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JEE Main 2024 · Physics · Electrostatics · hard

A small bob of mass 20 g and charge 2√3 muC hangs from a light insulating string of length 0.8 m in a uniform horizontal electric field of 10^5 V…

Single correct option

A small bob of mass 20 g20\ \mathrm{g} and charge 23 μC2\sqrt3\ \mu\mathrm{C} hangs from a light insulating string of length 0.8 m0.8\ \mathrm{m} in a uniform horizontal electric field of 105 V m−110^5\ \mathrm{V\,m^{-1}}. Taking g=10 m s−2g=10\ \mathrm{m\,s^{-2}}, the time period of small oscillations about its equilibrium position is
  1. A. 0.4π5 s0.4\pi\sqrt{5}\ \mathrm{s}
  2. B. 0.2π2 s0.2\pi\sqrt{2}\ \mathrm{s}
  3. C. 0.4π s0.4\pi\ \mathrm{s}
  4. D. 0.8π s0.8\pi\ \mathrm{s}
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Answer: (C) 0.4π s0.4\pi\ \mathrm{s}

Electric force qE=23×10−6×105=0.23 NqE=2\sqrt3\times10^{-6}\times10^{5}=0.2\sqrt3\ \mathrm{N} (horizontal) and weight mg=0.2 Nmg=0.2\ \mathrm{N}. The effective gravity is geff=g2+(qE/m)2=100+300=20 m s−2g_{eff}=\sqrt{g^2+(qE/m)^2}=\sqrt{100+300}=20\ \mathrm{m\,s^{-2}}, directed at 60∘60^\circ to the vertical. Small oscillations about the new equilibrium have T=2πLgeff=2π0.820=2π(0.2)=0.4π sT=2\pi\sqrt{\dfrac{L}{g_{eff}}}=2\pi\sqrt{\dfrac{0.8}{20}}=2\pi(0.2)=0.4\pi\ \mathrm{s}. Using g=10g=10 alone gives 0.4π20.4\pi\sqrt2.
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