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PHXI04:MOTION IN A PLANE

361840 A particle covers equal distance around a circular path, in equal intervals of time. Which of the following quantities connected with the motion of the particle remains constant with time?

1 Displacement

2 Velocity

3 Speed

4 Acceleration

PHXI04:MOTION IN A PLANE

361841 A man carrying a monkey on his shoulder does cycling smoothly on a circular track of radius $9 m$ and completes 120 revolutions in 3 minutes. The magnitude of centripetal acceleration of monkey is (in $m / s^{2}$ )

1

16 π^{2} m s^{- 2}

2 zero

3

57600 π^{2} m s^{- 2}

4

4 π^{2} m s^{- 2}

Explanation:

Radius of circular track, $r = 9 m$
Angular velocity, $ω = \frac{120 \times 2 π}{3 \times 60} = \frac{4 π}{3} r a d / s$
As we know,
Centripetal acceleration, $a_{c} = \frac{v^{2}}{r} = ω^{2} r$
$∴ a_{c} = {(\frac{4 R}{3})}^{2} \times 9$
$∴ a_{c} = 16 π^{2} m / s^{2}$

JEE - 2024

PHXI04:MOTION IN A PLANE

361842 In a non - uniform circular motion, the ratio of tangential to radial acceleration is ( where $r =$ radius of circle, $v =$ speed of the particle, $α =$ angular acceleration)

1

\frac{α^{2} r^{2}}{v}

2

\frac{α^{2} r}{v^{2}}

3

\frac{α r^{2}}{v^{2}}

4

\frac{v^{2}}{r^{2} α}

Explanation:

Given, radius of circle $= r$
Speed of particle $= v$
Angular acceleration $= α$
$∴$ Tangential acceleration $= α r (1)$
Radial acceleration $= \frac{v^{2}}{r} (2)$ {\mkern 1mu} {\mkern 1mu} (2)\]
Dividing eq.(1) by eq.(2),
$\frac{T a n g e n t i a l a c c e l e r a t i o n}{R a d i a l a c c e l e r a t i o n} = \frac{α r^{2}}{v^{2}}$

MHTCET - 2018

PHXI04:MOTION IN A PLANE

361843 A particle moving with uniform speed in a circular path maintains:

1 Constant velocity

2 Constant acceleration

3 Constant velocity but varying acceleration

4 Varying velocity and varying acceleration

NEET - 2024

PHXI04:MOTION IN A PLANE

361840 A particle covers equal distance around a circular path, in equal intervals of time. Which of the following quantities connected with the motion of the particle remains constant with time?

1 Displacement

2 Velocity

3 Speed

4 Acceleration

PHXI04:MOTION IN A PLANE

361841 A man carrying a monkey on his shoulder does cycling smoothly on a circular track of radius $9 m$ and completes 120 revolutions in 3 minutes. The magnitude of centripetal acceleration of monkey is (in $m / s^{2}$ )

1

16 π^{2} m s^{- 2}

2 zero

3

57600 π^{2} m s^{- 2}

4

4 π^{2} m s^{- 2}

Explanation:

Radius of circular track, $r = 9 m$
Angular velocity, $ω = \frac{120 \times 2 π}{3 \times 60} = \frac{4 π}{3} r a d / s$
As we know,
Centripetal acceleration, $a_{c} = \frac{v^{2}}{r} = ω^{2} r$
$∴ a_{c} = {(\frac{4 R}{3})}^{2} \times 9$
$∴ a_{c} = 16 π^{2} m / s^{2}$

JEE - 2024

PHXI04:MOTION IN A PLANE

361842 In a non - uniform circular motion, the ratio of tangential to radial acceleration is ( where $r =$ radius of circle, $v =$ speed of the particle, $α =$ angular acceleration)

1

\frac{α^{2} r^{2}}{v}

2

\frac{α^{2} r}{v^{2}}

3

\frac{α r^{2}}{v^{2}}

4

\frac{v^{2}}{r^{2} α}

Explanation:

Given, radius of circle $= r$
Speed of particle $= v$
Angular acceleration $= α$
$∴$ Tangential acceleration $= α r (1)$
Radial acceleration $= \frac{v^{2}}{r} (2)$ {\mkern 1mu} {\mkern 1mu} (2)\]
Dividing eq.(1) by eq.(2),
$\frac{T a n g e n t i a l a c c e l e r a t i o n}{R a d i a l a c c e l e r a t i o n} = \frac{α r^{2}}{v^{2}}$

MHTCET - 2018

PHXI04:MOTION IN A PLANE

361843 A particle moving with uniform speed in a circular path maintains:

1 Constant velocity

2 Constant acceleration

3 Constant velocity but varying acceleration

4 Varying velocity and varying acceleration

NEET - 2024

PHXI04:MOTION IN A PLANE

361840 A particle covers equal distance around a circular path, in equal intervals of time. Which of the following quantities connected with the motion of the particle remains constant with time?

