A bob of a simple pendulum of mass 40gm with a positive charge 4×10–6C is oscillating with time period ‘T1’. An electric field of intensity 3.6×104 N/c is applied vertically upwards now time period is T2. The value of 'T2′T1 is (g = 10m/s2)

T1∝1g;T2∝1g1;g=10 and g1=g−Eqm=10−3.6×104×4×10−640×10−3=10−3.6=6.4T2T1=gg1⇒106.4=10064=108=1.25

A bob of a simple pendulum of mass 40gm with a positive charge 4×10–6C is oscillating with time period ‘T1’. An electric field of intensity 3.6×104 N/c is applied vertically upwards now time period is T2. The value of 'T2′T1 is (g = 10m/s2)

T1∝1g;T2∝1g1;g=10 and g1=g−Eqm=10−3.6×104×4×10−640×10−3=10−3.6=6.4T2T1=gg1⇒106.4=10064=108=1.25

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A bob of a simple pendulum of mass 40gm with a positive charge 4×10^–6C is oscillating with time period ‘T₁’. An electric field of intensity 3.6×10⁴ N/c is applied vertically upwards now time period is T₂. The value of $\frac{^{'} {T_{2}}^{'}}{T_{1}}$ is (g = 10m/s²)

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0.64

1.25

0.16

0.8

answer is C.

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Detailed Solution

$\begin{aligned} T_{1} \propto \frac{1}{\sqrt{g}}; T_{2} \propto \frac{1}{\sqrt{g^{1}}}; g = 10 and g^{1} = g - \frac{Eq}{m} \\ = 10 - \frac{3.6 \times 10^{4} \times 4 \times 10^{- 6}}{40 \times 10^{- 3}} = 10 - 3.6 = 6.4 \\ \frac{T_{2}}{T_{1}} = \sqrt{\frac{g}{g_{1}}} \Rightarrow \sqrt{\frac{10}{6.4}} = \sqrt{\frac{100}{64}} = \frac{10}{8} = 1.25 \end{aligned}$