1. The current voltage relation of diode is given by I=(e1000V/T−1)mAI=(e1000V/T−1)mA , where the applied VV is in volts and the temperature TT is in degree kelvin. If a student makes an error measuring ±0.01V±0.01V while measuring the current of 5mA5mA at 300K300K , what will be the error in the value of current in mA?
(1) 0.2mA0.2mA
(2) 0.02mA0.02mA
(3) 0.5mA0.5mA
(4) 0.05mA0.05mA
Answer (1)
Sol. I=(e1000V/T−1)mAI=(e1000V/T−1)mA
When I=5mAI=5mA e1000V/T=6mAe1000V/T=6mA
Also,dI=(e1000V/T)×1000T⋅dVAlso,dI=(e1000V/T)×T1000⋅dV=(6mA)×1000300×(0.01)=(6mA)×3001000×(0.01)=0.2mA=0.2mA
2. From a tower of height HH a particle is thrown vertically upwards with a speed uu . The time taken by the particle, to hit the ground, is nn times that taken by it to reach the highest point of its path. The relation between HH , uu and nn is:
(1)2gH=n2u2(2)gH=(n−2)2u2(1)2gH=n2u2(2)gH=(n−2)2u2(3)2gH=nu2(n−2)(4)gH=(n−2)u2(3)2gH=nu2(n−2)(4)gH=(n−2)u2
Answer (3)
Sol. Time taken to reach highest point is t1=ugt1=gu
Speed on reaching ground =u2+2gh=u2+2gh
Now, v=u+atv=u+at
⇒u2+2gh=−u+gt⇒u2+2gh=−u+gt⇒t=u+u2+2ghg=nug⇒t=gu+u2+2gh=gnu⇒2gh=n(n−2)u2⇒2gh=n(n−2)u2
3. A mass mm is supported by a massless string wound around a uniform hollow cylinder of mass mm and radius RR . If the string does not slip on the cylinder, with what acceleration will the mass fall on release?
(1) 2g332g
(2) g22g
(3) 5g665g
(4) gg
Answer (2)
Sol. a=Rαa=Rα
mg−T=mamg−T=ma
T×R=mR2αT×R=mR2α
or T=maT=ma
⇒a=g2⇒a=2g
4. A block of mass mm is placed on a surface with a vertical cross-section given by y=x36y=6x3 . If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is
(1) 16m61m
(2) 23m32m
(3) 13m31m
(4) 12m21m
Answer (1)
Sol. tanθ=dydx=x22tanθ=dxdy=2x2
At limiting equilibrium,
μ=tanθμ=tanθ0.5=x220.5=2x2⇒x=±1⇒x=±1Now,y=16Now,y=61
5. When a rubber-band is stretched by a distance xx , it exerts a restoring force of magnitude F=ax+bx2F=ax+bx2 where aa and bb are constants. The work done in stretching the unstretched rubber-band by LL is:
(1)aL2+bL3(2)12(aL2+bL3)(1)aL2+bL3(2)21(aL2+bL3)(3)aL22+bL33(4)12(aL22+bL33)(3)2aL2+3bL3(4)21(2aL2+3bL3)
Answer (3)
6. A bob of mass mm attached to an inextensible string of length ll is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed ωω rad/s about the vertical. About the point of suspension
(1) Angular momentum is conserved
(2) Angular momentum changes in magnitude but not in direction
(3) Angular momentum changes in direction but not in magnitude
(4) Angular momentum changes both in direction and magnitude
Answer (3)
Sol. τ=mg×lτ=mg×l sin θθ . (Direction parallel to plane of rotation of particle)
as ττ is perpendicular to L⃗L , direction of L changes but magnitude remains same.
7. Four particles, each of mass MM and equidistant from each other, move along a circle of radius RR under the action of their mutual gravitational attraction. The speed of each particle is
(1) GMR (2) 22GMR (3) GMR(1+22) (4) 12GMR(1+22)(4) (1) RGM (2) 22RGM (3) RGM(1+22) (4) 21RGM(1+22)(4)
Answer (4)
Sol. F2+F2+F′=Mv2R2F+2F+F′=RMv2
2×GM22(R2)2+GM24R2=Mv2R2(R2)22×GM2+4R2GM2=RMv2GM2R[14+12]=Mv2RGM2[41+21]=Mv2v=GmR(2+442)=12GmR(1+22)v=RGm(422+4)=21RGm(1+22)
8. The pressure that has to be applied to the ends of a steel wire of length 10cm10cm to keep its length constant when its temperature is raised by 100∘C100∘C is:
(For steel Young's modulus is 2×1011Nm−22×1011Nm−2 and coefficient of thermal expansion is 1.1×10−5K−11.1×10−5K−1
(1)2.2×108Pa(3)2.2×107Pa(2)2.2×109Pa(4)2.2×106Pa(3)(1)(3)2.2×108Pa2.2×107Pa(2)2.2×109Pa(4)2.2×106Pa(3)
Answer (1)
Sol. As length is constant,
Strain=ΔLL=αΔQStrain=LΔL=αΔQ
Now pressure == stress =Y×=Y× strain
=2×1011×1.1×10−5×100=2×1011×1.1×10−5×100=2.2×108Pa=2.2×108Pa
9. There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d1d1 and d2d2 are filled in the tube. Each liquid subtends 90∘90∘ angle at centre. Radius joining their interface makes an angle αα with vertical. Ratio d1d2d2d1 is
11−sinα1−sinα11+tanα1−tanα1−tanα1+tanα
Answer (3)
Sol. Equating pressure at AA
(Rcosα+Rsinα)d2g=(Rcosα−Rsinα)d1g(Rcosα+Rsinα)d2g=(Rcosα−Rsinα)d1g⇒d1d2=cosα+sinαcosα−sinα=1+tanα1−tanα⇒d2d1=cosα−sinαcosα+sinα=1−tanα1+tanα
10. On heating water, bubbles being formed at the bottom of the vessel detatch and rise. Take the bubbles to be spheres of radius RR and making a circular contact of radius rr with the bottom of the vessel. If r≪Rr≪R , and the surface tension of water is TT , value of rr just before bubbles detatch is (Density of water is ρwρw )
R2ρwg3T(2)R2ρwg6TR23Tρwg(2)R26TρwgR2ρwgT(4)R23ρwgTR2Tρwg(4)R2T3ρwg
Answer (No answer)
Sol. When the bubble gets detached,
Buoyant force == force due to surface tension
∫T×dlsinθ=43πR3ρwg∫T×dlsinθ=34πR3ρwg⇒T×2πr×rR=43πR3ρwg⇒T×2πr×Rr=34πR3ρwg⇒r2=2R4ρwg3⇒r2=32R4ρwg⇒r=R22ρwg3T⇒r=R23T2ρwg
11. Three rods of copper, brass and steel are welded together to form a Y-shaped structure. Area of cross- section of each rod =4cm2=4cm2 . End of copper rod is maintained at 100∘C100∘C whereas ends of brass and steel are kept at 0∘C0∘C . Lengths of the copper, brass and steel rods are 46, 13 and 12 cm respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is
(1) 1.2 cal/s
(2) 2.4 cal/s
(3) 4.8 cal/s
(4) 6.0 cal/s
Answer (3)
Sol.
