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Let P be an arbitrary point having sum of the squares of the distance from the planes x + y + z = 0, l x – nz = 0 and x – 2y + z = 0, equal to 9. If the locus of the point P is x 2 + y 2 + z 2 = 9, then the value of l – n is equal to _______. JEE MAIN ONLINE PAPER 2021 Held on March 17, 2021 (Evening) Hints & Solutions PHYSICS SECTION-A 1.[4] v 0 = gh 2 v = gh 2 gh 2 e = ⇒ e = 0.9 S = h + 2e 2 h + 2e 4 h +.......... t = ......... g h 2 e 2 g h 2 e 2 g h 2 2 + + + s / m 5 . 2 t s v av = = 2.[2] f = 4 + 3 + 3 = 10 assuming non linear 2 . 1 10 12 f 2 1 C C V P = = + = = β 3.[Bonus] Official Ans. by NTA (NA) A = A 0 e – γ t ln2 = m 2 b × 120 b 120 1 2 693 . 0 = × × 1.16 × 10 –2 kg/sec. 4.[4] Conceptual 5.[4] P = h ρ g 2 – 3 3 10 36 . 1 8 . 9 10 10 2 V V p × × × × = Δ = β = 1.44 × 10 9 N/m 2 6.[3] T ∝ R 3/2 2 / 3 R 3 R 12 T 24 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⇒ T = 3hr 7.[4] w = 2 π f = 1.5 × 10 3 2 6 A = = 3 cm = 0.03 m 8.[2] Option (a) is wrong ; since in adiabatic process V ≠ constant. Option (b) is wrong, since in isothermal process T = constant Option (c) & (d) matches isothermes & adiabatic formula : TV γ –1 = constant & 1 – p T γ γ = constant 9.[2] v = v 0 + gt + Ft 2 dt ds = v 0 + gt + Ft 2 ∫ ∫ + + = 1 0 2 0 dt ) Ft gt v ( ds 1 0 2 2 0 3 Ft 2 gt t v s ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + + = s = 3 F 2 g v 0 + + 10.[4] Band width = 2 f m ω m = 1.57 × 10 8 = 2 π f m BW = 2f m = Hz 2 10 8 = 50 MHz 11.[2] E 2E R i 2 1 r r R E 3 i + + = TPD = 2E – ir 1 = 0 2E = ir 1 2 1 1 r r R r E 3 E 2 + + × = 2R + 2r 1 + 2r 2 = 3r 1 2 1 r – 2 r R = JEE-Main Online Paper [March, 2021] 2021 | 105 12.[2] B = 2 × B st.wire + B loop ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ π π + π × = 2 r 2 i μ r 4 i μ 2 B 0 0 ) 2 ( r 4 i μ B 0 π + π = 13.[3] 100 Ω 10 Ω G 60 Ω 5 Ω A C D B 10V 10 0 x y 0 10 0 – x 15 y – x 100 10 – x = + + 53x – 20y = 30 …

1. 0 5 0 – y 15 x – y 60 10 – y = + + 17 y – 4x = 10 ......
2. on solving
3. &
4. x = 0.865 y = 0.792 Δ V = 0.073 R = 15W i = 4.87 mA 14.[4] A 1 ω 1 = A 2 ω 2 m k A m k A 2 2 1 1 = 1 2 2 1 k k A A = 15.[4] (a) I V= V R (b) V L I (c) V C I (d) R X – X V V – V tan C L R C L = = φ 16.[1] C comes to rest V cm of A & B = 2 v μ = m 2 m m × = 2 m ⇒ 2 1 μ 2 2 m kx 2 1 v = v k 2 m k v μ

Correct answer: on solving

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