Exercises — SI units — seven base units and all derived units

Base unit woh hai jo saat fundamental bricks mein se ek hai; derived unit unse equation ke zariye bana hota hai.

• $\text{kelvin (K)}$ → base (temperature).
• $\text{mole (mol)}$ → base (amount of substance).
• $\text{candela (cd)}$ → base (luminous intensity).
• $\text{newton (N)}$ → derived, $F=ma$ se, toh ${\text{kg⋅m⋅s}}^{-2}$.
• $\text{watt (W)}$ → derived, $P=E_W/t$ se (power = work per time), toh ${\text{kg⋅m}}^2{\text{⋅s}}^{-3}$.

Work $E_W=Fd$ se (work = force along a distance): newton ke bricks ko ek metre se multiply karo. Yahan $\text{J}$ bas woh naam hai jo hum resulting brick-pattern ko dete hain. $[\text{J}]=[\text{N}]\cdot [\text{m}]=({\text{kg⋅m⋅s}}^{-2})\cdot \text{m}={\text{kg⋅m}}^2{\text{⋅s}}^{-2}.$

Kya hai: pressure force ko ek area mein spread karna hai. Kyun substitute karte hain: pascal ka koi independent standard nahi — yeh is equation se generate hota hai. $[p]=\frac{[F]}{[A]}=\frac{{\text{kg⋅m⋅s}}^{-2}}{{\text{m}}^2}={\text{kg⋅m}}^{-1}{\text{⋅s}}^{-2}.$ Numerator mein $\text{m}$ neeche ${\text{m}}^2$ se milta hai: ${\text{m}}^{1-2}={\text{m}}^{-1}$.

Step 1 (acceleration): $a=\frac{\Delta v}{\Delta t}=\frac{30-0}{10}=3{\text{m/s}}^2$. Step 2 (force): $F=ma=1200\times 3=3600\text{N}$. Unit check: $[F]=\text{kg}\cdot \frac{\text{m}}{{\text{s}}^2}={\text{kg⋅m⋅s}}^{-2}=\text{N}$, jaise hona chahiye. $a=\Delta v/\Delta t$ step ke liye Kinematics — velocity and acceleration aur $F=ma$ ke liye Newton's Laws of Motion dekho.

Charge: $[Q]=[I][t]=\text{A⋅s}=\text{C}$. Voltage: energy per charge, toh $[V]=\frac{[E_W]}{[Q]}=\frac{{\text{kg⋅m}}^2{\text{⋅s}}^{-2}}{\text{A⋅s}}={\text{kg⋅m}}^2{\text{⋅s}}^{-3}{\text{⋅A}}^{-1}.$ ${\text{s}}^{-2}$ over ${\text{s}}^{+1}$ deta hai ${\text{s}}^{-3}$; denominator mein $\text{A}$ ban jaata hai ${\text{A}}^{-1}$.

Hum kya karte hain: charge bricks ko voltage bricks se divide karo. Kyun: capacitance hai "charge stored per volt applied", toh iski recipe exactly woh ratio $Q/V$ hai. $[\mathcal{C}]=\frac{[Q]}{[V]}=\frac{\text{A⋅s}}{{\text{kg⋅m}}^2{\text{⋅s}}^{-3}{\text{⋅A}}^{-1}}.$ Dividing matlab exponents subtract karo aur denominator ke signs flip karo:

Step 1 — ohm ke bricks nikalo. Kyun? Hum $I^{2}R$ check nahi kar sakte jab tak hume pata nahi ki $R$ kya contribute karta hai, aur $R=V/I$ batata hai: volt ke bricks ko ek ampere se divide karo. $[R]=\frac{[V]}{[A]}=\frac{{\text{kg⋅m}}^2{\text{⋅s}}^{-3}{\text{⋅A}}^{-1}}{\text{A}}={\text{kg⋅m}}^2{\text{⋅s}}^{-3}{\text{⋅A}}^{-2}.$ (Neeche akela $\text{A}$ ampere power ko ek se ghata deta hai: ${\text{A}}^{-1-1}={\text{A}}^{-2}$.)