Q=Q1+Q2Q=Q1+Q20.92×4(100−T)46=0.26×4×(T−0)13+0.12×4×T12460.92×4(100−T)=130.26×4×(T−0)+120.12×4×T⇒200−2T=2T+T⇒200−2T=2T+T⇒T=40∘C⇒T=40∘C⇒Q=0.92×4×6046=4.8cal/s⇒Q=460.92×4×60=4.8cal/s
12. One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperatures at A,BA,B and CC are 400K400K , 800K800K and 600K600K respectively. Choose the correct statement
(1) The change in internal energy in whole cyclic process is 250R250R
(2) The change in internal energy in the process CA is 700R700R
(3) The change in internal energy in the process AB is −350R−350R
(4) The change in internal energy in the process BC is −500R−500R
Answer (4)
ΔU=nCVΔT=1×5R2ΔTΔU=nCVΔT=1×25RΔTFor BC,ΔT=−200KFor BC,ΔT=−200K⇒ΔU=−500R⇒ΔU=−500R
13. An open glass tube is immersed in mercury in such a way that a length of 8cm8cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46cm46cm . What will be length of the air column above mercury in the tube now?
(Atmospheric pressure =76cm=76cm of Hg)
(1) 16cm16cm
(2) 22cm22cm
(3) 38cm38cm
(4) 6cm6cm
15. A pipe of length 85 cm85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250Hz1250Hz . The velocity of sound in air is 340 m/s340 m/s .
(1) 12
(2) 8
(3) 6
(4) 4
Answer (3)
f=(2n−1)v4L≤1250f=4L(2n−1)v≤1250⇒(2n−1)×3400.85×4≤1250⇒0.85×4(2n−1)×340≤1250⇒2n−1≤12.5⇒2n−1≤12.5
Answer is 6.
P+x=P0P+x=P0
P=(76−x)P=(76−x)
8×A×76=(76−x)×A×(54−x)8×A×76=(76−x)×A×(54−x)
x=38x=38
Length of air column =54−38=16cm=54−38=16cm
14. A particle moves with simple harmonic motion in a straight line. In first ττ s, after starting from rest it travels a distance aa and in next ττ s it travels 2a2a in same direction then
(1) Amplitude of motion is 3a3a
(2) Time period of oscillations is 8τ8τ
(3) Amplitude of motion is 4a4a
(4) Time period of oscillations is 6τ6τ
Answer (4)
Sol. As it starts from rest, we have
x=Acosωt.Att=0,x=Ax=Acosωt.Att=0,x=A
when t=τ,x=A−at=τ,x=A−a
when t=2τ,x=A−3at=2τ,x=A−3a
⇒A−a=Acosωτ⇒A−a=AcosωτA−3a=Acos2ωτA−3a=Acos2ωτAscos2ωτ=2cos2ωτ−1Ascos2ωτ=2cos2ωτ−1⇒A−3aA=2(A−aA)2−1⇒AA−3a=2(AA−a)2−1A−3aA=2A2+2a2−4Aa−A2A2AA−3a=A22A2+2a2−4Aa−A2A2−3aA=A2+2a2−4AaA2−3aA=A2+2a2−4Aaa2=2aAa2=2aAA=2aA=2a
Now, A−a=AcosωτA−a=Acosωτ
⇒cosωτ=12⇒cosωτ=212πTτ=π3T2πτ=3π⇒T=6τ⇒T=6τ
16. Assume that an electric field E⃗=30x2i⃗E=30x2i exists in space. Then the potential difference VA−VOVA−VO where VOVO is the potential at the origin and VAVA the potential at x=2mx=2m is
(1)120 J (2)−120 J (1)120 J (2)−120 J (3)−80 J (4)80 J (3)−80 J (4)80 J
Answer (3)
Vol=V⃗=−E⃗⋅d⃗xVol=V=−E⋅dx∫V0VAdV=−∫0230x2dx∫V0VAdV=−∫0230x2dxVA−VO=−[10x3]02=−80 J VA−VO=−[10x3]02=−80 J
17. A parallel plate capacitor is made of two circular plates separated by a distance 5mm5mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3×1043×104 V/mV/m , the charge density of the positive plate will be close to
(1)6×10−7C/m2(2)3×10−7C/m2(1)6×10−7C/m2(2)3×10−7C/m2(3)3×104C/m2(4)6×104C/m2(3)3×104C/m2(4)6×104C/m2
Answer (1)
Vol:E=σKϵ0Vol:E=Kϵ0σσ=Kϵ0Eσ=Kϵ0E=2.2×8.85×10−12×3×104≈6×10−7C/m2=2.2×8.85×10−12×3×104≈6×10−7C/m2
18. In a large building, there are 15 bulbs of 40W40W 5 bulbs of 100W100W 5 fans of 80W80W and 1 heater of 1kW1kW . The voltage of the electric mains is 220V220V . The minimum capacity of the main fuse of the building will be :
(1) 8A
(2) 10A
(3) 12A
(4) 14A
21. In the circuit shown here, the point 'C' is kept connected to point 'A' till the current flowing through the circuit becomes constant. Afterward, suddenly, point 'C' is disconnected from point 'A' and connected to point 'B' at time t=0t=0 . Ratio of the voltage across resistance and the inductor at t=L/Rt=L/R will be equal to
Answer (3)
Sol. 15×40+5×100+5×80+1000=V×I15×40+5×100+5×80+1000=V×I
600+500+400+1000=220I600+500+400+1000=220I
I=2500220=11.36I=2202500=11.36
I=12AI=12A
19. A conductor lies along the zz -axis at −1.5≤z<1.5m−1.5≤z<1.5m and carries a fixed current of 10.0A10.0A in −az−az direction (see figure). For a field B⃗=3.0×10−4e−0.2xa^yT,B=3.0×10−4e−0.2xa^yT, find the power required to move the conductor at constant speed to x=2.0mx=2.0m y=0my=0m in 5×10−3s5×10−3s Assume parallel motion along the xx -axis
(1) 1.57 W
(2) 2.97 W
(3) 14.85 W
(4) 29.7 W
Answer (2)
Sol. Average Power == work time
W=∫02FdxW=∫02Fdx=∫023.0×10−4e−0.2x×10×3dx=∫023.0×10−4e−0.2x×10×3dx=9×10−3∫02e−0.2xdx=9×10−3∫02e−0.2xdx=9×10−30.2[−e−0.2×2+1]B=3.0×10−4e−0.2x=0.29×10−3[−e−0.2×2+1]B=3.0×10−4e−0.2x=9×10−30.2×[1−e−0.4]I=10AI=3m=0.29×10−3×[1−e−0.4]I=10AI=3m=9×10−3×(0.33)=9×10−3×(0.33)=2.97×10−3J=2.97×10−3JP=2.97×10−3(0.2)×5×10−3=2.97WP=(0.2)×5×10−32.97×10−3=2.97W
20. The coercivity of a small magnet where the ferromagnet gets demagnetized is 3×103Am−13×103Am−1 The current required to be passed in a solenoid of length 10cm10cm and number of turns 100, so that the magnet gets demagnetized when inside the solenoid, is
(1) 30mA30mA
(2) 60mA60mA
(3) 3A
(4) 6A
Answer (3)
Sol. B=μ0niB=μ0ni
Bμ0=niμ0B=ni3×103=NIL=100×i10×10−23×103=LNI=10×10−2100×iI=3A.I=3A.
21. In the circuit shown here, the point 'C' is kept connected to point 'A' till the current flowing through the circuit becomes constant. Afterward, suddenly, point 'C' is disconnected from point 'A' and connected to point 'B' at time t=0t=0 . Ratio of the voltage across resistance and the inductor at t=L/Rt=L/R will be equal to
Answer (3)
Sol. Applying Kirchhoff's law in closed loop, −VR−VC=0−VR−VC=0
⇒VR/VC=−1⇒VR/VC=−1
Note : The sense of voltage drop has not been defined. The answer could have been 1.
