How to Check CBSE Class 12 Physics Answer Key 2025?
The CBSE Class 12 Physics Exam 2025 is a crucial paper for students, especially those aiming for careers in engineering, medicine, and other science-related fields. After appearing for the exam, students eagerly look for the CBSE Class 12 Physics Answer Key 2025 to verify their answers and estimate their scores.
The CBSE Class 12 Physics Exam 2025 is a crucial paper for students, especially those aiming for careers in engineering, medicine, and other science-related fields. After appearing for the exam, students eagerly look for the CBSE Class 12 Physics Answer Key 2025 to verify their answers and estimate their scores.
At Reliable Scholar Academy, we understand students’ concerns and provide an accurate and detailed answer key soon after the exam. In this article, we will guide you on how to check both the unofficial and official answer keys, how to use them effectively, and how our coaching can help you succeed in future competitive exams.
Unofficial CBSE Class 12 Physics Answer Key 2025 – Reliable Scholar Academy
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SECTION – B
17. n identical cells, each of e.m.f E and internal resistance r, are connected in series. Later on, it was found out that two cells ‘X’ and ‘Y’ are connected in reverse polarities. Calculate the potential difference across the cell ‘X’.
Solution:
When n identical cells are in series, the total e.m.f. is nEnEnE. If two cells are reversed, their contribution is negative, so the net e.m.f. becomes:
Enet=(n−2)EE_{text{net}} = (n – 2)EEnet=(n−2)E
The total resistance is nrnrnr, so the current is:
I=(n−2)Enr=(n−2)EnrI = frac{(n – 2)E}{nr} = frac{(n – 2)E}{n r}I=nr(n−2)E=nr(n−2)E
The potential difference across cell X is:
VX=E−Ir=E−((n−2)Enr⋅r)=E(2n)V_X = E – Ir = E – left(frac{(n – 2)E}{n r} cdot rright) = E left( frac{2}{n} right)VX=E−Ir=E−(nr(n−2)E⋅r)=E(n2)
Thus, VX=2EnV_X = frac{2E}{n}VX=n2E.
18. (a) In a diffraction experiment, the slit is illuminated by light of wavelength 600 nm. The first minimum of the pattern falls at θ=30∘theta = 30^circθ=30∘. Calculate the width of the slit.
Solution:
For single-slit diffraction, the condition for the first minimum is given by:
asinθ=mλa sin theta = m lambdaasinθ=mλ
where:
- aaa = width of the slit
- λ=600lambda = 600λ=600 nm = 600×10−9600 times 10^{-9}600×10−9 m
- θ=30∘theta = 30^circθ=30∘
- m=1m = 1m=1 (for first minimum)
Substituting the values:
asin30∘=1×(600×10−9)a sin 30^circ = 1 times (600 times 10^{-9})asin30∘=1×(600×10−9)
Since sin30∘=0.5sin 30^circ = 0.5sin30∘=0.5:
a×0.5=600×10−9a times 0.5 = 600 times 10^{-9}a×0.5=600×10−9 a=600×10−90.5a = frac{600 times 10^{-9}}{0.5}a=0.5600×10−9 a=1.2×10−6 m=1.2 mma = 1.2 times 10^{-6} text{ m} = 1.2 text{ mm}a=1.2×10−6 m=1.2 mm
Thus, the width of the slit is 1.2 mm.
(b) In a Young’s double-slit experiment, two light waves, each of intensity I0I_0I0, interfere at a point, having a path difference λ8frac{lambda}{8}8λ on the screen. Find the intensity at this point.
Solution:
The intensity at any point in an interference pattern is given by:
I=I1+I2+2I1I2cosϕI = I_1 + I_2 + 2sqrt{I_1 I_2} cos phiI=I1+I2+2I1I2cosϕ
Since both waves have the same intensity I0I_0I0, we set I1=I2=I0I_1 = I_2 = I_0I1=I2=I0:
I=2I0(1+cosϕ)I = 2I_0 (1 + cos phi)I=2I0(1+cosϕ)
The phase difference ϕphiϕ is related to the path difference ΔxDelta xΔx by:
ϕ=2πλ×Δxphi = frac{2pi}{lambda} times Delta xϕ=λ2π×Δx
Given Δx=λ8Delta x = frac{lambda}{8}Δx=8λ, we substitute:
ϕ=2πλ×λ8=2π8=π4phi = frac{2pi}{lambda} times frac{lambda}{8} = frac{2pi}{8} = frac{pi}{4}ϕ=λ2π×8λ=82π=4π
Using cosπ4=12cos frac{pi}{4} = frac{1}{sqrt{2}}cos4π=21, we find:
I=2I0(1+12)I = 2I_0 left(1 + frac{1}{sqrt{2}}right)I=2I0(1+21)
Approximating 12≈0.707frac{1}{sqrt{2}} approx 0.70721≈0.707:
I≈2I0(1+0.707)=2I0×1.707I approx 2I_0 (1 + 0.707) = 2I_0 times 1.707I≈2I0(1+0.707)=2I0×1.707 I≈3.414I0I approx 3.414 I_0I≈3.414I0
Thus, the intensity at the given point is approximately 3.414I03.414 I_03.414I0.
19. A double convex lens of glass has both faces of the same radius of curvature 17 cm. Find its focal length if it is immersed in water. The refractive indices of glass and water are 1.5 and 1.33 respectively.
