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PHXI04:MOTION IN A PLANE

361835 A particle moving with a constant speed of $9 k m / h$ on a circle reaches from end $A$ to the diametrically opposite end $B$ in time $2 s$ . The average acceleration from $A$ to $B$ is
supporting img

1

9 k m / h^{2}

2

18 k m / s^{2}

3

5 k m / h^{2}

4

2.5 m / s^{2}

Explanation:

Average acceleration
$\begin{aligned} = \frac{Δ v}{Δ t} = \frac{[9 - (- 9)] k m / h}{2 s} = \frac{18}{2} \times \frac{\frac{5}{18} m / s}{5} \\ = 2.5 m / s^{2} \end{aligned}$

PHXI04:MOTION IN A PLANE

361836 A particle is moving in a circle of radius $R$ with constant speed $v$ , if radius is doubled then its centripetal force to keep the same speed should be

1 Doubled

2 Halved

3 Quadrupled

4 Unchanged

Explanation:

$F = \frac{m v^{2}}{r} .$ For same mass and same speed if radius is doubled then force should be halved.

PHXI04:MOTION IN A PLANE

361837 A particle is moving with constant speed $v$ in $x - y$ plane, as shown in the figure. The magnitude of its angular velocity about point $O$ is
supporting img

1

\frac{v}{\sqrt{a^{2} + b^{2}}}

2

\frac{v}{b}

3

\frac{v b}{(a^{2} + b^{2})}

4

\frac{v}{a}

Explanation:

supporting img
$v_{1} = r ω$
$v \cos θ = R ω$
$ω = \frac{v \cos θ}{\sqrt{a^{2} + b^{2}}}$
$ω = \frac{v b}{(a^{2} + b^{2})} [∵ R = \sqrt{a^{2} + b^{2}}]$

PHXI04:MOTION IN A PLANE

361838 The angle turned by a body undergoing circular motion depends on time as $θ = θ_{0} + θ_{1} t + θ_{2} t^{2} .$ Then the angular acceleration of the body is

1

θ_{1}

2

θ_{2}

3

2 θ_{1}

4

2 θ_{2}

Explanation:

Angular acceleration $= \frac{d^{2} θ}{d t^{2}} = 2 θ_{2}$

PHXI04:MOTION IN A PLANE

361839 A point $P$ moves in counter-clockwise direction on a circular path as shown in the figure. The movement of $^{'} P^{'}$ is such that it sweeps out a length $s = t^{3} + 5$ , where $s$ is in metres and $t$ is in seconds. The radius of the path is $27 m .$ The acceleration of ' $P$ ' when $t = 3 s$ is
(Take $\sqrt{13} = 3.6$ )
supporting img

1

42.2 m / s^{2}

2

51.4 m / s^{2}

3

63.1 m / s^{2}

4

32.4 m / s^{2}

Explanation:

As $s = t^{3} + 5$
$\frac{d s}{d t} = 3 t^{2} = v$
$∴ a_{t} = \frac{d v}{d t} = 6 t$
$| \vec{a} | = \sqrt{a_{c}^{2} + a_{t}^{2}}$
At $t = 3 s$
$| \vec{a} | = \sqrt{{(\frac{v^{2}}{R})}^{2} + a_{t}^{2}} = \sqrt{{(\frac{9 t^{4}}{R})}^{2} + {(6 t)}^{2}}$
$= \sqrt{{(27)}^{2} + 324} = \sqrt{1053} = 9 \sqrt{13}$
$∴ | \vec{a} | = 32.4 m / s^{2}$

PHXI04:MOTION IN A PLANE

361835 A particle moving with a constant speed of $9 k m / h$ on a circle reaches from end $A$ to the diametrically opposite end $B$ in time $2 s$ . The average acceleration from $A$ to $B$ is
supporting img

1

9 k m / h^{2}

2

18 k m / s^{2}

3

5 k m / h^{2}

4

2.5 m / s^{2}

Explanation:

Average acceleration
$\begin{aligned} = \frac{Δ v}{Δ t} = \frac{[9 - (- 9)] k m / h}{2 s} = \frac{18}{2} \times \frac{\frac{5}{18} m / s}{5} \\ = 2.5 m / s^{2} \end{aligned}$

PHXI04:MOTION IN A PLANE

361836 A particle is moving in a circle of radius $R$ with constant speed $v$ , if radius is doubled then its centripetal force to keep the same speed should be

1 Doubled

2 Halved

3 Quadrupled

4 Unchanged

Explanation:

$F = \frac{m v^{2}}{r} .$ For same mass and same speed if radius is doubled then force should be halved.

