.00 g l An increase in temperature means an increase in the average
Mol. wt. (M) of NaCl = 23 + 35.5 = 58.5
kinetic energy of the gas molecules, thus an increase in Cr ms.
Number of unit cell present in a cube shaped crystal of NaCl of l There will be more collisions per unit time, furthermore,
mass 1.00 g = r ´ a3 ´ N A = m ´ N A the momentum of each collision increases (molecules strike
M ´ Z M ´ Z the wall harder).
l Therefore, there will be an increase in pressure.
= 1 ´ 6.023 ´ 10 23 l If we allow the volume to change to maintain constant
58.5 ´ 4 pressure, the volume will increase with increasing
temperature (Charles law).
(In NaCl each unit cell has 4 NaCl units. Hence Z = 4). (iii) Avogadro’s Law
\ Number of unit cells = 0.02573 × 102 3 It states that under similar conditions of pressure and
temperature, equal volume of all gases contain equal number
= 2.57 × 102 1 unit cells of molecules. Considering two gases, we have
22. (d) : bcc Points are at corners and one in the centre of the unit P1V1 = 2 kT1 and P2V2 = 2 kT2
3 3
cell. Since P1 = P2 and T 1 = T 2, therefore
Number of atoms per unit cell = 8 ´ 81 + 1 = 2 P1V1 = (2 / 3) kT1 Þ V1 = n1
fcc Points are at the corners and also centre of the six faces P2V2 (2 / 3) kT2 V2 n2
of each cell. If volumes are identical, obviously n1 = n2 .
1 12 = 4 25. (c) : Units of R
8
Number of atoms per unit cell = 8 ´ + 6 ´ (i) in L atm Þ 0.082 L atm mol– 1 K– 1
23. (c) : From ideal gas equation, PV = nRT (ii) in C.G.S. system Þ 8.314 × 107 erg mol– 1 K– 1
\ n/V = P/RT (number of moles = n/V)
(iii) in M.K.S. system Þ 8.314 J mol– 1 K– 1
(iv) in calories Þ 1.987 cal mol– 1 K– 1
Atomic Structure 7
CHAPTER ATOMIC STRUCTURE
3
1. Energy of an electron is given by E = -2.178 ´ 10-18 J æ Zn2 2 øö÷ . (a) 1.52 × 10 –4 m (b) 5.10 × 10 –3 m (2009)
ç (c) 1.92 × 10 –3 m (d) 3.84 × 10 –3 m
è
Wavelength of light required to excite an electron in an 8. The ionization enthalpy of hydrogen atom is
hydrogen atom from level n = 1 to n = 2 will be 1.312 × 10 6 J mol –1 . The energy required to excite the electron
(h = 6.62 × 10 –34 J s and c = 3.0 × 10 8 m s –1 ) in the atom from n = 1 to n = 2 is
(a) 8.500 × 10 –7 m (b) 1.214 × 10 –7 m (a) 9.84 × 10 5 J mol –1
(c) 2.816 × 10 –7 m (d) 6.500 × 10 –7 m (2013) (b) 8.51 × 10 5 J mol –1
2. The electrons identified by quantum numbers n and l: (c) 6.56 × 10 5 J mol –1 (2008)
(1) n = 4, l = 1 (2) n = 4, l = 0 (d) 7.56 × 10 5 J mol –1
(3) n = 3, l = 2 (4) n = 3, l = 1 9. Which of the following sets of quantum numbers represents the
can be placed in order of increasing energy as highest energy of an atom?
(a) (4) < (2) < (3) < (1) (a) n = 3, l = 0, m = 0, s = + 1
(b) (2) < (4) < (1) < (3) 2
(c) (1) < (3) < (2) < (4)
(d) (3) < (4) < (2) < (1) (2012) (b) n = 3, l = 1, m = 1, s = + 1
2
3. A gas absorbs a photon of 355 nm and emits at two wavelengths.
+ 1
If one of the emission is at 680 nm, the other is at (c) n = 3, l = 2, m = 1, s = 2
(a) 1035 nm (b) 325 nm (d) n = 4, l = 0, m = 0, s = + 1
2
(c) 743 nm (d) 518 nm (2011) (2007)
4. The energy required to break one mole of Cl—Cl bonds in
Cl2 is 242 kJ mol –1 . The longest wavelength of light capable 10. Uncertainty in the position of an electron
(mass = 9.1 × 10 –31 kg) moving with a velocity 300 m s –1, accurate
of breaking a single Cl—Cl bond is (c = 3 × 10 8 m s –1
and NA = 6.02 × 10 23 mol –1 ) upto 0.001% will be (h = 6.6 × 10– 34 J s)
(a) 19.2 × 10 –2 m (b) 5.76 × 10– 2 m
(a) 494 nm (b) 594 nm
(c) 640 nm (d) 700 nm (2010) (c) 1.92 × 10 –2 m (d) 3.84 × 10– 2 m
5. Ionisation energy of He+ is 19.6 × 10 –18 J atom –1 . The energy (2006)
of the first stationary state (n = 1) of Li 2+ is 11. According to Bohr’s theory, the angular momentum of an
(a) 8.82 × 10 –17 J atom –1 electron in 5t h orbit is
(b) 4.41 × 10 –16 J atom –1 (a) 25 h (b) 1.0 h
(c) – 4.41 × 10 –17 J atom –1 p p
(d) – 2.2 × 10 –15 J atom –1 (2010) (c) 10 h (d) 2.5 h (2006)
p p
6. Calculate the wavelength (in nanometre) associated with a
proton moving at 1.0 × 10 3 m s –1 .
(Mass of proton = 1.67 × 10 –27 kg and h = 6.63 × 10 –34 J s) 12. Which of the following statements in relation to the hydrogen
atom is correct?
(a) 0.032 nm (b) 0.40 nm
(a) 3s orbital is lower in energy than 3p orbital.
(c) 2.5 nm (d) 14.0 nm (2009)
(b) 3p orbital is lower in energy than 3d orbital.
7. In an atom, an electron is moving with a speed of 600 m/s
(c) 3s and 3p orbitals are of lower energy than 3d orbital.
with an accuracy of 0.005%. Certainty with which the position
(d) 3s, 3p and 3d orbitals all have the same energy.
of the electron can be located is (h = 6.6 × 10 –34 kg m 2 s –1 ,
(2005)
mass of electron, e m = 9.1 × 10 –31 kg)
8 JEE MAIN CHAPTERWISE EXPLORER
13. In a multielectron atom, which of the following orbitals 17. The orbital angular momentum for an electron revolving in an
described by the three quantum numbers will have the same
energy in the absence of magnetic and electric fields? orbit is given by l( l + 1) × h . This momentum for an
(i) n = 1, l = 0, m = 0
(ii) n = 2, l = 0, m = 0 2 p
(iii) n = 2, l = 1, m = 1
(iv) n = 3, l = 2, m = 1 selectron will be given by
(v) n = 3, l = 2, m = 0
(a) (i) and (ii) (b) (ii) and (iii) (a) + 1 × h (b) zero
(c) (iii) and (iv) (d) (iv) and (v) 2 2p
(2005)
(c) h (d) 2 × h
2 p 2 p
(2003)
14. The wavelength of the radiation emitted, when in a hydrogen 18. The de Broglie wavelength of a tennis ball of mass 60 g moving
with a velocity of 10 metres per second is approximately
atom electron falls from infinity to stationary state 1, would be (Planck's constant, h = 6.63 × 10 –34 J s)
(a) 10 –33 metres (b) 10– 31 metres
(Rydberg constant = 1.097 × 107 m– 1) (c) 10– 16 metres (d) 10 –25 metres.