Solution:
The focal length of a lens in a medium is given by the lens-maker’s formula:
1fm=(nlensnmedium−1)(1R1−1R2)frac{1}{f_m} = left( frac{n_{text{lens}}}{n_{text{medium}}} – 1 right) left( frac{1}{R_1} – frac{1}{R_2} right)fm1=(nmediumnlens−1)(R11−R21)
Given Data:
- Refractive index of glass: nlens=1.5n_{text{lens}} = 1.5nlens=1.5
- Refractive index of water: nmedium=1.33n_{text{medium}} = 1.33nmedium=1.33
- Radius of curvature: R1=17R_1 = 17R1=17 cm, R2=−17R_2 = -17R2=−17 cm (convex lens)
Step 1: Calculate the focal length in water
1fm=(1.51.33−1)(117−1−17)frac{1}{f_m} = left( frac{1.5}{1.33} – 1 right) left( frac{1}{17} – frac{1}{-17} right)fm1=(1.331.5−1)(171−−171) =(1.51.33−1)(217)= left( frac{1.5}{1.33} – 1 right) left( frac{2}{17} right)=(1.331.5−1)(172) =(1.1278−1)×217= left( 1.1278 – 1 right) times frac{2}{17}=(1.1278−1)×172 =(0.1278)×217= (0.1278) times frac{2}{17}=(0.1278)×172 =0.255617= frac{0.2556}{17}=170.2556 =0.01503= 0.01503=0.01503 fm=10.01503≈66.5 cmf_m = frac{1}{0.01503} approx 66.5 text{ cm}fm=0.015031≈66.5 cm
Final Answer:
The focal length of the lens in water is approximately 66.5 cm.
20. An electron in the Bohr model of the hydrogen atom makes a transition from energy level −1.51 eV to −3.40 eV. Calculate the change in the radius of its orbit. The radius of the orbit of the electron in its ground state is 0.53 Å.
Solution:
The radius of an electron’s orbit in the Bohr model is given by:
rn=n2r1r_n = n^2 r_1rn=n2r1
where:
- rnr_nrn = radius of the nnnth orbit,
- r1=0.53r_1 = 0.53r1=0.53 Å (Bohr radius),
- nnn is the principal quantum number.
Step 1: Determine the quantum numbers
The energy levels in the Bohr model are given by:
En=−13.6n2 eVE_n = frac{-13.6}{n^2} text{ eV}En=n2−13.6 eV
For −3.40 eV:
−13.6n2=−3.40frac{-13.6}{n^2} = -3.40n2−13.6=−3.40 n2=13.63.40=4n^2 = frac{13.6}{3.40} = 4n2=3.4013.6=4 n=2n = 2n=2
For −1.51 eV:
−13.6n2=−1.51frac{-13.6}{n^2} = -1.51n2−13.6=−1.51 n2=13.61.51≈9n^2 = frac{13.6}{1.51} approx 9n2=1.5113.6≈9 n=3n = 3n=3
Step 2: Calculate the radii of the orbits
r2=22×0.53=4×0.53=2.12 A˚r_2 = 2^2 times 0.53 = 4 times 0.53 = 2.12 text{ Å}r2=22×0.53=4×0.53=2.12 A˚ r3=32×0.53=9×0.53=4.77 A˚r_3 = 3^2 times 0.53 = 9 times 0.53 = 4.77 text{ Å}r3=32×0.53=9×0.53=4.77 A˚
Step 3: Find the change in radius
Δr=r3−r2=4.77−2.12=2.65 A˚Delta r = r_3 – r_2 = 4.77 – 2.12 = 2.65 text{ Å}Δr=r3−r2=4.77−2.12=2.65 A˚
Final Answer:
The change in the radius of the orbit is 2.65 Å.
21. A p-type Si semiconductor is made by doping an average of one dopant atom per 5×1075 times 10^75×107 silicon atoms. If the number density of silicon atoms in the specimen is 5×10285 times 10^{28}5×1028 atoms m−3m^{-3}m−3, find the number of holes created per cubic centimetre in the specimen due to doping. Also, give one example of such dopants.
Solution:
Step 1: Calculate the number of dopant atoms per unit volume
- Number density of silicon atoms = 5×10285 times 10^{28}5×1028 atoms m−3m^{-3}m−3
- Each dopant atom is added per 5×1075 times 10^75×107 silicon atoms.
So, the number of dopant atoms per cubic meter is:
5×10285×107frac{5 times 10^{28}}{5 times 10^7}5×1075×1028 =1×1021 dopant atoms per m3= 1 times 10^{21} text{ dopant atoms per } m^3=1×1021 dopant atoms per m3
Step 2: Convert to per cubic centimetre
Since 1m3=106cm31 m^3 = 10^6 cm^31m3=106cm3,
Number of dopant atoms per cm3=1021106=1015 atoms per cm3text{Number of dopant atoms per } cm^3 = frac{10^{21}}{10^6} = 10^{15} text{ atoms per } cm^3Number of dopant atoms per cm3=1061021=1015 atoms per cm3
Step 3: Determine the number of holes created
- Each dopant atom contributes one hole in a p-type semiconductor.
- Therefore, the number of holes created per cubic centimetre is:
1015 holes/cm310^{15} text{ holes/cm}^31015 holes/cm3
Step 4: Example of a Dopant
A common p-type dopant for silicon is Boron (B).
Final Answer:
- The number of holes created per cubic centimetre = 101510^{15}1015 holes/cm³.
- Example of a p-type dopant: Boron (B).
