The bob of a pendulum of mass m and length l and a charge of q is in the rest position in a uniform horizontal electric field of E. The tension in the string of the pendulum is :

m g

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q E

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{[(m g)^{2} + (q E)^{2}]}^{1 / 2}

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{[(m g)^{2} + (q E)^{2}]}^{1 / 4}

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Solution

The correct option is B ${[(m g)^{2} + (q E)^{2}]}^{1 / 2}$
From FBD,
$T s i n θ = q E$ ............... (1) where $T$ is the tension in string.
$T c o s θ = m g$ ............... (2)
$∴ t a n θ = \frac{q E}{m g}$
$θ = t a n^{- 1} (\frac{q E}{m g})$
Doing $(1)^{2} + (2)^{2},$ we will get,
$T^{2} s i n^{2} θ + T^{2} c o s^{2} θ = (m g)^{2} + (q E)^{2}$
$T = \sqrt{(m g)^{2} + (q E)^{2}}$
Ans:(C)

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The bob of a pendulum of mass m and length l and a charge of q is in the rest position in a uniform horizontal electric field of E. The tension in the string of the pendulum is :

The correct option is B [(mg)2+(qE)2]1/2From FBD,Tsinθ=qE ............... (1) where T is the tension in string.Tcosθ=mg ............... (2)∴tanθ=qEmgθ=tan−1(qEmg)Doing (1)2+(2)2, we will get,T2sin2θ+T2cos2θ=(mg)2+(qE)2T=√(mg)2+(qE)2Ans:(C)