22. During the propagation of electromagnetic waves in a medium
(1) Electric energy density is double of the magnetic energy density
(2) Electric energy density is half of the magnetic energy density
(3) Electric energy density is equal to the magnetic energy density
(4) Both electric and magnetic energy densities are zero
Answer (3)
Sol. Energy is equally divided between electric and magnetic field
23. A thin convex lens made from crown glass (μ=32)(μ=23) has focal length ff . When it is measured in two different liquids having refractive indices 4334 and 5335 , it has the focal lengths f1f1 and f2f2 respectively. The correct relation between the focal lengths is
(1) f1=f2 (2) f1>ff1>f and f2f2 becomes negative
(3) f2>ff2>f and f1f1 becomes negative
(4) f1f1 and f2f2 both become negative
Answer (2)
Sol. By Lens maker's formula
1f1=(3/24/3−1)(1R1−1R2)f11=(4/33/2−1)(R11−R21)1f2=(3/25/3−1)(1R1−1R2)f21=(5/33/2−1)(R11−R21)1f=(32−1)(1R1−1R2)f1=(23−1)(R11−R21)⇒f1=4f&f2=−5f⇒f1=4f&f2=−5f
24. A green light is incident from the water to the air - water interface at the critical angle(0). Select the correct statement
(1) The entire spectrum of visible light will come out of the water at an angle of 90∘90∘ to the normal
(2) The spectrum of visible light whose frequency is less than that of green light will come out to the air medium
(3) The spectrum of visible light whose frequency is more than that of green light will come out to the air medium
(4) The entire spectrum of visible light will come out of the water at various angles to the normal
Answer (2)
Sol. sin θc=1μθc=μ1
For greater wavelength (i.e. lesser frequency) μμ is less So, θcθc would be more. So, they will not suffer reflection and come out at angles less then 90∘90∘
25. Two beams, AA and BB , of plane polarized light with mutually perpendicular planes of polarization are seen through a polaroid. From the position when the beam AA has maximum intensity (and beam BB has zero intensity), a rotation of polaroid through 30∘30∘ makes the two beams appear equally bright. If the initial intensities of the two beams are IAIA and IBIB respectively, then IAIBIBIA equals
(1) 3
(2) 3223
(3) 1
(4) 1331
Answer (4)
Sol. By law of Malus, I=I0cos2θI=I0cos2θ
Now, IA′=IAcos230IA′=IAcos230
IB′=IBcos260IB′=IBcos260
As IA′=IB′IA′=IB′
⇒IA×34=IB×14⇒IA×43=IB×41
IAIB=13IBIA=31
26. The radiation corresponding to 3→23→2 transition of hydrogen atoms falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3×10−43×10−4 T. If the radius of the largest circular path followed by these electrons is 10.0mm10.0mm , the work function of the metal is close to
(1) 1.8eV1.8eV
(2) 1.1eV1.1eV
(3) 0.8eV0.8eV
(4) 1.6eV1.6eV
Answer (2)
r=mvqBr=qBmv=2meVeB=eB2meV=1B2meV=B1e2mV⇒V=B2r2e2m=0.8 V ⇒V=2mB2r2e=0.8 V
29. Match List-I (Electromagnetic wave type) with List - II (Its association/application) and select the correct option from the choices given below the lists:
For transition between 3 to 2,
E=13.6(14−19)E=13.6(41−91)=13.6×536=1.88eV=3613.6×5=1.88eV
Work function =1.88eV−0.8eV=1.88eV−0.8eV
=1.08eV=1.1eV=1.08eV=1.1eV
27. Hydrogen (1H1)(1H1) , Deuterium (1H2)(1H2) , singly ionised Helium (2He4)+(2He4)+ and doubly ionised lithium (3Li6)++(3Li6)++ all have one electron around the nucleus. Consider an electron transition from n=2n=2 to n=1n=1 . If the wave lengths of emitted radiation are λ1λ1 , λ2λ2 , λ3λ3 and λ4λ4 respectively then approximately which one of the following is correct?
(1)4λ1=2λ2=2λ3=λ4(1)4λ1=2λ2=2λ3=λ4(2)λ1=2λ2=2λ3=Λ4(2)λ1=2λ2=2λ3=Λ4(3)λ1=λ2=4λ3=9λ4(3)λ1=λ2=4λ3=9λ4(4)λ1=2λ2=3λ3=4λ4(4)λ1=2λ2=3λ3=4λ4
Answer (3)
Sol.1λ=RZ2[1n12−1n2]Sol.λ1=RZ2[n121−n21]⇒λ∝1Z2for given n1&n2⇒λ∝Z21for given n1&n2⇒λ1=λ2=4λ3=9λ4⇒λ1=λ2=4λ3=9λ4
28. The forward biased diode connection is
Answer (1)
For forward Bias, pp - side must be at higher potential than nn - side.
List-I
(a) Infrared waves
(b) Radio waves
(c) X-rays
(d) Ultraviolet rays
List-II
(i) To treat muscular strain
(ii) For broadcasting
(iii) To detect fracture of bones
(iv) Absorbed by the ozone layer of the atmosphere
(a) (b) (c) (d)
(1) (iv) (iii) (ii) (i)
(2) (i) (ii) (iv) (iii)
(3) (iii) (ii) (i) (iv)
(4) (i) (ii) (iii) (iv)
Answer (4)
Sol. (a) Infrared rays are used to treat muscular strain
(b) Radiowaves are used for broadcasting
(c) X-rays are used to detect fracture of bones
(d) Ultraviolet rays are absorbed by ozone
30. A student measured the length of a rod and wrote it as 3.50cm3.50cm . Which instrument did he use to measure it?
(1) A meter scale
(2) A vernier calliper where the 10 divisions in vernier scale matches with 9 division in main scale and main scale has 10 divisions in 1 cm
(3) A screw gauge having 100 divisions in the circular scale and pitch as 1 mm
(4) A screw gauge having 50 divisions in the circular scale and pitch as 1 mm
Answer (2)
Sol. As measured value is 3.50cm3.50cm , the least count must be 0.01cm=0.1mm0.01cm=0.1mm
For vernier scale with 1MSD=1mm1MSD=1mm and 9MSD=10VS9MSD=10VS
Least count=1MSD−1VSDLeast count=1MSD−1VSD=0.1mm=0.1mm
PART-B : CHEMISTRY
31. The correct set of four quantum numbers for the valence electrons of rubidium atom (Z = 37) is
(1) 5,0,0,+125,0,0,+21
(2) 5,1,0,+125,1,0,+21
(3) 5,1,1,+125,1,1,+21
(4) 5,0,1,+125,0,1,+21
Answer (1)
Sol. 37 → 1s22s22p63s23p63d104s24p65s11s22s22p63s23p63d104s24p65s1
So last electron enters 5s orbital
Hence n = 5, l = 0, m_l = 0, m_s = ±1/2
32. If Z is a compressibility factor, van der Waals equation at low pressure can be written as
(1) Z=1+RTPbZ=1+PbRT
(2) Z=1−aVRTZ=1−VRTa
(3) Z=1−PbRTZ=1−RTPb
(4) Z=1+PbRTZ=1+RTPb
Answer (2)
Sol. Compressibility factor (Z) = PVRTRTPV
(for one mole of real gas)
van der Waal equation
(P+aV2)(V−b)=RT(P+V2a)(V−b)=RT
At low pressure
V−b≈VV−b≈V
(P+aV2)V=RT(P+V2a)V=RT
PV+aV=RTPV+Va=RT
PV=RT−aVPV=RT−Va
PVRT=1−aVRTRTPV=1−VRTa
So, Z=1−aVRTZ=1−VRTa
33. CsCl crystallises in body centred cubic lattice. If ‘a’ is its edge length then which of the following expressions is correct?