22. (a) Two batteries of emf’s 3V & 6V and internal resistances 0.2Ω & 0.4Ω are connected in parallel. This combination is connected to a 4Ω resistor. Find:
(i) the equivalent emf of the combination
(ii) the equivalent internal resistance of the combination
(iii) the current drawn from the combination
Solution:
Step 1: Formula for Equivalent EMF (Eeq) in Parallel Combination
Eeq=E1r2+E2r1r1+r2E_{text{eq}} = frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}Eeq=r1+r2E1r2+E2r1
where:
- E1=3VE_1 = 3VE1=3V, r1=0.2Ωr_1 = 0.2Ωr1=0.2Ω
- E2=6VE_2 = 6VE2=6V, r2=0.4Ωr_2 = 0.4Ωr2=0.4Ω
Substituting the values:
Eeq=(3×0.4)+(6×0.2)0.2+0.4E_{text{eq}} = frac{(3 times 0.4) + (6 times 0.2)}{0.2 + 0.4}Eeq=0.2+0.4(3×0.4)+(6×0.2) =1.2+1.20.6=2.40.6=4V= frac{1.2 + 1.2}{0.6} = frac{2.4}{0.6} = 4V=0.61.2+1.2=0.62.4=4V
Step 2: Equivalent Internal Resistance (Req)
For parallel internal resistances:
1Req=1r1+1r2frac{1}{R_{text{eq}}} = frac{1}{r_1} + frac{1}{r_2}Req1=r11+r21 =10.2+10.4=5+2.5=7.5= frac{1}{0.2} + frac{1}{0.4} = 5 + 2.5 = 7.5=0.21+0.41=5+2.5=7.5 Req=17.5=0.133ΩR_{text{eq}} = frac{1}{7.5} = 0.133ΩReq=7.51=0.133Ω
Step 3: Total Resistance in Circuit
Total resistance RtotalR_{text{total}}Rtotal is:
Rtotal=Req+4Ω=0.133Ω+4Ω=4.133ΩR_{text{total}} = R_{text{eq}} + 4Ω = 0.133Ω + 4Ω = 4.133ΩRtotal=Req+4Ω=0.133Ω+4Ω=4.133Ω
Step 4: Total Current Drawn from the Combination
Using Ohm’s Law:
I=EeqRtotal=44.133I = frac{E_{text{eq}}}{R_{text{total}}} = frac{4}{4.133}I=RtotalEeq=4.1334 I≈0.968AI approx 0.968AI≈0.968A
Final Answers:
(i) Equivalent emf of the combination: 4V
(ii) Equivalent internal resistance: 0.133Ω
(iii) Current drawn from the combination: 0.968A
(b) (i) A conductor of length lll is connected across an ideal cell of emf EEE. Keeping the cell connected, the length of the conductor is increased to 2l2l2l by gradually stretching it. If RRR and R′R’R′ are initial and final values of resistance and vdv_dvd and vd′v_d’vd′ are initial and final values of drift velocity, find the relation between:
(i) R′R’R′ and RRR
(ii) vd′v_d’vd′ and vdv_dvd
Solution:
Step 1: Relation Between R′R’R′ and RRR
We know that the resistance of a conductor is given by:
R=ρlAR = rho frac{l}{A}R=ρAl
where:
- ρrhoρ = resistivity of the material (constant)
- lll = length of the conductor
- AAA = cross-sectional area
Since the conductor is stretched to twice its length (2l2l2l), its volume remains constant:
A1l1=A2l2A_1 l_1 = A_2 l_2A1l1=A2l2
Since l2=2l1l_2 = 2l_1l2=2l1, we get:
A2=A12A_2 = frac{A_1}{2}A2=2A1
Now, the new resistance is:
R′=ρ2lA/2=ρ2l×2A=4RR’ = rho frac{2l}{A/2} = rho frac{2l times 2}{A} = 4RR′=ρA/22l=ρA2l×2=4R
Thus, the relation between R′R’R′ and RRR is:
R′=4RR’ = 4RR′=4R
Step 2: Relation Between vd′v_d’vd′ and vdv_dvd
Drift velocity (vdv_dvd) is given by:
vd=InAev_d = frac{I}{n A e}vd=nAeI
where:
- III = current
- nnn = number density of charge carriers
- AAA = cross-sectional area
- eee = charge of an electron
Since the conductor is connected across an ideal cell of constant emf EEE, the current is:
I=ERI = frac{E}{R}I=RE
For the new case:
I′=ER′I’ = frac{E}{R’}I′=R′E
Using R′=4RR’ = 4RR′=4R:
I′=E4R=I4I’ = frac{E}{4R} = frac{I}{4}I′=4RE=4I
Since vd∝Iv_d propto Ivd∝I, we get:
vd′=vd4v_d’ = frac{v_d}{4}vd′=4vd
Thus, the relation between vd′v_d’vd′ and vdv_dvd is:
vd′=vd4v_d’ = frac{v_d}{4}vd′=4vd
Final Answers:
(i) R′=4RR’ = 4RR′=4R
(ii) vd′=vd4v_d’ = frac{v_d}{4}vd′=4vd
Solution:
No, not all free electrons in the conductor move in the same direction at any given instant.
Explanation:
- Free electrons in a conductor exhibit random motion due to thermal energy, similar to gas molecules.
- When an electric field is applied, electrons experience a force opposite to the field (since electrons are negatively charged).
- This results in a net drift velocity in the direction opposite to the applied electric field.
- However, the individual electrons still continue their random motion, colliding with atoms and other electrons.
Thus, while there is an average drift of electrons from lower to higher potential (opposite to conventional current), the motion of individual electrons remains random.
Question 23
Solution:
Given Data:
- A long straight wire carrying current III along the positive x-axis.
- A charge qqq is moving with velocity vmathbf{v}v along the x-axis.
- The charge is at a perpendicular distance ddd along the y-axis from the wire.
- The charge remains undeviated, meaning the net force acting on it must be zero.
Using the right-hand rule and Biot-Savart law, we analyze the situation:
(a) Magnetic Field Bmathbf{B}B at distance ddd:
For an infinite straight current-carrying wire, the magnetic field at a perpendicular distance ddd is given by:
B=μ0I2πdB = frac{mu_0 I}{2pi d}B=2πdμ0I
The direction of Bmathbf{B}B is given by the right-hand rule:
- Thumb in the direction of current (+i^+hat{i}+i^).
