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PHXI04:MOTION IN A PLANE

361878 A particle moves in a circle of radius 25 $c m$ at two revolutions per sec. The acceleration of the particle in $m / s^{2}$ is :

1

π^{2}

2

8 π^{2}

3

4 π^{2}

4

2 π^{2}

Explanation:

Since, $n = 2, ω = 2 π \times 2 = 4 π r a d / s^{2}$
So acceleration $= ω^{2} r = (4 π)^{2} \times \frac{25}{100} m / s^{2} = 4 π^{2}$

PHXI04:MOTION IN A PLANE

361879 A particle is moving along a circular path of radius $6 m$ with uniform speed of $8 m s^{- 1}$ . The average acceleration when the particle completes one half of the revolution is-

1

\frac{16}{3 π} m / s^{2}

2

\frac{32}{3 π} m / s^{2}

3

\frac{64}{3 π} m / s^{2}

4 None of these

Explanation:

Average acceleration $= \frac{change in velocity}{change in time}$
supporting img
$= \frac{\vec{v} - (\vec{v})}{Δ t} = \frac{2 \vec{v}}{Δ t}$
$= \frac{64}{3 π} m / s^{2}$

PHXI04:MOTION IN A PLANE

361880 A small ball describes a horizontal circle on the smooth inner surface of a cone. Speed of the ball at a height of 0.1 $m$ above the vertex is
supporting img

1

4 m s^{- 1}

2

3 m s^{- 1}

3

2 m s^{- 1}

4

1 m s^{- 1}

Explanation:

Forces acting on the ball are shown in the figure.
Using $\frac{m v^{2}}{R} = N \cos θ a n d m g = N \sin θ$
we get $\tan θ = \frac{g R}{v^{2}}$
or $\frac{R}{h} = \frac{g R}{v^{2}} i . e ., v = \sqrt{g h}$
$\Rightarrow v = \sqrt{10 \times 0.1} = 1 m s^{- 1}$
supporting img

PHXI04:MOTION IN A PLANE

361881 A cylindrical vessel partially filled with water is rotated about its vertical central axis. Its surface will

1 Rise equally

2 Rise from the sides

3 Rise from the middle

4 Lowered equally

PHXI04:MOTION IN A PLANE

361882 A particle describes a horizontal circle in a conical funnel whose inner surface is smooth with speed of 0.5 $m$ / $s$ . What is height of the plane of circle from vertex of the funnel?

1

0.25 m

2

2 c m

3

4 c m

4

2.5 c m

Explanation:

The particle is moving in circular path
From the figure $m g = R \sin θ$
$\frac{m v^{2}}{r} = R \cos θ$
From Eqs. (1) and (2), we get
$\tan θ = \frac{r g}{v^{2}}$ and $\tan θ = \frac{r}{h}$
$h = \frac{v^{2}}{g} = \frac{{(0.5)}^{2}}{10} = 0.25 m = 2.5 c m$
supporting img

PHXI04:MOTION IN A PLANE

361878 A particle moves in a circle of radius 25 $c m$ at two revolutions per sec. The acceleration of the particle in $m / s^{2}$ is :

1

π^{2}

2

8 π^{2}

3

4 π^{2}

4

2 π^{2}

Explanation:

Since, $n = 2, ω = 2 π \times 2 = 4 π r a d / s^{2}$
So acceleration $= ω^{2} r = (4 π)^{2} \times \frac{25}{100} m / s^{2} = 4 π^{2}$

PHXI04:MOTION IN A PLANE

361879 A particle is moving along a circular path of radius $6 m$ with uniform speed of $8 m s^{- 1}$ . The average acceleration when the particle completes one half of the revolution is-

1

\frac{16}{3 π} m / s^{2}

2

\frac{32}{3 π} m / s^{2}

3

\frac{64}{3 π} m / s^{2}

4 None of these

Explanation:

Average acceleration $= \frac{change in velocity}{change in time}$
supporting img
$= \frac{\vec{v} - (\vec{v})}{Δ t} = \frac{2 \vec{v}}{Δ t}$
$= \frac{64}{3 π} m / s^{2}$

PHXI04:MOTION IN A PLANE

361880 A small ball describes a horizontal circle on the smooth inner surface of a cone. Speed of the ball at a height of 0.1 $m$ above the vertex is
supporting img

1

4 m s^{- 1}

2

3 m s^{- 1}

3

2 m s^{- 1}

4

1 m s^{- 1}

Explanation:

Forces acting on the ball are shown in the figure.
Using $\frac{m v^{2}}{R} = N \cos θ a n d m g = N \sin θ$
we get $\tan θ = \frac{g R}{v^{2}}$
or $\frac{R}{h} = \frac{g R}{v^{2}} i . e ., v = \sqrt{g h}$
$\Rightarrow v = \sqrt{10 \times 0.1} = 1 m s^{- 1}$
supporting img

PHXI04:MOTION IN A PLANE

361881 A cylindrical vessel partially filled with water is rotated about its vertical central axis. Its surface will

1 Rise equally

2 Rise from the sides

3 Rise from the middle

4 Lowered equally

PHXI04:MOTION IN A PLANE

361882 A particle describes a horizontal circle in a conical funnel whose inner surface is smooth with speed of 0.5 $m$ / $s$ . What is height of the plane of circle from vertex of the funnel?