PHXI04:MOTION IN A PLANE

361837 A particle is moving with constant speed $v$ in $x - y$ plane, as shown in the figure. The magnitude of its angular velocity about point $O$ is
supporting img

1

\frac{v}{\sqrt{a^{2} + b^{2}}}

2

\frac{v}{b}

3

\frac{v b}{(a^{2} + b^{2})}

4

\frac{v}{a}

Explanation:

supporting img
$v_{1} = r ω$
$v \cos θ = R ω$
$ω = \frac{v \cos θ}{\sqrt{a^{2} + b^{2}}}$
$ω = \frac{v b}{(a^{2} + b^{2})} [∵ R = \sqrt{a^{2} + b^{2}}]$

PHXI04:MOTION IN A PLANE

361838 The angle turned by a body undergoing circular motion depends on time as $θ = θ_{0} + θ_{1} t + θ_{2} t^{2} .$ Then the angular acceleration of the body is

1

θ_{1}

2

θ_{2}

3

2 θ_{1}

4

2 θ_{2}

Explanation:

Angular acceleration $= \frac{d^{2} θ}{d t^{2}} = 2 θ_{2}$

PHXI04:MOTION IN A PLANE

361839 A point $P$ moves in counter-clockwise direction on a circular path as shown in the figure. The movement of $^{'} P^{'}$ is such that it sweeps out a length $s = t^{3} + 5$ , where $s$ is in metres and $t$ is in seconds. The radius of the path is $27 m .$ The acceleration of ' $P$ ' when $t = 3 s$ is
(Take $\sqrt{13} = 3.6$ )
supporting img

1

42.2 m / s^{2}

2

51.4 m / s^{2}

3

63.1 m / s^{2}

4

32.4 m / s^{2}

Explanation:

As $s = t^{3} + 5$
$\frac{d s}{d t} = 3 t^{2} = v$
$∴ a_{t} = \frac{d v}{d t} = 6 t$
$| \vec{a} | = \sqrt{a_{c}^{2} + a_{t}^{2}}$
At $t = 3 s$
$| \vec{a} | = \sqrt{{(\frac{v^{2}}{R})}^{2} + a_{t}^{2}} = \sqrt{{(\frac{9 t^{4}}{R})}^{2} + {(6 t)}^{2}}$
$= \sqrt{{(27)}^{2} + 324} = \sqrt{1053} = 9 \sqrt{13}$
$∴ | \vec{a} | = 32.4 m / s^{2}$

PHXI04:MOTION IN A PLANE

361835 A particle moving with a constant speed of $9 k m / h$ on a circle reaches from end $A$ to the diametrically opposite end $B$ in time $2 s$ . The average acceleration from $A$ to $B$ is
supporting img

1

9 k m / h^{2}

2

18 k m / s^{2}

3

5 k m / h^{2}

4

2.5 m / s^{2}

Explanation:

Average acceleration
$\begin{aligned} = \frac{Δ v}{Δ t} = \frac{[9 - (- 9)] k m / h}{2 s} = \frac{18}{2} \times \frac{\frac{5}{18} m / s}{5} \\ = 2.5 m / s^{2} \end{aligned}$

PHXI04:MOTION IN A PLANE

361836 A particle is moving in a circle of radius $R$ with constant speed $v$ , if radius is doubled then its centripetal force to keep the same speed should be

1 Doubled

2 Halved

3 Quadrupled

4 Unchanged

Explanation:

$F = \frac{m v^{2}}{r} .$ For same mass and same speed if radius is doubled then force should be halved.

PHXI04:MOTION IN A PLANE

361837 A particle is moving with constant speed $v$ in $x - y$ plane, as shown in the figure. The magnitude of its angular velocity about point $O$ is
supporting img

1

\frac{v}{\sqrt{a^{2} + b^{2}}}

2

\frac{v}{b}

3

\frac{v b}{(a^{2} + b^{2})}

4

\frac{v}{a}

Explanation:

supporting img
$v_{1} = r ω$
$v \cos θ = R ω$
$ω = \frac{v \cos θ}{\sqrt{a^{2} + b^{2}}}$
$ω = \frac{v b}{(a^{2} + b^{2})} [∵ R = \sqrt{a^{2} + b^{2}}]$

