A bob of a simple pendulum of mass 40 g with a positive charge of 4×10−6 C is oscillating with time period T1. An electric field of intensity 3.6×104 N/C is applied vertically upwards, now the time period is T2. The value of T2T1 is(g=10 m/s2)

A bob of a simple pendulum of mass 40 g with a positive charge of 4×1

1.	A bob of a simple pendulum of mass 40 g with a positive charge of 4×10−6 C is oscillating with time period T1. An electric field of intensity 3.6×104 N/C is applied vertically upwards, now the time period is T2. The value of T2T1 is(g=10 m/s2)
Answer» A bob of a simple pendulum of mass $40 g$ with a positive charge of $4 \times 10^{- 6} C$ is oscillating with time period $T_{1}$ . An electric field of intensity $3.6 \times 10^{4} N/C$ is applied vertically upwards, now the time period is $T_{2}$ . The value of $\frac{T_{2}}{T_{1}}$ is $(g = 10 m / s^{2})$

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A bob of a simple pendulum of mass 40 g with a positive charge of 4×10−6 C is oscillating with time period T1. An electric field of intensity 3.6×104 N/C is applied vertically upwards, now the time period is T2. The value of T2T1 is(g=10 m/s2)

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