Class 12 Science – GenAI Prompt Engineering Academy

Maths

CBSE Class 12 Mathematics: Continuity and Differentiability

50 MCQs with Answer Key

Questions

A function f(x)f(x) is continuous at x=ax=a if:
a) lim⁡x→a−f(x)limx→a−f(x) exists
b) lim⁡x→a+f(x)limx→a+f(x) exists
c) lim⁡x→af(x)=f(a)limx→af(x)=f(a)
d) f(a)f(a) is defined
The function f(x)=∣x−3∣f(x)=∣x−3∣ is not differentiable at:
a) x=0x=0
b) x=1x=1
c) x=3x=3
d) x=5x=5
If f(x)=x2sin⁡(1x)f(x)=x2sin(x1) for x≠0x=0 and f(0)=0f(0)=0, then f(x)f(x) at x=0x=0 is:
a) Continuous but not differentiable
b) Differentiable but not continuous
c) Both continuous and differentiable
d) Neither continuous nor differentiable
The derivative of log⁡(sec⁡x+tan⁡x)log(secx+tanx) w.r.t. xx is:
a) sec⁡xsecx
b) tan⁡xtanx
c) sec⁡2xsec2x
d) 1sec⁡x+tan⁡xsecx+tanx1
If y=exxy=exx, then dydxdxdy is:
a) exx⋅xx(1+ln⁡x)exx⋅xx(1+lnx)
b) exx⋅xxln⁡xexx⋅xxlnx
c) exx⋅xx−1exx⋅xx−1
d) exx⋅xxexx⋅xx
The function f(x)=x2−4x−2f(x)=x−2x2−4 at x=2x=2:
a) Has a removable discontinuity
b) Has an infinite discontinuity
c) Is continuous
d) Is differentiable
If y=tan⁡−1(1+x2−1x)y=tan−1(x1+x2−1), then dydx=dxdy=
a) 12(1+x2)2(1+x2)1
b) 11+x21+x21
c) 121+x221+x21
d) x1+x21+x2x
Rolle’s Theorem is applicable for f(x)=x2−4x+3f(x)=x2−4x+3 in:
a) [0,2][0,2]
b) [1,3][1,3]
c) [−1,1][−1,1]
d) [2,4][2,4]
The second derivative of ex3ex3 w.r.t. xx is:
a) ex3(9×4+6x)ex3(9x4+6x)
b) ex3(3×2)ex3(3x2)
c) ex3(6x+9×4)ex3(6x+9x4)
d) ex3(3×2+6x)ex3(3x2+6x)
If f(x)={x2sin⁡(1/x)x≠00x=0f(x)={x2sin(1/x)0x=0x=0, then f′(0)=f′(0)=
a) 1
b) 0
c) -1
d) Does not exist
The derivative of sin⁡−1(2×1−x2)sin−1(2x1−x2) w.r.t. sin⁡−1xsin−1x is:
a) 0
b) 1
c) 2
d) 1221
For the function f(x)=x2/3f(x)=x2/3, which statement is true?
a) Differentiable at x=0x=0
b) Continuous at x=0x=0 but not differentiable
c) Discontinuous at x=0x=0
d) f′(0)=1f′(0)=1
If y=ln⁡∣x−1x+1∣y=lnx+1x−1, then dydx=dxdy=
a) 1x−1−1x+1x−11−x+11
b) 2×2−1x2−12
c) 1(x−1)2−1(x+1)2(x−1)21−(x+1)21
d) −2×2−1x2−1−2
The number of points where f(x)=∣x∣+∣x−1∣f(x)=∣x∣+∣x−1∣ is not differentiable in [0,2][0,2] is:
a) 0
b) 1
c) 2
d) 3
If x=a(θ−sin⁡θ)x=a(θ−sinθ), y=a(1−cos⁡θ)y=a(1−cosθ), then d2ydx2dx2d2y at θ=π2θ=2π is:
a) 1aa1
b) −1a−a1
c) 14a4a1
d) −14a−4a1
The derivative of sec⁡−1(12×2−1)sec−1(2x2−11) w.r.t. 1−x21−x2 at x=12x=21 is:
a) 0
b) 1
c) -1
d) 2
If y=tan⁡−1(3x−x31−3×2)y=tan−1(1−3x23x−x3), then dydx=dxdy=
a) 31+x21+x23
b) 31+9×21+9x23
c) 11+9×21+9x21
d) 31−3×21−3x23
The function f(x)=[x]f(x)=[x] (greatest integer function) is discontinuous at:
a) Only integers
b) Only non-integers
c) All real numbers
d) x=0x=0
If f(x)=∣x−1∣+∣x−3∣f(x)=∣x−1∣+∣x−3∣, then f′(x)f′(x) for x∈(1,3)x∈(1,3) is:
a) -2
b) 0
c) 2
d) Undefined
The derivative of xx+xa+ax+aaxx+xa+ax+aa w.r.t. xx is:
a) xx(1+ln⁡x)+axa−1+axln⁡axx(1+lnx)+axa−1+axlna
b) xx(1+ln⁡x)+xaln⁡x+axln⁡axx(1+lnx)+xalnx+axlna
c) xx(1+ln⁡x)+axa−1+axln⁡a+aaln⁡axx(1+lnx)+axa−1+axlna+aalna
d) xx(1+ln⁡x)+axa−1+axln⁡axx(1+lnx)+axa−1+axlna
If y=sin⁡(xy)y=sin(xy), then dydx=dxdy=
a) ysin⁡(xy)ln⁡xx(1−xycos⁡(xy)ln⁡x)x(1−xycos(xy)lnx)ysin(xy)lnx
b) y2cos⁡(xy)x(1−ysin⁡(xy)ln⁡x)x(1−ysin(xy)lnx)y2cos(xy)
c) ycos⁡(xy)x(1−ysin⁡(xy)ln⁡x)x(1−ysin(xy)lnx)ycos(xy)
d) ysin⁡(xy)ln⁡xx(1−ycos⁡(xy)ln⁡x)x(1−ycos(xy)lnx)ysin(xy)lnx
The function f(x)=x2−1x−1f(x)=x−1x2−1 at x=1x=1 has:
a) Removable discontinuity
b) Jump discontinuity
c) Infinite discontinuity
d) Continuity
If f(x)=x∣x∣f(x)=x∣x∣, then f′(x)f′(x) for x>0x>0 is:
a) 2×2x
b) −2x−2x
c) 00
d) xx
The derivative of cos⁡−1(1−x21+x2)cos−1(1+x21−x2) w.r.t. tan⁡−1xtan−1x is:
a) 1
b) -1
c) 2
d) -2
The value of cc in Rolle’s Theorem for f(x)=x3−6×2+11x−6f(x)=x3−6x2+11x−6 on [1,3][1,3] is:
a) 113311
b) 5225
c) 7337
d) 8338
If y=ln⁡(ln⁡(ln⁡x))y=ln(ln(lnx)), then dydx=dxdy=
a) 1xln⁡xln⁡(ln⁡x)xlnxln(lnx)1
b) 1ln⁡(ln⁡x)ln(lnx)1
c) 1xln⁡xxlnx1
