ISAT-2011 TEST PAPER 1.
A projectile is fired at an angle 60° with some velocity u. If the angle is changed infinitesimally, let the corresponding fractional changes in the range and the time of flight be x and y, y, respectively, respectively, Then y is (A)
Sol.
2 x 3
(B)
2 x 3
(C) 2x
(D) –2x
u 2 sin 2 R= g u2 cos 2d dR = 2 g 2u 2 cos 2 dR d = g u 2 sin 2 R g dR = 2cot2d = x R T=
2u sin g
2u dT = g cos d dT cot d = y T y cot cot 60 x 2 cot 2 2 cot 120
1 =
1 3 1 2 2 3
x 2 Correct Answer is not in the options.
y=
ISAT- 1
2.
A ball is dropped down vertically from a tall building. After falling a height h it bounces elastically from a table inclined at an angle and hits a wall at a distance d from the point of earlier impact horizontally, horizontally, then
(A) = (1/2) sin 1 (2d/h) (C*) = (1/4) sin 1 (d/h)
(B) = (1/2) tan 1 (d/h) (D) = (1/2) tan 1 (2d/h)
–
–
–
Sol.
u=
–
2gh
u 2 sin 2(90 º 2) d= 2g
d=
2gh sin 4 2g
1 1 d = 4 sin h
3.
A photon with an initial frequency 1011 Hz scatters of an electron at rest. Its final f inal frequency is 0.9×1011 Hz. The speed of the scattered electron is close to (h = 6.63 × 10 34 Js, me = 9.1×10 31 kg) (A*) 4 × 103 ms-1 (B) 3 × 102 ms-1 (C) 2 × 106 ms-1 (D) 30 ms-1 –
Sol.
h0 = h +
–
1 me v 2 2
h(1011 – 0.9 × 1011) =
1 mev 2 2
2 h 1010 v me v = 3.8 × 103 4 × 103 m/s Comprehensive (4 - 5) An -particle experiences the following force due to a nucleus (k > 0)
F
= 4.
k r2
k R3
rˆ
if
r>R
if
r< R
r
The correct potential energy diagram for the above force is
(A)
(B)
(C)
(D*)
ISAT- 2
Sol.
k
F
r2
rˆ
k
=
if
r>R
if
r
r
R3
R
r
Uf – Ui = F . d r = –
k
r
2
dr = k r
assuming U = 0 k r
for r > R, U = r
Ur – UR = F.d r
r
kr
k
F.dr R dr 2R
=
3
3
(r 2 R 2 )
R
k
Ur = U R + =
k 2R
Ur 0
3
2R
3
(r 2 R 2 )
[R 2 r 2 ]
k 2R
U= U= 5.
Sol.
k r
r >R
k 2R
3
(R2 + r2)
r
Suppose the particle starts from r = with a kinetic energy just enough to reach r = R. Its kinetic energy at r = R/2 will be (A) k/R (B) (5/8) (k/R) (C*) (3/8) (k/R) (D) 0 The particle is projected from fr om infinity such that is just reaches r = R i.e. its speed at r = R is zero. Now applying energy conservation from r = R to r = R/2. UR = U R/2 + K.E.
k 2 k R = 2R 3 R
R2 4 + KE
k 5k = K.E. R 8R 3k K.E. 8R
6.
Let a particle have an instantaneous position r ( t ) , velocity v ( t ) and acceleration a ( t ) . Necessary conditions for it to be considered as an instantaneous circular motion about the origin are
(A) r .v 0 ; a.v 0 ; a.r 0 (B) r .v 0 ; a.v 0 ; a.r 0 (C) r .v 0 ; a.v 0 ; a.r 0
(D*) r .v 0 ; a.v 0 ; a.r 0
ISAT- 3
Sol.
v.r 0 , a . v 0 , a . r 0
7.
A large parallel plate capacitor is made of two metal plates of size 2m × 1m. It has a dielectric slab made of two dielectrics of permeability K1 and K2 as shown in the figure, the distance between them being 0.1m . It is charged by a battery of 1000V after which the battery is disconnected. Now the dielectric slab is pulled out by 10cm. The work done in doing so is (ignore ( ignore the gap between the plates and the dielectric slab)
(A) 2.5 × 105 0 J (C) 5 × 1050 J
(B) -2.5 × 1050 J (D*) -5 × 1050 J
Sol.
Ceq = C1 + C2 =
3 0 2 0 0.1 0.1
Ceq = 50 0 Q = 500 × 1000 1 1 CeqV2 = 500 (1000)2 2 2 6 = 250 × 10 J Finally Ceq = C1’ + C2’
energy stored =
=
3 0 1.1 2 0 (0.9 ) = 51 0. 1 0 .1
0
1 50 2 02 10 6 Energy final = 2 510 1 50 2 0 10 6 Energy final = 2 51 1 50 2 0 10 6 2 51 = –5 × 105 0 J.
E =
8.
