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PHXI04:MOTION IN A PLANE

361861 A particle completes 3 revolutions per second on a circular path of radius $8 c m$ . Find the value of angular velocity and centripetal acceleration of the particle

1

6 π \frac{r a d}{s}; 288 π^{2} \frac{c m}{s^{2}}

2

π \frac{r a d}{s}; 275 π^{2} \frac{c m}{s^{2}}

3

6 π \frac{r a d}{s}; 288 \frac{c m}{s^{2}}

4 None

Explanation:

To find the angular velocity $(ω)$ and centripetal acceleration $(a_{r})$ of a particle moving in a circle with 3 revolutions per second (frequency $f = 3 H z$ ) and radius $r = 8 c m$ .
Angular velocity $ω$ is given by $ω = 2 π f$ , so $ω = 2 π \times 3 r a d / s = 6 π r a d / s$ .
Centripetal acceleration $a_{r}$ is given by $a_{r} = ω^{2} r$ ,
so $a_{r} = (6 π)^{2} \times 8 = 288 π^{2} c m / s^{2}$
Therefore, the correct option is (1).

PHXI04:MOTION IN A PLANE

361862 If $a_{r}$ and $a_{t}$ represent radial and tangential accelerations respectively, the motion of a particle will be uniformly circular if

1

a_{r} = 0

and

a_{t} = 0

2

a_{r} = 0

but

a_{t} \neq 0

3

a_{r} \neq 0

but

a_{t} = 0

4

a_{r} \neq 0

and

a_{t} \neq 0

Explanation:

If $a_{r} \neq 0$ but $a_{t} = 0$ , then motion is a
uniform circular.
If $a_{r} \neq 0$ and $a_{t} \neq 0$ , then motion is a non-
uniform circular.

PHXI04:MOTION IN A PLANE

361863 Assertion :
A uniform circular motion is an accelerated motion.
Reason :
Direction of acceleration is parallel to the velocity vector.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI04:MOTION IN A PLANE

361864 A Particle moves along a circle of radius $r$ with constant tangential acceleration. If the velocity of the particle is $v$ at the end of second revolution, after the revolution has started, then the tangential acceleration is

1

\frac{v^{2}}{8 π r}

2

\frac{v^{2}}{6 π r}

3

\frac{v^{2}}{4 π r}

4

\frac{v^{2}}{2 π r}

Explanation:

Applying equation of motion in tangential direction (along the circular path),
$v^{2} = u^{2} + 2 a s$
[Symbols have their usual meanings]
Given, $u = 0$ and $s = 2 \times 2 π r = 4 π r$
$∴ v^{2} = 2 a \times 4 π r \Rightarrow a = \frac{v^{2}}{8 π r}$

MHTCET - 2016

PHXI04:MOTION IN A PLANE

361861 A particle completes 3 revolutions per second on a circular path of radius $8 c m$ . Find the value of angular velocity and centripetal acceleration of the particle

1

6 π \frac{r a d}{s}; 288 π^{2} \frac{c m}{s^{2}}

2

π \frac{r a d}{s}; 275 π^{2} \frac{c m}{s^{2}}

3

6 π \frac{r a d}{s}; 288 \frac{c m}{s^{2}}

4 None

Explanation:

To find the angular velocity $(ω)$ and centripetal acceleration $(a_{r})$ of a particle moving in a circle with 3 revolutions per second (frequency $f = 3 H z$ ) and radius $r = 8 c m$ .
Angular velocity $ω$ is given by $ω = 2 π f$ , so $ω = 2 π \times 3 r a d / s = 6 π r a d / s$ .
Centripetal acceleration $a_{r}$ is given by $a_{r} = ω^{2} r$ ,
so $a_{r} = (6 π)^{2} \times 8 = 288 π^{2} c m / s^{2}$
Therefore, the correct option is (1).

PHXI04:MOTION IN A PLANE

361862 If $a_{r}$ and $a_{t}$ represent radial and tangential accelerations respectively, the motion of a particle will be uniformly circular if

1

a_{r} = 0

and

a_{t} = 0

2

a_{r} = 0

but

a_{t} \neq 0

3

a_{r} \neq 0

but

a_{t} = 0

4

a_{r} \neq 0

and

a_{t} \neq 0

Explanation:

If $a_{r} \neq 0$ but $a_{t} = 0$ , then motion is a
uniform circular.
If $a_{r} \neq 0$ and $a_{t} \neq 0$ , then motion is a non-
uniform circular.

PHXI04:MOTION IN A PLANE

361863 Assertion :
A uniform circular motion is an accelerated motion.
Reason :
Direction of acceleration is parallel to the velocity vector.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI04:MOTION IN A PLANE

361864 A Particle moves along a circle of radius $r$ with constant tangential acceleration. If the velocity of the particle is $v$ at the end of second revolution, after the revolution has started, then the tangential acceleration is

1

\frac{v^{2}}{8 π r}

2

\frac{v^{2}}{6 π r}

3

\frac{v^{2}}{4 π r}

4

\frac{v^{2}}{2 π r}

Explanation:

Applying equation of motion in tangential direction (along the circular path),
$v^{2} = u^{2} + 2 a s$
[Symbols have their usual meanings]
Given, $u = 0$ and $s = 2 \times 2 π r = 4 π r$
$∴ v^{2} = 2 a \times 4 π r \Rightarrow a = \frac{v^{2}}{8 π r}$

