If 10 x sin x e x ) x 1 ( log xe lim 2 x 2 e x 0 x = γ + + β − α − → , α, β, γ∈ R then the value of α + β + γ is _____ . JEE MAIN ONLINE PAPER 2021 Held on JULY 20, 2021 (Evening) Hints & Solutions PHYSICS Section -A 1.[1] K 2 = 4K 1 2 1 2 2 mv 2 1 4 mv 2 1 = v 2 = 2v 1 p = mv p 2 = mv 2 =

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If 10 x sin x e x ) x 1 ( log xe lim 2 x 2 e x 0 x = γ + + β − α − → , α, β, γ∈ R then the value of α + β + γ is _____ . JEE MAIN ONLINE PAPER 2021 Held on JULY 20, 2021 (Evening) Hints & Solutions PHYSICS Section -A 1.[1] K 2 = 4K 1 2 1 2 2 mv 2 1 4 mv 2 1 = v 2 = 2v 1 p = mv p 2 = mv 2 = 2mv 1 p 1 = mv 1 % change = 100 P P 1 × Δ = 1 1 1 mv mv mv 2 − × 100 = 100% 2.[3] L = Length of escalator 1 esc / b t L V = When only escalator is moving 2 esc t L V = when both are moving V b/g = V b/esc + V esc 2 1 g / b t L t L V + = ⇒ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + = = 2 1 2 1 g / b t t t t v L t 3.[4] T 2 ∝ R 3 T = kR 3/2 R dR 2 3 T dT = = 03 . 0 02 . 0 2 3 = × % Change = 3% 4.[1] β − β + = 1 1 f f 0 c v = β β − β + 1 1 f f 0 1 2 0 ) 1 ( ) 1 ( f f 1 − β − β + = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ Δ + β is small compared to 1 ) 2 1 ( f f 2 1 0 β + = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ Δ + c v f f 0 = Δ = β s / km 6 . 305 5890 c 6 v = × = 5.[2] T 1 = k( A 1 – A 0 ) T 2 = k( A 2 – A 0 ) 0 2 0 1 2 1 T T A A A A − − = 0 2 1 1 2 1 1 T T T T A A A = − − 6.[4] Direction of propagation = j ˆ k ˆ i ˆ B E − = × = × → → 7.[4] R = 100 Ω X L = ω L = 50 π × 10 –3 π = ω = 100 10 C 1 X 11 C X C >> X L & |X C –X L | >> R 8.[3] PV = nRT PV ∝ T Straight line with positive slope(nR) 9.[4] v 2 = ω 2 (A 2 –x 2 ) 2 2 2 2 2 2 2 1 2 1 2 v x v x A ω + = ω + = 2 2 2 1 2 1 2 2 2 x x v v − − = ω 2 1 2 2 2 2 2 1 v v x x 2 T − − = π = 10.[3] mv h = λ Kinetic energy, C 2 2 2 hc m 2 h m 2 P λ = λ = h c m 2 2 C λ = λ 2 | 2021 JEE-Main Online Paper [July, 2021] 11.[2] 2 2 mv 2 1 2 1 I 2 1 × = ω 2 mR 2 1 I = Body is solid cylinder 12.[4] m ∝ t a v b A c m ∝ [Τ] a [LT –1 ] b [ML 2 T –1 ] c M 1 L 0 T 0 = M C L b+2c T a–b–c comparing powers c = 1, b = –2, a = –1 m ∝ t –1 v –2 A 1 13.[1] f 2 1 + = γ 1 2 f − γ = 14.[4] R = R 0 e – λ t A nR = A nR 0 – λ t – λ is slope of straight line 20 3 = λ 62 . 4 2 n t 2 / 1 = λ = A 15.[2] T A = T B (since ω A – ω B ) 16.[4] A tan δ = tan δ cos θ = tan45° cos 30° 2 3 1 tan × = δ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = δ − 2 3 tan 1 17.[1] P = constant Pt mv 2 1 2 = ⇒ t v ∝ t C dt dx = C = constant by integration 1 2 1 t t C x 1 2 + + = x ∝ t 3/2 18.[4] x | Q | | P | = = → → ...(i) | Q P | n | Q P | → → → → − = + P 2 + Q 2 + 2PQ cos θ = n 2 (P 2 + Q 2 –2PQcos θ ) Using (i) in above equation 2 2 n 1 1 n cos + − = θ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − = = θ − 1 n 1 n cos 2 2 1 19.