- Fingers curl around the wire.
- At distance ddd above the wire, Bmathbf{B}B is along the negative k^hat{k}k^-direction.
B=μ0I2πd(−k^)mathbf{B} = frac{mu_0 I}{2pi d} (-hat{k})B=2πdμ0I(−k^)
(b) Magnetic Force Fmmathbf{F_m}Fm on charge qqq:
The force on a moving charge due to a magnetic field is:
Fm=q(v×B)mathbf{F_m} = q (mathbf{v} times mathbf{B})Fm=q(v×B)
- vmathbf{v}v is along i^hat{i}i^
- Bmathbf{B}B is along −k^-hat{k}−k^
Using the cross-product:
i^×(−k^)=j^hat{i} times (-hat{k}) = hat{j}i^×(−k^)=j^
Thus,
Fm=qv(μ0I2πd)j^mathbf{F_m} = q v left(frac{mu_0 I}{2pi d} right) hat{j}Fm=qv(2πdμ0I)j^
(c) Electric Field Emathbf{E}E:
For the particle to remain undeviated, the net force must be zero. That means the electric force Fe=qEmathbf{F_e} = qmathbf{E}Fe=qE must cancel Fmmathbf{F_m}Fm.
Since Fmmathbf{F_m}Fm is in the j^hat{j}j^-direction, the required electric field must be in the negative j^hat{j}j^-direction:
E=−μ0Iv2πdj^mathbf{E} = -frac{mu_0 I v}{2pi d} hat{j}E=−2πdμ0Ivj^
Final Answers:
(a) Magnetic Field:
B=μ0I2πd(−k^)mathbf{B} = frac{mu_0 I}{2pi d} (-hat{k})B=2πdμ0I(−k^)
(b) Magnetic Force:
Fm=qv(μ0I2πd)j^mathbf{F_m} = q v left(frac{mu_0 I}{2pi d} right) hat{j}Fm=qv(2πdμ0I)j^
(c) Electric Field:
E=−μ0Iv2πdj^mathbf{E} = -frac{mu_0 I v}{2pi d} hat{j}E=−2πdμ0Ivj^
Question (Converted to Text):
24. An AC source of voltage v=vmsinωtv = v_m sin omega tv=vmsinωt is connected to a series combination of an LCR circuit. Draw the phasor diagram. Using it, obtain an expression for the impedance of the circuit and the phase difference between the applied voltage and the current.
Answer:
In a series LCR circuit, an AC voltage v=vmsinωtv = v_m sin omega tv=vmsinωt is applied across an inductor (L), capacitor (C), and resistor (R). The current (I) remains the same in all components but the voltages across them differ in phase.
Phasor Diagram:
- The resistor’s voltage VRV_RVR is in phase with the current.
- The inductor’s voltage VLV_LVL leads the current by 90∘90^circ90∘.
- The capacitor’s voltage VCV_CVC lags the current by 90∘90^circ90∘.
- The net voltage is the phasor sum: V=VR2+(VL−VC)2V = sqrt{V_R^2 + (V_L – V_C)^2}V=VR2+(VL−VC)2
Impedance ZZZ:
Z = sqrt{R^2 + (X_L – X_C)^
Question (Converted to Text):
25.
(a) A parallel plate capacitor is charged by an AC source. Show that the sum of conduction current (IcI_cIc) and the displacement current (IdI_dId) has the same value at all points of the circuit.
(b) In case (a) above, is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.
Answer:
(a) Conduction and Displacement Current Conservation
In an AC circuit with a capacitor, a conduction current IcI_cIc flows in the wires, but no actual charge crosses the capacitor gap. Instead, a displacement current IdI_dId is created in the capacitor due to the changing electric field.
From Maxwell’s equations, the displacement current is:
Id=ϵ0dΦEdtI_d = epsilon_0 frac{dPhi_E}{dt}Id=ϵ0dtdΦE
where ΦEPhi_EΦE is the electric flux.
Since the capacitor is connected in series with the circuit, the conduction current IcI_cIc in the wires must match the displacement current IdI_dId in the capacitor to ensure continuity. Thus, at all points in the circuit:
Ic=IdI_c = I_dIc=Id
This proves that the total current (conduction + displacement) remains constant throughout the circuit.
(b) Validity of Kirchhoff’s First Rule at the Capacitor Plates
Kirchhoff’s first rule (junction rule) states that the sum of currents entering a junction equals the sum leaving it. However, at the capacitor plates:
- The conduction current stops at the plate.
- The displacement current continues in the gap.
Since displacement current behaves like a real current in maintaining continuity, Kirchhoff’s junction rule remains valid only if we include displacement current as a real current. Without this inclusion, it would seem like charge is accumulating, violating the rule.
Thus, Kirchhoff’s junction rule holds true when displacement current is considered.
25.
(a) A parallel plate capacitor is charged by an AC source. Show that the sum of conduction current (IcI_cIc) and the displacement current (IdI_dId) has the same value at all points of the circuit.
(b) In case (a) above, is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.
Answer:
(a) Conservation of Conduction and Displacement Current
A parallel plate capacitor connected to an AC source experiences a time-varying voltage, leading to a time-dependent charge accumulation on the plates. The conduction current IcI_cIc flows in the circuit wires, while inside the capacitor, a displacement current IdI_dId arises due to the changing electric field.
Maxwell’s displacement current is given by:
Id=ϵ0dΦEdtI_d = epsilon_0 frac{dPhi_E}{dt}Id=ϵ0dtdΦE
where ΦEPhi_EΦE is the electric flux.
