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The compound(s) having peroxide linkage is(are) (A) H 2 S 2 O 7 𝐻 2 𝑆 2 𝑂 7 (B) H 2 S 2 O 8 𝐻 2 𝑆 2 𝑂 8 (C) H 2 S 2 O 5 𝐻 2 𝑆 2 𝑂 5 (D) H 2 S O 5

on July 07, 2025

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Math JEE Adv. 2025 paper 1

Problem Statement: We are given: An ellipse E : x 2 6 + y 2 3 = 1 E : 6 x 2 + 3 y 2 = 1 . A parabola P : y 2 = 12 x P : y 2 = 12 x . Two distinct common tangents T 1 T 1 and T 2 T 2 to both E E and P P . T 1 T 1 touches P P at A 1 A 1 and E E at A 2 A 2 . T 2 T 2 touches P P at A 4 A 4 and E E at A 3 A 3 . We need to determine which of the given statements (A)-(D) are true. Step 1: Find the condition for a line to be tangent to the parabola P : y 2 = 12 x P : y 2 = 12 x . A general line can be written as y = m x + c y = m x + c . For this line to be tangent to the parabola y 2 = 12 x y 2 = 12 x , the condition is: c = 3 m c = m 3 Thus, the equation of the tangent to the parabola is: y = m x + 3 m y = m x + m 3 Step 2: Find the condition for the same line to be tangent to the ellipse E ...

The firing order in a six cylinder vertical two stroke in-line engine is 1-4-2-6-3-5. The piston stroke is 100 mm and length of each connecting rod is 200 mm. The pitch distances between the cylinder centre lines are 100 mm, 100 mm, 150 mm, 100 mm and 100 mm respectively. The reciprocating mass per cylinder is 1 kg and the engine runs at 3000 rpm. Determine the out of balance primary and secondary forces and couples on this engine, taking the plane of cylinder 2 as the reference plane.

Six-Cylinder Two-Stroke Engine Balance Analysis Let's analyze the out-of-balance forces and couples for this six-cylinder two-stroke engine. Given Data: Engine type: 6-cylinder vertical two-stroke in-line Firing order: 1-4-2-6-3-5 Piston stroke: 100 mm → Crank radius (R) = 50 mm = 0.05 m Connecting rod length (L) = 200 mm = 0.2 m Cylinder spacing: 100, 100, 150, 100, 100 mm (from cylinder 1 to 6) Reciprocating mass per cylinder (m) = 1 kg Engine speed = 3000 rpm → ω = 2π × 3000/60 = 314.16 rad/s Reference plane: Cylinder 2 Crank Angle Determination: For a two-stroke engine with firing order 1-4-2-6-3-5, the crank angles are spaced at: θ₁ = 0° (cylinder 1) θ₄ = 60° θ₂ = 120° θ₆ = 180° θ₃ = 240° θ₅ = 300° Primary Forces and Couples: Primary force for each cylinder: Fₚ = mω²Rcosθᵢ Primary Force Calculation: Sum of primary forces: ΣFₚ = mω²R[cos(0°) + cos(60°) + cos(120°) + cos(180°) + cos(240°) + cos(300°)] = 0 Primary Couple Calculation: Taking cylinder 2 as reference (distance = 0...

A table tennis ball has radius (3/2) × 10−2 m and mass (22/7) × 10−3 kg. It is slowly pushed down into a swimming pool to a depth of 𝑑 = 0.7 m below the water surface and then released from rest. It emerges from the water surface at speed 𝑣, without getting wet, and rises up to a height 𝐻. Which of the following option(s) is(are) correct? [Given: 𝜋 = 22/7, 𝑔 = 10 m s −2 , density of water = 1 × 103 kg m−3 , viscosity of water = 1 × 10−3 Pa-s.] (A) The work done in pushing the ball to the depth 𝑑 is 0.077 J. (B) If we neglect the viscous force in water, then the speed 𝑣 = 7 m/s. (C) If we neglect the viscous force in water, then the height 𝐻 = 1.4 m. (D) The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is 500/9.

To solve this problem, we need to analyze the motion of the table tennis ball as it is pushed into the water, released, and rises back to the surface. We will consider the forces acting on the ball, including buoyancy, gravity, and viscous drag, and calculate the work done, speed, height, and force ratios. Step 1: Work Done in Pushing the Ball to Depth d d The work done in pushing the ball to a depth d d is equal to the change in potential energy of the ball-water system. The ball displaces a volume of water equal to its own volume, and the buoyant force acts upward. The work done is: W = Buoyant Force × d = ( ρ w V g ) ⋅ d W = Buoyant Force × d = ( ρ w V g ) ⋅ d where: ρ w = 1 × 1 0 3 kg/m 3 ρ w = 1 × 1 0 3 kg/m 3 is the density of water, V = 4 3 π r 3 V = 3 4 π r 3 is the volume of the ball, r = 3 2 × 1 0 − 2 m r = 2 3 × 1 0 − 2 m is the radius of the ball, g = 10 m/s 2 g = 10 m/s 2 is the acceleration due to gra...

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