(1) rCs++rCl−=3arCs++rCl−=3a
(2) rCs++rCl−=3a2rCs++rCl−=23a
(3) rCs++rCl−=32arCs++rCl−=23a
(4) rCs++rCl−=3arCs++rCl−=3a
Answer (3)
Sol.
2rCl−+2rCs+=3a2rCl−+2rCs+=3a
rCl−+rCs+=3a2rCl−+rCs+=23a
34. For the estimation of nitrogen, 1.4 g of an organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of M1010M sulphuric acid. The unreacted acid required 20 mL of M1010M sodium hydroxide for complete neutralization. The percentage of nitrogen in the compound is
(1) 6%
(2) 10%
(3) 3%
(4) 5%
Answer (2)
Sol. As per question
| |
Normality |
Volume |
| H₂SO₄ |
N/5 |
60mL |
| NaOH |
N/10 |
20mL |
(ngeq)H2SO4=(ngeq)NaOH+(ngeq)NH3(ngeq)H2SO4=(ngeq)NaOH+(ngeq)NH3
15×601000=110×201000+(ngeq)NH351×100060=101×100020+(ngeq)NH3
6500=1500+(ngeq)NH35006=5001+(ngeq)NH3
(ngeq)NH3=5500=1100(ngeq)NH3=5005=1001
(nmol)N=(nmol)NH3=(ngeq)NH3=1100(nmol)N=(nmol)NH3=(ngeq)NH3=1001
(Mass)N=14100=0.14g(Mass)N=10014=0.14g
Percentage of "N" = 0.141.4×100=10%1.40.14×100=10%
35. Resistance of 0.2M0.2M solution of an electrolyte is 50Ω50Ω . The specific conductance of the solution is 1.4Sm−11.4Sm−1 . The resistance of 0.5M0.5M solution of the same electrolyte is 280Ω280Ω . The molar conductivity of 0.5M0.5M solution of the electrolyte in Sm2mol−1Sm2mol−1 is
(1)5×10−4(2)5×10−3(1)5×10−4(2)5×10−3(3)5×103(4)5×102(3)5×103(4)5×102
Answer (1)
Sol. For 0.2M0.2M solution
R=50ΩR=50Ω
σ=1.4Sm−1=1.4×10−2Scm−1σ=1.4Sm−1=1.4×10−2Scm−1
⇒ρ=1σ=11.4×10−2Ωcm⇒ρ=σ1=1.4×10−21Ωcm
Now,R=ρlaNow,R=ρal
⇒la=Rρ=50×1.4×10−2⇒al=ρR=50×1.4×10−2
For 0.5M0.5M solution
R=280ΩR=280Ω
σ=?σ=?
la=50×1.4×10−2al=50×1.4×10−2
⇒R=ρla⇒R=ρal
⇒1ρ=1R×la⇒ρ1=R1×al
⇒σ=1280×50×1.4×10−2⇒σ=2801×50×1.4×10−2
=1280×70×10−2=2801×70×10−2
=2.5×10−3Scm−1=2.5×10−3Scm−1
Now,λm=σ×1000MNow,λm=Mσ×1000
=2.5×10−3×10000.5=0.52.5×10−3×1000
=5Scm2mol−1=5Scm2mol−1
=5×10−4Sm2mol−1=5×10−4Sm2mol−1
36. For complete combustion of ethanol,
C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l),C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l),
the amount of heat produced as measured in bomb calorimeter, is 1364.47kJmol−11364.47kJmol−1 at 25∘C25∘C . Assuming ideality the enthalpy of combustion, ΔcHΔcH for the reaction will be
(R=8.314kJmol−1)(R=8.314kJmol−1)(1)−1366.95kJmol−1(2)−1361.95kJmol−1(1)−1366.95kJmol−1(2)−1361.95kJmol−1(3)−1460.50kJmol−1(4)−1350.50kJmol−1(3)−1460.50kJmol−1(4)−1350.50kJmol−1
Answer (1)
Sol. C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l)C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l)
Bomb calorimeter gives ΔUΔU of the reaction
So, as per question
ΔU=−1364.47kJmol−1ΔU=−1364.47kJmol−1
Δng=−1Δng=−1
ΔH=ΔU+ΔngRTΔH=ΔU+ΔngRT
=−1364.47−1×8.314×2981000=−1364.47−10001×8.314×298
=−1366.93kJmol−1=−1366.93kJmol−1
37. The equivalent conductance of NaCl at concentration C and at infinite dilution are λcλc and λscλsc respectively. The correct relationship between λcλc and λscλsc is given as
(Where the constant B is positive)
(1)λc=λsc+(B)C(2)λc=λsc−(B)C(1)λc=λsc+(B)C(2)λc=λsc−(B)C(3)λc=λsc−(B)C(4)λc=λsc+(B)C(3)λc=λsc−(B)C(4)λc=λsc+(B)C
Answer (3)
Sol. According to Debye Huckle onsager equation,
λc=λsc−ACλc=λsc−AC
Here A=BA=B
∴λc=λsc−BC∴λc=λsc−BC
38. Consider separate solutions of 0.500MC2H5OH(aq)0.500MC2H5OH(aq) 0.100MMg3(PO4)2(aq)0.100MMg3(PO4)2(aq) 0.250MKBr(aq)0.250MKBr(aq) and 0.125MNa3PO4(aq)0.125MNa3PO4(aq) at 25∘C25∘C . Which statement is true about these solutions, assuming all salts to be strong electrolytes?
(1) They all have the same osmotic pressure.
(2) 0.100MMg3(PO4)2(aq)0.100MMg3(PO4)2(aq) has the highest osmotic pressure.
(3) 0.125MNa3PO4(aq)0.125MNa3PO4(aq) has the highest osmotic pressure.
(4) 0.500MC2H5OH(aq)0.500MC2H5OH(aq) has the highest osmotic pressure.