Since no actual charge moves across the capacitor gap, for current continuity in the circuit:
Ic=IdI_c = I_dIc=Id
Thus, the sum of conduction and displacement currents remains the same at all points in the circuit, ensuring there is no discontinuity in current flow.
(b) Validity of Kirchhoff’s First Rule at Capacitor Plates
Kirchhoff’s first rule states that the algebraic sum of currents entering and leaving a junction is zero. However, at the capacitor plates:
- The conduction current IcI_cIc brings charge to the plate but does not cross the gap.
- The displacement current IdI_dId allows the effect of current continuity across the capacitor.
Since IdI_dId behaves like a real current, Kirchhoff’s junction rule remains valid when displacement current is included in the current balance equation. If displacement current were ignored, it would seem like charge is accumulating, violating Kirchhoff’s law.
Thus, Kirchhoff’s first rule is applicable if we consider displacement current as part of the total current.
Bottom of Form
26.
(a) Mention any three features of results of experiment on photoelectric effect which cannot be explained using the wave theory of light.
(b) In his experiment on photoelectric effect, Robert A. Millikan found the slope of the cut-off voltage versus frequency of incident light plot to be 4.12×10−154.12 times 10^{-15}4.12×10−15 Vs. Calculate the value of Planck’s constant from it.
Answer:
(a) Three Features of the Photoelectric Effect Unexplained by Wave Theory
The wave theory of light fails to explain the following results of the photoelectric effect experiment:
- Instantaneous Emission of Photoelectrons:
- According to wave theory, energy is distributed over time, so there should be a delay before electrons are ejected.
- However, in the photoelectric effect, electrons are emitted instantly when light of sufficient frequency is incident.
- Threshold Frequency:
- Wave theory suggests that light of any frequency, if intense enough, should eventually eject electrons.
- But, experiments show that below a certain threshold frequency (f0f_0f0), no photoelectrons are emitted, no matter how intense the light is.
- Dependence on Frequency, Not Intensity:
- Wave theory predicts that higher intensity (brighter light) should increase the kinetic energy of emitted electrons.
- In reality, increasing the intensity increases the number of electrons emitted, but their energy depends only on the frequency of incident light.
These observations support Einstein’s quantum theory of light, where photons carry energy E=hfE = h fE=hf.
(b) Calculation of Planck’s Constant
Einstein’s photoelectric equation is:
eV0=hf−ϕeV_0 = h f – phieV0=hf−ϕ
The slope of the stopping potential (V0V_0V0) vs. frequency (fff) graph is given by:
Slope=hetext{Slope} = frac{h}{e}Slope=eh
Given:
he=4.12×10−15 Vsfrac{h}{e} = 4.12 times 10^{-15} text{ Vs}eh=4.12×10−15 Vs
Charge of an electron, e=1.6×10−19e = 1.6 times 10^{-19}e=1.6×10−19 C.
Calculating hhh:
h=(4.12×10−15)×(1.6×10−19)h = (4.12 times 10^{-15}) times (1.6 times 10^{-19})h=(4.12×10−15)×(1.6×10−19) h=6.592×10−34 Jsh = 6.592 times 10^{-34} text{ Js}h=6.592×10−34 Js
Approximating,
h≈6.63×10−34 Jsh approx 6.63 times 10^{-34} text{ Js}h≈6.63×10−34 Js
This matches the accepted value of Planck’s constant.
27.
(a) Draw circuit arrangement for studying V-I characteristics of a p-n junction diode.
(b) Show the shape of the characteristics of a diode.
(c) Mention two information that you can get from these characteristics.
Answer:
(a) Circuit Arrangement for Studying V-I Characteristics of a p-n Junction Diode
The V-I characteristics of a p-n junction diode can be studied using the following circuit arrangement:
Components Required:
- p-n junction diode
- Variable DC power supply
- Voltmeter (for measuring voltage across the diode)
- Ammeter (for measuring current through the diode)
- Resistor (for current limiting)
- Connecting wires
Circuit Setup:
- The diode is connected in forward bias by connecting its anode to the positive terminal of the power supply and its cathode to the negative terminal.
- The voltage across the diode and the current through it are measured.
- The same setup is used for reverse bias, but the diode’s terminals are reversed.
- The observations are used to plot the V-I characteristics.
(b) Shape of the Characteristics of a Diode
The V-I characteristics of a diode consist of two regions:
- Forward Bias Region:
- Initially, the current is very small until the diode reaches a threshold voltage (around 0.7V for silicon and 0.3V for germanium).
- Beyond this, the current increases rapidly with a small increase in voltage.
- Reverse Bias Region:
- In reverse bias, a very small leakage current flows until breakdown occurs.
- At breakdown voltage, a large current suddenly flows.
Graph:
The V-I characteristic curve of a diode shows:
- An exponential increase in current in forward bias after threshold voltage.
- A small leakage current in reverse bias until breakdown.
(c) Two Important Information from V-I Characteristics
- Threshold Voltage (Cut-in Voltage):
- The voltage at which the diode starts conducting significantly in forward bias.
- Example: For silicon, it is 0.7V; for germanium, it is 0.3V.
- Breakdown Voltage:
- The reverse voltage at which the diode starts conducting heavily due to breakdown.
- Useful for understanding Zener diodes and voltage regulation applications.
These characteristics help in designing circuits using diodes for rectification, switching, and voltage regulation applications.
28.
(a) Define ‘Mass defect’ and ‘Binding energy’ of a nucleus. Describe ‘Fission process’ on the basis of binding energy per nucleon.