–
250 × 106J
A current I is flowing in a long straight wire along the z-axis. z-ax is. A particle with mass m and charge q has an initial position x0 ˆi and velocity v0 kˆ The z-component of its velocity after a very short time interval t is (A*) z (t) = 0 [1 –
(C) z (t) = 0 [1 –
1 2
3 2
2
o I q2 (t)2] 2x 0 m2
1 (B) z(t) = 0[1 + 2
2
o I q2 (t)2] 2x 0 m2
2
o I q2 (Dt)2] 2 2 x 0 m
(D) z (t) = 0
Sol.
ISAT- 4
mv 0 mv 0 r = qB q (2x 0 ) 0
=
v 0 t r
v z v 0 cos( )
( )2 = v0 1 2
2
( for small angle cos = 1 –
) 2
v 02 (t )2 = v 0 1 2r 2 v 2 ( t )2 2 q2 0 0 2 2 = v 0 1 2 2x 0 m v 0 1 2 q2 2 0 v z v0 1 t 2 2x 0 m 2 9.
A non-conducting sphere of radius R has a charge Q distributed uniformly over its volume. The T he sphere is surrounded by a thin metal shell of radius b (b > R) with a charge –Q. The space between the shell and the sphere is filled with air. Which of the following graphs correctly represents the corresponding electric field.
(A*)
(B)
(C)
(D)
Sol.
E
E
kQr R3 kQ r2
rR R
E =0
r>b
ISAT- 5
10.
The magnetic field at the center of a loop carrying a current I in the circuit shown is given by
(A*)
Sol.
B
=
oI 7 ˆ 3R 8
k
(B)
oI 5 ˆ 3R 8
k
(C)
o I 7 ˆ 3R 8
k
(D)
oI 5 ˆ 3R 8
k
0 ˆ 0 5 ˆ k k 4 R 3 4(2R) 3 0 5 ˆ 1 k 2R 2
7 0 ˆ k 24R A current I is flowing in a wire of length l . The total momentum carried by the charge carrier of mass m and charge q is =
11.
(A*) Sol.
m I l I l q
(B)
2m I l q
(C)
q I l I l m
(D)
2q I l I l m
= nqAvd
P = nAmvd P= 12.
q
m
In an oil drop experiment, charged charge d oil drops of mass m and charge q are released relea sed at a height h, one at a time, at intervals t >
2h / g . The drops are collected in a large m etal sphere of radius R with a small opening at
the top. The total number of drops dr ops that are able to enter the sphere will be. (A)
(C) Sol.
mg4 0 (h R)
(B*)
q2 mg 4 0 (h R )2 q
(D)
mg40 (h R) 2 q2 mg40 (h R)q
Let n number of drops have fallen f allen into the sphere and the charge of the sphere has become Q = nq, and the next drops just reaches the top of the sphere, applying energy conservation kQq kqQ mgh mg(2R) (h R ) R mg(h – 2R) =
kqQ kQq R h R
mg(h – 2R) =
kQq(h 2R) R(h R)
4 0 mgR(h R ) q2
n
None of the answers is matching. Dimensionally B is correct.
ISAT- 6
13.
Two lenses, one biconvex of focal length f1 and another biconcave of focal length f2 are placed along the same axis. They are separated by a certain distance such that a parallel beam of light incident on the convex lens also emerges parallel from the concave lens subsequently. subsequently. The magnification m agnification of the combination is given by (A) M = f12 / f22
(B*) M = f2 / f1
(C) M = f1 / f2
(D) M = (f1 f2)( f12
f22 )
Sol.
f2 m= f 1 14.
Sol.
The central fringe in a Young’s double slit experiment with the He-Ne laser ( = 632.8 nm) has intensity I0. If one of the slits is covered by a 5m thick film of plastic (refractive index = 1.4), the intensity becomes I1. The ratio I1 /I0 is close to (A) 3/16 (B) 1/4 (C) 1/2 (D*) 3/4 Path difference = ( – 1) t = (1.4 – 1) × 5 × 10 6 = 2 × 10 6 m –
–
Phase difference =
2
x
= 19.86 rad
2
I1 = I0 cos2 I1 3 I0 4 15.
Sol.
16.
A polarizer is introduced in the path of a beam of unpolarized unpolarized light incident on a block of transparent material (refractive (refract ive index = 3 ). The polarizer can be placed such that its axis is parallel (P) or normal (N) to the plane of incidence. The incident beam makes an angle with the surface of the block. The light will be completely transmitted if (A*) = 30° and the polarizer is placed in P (B) = 30° and the polarizer is placed in N (C) = 60° and the polarizer is placed in P (D) = 60° and the polarizer is placed in N 1 = tan () = 60º = 30º = 30º & polarizer is placed in P. –
–
A submarine traveling at 10 ms 1 is chasing another one in front of it. It locates its position and speed by sending Sonar (ultrasonic sound) towards it and recording the time of its travel and return frequency. frequency. The frequency of the Sonar is 25000 Hz and the frequency of the reflected r eflected signal is 24900 Hz. If the speed of sound in water is 1500 ms 1 , the speed of the submarine subm arine being chased is (A) 16 ms 1 (B) 14 ms 1 (C*) 13 ms 1 (D) 11 ms 1 –
–
Sol.
v s v v s 10 v 10 v v s s
f ’ = f
–
–
–
vs – velocity of sound
v = 13 m/s
ISAT- 7
17.
