1.3.10 · D3Efficiency
Yeh Efficiency ka problem-drill child hai. Parent ne idea build kiya; yahan hum har tarah ke question dhundte hain jo efficiency puch sakti hai aur unme se ek-ek solve karte hain — including weird wale (zero output, 100% ideal, chained losses, "which is useful?" traps, aur ek exam twist). Kuch bhi naya assume nahi kiya: agar koi symbol aaya, toh woh parent mein earn hua tha ya yahan re-earn hoga.
Recall Woh do formulas jinpar hum rely karte hain (parent se)Efficiency woh useful fraction hai jo andar gayi cheez ki hai:
η = E total in E useful out = P in P useful out = 1 − E in E wasted E J P W η 0 1 100 \%).
Solve karne se pehle, chaliye har case class list karte hain jo ek question mein ho sakti hai. Neeche har worked example us cell ke saath tagged hai jise woh fill karta hai.
Cell
Case class
Kyun tricky hai
Example
C1 Standard energy input/output
Useful energy pick karoEx 1
C2 Power form (P E
Same-unit division
Ex 2
C3 Reverse solve (input/output/waste dhundho)
Formula rearrange karo
Ex 3
C4 Zero useful output (degenerate)
η = 0 Ex 4
C5 Ideal / limiting (η → 1
Frictionless bound; kyun real η < 1
Ex 5
C6 Chained machines (series)
Efficiencies multiply hoti hain
Ex 6
C7 "Which output is useful?" trap
Total ≠ useful
Ex 7
C8 Real-world word problem + unit clash
kJ vs J, per-second rates
Ex 8
C9 Exam twist (efficiency ek badi quantity ke andar hidden)
Ek non-η
Ex 9
Hum inhe upar se neeche chalenge. Ex 1 master figure bhi carry karta hai.
Worked example Ex 1 — Crane ek crate uthata haiEk crane 8000 J 40 kg 15 m g = 10 m/s 2 η
Forecast: Guess karo — kya η 80% (pause)
Step 1 — Useful output identify karo. Wanted kaam hai crate ko uthana , toh useful energy = Gravitational Potential Energy mein gain:
E useful = m g h = 40 × 10 × 15 = 6000 J . Yeh step kyun? Efficiency ke numerator mein sirf woh energy aati hai jo intended task ne ki — yahan, lifting.
Step 2 — Definition apply karo.
η = E in E useful = 8000 6000 = 0.75 = 75%. Yeh step kyun? Input poori supplied electrical energy hai; definition baaki kaam kar deti hai.
Verify: Waste = 8000 − 6000 = 2000 J 1 − 2000/8000 = 0.75 80%
Figure dekho: burnt-orange bar poora 8000 J
Worked example Ex 2 — Ceiling fanEk fan 60 W 27 W η
Forecast: Roughly aadha? Ek-chaurthai? Divide karne se pehle guess karo.
Step 1 — Powers ko directly divide karo. Dono watts mein hain, toh koi conversion nahi:
η = P in P useful = 60 27 = 0.45 = 45%. Yeh step kyun? Work and Power se, P = E / t t
Step 2 — Wasted power.
P waste = P in − P useful = 60 − 27 = 33 W . Yeh step kyun? Conservation input power ko useful + wasted mein split karta hai.
Verify: 27 + 33 = 60 1 − 33/60 = 0.45
Worked example Ex 3 — Input dhundhoEk machine 60% 300 J
Forecast: 300 J 400 500
Step 1 — Input ke liye definition rearrange karo. η = E useful / E in E in = η E useful = 0.60 300 = 500 J . Yeh step kyun? Hume output aur η E in 1 badhata hai, jo "zyada andar gaya" se match karta hai.
Step 2 — Wasted energy.
E waste = E in − E useful = 500 − 300 = 200 J . Yeh step kyun? Jo bhi useful nahi woh waste hai (heat, sound).
Verify: η = 300/500 = 0.60 200/500 = 0.40 = 1 − η
Worked example Ex 4 — Ek locked wall ko push karnaTum 250 J nahi hilti . Is "lifting" attempt ki efficiency kya hai?
Forecast: Lagta hai effort ka kuch toh count hona chahiye — par kya hota hai?
Step 1 — Useful output. Useful work = force × 0 E useful = 0 J . Yeh step kyun? Work ke liye motion chahiye (dekho Work and Power ). Koi displacement nahi ⇒ koi useful energy deliver nahi, chahe tum kitna bhi thak jao.
Step 2 — Efficiency.
η = 250 0 = 0 = 0%. Yeh step kyun? Numerator mein zero hone se η = 0 0 1
Verify: Saare 250 J Heat and Internal Energy ): waste = 250 − 0 = 250 J 1 − 250/250 = 0 degenerate floor hai — efficiency ka minimum.
