Question

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

1. 0.8
2. 1.25
3. 0.64
4. 0.16​

Answer 1

the answer of this question 1.25

Answer 2

Given: A bob of a simple pendulum of mass 40 gm with a positive charge 40×10^-6 is oscillating with time period T1. An electric field of intensity 3.6×10^4 N/C is applied vertically upwards now time period is T2.

To find: The value of T1/T2

Solution:

• Now we know the formula for time period as:

               T = 2π√ ( l / g)

• Here g will be g - Eq/m as per the question.

               T2 = 2π √( l / (g - (Eq/m)) )

• Now dividing T2 and T1, we get:

               T2 / T1 = √g / (g-Eq/m )

               T2 / T1 = √10 / ( 10 - (3.6 x 10^4 x 4 x 10^-6 / 40 x 10^-3))

• After solving this, we get:

               T2 / T1 = √10 / 6.4

• Multiplying and dividing by √10 in numerator and denominator, we get:

               T2 / T1 = √10 / 6.4 x √10/10

               T2 / T1 = √100/64

               T2 / T1 = 10/8

• Reciprocating it, we get:

               T1/T2 = 8/10

               T1/T2 = 0.8

Answer:

                So the value of T1/T2 is 0.8