(2003)
(a) 91 nm (b) 192 nm
(c) 406 nm (d) 9.1 × 10– 8 nm
(2004) 19. In Bohr series of lines of hydrogen spectrum, the third line from
15. Consider the ground state of Cr atom (Z = 24). The numbers of the red end corresponds to which one of the following inter
electrons with the azimuthal quantum numbers, l = 1 and 2 are, orbit jumps of the electron for Bohr orbits in an atom of
respectively hydrogen?
(a) 12 and 4 (b) 12 and 5 (a) 3 ® 2 (b) 5 ® 2
(c) 16 and 4 (d) 16 and 5 (c) 4 ® 1 (d) 2 ® 5
(2004) (2003)
16. Which of the following sets of quantum numbers is correct for 20. Uncertainty in position of a minute particle of mass 25 g in
an electron in 4f orbital?
space is 10 –5 m. What is the uncertainty in its velocity
1 (in m s –1) ? (h = 6.6 × 10 –34 J s)
+
(a) n = 4, l = 3, m = +4, s = 2 (a) 2.1 × 10– 34 (b) 0.5 × 10 –34
1 (c) 2.1 × 10 –28 (d) 0.5 × 10 –23 (2002)
(b) n = 4, l = 4, m = –4, s = - 21. In a hydrogen atom, if energy of an electron in ground state is
2
1 13.6 eV, then that in the 2 nd excited state is
(c) n = 4, l = 3, m = +1, s = +
2 (a) 1.51 eV (b) 3.4 eV
1 (c) 6.04 eV (d) 13.6 eV
(d) n = 3, l = 2, m = –2, s = + (2004)
2 (2002)
Answer Key
1. (b) 2. (a) 3. (c) 4. (a) 5. (c) 6. (b)
11. (d) 12. (d)
7. (c) 8. (a) 9. (c) 10. (c) 17. (b) 18. (a)
13. (d) 14. (a) 15. (b) 16. (c)
19. (b) 20. (c) 21. (a)
Atomic Structure 9
1. (b) : E = -2.178 ´ 10 -18 Z 2 çæ 1 - 1 ö Þ 242 ´ 103 = 6.63 ´ 10-34 ´ 3 ´ 108
è n12 n22 ÷ø 6.02 ´ 1023 l
E = -2.178 ´ 10 -18 éê 1 - 1 ù \ l = 6.63 ´ 10-34 ´ 3 ´ 108 ´ 6.02 ´ 10 23
ëê (2)2 (1) 2 úûú 242 ´ 103
E = +2.178 ´ 10-18 ´ 3 = 1.6335 ´ 10-18 J = 0.494 × 10 –6 m = 494 nm
4
5. (c) : I.E.(He +) = 19.6 × 10 –18 J atom –1
E = hc E 1 (for H) × Z 2 = I.E.
l E 1 × 4 = –19.6 × 10 –18
E1 (for Li 2+) = E1 (for H) × 9
Þ l = hc = 6.62 ´ 10-34 J s ´ 3 ´ 108 m
E 1.6335 ´ 10-18 J = -19.6 ´ 10-18 ´ 9 = - 44.1 ´ 10-18 J atom -1
4
l = 12.14 × 10 –8 m = – 4.41 × 10 –17 J atom –1
or l = 1.214 × 10 –7 m
2. (a) : (1) n = 4, l = 1 Þ 4p 6. (b) : According to deBroglie’s equation,
(2) n = 4, l = 0 Þ 4s h
mv
(3) n = 3, l = 2 Þ 3d l =
(4) n = 3, l = 1 Þ 3p Given, v = 1.0 × 10 3 m s –1
Increasing order of energy is 3p < 4s < 3d < 4p 6.63 ´ 10 -34
´ 10-27 ´ 1.0 ´ 103
(4) < (2) < (3) < (1) \ l = 1.67 = 3.9 ´ 10-10 m
Alternatively, or l » 0.4 nm
for (1) n + l = 5 ; n = 4
(2) n + l = 4 ; n = 4 7. (c) : Given, velocity of e – , v = 600 m s –1
(3) n + l = 5 ; n = 3 Accuracy of velocity = 0.005%
(4) n + l = 4 ; n = 3 \ Dv = 6001´000.005 = 0.03
Lower n + l means less energy and if for two subshells According to Heisenberg’s uncertainty principle,
n + l is same than lower n, lower will be the energy.
Thus correct order is (4) < (2) < (3) < (1). Dx × mDv ³ h
4p
3. (c) : We know that
E = hu = hc/l Þ Dx = 4 ´ 6.6 ´ 10 -34
3.14 ´ 9.1 ´10-31 ´ 0.03
E = E 1 + E 2 or hc = hc + hc = 1.92 × 10 –3 m
l l1 l 2
8. (a) : The ionisation of Hatom is the energy absorbed when
Þ 1 = 1 + 1 Þ 1 = 1 + 1 the electron in an atom gets excited from first shell (E 1) to
l l1 l2 355 680 l 2
infinity (i.e., E¥ )
\ l2 = 355 ´ 680 = 742.769 nm » 743 nm I.E = E¥ – E 1
680 - 355 1.312 × 10 6 = 0 – E 1
E1 = –1.312 × 10 6 J mol –1
4. (a) : Energy required to break 1 mol of bonds = 242 kJ mol– 1 E2 = - 1.312 ´106 = - 1.3124´ 106
(2) 2
242 ´ 10 3
\ Energy required to break 1 bond = 6.02 ´ 1023 J Energy of electron in second orbit (n = 2)
hc \ Energy required when an electron makes transition from
l
We know that, E= n = 1 to n = 2
Given, c = 3 × 10 8 m s –1 DE = E 2 – E1 = -1.3124´10 6 - (-1.312 ´106 )
10 JEE MAIN CHAPTERWISE EXPLORER
= -1.312 ´ 106 + 5.248 ´ 106 = 0.984 × 10 6 For l = 1, total number of electrons = 12
4 [2p 6 and 3p 6 ]
DE = 9.84 ×10 5 J mol –1 For l = 2, total number of electrons = 5 [3d 5 ]
9. (c) : n = 3, l = 0 represents 3s orbital 16. (c) : For 4 f orbital electrons, n = 4
n = 3, l = 1 represents 3p orbital
n = 3, l = 2 represents 3d orbital s p d f
n = 4, l = 0 represents 4s orbital l = 3 (because 0 1 2 3 )
m = +3, +2, +1, 0, –1, –2, –3
The order of increasing energy of the orbitals is s = ±1/2
3s < 3p < 4s < 3d.
