PHYSICS

PHYSICS

The dimensional formula of energy density is [M L^{-1 T^{-2] . What is the dimensional formula of energy flux (energy pe

Given: The dimensional formula of energy density is [M L^{-1 T^{-2] . What is the dimensional formula of energy flux (energy per unit area per unit time)? Formula: Energy flux = Energy / Area / Time. Substitution & Calculation: [M L² T^{-2] / [L²] / [T] = [M L² T^{-2] [L^{-2] [T^{-1] = [M T^{-3] . 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

An electron moves at 6.5 × 10⁶m/s perpendicular to a field of 0.2 T . What is the radius of its path? (Mass = 9.1 × 10⁻³

Given: An electron moves at 6.5 × 10⁶m/s perpendicular to a field of 0.2 T . What is the radius of its path? (Mass = 9.1 × 10⁻³¹kg, charge = 1.6 × 10⁻¹⁹C ) Formula: r = mv/qB. Substitution & Calculation: r = frac9.1 × 10⁻³¹ × 6.5 × 10⁶¹.6 × 10⁻¹⁹ × 0.2 = frac5.915 × 10⁻²⁴³.2 × 10⁻²⁰= 1.848 × 10⁻⁴m approx 0.0185 cm . 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 spring of k = 250 N/m has a 2.5 kg mass. If E = 1.25 J, what is the amplitude?

Given: A spring of k = 250 N/m has a 2.5 kg mass. If E = 1.25 J, what is the amplitude? Formula: Total energy: E = 1/2 k A². Substitution & Calculation: 1.25 = 0.5 × 250 × A² Rightarrow 1.25 = 125 A² Rightarrow A² = 0.01 Rightarrow A = 0.1 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 gas occupies 5.6 litres at STP. How many moles of the gas are present? (Molar volume at STP = 22.4 litres)

Given: A gas occupies 5.6 litres at STP. How many moles of the gas are present? (Molar volume at STP = 22.4 litres) Formula: Number of moles (μ) = fracVolumeMolar volume. Substitution & Calculation: μ = 5.6/22.4 = 0.25 mol. 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

An electron is accelerated through 50 V . What is its de Broglie wavelength? (Take h = 6.63 × 10⁻³⁴J s, m_e = 9.11 × 10⁻

Given: An electron is accelerated through 50 V . What is its de Broglie wavelength? (Take h = 6.63 × 10⁻³⁴J s, m_e = 9.11 × 10⁻³¹kg, e = 1.6 × 10⁻¹⁹C ) Formula: K = e V = 1.6 × 10⁻¹⁹ × 50 = 8.0 × 10⁻¹⁸J. Substitution & Calculation: p = √2 m K = √2 × 9.11 × 10⁻³¹ × 8.0 × 10⁻¹⁸approx 3.816 × 10⁻²⁴kg m/s . lambda = h/p = frac6.63 × 10⁻³⁴³.816 × 10⁻²⁴approx 1.737 × 10⁻¹⁰m = 0.1737 nm . 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 square loop of side 26 cm rotates at 12 rad/s in a 0.3 T field. What is the maximum emf induced?

Given: A square loop of side 26 cm rotates at 12 rad/s in a 0.3 T field. What is the maximum emf induced? Formula: A = (0.26)² = 0.0676 m². Substitution & Calculation: ε_0 = N B A omega = 1 × 0.3 × 0.0676 × 12 = 0.24336 V approx 0.243 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), Chapter: Electromagnetic Induction (Latest NCERT 2026-27), Topic: Rotating coil, maximum emf ε₀ = NBAω, N = 300, B = 0.07 T, A = 0.012 m². The section explains governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, SI.

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PHYSICS

A wire of length 2.2 m carrying 3.5 A is at 60° to a magnetic field of 0.2 T . What is the force on the wire?

