Home
Class 11
PHYSICS
[" The frequency heard by the divere "3"...

[" The frequency heard by the divere "3" ) "411Hz" .Af "448Hz],[" 1) "350Hz],[" The bob of a simple pendulum of mass "40grams" with a positive charge "4times10^(-6)C" is "],[" The bob of a simple pendulum of mass "40grams" with a positity "3.6times10^(4)N/c" is applied "],[" oscillating with a time period.T_1 .~ A n ~ e l e c t r i c ~ f i e l d ~ o f ~ i n t e n s i t y ~ "3.6times10^(4)N/c" is "10m/s^(2)" )."],[" vertically upwards.Now the time period is "T_(2)" ,the value of "(T_(2))/(T_(1))" is "(g=10m/s^(2)).]

Promotional Banner

Similar Questions

Explore conceptually related problems

A bob of a simple pendulum of mass 40gm with a positive charge 4xx10^(-6)C is oscillating with a time period T_(1) .An electric field of intensity 3.6xx10^(4)N//C is applied vertically upwards.Now the time period is T_(2) the value of (T_(2))/(T_(1)) is (g=10//s^(2))

A bob of a simple pendulum of mass 40 mg with a positive charge 4 xx 10^(-6) C is oscilliating with time period "T_(1). An electric field of intensity 3.6 xx 10^(4) N/c is applied vertically upwards now time period is T_(2) . The value of (T_(2))/(T_(1)) is (g = 10 m/s^(2))

A bob of a simple pendulum of mass 40 mg with a positive charge 4 xx 10^(-6) C is oscilliating with time period "T_(1). An electric field of intensity 3.6 xx 10^(4) N/c is applied vertically upwards now time period is T_(2) . The value of (T_(2))/(T_(1)) is (g = 10 m/s^(2))

The bob of a simple pendulum of mass 100g is oscillating with a time period of 1.42s . If the bob is replaced by another bob of mass 150g but of same radius, the new time period of oscillation

At which position the bob of simple pendulum has maximum K.E.?

The bob of a simple pendulum of mass 100 g is oscillating with a time period of 1.42s. If the bob is replaced by another bob of mass 150 g but of same radius what be the new time period of oscillation ?