📝 Numerical 1.1 – Light Year in Metres

Is numerical mein humein light year ko metres mein convert karna hai. Light year wo distance hai jo light 1 year mein travel karti hai. Humein speed of light di gayi hai (3 × 10⁸ m/s). Pehle hum 1 year ko seconds mein convert karenge (365 days × 24 hours × 60 minutes × 60 seconds). Phir hum distance ka formula use karenge: Distance = Speed × Time. Is tarah humein light year ki value metres mein mil jayegi. Answer ko scientific notation mein likhna hai.

📝 Numerical 1.2 – Seconds in a Year and Years in a Second

Is numerical ke do parts hain. Part (a) mein humein yeh batana hai ke 1 year mein kitne seconds hote hain. Iske liye hum 365 days, 24 hours, 60 minutes, aur 60 seconds ko multiply karenge. Part (b) mein humein yeh batana hai ke 1 second mein kitne years hote hain. Iske liye hum part (a) ke answer ka reciprocal lein ge, yani 1 ko seconds ki total number se divide karenge. Dono answers ko standard form (scientific notation) mein likhna hai.

📝 Numerical 1.3 – Area of a Rectangle (Significant Figures)

Is numerical mein humein ek rectangular plate ka area nikalna hai. Length aur width di gayi hain. Humein area nikalne ke liye dono ko multiply karna hai. Lekin answer ko sahih significant figures mein dena hai. Is ke liye hum dekhte hain ke length aur width mein se kis mein significant figures ki number kam hai. Answer mein utni hi significant figures rakhni hain. Phir area ko round off karke final answer likhna hai.

📝 Numerical 1.4 – Sum of Masses (Precision)

Is numerical mein humein 4 different masses ko add karna hai. Sab masses kg mein di gayi hain. Humein sum nikalna hai aur answer ko sahih precision (decimal places) mein dena hai. Is ke liye hum dekhte hain ke di gayi masses mein se kis ki decimal places sab se kam hain. Answer mein utni hi decimal places rakhni hain. Phir sum ko round off karke final answer likhna hai.

📝 Numerical 1.5 – Volume of Cylinder with Uncertainty

Is numerical mein humein ek metal cylinder ka volume nikalna hai aur uski uncertainty bhi nikalni hai. Diameter aur length di gayi hain, aur Vernier Callipers ki least count bhi di gayi hai (0.01 cm). Pehle hum diameter se radius nikalte hain (radius = diameter/2). Phir volume ka formula use karte hain: Volume = π × r² × h. Is se volume aa jayega. Phir uncertainty nikalne ke liye, hum percentage uncertainty nikalte hain radius aur length mein, phir unko add karte hain (kyunki volume mein radius square hai, is liye radius ki uncertainty ko 2 se multiply karte hain). Phir absolute uncertainty nikal kar final answer likhte hain.

📝 Numerical 1.6 – Dimensional Analysis of vf² – vi² = 2aS

Is numerical mein humein yeh dikhana hai ke equation vf² – vi² = 2aS dimensionally correct hai. Is ke liye hum har term ke dimensions nikalte hain. vf aur vi velocity hain, unka dimension [LT⁻¹] hai. Square karne par [L²T⁻²] aa jata hai. a acceleration hai, dimension [LT⁻²] hai. S distance hai, dimension [L] hai. 2aS ka dimension [LT⁻²] × [L] = [L²T⁻²] aa jata hai. Dono sides ke dimensions same hain, is liye equation dimensionally correct hai.

📝 Numerical 1.7 – Dimensional Analysis of E = mc²

Is numerical mein humein Einstein ki famous equation E = mc² ko dimensionally check karna hai. E energy hai, is ka dimension [ML²T⁻²] hai. m mass hai, dimension [M] hai. c speed of light hai, dimension [LT⁻¹] hai. c² ka dimension [L²T⁻²] hai. Ab mc² ka dimension [M] × [L²T⁻²] = [ML²T⁻²] aa jata hai. Dono sides ke dimensions same hain, is liye equation dimensionally correct hai.

📝 Numerical 1.8 – Derive Formula for Time Period of Pendulum

Is numerical mein humein simple pendulum ke time period (T) ka formula derive karna hai using dimensional analysis. Humein bataya gaya hai ke T ka depend karta hai:

• Length of pendulum (l)

• Mass of bob (m)

• Angle (θ) – jo dimensionless hai

• Acceleration due to gravity (g)

Hum maan lete hain ke T = k × lᵃ × mᵇ × gᶜ (jahan k constant hai aur θ ko ignore kar dete hain kyunki dimensionless hai). Phir hum dimensions likhte hain: T ka dimension [T], l ka [L], m ka [M], g ka [LT⁻²]. Dono sides ke dimensions compare karte hain. M ki power zero aati hai, is liye mass ka koi effect nahi. L aur T ki powers ko solve karte hain. Aakhir mein formula aata hai: T = k × √(l/g). Yehi simple pendulum ka time period ka formula hai.