d) 1ln⁡xlnx1
The function f(x)={sin⁡5xxx≠0kx=0f(x)={xsin5xkx=0x=0 is continuous at x=0x=0 if k=k=
a) 0
b) 5
c) 1
d) 1551
If f(x)=x3f(x)=x3, g(x)=sin⁡2xg(x)=sin2x, then ddx[f(g(x))]=dxd[f(g(x))]=
a) 3sin⁡22x⋅cos⁡2x3sin22x⋅cos2x
b) 3sin⁡22x⋅2cos⁡2x3sin22x⋅2cos2x
c) 3sin⁡22x3sin22x
d) 6sin⁡22xcos⁡2x6sin22xcos2x
The derivative of sec⁡(tan⁡x)sec(tanx) w.r.t. xx is:
a) sec⁡(tan⁡x)tan⁡(tan⁡x)⋅sec⁡2x⋅12xsec(tanx)tan(tanx)⋅sec2x⋅2x1
b) sec⁡(tan⁡x)tan⁡(tan⁡x)⋅sec⁡2xsec(tanx)tan(tanx)⋅sec2x
c) sec⁡(tan⁡x)⋅12xsec(tanx)⋅2x1
d) sec⁡(tan⁡x)tan⁡(tan⁡x)⋅12xsec(tanx)tan(tanx)⋅2x1
If y=eacos⁡−1xy=eacos−1x, then (1−x2)y2−xy1=(1−x2)y2−xy1= (where y1,y2y1,y2 are derivatives)
a) a2ya2y
b) −a2y−a2y
c) ayay
d) −ay−ay
The function f(x)=1x−2f(x)=x−21 has:
a) Discontinuity at x=2x=2 only
b) Differentiability at x=2x=2
c) Continuity for all xx
d) Removable discontinuity at x=2x=2
If y=sin⁡x+sin⁡x+sin⁡x+⋯∞y=sinx+sinx+sinx+⋯∞, then dydx=dxdy=
a) cos⁡x2y−12y−1cosx
b) sin⁡x2y−12y−1sinx
c) cos⁡x2y+12y+1cosx
d) sin⁡x2y+12y+1sinx
The left-hand derivative of f(x)=cos⁡∣x∣f(x)=cos∣x∣ at x=0x=0 is:
a) 0
b) 1
c) -1
d) Does not exist
If xy=ex−yxy=ex−y, then dydx=dxdy=
a) 1+ln⁡x(1+ln⁡x)2(1+lnx)21+lnx
b) ln⁡x(1+ln⁡x)2(1+lnx)2lnx
c) 1(1+ln⁡x)2(1+lnx)21
d) ln⁡x1+ln⁡x1+lnxlnx
The derivative of log⁡x25logx25 w.r.t. xx is:
a) −ln⁡5x(ln⁡x2)2x(lnx2)2−ln5
b) −ln⁡5xln⁡xxlnx−ln5
c) −ln⁡52x(ln⁡x)22x(lnx)2−ln5
d) −ln⁡52xln⁡x2xlnx−ln5
The function f(x)=x3−3×2+3x−1f(x)=x3−3x2+3x−1 is:
a) Strictly increasing on RR
b) Strictly decreasing on RR
c) Increasing in (−∞,1)(−∞,1) and decreasing in (1,∞)(1,∞)
d) Decreasing in (−∞,1)(−∞,1) and increasing in (1,∞)(1,∞)
If y=sin⁡−1(2×1+x2)y=sin−1(1+x22x), then dydx=dxdy=
a) 21+x21+x22 for ∣x∣>1∣x∣>1
b) 21+x21+x22 for ∣x∣<1∣x∣<1
c) −21+x21+x2−2 for ∣x∣<1∣x∣<1
d) 21−x21−x22 for ∣x∣<1∣x∣<1
The derivative of tan⁡−1(sin⁡x+cos⁡xcos⁡x−sin⁡x)tan−1(cosx−sinxsinx+cosx) w.r.t. xx is:
a) 1
b) -1
c) 1221
d) 0
If f(x)=x2+2x+3f(x)=x2+2x+3 and g(x)=f(f(x))g(x)=f(f(x)), then g′(1)=g′(1)=
a) 20
b) 40
c) 60
d) 80
The function f(x)=sin⁡x−cos⁡xf(x)=sinx−cosx is decreasing in:
a) (0,π2)(0,2π)
b) (π2,π)(2π,π)
c) (π,3π2)(π,23π)
d) (3π2,2π)(23π,2π)
If y=cos⁡(2tan⁡−11−x1+x)y=cos(2tan−11+x1−x), then dydx=dxdy=
a) 11−x21−x21
b) −11−x21−x2−1
c) 11+x21+x21
d) −11+x21+x2−1
The second derivative of log⁡(log⁡x)log(logx) w.r.t. xx is:
a) −1x2ln⁡x−1×2(ln⁡x)2x2lnx−1−x2(lnx)21
b) −1x2ln⁡xx2lnx−1
c) 1×2(ln⁡x)2x2(lnx)21
d) −1×2(1ln⁡x+1(ln⁡x)2)x2−1(lnx1+(lnx)21)
The set of points where f(x)=x∣x∣f(x)=x∣x∣ is differentiable is:
a) R−{0}R−{0}
b) RR
c) (0,∞)(0,∞)
d) (−∞,0)(−∞,0)
If f′(x)=g(x)f′(x)=g(x) and g′(x)=−f(x)g′(x)=−f(x) for all xx, and f(0)=0f(0)=0, g(0)=1g(0)=1, then f2(x)+g2(x)=f2(x)+g2(x)=
a) 0
b) 1
c) xx
d) 2×2x
The derivative of xsin⁡x+(sin⁡x)xxsinx+(sinx)x w.r.t. xx is:
a) xsin⁡x(sin⁡xx+cos⁡xln⁡x)+(sin⁡x)x(cot⁡x+ln⁡(sin⁡x))xsinx(xsinx+cosxlnx)+(sinx)x(cotx+ln(sinx))
b) xsin⁡x(cos⁡xln⁡x+sin⁡xx)+(sin⁡x)x(xcot⁡x+ln⁡(sin⁡x))xsinx(cosxlnx+xsinx)+(sinx)x(xcotx+ln(sinx))
c) xsin⁡xcos⁡xln⁡x+(sin⁡x)xln⁡(sin⁡x)xsinxcosxlnx+(sinx)xln(sinx)
d) xsin⁡x(sin⁡xx+cos⁡xln⁡x)+(sin⁡x)x(cot⁡x+xsin⁡xcos⁡x)xsinx(xsinx+cosxlnx)+(sinx)x(cotx+sinxxcosx)
The function f(x)={x2−1x≤14x−4x>1f(x)={x2−14x−4x≤1x>1 at x=1x=1 is:
a) Continuous and differentiable
b) Continuous but not differentiable
c) Discontinuous
d) Differentiable but not continuous
If y=tan⁡−1(a+btan⁡xb−atan⁡x)y=tan−1(b−atanxa+btanx), then dydx=dxdy=
a) abba
b) baab
c) 1
d) 0
The derivative of tan⁡xtanx w.r.t. xx is:
a) sec⁡2x4xtan⁡x4xtanxsec2x
b) sec⁡2x2xtan⁡x2xtanxsec2x
c) sec⁡2x4xtan⁡x4xtanxsec2x
d) sec⁡xtan⁡x4xtan⁡x4xtanxsecxtanx
If f(x)=kcos⁡xπ−2xf(x)=π−2xkcosx for x≠π2x=2π, f(π2)=3f(2π)=3, and ff is continuous at x=π2x=2π, then k=k=
a) 6
b) 3
c) 12
d) 9
The value of ddx(sin⁡−12×1+x2)dxd(sin−11+x22x) at x=12x=21 is:
a) 4554
b) 3553
c) 8558
d) 6556