(b) A deuteron contains a proton and a neutron and has a mass of 2.013553 u. Calculate the mass defect for it in u and its energy equivalence in MeV. (mₚ = 1.007277 u, mₙ = 1.008665 u, 1u = 931.5 MeV/c²)
Answer:
(a) Definitions and Explanation
1. Mass Defect:
Mass defect (ΔmDelta mΔm) is the difference between the total mass of the individual nucleons (protons and neutrons) and the actual mass of the nucleus.
Δm=(Zmp+Nmn)−MnucleusDelta m = (Z m_p + N m_n) – M_{text{nucleus}}Δm=(Zmp+Nmn)−Mnucleus
where:
- ZZZ = number of protons
- NNN = number of neutrons
- mpm_pmp = mass of proton
- mnm_nmn = mass of neutron
- MnucleusM_{text{nucleus}}Mnucleus = actual mass of nucleus
The missing mass is converted into binding energy according to Einstein’s equation:
E=Δm⋅c2E = Delta m cdot c^2E=Δm⋅c2
2. Binding Energy:
Binding energy (EBE_BEB) is the energy required to completely separate a nucleus into its individual protons and neutrons. It is given by:
EB=Δm×931.5 MeVE_B = Delta m times 931.5 text{ MeV}EB=Δm×931.5 MeV
where 931.5 MeV is the energy equivalent of 1 atomic mass unit (u).
3. Fission Process and Binding Energy per Nucleon:
- Nuclear fission occurs when a heavy nucleus (e.g., Uranium-235) splits into smaller nuclei, releasing energy.
- The binding energy per nucleon (EBAfrac{E_B}{A}AEB) is the binding energy divided by the total number of nucleons.
- Heavier nuclei have lower binding energy per nucleon, while medium-mass nuclei (e.g., Iron-56) have the highest binding energy per nucleon.
- Fission increases the total binding energy per nucleon, releasing a large amount of energy.
(b) Calculation of Mass Defect and Energy Equivalence
Step 1: Calculate Mass Defect
The given data:
- Mass of proton, mp=1.007277m_p = 1.007277mp=1.007277 u
- Mass of neutron, mn=1.008665m_n = 1.008665mn=1.008665 u
- Actual mass of deuteron, MD=2.013553M_D = 2.013553MD=2.013553 u
Total mass of nucleons (if not bound in a nucleus):
mp+mn=1.007277+1.008665=2.015942 um_p + m_n = 1.007277 + 1.008665 = 2.015942 text{ u}mp+mn=1.007277+1.008665=2.015942 u
Mass defect (ΔmDelta mΔm):
Δm=(mp+mn)−MDDelta m = (m_p + m_n) – M_DΔm=(mp+mn)−MD Δm=2.015942−2.013553Delta m = 2.015942 – 2.013553Δm=2.015942−2.013553 Δm=0.002389 uDelta m = 0.002389 text{ u}Δm=0.002389 u
Step 2: Calculate Energy Equivalence
Using the energy-mass equivalence formula:
E=Δm×931.5 MeVE = Delta m times 931.5 text{ MeV}E=Δm×931.5 MeV E=0.002389×931.5E = 0.002389 times 931.5E=0.002389×931.5 E≈2.23 MeVE approx 2.23 text{ MeV}E≈2.23 MeV
Final Answer:
- Mass defect for deuteron: 0.002389 u
- Binding energy: 2.23 MeV
This energy represents the stability of the deuteron and is responsible for holding the nucleons together.
SECTION – E
31. (a) (i)
(1) What are coherent sources? Why are they necessary for observing a sustained interference pattern?
(2) Lights from two independent sources are not coherent. Explain.
(ii) Two slits 0.1 mm apart are arranged 1.20 m from a screen. Light of wavelength 600 nm from a distant source is incident on the slits.
(1) How far apart will adjacent bright interference fringes be on the screen?
(2) Find the angular width (in degrees) of the first bright fringe.
(b) (i) Define a wavefront.
Answer Explanation
(a) (i) (1) What are Coherent Sources?
Coherent sources are two or more wave sources that emit waves having:
- The same frequency and wavelength
- A constant phase difference
They are essential for producing sustained interference patterns because, without coherence, the phase difference between the waves would change randomly over time, destroying the interference pattern.
(a) (i) (2) Why are Lights from Independent Sources Not Coherent?
- Two independent sources emit light waves randomly, leading to a varying phase difference over time.
- Natural light sources (e.g., bulbs, the Sun) emit light due to random atomic transitions, making them incoherent with different phase relations.
- Coherent light is produced using laser sources or division of wavefront methods (Young’s Double Slit Experiment).
(a) (ii) (1) Calculation of Fringe Width
Given Data:
- Distance between slits: d=0.1 mm=1.0×10−4 md = 0.1 text{ mm} = 1.0 times 10^{-4} text{ m}d=0.1 mm=1.0×10−4 m
- Wavelength of light: λ=600 nm=600×10−9 mlambda = 600 text{ nm} = 600 times 10^{-9} text{ m}λ=600 nm=600×10−9 m
- Distance between slits and screen: D=1.20 mD = 1.20 text{ m}D=1.20 m
Formula for Fringe Width (βbetaβ):
β=λDdbeta = frac{lambda D}{d}β=dλD β=(600×10−9)(1.2)1.0×10−4beta = frac{(600 times 10^{-9}) (1.2)}{1.0 times 10^{-4}}β=1.0×10−4(600×10−9)(1.2) β=7.2×10−710−4beta = frac{7.2 times 10^{-7}}{10^{-4}}β=10−47.2×10−7 β=7.2×10−3 m=7.2 mmbeta = 7.2 times 10^{-3} text{ m} = 7.2 text{ mm}β=7.2×10−3 m=7.2 mm
Answer: Adjacent bright fringes will be 7.2 mm apart on the screen.