Given: A wire of length 2.2 m carrying 3.5 A is at 60° to a magnetic field of 0.2 T . What is the force on the wire? Formula: Force F = I l B sin θ. Substitution & Calculation: F = 3.5 × 2.2 × 0.2 × sin 60° = 7.7 × 0.2 × 0.866 = 1.3336 approx 1.33 N . 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 bubble of radius 5.5 mm is blown at 45 cm depth in water ( rho = 1000 kg/m³, S = 0.0727 N/m ). What is the total press

Given: A bubble of radius 5.5 mm is blown at 45 cm depth in water ( rho = 1000 kg/m³, S = 0.0727 N/m ). What is the total pressure inside? (Take P_a = 1.01 × 10⁵Pa, g = 10 m/s² ) Formula: P_i = P_a + rho g h + 2 S/r. Substitution & Calculation: P_a = 1.01 × 10⁵Pa, rho g h = 1000 × 10 × 0.45 = 4500 Pa . 2 S/r = frac2 × 0.07275.5 × 10⁻³= 26.44 Pa . P_i = 1.01 × 10⁵+ 4500 + 26.44 = 1.05526 × 10⁵Pa . 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

Three particles of masses 2 kg, 4 kg, and 6 kg are at (0, 2), (3, 0), and (1, 4) respectively. What is the x-coordinate

Given: Three particles of masses 2 kg, 4 kg, and 6 kg are at (0, 2), (3, 0), and (1, 4) respectively. What is the x-coordinate of their nter of mass? Formula: Formula: X = m_1 x_1 + m_2 x_2 + m_3 x_3/m_1 + m_2 + m_3. Substitution & Calculation: Masses: 2, 4, 6 kg ; x-coordinates: 0, 3, 1 . X = (2 × 0) + (4 × 3) + (6 × 1)/2 + 4 + 6 = 0 + 12 + 6/12 = 18/12 = 1.5 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

What is the orbital period of an electron in the n = 4 orbit if v_1 = 2.2 × 10⁶m/s and r_1 = 5.3 × 10⁻¹¹m ?

Given: What is the orbital period of an electron in the n = 4 orbit if v_1 = 2.2 × 10⁶m/s and r_1 = 5.3 × 10⁻¹¹m ? Formula: v_4 = 2.2 × 10⁶/4 = 5.5 × 10⁵m/s. Substitution & Calculation: r_4 = 16 × 5.3 × 10⁻¹¹= 8.48 × 10⁻¹⁰m . T = 2π r_4/v_4 = frac2 × 3.14 × 8.48 × 10⁻¹⁰⁵.5 × 10⁵approx 9.68 × 10⁻¹⁵s . 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 solenoid produces B = 1.2 T with a core of μ_r = 400 and n = 1500 m^{-1 . What is the current I ? (Take μ_0 = 4π × 10⁻

Given: A solenoid produces B = 1.2 T with a core of μ_r = 400 and n = 1500 m^{-1 . What is the current I ? (Take μ_0 = 4π × 10⁻⁷T m A^{-1 ). Formula: B = μ_0 μ_r n I, so I = B/μ_0 μ_r n. Substitution & Calculation: Given: B = 1.2 T, μ_r = 400, n = 1500 m^{-1, μ_0 = 4π × 10⁻⁷. I = frac1.24π × 10⁻⁷ × 400 × 1500 = frac1.27.539 × 10⁻¹approx 1.592 A approx 1.6 A . 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 conductor has a surface charge density of 1.5 × 10⁻⁶C/m² . What is the electric field just outside it? (Take ε_0 = 8.8

Given: A conductor has a surface charge density of 1.5 × 10⁻⁶C/m² . What is the electric field just outside it? (Take ε_0 = 8.85 × 10⁻¹²C² N^{-1 m^{-2 ). Formula: E = sigma/ε_0 = frac1.5 × 10⁻⁶⁸.85 × 10⁻¹²approx 1.695 × 10⁵N/C .. 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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