Answer Key

*All questions are aligned with NCERT Class 12 Mathematics (Chapter 5: Continuity and Differentiability) and CBSE exam patterns. Concepts covered include continuity, differentiability, derivatives of various functions, chain rule, parametric derivatives, implicit differentiation, logarithmic differentiation, and second-order derivatives.*

Physics

Class: 12
Subject: Physics
Unit: 2 – Current Electricity
NCERT Reference: Chapters 3 (Current Electricity)

Multiple Choice Questions (MCQs)

The SI unit of electric current is:
a) Coulomb (C)
b) Volt (V)
c) Ampere (A)
d) Ohm (Ω)
The net charge flowing through any cross-section of a conductor in 10 seconds is 50 C. The current is:
a) 0.2 A
b) 5 A
c) 50 A
d) 500 A
Drift velocity (vdvd) of electrons in a conductor is related to the electric field (EE) applied as:
a) vd∝Evd∝E
b) vd∝1Evd∝E1
c) vd∝E2vd∝E2
d) vdvd is independent of EE
The drift velocity of electrons in a metallic conductor is of the order of:
a) 108108 m/s
b) 104104 m/s
c) 10−210−2 m/s
d) 10−410−4 m/s
Ohm’s law states the relationship between:
a) Charge and Time
b) Current and Potential Difference
c) Resistance and Length
d) Power and Energy
A conductor obeys Ohm’s law when:
a) Its resistance changes with temperature
b) The V-I graph is a straight line passing through the origin
c) The current is directly proportional to resistance
d) It is made of semiconductor material
The resistance (RR) of a uniform conductor depends on its length (ll), cross-sectional area (AA), and resistivity (ρρ) as:
a) R=ρlAR=ρAl
b) R=ρAlR=ρlA
c) R=ρlAR=Aρl
d) R=ρAlR=lρA
The SI unit of resistivity (ρρ) is:
a) Ohm (Ω)
b) Ohm-meter (Ω m)
c) Ohm per meter (Ω/m)
d) Siemens (S)
Resistivity of a material is a measure of its:
a) Ability to conduct electric current
b) Opposition to the flow of electric current
c) Internal resistance per unit length
d) Intrinsic property independent of dimensions
Which of the following has the highest resistivity at room temperature?
a) Copper
b) Silicon
c) Glass
d) Silver
The resistance of a metallic wire increases with temperature primarily because:
a) Length of the wire increases
b) Area of cross-section decreases
c) Relaxation time of electrons decreases
d) Number density of electrons increases
The temperature coefficient of resistance (αα) for a conductor is defined as:
a) α=Rt−R0R0tα=R0tRt−R0
b) α=Rt−R0R0×1tα=R0Rt−R0×t1
c) α=R0−RtR0tα=R0tR0−Rt
d) α=Rt−R0R0×1(T−T0)α=R0Rt−R0×(T−T0)1
Two resistors R1R1 and R2R2 are connected in series. Their equivalent resistance is:
a) Always less than R1R1 or R2R2
b) Always greater than R1R1 or R2R2
c) Equal to R1+R2R1+R2
d) Equal to 1R1+1R2R11+R21
Two resistors of 2 Ω and 4 Ω are connected in parallel. Their equivalent resistance is:
a) 4334 Ω
b) 3443 Ω
c) 6 Ω
d) 8 Ω
Kirchhoff’s first rule (Junction rule) is based on the conservation of:
a) Energy
b) Charge
c) Momentum
d) Current
Kirchhoff’s second rule (Loop rule) is based on the conservation of:
a) Energy
b) Charge
c) Momentum
d) Current
In the Wheatstone bridge circuit shown below, the bridge is balanced when:
(Assume standard arrangement: P, Q, R, S resistors)
a) PQ=RSQP=SR
b) PR=QSRP=SQ
c) P×Q=R×SP×Q=R×S
d) P×S=Q×RP×S=Q×R
The Meter Bridge works on the principle of:
a) Ohm’s Law
b) Kirchhoff’s Laws
c) Faraday’s Law
d) Wheatstone Bridge
In a Meter Bridge experiment, the null point is obtained at 40 cm from end A when resistance X is in the left gap and resistance Y is in the right gap. If X = 10 Ω, then Y is:
a) 10066100 Ω ≈ 16.67 Ω
b) 40101040 Ω = 4 Ω
c) 60101060 Ω = 6 Ω
d) 10×60404010×60 Ω = 15 Ω
The potentiometer is preferred over a voltmeter for measuring the emf of a cell because:
a) Potentiometer is more sensitive
b) Potentiometer draws no current from the cell
c) Voltmeter has a very high resistance
d) Potentiometer can measure larger emfs
The emf (E) of a cell is defined as:
a) The potential difference across its terminals when no current flows
b) The work done per unit charge to move charge through the cell
c) Both (a) and (b)
d) The potential difference across its terminals when current flows
The terminal potential difference (V) of a cell of emf (E) and internal resistance (r), when delivering a current (I), is given by:
a) V=EV=E
b) V=E+IrV=E+Ir
c) V=E−IrV=E−Ir
d) V=IrV=Ir
A cell of emf 1.5 V and internal resistance 0.5 Ω is connected to an external resistance of 2.5 Ω. The current in the circuit is:
a) 0.5 A
b) 1.0 A
c) 1.5 A
d) 3.0 A
For the circuit in Q23, the terminal voltage of the cell is:
a) 1.25 V
b) 1.5 V
c) 1.0 V
d) 1.75 V
Two identical cells, each of emf E and internal resistance r, are connected in series across an external resistance R. The current through R is:
a) 2E2r+R2r+R2E
b) 2Er+Rr+R2E
c) E2r+R2r+RE
d) Er+Rr+RE
Two identical cells, each of emf E and internal resistance r, are connected in parallel across an external resistance R. The current through R is:
a) 2E2r+R2r+R2E
b) 2Er+2Rr+2R2E
c) E2r+R2r+RE
d) Er+Rr+RE
The maximum current that can be drawn from a cell is:
a) Infinite
b) ErrE
c) ERRE
d) E×rE×r
The condition for maximum power transfer from a source (emf E, internal resistance r) to an external load resistance (R) is:
a) R = 0
b) R = r
c) R = 2r
d) R = r/2
The heating effect of electric current is due to:
a) Drift of electrons
b) Collisions of electrons with ions/atoms
c) Flow of protons
d) High electric field
The power dissipated as heat in a resistor R carrying a current I is given by:
a) I2RI2R
b) IR2IR2
c) V2/RV2/R
d) Both (a) and (c)
A 100 W, 220 V electric bulb is operated at 110 V. Its power consumption will be approximately:
a) 100 W
b) 50 W
c) 25 W
d) 12.5 W
The commercial unit of electrical energy is:
a) Joule (J)
b) Watt (W)
c) Kilowatt-hour (kWh)
d) Volt-Ampere (VA)
How much energy in kWh is consumed by a 2 kW electric heater used for 3 hours?
a) 6 kWh
b) 1.5 kWh
c) 0.67 kWh
d) 5 kWh
The colour code on a resistor is: Brown, Black, Red, Gold. Its resistance and tolerance are:
a) 10 Ω ± 5%
b) 100 Ω ± 5%
c) 1 kΩ ± 5%
d) 10 kΩ ± 5%
Mobility (μμ) of a charge carrier is defined as:
a) Drift velocity per unit electric field (μ=vd/Eμ=vd/E)
b) Charge per unit mass
c) Current density per unit electric field
d) Resistance per unit length
For a conductor carrying steady current, the current density (J) is:
a) Directly proportional to the electric field (E)
b) Inversely proportional to E
c) Independent of E
d) Proportional to E²
A wire of resistance 12 Ω is stretched to double its length. Assuming uniform cross-section and constant volume, its new resistance is:
a) 12 Ω
b) 24 Ω
c) 36 Ω
d) 48 Ω
The resistivity of a conductor depends on:
a) Length
b) Area of cross-section
c) Material and Temperature
d) Current
A carbon resistor and a copper wire of the same resistance at room temperature are heated to the same higher temperature. Which one will have higher resistance now?
a) Carbon resistor
b) Copper wire
c) Both will have the same resistance
d) Cannot be determined
In the circuit shown below (Two resistors R1 and R2 in series with a battery V), the potential difference across R1 is:
a) VR1R1+R2VR1+R2R1
b) VR2R1+R2VR1+R2R2
c) V(R1+R2)V(R1+R2)
d) V(R1−R2)V(R1−R2)
Three resistors, 2 Ω, 3 Ω, and 6 Ω, are connected in parallel. Their equivalent resistance is:
a) 1 Ω
b) 11 Ω
c) 0.917 Ω ≈ 1/1.09 Ω
d) 1.5 Ω
A 6 V battery is connected across a 3 Ω resistor. The charge passing through the resistor in 10 seconds is:
a) 2 C
b) 20 C
c) 60 C
d) 180 C
The internal resistance of a cell is determined using a potentiometer by:
a) Comparing EMFs of two cells
b) Finding the balancing length for the cell alone and then with a known resistance across it
c) Measuring the terminal voltage directly
d) Using Ohm’s law with a known external resistance
The potential gradient (kk) along a potentiometer wire of length LL and resistance RR, connected to a driver cell of emf EdEd, is:
a) k=EdLk=LEd
b) k=EdRLk=LEdR
c) k=EdRLk=RLEd
d) k=EdL×RR+rdk=LEd×R+rdR (where rdrd is driver cell internal resistance)
The resistivity of a semiconductor:
a) Increases with temperature
b) Decreases with temperature
c) Is independent of temperature
d) First increases then decreases with temperature
A battery is charged. During charging:
a) The terminal voltage is less than its EMF
b) The terminal voltage is equal to its EMF
c) The terminal voltage is greater than its EMF
d) Current flows from the positive to the negative terminal inside the battery
In the circuit shown, the equivalent resistance between points A and B is:
(Diagram Description: Two 6 Ω resistors in parallel, connected in series with a 3 Ω resistor.)
a) 3 Ω
b) 6 Ω
c) 9 Ω
d) 12 Ω
Kirchhoff’s loop rule applied to a closed loop in a DC circuit states that:
a) The sum of currents entering a junction is zero
b) The algebraic sum of changes in potential around any closed loop is zero
c) The sum of EMFs is equal to the sum of IR drops
d) Both (b) and (c)
The resistance of a superconductor below its critical temperature is:
a) Very high
b) Very low but finite
c) Exactly zero
d) Negative
The purpose of using thick copper strips in a meter bridge is to:
a) Increase the resistance of the bridge wire
b) Decrease the resistance of the connections
c) Increase the sensitivity
d) Make the bridge look stronger