(a) (ii) (2) Calculation of Angular Width
Formula for Angular Width (θthetaθ):
θ=βDtheta = frac{beta}{D}θ=Dβ θ=7.2×10−31.2theta = frac{7.2 times 10^{-3}}{1.2}θ=1.27.2×10−3 θ=6×10−3 radtheta = 6 times 10^{-3} text{ rad}θ=6×10−3 rad
To convert to degrees:
θ=6×10−3×180πtheta = 6 times 10^{-3} times frac{180}{pi}θ=6×10−3×π180 θ≈0.344 degreestheta approx 0.344 text{ degrees}θ≈0.344 degrees
Answer: Angular width of the first bright fringe is 0.344 degrees.
(b) (i) Definition of Wavefront
A wavefront is a surface of constant phase representing points where the wave has the same phase at a given instant.
Types of Wavefronts:
- Plane Wavefront – Produced by distant sources or lasers.
- Spherical Wavefront – Produced by point sources like a bulb.
- Cylindrical Wavefront – Produced by linear sources like a laser beam passing through a slit.
Final Answers:
- Coherent sources emit waves of the same frequency and a constant phase difference, necessary for sustained interference patterns.
- Independent sources are not coherent because they emit waves with random phase differences over time.
- Fringe width = 7.2 mm
- Angular width = 0.344 degrees
- Wavefront is a surface of constant phase representing wave motion.
(b) (i) Define a wavefront. An incident plane wave falls on a convex lens and gets refracted through it. Draw a diagram to show the incident and refracted wavefront.
(ii) A beam of light coming from a distant source is refracted by a spherical glass ball (refractive index 1.5) of radius 15 cm. Draw the ray diagram and obtain the position of the final image formed.
Answer Explanation
(b) (i) Definition of Wavefront
A wavefront is a surface over which all points of a wave oscillate in phase. It represents the locus of points having the same phase in a propagating wave.
Types of Wavefronts:
- Plane Wavefront – When light originates from a distant source, the wavefronts appear parallel.
- Spherical Wavefront – Produced by a point source, expanding outward in spherical layers.
- Cylindrical Wavefront – Produced when light is confined in a cylindrical shape, like a laser beam passing through a narrow slit.
Wavefront Refraction through a Convex Lens
- A plane wavefront from a distant source falls on a convex lens.
- The lens converges the wavefront, forming a spherical wavefront after refraction.
- The new wavefront focuses at the focal point of the lens.
- Diagram: The diagram should show an incident plane wavefront, a convex lens, and the refracted spherical wavefront converging to a point.
(b) (ii) Refraction of Light through a Spherical Glass Ball
Given Data:
- Refractive index of the glass ball: n=1.5n = 1.5n=1.5
- Radius of the sphere: R=15R = 15R=15 cm
Explanation:
- A distant light source emits parallel rays.
- When these rays enter the spherical glass ball, they bend towards the normal due to refraction.
- Inside the sphere, light travels and undergoes a second refraction at the other side.
- The final image position can be found using the lens maker’s formula and the formula for refraction at a curved surface.
- Diagram: The diagram should illustrate parallel incident rays, their refraction inside the glass ball, and the formation of the final image.
Final Answer Summary:
- Wavefront is a surface where all points have the same phase in a propagating wave.
- Plane wavefronts become spherical when passing through a convex lens, converging at the focal point.
- A spherical glass ball refracts parallel light rays, forming a focused image depending on its refractive index and radius.
33. (a) (i) A proton moving with velocity V⃗vec{V}V in a non-uniform magnetic field traces a path as shown in the figure.
The path followed by the proton is always in the plane of the paper. What is the direction of the magnetic field in the region near points P, Q, and R? What can you say about the relative magnitude of magnetic fields at these points?
(ii) A current-carrying circular loop of area AAA produces a magnetic field BBB at its center. Show that the magnetic moment of the loop is
2BAμ0Aπfrac{2BA}{mu_0} sqrt{frac{A}{pi}}μ02BAπA
Answer Explanation
(a) (i) Direction of Magnetic Field and its Relative Magnitude
Concept Used: Motion of a Charged Particle in a Magnetic Field
- A charged particle like a proton moves in a curved path in the presence of a magnetic field due to the Lorentz force: F⃗=q(V⃗×B⃗)vec{F} = q (vec{V} times vec{B})F=q(V×B) where qqq is the charge of the proton, V⃗vec{V}V is its velocity, and B⃗vec{B}B is the magnetic field.
- The curvature of the path is determined by the strength of the magnetic field at different points.
Direction of Magnetic Field at Points P, Q, and R
- Since the proton is moving in the plane of the paper, and the force due to the magnetic field is always perpendicular to its velocity, the magnetic field must be perpendicular to the plane of the paper (either into or out of the plane).
- Using the right-hand rule (for positive charges):
- If the force is causing counterclockwise motion, the magnetic field is out of the plane.
- If the force is causing clockwise motion, the magnetic field is into the plane.
Relative Magnitude of Magnetic Fields
- The radius of curvature of the proton’s path is given by: r=mVqBr = frac{mV}{qB}r=qBmV where mmm is the proton’s mass, VVV is velocity, qqq is charge, and BBB is magnetic field strength.
- Larger curvature radius → Weaker magnetic field (lower BBB).
- Smaller curvature radius → Stronger magnetic field (higher BBB).
- Comparing points P, Q, and R:
- If R has the smallest radius, it has the strongest magnetic field.
- If P has the largest radius, it has the weakest magnetic field.
(a) (ii) Magnetic Moment of a Current-Carrying Circular Loop
Concept Used: Magnetic Dipole Moment
- A current-carrying circular loop produces a magnetic field at its center. The magnetic moment of the loop is given by: M=IAM = IAM=IA where III is the current and AAA is the area of the loop.
