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PHXI04:MOTION IN A PLANE

361844 A particle is performing $U$ . $C$ . $M$ along the circumference of a circle of diameter 50 $c m$ with frequency 2 $H z$ . The acceleration of the particle in $m / s^{2}$ is

1

2 π^{2}

2

8 π^{2}

3

π^{2}

4

4 π^{2}

Explanation:

Given, diameter of circle, $d = 50 c m$
$= 50 \times 10^{- 2} m$
and frequency , $f = 2 H z$
The acceleration of particle in a uniform circular motion can be given as
$a = ω^{2} x$
where, $ω =$ angular frequency $= 2 π f$ , and $x =$ distance from centre $= \frac{d}{2}$ .
$\Rightarrow a = 4 π^{2} f^{2} \times \frac{d}{2} (1)$
$\Rightarrow a = 4 π^{2} \times 4 \times \frac{50 \times 10^{- 2}}{2} = 4 π^{2}$

MHTCET - 2019

PHXI04:MOTION IN A PLANE

361845 A body is moving in a circular path with acceleration $a$ . If its velocity gets doubled, find the ratio of acceleration after and before the change

1

1 : 4

2

\frac{1}{4} : 2

3

2 : 1

4

4 : 1

Explanation:

In a circular motion
Centripetal acceleration $\frac{v^{2}}{r} = A$
Velocity is doubled acceleration becomes
$\frac{4. v^{2}}{r} = 4 A$
Ratio of after and before change
$= \frac{4 A}{A} = \frac{4}{1}$
$\Rightarrow 4 : 1$

PHXI04:MOTION IN A PLANE

361846 In 1.0 $s$ , a particle goes from point $A$ to point $B$ , moving in a semicircle of radius 1.0 $m$ (see figure). The magnitude of the average velocity is
supporting img

1

2.0 m / s

2

3.14 m / s

3

Zero

4

1.0 m / s

Explanation:

$A v e r a g e v e l o c i t y = \frac{T o t a l d i s p l a c e m e n t}{t i m e}$

PHXI04:MOTION IN A PLANE

361847 A particle of mass ' $m$ ' is revolving in a horizontal circle of radius $r$ with a constant angular speed $w$ . The area velocity of the particle is

1

r^{2} ω

2

r^{2} θ

3

\frac{r^{2} ω}{2}

4

\frac{r ω^{2}}{2}

Explanation:

Particle is moving in a horizontal circle of radius $r$ . Let $θ$ be the angular displacement of the particle after time $d t$ as shown in the figure.
supporting img
Area covered in time $d t$ is given by
$d A = \frac{π r^{2} θ}{2 π}$
Areal velocity,
$v_{A} = \frac{d A}{d t} = \frac{1}{2} r^{2} ω (∵ \frac{d θ}{d t} = ω)$ .
So correct option is (3)

PHXI04:MOTION IN A PLANE

361844 A particle is performing $U$ . $C$ . $M$ along the circumference of a circle of diameter 50 $c m$ with frequency 2 $H z$ . The acceleration of the particle in $m / s^{2}$ is

1

2 π^{2}

2

8 π^{2}

3

π^{2}

4

4 π^{2}

Explanation:

Given, diameter of circle, $d = 50 c m$
$= 50 \times 10^{- 2} m$
and frequency , $f = 2 H z$
The acceleration of particle in a uniform circular motion can be given as
$a = ω^{2} x$
where, $ω =$ angular frequency $= 2 π f$ , and $x =$ distance from centre $= \frac{d}{2}$ .
$\Rightarrow a = 4 π^{2} f^{2} \times \frac{d}{2} (1)$
$\Rightarrow a = 4 π^{2} \times 4 \times \frac{50 \times 10^{- 2}}{2} = 4 π^{2}$

MHTCET - 2019

PHXI04:MOTION IN A PLANE

361845 A body is moving in a circular path with acceleration $a$ . If its velocity gets doubled, find the ratio of acceleration after and before the change

1

1 : 4

2

\frac{1}{4} : 2

3

2 : 1

4

4 : 1

Explanation:

In a circular motion
Centripetal acceleration $\frac{v^{2}}{r} = A$
Velocity is doubled acceleration becomes
$\frac{4. v^{2}}{r} = 4 A$
Ratio of after and before change
$= \frac{4 A}{A} = \frac{4}{1}$
$\Rightarrow 4 : 1$

PHXI04:MOTION IN A PLANE

361846 In 1.0 $s$ , a particle goes from point $A$ to point $B$ , moving in a semicircle of radius 1.0 $m$ (see figure). The magnitude of the average velocity is
supporting img

1

2.0 m / s

2

3.14 m / s

3

Zero

4

1.0 m / s

Explanation:

$A v e r a g e v e l o c i t y = \frac{T o t a l d i s p l a c e m e n t}{t i m e}$

PHXI04:MOTION IN A PLANE

361847 A particle of mass ' $m$ ' is revolving in a horizontal circle of radius $r$ with a constant angular speed $w$ . The area velocity of the particle is

1

r^{2} ω

2

r^{2} θ

3

\frac{r^{2} ω}{2}

4

\frac{r ω^{2}}{2}

Explanation:

Particle is moving in a horizontal circle of radius $r$ . Let $θ$ be the angular displacement of the particle after time $d t$ as shown in the figure.
supporting img
Area covered in time $d t$ is given by
$d A = \frac{π r^{2} θ}{2 π}$
Areal velocity,
$v_{A} = \frac{d A}{d t} = \frac{1}{2} r^{2} ω (∵ \frac{d θ}{d t} = ω)$ .
So correct option is (3)