Answer Key

c) Ampere (A)
b) 5 A (I = Q/t = 50 C / 10 s)
a) vd∝Evd∝E
d) 10−410−4 m/s
b) Current and Potential Difference
b) The V-I graph is a straight line passing through the origin
a) R=ρlAR=ρAl
b) Ohm-meter (Ω m)
d) Intrinsic property independent of dimensions
c) Glass
c) Relaxation time of electrons decreases
d) α=Rt−R0R0×1(T−T0)α=R0Rt−R0×(T−T0)1
c) Equal to R1+R2R1+R2
a) 4334 Ω (Req = (2*4)/(2+4) = 8/6 = 4/3 Ω)
b) Charge
a) Energy
d) P×S=Q×RP×S=Q×R (Standard Balance Condition: P/Q = R/S implies PS = QR)
d) Wheatstone Bridge
d) 10×60404010×60 Ω = 15 Ω (Balance: X/l = Y/(100-l) => 10/40 = Y/60 => Y = (10*60)/40)
b) Potentiometer draws no current from the cell
c) Both (a) and (b)
c) V=E−IrV=E−Ir
a) 0.5 A (I = E/(R+r) = 1.5/(2.5+0.5) = 1.5/3)
a) 1.25 V (V = E – Ir = 1.5 – (0.5 * 0.5) = 1.5 – 0.25)
a) 2E2r+R2r+R2E (Total EMF = 2E, Total Resistance = 2r + R)
d) Er+Rr+RE (Parallel Cells: E_eq = E, r_eq = r/2; I = E/(r/2 + R) = E/((r + 2R)/2) = 2E/(r + 2R))
b) ErrE (When R=0, short circuit)
b) R = r
b) Collisions of electrons with ions/atoms
d) Both (a) and (c) (P = I²R = V²/R)
c) 25 W (Resistance R = V²/P = 220²/100 = 484 Ω. At 110V, P’ = V’²/R = 110²/484 = 12100/484 ≈ 25 W)
c) Kilowatt-hour (kWh)
a) 6 kWh (Energy = Power * Time = 2 kW * 3 h)
c) 1 kΩ ± 5% (Brown=1, Black=0, Red=2 (multiplier), Gold=±5% => 10 * 10² Ω = 1000 Ω = 1kΩ)
a) Drift velocity per unit electric field (μ=vd/Eμ=vd/E)
a) Directly proportional to the electric field (E) (J = σE, σ constant for Ohmic material)
d) 48 Ω (R ∝ l², since Volume constant => l doubles, A halves => R_new = R_old * (2)² * (1/ (1/2)) wait: R ∝ l² => R_new / R_old = (l_new / l_old)² = 4 => R_new = 12 * 4)
c) Material and Temperature
a) Carbon resistor (Carbon has negative α, Copper has positive α)
a) VR1R1+R2VR1+R2R1 (Potential divider rule)
a) 1 Ω (1/Req = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 => Req=1Ω)
b) 20 C (I = V/R = 6/3 = 2A; Q = I*t = 2 * 10)
b) Finding the balancing length for the cell alone (L1) and then with a known resistance (R) across it (L2); r = R*(L1/L2 – 1)
d) k=EdL×RR+rdk=LEd×R+rdR (Potential drop per unit length = (E_d * R_wire / R_total) / L)
b) Decreases with temperature
c) The terminal voltage is greater than its EMF (V = E + I*r during charging)
b) 6 Ω (R_parallel = (6*6)/(6+6) = 36/12 = 3Ω; R_AB = 3Ω + 3Ω = 6Ω)
d) Both (b) and (c) (ΣΔV = 0 is fundamental; ΣEMF = ΣIR is a consequence for loops with EMFs and resistors only)
c) Exactly zero
b) Decrease the resistance of the connections (To make contact resistance negligible)

Chemistry

MCQs: Coordination Compounds (Class 12 CBSE)

