PHYSICS

PHYSICS

A hydrogen atom absorbs a photon of energy 12.75 eV from the ground state. To which energy level does it jump? (Use E_n

Given: A hydrogen atom absorbs a photon of energy 12.75 eV from the ground state. To which energy level does it jump? (Use E_n = -13.6/n² eV ) These values define the system as per NCERT data. Formula: E_1 = -13.6 eV. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: E_n = -13.6 + 12.75 = -0.85 eV . -0.85 = -13.6/n² Rightarrow n² = 16 Rightarrow n = 4 . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Laws of Motion, Work Energy Power, Gravitation and System of Particles, Topic: Newton's laws, work-energy theorem and rotational dynamics.

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PHYSICS

The number of valence electrons in Si and Ge atoms is:

Si (third orbit) and Ge (fourth orbit) are group IV elements, each with four valence electrons (2s and 2p for Si, 4s and 4p for Ge), forming covalent bonds in their lattice.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Dual Nature, Atoms, Nuclei and Electronic Devices, Topic: Atomic structure, nuclear binding and semiconductor physics.

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PHYSICS

A compound microscope has an objective of focal length 2 cm and eyepiece of focal length 5 cm with a tube length of 18 c

Given: A compound microscope has an objective of focal length 2 cm and eyepiece of focal length 5 cm with a tube length of 18 cm . What is the magnification at infinity? These values define the system as per NCERT data. Formula: Objective magnification: m_o = L/f_o = 18/2 = 9. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Eyepiece magnification: m_e = D/f_e = 25/5 = 5 . Total magnification: m = m_o × m_e = 9 × 5 = 45 . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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PHYSICS

A circular loop of radius 18 cm is deformed into a straight wire in a 0.14 T field in 0.7 s. What is the induced emf?

Given: A circular loop of radius 18 cm is deformed into a straight wire in a 0.14 T field in 0.7 s. What is the induced emf? These values define the system as per NCERT data. Formula: Initial flux: Phi = B A = 0.14 × π × (0.18)² = 0.01425 Wb. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Final flux = 0. varepsilon = Δ Phi/Δ t = 0.01425/0.7 = 0.02036 V approx 0.02 V . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Electromagnetic Induction, Alternating Current and Electromagnetic Waves, Topic: Induced emf, inductance and EM wave properties.

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PHYSICS

A sphere of radius 0.025 m moves at 0.08 m/s through blood ( eta = 2.7 × 10⁻³ Pa s ). What is the viscous drag force

Given: A sphere of radius 0.025 m moves at 0.08 m/s through blood ( eta = 2.7 × 10⁻³ Pa s ). What is the viscous drag force? These values define the system as per NCERT data. Formula: Stokes’ law: F = 6 π eta a v. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: eta = 2.7 × 10⁻³ Pa s, a = 0.025 m, v = 0.08 m/s . F = 6 × 3.14 × 2.7 × 10⁻³ × 0.025 × 0.08 = 1.017 × 10⁻⁴ N . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Laws of Motion, Work Energy Power, Gravitation and System of Particles, Topic: Newton's laws, work-energy theorem and rotational dynamics.

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PHYSICS

What is the volume of a nucleus with radius 3.0 × 10⁻¹⁵ m ? (Use π = 3.14 )

Given: What is the volume of a nucleus with radius 3.0 × 10⁻¹⁵ m ? (Use π = 3.14 ) These values define the system as per NCERT data. Formula: Volume = 4/3 π R³. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: R³ = (3.0 × 10⁻¹⁵)³ = 2.7 × 10⁻⁴⁴ m³ . Volume = 4/3 × 3.14 × 2.7 × 10⁻⁴⁴ approx 1.13 × 10⁻⁴³ m³ . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Dual Nature, Atoms, Nuclei and Electronic Devices, Topic: Atomic structure, nuclear binding and semiconductor physics.

