PHYSICS

PHYSICS

The SI unit of energy is kg m² s^{-2 . What is its dimensional formula?

Given: The SI unit of energy is kg m² s^{-2 . What is its dimensional formula? Formula: kg = [M], m² = [L²], s^{-2 = [T^{-2]. Substitution & Calculation: kg m² s^{-2 = [M L² T^{-2] . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

A soap film supports 3.0 × 10⁻²N over a 50 cm slider. What is the surface tension?

Given: A soap film supports 3.0 × 10⁻²N over a 50 cm slider. What is the surface tension? Formula: S = F/2 l. Substitution & Calculation: F = 3.0 × 10⁻²N, l = 0.5 m . S = frac3.0 × 10⁻²² × 0.5 = 0.03 N/m . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

A 4 μF capacitor charged to 300 V is connected to an uncharged 8 μF capacitor. What is the energy lost?

Given: A 4 μF capacitor charged to 300 V is connected to an uncharged 8 μF capacitor. What is the energy lost? Formula: Initial energy: U_i = 1/2 × 4 × 10⁻⁶ × (300)² = 0.18 J. Substitution & Calculation: Charge: Q = 4 × 10⁻⁶ × 300 = 1.2 × 10⁻³C . Total C = 4 + 8 = 12 μF, V = frac1.2 × 10⁻³¹² × 10⁻⁶= 100 V . Final energy: U_f = 1/2 × 12 × 10⁻⁶ × (100)² = 0.06 J . Loss: U_i - U_f = 0.18 - 0.06 = 0.12 J . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

In a Wheatstone bridge with R_1 = 30 Ω, R_2 = 60 Ω, R_3 = 15 Ω, and R_4 = 32 Ω, a 10 V battery is connected across AC. W

Given: In a Wheatstone bridge with R_1 = 30 Ω, R_2 = 60 Ω, R_3 = 15 Ω, and R_4 = 32 Ω, a 10 V battery is connected across AC. What is the current through the galvanometer ( R_g = 10 Ω )? Formula: Junction B: I_1 = I_g + I_4, Junction D: I_2 = I_g + I_3. Substitution & Calculation: Apply Kirchhoff’s rules. Let currents be I_1 (AB), I_2 (AD), I_g (BD). . Loop BADB: 30 I_1 + 10 I_g - 60 I_2 = 0 Rightarrow 3 I_1 + I_g - 6 I_2 = 0 . Loop BCDB: 60 (I_1 - I_g) - 10 I_g - 15 (I_2 + I_g) = 0 Rightarrow 4 I_1 - 2 I_2 - 2 I_g = 0 . Loop ADCEA: 60 I_2 + 15 (I_2 + I_g) = 10 Rightarrow 75 I_2 + 15

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

A dipole with m = 0.7 A m² in B = 0.6 T at 30° has torque:

Given: A dipole with m = 0.7 A m² in B = 0.6 T at 30° has torque: Formula: tau = m B sinθ. Substitution & Calculation: Given: m = 0.7 A m², B = 0.6 T, θ = 30°, sin 30° = 0.5 . tau = 0.7 × 0.6 × 0.5 = 0.21 N m . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

A neutron ( 1 u ) at 10⁶m/s collides elastically with a carbon ( 12 u ). What fraction of its kinetic energy is transfer

Given: A neutron ( 1 u ) at 10⁶m/s collides elastically with a carbon ( 12 u ). What fraction of its kinetic energy is transferred? Formula: Fraction transferred f_2 = 4 m_1 m_2/(m_1 + m_2)² = 4 × 1 × 12/(1 + 12)² = 48/169 approx 0.284 .. Substitution: Substituting given values into formula as per NCERT 2026-27 method. Calculation: Simplifying step by step with proper SI units like m/s², J kg⁻¹ K⁻¹, 10⁻⁵, A m⁻¹ etc. Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

A solid sphere of mass 5 kg and radius 0.2 m rotates at 10 rad/s . What is its angular momentum about its axis?

Given: A solid sphere of mass 5 kg and radius 0.2 m rotates at 10 rad/s . What is its angular momentum about its axis? Formula: I = 2/5 M R² = 2/5 × 5 × (0.2)² = 0.08 kg m². Substitution & Calculation: L = I omega = 0.08 × 10 = 0.8 kg m²/s . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

The maximum speed of photoelectrons emitted from a surface is 5.0 × 10⁵m/s . What is the stopping potential? (Take m_e =

Given: The maximum speed of photoelectrons emitted from a surface is 5.0 × 10⁵m/s . What is the stopping potential? (Take m_e = 9.11 × 10⁻³¹kg, e = 1.6 × 10⁻¹⁹C ) Formula: K_{max = 1/2 m v_{max² = 1/2 × 9.11 × 10⁻³¹ × (5.0 × 10⁵)² = 1.13875 × 10⁻¹⁹J. Substitution & Calculation: V_0 = fracK_{maxe = frac1.13875 × 10⁻¹⁹¹.6 × 10⁻¹⁹approx 0.71 V . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

What is the total energy of a 100 kg satellite orbiting Earth at 3 R_E from the nter? ( M_E = 6 × 10²⁴kg, R_E = 6.4 × 10

Given: What is the total energy of a 100 kg satellite orbiting Earth at 3 R_E from the nter? ( M_E = 6 × 10²⁴kg, R_E = 6.4 × 10⁶m, G = 6.67 × 10⁻¹¹N m²/kg² ) Formula: E = -G M_E m/2 r. Substitution & Calculation: r = 3 R_E = 3 × 6.4 × 10⁶= 1.92 × 10⁷m . E = -frac6.67 × 10⁻¹¹ × 6 × 10²⁴ × 1002 × 1.92 × 10⁷. E = -frac4.002 × 10¹⁶³.84 × 10⁷approx -1.04 × 10⁹J . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

What is the approximate wavelength range of radio waves as per the document?

The document states that radio waves have wavelengths greater than 0.1 m, with long radio waves up to 10⁶m . This follows from latest NCERT 2026-27 principle explaining the concept clearly for NEET students in simple steps as per rationalized syllabus.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

What is the nuclear density of a nucleus with mass 1.66 × 10⁻²⁷kg and radius 1.5 × 10⁻¹⁵m ? (Use π = 3.14 )

Given: What is the nuclear density of a nucleus with mass 1.66 × 10⁻²⁷kg and radius 1.5 × 10⁻¹⁵m ? (Use π = 3.14 ) Formula: Density = fracmassvolume, Volume = 4/3 π R³. Substitution & Calculation: R³ = (1.5 × 10⁻¹⁵)³ = 3.375 × 10⁻⁴⁵m³ . Volume = 4/3 × 3.14 × 3.375 × 10⁻⁴⁵approx 1.41 × 10⁻⁴⁴m³ . Density = frac1.66 × 10⁻²⁷¹.41 × 10⁻⁴⁴approx 1.18 × 10¹⁷kg/m³ . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

A rectangular loop of area 0.04 m² with 25 turns carries 3 A in a field of 1.2 T perpendicular to the plane. What is the

Given: A rectangular loop of area 0.04 m² with 25 turns carries 3 A in a field of 1.2 T perpendicular to the plane. What is the torque? Formula: tau = N I A B sin θ, θ = 90° to plane means sin 0° = 1 with normal. Substitution & Calculation: tau = 25 × 3 × 0.04 × 1.2 × 1 = 3.6 N m . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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