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A particle of mass m falls through a tunnel drilled along the diameter of Earth starting from point A, which is at a height 2R above Earth's surface. What is the velocity of the particle when it crosses point B, located at a distance R/2 below the Earth's centre? (M = mass of Earth, R = radius of Earth)

1. (5/2) * sqrt(GM/(3R))
2. sqrt(GM/(3R))
3. 2 * sqrt(GM/(3R))
4. (3/2) * sqrt(GM/(3R))

Correct answer: (5/2) * sqrt(GM/(3R))

Solution

Particle starts at rest at height 2R above surface => distance from centre = 3R. Point B: R/2 below centre => distance from centre = R/2 (inside Earth). Potential energy at r=3R (outside): U_A = -GMm/(3R). Potential energy at r=R/2 (inside uniform sphere): U_B = -GMm(3R² - r²)/(2R³) at r=R/2. U_B = -GMm(3R² - R²/4)/(2R³) = -GMm*(11R²/4)/(2R³) = -11GMm/(8R). Energy conservation (KE_A = 0): U_A + 0 = U_B + (1/2)mv². -GMm/3R = -11GMm/8R + (1/2)mv². (1/2)v² = GMm/m * (11/8R - 1/3R) = GM(33-8)/(24R) = 25GM/(24R). v² = 25GM/(12R). v = 5/2 * sqrt(GM/(3R)). [Since sqrt(25GM/12R) = 5*sqrt(GM/12R) = 5/2 * sqrt(GM/3R)]

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