Answer (1)
Sol. π=iCRTπ=iCRT
πC2H5OH=1×0.500×R×T=0.5RTπC2H5OH=1×0.500×R×T=0.5RTπMg3(PO4)2=5×0.100×R×T=0.5RTπMg3(PO4)2=5×0.100×R×T=0.5RTπKBr=2×0.250×R×T=0.5RTπKBr=2×0.250×R×T=0.5RTπNa3PO4=4×0.125×RT=0.5RTπNa3PO4=4×0.125×RT=0.5RT
39. For the reaction SO2(g)+12O2(g)⇌SO3(g)′SO2(g)+21O2(g)⇌SO3(g)′ if Kp=Kc(RT)xKp=Kc(RT)x where the symbols have usual meaning then the value of xx is (assuming ideality)
(1)−1(2)−12(1)−1(2)−21(3)12(4)1(3)21(4)1
Answer (2)
SO2(g)+12O2(g)⇌SO3(g)SO2(g)+21O2(g)⇌SO3(g)Kp=Kc(RT)xKp=Kc(RT)xx=Δng=no. of gaseous moles in productx=Δng=no. of gaseous moles in product − no. of gaseous moles in reactant − no. of gaseous moles in reactant − 1−(1+12)=1−32=−12 − 1−(1+21)=1−23=2−1
40. For the non-stoichiometre reaction 2A+B→C+D2A+B→C+D the following kinetic data were obtained in three separate experiments, all at 298 K298 K
| Initial Concentration (A) |
Initial Concentration (B) |
Initial rate of formation of C (mol L-1s-1) |
| 0.1 M |
0.1 M |
1.2 × 10-3 |
| 0.1 M |
0.2 M |
1.2 × 10-3 |
| 0.2 M |
0.1 M |
2.4 × 10-3 |
The rate law for the formation of C is
(1)dCdt=k[A][B](2)dCdt=kA[A]2[B](1)dtdC=k[A][B](2)dtdC=kA[A]2[B](3)dCdt=k[A][B]2(4)dCdt=k[A](3)dtdC=k[A][B]2(4)dtdC=k[A]
Answer (4)
Sol. 2A+B⟶C+D2A+B⟶C+D
Rate of Reaction= −1 d[A] ˙2 d t=−d[B]d tRate of Reaction=2 d t −1 d[A] ˙=−d td[B] −d[C]dt=d[D]dt −dtd[C]=dtd[D]
Let rate of Reaction =k[A]x[B]y=k[A]x[B]y
Or,d[C]dt=k[A]x[B]yOr,dtd[C]=k[A]x[B]y
Now from table,
1.2×10−3=k[0.1]x[0.1]y(ii)1.2×10−3=k[0.1]x[0.1]y(ii)1.2×10−3=k[0.1]x[0.2]y(iii)1.2×10−3=k[0.1]x[0.2]y(iii)
Dividing equation (i) by (ii)
⇒1.2×10−31.2×10−3=k[0.1]x[0.1]yk[0.1]x[0.2]y⇒1.2×10−31.2×10−3=k[0.1]x[0.2]yk[0.1]x[0.1]y⇒1=[12]y⇒1=[21]y⇒[v=0v=0]⇒[v=0v=0]
Now Dividing equation (i) by (iii)
⇒1.2×10−32.4×10−3=k[0.1]x[0.1]yk[0.2]x[0.1]y⇒2.4×10−31.2×10−3=k[0.2]x[0.1]yk[0.1]x[0.1]y⇒[12]x=[12]x⇒[21]x=[21]x⇒x=1x=1⇒x=1x=1Henced[C]dt=k[A]i[B]o.Hencedtd[C]=k[A]i[B]o.
41. Among the following oxoacids, the correct decreasing order of acid strength is
(1)HOCl>HClO2>HClO3>HClO4(1)HOCl>HClO2>HClO3>HClO4(2)HClO4>HClO>HClO2>HClO3(2)HClO4>HClO>HClO2>HClO3(3)HClO4>HClO3>HClO2>HClO(3)HClO4>HClO3>HClO2>HClO(4)HClO2>HClO4>HClO3>HClO(4)HClO2>HClO4>HClO3>HClO
Answer (3)
SO4HClO4⇌ClO4−+H+SO4HClO4⇌ClO4−+H+HClO3⇌ClO3−+H+HClO3⇌ClO3−+H+HClO2⇌ClO2−+H+HClO2⇌ClO2−+H+HClO⇌ClO−+H+HClO⇌ClO−+H+Resonance produced conjugate base.Resonance produced conjugate base.
Resonance produced conjugate base.
As per resonance stability order of conjugate base is
ClO4−>ClO3−>ClO2−>ClO−ClO4−>ClO3−>ClO2−>ClO−
Hence acidic strength order is
HClO4>HClO3>HClO2>HClOHClO4>HClO3>HClO2>HClO
42. The metal that cannot be obtained by electrolysis of an aqueous solution of its salts is
(1) Ag
(2) Ca
(3) Cu
(4) Cr
Answer (2)
Sol. On electrolysis only in case of Ca2+Ca2+ salt aqueous solution H2H2 gas discharge at Cathode.
Case of Cr
At cathode: Cr3++2e−⟶CrCr3++2e−⟶Cr
So, Cr is deposited.
Case of Ag
At cathode: Ag++e−⟶AgAg++e−⟶Ag
So, Ag is deposited.
Case of Cu
At cathode: Cu2++2e−⟶CuCu2++2e−⟶Cu
Case of Ca2+Ca2+
At cathode:H2O+e−⟶12H2+OH−At cathode:H2O+e−⟶21H2+OH−
43. The octahedral complex of a metal ion M3+M3+ with four monodentate ligands L1, L2, L3L1, L2, L3 and L4L4 absorb wavelengths in the region of red, green, yellow and blue, respectively. The increasing order of ligand strength of the four ligands is
L4
Answer (2)
Sol.
The energy of red light is less than that of violet light.
So energy order is
Red The complex absorbs lower energy light lower will be its strength. So order of ligand strength is
L1
44. Which one of the following properties is not shown by NO?
(1) It is diamagnetic in gaseous state
(2) It is a neutral oxide
(3) It combines with oxygen to form nitrogen dioxide
(4) It's bond order is 2.5
Answer (1)
Sol. Nitric oxide is paramagnetic in the gaseous state as it has one unpaired electron in its outermost shell. The electronic configuration of NO is
σ1s2σ1s2σ2s2σ2p2π2px2=π2py2π2px2σ1s2σ1s2σ2s2σ2p2π2px2=π2py2π2px2
However, it dimerises at low temperature to become diamagnetic.
2NO⇌N2O22NO⇌N2O2
Its bond order is 2.5 and it combines with O2O2 to give nitrogen dioxide.
45. In which of the following reactions H2O2H2O2 acts as a reducing agent?
H2O2+2H++2e−⟶2H2OH2O2+2H++2e−⟶2H2OH2O2−2e−⟶O2+2H+H2O2−2e−⟶O2+2H+H2O2+2e−⟶2OH−H2O2+2e−⟶2OH−H2O2+2OH−−2e−⟶O2+2H2OH2O2+2OH−−2e−⟶O2+2H2O
(1) (a), (b)
(2) (c), (d)
(3) (a), (c)
(4) (b), (d)
Answer (4)
Sol. The reducing agent oxidises itself.
H2O2−1+2H++2e−⟶2H2O−2H2O2−1+2H++2e−⟶2H2O−2H2O2−1−2e−⟶O2+2H+H2O2−1−2e−⟶O2+2H+H2O2−1+2e−⟶2OH−H2O2−1+2e−⟶2OH−H2O2−1+2OH−−2e−⟶O2+H2OH2O2−1+2OH−−2e−⟶O2+H2O
Note: Powers of 1O′1O′ are oxidation number of 1O′1O′ in the compound.
46. The correct statement for the molecule, CsI₃, is
(1) It is a covalent molecule
(2) It contains Cs⁺ and I₃⁻ ions
(3) It contains Cs³⁺ and I⁻ ions
(4) It contains Cs⁺, I⁻ and lattice I₂ molecule
Answer (2)
Sol. It contains Cs⁺ and I₃⁻ ions.