[1] R R R' + = 3 3 3 ' R 3 4 R 3 4 R 3 4 π = π + π R 2 ' R 3 1 = ... (i) ] R 4 [ 2 A 2 i π = 2 f ' R 4 A π = 3 / 1 2 3 / 2 2 f i f i 2 R 2 R 2 A A U U = = = 20.[2] μ = μ 0 (1 + x m ) = 4 π × 10 –7 × 500 = 2 π × 10 –4 H/m Section -B 21.[192] P = Vi 0.5 = 8i A 16 1 i = E = 20 = 8 + iR P R P = 12 × 16 = 192 Ω 22.[3] d a t 2 1 t = d a a s 2 2 1 a s 2 = .... (i) a a = gsin θ + μ gcos θ g 2 3 2 g μ + = a d = gsin θ – μ gcos θ JEE-Main Online Paper [July, 2021] 2021 | 3 = g 2 3 2 g μ − using the above values of a a and a d and putting in equation (i) we will gate 5 3 = μ 23.[10] When switch S 1 and S 2 are closed 12 6 12 6 2 6 12 6 12 + × + + + × Ω = + + = + + 10 4 2 4 18 72 2 18 72 24.[3] I in both cases in about point of contact Ring 2 I 2 1 mgh ω = 2 2 R 2 R v ) mR 2 ( 2 1 mgh = gh v R = Solid cylinder 2 I 2 1 mgh ω = 2 2 C 2 R v mR 2 3 2 1 mgh ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = 3 gh 4 v C = 2 3 v v C R = 25.[25] 65 . 0 75 . 0 10 1 5 1 di 1 di dV R 3 d − × − = = = − Ω = 25 4 100 26.[125] X L = X C (due to resonance) Z = R so R V Z V i rms = = W 10 125 5 250 250 R V 2 2 × = × = 27.[17258] Process of isothermal ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 1 2 v v n nRT W A = 1 × 8.3 × 300 × A n2 = 17258 × 10 –1 J 28.[4] hc hc KE φ − λ = φ − λ = 0 0 hc ) V 3 ( e .... (i) φ − λ = 0 0 2 hc eV ....(iii) using (i) & (ii) t 0 hc 4 hc λ = λ = φ λ t = 4 λ 0 29.[200] ω f = ω 0 + α t α = 1200 × 6 2 0 t 2 1 t α + ω = θ = 36 1 6 1200 2 1 60 10 600 × × × + × θ = 200 30.[20] 16 N 8 N 4 N 2 N N 0 t 0 t 0 t 0 t 0 2 1 2 1 2 1 2 1 ⎯→ ⎯ ⎯→ ⎯ ⎯→ ⎯ ⎯→ ⎯ 4×t 1/2 = 80 t 1/2 = 20 days CHEMISTRY Section -A 31.[4] and || O || O are metamers 32.[4] 4 | 2021 JEE-Main Online Paper [July, 2021] 33.[2] 34.[3] Informative, according to ncert uses of di hydrogen. In fact NH 3 largest production in used to manufacture nitrogenous fertilisers. 35.[1] (A) ) g

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s ( 3 CO ZnO ) S ( ZnCO + ⎯→ ⎯ Δ Heating in absence of oxygen in calcination. (B) 2Zns(s) + 3O 2 (g) → 2ZnO(g) + 2SO 2 (g) heating in presence of oxygen in roasting Hence (A) is calcination while (B) in roasting. 36.[4] (i) [Ag + ] required to ppt AgCl(s) K sp = IP = [Ag + ] [Cl – ] = 1.7 × 10 –10 [Ag + ] = 1.7 × 10 –9 (ii) [Ag + ] required to ppt Ag 2 CrO 4 (s) K sp = IP = [Ag + ] 2 [CrO 4 –2 ] = 1.9 × 10 –12 [Ag + ] = 4.3 × 10 –5 [Ag + ] required to ppt AgCl is low so AgCl will ppt 1 st . 37.[4] The element E is Ga and the diagonal element of 5 th period is 50 Sn having outer electronic configuration will be [Kr] 5s 2 4d 10 5p 2 . 38.[2] Metallic sodium does not react with 2-butyne because

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