17. (b) : The value of l (azimuthal quantum number) for s electron
10. (c) : According to Heisenberg’s uncertainty principle, is equal to zero.
Dx ´ Dp = h Orbital angular momentum = l (l + 1) × h
4 p 2 p
Dx × (m × Dv) = h Þ Dx = h Substituting the value of l for selectron
4p 4 pm × D v
= 0(0 + 1) × 2h p = 0
0.001 3 ´10-3 m s -1
Here Dv = ´ 300 =
100
18. (a) : l = h = 6.63 ´ 10-34 ´1 000 m
6.63 ´ 10 -34 mv 60 ´ 10
\ Dx = 4 ´ 3.14 ´ 9.1´10-31 ´ 3´ 10 -3
= 11.05 × 10 –34 m = 1.105 × 10 –33 metres.
= 1.92 ´ 10-2 m 19. (b) : The electron has minimum energy in the first orbit and
its energy increases as n increases. Here n represents number
11. (d) : Angular momentum of the electron, mvr = nh of orbit, i.e. 1 st , 2 nd , 3 rd .... The third line from the red end
2p corresponds to yellow region i.e. 5. In order to obtain less
energy electron tends to come in 1 st or 2 nd orbit. So jump
when n = 5 (given) may be involved either 5 ® 1 or 5 ® 2. Thus option (b) is
correct here.
\ Angular momentum = 5 h = 2.5 h
2 p p
12. (d) : For hydrogen the energy order of orbital is 20. (c) : According to Heisenberg uncertainty principle,
1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f
h
13. (d) : Orbitals having same (n + l) value in the absence of Dx × mDv = 4p
electric and magnetic field will have same energy.
6.6 ´10 -34
1 æ 1 1 ö 1 m -1 çèæ 1 1 ö Dv = 4 ´ 3.14 ´ 25´10 -5
l R ççè n12 n2 2 ÷÷ø l 12 ¥2 ÷ø
14. (a) : = - Þ = 1.097 ´107 -
\ l = 91 × 10 –9 m = 91 nm \ Dv = 2.1 ´ 10 -28 m s -1
15. (b) : 24 Cr ® 1s 2 2s2 2p 6 3s 2 3p 6 3d 5 4s 1 21. (a) : 2n d excited state will be the 3rd energy level.
we know for p, l = 1 and for d, l = 2.
E n = 1n32.6 eV or E = 139. 6 = 1.51 eV
Chemical Bonding and Molecular Structure 11
CHAPTER CHEMICAL BONDING AND
4 MOLECULAR STRUCTURE
1. Stability of the species Li 2 , Li 2– and Li 2+ increases in the order (a) N 2, O 2– , NO +, CO
(b) C2 2 – , O 2– , CO, NO
of (c) NO +, C 22 – , CN – , N 2
(d) CN –, N 2 , O 22 – , C 22 –
(a) Li2 – < Li 2 < Li+ 2 (2008)
(b) Li2 < Li+ 2 < Li 2–
(c) Li2 – < Li+ 2 < Li 2 (2013) 11. Which one of the following pairs of species have the same
(d) Li2 < Li 2– < Li+ 2
bond order?
2. In which of the following pairs of molecules/ions, both the
(a) NO + and CN + (b) CN – and NO +
species are not likely to exist? (c) CN – and CN + (d) O 2– and CN – (2008)
(a) H2 – , He2 2 + (b) H2 + , He2 2 – 12. Which of the following hydrogen bonds is the strongest?
(c) H2 – , He2 2 – (d) H2 2 + , He2 (2013) (a) O — H F
3. Which one of the following molecules is expected to exhibit (b) O — H H
diamagnetic behaviour? (c) F — H F
(a) S 2 (b) C2 (d) O — H O (2007)
(c) N 2 (d) O 2
(2013) 13. In which of the following ionization processes, the bond order
4. The molecule having smallest bond angle is has increased and the magnetic behaviour has changed?
(a) AsCl3 (b) SbCl 3 (a) N 2 ® N 2+ (b) C 2 ® C2 +
(c) PCl 3 (d) NCl3
(2012) (c) NO ® NO + (d) O 2 ® O 2+ (2007)
5. In which of the following pairs the two species are not 14. The charge/size ratio of a cation determines its polarizing power.
isostructural? Which one of the following sequences represents the increasing
(a) PCl 4+ and SiCl 4 (b) PF5 and BrF 5 order of the polarizing power of the cationic species, K+ , Ca 2+ ,
(c) AlF 63 – and SF 6 (d) CO 32 – and NO 3–
(2012) Mg2 + , Be2 + ?
6. The structure of IF 7 is (a) Ca 2+ < Mg 2+ < Be+ < K+
(a) square pyramid (b) trigonal bipyramid
(b) Mg2 + < Be2 + < K+ < Ca 2+
(c) octahedral (d) pentagonal bipyramid. (2011) (c) Be2 + < K+ < Ca 2+ < Mg 2+
(d) K+ < Ca 2+ < Mg 2+ < Be2 + (2007)
7. The hybridisation of orbitals of N atom in NO 3– , NO 2+ and NH 4+ 15. Which of the following species exhibits the diamagnetic
are respectively
behaviour?
(a) sp, sp 2 , sp 3 (b) sp2 , sp, sp 3 (b) O 22 –
(a) NO (d) O2
(c) sp, sp 3 , sp 2 (d) sp2 , sp 3 , sp (2011) (c) O 2+ (2007)
8. Among the following the maximum covalent character is shown 16. In which of the following molecules/ions are all the bonds not
by the compound
equal?
(a) FeCl 2 (b) SnCl 2
(c) AlCl 3 (d) MgCl2 (2011) (a) SF4 (b) SiF4 (2006)
(c) XeF4 (d) BF4 –
9. Using MO theory predict which of the following species has 17. Among the following mixtures, dipoledipole as the major
the shortest bond length? interaction, is present in
(a) O 22+ (b) O +2 (2009) (a) benzene and ethanol
(c) O –2 (d) O 22–
(b) acetonitrile and acetone
10. Which one of the following constitutes a group of the (c) KCl and water
isoelectronic species?
(d) benzene and carbon tetrachloride. (2006)
12 JEE MAIN CHAPTERWISE EXPLORER
18. Which of the following molecules/ions does not contain unpaired 25. The pair of species having identical shapes for molecules of
electrons? both species is
(a) O 22 – (b) B2 (2006) (a) CF4 , SF4 (b) XeF2 , CO 2
(c) N 2+ (d) O2 (c) BF3 , PCl3 (d) PF5 , IF5
19. Of the following sets which one does NOT contain isoelectronic (2003)
species? 26. Which one of the following compounds has the smallest bond
(a) PO4 3 – , SO4 2 –, ClO 4– angle in its molecule?
(b) CN – , N 2 , C2 2 –
(c) SO3 2 –, CO 32 –, NO 3– (a) SO2 (b) OH2 (2003)
(d) BO 33 –, CO 32 –, NO 3– (c) SH2 (d) NH3
(2005)
20. Which one of the following species is diamagnetic in nature? 27. Which of the following are arranged in an increasing order of
(a) He2 + (b) H2 (2005) their bond strengths?