47. The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of their molecule is
(1) 1 : 4
(2) 7 : 32
(3) 1 : 8
(4) 3 : 16
Answer (2)
Sol. Let the mass of O₂ = x
Mass of N₂ = 4x
Number of moles of O₂ = x/32
Number of moles of N₂ = 4x/28 = x/7
∴ Ratio = x/32 : x/7 = 7 : 32
48. Given below are the half-cell reactions
Mn²⁺ + 2e⁻ → Mn; E° = − 1.18 V
(Mn³⁺ + e⁻ → Mn²⁺); E° = + 1.51 V
The E° for 3 Mn²⁺ → Mn + 2Mn³⁺ will be
(1) −2.69 V; the reaction will not occur
(2) −2.69 V; the reaction will occur
(3) −0.33 V; the reaction will not occur
(4) −0.33 V; the reaction will occur
Answer (1)
Sol. (1) Mn²⁺ + 2e → Mn; E° = −1.18V;
ΔG1∘=−2F(−1.18)=2.36FΔG1∘=−2F(−1.18)=2.36F
(2) Mn³⁺ + e → Mn²⁺; E° = +1.51V;
ΔG2∘=−F(1.51)=−1.51FΔG2∘=−F(1.51)=−1.51F
(1) − 2 × (2)
3Mn²⁺ → Mn + 2Mn³⁺;
ΔG3∘=ΔG1∘−2ΔG2∘ΔG3∘=ΔG1∘−2ΔG2∘
= [2.36 − 2(−1.51)] F
= (2.36 + 3.02) F
= 5.38 F
But ΔG3∘=−2FE∘ΔG3∘=−2FE∘
⇒ 5.38F = −2FE°
⇒ E° = −2.69 V
As E° value is negative reaction is non spontaneous.
49. Which series of reactions correctly represents chemical reactions related to iron and its compound?
(1) Fe →dil. H2SO4dil. H2SO4 FeSO₄ →H2SO4,O2H2SO4,O2 Fe₂(SO₄)₃ →heatheat Fe
(2) Fe →O2,heatO2,heat FeO →dil. H2SO4dil. H2SO4 FeSO₄ →heatheat Fe
(3) Fe →Cl2,heatCl2,heat FeCl₃ →heat, airheat, air FeCl₂ →ZnZn Fe
(4) Fe →O2,heatO2,heat Fe₃O₄ →CO,600∘CCO,600∘C FeO →CO,700∘CCO,700∘C Fe
Answer (4)
Sol. Fe →O2,HeatO2,Heat Fe₃O₄
This reaction is corresponding to the combustion of Fe.
FeO →CO,600∘CCO,600∘C FeO →CO,700∘CCO,700∘C Fe
These reactions correspond to the production of Fe by reduction of Fe₃O₄ in blast furnace.
50. The equation which is balanced and represents the correct product(s) is
(1) Li₂O + 2KCl → 2LiCl + K₂O
(2) [CoCl(NH₃)₅]⁺ + 5H⁺ → Co²⁺ + 5NH₄⁺ + Cl⁻
(3) [Mg(H₂O)₆]²⁺ + (EDTA)⁴⁻ →excess NaOHexcess NaOH [Mg(EDTA)]²⁺ + 6H₂O
(4) CuSO₄ + 4KCN → K₂[Cu(CN)₄] + K₂SO₄
Answer (2)
Sol. The complex [CoCl(NH₃)₅]⁺ decomposes under acidic medium, so
[CoCl(NH₃)₅]⁺ + 5H⁺ → Co²⁺ + 5NH₄⁺ + Cl⁻.
51. In Sₙ2 reactions, the correct order of reactivity for the following compounds CH₃Cl, CH₃CH₂Cl, (CH₃)₂CHCl and (CH₃)₃CCl is
(1) CH₃Cl > (CH₃)₂CHCl > CH₃CH₂Cl > (CH₃)₃CCl
(2) CH₃Cl > CH₃CH₂Cl > (CH₃)₂CHCl > (CH₃)₃CCl
(3) CH₃CH₂Cl > CH₃Cl > (CH₃)₂CHCl > (CH₃)₃CCl
(4) (CH₃)₂CHCl > CH₃CH₂Cl > CH₃Cl > (CH₃)₃CCl
52. On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, the organic compound formed is
(1) An alkanol
(2) An alkanediol
(3) An alkyl cyanide
(4) An alkyl isocyanide
Answer (4)
Sol. R−CH2−NH2→CHCl3/KOHC2H5OH→R−CH2−NCSol. R−CH2−NH2CHCl3/KOHC2H5OH→R−CH2−NC
53. The most suitable reagent for the conversion of R−CH2−OH→R−CHOR−CH2−OH→R−CHO is
(1) KMnO4KMnO4
(2) K2Cr2O7K2Cr2O7
(3) CrO3CrO3
(4) PCC (Pyridinium Chlorochromate)
Answer (4)
Sol. PCC is mild oxidising agent, it will convert R−CH2−OH⟶R−CHOR−CH2−OH⟶R−CHO
54. The major organic compound formed by the reaction of 1, 1, 1-trichloroethane with silver powder is
(1) Acetylene
(2) Ethene
(3) 2-Butyne
(4) 2-Butene
Answer (3)
Sol. 2Cl−C−CH3→AgCH3C≡CCH3+6AgClSol. 2Cl−C−CH3AgCH3C≡CCH3+6AgClClCl1,1,1−trichloroethane1,1,1−trichloroethane
55. Sodium phenoxide when heated with CO2CO2 under pressure at 125∘C125∘C yields a product which on acetylation produces C.
The major product C would be
Answer (1)
Sol.
[Reaction scheme as in PDF]
56. Considering the basic strength of amines in aqueous solution, which one has the smallest pKbpKb value?
(1) (CH3)2NH(CH3)2NH
(2) CH3NH2CH3NH2
(3) (CH3)3N(CH3)3N
(4) C6H5NH2C6H5NH2
Answer (1)
Sol. Among C6H5NH2C6H5NH2 CH3NH2CH3NH2 (CH3)2NH(CH3)2NH
(CH3)3N⋅C6H5NH2(CH3)3N⋅C6H5NH2 is least basic due to resonance.
Out of (CH3)3N(CH3)3N CH3NH2CH3NH2 (CH3)2NH(CH3)2NH (CH3)2NH(CH3)2NH is most basic due to +1+1 effect and hydrogen bonding in H2OH2O
57. For which of the following molecule significant μ≠0μ=0 ?
(a)
(b)
(c)
(d)
(1) Only (a)
(2) (a) and
(3) Only (c)
(4) (c) and
Answer (4)
Sol. (a) μ=0μ=0
(b) μ=0μ=0
(c) μ≠0μ=0
(d) μ≠0μ=0
58. Which one is classified as a condensation polymer?
(1) Dacron
(2) Neoprene
(3) Teflon
(4) Acrylonitrile
Answer (1)
Sol. Dacron is polyester formed by condensation polymerisation of terephthalic acid and ethylene glycol.
Acrylonitrile, Neoprene and Teflon are addition polymers of acrylonitrile, isoprene and tetrafluoro ethylene respectively.
59. Which one of the following bases is not present in DNA?
(1) Quinoline
(2) Adenine
(3) Cytosine
(4) Thymine
Answer (1)
Sol. DNA contains ATGC bases
A - Adenine T - Thymine G - Guanine C - Cytocine So quinoline is not present.