PHXI04:MOTION IN A PLANE

361838 The angle turned by a body undergoing circular motion depends on time as $θ = θ_{0} + θ_{1} t + θ_{2} t^{2} .$ Then the angular acceleration of the body is

1

θ_{1}

2

θ_{2}

3

2 θ_{1}

4

2 θ_{2}

Explanation:

Angular acceleration $= \frac{d^{2} θ}{d t^{2}} = 2 θ_{2}$

PHXI04:MOTION IN A PLANE

361839 A point $P$ moves in counter-clockwise direction on a circular path as shown in the figure. The movement of $^{'} P^{'}$ is such that it sweeps out a length $s = t^{3} + 5$ , where $s$ is in metres and $t$ is in seconds. The radius of the path is $27 m .$ The acceleration of ' $P$ ' when $t = 3 s$ is
(Take $\sqrt{13} = 3.6$ )
supporting img

1

42.2 m / s^{2}

2

51.4 m / s^{2}

3

63.1 m / s^{2}

4

32.4 m / s^{2}

Explanation:

As $s = t^{3} + 5$
$\frac{d s}{d t} = 3 t^{2} = v$
$∴ a_{t} = \frac{d v}{d t} = 6 t$
$| \vec{a} | = \sqrt{a_{c}^{2} + a_{t}^{2}}$
At $t = 3 s$
$| \vec{a} | = \sqrt{{(\frac{v^{2}}{R})}^{2} + a_{t}^{2}} = \sqrt{{(\frac{9 t^{4}}{R})}^{2} + {(6 t)}^{2}}$
$= \sqrt{{(27)}^{2} + 324} = \sqrt{1053} = 9 \sqrt{13}$
$∴ | \vec{a} | = 32.4 m / s^{2}$

PHXI04:MOTION IN A PLANE

361835 A particle moving with a constant speed of $9 k m / h$ on a circle reaches from end $A$ to the diametrically opposite end $B$ in time $2 s$ . The average acceleration from $A$ to $B$ is
supporting img

1

9 k m / h^{2}

2

18 k m / s^{2}

3

5 k m / h^{2}

4

2.5 m / s^{2}

Explanation:

Average acceleration
$\begin{aligned} = \frac{Δ v}{Δ t} = \frac{[9 - (- 9)] k m / h}{2 s} = \frac{18}{2} \times \frac{\frac{5}{18} m / s}{5} \\ = 2.5 m / s^{2} \end{aligned}$

PHXI04:MOTION IN A PLANE

361836 A particle is moving in a circle of radius $R$ with constant speed $v$ , if radius is doubled then its centripetal force to keep the same speed should be

1 Doubled

2 Halved

3 Quadrupled

4 Unchanged

Explanation:

$F = \frac{m v^{2}}{r} .$ For same mass and same speed if radius is doubled then force should be halved.

PHXI04:MOTION IN A PLANE

361837 A particle is moving with constant speed $v$ in $x - y$ plane, as shown in the figure. The magnitude of its angular velocity about point $O$ is
supporting img

1

\frac{v}{\sqrt{a^{2} + b^{2}}}

2

\frac{v}{b}

3

\frac{v b}{(a^{2} + b^{2})}

4

\frac{v}{a}

Explanation:

supporting img
$v_{1} = r ω$
$v \cos θ = R ω$
$ω = \frac{v \cos θ}{\sqrt{a^{2} + b^{2}}}$
$ω = \frac{v b}{(a^{2} + b^{2})} [∵ R = \sqrt{a^{2} + b^{2}}]$

PHXI04:MOTION IN A PLANE

361838 The angle turned by a body undergoing circular motion depends on time as $θ = θ_{0} + θ_{1} t + θ_{2} t^{2} .$ Then the angular acceleration of the body is

1

θ_{1}

2

θ_{2}

3

2 θ_{1}

4

2 θ_{2}

Explanation:

Angular acceleration $= \frac{d^{2} θ}{d t^{2}} = 2 θ_{2}$

PHXI04:MOTION IN A PLANE

361839 A point $P$ moves in counter-clockwise direction on a circular path as shown in the figure. The movement of $^{'} P^{'}$ is such that it sweeps out a length $s = t^{3} + 5$ , where $s$ is in metres and $t$ is in seconds. The radius of the path is $27 m .$ The acceleration of ' $P$ ' when $t = 3 s$ is
(Take $\sqrt{13} = 3.6$ )
supporting img