(c) H2 + (d) H2 –
(a) O 2– < O 2 < O 2+ < O 22 –
21. The maximum number of 90° angles between bond pairbond (b) O 22 – < O 2– < O 2 < O 2+
(c) O 2– < O 22 – < O 2 < O 2+
pair of electrons is observed in (d) O 2+ < O 2 < O 2– < O 22 –
(a) dsp 3 hybridisation (2002)
(b) sp 3d hybridisation 28. A square planar complex is formed by hybridisation of which
(c) dsp 2 hybridisation atomic orbitals?
(d) sp 3d 2 hybridisation. (2004) (a) s, px , py , dy z (b) s, p x , py , dx 2 – y2
(c) s, px , p y , dz 2 (d) s, p y , p z , dx y
22. Which one of the following has the regular tetrahedral structure?
(a) XeF4 (b) SF4 (2002)
(c) BF4 – (d) [Ni(CN)4 ]2 – 29. Number of sigma bonds in P4 O 10 is
(Atomic nos.: B = 5, S = 16, Ni = 28, Xe = 54)
(2004) (a) 6 (b) 7
23. The bond order in NO is 2.5 while that in NO + is 3. Which of (c) 17 (d) 16
the following statements is true for these two species? (2002)
(a) Bond length in NO + is greater than in NO. 30. In which of the following species is the underlined carbon having
(b) Bond length in NO is greater than in NO +. sp 3 hybridisation?
(c) Bond length in NO + is equal to that in NO. (a) CH3 COOH (b) CH3 CH2 OH
(c) CH3 C OCH3 (d) CH2 CH – CH3
(d) Bond length is unpredictable. (2004)
24. The correct order of bond angles (smallest first) in H2 S , NH3 , (2002)
BF3 and SiH4 is 31. In which of the following species the interatomic bond angle is
(a) H2 S < SiH4 < NH3 < BF3 109°28¢?
(b) NH3 < H2 S < SiH4 < BF3 (a) NH3 , (BF4 ) – 1 (b) (NH4 ) + , BF3
(c) H2 S < NH3 < SiH4 < BF3 (c) NH3 , BF3 (d) (NH2 ) –1 , BF3 (2002)
(d) H2 S < NH3 < BF3 < SiH4 . (2004)
Answer Key
1. (c) 2. (d) 3. (d) 4. (b) 5. (b) 6. (d)
7. (b) 8. (c) 9. (a) 10. (c) 11. (b) 12. (c)
13. (c) 14. (d) 15. (b) 16. (a) 17. (b) 18. (a)
19. (c) 20. (b) 21. (d) 22. (c) 23. (b) 24. (c)
25. (b) 26. (c) 27. (b) 28. (b) 29. (d) 30. (b)
31. (a)
Chemical Bonding and Molecular Structure 13
1. (c) : Species Bond order 8. (c) : We know that, extent of polarisation µ covalent character
Li 2 1 in ionic bond.
Fajan’s rule states that
Li 2 – 0.5 (i) the polarising power of cation increases, with increase in
magnitude of positive charge on the cation
Li2 + 0.5
Higher the bond order, greater is the stability.
2. (d) : Species with zero bond order does not exist. \ polarising power µ charge of cation
H2 2 + : s(1s) 0
(ii) the polarising power of cation increases with the decrease
\ Bond order = 0 in the size of a cation.
He 2 : s(1s) 2 s*(1s) 2 \ polarising power µ 1
size of cation
Bond order = 2 - 2 = 0 Here the AlCl 3 is satisfying the above two conditions i.e., Al
2 is in +3 oxidation state and also has small size. So it has
more covalent character.
3. (d) : O 2 is expected to be diamagnetic in nature but actually 9. (a) : According to MOT, the molecular orbital electronic
it is paramagnetic. configuration of
4. (b) : As we move down the group the size of atom increases O22 + : (s1s) 2 (s *1 s) 2 (s2s) 2 (s *2 s) 2 (s2pz) 2 (p2px) 2 = (p2py) 2
and as size of central atom increases, lone pairbond pair
repulsion also increases. Thus bond angle decreases. \ B.O. = 10 - 4 = 3
Increasing order of atomic radius : 2
N < P < As < Sb O+2 : (s1s) 2( s * 1s) 2 (s2s) 2( s * 2s) 2 (s2pz) 2 (p2px) 2
Decreasing order of bond angle :
= (p2py) 2 (p * 2px) 1
NCl 3 > PCl 3 > AsCl 3 > SbCl 3 \ B.O . = 10 - 5 = 2.5
5. (b) : PCl 4+ and SiCl 4 Þ both tetrahedral 2
PF 5 Þ trigonal bipyramidal O2– : (s1s) 2( s *1 s) 2 (s2s) 2( s * 2s) 2 (s2pz)2 (p2px) 2
BrF 5 Þ square pyramidal
AlF 63 – and SF 6 both are octahedral, CO 32 – and NO 3– both are = (p2py) 2 (p *2 px) 2 = (p * 2py) 1
trigonal planar.
\ B.O. = 10 - 7 = 1.5
6. (d) : The structure is pentagonal bipyramidal having sp 3d 3 2
hybridisation as given below: O22 – : (s1s) 2( s *1 s) 2 (s2s) 2( s *2 s) 2 (s2pz) 2 (p2px) 2
= (p2py) 2( p *2 px) 2 = (p * 2py) 2
\ B.O. = 10 - 8 = 1.0
2
Q B.O. µ Bond 1l ength ,
\ O22 + has the shortest bond length.
10. (c) : Number of electrons in each species are given below
7. (b) : The structures of NO 3– , NO 2+ and NH 4+ is N 2 = 14 CN – = 14
sp 2 hybridisation O 2– = 17
sp hybridisation C2 2 – = 14
NO + = 14 O 22 – = 18
CO = 14 NO = 15
It is quite evident from the above that NO + ,
C 22 – , CN –, N 2 and CO are isoelectronic in nature. Hence option
(c) is correct.
sp 3 hybridisation 11. (b) : In the given pair of species, number of electron in NO +
= number of electron in CN – = 14 electrons.
So they are isoelectronic in nature.
14 JEE MAIN CHAPTERWISE EXPLORER
Hence bond order of these two species will be also F F ..
16. (a) :
similar which is shown below. S
NO + ® s1s 2 s*1s 2 s2s 2 s*2s 2 s2p z2 p2p x 2 p2p y 2
F F
B.O = 1/2 [N b – N a ] = 1/2 [10 – 4] SF4 molecule shows sp 3d hybridisation but its expected trigonal
or B.O = 3 bipyramidal geometry gets distorted due to presence of a
CN – ® s1s 2 s*1s 2 s2s 2 s*2s 2 p2p x2 p2p y 2 s2p z 2
lone pair of electrons and it becomes distorted tetrahedral
B.O =1/2 [10 – 4] or B.O = 3
or seesaw with the bond angles equal to 89° and 177° instead
12. (c) : Because of highest electronegativity of F, hydrogen of the expected angles of 90° and 180° respectively.
