Exams › JEE Advanced › Physics
Correct answer: 2*sqrt(10) m/s
Net force on bead = buoyancy - weight = (rho_water - rho_bead)*V*g = 0.5*rho_water*V*g upward. Effective g_eff = g upward (since mass = 0.5*rho_water*V). With g_eff upward, the effective potential energy is maximum at the bottom (h=0) and minimum at the top (h=2R=4m). The bead naturally accelerates toward the top. For a bead threaded on a wire, it can only fail to complete the circle if it stops and reverses. The critical point is the bottom of the circle (lowest PE in terms of motion). However, starting at the bottom with V0, the bead gains KE as it rises (since g_eff is upward). It loses KE coming back down from the top. The critical condition is that the bead must not stop at the bottom on the return. Since energy is conserved and the path is a closed circle, the bead returns to the bottom with the same speed V0. So any V0>0 allows completion — unless the question refers to a specific constraint from the wire (no pull constraint, wire can only push). In that case the normal force condition at the top (where g_eff pushes the bead away from center) requires: mv²/R >= m*g_eff (outward), i.e., v² >= g_eff*R = g*R = 10*2 = 20 m²/s² at the top. Energy conservation from bottom to top: (1/2)*V0² = (1/2)*v_top² - g*2R (g_eff does positive work going up 2R). V0² = v_top² - 4gR. For v_top² = gR (minimum): V0² = gR - 4gR = -3gR < 0 — impossible. Using v_top² = g*R: V0² = g*R - 2*g*(2R)... Let me be careful. Going from bottom (h=0) to top (h=4): work by g_eff (upward) = m*g*4 (positive, since displacement is upward). So KE_top = KE_bottom + m*g*(2R)*2... using work-energy: (1/2)*m*V_top² = (1/2)*m*V0² + m*g*(2R). V_top² = V0² + 4gR. Min V_top² from normal force at top (wire can only push inward, i.e., cannot pull, meaning N>=0 and N is inward): at top, centripetal = N + m*g (both inward since g_eff is upward = inward at top). Min when N=0: V_top²/R = g => V_top² = gR. But V_top² = V0² + 4gR > gR always. So wire always pushes. For the bottom: centripetal is upward = inward (bottom of circle). N - m*g (N upward from wire, g_eff upward): N - mg = mV²/R => N = mg + mV²/R > 0 always. No constraint at bottom either. Hmm. Perhaps there is a side point (h=R, the 3 o'clock or 9 o'clock position) where the wire must be able to maintain contact. At the side (h=R above bottom), g_eff is upward and the centripetal direction is horizontal (toward center). N is horizontal. mg_eff is upward = tangential at this point. So N = mV_side²/R (purely horizontal). This is always positive as long as V_side > 0. V_side² = V0² + 2*g*R. Always > 0. So completion requires V0 > 0? That gives V0_min approaching 0, which contradicts the answer choices. There must be a subtlety I'm missing. Most likely: the question considers the bead at the ORIGIN (center of frame, not the bottom of the circle), and the bead enters the circular wire at a point at the same height as the center. This gives a very different geometry. Taking the origin at the center, bead starts at (R,0) = (2,0) (rightmost point, height = center height = R above the bottom). Then the circle goes up to (0,2R) and down to (0,0). With g_eff upward, the highest PE point for motion is the bottom of the circle (0,0). Energy conservation from (R,0) to (0,0): (1/2)*V0² = (1/2)*V_bottom² - g*R (moving down by R, g_eff upward does negative work going down). V_bottom² = V0² + 2gR... no: going from height R to height 0 (downward), g_eff (upward) does negative work = -m*g*R. KE_bottom = KE_start - m*g*R. V_bottom² = V0² - 2gR. For V_bottom > 0: V0² > 2gR = 2*10*2 = 40 => V0 > 2*sqrt(10) m/s. This matches option C! The critical point is the bottom of the circle, and the bead starts at the side. Minimum V0 = sqrt(2gR) = sqrt(40) = 2*sqrt(10) m/s.