When light of intensity I reflects from a surface separating two media with refractive index 1 and 2 (2 > 1), the intensity of the reflected light is (2 –1)2 /(2 +1)2 . T To o make reflection r eflection zero a thin layer of a material of refractive index of thickness t is inserted between the two media. The value value of and t such that wavelength of light is not reflected at all is (A) = (2 + 1)/ 2; 2t = (2n+1)/2 (B) = (1 + 2)/ 2; 2t = 2n (C*) =
1 2 ;
2t = (2n+1)/2
(D) =
1 2 ;
2t = 2n
Sol.
For destructive interference of ray 1 and 2 2t = (2n + 1)
2 for zero intensity of reflected light 1 = 2 2
2 – – 1 1 2 2 – – 1 2 1 18.
2
= 1 2
A point object is placed below belo w a wide glass plate of refractive ref ractive index n. As an observer moves from left le ft to right above the glass plate, the angle subtended by the apparent object is n
–
(A) 2 tan
1
2
n 1
–
(C*) 2 tan
1
n n2 1
–
(B) tan
1
n n2 1
–
(D) tan
1
2n n2 1
Sol.
(C) (C)
19.
A light sensor is fixed at one corner of the bottom of a rectangular tank of depth 10 m full of a liquid of refractive index 2 / 3 . The illuminated area through which the light can exit at the top of the tank is (A) (A) a quar quarte terr of of a circ circlle of of rad radiius > 10 m (B) (B) a quar quarte terr of of a circ circlle of of rad radiu ius s < 10 m (C*) a quarter of a circle of radius = 10 3 m
(D) a quarter of of a circl rcle of of ra radius > 10 3 m
Sol.
sinc =
3 2
r tan 60 10
r = 10 3 Area =
r 2 4 ISAT- 8
20.
The average pressure on a sphere submerged in water is the pressure at the depth of its center. center. A sphere sphere of radius 10 cm made of steel is held in water as shown in the figure. The force that water applies on the surface surfa ce 3 2 3 of the shaded hemisphere is (g = 10ms ; water = 10 kgm ) –
Sol.
(A) 0 N (B*) 63 N 2 FHz = P R = g (0.2) × (0.1)2 = 63 Nt Fvertical = Vg
–
(C) 126 N
(D) 252 N
3
2 1 = 10 × 10 3 10 3
=
20 = 21Nt 3
Fres = 21 10 None of the options is matching m atching so the most appropriate answer is corresponding to the Horizontal thrust force i.e. 63 Newton. 21.
Sol. 22.
Laplace correction to the speed of sound is made only for gases and not for solids and liquids. This is because, in comparison to gases, liquids and solids have (A) larger thermal therm al conductivity (B) much smaller compressibility (C) much smaller coefficient of thermal expansion (D*) much smaller relative pressure change when the wave is passing through them. (D) (D) Three rods of equal lengths and cross sectional areas are joined as shown in the figure, with respective thermal conductivities K, 2Kand K. The left end is at a temperature TA and the right end at a temperature tem perature TB. In steady state, the temperatures T1 and T2 at the junctions j unctions are given by.
(A*) T1 = (C) T1 =
3 2 2 3 T A + TB ; T2 = TA + T B 5 5 5 5 3 2 1 4 TA + TB ; T2 = TA + TB 5 5 5 5
(B) T1 =
4 1 1 4 TA + TB ; T2 = T A + TB 5 5 5 5
(D) T1 =
4 1 2 3 TA + TB ; T2 = TA + TB 5 5 5 5
Sol. T1 TA
KA T1 =
+
T1 TB =0 3 2KA
3 2 TA TB 5 5
ISAT- 9
T2 TA T2 TB 0 3 2KA KA T2 = 23.
2 3 TA TB 5 5
The diameter of a metal wire is measured using a screw gauge, whose circular scale s cale has 50 divisions. Two Two full rotations of the circular scale move two main scale divisions of 0.5 mm each. W hen it is used to measure the diameter of a wire of length 3.14 m, and resistance 10 , its reading is as shown in the figure. The resistivity of the wire is
(A) 4.84 × 10 5 m Pitch = 0.5 mm –
Sol.
(B) 2.42 × 10 5 m –
(C*) 1.21 × 10 5m –
(D) 2.42 × 10 10 6 m –
0.5 = 0.01 mm 50 P.S.R. = 4 × 0.5 = 2mm C.S.R. = 20 × L.C. = 0.2 m diameter = 2.2 mm
L.C. =
= 24. Sol. 25.
Sol.