SiF 4 : sp 3 hybridisation and tetrahedral geometry.
bonding in F — H F is strongest.
F
13. (c) : Molecular orbital configuration of
O 2 Þ s1s 2 s*1s 2 s2s 2 s*2s 2 s2pz 2 p2p x 2 p2p y 2 p*2px 1 p*2p y 1
Þ paramagnetic
Bond order = 10 - 6 = 2
2
O 2+ Þ s1s 2 s*1s 2 s2s 2 s*2s 2 s2pz 2 p2p x2 p2p y 2 p*2p x 1 109.5°
Si
Þ paramagnetic F F
F
Bond order = 10 - 5 = 2.5
2
XeF 4 : sp 3 d 2 hybridisation, shape is square planar instead of
N 2 Þ s1s 2 s*1s 2 s2s 2 s*2s 2 p2p x 2 p2p y 2 p2p Z 2 octahedral due to pres ence of two lone pair of electrons on
Þ paramagnetic
Xe atom.
Bond order = 10 - 4 = 3 F F
2
Xe
N 2+ Þ s1s 2 s*1s 2 s2s 2 s*2s 2 p2p x 2 p2p y 2 p2p z 1
Þ paramagnetic
Bond order = 9 - 4 = 2.5 F F
2
BF 4 – : sp 3 hybridisation and tetrahedral geometry.
C 2 Þ s1s 2 s*1s 2 s2s 2 s*2s 2 p2p x 2 p2p y 2
Þ diamagnetic 17. (b) : Dipoledipole interactions occur among the polar
molecules. Polar molecules have permanent dipoles. The
Bond order = 8 - 4 = 2
2 positive pole of one molecule is thus attracted by the negative
C 2+ Þ s1s 2 s*1s 2 s2s 2 s*2s 2 p2p x 2 p2p y 1 pole of the other molecule. The magnitude of dipoledipole
Þ paramagnetic
forces in different polar molecules is predicted on the basis
Bond order = 7 - 4 = 1.5 of the polarity of the molecules, which in turn depends upon
2
NO Þ s1s 2 s*1s 2 s2s 2 s*2s2 s2p z2 p2p x 2 p2py 2 p*2p x 1 the electronegativities of the atoms present in the molecule
Þ paramagnetic and the geometry of the molecule (in case of polyatomic
Bond order = 10 - 5 = 2.5 molecules, containing more than two atoms in a molecule).
2
18. (a) : The molecular orbital configuration of O 22 – ion is
NO + Þ s1s 2 s*1s 2 s2s 2 s*2s2 s2p z 2 p2p x 2 p2p y 2 KK s(2s)2 s*(2s) 2 s(2p z) 2 p(2p x )2 p(2py ) 2 p*(2px ) 2 p*(2py ) 2
Þ diamagnetic Here KK represents nonbonding molecular orbital of 1s orbital.
O 22 – contains no unpaired electrons.
Bond order = 10 - 4 = 3 The molecular orbital configuration of B2 molecule is
2 KK s(2s) 2 s*(2s) 2 p(2p x) 1 p(2py ) 1
It contains 2 unpaired electrons.
14. (d) : High charge and small size of the cations increases The molecular orbital configuration of N +2 ion is
KK s(2s) 2 s*(2s) 2 s(2p z) 2 p(2p x) 2 p(2p y )2 s(2p z ) 1
polarisation. It contains one unpaired electron.
As the size of the given cations decreases as The molecular orbital configuration of O 2 molecule is
KK s(2s) 2 s*(2s) 2 p(2p x) 2 p(2p y )2 s(2p z ) 2 p*(2px ) 1 p*(2p y ) 1
K+ > Ca 2+ > Mg2 + > Be2 + It contains 2 unpaired electrons.
Hence, polarising power decreases as 19. (c) : Number of electrons in SO 32 –
= 16 + 8 × 3 + 2 = 42
K+ < Ca 2+ < Mg2 + < Be2 +
Number of electrons in CO 32 – = 6 + 8 × 3 + 2= 32
15. (b) : Molecular orbital configuration is Number of electrons in NO 3– = 7 + 8 × 3 + 1= 32
These are not isoelectronic species as number of electrons
NO Þ s1s2 s*1s2 s2s 2 s*2s 2 s2p z 2 p2p x 2 p2p y2 p*2p x 1
Þ paramagnetic are not same.
O 2 Þ s1s 2 s*1s 2 s2s 2 s*2s 2 s2p z 2 p2px 2 p2p y 2 p*2p x1 p*2p y 1
Þ paramagnetic
O 22 – Þ s1s2 s*1s 2 s2s 2 s*2s 2 s2p z 2 p2p x2 p2p y 2 p*2p x2 p*2p y2
Þ diamagnetic
O 2+ Þ s1s 2 s*1s 2 s2s2 s*2s 2 s2p z 2 p2p x 2 p2p y2 p*2p x 1
Þ paramagnetic
Chemical Bonding and Molecular Structure 15
20. (b) : He 2 + ® s(1s) 2 s*(1s) 1 , one unpaired electron 25. (b) : Central atom in each being sp hybridised shows linear
H2 ® s(1s) 2 s*(1s)0 , no unpaired electron
H2 + ® s(1s) 1 s*(1s) 0, one unpaired electron shape.
H2 – ® s(1s) 2 s*(1s) 1 , one unpaired electron.
Due to absence of unpaired electrons, H 2 will be diamagnetic. F — Xe — F O C O
21. (d) : 26. (c) : SO 2 OH 2 SH 2 NH 3
Bond angle : 119.5° 104.5° 92.5° 106.5°
27. (b) : Molecular orbital configuration of
O 2 Þ s(1s) 2s*(1s)2 s(2s)2 s*(2s)2 s(2p z ) 2p(2px ) 2
dsp 2 hybridisation p(2p y )2 p*(2p x) 1p*(2p y) 1; B.O. = 10 - 6 = 2
dsp 2 hybridisation 2
(four 90° angles between sp 3 d or dsp 3 hybridisation
bond pair and bond pair) (six 90° angles between O 2+ Þ s(1s) 2s*(1s) 2s(2s) 2s*(2s) 2s(2p z ) 2p(2px ) 2
bond pair and bond pair)
10 - 5 = 2.5
p(2p y ) 2p*(2p x ) 1; B.O. = 2
O 2– Þ s(1s) 2s*(1s) 2s(2s) 2s*(2s) 2s(2p z ) 2p(2p x ) 2
p(2p y ) 2p*(2p x) 2p*(2p y) 1; B.O. = 10 - 7 = 1.5
2
sp 3 d 2 hybridisation O 22 – Þ s(1s) 2s*(1s)2 s(2s)2 s*(2s) 2s(2p z) 2p(2px ) 2
(twelve 90° angle between bond pair and bond pair)
2– F – p(2p y )2 p*(2p x ) 2p*(2p y) 2 ; B.O. = 10 - 8 = 1
NC CN 2
22. (c) : Ni B Hence increasing order of bond order is
NC CN F O 22 – < O 2– < O 2 < O 2+
square planar F F 28. (b) : dsp 2 hybridisation gives square planar structure with s,
px , p y and d x2 – y 2 orbitals with bond angles of 90°.
regular tetrahedral
O
F ..