60. In the reaction, CH3COOH→LiAlH4A→PCl5B→AlC.KOHC,CH3COOHLiAlH4APCl5BAlC.KOHC, the product C is
(1) Acetaldehyde
(2) Acetylene
(3) Ethylene
(4) Acetyl chloride
Answer (3)
Sol. Ethylene
61. If X={4n−3n−1:n∈N}X={4n−3n−1:n∈N} and Y={9(n−1):n∈N}Y={9(n−1):n∈N} where NN is the set of natural numbers, then X∪YX∪Y is equal to
(1)X(2)Y(3)N(4)Y−X(4)(1)X(3)N(2)Y(4)Y−X(4)
Answer (2)
X={(1+3)n−3n−1,n∈N}=32(nC2+nC3.3+…+3n−2),n∈N}={Divisible by 9}Y={9(n−1),n∈N}={All multiples of 9}S0,X⊆Yi.e.,X∪Y=YX∪Y=Y(4)X={(1+3)n−3n−1,n∈N}=32(nC2+nC3.3+…+3n−2),n∈N}={Divisible by 9}Y={9(n−1),n∈N}={All multiples of 9}S0,X⊆Yi.e.,X∪Y=YX∪Y=Y(4)
62. If ZZ is a complex number such that ∣z∣≥2∣z∣≥2 then the minimum value of ∣z+12∣z+21
(1) Is strictly greater than 5225
(2) Is strictly greater than 3223 but less than 5225
(3) Is equal to 5225
(4) Lies in the interval (1, 2)
Answer (4)
Sol.
So, ∣z+12∣≥∣∣z∣−12∣z+21≥∣z∣−21
⇒ ∣z+12∣≥∣2−12∣z+21≥2−21
⇒ ∣zmin.=32∣∣zmin.=23∣
63. If a∈Ra∈R and the equation
−3(x−[x])2+2(x−[x])+a2=0−3(x−[x])2+2(x−[x])+a2=0
(where [x][x] denotes the greatest integer ≤x≤x ) has no integral solution, then all possible values of aa lie in the interval
(1)(−2,−1)(1)(−2,−1)(2)(−∞,−2)∪(2,∞)(2)(−∞,−2)∪(2,∞)(3)(−1,0)∪(0,1)(3)(−1,0)∪(0,1)(4)(1,2)(4)(1,2)
Answer (3)
−3(x−[x])2+2[x−[x])+a2=0−3(x−[x])2+2[x−[x])+a2=03[x]2−2[x]−a2=03[x]2−2[x]−a2=0a≠0,3([x]2−23[x])=a2a=0,3([x]2−32[x])=a2a2=3([x]−13)2−13a2=3([x]−31)2−310≤[x]<1and−13≤[x]−13<230≤[x]<1and−31≤[x]−31<320≤3([x]−13)2<430≤3([x]−31)2<34−13≤3([x]−13)2−13<1−31≤3([x]−31)2−31<10
For non- integral solution
0
Alternative
−3[x]2+2[x]+a2=0−3[x]2+2[x]+a2=0
Now, −3[x]2+2[x]−3[x]2+2[x]
to have no integral roots 0
∴a∈(−1,0)∪(0,1)∴a∈(−1,0)∪(0,1)
64. Let αα and ββ be the roots of equation px2+qx+r=0,p≠0px2+qx+r=0,p=0. If p, q, r are in A.P. and 1α+1β=4α1+β1=4, then the value of ∣α−β∣∣α−β∣ is
(1) 349934
(2) 21399213
(3) 619961
(4) 21799217
Answer (2)
Sol.
∵ p, q, r are in AP
2q = p + r ...(i)
Also 1α+1β=4α1+β1=4
⇒ α+βαβ=4αβα+β=4
−qr=4⇒q=−4rr−q=4⇒q=−4r ...(ii)
From (i)
2(−4r) = p + r
p = − 9r
q = − 4r
r = r
Now ∣α−β∣=(α+β)2−4αβ∣α−β∣=(α+β)2−4αβ
= (−qp)2−4rp(p−q)2−p4r
= q2−4pr∣p∣∣p∣q2−4pr
= 16r2+36r2∣−9r∣∣−9r∣16r2+36r2
= 21399213
65. If α,β≠0α,β=0, and f(n)=αn+βnf(n)=αn+βn and
∣31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)∣=K(1−α)2(1−β)2(α−β)2, then K is equal to31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)=K(1−α)2(1−β)2(α−β)2, then K is equal to
(1) 1
(2) −1
(3) αβ
(4) 1αβαβ1
Answer (1)
Sol.
∣1+1+11+α+β1+α2+β21+α+β1+α2+β21+α3+β31+α2+β21+α3+β31+α4+β4∣=∣1111αβ1α2β2∣×∣1111αα21ββ2∣=[(1−α)(1−β)(1−β)]21+1+11+α+β1+α2+β21+α+β1+α2+β21+α3+β31+α2+β21+α3+β31+α4+β4=1111αα21ββ2×1111αβ1α2β2=[(1−α)(1−β)(1−β)]2
So, K=1K=1
66. If A is an 3 × 3 non-singular matrix such that AA′=A′AAA′=A′A and B=A−1A′B=A−1A′, then BB' equals
(1) B−1B−1
(2) (B−1)′(B−1)′
(3) I+BI+B
(4) II
Answer (4)
Sol. BB′=(A−1A′)(A−1A′)′BB′=(A−1A′)(A−1A′)′
= A−1.A.A′.(A−1)′A−1.A.A′.(A−1)′ {as AA′=A′AAA′=A′A}
= I(A−1A)′I(A−1A)′
= I.I=I2=II.I=I2=I
67. If the coefficients of x3x3 and x4x4 in the expansion of (1+ax+bx2)(1−2x)18(1+ax+bx2)(1−2x)18 in powers of x are both zero, then (a, b) is equal to
(1) (14,2723)(14,3272)
(2) (16,2723)(16,3272)
(3) (16,2513)(16,3251)
(4) (14,2513)(14,3251)
Answer (2)
Sol. (1+ax+bx2)(1−2x)18(1+ax+bx2)(1−2x)18
(1+ax+bx2)[18C0−18C1(2x)+18C2(2x)2−18C3(2x)3+18C4(2x)4−......](1+ax+bx2)[18C0−18C1(2x)+18C2(2x)2−18C3(2x)3+18C4(2x)4−......]
Coeff. of x3=−18C38+a×4.18C2−2b×18=0x3=−18C38+a×4.18C2−2b×18=0
= −18×17×1668+4a+18×172−36b=0−618×17×168+24a+18×17−36b=0
= −51×16×8+a×36×17−36b=0−51×16×8+a×36×17−36b=0
= −34×16+51a−3b=0−34×16+51a−3b=0
= 51a−3b=34×16=54451a−3b=34×16=544
= 51a−3b=54451a−3b=544 ...(i)
Only option number (2) satisfies the equation number (i).
68. If (10)9+2(11)1(10)8+3(11)2(10)7+…+10(11)9=k(10)9(10)9+2(11)1(10)8+3(11)2(10)7+…+10(11)9=k(10)9 , then kk is equal to
(1)100(2)110(1)100(2)110(3)12110(4)441100(3)10121(4)100441
Answer (1)
Sol.109+2⋅(11)(10)8+3(11)2(10)7+…+10(11)9=k(10)9Sol.109+2⋅(11)(10)8+3(11)2(10)7+…+10(11)9=k(10)9x=109+2⋅(11)(10)8+3(11)2(10)7+…+10(11)9x=109+2⋅(11)(10)8+3(11)2(10)7+…+10(11)91110x=11⋅108+2⋅(11)2⋅(10)7+…+9(11)9+11101011x=11⋅108+2⋅(11)2⋅(10)7+…+9(11)9+1110x(1−1110)10=109+11(10)8+112x(10)7+…+119−1110⇒x10=109((1110)10−11110−1)−1110⇒x10=x10=(1110−1010)−1110=−1010⇒x=1011=k⋅109⇒k=100(1)10x(1−1011)=109+11(10)8+112x(10)7+…+119−1110⇒10x=109(1011−1(1011)10−1)−1110⇒10x=10x=(1110−1010)−1110=−1010⇒x=1011=k⋅109⇒k=100(1)
69. Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. Then the common ratio of the G.P. is
(1)2−3(2)2+3(1)2−3(2)2+3(3)2+3(4)3+2(3)2+3(4)3+2
Answer (2)
Sol. a,ar,ar2→G.P.a,ar,ar2→G.P.
a,2ar,ar2→A.P.a,2ar,ar2→A.P.