RA
10 (1.1 10 3 ) 2 = 1.21 × 10 3.14
–
5
Which of the following quantities has the least number of significant digits? (A) 0.80760 (B*) 0.08765 (C) 5.7423 × 102 (D) 80.760 0.08765 In an experiment designed to determine the universal gravitational constant, G, the percentage errors in measuring the appropriate a ppropriate mass, mas s, length and time variables are given by a, b, c respectively respecti vely.. The total error in determining G is then (A*) (a + 3b + 2c) (B) ( – a + 3b – 2c) (C) (2a + 3b + 2c) (D) (a + 9b + 4c) 1 3 2 [G] = M L T –
–
G M 3L 2T 100 100 L T G M G 100 = a + 3b + 2c G 26.
Ans. 27. Ans.
The relative stability of the octahedral complexes of Fe(III) over Fe(II) with the bidentate ligands, (i) HO-CH2-CH2-OH, (ii) HO-CH2-CH2-NH2, (iii) H2N-CH2-CH2-NH2, (iv) H2N-CH2-CH2-SH, follows the order (A) (i) > (ii) > (iii) > (iv) (B) (ii) > (i) > (iv) > (iii) (C*) (iii) > (ii) > (iv) > (i) (D) (iv) > (i ) > (iii ) > (ii) It is fact (A/c to spectrochemical series of ligand strength). Number of isomers that [Pt(Cl)(Br)(NO)(NH3)] exhibit, (A*) 3 (B) 6 (C) 4 In Mabcd compounds total geometical isomer are three.
(D) 8
ISAT- 10
28.
Ans. 29.
When a metal is in its low oxidation state, the metal-carbon bond in M-CO is stronger than the metal-ch metal-chloride loride bond in M-Cl, because, (A) chloride is a donor and carbon monoxide is a acceptor (B) chloride is a acceptor and carbon monoxide is a donor (C*) chloride is a donor and the carbon monoxide is both a donor as well as a acceptor (D) chloride is both the donor as well as acceptor and the carbon monoxide is both and accepter Due to synergic effect in metal carbonyl car bonyl complexes. Freshly prepared, bright blue coloured, colour ed, dilute solution of sodium in liquid ammonia amm onia can be used to reduce the organic functional moieties. In this, the actual reducing species is, (‘e’ is an electron) (A) [Na(NH3)n]+ (B) [H2(NH3)n] (C) [NaNH2(NH3)n] (D*) [e(NH3)n] Dilute solution of Na & NH3 forms ammonated e which act as reducing agent. Na + (x + y) NH 3 [Na(NH3)x]+ + [e(NH3)y]or [e(N [e(NH H3)n] ammonated e The statement that is NOT correct corr ect in case of silicates is, (A) cement is a silicate (B) the Si-O bond is 50% covalent and 50% ionic (C) silicate structures could have holes to fit cations in tetrahedral and octahedral geometries (D*) silicates are mainly m ainly built through ‘SiO2’ units Silicate cantions SiO44 units in their lattice. –
Sol.
–
–
30.
Sol. 31. Sol. 32.
Sol.
33.
Sol.
34.
–
The (SiO32 )n are (A) pyrosilicates (B) orthosilicates (C) cyclic silicates (D) sheet silicates 2n The normal form ula of cyclic silicate anion is (SinO3n )n. So, the ans is cyclic silicate. –
–
The oxo-acid of sulphur that does NOT contain S-S bond is, (A) pyrosulphurous acid (H2S2O5) (B) dithionus acid (H2S2O4) (C) dithionic acid (H2S2O6) (D) pyrosulphuric acid (H2S2O7) Structure of oxyacid are follows f ollows
The reason for the formation of H+ when B(OH)3 is dissolved in water is, (A) acidic nature of B(OH)3 (B) high polarizing power of B3+ (C) hydrogen bonding between B(OH)3 and water (D) high electronegativity electronegativity of oxygen Boric acid is behave as levies acid with H2O
An optically active alcohol (X) on catalytic hydrogenation hydrogenation gives an optically inactive alcohol. The alcohol (X) is (A) 3-ethyl-3-buten-2-ol (B) 3-methyl-3-penten-2-ol (C*) 2-ethyl-3-buten-1-ol (D) 4-methyl-4-penten-2-ol
Sol.
ISAT- 11
35.
The major product for formed med in the following reaction is i) O3 / Zn ii) aq, NaOH
(A*)
(B)
(C)
(D)
O3 / Zn
Sol.
H2O
36.
The following transformation is effected by
Sol.
(A) alkaline KMnO4 (B) NaOH/CHCl3 (C*) NaOH/ 2 (D) peracetic acid I2 in alkaline medium doesnot add to C = C but removes C1 by iodoform reaction.
37.
Among the following halides, the one that is least reactive towards solvolysis
(I) (A) I
(II) (B) II
(III) (C*) III
(IV) (D) IV
Sol.
All solvolysis reaction are SN1 and passes, through carbocation formation. The carbocation c arbocation formed in III is antiaromatic hence it is least reactive.
38.