F F F
Xe O PO
S F F .. F 29. (d) : O P OP O
.. O
square planar OP
O
F
seesaw shaped O
No. of s bonds = 16
23. (b) : Higher the bond order, shorter will be the bond length. No. of p bonds = 4
Thus NO + is having higher bond order than that of NO so
NO + has shorter bond length. O O
24. (c) : The correct order of bond angle (smallest first) is 30. (b) : In molecules (a) CH 3 C OH , (c) CH 3 C CH 3
H 2 S < NH 3 < SiH 4 < BF 3 and (d) (CH 2 CH — CH 3 ), the carbon atom has a multiple
92.6° < 107° < 109°28¢ < 120° bond, only (b) has sp 3 hybridization.
×× H F 31. (a) : Both undergoes sp 3 hybridization. The expected bond angle
should be 109º28¢ but actual bond angle is less than 109º28¢
H 109°28¢ 120° because of the repulsion between lone pair and bonded pairs
107° N due to which contraction occurs.
Si B
92.6°
S H H H H H H F F
16 JEE MAIN CHAPTERWISE EXPLORER
CHAPTER CHEMICAL
THERMODYNAMICS
5
1. A piston filled with 0.04 mol of an ideal gas expands reversibly (a) – 22.88 kJ (b) – 228.88 kJ
from 50.0 mL to 375 mL at a constant temperature of 37.0°C. (c) + 228.88 kJ (d) – 343.52 kJ (2009)
As it does so, it absorbs 208 J of heat. The values of q and 7. In a fuel cell methanol is used as fuel and oxygen gas is used
w for the process will be (R = 8.314 J/mol K) (ln 7.5 = 2.01) as an oxidizer. The reaction is
(a) q = + 208 J, w = + 208 J
(b) q = + 208 J, w = – 208 J CH 3 OH ( l ) + 23 O 2( g ) CO 2( g ) + 2 H 2 O ( l ) .
(c) q = – 208 J, w = – 208 J At 298 K standard Gibb’s energies of formation for CH 3 OH (l) ,
(d) q = – 208 J, w = + 208 J (2013) H2 O (l) and CO 2(g) are –166.2, –237.2 and –394.4 kJ mol –1
2. The incorrect expression among the following is respectively. If standard enthalpy of combustion of methanol
wreversible = - nRT ln Vf is –726 kJ mol –1, efficiency of the fuel cell will be
Vi
(a) in isothermal process, (a) 80 % (b) 87%
(b) ln K = DH° - TDS° (c) 90% (d) 97% (2009)
RT
8. Standard entropy of X2 , Y2 and XY3 are 60, 40 and 50 J K– 1 mol– 1,
(c) K = e– DG°/RT respectively. For the reaction, 1/2 X 2 + 3/2 Y 2 ® XY3 ,
DGsystem = -T DH = –30 kJ, to be at equilibrium, the temperature will be
DS total
(d) (2012) (a) 1000 K (b) 1250 K
3. The entropy change involved in the isothermal reversible (c) 500 K (d) 750 K (2008)
expansion of 2 moles of an ideal gas from a volume of 10 9. Oxidising power of chlorine in aqueous solution can be
dm3 to a volume of 100 dm 3 at 27°C is determined by the parameters indicated below:
(a) 38.3 J mol –1 K –1 (b) 35.8 J mol –1 K –1 (2011) – Cl (g ) – Cl – (g ) D hydH – Cl – (aq )
(c) 32.3 J mol –1 K –1 (d) 42.3 J mol –1 K –1 D eg H
1/2 Cl2 (g) 1/2 D dissH
4. The standard enthalpy of formation of NH3 is – 46 kJ mol– 1 . The energy involved in the conversion of 1/2 Cl 2(g) to Cl –( aq)
If the enthalpy of formation of H 2 from its atoms is (using data, D dissH –Cl2 = 240 kJ mol– 1, D egH –Cl = –349 kJ mol –1,
– 436 kJ mol –1 and that of N 2 is –712 kJ mol –1 , the average D hyd H–C l – = –381 kJ mol– 1) will be
bond enthalpy of N—H bond in NH 3 is
(a) – 1102 kJ mol –1 (b) – 964 kJ mol –1 (a) +120 kJ mol –1 (b) +152 kJ mol –1
(c) + 352 kJ mol –1 (d) + 1056 kJ mol –1 (2010) (c) –610 kJ mol –1 (d) –850 kJ mol –1 (2008)
5. For a particular reversible reaction at temperature T, DH and 10. Identify the correct statement regarding a spontaneous process:
DS were found to be both +ve. If T e is the temperature at (a) Lowering of energy in the reaction process is the only
equilibrium, the reaction would be spontaneous when criterion for spontaneity.
(a) T = T e (b) Te > T (b) For a spontaneous process in an isolated system, the change
(c) T > T e (d) Te is 5 times T (2010) in entropy is positive.
6. On the basis of the following thermochemical data : (D f G° (c) Endothermic processes are never spontaneous.
H+ ( aq) = 0). (d) Exothermic processes are always spontaneous. (2007)
H2 O (l) ® H +( aq) + OH – (aq) ; DH = 57.32 kJ 11. Assuming that water vapour is an ideal gas, the internal energy
H2(g) + 21 O2(g) ® H2O(l ) ; DH = - 286.2 kJ change (DU) when 1 mol of water is vapourised at 1 bar pressure
and 100°C, (given : molar enthalpy of vapourisation of water at
The value of enthalpy of formation of OH – ion at 25°C is 1 bar and 373 K = 41 kJ mol– 1 and R = 8.3 J mol– 1 K– 1) will be
Chemical Thermodynamics 17
(a) 41.00 kJ mol– 1 (b) 4.100 kJ mol– 1 (a) 100 kJ mol– 1 (b) 200 kJ mol– 1 (2005)
(c) 3.7904 kJ mol– 1 (c) 800 kJ mol– 1 (d) 400 kJ mol– 1
(d) 37.904 kJ mol– 1
(2007) 18. A schematic plot of ln Ke q versus inverse of temperature for a
reaction is shown in the figure. The reaction must be
12. In conversion of limestone to lime,
CaCO 3(s) ® CaO( s) + CO 2(g) 6.0
the values of DH° and DS° are +179.1 kJ mol –1 and
160.2 J/K respectively at 298 K and 1 bar. Assuming that DH°
and DS° do not change with temperature, temperature above ln K eq
which conversion of limestone to lime will be spontaneous is
(a) 1118 K (b) 1008 K 2.0
1.5 × 10 –3
(c) 1200 K (d) 845 K (2007) 2.0 × 10 –3
13. (DH – DU) for the formation of carbon monoxide (CO) from its 1/T (K –1 )
elements at 298 K is (a) exothermic
(R = 8.314 JK –1 mol –1 ) (b) endothermic
(a) –1238.78 J mol– 1 (b) 1238.78 J mol– 1 (c) one with negligible enthalpy change
(c) –2477.57 J mol– 1 (d) 2477.57 J mol– 1 (2006) (d) highly spontaneous at ordinary temperature. (2005)
14. The enthalpy changes for the following processes are listed 19. Consider the reaction: N 2 + 3H2 ® 2NH3 carried out at constant
temperature and pressure. If DH and DU are the enthalpy and
below:
internal energy changes for the reaction, which of the following
Cl2 (g) = 2Cl( g) , 242.3 kJ mol– 1
I2 (g) = 2I (g) , 151.0 kJ mol– 1 expressions is true?
