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PhysicsClass 11CBSE

λ₁ = 350 nm, λ₂ = 450 nm; maximum velocities differ by factor 2 (v₁ = 2v₂). Find work function φ.

Einstein’s equations:hc/λ₁ = φ + ½mᵉv₁² ...(1)hc/λ₂ = φ + ½mᵉv₂² ...(2)Substitute v₁ = 2v₂ ⇒ v₁² = 4v₂² and subtract (2) from (1):hc(1/λ₁ − 1/λ₂) = (3/2)mᵉv₂²Calculate left side:1/350nm − 1/450nm = 100/(350×450) nm⁻¹ = 6.349×10⁵ m⁻¹hc × 6.349×10⁵ = 1.262×10⁻¹⁹ JSolve for v₂²:v₂² = (2/3) × 1.262×10⁻¹⁹ / 9.1×10⁻³¹ = 9.245×10¹⁰ m²/s²Substitute into equation (2):hc/λ₂ = (6.626×10⁻³⁴×3×10⁸) / (450×10⁻⁹) = 4.417×10⁻¹⁹ J½mᵉv₂² = 0.5 × 9.1×10⁻³¹ × 9.245×10¹⁰ = 4.206×10⁻²⁰ Jφ = 4.417×10⁻¹⁹ − 0.4206×10⁻¹⁹ ≈ 3.996×10⁻¹⁹ J ≈ 2.5 eVAnswer: Work function φ ≈ 2.5 eV

PhysicsClass 11CBSE

Charges −q, +q, −q at corners of equilateral triangle (side a). Find net force on +q at centroid O.

Distance from centroid to vertex: r = a/√3Force magnitude from each corner on +q at O:F = kq²/r² = kq²/(a²/3) = 3kq²/a²Directions (place +q corner at top, −q corners at bottom-left and bottom-right):From +q at top: repels downward (0, −F)From −q at bottom-L: attracts toward B-L (−F√3/2, −F/2)From −q at bottom-R: attracts toward B-R (+F√3/2, −F/2)Sum of components:Ex = 0 − F√3/2 + F√3/2 = 0Ey = −F − F/2 − F/2 = −2FF_net = 2F = 6kq²/a²Answer: F_net = 6kq²/a², directed away from the +q corner (toward midpoint of opposite side)Symmetry only guarantees x-cancellation here. The y-components all add never assume the force is zero without resolving components.

PhysicsClass 11CBSE

Two identical charges q₁ = q₂ = 5 μC. Find largest charge q to transfer so force decreases to 1/2.5 of original.

Step-1: After transferring charge q: new charges = (5−q) and (5+q) μCStep-2: New force = F₀ / 2.5:k(5−q)(5+q)/r² = k×25/(2.5×r²) = 10k/r²(5−q)(5+q) = 1025 − q² = 10 ⇒ q² = 15 ⇒ q = √15 μCAnswer: q = √15 μC ≈ 3.87 μCNote: (5−q)(5+q) = 25−q² is the difference-of-squares identity. Transferring charge always decreases the product q₁q₂ for charges of the same sign.

PhysicsClass 11CBSE

Charges +Q, −Q, +Q, −Q at corners of a square (side 5 cm). Find field at centre. (Q = 1 μC)

With alternating charges, the two +Q charges are at diagonally opposite corners, and the two −Q charges are at diagonally opposite corners.Field at centre from each pair:Both +Q fields: exactly opposite directions ⇒ cancelBoth −Q fields: exactly opposite directions ⇒ cancelEvery field vector has an equal and opposite partner by symmetry.Answer: Electric field at centre = 0 (zero)

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