Isopropanol can be converted to methylacetate using (A*) pyridinium chlorochromate followed by peracetic acid (B) peracetic acid followed by pyridinium pyridinium chlorochromate (C) pyridinium chlorochromate followed by NaOH/ 2 (D) NaOH/ 2 followed by pyridinium chlorochromate
Sol. 39.
Among the isomeric butylbenzenes, butylbenzenes, the one that is NOT oxidized by alkaline KMnO4 to benzoic acid is
(A) Sol.
(B)
(C*)
(D)
Only alkyl benzene with at least one benzylic benzylic H is oxidised.
ISAT- 12
40.
The following reaction is effected by
(A*) i. (CH3)2CHCOCl/AlCl 3; ii. Br2 /FeBr3; iii. NH2.NH2 /KOH (B) i. (CH3)2CH-CH2Cl/AlCl3; ii. Br2 /FeBr3 (C) i. (CH3)2CHCOCl/AlCl 3; ii. NH2.NH2 /KOH; iii. Br 2 /FeBr3 (D) i. Br2 /FeBr3; ii. (CH3)2CH-CH2Cl/AlCl3
Sol.
41.
The major product of the following reaction is
(A)
(B*)
(C)
(D)
Sol.
42.
Conversion of benzene into 1,3-dibromobenzene is accomplished through (A) i. Br2 /FeBr3; ii. HNO3 /conc. H2SO4; iii. Sn/HCl; iv. NaNO 2 /HCl, 0-5 °C ; v. CuBr (B) i. Br2 /FeBr3; ii. HNO3 /conc. H 2SO4; iii. Sn/HCl; iv. NaNO 2 /HCl, 0-5 ºC ; v. CuBr 2 (C*) i. HNO3 /conc. H 2SO4; ii. Br 2 /FeBr3; iii. Sn/HCl; iv. NaNO2 /HCl, 0-5 ºC ; v. CuBr (D) i. HNO3 /conc. H 2SO4; ii. Br2 /FeBr3; iii. Sn/HCl; iv. NaNO2 /HCl, 0-5 ºC ; v. CuBr 2
Sol.
43.
Sol.
44.
Liquid oxygen and liquid nitrogen are allowed to flow between the poles of an electromagnet. Choose the correct observation (A) Both will be attracted but to opposing pole pieces (B) Both will be attracted to the same sam e pole (C) Liquid oxygen will be attracted and liquid nitrogen will be repelled to the same degree (D*) Liquid oxygen will be attracted but liquid nitrogen unaffected On the basic of MOT, O2 is paramagnetic while N2 is diamagnetic. So, liquid oxygen will be attracted but liquid nitrogen unattracted. The highest transition energy in the Balmer series in the emission spectra of hydrogen is (RH =109737 cm 1) (A) 4389.48 cm 1 (B) 2194.74 cm 1 (C) 5486.85 cm 1 (D*) 27434.25 cm For balmer series of hydrogen n1 = 2, z = 1 and for highest energy, energy, n2 = –
–
Sol.
–
–
–
1
ISAT- 13
1 1 = RH × Z 2 2 n1 n 2 1
1
45.
2
= 109737 × (1)2 ×
2
46.
2
–
= 27434.25 cm
1
A one litre glass bulb is evacuated and weighed. The weight is 500 g. It is then filled with an ideal gas at 1 atm pressure at 312.5 K. The weight of the filled bulb is 501.2 g. The molar weight of the gas is (R = 8 × 10 2L.atm.K 1.mol 1) (A) 28 (B) 34 (C*) 30 (D) 24 V = 1, wgas = 501.2 – 500 = 1.2 gm –
Sol.
1
–
–
using
Pv=nRT 1 × 1 =
M = 30
1.2 × 8 × 10 2 × 312.5 M –
The van der Waals coeffcient of the inert gases He, Ar and Xe are given below Inert gas a (atm.dm 6 .mol 2 ) b (10 2 dm3 mol 1
Sol.
47.
He
0.34
2.38
Ar
1.337
3.20
Xe
4.137
5.16
Choose the appropriate pair to complete the following statement. “The increase in the value of a signifies the increasing importance of ——- interaction while increase in value in b is due to ———————-” (A) Ion-ion; increased atomic volume (B) Induced dipole-induced dipole; increased atomic volume (C) Induced dipole-dipole; dipole-dipole interaction (D) Dipole-dipole; decreasing ionization energies On going down the group, size increases increases due to which atomic volume increases and with with increases of size vanderwall force increases which is induced dipole - induced dipole. Assuming H0 and S0 do not change with temperature, calculate the boiling point of liquid A using the thermodynamic data given below Thermodynamic data 0
H (kJ/mol) 0 S (J/K/mol)
Sol.
48.
130 100
—
(A*) 300 K (B) 130 K G° = H° – TS°, Aliq Agas. H° = –100 –( –130) = 30 kJ/mole S° = 200 – 100 = 100 J/K /mole 0 = 30,000 – T × 100 T = 300 K
A(gas) 100 200
—
(C) 150 K
(D) 50 K
–
A solution of CaCl2 was prepared by dissolving 0.0112 g in 1 kg of distilled water (molar (m olar mass of Ca = 41 g mol 1 and Cl = 35.5 g mol 1). The freezing point constant of water is 2 K.kg mol 1. The depression in the freezing point of the solution is (A) 0.0002 (B) 0.002 (C) 0.003 (D*) 0.0006 For CaCl2, i = 3 Tf = i × Kf × m –
Sol.