Werner’s theory successfully explained:
a) Only the primary valency of central metal ions
b) Only the secondary valency of central metal ions
c) Both primary and secondary valencies of central metal ions
d) Only the colour of coordination compounds
Answer: c)
The primary valency of a metal ion in a coordination compound corresponds to its:
a) Coordination number
b) Oxidation state
c) Number of ligands attached
d) Charge on the complex ion
Answer: b)
In the coordination entity [Co(NH₃)₆]³⁺, the secondary valency of Co³⁺ is:
a) 3
b) 6
c) 0
d) 4
Answer: b)
Which of the following is NOT a coordination entity?
a) [Fe(CN)₆]⁴⁻
b) Ni(CO)₄
c) K₄[Fe(CN)₆]
d) [Co(NH₃)₆]Cl₃
Answer: c) (K₄[Fe(CN)₆] is the compound, [Fe(CN)₆]⁴⁻ is the coordination entity)
The oxidation state of chromium in [Cr(H₂O)₆]³⁺ is:
a) +6
b) +3
c) +2
d) 0
Answer: b)
The coordination number of the central metal ion in [Fe(C₂O₄)₃]³⁻ is:
a) 3
b) 4
c) 6
d) 2
Answer: c) (C₂O₄²⁻ is a bidentate ligand)
The IUPAC name of [Co(NH₃)₅Cl]Cl₂ is:
a) Pentaamminechlorocobalt(III) chloride
b) Pentaamminechlorocobalt(II) chloride
c) Chloropentaamminecobalt(III) chloride
d) Cobalt(III) pentaamminechloride chloride
Answer: c)
The IUPAC name of K₃[Fe(CN)₆] is:
a) Potassium hexacyanoferrate(II)
b) Potassium hexacyanoferrate(III)
c) Tripotassium hexacyanoiron(III)
d) Potassium ferricyanide
Answer: b)
The IUPAC name of [Ni(CO)₄] is:
a) Tetracarbonylnickelate(0)
b) Tetracarbonylnickel(II)
c) Tetracarbonylnickel(0)
d) Nickel tetracarbonyl
Answer: c)
Which ligand is unidentate?
a) Ethylenediamine (en)
b) Oxalate (ox)
c) Carbon monoxide (CO)
d) Ethylenediaminetetraacetate (EDTA)
Answer: c)
The ligand EDTA is:
a) Unidentate
b) Bidentate
c) Tridentate
d) Hexadentate
Answer: d)
Which of the following can act as a chelating ligand?
a) NH₃
b) Cl⁻
c) H₂NCH₂CH₂NH₂ (en)
d) CN⁻
Answer: c)
Geometrical isomerism is possible in:
a) [Pt(NH₃)₂Cl₂] (Square planar)
b) [Co(NH₃)₆]³⁺ (Octahedral)
c) [ZnCl₄]²⁻ (Tetrahedral)
d) [Ni(CO)₄] (Tetrahedral)
Answer: a)
The complex [CoCl₂(en)₂]⁺ exhibits:
a) Optical isomerism only
b) Geometrical isomerism only
c) Both geometrical and optical isomerism
d) Linkage isomerism
Answer: c) (Geometrical cis/trans; Optical isomers possible for cis form)
Which type of isomerism is shown by [Co(NH₃)₅(NO₂)]Cl₂ and [Co(NH₃)₅(ONO)]Cl₂?
a) Ionisation isomerism
b) Linkage isomerism
c) Coordination isomerism
d) Geometrical isomerism
Answer: b)
Ionisation isomers have:
a) Same molecular formula but different ligands
b) Same molecular formula but give different ions in solution
c) Different molecular formulas
d) The same central metal oxidation state but different coordination numbers
Answer: b)
An example of coordination isomerism is:
a) [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br
b) [Cr(H₂O)₆]Cl₃ and [Cr(H₂O)₅Cl]Cl₂·H₂O
c) [Co(NH₃)₅NO₂]Cl₂ and [Co(NH₃)₅ONO]Cl₂
d) [Co(en)₃][Cr(ox)₃] and [Cr(en)₃][Co(ox)₃]
Answer: d)
Valence Bond Theory explains the magnetic properties of coordination compounds based on:
a) Crystal field splitting
b) Hybridization and presence of unpaired electrons
c) Ligand field strength
d) Electron pairing energy
Answer: b)
In an octahedral complex, d²sp³ hybridization results in:
a) Paramagnetic complex
b) Diamagnetic complex
c) Either paramagnetic or diamagnetic depending on ligands
d) Always high spin complex
Answer: b) (Involves pairing of electrons)
The complex [FeF₆]³⁻ is paramagnetic. The hybridisation of Fe³⁺ in this complex is:
a) d²sp³
b) sp³d²
c) dsp²
d) sp³
Answer: b) (Weak field ligand, outer orbital complex, high spin, sp³d² hybridization)
Crystal Field Theory explains the colour of coordination compounds due to:
a) Charge transfer
b) d-d transitions
c) σ-bond formation
d) π-bond formation
Answer: b)
In an octahedral complex, the t₂g orbitals are:
a) Lower in energy than eg orbitals
b) Higher in energy than eg orbitals
c) Degenerate with eg orbitals
d) Not involved in splitting
Answer: a)
The magnitude of crystal field splitting energy (Δ₀) is maximum for which ligand?
a) I⁻
b) F⁻
c) CN⁻
d) H₂O
Answer: c) (CN⁻ is strongest field ligand)
A complex appears green. It absorbs light in the region of:
a) Red
b) Orange
c) Yellow
d) Violet
Answer: a) (Complementary colour)
For a d⁴ metal ion in an octahedral field, the high spin complex has:
a) 0 unpaired electrons
b) 2 unpaired electrons
c) 4 unpaired electrons
d) 3 unpaired electrons