Step 2 — $I^2$ se multiply karo. Kyun? Resistor mein dissipated power $P=I^{2}R$ hai, toh iski unit test karne ke liye hum $[I]=\text{A}$ squared aur abhi nikale ohm bricks substitute karte hain. $[I^{2}R]={\text{A}}^2\cdot ({\text{kg⋅m}}^2{\text{⋅s}}^{-3}{\text{⋅A}}^{-2})={\text{kg⋅m}}^2{\text{⋅s}}^{-3}\underset{={\text{A}}^0=1}{\underbrace{{\text{A}}^{2-2}}}={\text{kg⋅m}}^2{\text{⋅s}}^{-3}=\text{W}.$ Kyun amperes gayab hote hain: current squared appear hota hai ($+2$) aur ohm ${\text{A}}^{-2}$ carry karta hai, toh woh add hokar ${\text{A}}^0=1$ ho jaate hain, exactly watt chhodke.

Hum magnetic flux ko $\Phi$ likhte hain (Greek capital "phi", flux ke liye ek common symbol — yeh measure karta hai ki kitna magnetic field ek loop se guzarta hai). Bricks ko table se padhte hue, ${\text{kg⋅m}}^2{\text{⋅s}}^{-2}{\text{⋅A}}^{-1}$ weber (Wb) hai, us flux $\Phi$ ki unit. Yeh $\Phi =V\cdot t$ se aata hai (flux = voltage × time), kyunki $[V][t]=({\text{kg⋅m}}^2{\text{⋅s}}^{-3}{\text{⋅A}}^{-1})\cdot \text{s}={\text{kg⋅m}}^2{\text{⋅s}}^{-2}{\text{⋅A}}^{-1}$, exactly match karta hai.

Step 1 — $G$ isolate karo. Kyun? $G$ ki unit chhupi rehti hai jab tak $G$ ek taraf akela nahi hota, toh hum law rearrange karte hain: $G=\frac{F{r}^2}{m_1{m}_2}$. Kyunki $[r]=\text{m}$ hai, $[r^2]={\text{m}}^2$ hai. Base bricks substitute karo: $[G]=\frac{[F][r^2]}{[m{]}^2}=\frac{({\text{kg⋅m⋅s}}^{-2})\cdot {\text{m}}^2}{{\text{kg}}^2}={\text{m}}^3{\text{⋅kg}}^{-1}{\text{⋅s}}^{-2}.$ Step 2 — density. $[\rho ]=\frac{\text{kg}}{{\text{m}}^3}={\text{kg⋅m}}^{-3}$. Step 3 — product $G\rho$. Kyun multiply karte hain? Expression $G\rho$ maangta hai, toh hum bricks combine karte hain exponents add karke: $[G\rho ]=({\text{m}}^3{\text{⋅kg}}^{-1}{\text{⋅s}}^{-2})({\text{kg⋅m}}^{-3})=\underset{1}{\underbrace{{\text{m}}^{3-3}}}\underset{1}{\underbrace{{\text{kg}}^{-1+1}}}{\text{s}}^{-2}={\text{s}}^{-2}.$ Step 4 — square root. Kyun? Expression $\sqrt{G\rho }$ hai, aur square root har exponent ko half karta hai: $[\sqrt{G\rho }]={\text{s}}^{-1}$ = ek frequency (hertz). Khoobsurat: $\sqrt{G\rho }$ ek inverse-time hai — yeh ek natural gravitational timescale set karta hai.