A(liq)
= 3 × 2x
–
.0112 / 111 1
= .0006
ISAT- 14
49.
Of the four values of pH given below which is the closet to the pH of 0.004 M carbonic acid (Ka1 = 4 × 10 7 ; Ka2 = 2 × 10 12) (A*) 4.4 (B) 5.0 (C) 5.4 (D) 4.0 Due to large difference in Ka1 and Ka2 values pH is only due to Ka1 –
Sol.
50.
Sol.
–
Ka1 C
4 10 7
=
[H+] = C = 4 × 10 3 × 10 2 = 4 × 10 pH = 5 – log4 = 4.4
4 10 3 –
= .01 (1%) –
–
5
The Haber’s process for the production of ammonia involves the equilibrium N2(g) + 3H 2(g) 2NH3(g) Assuming that H° and S° for the reaction does not change with temperature, which of the statements is true (H° = – 95 kJ and S° = – 190 J/K) (A) Ammonia dissociates spontaneously below 500 K (B*) Ammonia dissociates spontaneously above 500 K (C) Ammonia dissociates at all temperatures (D) Ammonia does not dissociates at any temperature G° = H° – TS° Spontaneous, G° < 0 G° = 95 × 103 – T × ( –190) = 95,000 × 190 T < 0 –95000 < –190 T 190 T > 95000 T > 500 K
51.
Martin throws two dice simultaneously. simultaneously. If the sum of the outcomes is 12, he offers lunch at a five star hotel with probability 2/3. 2/3. If the sum is 7, he offers lunch with probability 1/2. 1/2. In all the other cases, he offers lunch with probability 1/3. Given that the lunch was offered, the probability that the sum of the outcomes equals 12 is (A) 1/18 (B*) 1/20 (C) 1/24 (D) 1/36
Sol.
P(12) = P(7) =
1 36
1 6
P(not 7 not 12) = 1 –
1 1 29 – = 36 6 36
1 2 . 36 3 P= 1 2 1 1 29 1 . . . 36 3 6 2 36 3 =
2 2 1 = 2 9 29 40 20
52.
A species has an initial population 4 10. At the end of the first day, the population increases by 50% At the end of the second day, itit decreases by the same percentage. If the process continues in the same sam e pattern, the number of days for the population to reach 310 is (A) 10 (B*) 20 (C) 50 (D) 100
Sol.
end of first day = (1.5) . 410 = end of second day = 3.219.
3 . 220 = 3.219 2
1 = 3.218 2
ISAT- 15
3.219 + 3.218 + 9.217 + 9.216 + 27.215 + . . . . . . . . at the end of 20 trial it will become 310 53.
If 4 squares are chosen at random on a chessboard (there are ar e 64 square arranged in 8 rows and 8 columns in a chessboard), then the probability that all the four squares are in the same sam e main diagonal is 8
(A)
8
C4
64
(B*) 2
C4
8
C4
64
(C) 3
C4
8
C4
64
(D) 4
C4
C4
64
C4
Sol.
2 . 8C 4
54.
A student was calculating the variance of a data that consists of ten observations. By mistake, he used
64
C4
Two main diagonal have 8 element.
one of the observations as 1 instead of 10 and found the variance as
744 46 and mean as . The actual 100 10
variance of the data is 825 100
(A*) Sol.
x0
46 10
(C)
625 100
(D)
525 100
(given)
Now, Now, actual mean x variance =
725 100
(B)
55 10
744 100
2 2 2 ( x1 – x ) ( x 2 – x ) ........ ( x10 – x ) 744
(10 )
100
744 so actual variance = 100 744 1 55 = 100 10 10 – 10
2
55.
–
(1 – x 0 )2 10
46 – 1 – 10
(10 – x )2 10
825 100
2
A fair coin is tossed 6 times. The probability that the head appears in the sixth trial for the third time is (A)
5 16
(B*)
5 32
(C)
5 36
(D)
3 64
5
Sol.
56. Sol.
1 1 10 5 5 C2 . = 2 2 64 32 –
The sum of the roots of the equation x + 1 – 2log2(2x+ 3) + 2log4(10 – 2 x) = 0 is (A*) log2 11 (B) log2 12 (C) log2 13 (D) log2 14 x 2 x x + 1 – log2 (2 + 3) + log2 (10 – 2 ) = 0 –
(2x 3)2 10 – 2 x
x + 1 = log2
2x.2 =
–
( 2 x 3 )2 –
10 – 2
x
20.2x – 2 = (2x)2 + 9 + 6.2x , put 2x = y y2 – 14y + 11 = 0
ISAT- 16
y1 y2 = 11 2 x1 x 2 = 11 x1 + x2 = log2 11
57.
