Length of a simple pendulum is increased by 2% . Its time period will

English

Gujarati

Hindi

13.Oscillations

easy

The length of a simple pendulum is increased by $2\%$. Its time period will

A

$2$

B

$1$

C

$4$

D

$8$

(AIPMT-1997)

Solution

$l_{2}=1.02 l_{1} .$

Time period $(T)=2 \pi \times \sqrt{\frac{l}{g}} \propto \sqrt{l}$

Therefore $\frac{T_{2}}{T_{1}}=\sqrt{\frac{l_{2}}{l_{1}}}=\sqrt{\frac{1.02 l_{1}}{l_{1}}}=1.01$

Thus time period increased by $1 \%$.

Standard 11

Physics

Share

Similar Questions

A simple pendulum is attached to the roof of a lift. If time period of oscillation, when the lift is stationary is $T$. Then frequency of oscillation, when the lift falls freely, will be

easy

A pendulum has time period $T$. If it is taken on to another planet having acceleration due to gravity half and mass $ 9 $ times that of the earth then its time period on the other planet will be

easy

A chimpanzee swinging on a swing in a sitting position, stands up suddenly, the time period will

easy

(AIIMS-2012)

In the following table relation of graph in column$-I$ and shape of graph in column$-II$ is shown match them appropriately.

column$-I$ column $-II$

$(a)$ ${T^2} \to l$ $(i)$ Linear

$(b)$ ${T^2} \to g$ $(ii)$ Parabolic

$(c)$ ${T} \to l$ $(iii)$ Hyperbolic

medium

A simple pendulum of length $1\,m$ is allowed to oscillate with amplitude $2^o$. It collides elastically with a wall inclined at $1^o$ to the vertical. Its time period will be : (use $g = \pi ^2$ )

normal