Answer: c)
For a d⁶ metal ion in an octahedral field, the low spin complex has:
a) 0 unpaired electrons
b) 2 unpaired electrons
c) 4 unpaired electrons
d) 5 unpaired electrons
Answer: a) (All electrons paired in t2g orbitals)
The magnetic moment of a complex is 5.92 BM. The number of unpaired electrons in the metal ion is:
a) 2
b) 3
c) 4
d) 5
Answer: d) (μ ≈ √[n(n+2)]; n=5 gives √35 ≈ 5.92 BM)
Which pair represents coordination isomers?
a) [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br
b) [Cr(H₂O)₆]Cl₃ and [Cr(H₂O)₅Cl]Cl₂·H₂O
c) [Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆]
d) [Co(NH₃)₅NO₂]Cl₂ and [Co(NH₃)₅ONO]Cl₂
Answer: c)
How many geometrical isomers are possible for [Ma₃b₃] type octahedral complex?
a) 1
b) 2
c) 3
d) 4
Answer: b) (facial and meridional)
The effective atomic number (EAN) of Fe in [Fe(CN)₆]⁴⁻ is:
a) 24
b) 30
c) 34
d) 36
Answer: d) (Fe²⁺ Z=26, loses 2e⁻, gains 12e⁻ from 6 CN⁻ ligands: 26 – 2 + 12 = 36)
Which of the following is an ambidentate ligand?
a) NH₃
b) Cl⁻
c) SCN⁻
d) en
Answer: c) (Can bind through S or N)
The denticity of Ethylenediaminetetraacetate ion (EDTA⁴⁻) is:
a) 2
b) 4
c) 6
d) 8
Answer: c)
The correct IUPAC name of [NiCl₂(PPh₃)₂] is:
a) Dichloridobis(triphenylphosphine)nickel(II)
b) Bis(triphenylphosphine)dichloridonickel(II)
c) Dichloridobis(triphenylphosphine)nickelate(II)
d) Nickel(II) dichloride bis(triphenylphosphine)
Answer: a)
Crystal Field Stabilization Energy (CFSE) for a d³ octahedral complex in terms of Δ₀ is:
a) -0.4 Δ₀
b) -0.6 Δ₀
c) -1.2 Δ₀
d) -1.6 Δ₀
Answer: c) (3 electrons in t2g; 3 * -0.4Δ₀ = -1.2Δ₀)
The complex ion [CoF₆]³⁻ is high spin. The number of unpaired electrons is:
a) 0
b) 2
c) 3
d) 4
Answer: d) (Co³⁺ d⁶, weak field ligand F⁻, high spin: t2g⁴ eg², 4 unpaired)
The complex that can show optical activity is:
a) trans-[CoCl₂(en)₂]⁺
b) cis-[CoCl₂(en)₂]⁺
c) [Co(NH₃)₆]³⁺
d) [NiCl₄]²⁻
Answer: b) (cis form lacks plane of symmetry)
The magnetic moment value for a d⁷ ion in an octahedral complex depends on:
a) Only the metal ion
b) Only the ligands
c) Both metal ion and ligands
d) Neither metal ion nor ligands
Answer: c) (Ligands determine high/low spin, which affects number of unpaired electrons)
In Crystal Field Theory, the pairing energy (P) is:
a) The energy required to pair two electrons
b) The energy difference between t2g and eg orbitals
c) The energy released when electrons pair
d) The energy of d-d transition
Answer: a)
A complex is low spin if:
a) Δ₀ P
c) Δ₀ = P
d) Δ₀ = 0
Answer: b)
For a d⁵ octahedral complex, the high spin configuration has CFSE of:
a) 0 Δ₀
b) -0.4 Δ₀
c) -0.6 Δ₀
d) -2.0 Δ₀
Answer: a) (t2g³ eg²; symmetric, no net stabilization)
The complex [Fe(CN)₆]⁴⁻ is:
a) Paramagnetic with 4 unpaired electrons
b) Paramagnetic with 1 unpaired electron
c) Diamagnetic
d) Paramagnetic with 5 unpaired electrons
Answer: b) (Fe²⁺ d⁶, strong field CN⁻, low spin: t2g⁶, one unpaired electron due to pairing)
Which ligand is likely to form a high spin octahedral complex with Fe³⁺?
a) CN⁻
b) CO
c) H₂O
d) NO₂⁻ (N-bound)
Answer: c) (H₂O is weaker field than CN⁻, CO, or NO₂⁻)
The type of isomerism not shown by [Cr(NH₃)₄Cl₂]⁺ is:
a) Geometrical
b) Optical
c) Ionisation
d) Linkage
Answer: c) (No counter ion to exchange with in the coordination sphere)
The colour of a coordination compound depends on:
a) The magnitude of Δ₀
b) The oxidation state of the metal
c) The nature of the ligand
d) All of the above
Answer: d)
In the complex [Ag(NH₃)₂]⁺, the geometry and hybridisation are:
a) Tetrahedral, sp³
b) Square planar, dsp²
c) Linear, sp
d) Octahedral, d²sp³
Answer: c)
The IUPAC name of H₂[PtCl₆] is:
a) Hydrogen hexachloroplatinate(IV)
b) Hexachloroplatinic(IV) acid
c) Hydrogen hexachloroplatinate(II)
d) Dihydrogen hexachloroplatinum(IV)
Answer: b)
The coordination number of nickel in [Ni(CO)₄] is:
a) 0
b) 2
c) 4
d) 6
Answer: c)
Which of the following is NOT a central atom/ion in coordination compounds?
a) Co³⁺
b) Fe³⁺
c) Pt²⁺
d) SO₄²⁻
Answer: d)
The complex used in the estimation of hardness of water is:
a) [Co(NCS)₄]²⁻
b) [Fe(CN)₆]⁴⁻
c) [Fe(CN)₆]³⁻
d) [Ni(CN)₄]²⁻
Answer: a) (Disodium salt of [Co(NCS)₄]²⁻ used in Patton and Reeder’s indicator)
The coordination compound present in the pigment of chlorophyll is:
a) Haemoglobin
b) Vitamin B₁₂
c) Chlorophyll contains Mg²⁺
d) Cisplatin
Answer: c) (Mg²⁺ coordinated in a porphyrin ring)

