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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 ...

Question No. 14 A block with mass M M is connected by a massless spring with stiffness constant k k to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude A A about an equilibrium position x 0 x 0 . Consider two cases: (i) when the block is at x 0 x 0 ; and (ii) when the block is at x = x 0 + A x = x 0 + A . In both the cases, a particle with mass m m ( < M < M ) is softly placed on the block after which they stick to each other. Which of the following statement(s) is (are) true about the motion after the mass m m is placed on the mass M M ? (A) The amplitude of oscillation in the first case changes by a factor of M m + M m + M M , whereas in the second case it remains unchanged (B) The final time period of oscillation in both the cases is same (C) The total energy decreases in both the cases (D) The instantaneous s...

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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