1

42.2 m / s^{2}

2

51.4 m / s^{2}

3

63.1 m / s^{2}

4

32.4 m / s^{2}

Explanation:

As $s = t^{3} + 5$
$\frac{d s}{d t} = 3 t^{2} = v$
$∴ a_{t} = \frac{d v}{d t} = 6 t$
$| \vec{a} | = \sqrt{a_{c}^{2} + a_{t}^{2}}$
At $t = 3 s$
$| \vec{a} | = \sqrt{{(\frac{v^{2}}{R})}^{2} + a_{t}^{2}} = \sqrt{{(\frac{9 t^{4}}{R})}^{2} + {(6 t)}^{2}}$
$= \sqrt{{(27)}^{2} + 324} = \sqrt{1053} = 9 \sqrt{13}$
$∴ | \vec{a} | = 32.4 m / s^{2}$

PHXI04:MOTION IN A PLANE

361835 A particle moving with a constant speed of $9 k m / h$ on a circle reaches from end $A$ to the diametrically opposite end $B$ in time $2 s$ . The average acceleration from $A$ to $B$ is
supporting img

1

9 k m / h^{2}

2

18 k m / s^{2}

3

5 k m / h^{2}

4

2.5 m / s^{2}

Explanation:

Average acceleration
$\begin{aligned} = \frac{Δ v}{Δ t} = \frac{[9 - (- 9)] k m / h}{2 s} = \frac{18}{2} \times \frac{\frac{5}{18} m / s}{5} \\ = 2.5 m / s^{2} \end{aligned}$

PHXI04:MOTION IN A PLANE

361836 A particle is moving in a circle of radius $R$ with constant speed $v$ , if radius is doubled then its centripetal force to keep the same speed should be

1 Doubled

2 Halved

3 Quadrupled

4 Unchanged

Explanation:

$F = \frac{m v^{2}}{r} .$ For same mass and same speed if radius is doubled then force should be halved.

PHXI04:MOTION IN A PLANE

361837 A particle is moving with constant speed $v$ in $x - y$ plane, as shown in the figure. The magnitude of its angular velocity about point $O$ is
supporting img

1

\frac{v}{\sqrt{a^{2} + b^{2}}}

2

\frac{v}{b}

3

\frac{v b}{(a^{2} + b^{2})}

4

\frac{v}{a}

Explanation:

supporting img
$v_{1} = r ω$
$v \cos θ = R ω$
$ω = \frac{v \cos θ}{\sqrt{a^{2} + b^{2}}}$
$ω = \frac{v b}{(a^{2} + b^{2})} [∵ R = \sqrt{a^{2} + b^{2}}]$

PHXI04:MOTION IN A PLANE

361838 The angle turned by a body undergoing circular motion depends on time as $θ = θ_{0} + θ_{1} t + θ_{2} t^{2} .$ Then the angular acceleration of the body is

1

θ_{1}

2

θ_{2}

3

2 θ_{1}

4

2 θ_{2}

Explanation:

Angular acceleration $= \frac{d^{2} θ}{d t^{2}} = 2 θ_{2}$

PHXI04:MOTION IN A PLANE

361839 A point $P$ moves in counter-clockwise direction on a circular path as shown in the figure. The movement of $^{'} P^{'}$ is such that it sweeps out a length $s = t^{3} + 5$ , where $s$ is in metres and $t$ is in seconds. The radius of the path is $27 m .$ The acceleration of ' $P$ ' when $t = 3 s$ is
(Take $\sqrt{13} = 3.6$ )
supporting img

1

42.2 m / s^{2}

2

51.4 m / s^{2}

3

63.1 m / s^{2}

4

32.4 m / s^{2}

Explanation:

As $s = t^{3} + 5$
$\frac{d s}{d t} = 3 t^{2} = v$
$∴ a_{t} = \frac{d v}{d t} = 6 t$
$| \vec{a} | = \sqrt{a_{c}^{2} + a_{t}^{2}}$
At $t = 3 s$
$| \vec{a} | = \sqrt{{(\frac{v^{2}}{R})}^{2} + a_{t}^{2}} = \sqrt{{(\frac{9 t^{4}}{R})}^{2} + {(6 t)}^{2}}$
$= \sqrt{{(27)}^{2} + 324} = \sqrt{1053} = 9 \sqrt{13}$
$∴ | \vec{a} | = 32.4 m / s^{2}$