2×2ar=a+ar22×2ar=a+ar2
4r=1+r24r=1+r2
⇒r2−4r+1=0⇒r2−4r+1=0
r=4±16−42=2±3r=24±16−4=2±3
r=2+3r=2+3
r=2−3 is rejected r=2−3 is rejected
∴(r>1)∴(r>1)
G.P. is increasing.
70. limx→0sin(πcos2x)x2 is equal to limx→0x2sin(πcos2x) is equal to
(1) −π (2) π (1) −π (2) π (3) π2 (4) 1 (3) 2π (4) 1
Answer (2)
Sol.limx→0sin(πcos2x)x2Sol.x→0limx2sin(πcos2x)=limx→0sin(π(1−sin2x)x2=x→0limx2sin(π(1−sin2x)=limx→0sin(π−πsin2x)x2=x→0limsinx2(π−πsin2x)=limx→0sin(πsin2x)x2[∵sin(π−θ)=sinθ]=x→0limx2sin(πsin2x)[∵sin(π−θ)=sinθ]=limx→0sin(πsin2x)(πsin2x)×πsin2xx2=x→0limsin(πsin2x)(πsin2x)×x2πsin2x=limx→01×π(sinxx)2=π=x→0lim1×π(xsinx)2=π
71. If gg is the inverse of a function ff and f′(x)=11+x5f′(x)=1+x51 then g′(x)g′(x) is equal to
(1)11+{g(x)}5(2)1+{g(x)}5(1)1+{g(x)}51(2)1+{g(x)}5(3)1+x5(4)5x4(3)1+x5(4)5x4
Answer (2)
Sol.f′(x)=11+x5=f(g(x))=x→f′(g(x))g′(x)=1Sol.f′(x)=1+x51=f(g(x))=x→f′(g(x))g′(x)=1g′(x)=1f′(g(x))=1+(g(x))5g′(x)=f′(g(x))1=1+(g(x))5
72. If ff and gg are differentiable functions in [0, 1] satisfying f(0)=2=g(1)f(0)=2=g(1) , g(0)=0g(0)=0 and f(1)=6f(1)=6 , then for some c∈[0,1]c∈[0,1]
(1)f′(c)=g′(c)(2)f′(c)=2g′(c)(1)f′(c)=g′(c)(2)f′(c)=2g′(c)(3)2f′(c)=g′(c)(4)2f′(c)=3g′(c)(3)2f′(c)=g′(c)(4)2f′(c)=3g′(c)
Answer (2)
Sol. Using, mean value theorem
f′(c)=f(1)−f(0)1−0=4f′(c)=1−0f(1)−f(0)=4g′(c)=g(1)−g(0)1−0=2g′(c)=1−0g(1)−g(0)=2so,[f′(c)=2g′(c)]so,[f′(c)=2g′(c)]so,[f′(c)=2g′(c)]so,[f′(c)=2g′(c)]
73. If x=−1x=−1 and x=2x=2 are extreme points of f(x)=αlog∣x∣+βx2+xf(x)=αlog∣x∣+βx2+x then
(1) α=2,β=−12α=2,β=−21
(2) α=2,β=12α=2,β=21
(3) α=−6,β=12α=−6,β=21
(4) α=−6,β=−12α=−6,β=−21
Answer (1)
Sol. f(x)=αlog∣x∣+βx2+xf(x)=αlog∣x∣+βx2+x
f′(x)=αx+2βx+1=0f′(x)=xα+2βx+1=0 at x=−1,2x=−1,2
∴ −α−2β+1=0⇒α+2β=1−α−2β+1=0⇒α+2β=1 ...(i)
α2+4β+1=0⇒α+8β=−22α+4β+1=0⇒α+8β=−2 ...(ii)
6β=−3⇒β=−126β=−3⇒β=−21
∴ α=2α=2
74. The integral ∫(1+x−1x)ex+1xdx∫(1+x−x1)ex+x1dx is equal to
(1) (x+1)ex+1x+c(x+1)ex+x1+c
(2) −xex+1x+c−xex+x1+c
(3) (x−1)ex+1x+c(x−1)ex+x1+c
(4) xex+1x+cxex+x1+c
Answer (4)
Sol. I=∫{e(x+1x)+x(1−1x2)ex+1x}dxI=∫{e(x+x1)+x(1−x21)ex+x1}dx
= x.ex+1x+cx.ex+x1+c
As ∫(xf′(x)+f(x))dx=xf(x)+c∫(xf′(x)+f(x))dx=xf(x)+c
75. The integral ∫0π1+4sin2x2−4sinx2dx∫0π1+4sin22x−4sin2xdx equals
(1) 43−443−4
(2) 43−4−π343−4−3π
(3) π−4π−4
(4) 2π3−4−4332π−4−43
Answer (2)
Sol.
∫0π1+4sin2x2−4sinx2dx∫0π1+4sin22x−4sin2xdx
= ∫0π∣2sinx2−1∣dx∫0π2sin2x−1dx
[sinx2=12⇒x2=π6⇒x=π3;x2=5π6⇒x=5π3][sin2x=21⇒2x=6π⇒x=3π;2x=65π⇒x=35π]
= ∫0π/3(1−2sinx2)dx+∫π/3π(2sinx2−1)dx∫0π/3(1−2sin2x)dx+∫π/3π(2sin2x−1)dx
= [x+4cosx2]0π/3+[−4cosx2−x]π/3π[x+4cos2x]0π/3+[−4cos2x−x]π/3π
= π3+432−4+(0−π+432+π3)3π+423−4+(0−π+423+3π)
= 43−4−π343−4−3π
76. The area of the region described by A={(x,y):x2+y2≤1 and y2≤1−x}A={(x,y):x2+y2≤1 and y2≤1−x} is
(1) π2+232π+32
(2) π2+432π+34
(3) π2+432π+34
(4) π2−432π−34
Answer (3)
Sol.
Shaded area
= π(1)22+2∫011−xdx2π(1)2+2∫011−xdx
= π2+2(1−x)3/23/2(−1)∣012π+3/22(1−x)3/2(−1)01
= π2+43(0−(−1))2π+34(0−(−1))
= π2+432π+34
77. Let the population of rabbits surviving at a time tt be governed by the differential equation dp(t)dt=12p(t)−200.Ifp(0)=100,thenp(t)equalsdtdp(t)=21p(t)−200.Ifp(0)=100,thenp(t)equals
(1) 600−500et/2600−500et/2
(2) 400−300et/2400−300et/2
(3) 400&minu
4) 300−200et/2300−200et/2
Answer (3)
Sol.dp(t)dt=12p(t)−200Sol.dtdp(t)=21p(t)−200
⋅{d(p(t))2p(t)−200}=∫ttdt⋅{2d(p(t))p(t)−200}=∫ttdt
\cdot \left.\frac{1}{2}\log \left(\frac{p(t)}{2} -200\right) = t + c
⋅p(t)2−200=t22k⋅2p(t)−200=2t2k
Using given condition p(t)=400−300et/2p(t)=400−300et/2
78. Let PSPS be the median of the triangle with vertices
