Length of second's hand in a watch is 1 ,cm. change in velocity of its tip in 15, seconds is

English

Gujarati

Hindi

3-2.Motion in Plane

medium

The length of second's hand in a watch is $1 \,cm.$ The change in velocity of its tip in $15\, seconds$ is

AZero

B$\frac{\pi }{{30\sqrt 2 }}cm/\sec $

C$\frac{\pi }{{30}}cm/\sec $

D$\frac{{\pi \sqrt 2 }}{{30}}cm/\sec $

Solution

(d) In $15\, second's$ hand rotate through $90°$.
Change in velocity $\left| {\overrightarrow {\Delta v} } \right| = 2v\sin (\theta /2)$
$ = 2(r\omega )\sin (90^\circ /2)$$ = 2 \times 1 \times \frac{{2\pi }}{T} \times \frac{1}{{\sqrt 2 }}$
$ = \frac{{4\pi }}{{60\sqrt 2 }} = \frac{{\pi \sqrt 2 }}{{30}}\frac{{cm}}{{\sec }}$ [As $T = 60 \,sec$]

Standard 11

Physics

A particle is revolving in a circle of radius $2\ m$ with angular velocity $\omega = t^2 -4t + 8\ rad/s$ . The time when speed of the particle becomes $8\ m/s$ is ……… $\sec$

medium

Consider a circle of radius $42\ cm$. An insect crawls with uniform speed of $1.3\ cm/s$ along the chord $AB$ then along the circular arc $BCD$ to reach point $D$ and then following cord $DA$ to reach finally $A$. Time spend by the insect to crawl from $A$ to $A$ is closest to ……… $\sec$

hard

body moves with constant angular velocity on a circle. Magnitude of angular acceleration

easy

What happens to the centripetal acceleration of a revolving body if you double the orbital speed $v$ and half the angular velocity $\omega $

easy

The angle turned by a body undergoing circular motion depends on time as $\theta = {\theta _0} + {\theta _1}t + {\theta _2}{t^2}$. Then the angular acceleration of the body is

easy

The length of second's hand in a watch is $1 \,cm.$ The change in velocity of its tip in $15\, seconds$ is

Solution

Similar Questions

A particle is revolving in a circle of radius $2\ m$ with angular velocity $\omega = t^2 -4t + 8\ rad/s$ . The time when speed of the particle becomes $8\ m/s$ is ……… $\sec$

Consider a circle of radius $42\ cm$. An insect crawls with uniform speed of $1.3\ cm/s$ along the chord $AB$ then along the circular arc $BCD$ to reach point $D$ and then following cord $DA$ to reach finally $A$. Time spend by the insect to crawl from $A$ to $A$ is closest to ……… $\sec$

body moves with constant angular velocity on a circle. Magnitude of angular acceleration

What happens to the centripetal acceleration of a revolving body if you double the orbital speed $v$ and half the angular velocity $\omega $

The angle turned by a body undergoing circular motion depends on time as $\theta = {\theta _0} + {\theta _1}t + {\theta _2}{t^2}$. Then the angular acceleration of the body is

Start a Free Trial Now