ICl( g) = I (g) + Cl( g) , 211.3 kJ mol– 1
I 2(s) = I2 (g) , 62.76 kJ mol– 1 (a) DH = 0 (b) DH = DU
Given that the standard states for iodine and chlorine are I2 (s)
(c) DH < DU (d) DH > DU
and Cl2 (g) , the standard enthalpy of formation for ICl( g) is
(a) – 14.6 kJ mol– 1 (b) –16.8 kJ mol– 1 (2005)
20. For a spontaneous reaction the DG, equilibrium constant (K)
(c) +16.8 kJ mol– 1 (d) +244.8 kJ mol– 1 (2006) and E° cell will be respectively
(a) –ve, >1, +ve (b) +ve, >1, –ve
15. An ideal gas is allowed to expand both reversibly and irreversibly (c) –ve, <1, –ve (d) –ve, >1, –ve (2005)
in an isolated system. If Ti is the initial temperature and Tf is the
final temperature, which of the following statements is correct? 21. The enthalpies of combustion of carbon and carbon monoxide
(a) (Tf ) irrev > (Tf ) r ev are –393.5 and –283 kJ mol– 1 respectively. The enthalpy of
(b) Tf > Ti for reversible process but Tf = Ti for irreversible
formation of carbon monoxide per mole is
process
(a) 110.5 kJ (b) 676.5 kJ
(c) (Tf ) rev = (Tf ) irrev
(d) Tf = Ti for both reversible and irreversible processes. (c) –676.5 kJ (d) –110.5 kJ (2004)
(2006) 22. An ideal gas expands in volume from 1 × 10 –3 m 3 to
1 × 10– 2 m3 at 300 K against a constant pressure of 1 × 105 Nm– 2.
The work done is
16. The standard enthalpy of formation (DH f °) at (a) – 900 J (b) – 900 kJ
298 K for methane, CH4 (g) is –74.8 kJ mol– 1. The additional
information required to determine the average energy for (c) 270 kJ (d) 900 kJ (2004)
C – H bond formation would be
23. The internal energy change when a system goes from state A to
(a) the dissociation energy of H2 and enthalpy of sublimation
of carbon B is 40 kJ/mole. If the system goes from A to B by a reversible
(b) latent heat of vaporisation of methane path and returns to state A by an irreversible path what would
(c) the first four ionisation energies of carbon and electron be the net change in internal energy?
gain enthalpy of hydrogen
(a) 40 kJ (b) > 40 kJ
(d) the dissociation energy of hydrogen molecule, H2 .
(2006) (c) < 40 kJ (d) zero (2003)
17. If the bond dissociation energies of XY, X 2 and Y2 (all diatomic 24. In an irreversible process taking place at constant T and P and
molecules) are in the ratio of 1 : 1 : 0.5 and DH f for the formation in which only pressurevolume work is being done, the change
of XY is –200 kJ mol– 1. The bond dissociation energy of X 2 will in Gibbs free energy (dG) and change in entropy (dS), satisfy
be the criteria
(a) (dS)V , E < 0, (dG)T , P < 0
(b) (dS)V , E > 0, (dG)T , P < 0
18 JEE MAIN CHAPTERWISE EXPLORER
(c) (dS)V , E = 0, (dG)T , P = 0 (2003) (a) violates 1s t law of thermodynamics
(d) (dS)V , E = 0, (dG)T , P > 0
(b) violates 1 st law of thermodynamics if Q1 is –ve
25. The enthalpy change for a reaction does not depend upon the (c) violates 1 st law of thermodynamics if Q 2 is –ve
(a) physical states of reactants and products (d) does not violate 1 st law of thermodynamics. (2002)
(b) use of different reactants for the same product
(c) nature of intermediate reaction steps 28. If an endothermic reaction is nonspontaneous at freezing point
(d) difference in initial or final temperatures of involved
substances. of water and becomes feasible at its boiling point, then
(2003)
(a) DH is –ve, DS is +ve
(b) DH and DS both are +ve
(c) DH and DS both are –ve
26. If at 298 K the bond energies of C – H, C – C, (d) DH is +ve, DS is –ve (2002)
C C and H – H bonds are respectively 414, 347, 615 and 29. For the reactions,
C + O 2 ® CO 2 ; DH = –393 J
435 kJ mol– 1, the value of enthalpy change for the reaction 2Zn + O 2 ® 2ZnO ; DH = –412 J
(a) carbon can oxidise Zn
H2 C CH2 (g) + H2 (g) ® H3 C – CH3 (g) (b) oxidation of carbon is not feasible
at 298 K will be (c) oxidation of Zn is not feasible
(d) Zn can oxidise carbon.