Answer Key:

c) 2. b) 3. b) 4. c) 5. b) 6. c) 7. c) 8. b) 9. c) 10. c)
d) 12. c) 13. a) 14. c) 15. b) 16. b) 17. d) 18. b) 19. b) 20. b)
b) 22. a) 23. c) 24. a) 25. c) 26. a) 27. d) 28. c) 29. b) 30. d)
c) 32. c) 33. a) 34. c) 35. d) 36. b) 37. c) 38. a) 39. b) 40. a)
b) 42. c) 43. c) 44. d) 45. c) 46. b) 47. c) 48. d) 49. a) 50. c)

Key Features Aligned with CBSE:

Strict NCERT Focus: Questions based solely on concepts covered in NCERT Class 12 Chemistry textbook (Unit 9).
Balanced Mix: Includes factual recall (e.g., Q1, Q5, Q10), conceptual understanding (e.g., Q13, Q18, Q21, Q39), and application/reasoning (e.g., Q7, Q8, Q14, Q27, Q34).
Latest Pattern: Follows MCQ format, language style, and depth found in CBSE sample papers and previous years’ questions (e.g., IUPAC naming, isomerism identification, magnetic moment calculation, hybridization, CFT predictions).
Age-Appropriate Language: Uses clear, concise terminology suitable for Class 12 students preparing for board exams.
Comprehensive Coverage: Tests all major subtopics – Werner’s theory, definitions, nomenclature, isomerism, VBT, CFT, magnetic properties, stability, applications.

Biology

Class: 12
Subject: Biology
Topic: Molecular Basis of Inheritance (Chapter 6, NCERT)

Multiple Choice Questions (MCQs):