Let z = a cos
i sin , a R, |a| < 1. then z2010 + z20111 + . . . equals 5 5
z 2010 (A) 1 – a Sol.
z
2011
+z
a 2010 (D*) 1 – z
z 2010 a 2010 . ei402 a 2010 +.....+ = = = 1 – z 1 – z 1 – z
The locus of the point z satisfying arg(z + 1) = and arg(z – 1) = , when (0, ) vary subject to the condition
Sol.
za2010 (C) 1 – z
z = a ei /5 2010
58.
a 2010 (B) 1 – a
1 tan
–
1 = 2, is tan
(A) two parallel lines (C*) a line parallel to the x-axis arg(z + 1) = arg(z – 1) = y tan 1 x 1 = & tan –
–
tan = 1 tan
–
1
(B) a single point (D) two intersecting lines
y
x – 1 =
y
y & tan = x 1 x – 1 1 =2 tan
x 1 x – 1 – = 2 y = 1 y y
59.
For the equation, sin x + cos x =
1 1 a , a > 0, 2 a
(A) there is no solution, for any a > 0 (B) there is a solution, for infinitely many a > 0 (C) there is a solution, for two or more, but finitely f initely many a > 0 (D*) there is a solution, for exactly one a > 0 Sol.
sin x + cos x =
1 1 a 2 a
2
solution will only be there if a = 1 60. Sol.
The number of solution of the equation sin x + cos x = 1 – sin 2x in the interval [ –2, 3] is (A) 2 (B) 4 (C*) 6 (D) 8 sin x + cos x = 1 – sin 2x 1 + t = t2 – 2t + 1 t2 – 3t = 0 t = 0, 3 sin 2x = 0 x= x=
n 2
2
,
5 , 2
–
3 , 2 , 0, –2 2 ISAT- 17
61.
Consider the circles C1 : x2 + y2 = 64 and C2 with radius 10. If the center of C2 lies on the line y = x and C 2 intersects C1 such that the length of the common chord is 16, then the center of the circle C2 is
3 3 , 2 2
(A) Sol.
4 4 , 2 2
(B)
6 6 , 2 2
8 8 , 2 2
(C)
(D)
(x – a)2 + (y – a)2 = 102 x2 + y2 = 64 equation of common chord is 2 – 2ax – 2ay + 2a – 100 + 64 = 0 It passes through (0, 0)
6 6 a2 = 18, a = 3 2 , 2 2 62.
A line segment joining (1, 0, 1) and the origin (0, 0, 0) is revolved about the x-axis to form a right circular cone. If (x, y, z) is any point on the cone, other than the origin, then it satisfies the equation (A) x2 – 2y2 – z2 = 0 (B*) x2 – y2 – z2 = 0 (C) 2x2 – y2 – 2z2 = 0 (D) x2 – 2y2 – 2z2= 0
Sol
Equaiton of plane AB is x = 1 OA =
2
Cone OACB can be find by section of plane x = 1 & sphere x2 + y2 + z2 =
2
2
by homoginizing the equation of sphere by plane equation of required cone x2 + y2 + z2 = 2(x)2 x2 – y2 – z2 = 0
63.
Let (x, y, z) be any point on the line passing through (x0, y0, 0) and parallel to the vector i x0 + y0 = 2, then (x, y, z) lies on the plane normal to the vector
(A*) i Sol.
j – 2k
(B) i
j 2k
(C) i
j – k
(D) i
j k
x – x 0 y – y 0 z 1 1 1 Let point on this line is (r + x0, r + y0, r) But x0 + y0 = 2 Point is (r + x0, r + 2 – x0, r) by eleminating x0 & r by x = r + x0, y = r + 2 – x0, z = r we get x + y – 2z = 2 Equation of line
Normal vector to the plane ˆi jˆ – 2kˆ
j k . If
ISAT- 18
64.
x2 y2 A tangent to the ellipse = 1 meets the coordinate axes at A and B. If the points A, A, B and the origin 25 16 are vertices of an isosceles triangle, then the length of AB is (A)
Sol.
(B)
41 / 2
(C*)
41
82
(D) 164
x cos y sin =1 5 4
5 , 0 cos
A
4 sin
B 0 ,
5 4 cos sin tan =
4 5
AB =
25 sec 2 16 cos ec 2
16 25 16 1 25 16
25 1
=
25 16 16 25 =
65.
Let an =
82
1 [(2n + 1) (2n + 2) ... (2n + n)] 1/n, for n N and lim an = eL, then L is n n
3
(A*)
3
log x dx
(B)
2
Sol.
3
log 2x dx
(C)
2
1 2 3 n an = 2 2 2 ....... 2 n n n n
log(1 x) dx 2
3
(D)
log( 2 x) dx 2
1 / n
n
1 r log 2 log an = n r 1 n
1
log an =
log( 2 x) dx
Put x + 2 = t
0
3
=
log t dt 2
66.