(a) +250 kJ (b) –250 kJ
(c) +125 kJ (d) –125 kJ. (2003)
27. A heat engine absorbs heat Q 1 at temperature T1 and heat Q 2 at (2002)
temperature T 2. Work done by the engine is J (Q1 + Q 2) . This
data
Answer Key
1. (b) 2. (b) 3. (a) 4. (c) 5. (c) 6. (b)
7. (d) 8. (d) 11. (d) 12. (a)
13. (a) 14. (c) 9. (c) 10. (b) 17. (c) 18. (a)
19. (c) 20. (a) 23. (d) 24. (b)
25. (c) 26. (d) 15. (a) 16. (a) 29. (d)
21. (d) 22. (a)
27. (d) 28. (b)
Chemical Thermodynamics 19
1. (b) : As it absorbs heat, 7. (d) : For the given reaction,
\ q = + 208 J CH3OH(l) + 32 O2(g ) ¾¾® CO2(g) + 2H2O(l) ,
DH = –726 kJ mol –1
wr ev = æ VV 12 ø÷ö
-2.303 nRT log 10 èç also, DG°f [CH3OH(l)] = –166.2 kJ mol –1
DG°f [H2O(l )] = –237.2 kJ mol –1
wr ev = -2.303 ´ (0.04) ´ 8.314 ´ 310 log 10 æèç 375 ö and DG°f [CO2(g)] = –394.4 kJ mol –1
50 ÷ø
\ wr ev = – 207.76 » – 208 J
2. (b) : DG° = –RT lnK Now, DG°reaction = å DG o - å DG of reactants
f products
DG° = DH° – TDS° = [–394.4 + 2 × (–237.2)] – (–166.2)
= – 702.6 kJ mol –1
DH° – TDS° = –RT lnK
ln K = - æ DH° - TDS° ö % Efficiency = DDHG ´ 100 = --77 0226.6 ´100
èç RT ÷ø = 96.77%
3. (a) : Entropy change for an isothermal process is \ Efficiency » 97%
DS = 2.303nR log æ V2 ö 8. (d) : 1 X2 + 32 Y2 ® XY3
çè V1 ÷ø 2
DS = 2.303 ´ 2 ´ 8.314 ´ log æèç 11000 ÷öø DSr0eaction = DSp0roducts - DS r0e actants
= 38.294 J mol –1 K –1 » 38.3 J mol –1 K –1
4. (c) : 1/2N 2 + 3/2H 2 NH 3 \ DS 0 = DS 0 - 1 DS 0 2 - 3 DSY02
B.E. 712 436 reaction XY3 2 X 2
= 50 - 1 ´ 60 - 3 ´ 40 = – 40 J K –1 mol –1
2 2
é ù
\ (DH°f )NH3 = ëê 1 B.E.N2 + 32 B.E.H2 - 3B.E.N - H úû Using equation, DG = DH – TDS
2
We have DH = – 30 kJ, DS = – 40 J K– 1 mol– 1 and at equilibrium
é1 3 ´ 3B.E.N - H ùúû DG = 0. Therefore
ëê 2 2
- 46 = ´ 712 + 436 - D H -30-´4100 00 =
DS
T = = 750 K
– 46 = 356 + 654 – 3B.E. N—H 9. (c) : 1 Cl2( g )® Cl ( g ) ;
3B.E.N —H = 1056 2
B.E.N- H = 1056 = 352 kJ mol -1 DH1 = 1 D diss H Ce-l 2 = 2420 = 120 kJ mol -1
3 2
5. (c) : According to Gibb’s formula, Cl(g) ® Cl(-g) ; DH2 = Deg H Ce-l = -349 kJ mol-1
DG = DH – TDS Cl(-g ) + aq ® Cl(-aq ) ;
DH3 = Dhyd H e- = -381 kJ mol-1
Since DH and DS, both are +ve, for DG < 0, the value of
T > T e .
6. (b) : The reaction for the formation of OH – (aq) is The required reaction is 1 Cl2( g ) ® Cl(-aq ) ; D H
2
H 2( g ) + 12 O 2( g )
H (+ aq ) + OH (– a q ) 1 e- Deg H e- + Dhyd H e-
2
This is obtained by adding the two given equations. Then DH = D diss H +
\ DH for the above reaction = 57.32 + (–286.2) = 120 + (– 349) + (–381) = –610 kJ mol –1
= –228.88 kJ
20 JEE MAIN CHAPTERWISE EXPLORER
10. (b) : In an isolated system, there is neither exchange of energy 16. (a) : C + 2H2 ® CH4 ; DH° = –74.8 kJ mol– 1
nor matter between the system and surrounding. For a In order to calculate average energy for C – H bond formation
spontaneous process in an isolated system, the change in entropy we should know the following data.
is positive, i.e. DS > 0. C( graphite) ® C (g) ; DH°f = enthalpy of sublimation of carbon
Most of the spontaneous chemical reactions are exothermic. A H2 (g) ® 2H( g) ; DH° = bond dissociation energy of H2
number of endothermic reactions are spontaneous e.g. melting
of ice (an endothermic process) is a spontaneous reaction. 17. (c) : Let the bond dissociation energy of XY, X2 and Y2 be x kJ
The two factors which are responsible for the spontaneity of mol– 1 , x kJ mol– 1 and 0.5x kJ mol– 1 respectively.
a process are
(i) tendency to acquire minimum energy 1 + 1 XY ; DH f = -200 kJ mol -1
(ii) tendency to acquire maximum randomness. 2 X2 2 Y2 ®
DHr eaction = [(sum of bond dissociation energy of all reactants)
– (sum of bond dissociation energy of product)]
11. (d) : DU = DH – D nRT = é1 DH X2 + 1 - DH ù
êë 2 2 DHY2 XY úû
= 41000 – 1 × 8.314 × 373
= 41000 – 3101.122 = x + 0.5 x - x = -200
= 37898.878 J mol– 1 = 37.9 kJ mol– 1 2 2
12. (a) : For DG° = DH° – TDS° \ x = 200 = 800 kJ mol -1
For a spontaneous process DG° < 0 0.25
i.e. DH° – TDS° < 0
or DH° < TDS° or, TDS° > DH° ln K 2 = DH é 1 - 1 ù
(a) : K1 ê T2 úû
18. R ë T1
T DH ° i.e. T > 179.1 ´ 1000
or > ln 6 = DH [1.5 ´10-3 - 2 ´ 10-3 ]
D S° 160.2 2 R
or T > 1117.9 K » 1118 K
13. (a) : DH – DU = Dng RT or, ln 3 = DH ´ (-0.5 ´ 10-3 )
1 R
C + 2 O 2 ® CO DH of reaction comes out to be negative. Hence reaction is
exothermic.
Dng =1- æçè1 + 1ö = - 1
2ø÷ 2 19. (c) : N 2 + 3H2 ® 2NH3
Dn = 2 – 4 = –2
DH – DU = - 1 ´ 8.314 ´ 298 = - 1238.78 J mol -1 DH = DU + DnRT = DU – 2RT
2 \ DH < DU
1 1 20. (a) : For spontaneous process, DG = –ve
14. (c) : 2 I 2 (s) + 2 Cl2 (g) ® ICl (g) Now DG = –RT ln K
When K > 1, DG = –ve
DH ICl( g) = é 1 DH I2 (s)®I2 ( g ) + 1 DH I -I + 1 DH ù -[DH I-Cl ] Again DGº = – nFEº
ëê 2 2 2 Cl-Cl ûú When Eº = +ve, DGº = –ve
= é 1 ´ 62.76 + 1 ´151.0 + 1 ´ 242.3úùû - [211.3] 21. (d) : C( s) + O 2(g) ® CO 2(g) ; DH = –393.5 kJ mol– 1 .. (i)
ëê 2 2 2
= [31.38 + 75.5 + 121.15] – 211.3 CO (g) + 12 O 2(g) ® CO 2(g) ; DH = –283 kJ mol– 1 ... (ii)
= 228.03 – 211.3 = 16.73 kJ/mol
On subtraction equation (ii) from equation (i), we get
15. (a) : If a gas was to expand by a certain volume reversibly, then C( s) + O 2(g) ® CO (g) ; DH = –110.5 kJ mol– 1
it would do a certain amount of work on the surroundings. If The enthalpy of formation of carbon monoxide per mole
it was to expand irreversibly it would have to do the same amount
of work on the surroundings to expand in volume, but it would = –110.5 kJ mol– 1
also have to do work against frictional forces. Therefore the
amount of work have greater modulus but –ve sign. 22. (a) : W = – PDV
Wi rrev. > Wr eve. ; (Tf ) i rrev. > (Tf ) r ev.
= –1 × 105 (1 × 10– 2 – 1 × 10 –3)
= –1 × 10 5 × 9 × 10– 3 = –900 J
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