The experiment that proved DNA is the genetic material in bacteriophages was conducted by:
a) Griffith
b) Hershey and Chase
c) Meselson and Stahl
d) Avery, MacLeod, and McCarty
Answer: b) Hershey and Chase
In a DNA molecule, the nitrogenous bases are attached to the:
a) Phosphate group
b) Deoxyribose sugar
c) Hydrogen bond
d) Ribose sugar
Answer: b) Deoxyribose sugar
The length of DNA is usually defined in terms of:
a) Micrometers
b) Base pairs
c) Nanometers
d) Angstroms
Answer: b) Base pairs
Which of the following is NOT a pyrimidine base?
a) Thymine
b) Cytosine
c) Adenine
d) Uracil
Answer: c) Adenine
The structure of DNA double helix was proposed by:
a) Erwin Chargaff
b) Francis Crick and James Watson
c) Maurice Wilkins and Rosalind Franklin
d) Linus Pauling
Answer: b) Francis Crick and James Watson
According to Chargaff’s rule, in a double-stranded DNA:
a) A = T and G = C
b) A = G and T = C
c) A + T = G + C
d) A + C = T + G
Answer: a) A = T and G = C
The enzyme that catalyses the polymerisation of deoxynucleotides to form DNA is:
a) DNA polymerase
b) RNA polymerase
c) Helicase
d) Ligase
Answer: a) DNA polymerase
During DNA replication, the template strand is read by DNA polymerase in the:
a) 5′ → 3′ direction
b) 3′ → 5′ direction
c) Both directions simultaneously
d) Direction varies
Answer: b) 3′ → 5′ direction
Meselson and Stahl’s experiment demonstrated that DNA replication is:
a) Conservative
b) Dispersive
c) Semi-conservative
d) Non-conservative
Answer: c) Semi-conservative
The discontinuously synthesised fragments of DNA on the lagging strand are called:
a) Primers
b) Okazaki fragments
c) Replicons
d) Transcripts
Answer: b) Okazaki fragments
The enzyme that joins the Okazaki fragments is:
a) DNA polymerase I
b) DNA polymerase III
c) DNA ligase
d) Primase
Answer: c) DNA ligase
The process of copying genetic information from one strand of DNA into RNA is termed:
a) Translation
b) Transcription
c) Replication
d) Transformation
Answer: b) Transcription
In transcription, the DNA strand that acts as the template is called the:
a) Coding strand
b) Non-template strand
c) Sense strand
d) Antisense strand
Answer: d) Antisense strand
The enzyme responsible for transcription in bacteria is:
a) DNA-dependent RNA polymerase
b) DNA-dependent DNA polymerase
c) RNA-dependent RNA polymerase
d) RNA-dependent DNA polymerase
Answer: a) DNA-dependent RNA polymerase
A transcription unit in DNA is defined by three regions in the DNA sequence. These are:
a) Promoter, Structural gene, Terminator
b) Operator, Structural gene, Terminator
c) Promoter, Operator, Terminator
d) Exon, Intron, Terminator
Answer: a) Promoter, Structural gene, Terminator
The RNA polymerase binds to the ___________ region of a transcription unit to initiate transcription.
a) Structural gene
b) Terminator
c) Promoter
d) Operator
Answer: c) Promoter
In eukaryotes, the primary transcript (hnRNA) undergoes splicing to remove:
a) Exons
b) Introns
c) Codons
d) Anticodons
Answer: b) Introns
The genetic code is read in a contiguous manner without punctuation. This property is called:
a) Degeneracy
b) Universality
c) Non-overlapping
d) Commaless
Answer: d) Commaless
The codon AUG codes for Methionine and also serves as the:
a) Stop codon
b) Release factor codon
c) Initiator codon
d) Operator codon
Answer: c) Initiator codon
The genetic code is degenerate. This means:
a) One codon codes for multiple amino acids
b) One amino acid is coded by multiple codons
c) Codons do not code for any amino acid
d) Codons overlap each other
Answer: b) One amino acid is coded by multiple codons
The process of synthesis of proteins directed by mRNA sequence occurs on:
a) Nucleus
b) Mitochondria
c) Ribosomes
d) Golgi apparatus
Answer: c) Ribosomes
Transfer RNA (tRNA) molecules carry specific amino acids to the ribosome. The site on tRNA that recognises the codon on mRNA is the:
a) DHU loop
b) TψC loop
c) Anticodon loop
d) Amino acid acceptor end
Answer: c) Anticodon loop
Translation begins with the binding of the small ribosomal subunit to the mRNA near the:
a) Stop codon
b) Promoter
c) Start codon (AUG)
d) Operator
Answer: c) Start codon (AUG)
The central dogma of molecular biology states the flow of genetic information as:
a) DNA → RNA → Protein
b) RNA → DNA → Protein
c) Protein → RNA → DNA
d) DNA → Protein → RNA
Answer: a) DNA → RNA → Protein
In some viruses (retroviruses), the flow of genetic information is from RNA to DNA. This process is called:
a) Transcription
b) Reverse transcription
c) Translation
d) Replication
Answer: b) Reverse transcription
The lac operon in E. coli is regulated by:
a) Tryptophan
b) Glucose
c) Lactose
d) Galactose
Answer: c) Lactose
In the absence of lactose, the repressor protein binds to the ___________ of the lac operon.
a) Promoter
b) Structural gene
c) Operator
d) Terminator
Answer: c) Operator
Human Genome Project (HGP) was launched in the year:
a) 1980
b) 1990
c) 2000
d) 2003
Answer: b) 1990
One of the main goals of the Human Genome Project was:
a) To develop transgenic crops
b) To sequence the entire human genome
c) To cure genetic diseases
d) To clone humans
Answer: b) To sequence the entire human genome
Approximately how many base pairs are estimated to be present in the human haploid genome?
a) 3 million
b) 3 billion
c) 30 billion
d) 300 million
Answer: b) 3 billion
Satellite DNA forms the basis of a powerful tool for genome analysis called:
a) PCR
b) DNA fingerprinting
c) Gene therapy
d) Recombinant DNA technology
Answer: b) DNA fingerprinting
The core particle of a nucleosome consists of:
a) DNA only
b) RNA only
c) Histone proteins only
d) DNA wrapped around a histone octamer
Answer: d) DNA wrapped around a histone octamer
The histones associated with the linker DNA between nucleosomes are called:
a) H1
b) H2A
c) H2B
d) H3
Answer: a) H1
V.N. Khorana’s work contributed significantly to:
a) Discovering the structure of DNA
b) Developing DNA sequencing techniques
c) Deciphering the genetic code
d) Proving DNA is the genetic material
Answer: c) Deciphering the genetic code
Which of the following is a stop codon?
a) UGA
b) UAG
c) UAA
d) All of the above
Answer: d) All of the above
In sickle cell anaemia, the mutation from GAG to GUG in the beta-globin gene leads to the substitution of:
a) Glutamic acid by Valine
b) Valine by Glutamic acid
c) Glycine by Valine
d) Valine by Glycine
Answer: a) Glutamic acid by Valine
A point mutation involving the substitution of a purine by a pyrimidine or vice versa is called:
a) Transition
b) Transversion
c) Frameshift mutation
d) Silent mutation
Answer: b) Transversion
A mutation that changes a codon specifying an amino acid into a stop codon is a:
a) Missense mutation
b) Nonsense mutation
c) Silent mutation
d) Frameshift mutation
Answer: b) Nonsense mutation
The sequence of DNA from which a functional RNA is transcribed is called a:
a) Codon
b) Gene
c) Operon
d) Chromosome
Answer: b) Gene
In eukaryotes, RNA polymerase II transcribes:
a) rRNA
b) tRNA
c) mRNA
d) snRNA
Answer: c) mRNA
The TATA box is a component of the:
a) Terminator
b) Structural gene
c) Promoter
d) Operator
Answer: c) Promoter
The amino acid carried by a tRNA is determined by its:
a) Anticodon sequence
b) DHU loop sequence
c) Specific aminoacyl-tRNA synthetase
d) Size of the tRNA molecule
Answer: c) Specific aminoacyl-tRNA synthetase
During translation elongation, a new amino acid is added to the growing polypeptide chain at the:
a) P-site of the ribosome
b) A-site of the ribosome
c) E-site of the ribosome
d) Small ribosomal subunit
Answer: b) A-site of the ribosome
The ‘wobble’ hypothesis explains:
a) Degeneracy of the genetic code
b) Non-overlapping nature of the code
c) Commaless nature of the code
d) Universality of the code
Answer: a) Degeneracy of the genetic code
The inducer molecule for the lac operon is derived from:
a) Glucose
b) Allolactose
c) Galactose
d) Tryptophan
Answer: b) Allolactose
Which histone protein is NOT part of the core histone octamer?
a) H1
b) H2A
c) H2B
d) H3
Answer: a) H1
A key finding of the Human Genome Project was that less than 2% of the genome codes for proteins. The majority consists of:
a) Repetitive DNA sequences
b) Mitochondrial DNA
c) Viral DNA inserts
d) Regulatory genes
Answer: a) Repetitive DNA sequences
The beta-chain of human haemoglobin has approximately how many nucleotides?
a) 146
b) 438
c) 146 * 3
d) 438 * 3
Answer: c) 146 * 3 (Since 146 amino acids, each coded by 3 nucleotides)
Frederick Griffith’s experiment with Streptococcus pneumoniae demonstrated:
a) DNA is the genetic material
b) RNA can be genetic material
c) The ‘transforming principle’
d) Semi-conservative DNA replication
Answer: c) The ‘transforming principle’
Avery, MacLeod, and McCarty identified the ‘transforming principle’ as DNA by treating the heat-killed S-strain extract with:
a) Proteases
b) RNases
c) DNases
d) All of the above
Answer: d) All of the above (Transformation failed only when DNase was used)

Answer Key:

Alignment Notes:

Syllabus Coverage: Questions comprehensively cover DNA structure, replication, transcription, genetic code, translation, regulation (lac operon), human genome project, DNA packaging, and mutations as per NCERT Class 12 Biology Chapter 6.
Question Types: Includes factual recall (e.g., discoverers, enzymes, structures), conceptual understanding (e.g., Chargaff’s rule, central dogma, operon regulation, genetic code properties), and application-based reasoning (e.g., mutation consequences, interpreting experiments like Meselson-Stahl/Hershey-Chase, HGP findings).
Language & Level: Uses clear, precise language appropriate for Class 12 CBSE students. Avoids unnecessary jargon beyond the NCERT scope.
CBSE Pattern: Follows the standard MCQ format (4 options, single correct answer) prevalent in CBSE board exams. Distractors are plausible misconceptions.
NCERT Basis: All concepts, definitions, experiments, and facts are drawn directly from the NCERT Class 12 Biology textbook.

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