PHXI04:MOTION IN A PLANE

361844 A particle is performing $U$ . $C$ . $M$ along the circumference of a circle of diameter 50 $c m$ with frequency 2 $H z$ . The acceleration of the particle in $m / s^{2}$ is

1

2 π^{2}

2

8 π^{2}

3

π^{2}

4

4 π^{2}

Explanation:

Given, diameter of circle, $d = 50 c m$
$= 50 \times 10^{- 2} m$
and frequency , $f = 2 H z$
The acceleration of particle in a uniform circular motion can be given as
$a = ω^{2} x$
where, $ω =$ angular frequency $= 2 π f$ , and $x =$ distance from centre $= \frac{d}{2}$ .
$\Rightarrow a = 4 π^{2} f^{2} \times \frac{d}{2} (1)$
$\Rightarrow a = 4 π^{2} \times 4 \times \frac{50 \times 10^{- 2}}{2} = 4 π^{2}$

MHTCET - 2019

PHXI04:MOTION IN A PLANE

361845 A body is moving in a circular path with acceleration $a$ . If its velocity gets doubled, find the ratio of acceleration after and before the change

1

1 : 4

2

\frac{1}{4} : 2

3

2 : 1

4

4 : 1

Explanation:

In a circular motion
Centripetal acceleration $\frac{v^{2}}{r} = A$
Velocity is doubled acceleration becomes
$\frac{4. v^{2}}{r} = 4 A$
Ratio of after and before change
$= \frac{4 A}{A} = \frac{4}{1}$
$\Rightarrow 4 : 1$

PHXI04:MOTION IN A PLANE

361846 In 1.0 $s$ , a particle goes from point $A$ to point $B$ , moving in a semicircle of radius 1.0 $m$ (see figure). The magnitude of the average velocity is
supporting img

1

2.0 m / s

2

3.14 m / s

3

Zero

4

1.0 m / s

Explanation:

$A v e r a g e v e l o c i t y = \frac{T o t a l d i s p l a c e m e n t}{t i m e}$

PHXI04:MOTION IN A PLANE

361847 A particle of mass ' $m$ ' is revolving in a horizontal circle of radius $r$ with a constant angular speed $w$ . The area velocity of the particle is

1

r^{2} ω

2

r^{2} θ

3

\frac{r^{2} ω}{2}

4

\frac{r ω^{2}}{2}

Explanation:

Particle is moving in a horizontal circle of radius $r$ . Let $θ$ be the angular displacement of the particle after time $d t$ as shown in the figure.
supporting img
Area covered in time $d t$ is given by
$d A = \frac{π r^{2} θ}{2 π}$
Areal velocity,
$v_{A} = \frac{d A}{d t} = \frac{1}{2} r^{2} ω (∵ \frac{d θ}{d t} = ω)$ .
So correct option is (3)

PHXI04:MOTION IN A PLANE

361844 A particle is performing $U$ . $C$ . $M$ along the circumference of a circle of diameter 50 $c m$ with frequency 2 $H z$ . The acceleration of the particle in $m / s^{2}$ is

1

2 π^{2}

2

8 π^{2}

3

π^{2}

4

4 π^{2}

Explanation:

Given, diameter of circle, $d = 50 c m$
$= 50 \times 10^{- 2} m$
and frequency , $f = 2 H z$
The acceleration of particle in a uniform circular motion can be given as
$a = ω^{2} x$
where, $ω =$ angular frequency $= 2 π f$ , and $x =$ distance from centre $= \frac{d}{2}$ .
$\Rightarrow a = 4 π^{2} f^{2} \times \frac{d}{2} (1)$
$\Rightarrow a = 4 π^{2} \times 4 \times \frac{50 \times 10^{- 2}}{2} = 4 π^{2}$

MHTCET - 2019

PHXI04:MOTION IN A PLANE

361845 A body is moving in a circular path with acceleration $a$ . If its velocity gets doubled, find the ratio of acceleration after and before the change

1

1 : 4

2

\frac{1}{4} : 2

3

2 : 1

4

4 : 1

Explanation:

In a circular motion
Centripetal acceleration $\frac{v^{2}}{r} = A$
Velocity is doubled acceleration becomes
$\frac{4. v^{2}}{r} = 4 A$
Ratio of after and before change
$= \frac{4 A}{A} = \frac{4}{1}$
$\Rightarrow 4 : 1$

PHXI04:MOTION IN A PLANE

361846 In 1.0 $s$ , a particle goes from point $A$ to point $B$ , moving in a semicircle of radius 1.0 $m$ (see figure). The magnitude of the average velocity is
supporting img

1

2.0 m / s

2

3.14 m / s

3

Zero

4

1.0 m / s

Explanation:

$A v e r a g e v e l o c i t y = \frac{T o t a l d i s p l a c e m e n t}{t i m e}$

PHXI04:MOTION IN A PLANE

361847 A particle of mass ' $m$ ' is revolving in a horizontal circle of radius $r$ with a constant angular speed $w$ . The area velocity of the particle is

1

r^{2} ω

2

r^{2} θ

3

\frac{r^{2} ω}{2}

4

\frac{r ω^{2}}{2}

Explanation:

Particle is moving in a horizontal circle of radius $r$ . Let $θ$ be the angular displacement of the particle after time $d t$ as shown in the figure.
supporting img
Area covered in time $d t$ is given by
$d A = \frac{π r^{2} θ}{2 π}$
Areal velocity,
$v_{A} = \frac{d A}{d t} = \frac{1}{2} r^{2} ω (∵ \frac{d θ}{d t} = ω)$ .
So correct option is (3)

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