Kyun check karte hain solve nahi? Hum koi number nahi nikal rahe; hum test kar rahe hain ki dono sides ke bricks match karte hain ya nahi. Agar nahi karte, toh equation definitely galat hai (dekho Dimensional Analysis). $\left[\frac{GM}{r}\right]=\frac{({\text{m}}^3{\text{⋅kg}}^{-1}{\text{⋅s}}^{-2})\cdot \text{kg}}{\text{m}}={\text{m}}^{3-1}{\text{⋅kg}}^{-1+1}{\text{⋅s}}^{-2}={\text{m}}^2{\text{⋅s}}^{-2}.$ $2$ dimensionless hai aur drop ho jaata hai. Square root lo: $\sqrt{{\text{m}}^2{\text{⋅s}}^{-2}}={\text{m⋅s}}^{-1}=\text{speed}.$ Consistent. (Note: consistency yeh prove nahi karta ki factor $2$ sahi hai — dimension checks pure numbers nahi dekh sakte.)

Yeh route kyun? Figure watt ko newton ÷ second ke roop mein build karta hai ($P=E_W/t$ se). Yahan hum ise alag tarike se — force × velocity — banate hain, aur agar physics consistent hai toh bricks agree karni chahiye. Substitute karo ($[F]={\text{kg⋅m⋅s}}^{-2}$, $[v]={\text{m⋅s}}^{-1}$) aur exponents add karo kyunki hum multiply kar rahe hain: $[Fv]=({\text{kg⋅m⋅s}}^{-2})({\text{m⋅s}}^{-1})={\text{kg⋅m}}^{1+1}{\text{⋅s}}^{-2-1}={\text{kg⋅m}}^2{\text{⋅s}}^{-3}=\text{W}.$ Figure ke watt jaisa same box — dono routes agree karte hain. Work, Energy and Power dekho.

Step 1 — $B$ isolate karo. Kyun? Pehle jaisi wajah: field ki unit tabhi reveal hoti hai jab $B$ akela ho, toh $B=\frac{F}{qv}$. Step 2 — denominator ke bricks, har exponent carefully build karo. Kyun carefully? Stray $\text{s}$ powers yahan chhup jaate hain, toh hum cancel karne se pehle har base brick ka exponent explicitly likhte hain. With $[q]=\text{A⋅s}$ aur $[v]={\text{m⋅s}}^{-1}$:

Hum kya karte hain: Planck ke bricks ko charge ke bricks se divide karo. Kyun: $X$ defined hai ratio $h/e$ ke roop mein, toh iski unit-recipe exactly $[h]/[e]$ hai.

(a) $a^0=1$ (ek pure number), toh $[Y]=[m]\cdot [a^0]=\text{kg}\cdot 1=\text{kg}$. Bas ek mass. (b) Har exponent batata hai ki tum us brick ki kitni copies multiply karte ho. Power $0$ = "us brick ki zero copies" = empty product = dimensionless $1$. Toh $[\text{kuch bhi}{]}^0$ dimensionless hai — yeh recipe mein kuch nahi add karta. Edge case: agar quantity itself literally zero hai (jaise $a=0{\text{m⋅s}}^{-2}$), toh $a^0=0^0$ mathematically undefined hai. Physics mein hum isse avoid karte hain: unit analysis $a$ ko ek symbol treat karta hai jo bricks carry karta hai, kabhi number $0$ nahi, toh $[a{]}^0=1$ hamesha safe hai. Trap sirf numerical $0^0$ mein hai, dimensional mein nahi. (c) Radian (opening definition se) ek aur dimensionless quantity hai: $\text{m}/\text{m}=1$. Dono $a^0$ aur $\text{rad}$ same "no-brick" pattern $1$ par land karte hain — ek unit check unhe alag nahi kar sakta, yahi exactly wajah hai ki named-but-dimensionless units exist karte hain taaki humans yaad rakhein ratio ka matlab kya tha.

Har rung ne same move use kiya: $[X]$ likho, saat base bricks substitute karo, multiply karte waqt exponents add karo / divide karte waqt subtract karo, aur dimensionless numbers ko girane do.