Magnetic Field at the Center of the Loop
- The magnetic field at the center of a circular current loop is: B=μ0I2RB = frac{mu_0 I}{2R}B=2Rμ0I where RRR is the radius of the loop, and μ0mu_0μ0 is the permeability of free space.
Derivation of Magnetic Moment
- Expressing Current in Terms of B: I=2BRμ0I = frac{2BR}{mu_0}I=μ02BR
- Area of the Loop: A=πR2A = pi R^2A=πR2
- Magnetic Moment Expression: M=IA=(2BRμ0)(πR2)M = IA = left(frac{2BR}{mu_0}right) (pi R^2)M=IA=(μ02BR)(πR2) M=2BπR3μ0M = frac{2B pi R^3}{mu_0}M=μ02BπR3
- Substituting R=AπR = sqrt{frac{A}{pi}}R=πA: M=2BAμ0AπM = frac{2B A}{mu_0} sqrt{frac{A}{pi}}M=μ02BAπA
Thus, the magnetic moment of the loop is:
2BAμ0Aπfrac{2BA}{mu_0} sqrt{frac{A}{pi}}μ02BAπA
Final Answer Summary:
- The magnetic field at P, Q, and R is perpendicular to the plane of the paper (determined using the right-hand rule).
- The relative magnitude of the magnetic field at different points depends on the curvature of the path:
- Stronger magnetic field → Smaller radius of curvature.
- Weaker magnetic field → Larger radius of curvature.
- The magnetic moment of a circular current loop is derived as: 2BAμ0Aπfrac{2BA}{mu_0} sqrt{frac{A}{pi}}μ02BAπA using the relation between magnetic field, current, and loop area.
(b) OR
(i) Derive an expression for the torque acting on a rectangular current loop suspended in a uniform magnetic field.
(ii) A charged particle is moving in a circular path with velocity V⃗vec{V}V in a uniform magnetic field B⃗vec{B}B. It is made to pass through a sheet of lead and, as a consequence, it loses one-half of its kinetic energy without a change in its direction.
How will:
(1) the radius of its path change?
(2) its time period of revolution change?
Answer Explanation
(i) Torque on a Rectangular Current Loop in a Uniform Magnetic Field
Concept Used: Magnetic Torque on a Current-Carrying Loop
- A current-carrying rectangular loop in a uniform magnetic field experiences a torque due to the force acting on its sides.
- The magnetic dipole moment of the loop interacts with the magnetic field, producing a torque given by: τ=M⃗×B⃗tau = vec{M} times vec{B}τ=M×B where M⃗=IAn^vec{M} = I A hat{n}M=IAn^ (magnetic moment), and BBB is the uniform magnetic field.
Derivation of Torque
- Consider a rectangular loop of length lll and width www, carrying current III in a magnetic field BBB.
- The forces on opposite sides of the loop are equal and opposite, forming a couple that exerts a torque.
- The torque is maximum when the normal to the loop is perpendicular to BBB and zero when parallel.
- The torque is given by: τ=IABsinθtau = I A B sin thetaτ=IABsinθ where A=l×wA = l times wA=l×w is the area of the loop, and θthetaθ is the angle between the normal to the loop and the magnetic field.
Final Expression for Torque
τ=IABsinθtau = IAB sin thetaτ=IABsinθ
This expression shows that the torque depends on the current, area of the loop, magnetic field strength, and the orientation of the loop.
(ii) Motion of a Charged Particle in a Magnetic Field
Concept Used: Circular Motion of a Charged Particle in a Magnetic Field
- A charged particle moving perpendicular to a uniform magnetic field follows a circular path due to the Lorentz force: F=qvBF = q v BF=qvB
- The centripetal force required for circular motion is provided by this magnetic force: mv2r=qvBfrac{m v^2}{r} = q v Brmv2=qvB which gives the radius of the circular path as: r=mvqBr = frac{m v}{q B}r=qBmv
Effect of Energy Loss on Radius
- Kinetic energy relation: K=12mv2K = frac{1}{2} m v^2K=21mv2
- If the particle loses half of its kinetic energy, the new kinetic energy is: K′=K2K’ = frac{K}{2}K′=2K
- Since kinetic energy is proportional to velocity squared: 12mv′2=12×12mv2frac{1}{2} m v’^2 = frac{1}{2} times frac{1}{2} m v^221mv′2=21×21mv2 v′=v2v’ = frac{v}{sqrt{2}}v′=2v
- The new radius of the path is: r′=mv′qB=m(v/2)qB=r2r’ = frac{m v’}{q B} = frac{m (v/sqrt{2})}{q B} = frac{r}{sqrt{2}}r′=qBmv′=qBm(v/2)=2r So, the radius decreases by a factor of 12frac{1}{sqrt{2}}21.
Effect on Time Period of Revolution
- Time period of circular motion: T=2πrvT = frac{2 pi r}{v}T=v2πr Substituting r=mvqBr = frac{m v}{q B}r=qBmv, T=2πmqBT = frac{2 pi m}{q B}T=qB2πm
- The time period is independent of velocity and depends only on mass, charge, and magnetic field strength.
- Since none of these parameters change, the time period remains the same.
Final Answer Summary:
- Torque on a rectangular loop in a magnetic field is given by:
τ=IABsinθtau = IAB sin thetaτ=IABsinθ
where III is the current, AAA is the area, BBB is the magnetic field, and θthetaθ is the angle.
- For the charged particle moving in a circular path in a uniform magnetic field:
- The radius decreases by a factor of 12frac{1}{sqrt{2}}21 because velocity decreases due to kinetic energy loss.
- The time period remains unchanged because it does not depend on velocity.
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