Worked example Ex 5 — Frictionless dreamEk frictionless, resistance-free ramp system 2 kg 3 m m g h η 12 J Friction mein lose karta hai. Naya η
Forecast: Ideal case — kya η 100%
Step 1 — Ideal output = input. Koi loss nahi, toh har input joule PE ban jaata hai:
E useful = m g h = 2 × 10 × 3 = 60 J , E in = 60 J . η ideal = 60 60 = 1 = 100%. Yeh step kyun? η = 1 theoretical ceiling hai — sirf tab reachable jab E wasted = 0 limit hai, koi real machine nahi.
Step 2 — Real friction add karo. Ab input ko PE aur 12 J E in = 60 + 12 = 72 J , η = 72 60 = 0.8 3 ≈ 83.3%. Yeh step kyun? Friction dikhata hai kyun real η < 1 60 J
Verify: 60/72 = 5/6 ≈ 0.8333 1 − 12/72 = 1 − 1/6 = 5/6 Ceiling (100%
Worked example Ex 6 — Generator phir motorEk diesel generator 35% 80% 10 000 J
Forecast: Kya overall η chote stage (35%
Step 1 — Stage efficiencies multiply karo. Motor ka input hi generator ka output hai:
η total = η 1 × η 2 = 0.35 × 0.80 = 0.28 = 28%. Yeh step kyun? Fractions ke fractions multiply hote hain — stage one se jo 35% 80%
Step 2 — Joules trace karo confirm karne ke liye.
10 000 × 0.35 3500 J × 0.80 2800 J useful . η total = 10 000 2800 = 0.28 ✓ Yeh step kyun? Actual energy follow karna multiply rule prove karta hai aur yeh bhi ki result (28% sabse chhote stage se bhi neeche hai (35%
Verify: 0.35 × 0.80 = 0.28 2800/10000 = 0.28 35%
Worked example Ex 7 — Light bulbEk filament bulb 100 W 90 W 10 W lamp ke roop mein, uski efficiency kya hai?
Forecast: Tempting hai kehna "100 W 100%
Step 1 — Intended purpose pick karo. Lamp ka wanted output light hai, heat nahi:
P useful = 10 W ( sirf light ) . Yeh step kyun? Parent ka rule: sirf intended form count karta hai. Yahan heat waste hai (Heat and Internal Energy ), chahe woh real energy ho.
Step 2 — Efficiency.
η = 100 10 = 0.10 = 10%. Yeh step kyun? Total output (100 W 100%
Verify: Light + heat = 10 + 90 = 100 W = 10/100 = 10%
100 100% Fix: Total output ≠ useful output. Puchho mujhe kya chahiye tha? — light. η = 10% 100%
Worked example Ex 8 — Kettle paani ubaal raha haiEk kettle rated 2.0 kW 90 s 150 kJ η
Forecast: kW, seconds, kJ sab jumbled — kya yeh around 80%
Step 1 — Input energy matching units mein compute karo. Work and Power se, E = P × t E in = 2.0 kW × 90 s = 2000 W × 90 s = 180 000 J = 180 kJ . Yeh step kyun? Divide karne se pehle input aur output ko same unit mein express karna zaroori hai — parent ki unit-mismatch warning. Yahan dono kJ ban jaate hain.
Step 2 — Efficiency.
η = E in E useful = 180 kJ 150 kJ = 0.8 3 ≈ 83.3%. Yeh step kyun? Ab yeh ek clean pure ratio hai; kJ cancel ho jaate hain.
Verify: 150/180 = 5/6 ≈ 0.8333 = 180 − 150 = 30 kJ 1 − 30/180 = 5/6
η 70% 4200 J 10 kg height tak uthata hai? (g = 10 m/s 2
Forecast: Tumhe η h h
Step 1 — Pehle useful energy dhundho. Efficiency batati hai kitna input PE ban gaya:
E useful = η × E in = 0.70 × 4200 = 2940 J . Yeh step kyun? Hoist sirf apna 70%
Step 2 — Useful energy ko height mein convert karo. Useful energy PE = m g h h = m g E useful = 10 × 10 2940 = 100 2940 = 29.4 m . Yeh step kyun? m g h = E useful h andar hai, answer nahi.
Verify: Backward check: PE gained = 10 × 10 × 29.4 = 2940 J η = 2940/4200 = 0.70 4200/100 = 42 m 29.4 m
Recall Coverage tick-listWhich cell forces η = 0 η = 1 multiplying efficiencies? ::: C6 — machines in series.
Which cell is the "total ≠ useful" trap? ::: C7 — the light bulb.
Which cell needs unit conversion before dividing? ::: C8 — kettle (kW·s → kJ).
Which cell asks for a non-η h
"U over IN, watch the WIN." Useful ÷ In. Par pehle setup jeet lo : (1) kaun si energy useful hai? (2) same units? (3) chained ⇒ multiply.