MHTCET - 2016

PHXI04:MOTION IN A PLANE

361861 A particle completes 3 revolutions per second on a circular path of radius $8 c m$ . Find the value of angular velocity and centripetal acceleration of the particle

1

6 π \frac{r a d}{s}; 288 π^{2} \frac{c m}{s^{2}}

2

π \frac{r a d}{s}; 275 π^{2} \frac{c m}{s^{2}}

3

6 π \frac{r a d}{s}; 288 \frac{c m}{s^{2}}

4 None

Explanation:

To find the angular velocity $(ω)$ and centripetal acceleration $(a_{r})$ of a particle moving in a circle with 3 revolutions per second (frequency $f = 3 H z$ ) and radius $r = 8 c m$ .
Angular velocity $ω$ is given by $ω = 2 π f$ , so $ω = 2 π \times 3 r a d / s = 6 π r a d / s$ .
Centripetal acceleration $a_{r}$ is given by $a_{r} = ω^{2} r$ ,
so $a_{r} = (6 π)^{2} \times 8 = 288 π^{2} c m / s^{2}$
Therefore, the correct option is (1).

PHXI04:MOTION IN A PLANE

361862 If $a_{r}$ and $a_{t}$ represent radial and tangential accelerations respectively, the motion of a particle will be uniformly circular if

1

a_{r} = 0

and

a_{t} = 0

2

a_{r} = 0

but

a_{t} \neq 0

3

a_{r} \neq 0

but

a_{t} = 0

4

a_{r} \neq 0

and

a_{t} \neq 0

Explanation:

If $a_{r} \neq 0$ but $a_{t} = 0$ , then motion is a
uniform circular.
If $a_{r} \neq 0$ and $a_{t} \neq 0$ , then motion is a non-
uniform circular.

PHXI04:MOTION IN A PLANE

361863 Assertion :
A uniform circular motion is an accelerated motion.
Reason :
Direction of acceleration is parallel to the velocity vector.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI04:MOTION IN A PLANE

361864 A Particle moves along a circle of radius $r$ with constant tangential acceleration. If the velocity of the particle is $v$ at the end of second revolution, after the revolution has started, then the tangential acceleration is

1

\frac{v^{2}}{8 π r}

2

\frac{v^{2}}{6 π r}

3

\frac{v^{2}}{4 π r}

4

\frac{v^{2}}{2 π r}

Explanation:

Applying equation of motion in tangential direction (along the circular path),
$v^{2} = u^{2} + 2 a s$
[Symbols have their usual meanings]
Given, $u = 0$ and $s = 2 \times 2 π r = 4 π r$
$∴ v^{2} = 2 a \times 4 π r \Rightarrow a = \frac{v^{2}}{8 π r}$

MHTCET - 2016

PHXI04:MOTION IN A PLANE

361861 A particle completes 3 revolutions per second on a circular path of radius $8 c m$ . Find the value of angular velocity and centripetal acceleration of the particle

1

6 π \frac{r a d}{s}; 288 π^{2} \frac{c m}{s^{2}}

2

π \frac{r a d}{s}; 275 π^{2} \frac{c m}{s^{2}}

3

6 π \frac{r a d}{s}; 288 \frac{c m}{s^{2}}

4 None

Explanation:

To find the angular velocity $(ω)$ and centripetal acceleration $(a_{r})$ of a particle moving in a circle with 3 revolutions per second (frequency $f = 3 H z$ ) and radius $r = 8 c m$ .
Angular velocity $ω$ is given by $ω = 2 π f$ , so $ω = 2 π \times 3 r a d / s = 6 π r a d / s$ .
Centripetal acceleration $a_{r}$ is given by $a_{r} = ω^{2} r$ ,
so $a_{r} = (6 π)^{2} \times 8 = 288 π^{2} c m / s^{2}$
Therefore, the correct option is (1).

PHXI04:MOTION IN A PLANE

361862 If $a_{r}$ and $a_{t}$ represent radial and tangential accelerations respectively, the motion of a particle will be uniformly circular if

1

a_{r} = 0

and

a_{t} = 0

2

a_{r} = 0

but

a_{t} \neq 0

3

a_{r} \neq 0

but

a_{t} = 0

4

a_{r} \neq 0

and

a_{t} \neq 0

Explanation:

If $a_{r} \neq 0$ but $a_{t} = 0$ , then motion is a
uniform circular.
If $a_{r} \neq 0$ and $a_{t} \neq 0$ , then motion is a non-
uniform circular.

PHXI04:MOTION IN A PLANE

361863 Assertion :
A uniform circular motion is an accelerated motion.
Reason :
Direction of acceleration is parallel to the velocity vector.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI04:MOTION IN A PLANE

361864 A Particle moves along a circle of radius $r$ with constant tangential acceleration. If the velocity of the particle is $v$ at the end of second revolution, after the revolution has started, then the tangential acceleration is

1

\frac{v^{2}}{8 π r}

2

\frac{v^{2}}{6 π r}

3

\frac{v^{2}}{4 π r}

4

\frac{v^{2}}{2 π r}

Explanation:

Applying equation of motion in tangential direction (along the circular path),
$v^{2} = u^{2} + 2 a s$
[Symbols have their usual meanings]
Given, $u = 0$ and $s = 2 \times 2 π r = 4 π r$
$∴ v^{2} = 2 a \times 4 π r \Rightarrow a = \frac{v^{2}}{8 π r}$

MHTCET - 2016