The value of lim
13 23 ... (3n)3 3n 4 (B) 9/4
n
(A) 1/2 Sol.
lim
n
(3n)2 (3n 1)2 3n 4 . 4
=
is (C*) 27/4
(D) 81/4
81 27 = 3 .4 4
ISAT- 19
/ 2
67.
The value of
0
2 sin x x / 2 e dx is 1 cos x
(A) e /4
(B) e /2
(C) 2e /2
(D*) 2e /4
/ 2
Sol.
e sec (x / 2) tan(x / 2)dx x / 2
2
0
x =t 2
at
/ 4
e sec t
=
2
t tan t . 2dt
0
/ 2
= e x / 2 . 2 tan x / 2 0 = 2e /4 68.
Sol.
The differential equation satisfied by the family fam ily of curves which cut the curves y = x3, R at right angles is (A) xy' – 3y = 0 (B*) x + 3yy' = 0 (C) y' – 3x2 = 0 (D) 3x2y' + 1 = 0 y = x3 dy 3y 3y = 3x2 = 3 . x2 = dx x x
–
dx 3 y dy x
x dx + 3y dy = 0 x2 + 3y2 = C
69.
Sol.
70. Sol.
Let f(x) = x (|x – |) (2 + cos2x), x R. Then the function f : R R is (A) one-one but NOT onto (B*) onto but NOT one-one (C) both one-one and onto (D) neither one-one nor onto 2 f(x) = x(|x – |) (2 + cos x) It is onto Range R value at x = 0, is 0 so function is many one.
The equation 2x3 – 3x2 + p = 0 has three distinct real roots for all p belonging to (A*) (0, 1) (B) (2, ) (C) ( – , 1/2) (D) ( –, 0) (1, ) 3 2 – p = 2x – 3x dy = 6x2 – 6x dx = 6x(x – 1)
–
1 < – p < 0 0
ISAT- 20
71.
For a real number x, let [ x ] denote the greatest integer less than or equal to x. Let f : [1/2,] R be defined by f(x) = (x – [ x ]) [x] cos
x
. Then f is 2 (A*) Continuous at x = 1 but NOT continuous at x = 2 (B) Continuous at x = 2 but NOT continuous at x = 1 (C) Continuous at both x = 1 and x = 2 (D) Discontinuous at x = 1 and x = 2
Sol.
1 f : , R 2 f(x) = (x – [x])[x] cos
x 2
cont. at x = 1 but not at x = 2
72.
Let f : (0, ) R be defined by f(x) = 2xsin2x cos2x. Then lim f(x) is x 0
(A) 1 Sol.
(B*) 2
(C) e
(D) 2e
y = lim 2xsin2x cos 2x x0
log y = lim (log2 + sin 2x log x + log cos 2x) x0
log x + lim log cos 2x x0 x 0 cos ec 2 x
log y = log 2 + lim
1 x = log 2 + lim +0 x 0 – 2 cos ec 2x cot 2x y=2
73.
The distance of the point (1, 2, 3) from the plane r . (2 i j 2k ) =
–
5 measured parallel to the line
r ( –3 i 2 j ) ( i j k ) is (A*) 3 3 Sol.
(B) 5 2
(C) 3 5
(D) 5 6
x – 1 y – 2 z – 3 (x, y, z) ( + 1, + 2, + 3) 1 1 1
r . 2 ˆi jˆ 2kˆ = – 5 2( + 1) + ( + 2) + 2( + 3) = – 5 5 = – 15 = – 3 P ( –2, –1, 0) A (1, 2, 3) AP =
999 = 3 3
ISAT- 21
74.
If the vector 3 i
4 j 7k = v1 v 2 , where v1 is parallel to i – j k and v 2 is perpendicular to 2 i – k , then
| v1 |2 + | v 2 |2 is (A) 68 Sol.
(B) 70
(C) 88
(D*) 92
v 1 v 2 3 ˆi 4 jˆ 7kˆ
v1 =
ˆi – jˆ kˆ
v 2 = (3 – ) ˆi + (4 + ) jˆ + (7 – ) kˆ
v 2 . 2ˆi – kˆ = 0
2(3 – ) – (7 – ) = 0 6 – 2 – 7 + = 0 – – 1 = 0 = – 1
v 1 – ˆi jˆ – kˆ
v1 3
v 2 89
v 2 4ˆi 3 jˆ 8kˆ 2
v1 v 2 75.
2
= 92
A plane H passes through the intersection of the planes r .( i
j k ) = – 3 and r .( i – j k ) = 2. If H divides
the line segment joining (3, 0, 2) and (0, 3, –1) in the ratio 2 : 1 internally, internally, then the equation of H is
(A*) r .(3 i – j 3k ) = 1 (B) r .( i
j 3k ) = 3
(C) r .(5 i – 3 j k ) = – 1 (D) r .(2 i
j 3k ) = 4
Sol. P (1, 2, 0) (x + y + z + 3) + (x – y + z – 2) = 0 passes through (1, 2, 0) 6 + ( –3) = 0 =2 3x – y + 3z – 1 = 0
ISAT- 22