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PHYSICS

A bar magnet with m = 1.5 A m² produces a field at 0.5 m on its equatorial line. What is B ? (Take μ_0 = 4Ï€ × 10⁻â

Given: A bar magnet with m = 1.5 A m² produces a field at 0.5 m on its equatorial line. What is B ? (Take μ_0 = 4π × 10⁻⁷ T m A^{-1 ). These values define the system as per NCERT data. Formula: B = μ_0/4π m/r³. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: m = 1.5 A m², r = 0.5 m, μ_0/4π = 10⁻⁷. B = 10⁻⁷ × 1.5/(0.5)³ = 10⁻⁷ × 1.5/0.125 = 1.2 × 10⁻⁶ T . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Moving Charges and Magnetism and Magnetism and Matter, Topic: Magnetic field due to current loop, solenoid and magnetic dipole moment. Page number should be added only after verification from the.

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PHYSICS

A conductor has a resistivity of 1.0 × 10⁻⁷Ω m and α = 4 × 10⁻³°C^{-1 at 25° C . What is its resistivity at

Given: A conductor has a resistivity of 1.0 × 10⁻⁷Ω m and α = 4 × 10⁻³°C^{-1 at 25° C . What is its resistivity at 85° C ? These values define the system as per NCERT data. Formula: Use: rho_t = rho_0 [1 + α (T - T_0)]. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substitute: rho_t = 1.0 × 10⁻⁷[1 + 4 × 10⁻³(85 - 25)] . Calculate: rho_t = 1.0 × 10⁻⁷[1 + 0.24] = 1.0 × 10⁻⁷ × 1.24 = 1.24 × 10⁻⁷Ω m . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Current Electricity, Topic: Resistivity, temperature dependence and circuit analysis.

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PHYSICS

A coil of 150 turns and area 0.05 m² is in a field that increases from 0 to 0.06 T in 0.3 s. What is the induced emf?

Given: A coil of 150 turns and area 0.05 m² is in a field that increases from 0 to 0.06 T in 0.3 s. What is the induced emf? These values define the system as per NCERT data. Formula: Δ Phi = B A = 0.06 × 0.05 = 0.003 Wb. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: varepsilon = N Δ Phi/Δ t = 150 × 0.003/0.3 = 150 × 0.01 = 1.5 V . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Moving Charges and Magnetism and Magnetism and Matter, Topic: Magnetic field due to current loop, solenoid and magnetic dipole moment. Page number should be added only after verification from the.

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PHYSICS

A steel rod of radius 0.01 m and length 1.5 m is subjected to a tensile force producing a stress of 4 × 10⁷ N/m² . W

Given: A steel rod of radius 0.01 m and length 1.5 m is subjected to a tensile force producing a stress of 4 × 10⁷ N/m² . What is the force applied? (Take π approx 3.14 ) These values define the system as per NCERT data. Formula: Stress: Stress = F/A. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Area: A = π r² = 3.14 × (0.01)² = 3.14 × 10⁻⁴ m² . Force: F = Stress × A = 4 × 10⁷ × 3.14 × 10⁻⁴= 1.256 × 10⁴ N . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Laws of Motion, Work Energy Power, Gravitation and System of Particles, Topic: Newton's laws, work-energy theorem and rotational dynamics.

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PHYSICS

A thin spherical shell of radius 25 cm has a charge of 15 μC . What is the electric field at a point 30 cm from the nte

Given: A thin spherical shell of radius 25 cm has a charge of 15 μC . What is the electric field at a point 30 cm from the nter? These values define the system as per NCERT data. Formula: Outside shell ( r > R ): E = k q/r². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: k = 9 × 10⁹ Nm²/C², q = 15 × 10⁻⁶ C, r = 0.3 m . E = 9 × 10⁹ × frac15 × 10⁻⁶(0.3)² = 9 × 10⁹ × frac15 × 10⁻⁶⁰.09 = 1.5 × 10⁶ N/C . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Electric Charges and Fields and Electrostatic Potential, Topic: Electric field, potential and capacitance concepts.

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PHYSICS

A 0.3 g drop falls from 500 m and hits the ground at 15 m/s . What is the work done by air resistance? (Take g = 10 m/sÂ

Given: A 0.3 g drop falls from 500 m and hits the ground at 15 m/s . What is the work done by air resistance? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: Work by gravity W_g = mgh = 0.0003 × 10 × 500 = 1.5 J. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Final K = 1/2 × 0.0003 × 15² = 0.03375 J . K_f = W_g + W_r Rightarrow 0.03375 = 1.5 + W_r Rightarrow W_r = -1.46625 J approx -1.47 J . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Current Electricity, Topic: Resistivity, temperature dependence and circuit analysis.

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