1 Displacement

2 Velocity

3 Speed

4 Acceleration

PHXI04:MOTION IN A PLANE

361841 A man carrying a monkey on his shoulder does cycling smoothly on a circular track of radius $9 m$ and completes 120 revolutions in 3 minutes. The magnitude of centripetal acceleration of monkey is (in $m / s^{2}$ )

1

16 π^{2} m s^{- 2}

2 zero

3

57600 π^{2} m s^{- 2}

4

4 π^{2} m s^{- 2}

Explanation:

Radius of circular track, $r = 9 m$
Angular velocity, $ω = \frac{120 \times 2 π}{3 \times 60} = \frac{4 π}{3} r a d / s$
As we know,
Centripetal acceleration, $a_{c} = \frac{v^{2}}{r} = ω^{2} r$
$∴ a_{c} = {(\frac{4 R}{3})}^{2} \times 9$
$∴ a_{c} = 16 π^{2} m / s^{2}$

JEE - 2024

PHXI04:MOTION IN A PLANE

361842 In a non - uniform circular motion, the ratio of tangential to radial acceleration is ( where $r =$ radius of circle, $v =$ speed of the particle, $α =$ angular acceleration)

1

\frac{α^{2} r^{2}}{v}

2

\frac{α^{2} r}{v^{2}}

3

\frac{α r^{2}}{v^{2}}

4

\frac{v^{2}}{r^{2} α}

Explanation:

Given, radius of circle $= r$
Speed of particle $= v$
Angular acceleration $= α$
$∴$ Tangential acceleration $= α r (1)$
Radial acceleration $= \frac{v^{2}}{r} (2)$ {\mkern 1mu} {\mkern 1mu} (2)\]
Dividing eq.(1) by eq.(2),
$\frac{T a n g e n t i a l a c c e l e r a t i o n}{R a d i a l a c c e l e r a t i o n} = \frac{α r^{2}}{v^{2}}$

MHTCET - 2018

PHXI04:MOTION IN A PLANE

361843 A particle moving with uniform speed in a circular path maintains:

1 Constant velocity

2 Constant acceleration

3 Constant velocity but varying acceleration

4 Varying velocity and varying acceleration

NEET - 2024

PHXI04:MOTION IN A PLANE

361840 A particle covers equal distance around a circular path, in equal intervals of time. Which of the following quantities connected with the motion of the particle remains constant with time?

1 Displacement

2 Velocity

3 Speed

4 Acceleration

PHXI04:MOTION IN A PLANE

361841 A man carrying a monkey on his shoulder does cycling smoothly on a circular track of radius $9 m$ and completes 120 revolutions in 3 minutes. The magnitude of centripetal acceleration of monkey is (in $m / s^{2}$ )

1

16 π^{2} m s^{- 2}

2 zero

3

57600 π^{2} m s^{- 2}

4

4 π^{2} m s^{- 2}

Explanation:

Radius of circular track, $r = 9 m$
Angular velocity, $ω = \frac{120 \times 2 π}{3 \times 60} = \frac{4 π}{3} r a d / s$
As we know,
Centripetal acceleration, $a_{c} = \frac{v^{2}}{r} = ω^{2} r$
$∴ a_{c} = {(\frac{4 R}{3})}^{2} \times 9$
$∴ a_{c} = 16 π^{2} m / s^{2}$

JEE - 2024

PHXI04:MOTION IN A PLANE

361842 In a non - uniform circular motion, the ratio of tangential to radial acceleration is ( where $r =$ radius of circle, $v =$ speed of the particle, $α =$ angular acceleration)

1

\frac{α^{2} r^{2}}{v}

2

\frac{α^{2} r}{v^{2}}

3

\frac{α r^{2}}{v^{2}}

4

\frac{v^{2}}{r^{2} α}

Explanation:

Given, radius of circle $= r$
Speed of particle $= v$
Angular acceleration $= α$
$∴$ Tangential acceleration $= α r (1)$
Radial acceleration $= \frac{v^{2}}{r} (2)$ {\mkern 1mu} {\mkern 1mu} (2)\]
Dividing eq.(1) by eq.(2),
$\frac{T a n g e n t i a l a c c e l e r a t i o n}{R a d i a l a c c e l e r a t i o n} = \frac{α r^{2}}{v^{2}}$

MHTCET - 2018

PHXI04:MOTION IN A PLANE

361843 A particle moving with uniform speed in a circular path maintains:

1 Constant velocity

2 Constant acceleration

3 Constant velocity but varying acceleration

4 Varying velocity and varying acceleration

NEET - 2024

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