1

0.25 m

2

2 c m

3

4 c m

4

2.5 c m

Explanation:

The particle is moving in circular path
From the figure $m g = R \sin θ$
$\frac{m v^{2}}{r} = R \cos θ$
From Eqs. (1) and (2), we get
$\tan θ = \frac{r g}{v^{2}}$ and $\tan θ = \frac{r}{h}$
$h = \frac{v^{2}}{g} = \frac{{(0.5)}^{2}}{10} = 0.25 m = 2.5 c m$
supporting img

PHXI04:MOTION IN A PLANE

361878 A particle moves in a circle of radius 25 $c m$ at two revolutions per sec. The acceleration of the particle in $m / s^{2}$ is :

1

π^{2}

2

8 π^{2}

3

4 π^{2}

4

2 π^{2}

Explanation:

Since, $n = 2, ω = 2 π \times 2 = 4 π r a d / s^{2}$
So acceleration $= ω^{2} r = (4 π)^{2} \times \frac{25}{100} m / s^{2} = 4 π^{2}$

PHXI04:MOTION IN A PLANE

361879 A particle is moving along a circular path of radius $6 m$ with uniform speed of $8 m s^{- 1}$ . The average acceleration when the particle completes one half of the revolution is-

1

\frac{16}{3 π} m / s^{2}

2

\frac{32}{3 π} m / s^{2}

3

\frac{64}{3 π} m / s^{2}

4 None of these

Explanation:

Average acceleration $= \frac{change in velocity}{change in time}$
supporting img
$= \frac{\vec{v} - (\vec{v})}{Δ t} = \frac{2 \vec{v}}{Δ t}$
$= \frac{64}{3 π} m / s^{2}$

PHXI04:MOTION IN A PLANE

361880 A small ball describes a horizontal circle on the smooth inner surface of a cone. Speed of the ball at a height of 0.1 $m$ above the vertex is
supporting img

1

4 m s^{- 1}

2

3 m s^{- 1}

3

2 m s^{- 1}

4

1 m s^{- 1}

Explanation:

Forces acting on the ball are shown in the figure.
Using $\frac{m v^{2}}{R} = N \cos θ a n d m g = N \sin θ$
we get $\tan θ = \frac{g R}{v^{2}}$
or $\frac{R}{h} = \frac{g R}{v^{2}} i . e ., v = \sqrt{g h}$
$\Rightarrow v = \sqrt{10 \times 0.1} = 1 m s^{- 1}$
supporting img

PHXI04:MOTION IN A PLANE

361881 A cylindrical vessel partially filled with water is rotated about its vertical central axis. Its surface will

1 Rise equally

2 Rise from the sides

3 Rise from the middle

4 Lowered equally

PHXI04:MOTION IN A PLANE

361882 A particle describes a horizontal circle in a conical funnel whose inner surface is smooth with speed of 0.5 $m$ / $s$ . What is height of the plane of circle from vertex of the funnel?

1

0.25 m

2

2 c m

3

4 c m

4

2.5 c m

Explanation:

The particle is moving in circular path
From the figure $m g = R \sin θ$
$\frac{m v^{2}}{r} = R \cos θ$
From Eqs. (1) and (2), we get
$\tan θ = \frac{r g}{v^{2}}$ and $\tan θ = \frac{r}{h}$
$h = \frac{v^{2}}{g} = \frac{{(0.5)}^{2}}{10} = 0.25 m = 2.5 c m$
supporting img

PHXI04:MOTION IN A PLANE

361878 A particle moves in a circle of radius 25 $c m$ at two revolutions per sec. The acceleration of the particle in $m / s^{2}$ is :

1

π^{2}

2

8 π^{2}

3

4 π^{2}

4

2 π^{2}

Explanation:

Since, $n = 2, ω = 2 π \times 2 = 4 π r a d / s^{2}$
So acceleration $= ω^{2} r = (4 π)^{2} \times \frac{25}{100} m / s^{2} = 4 π^{2}$

PHXI04:MOTION IN A PLANE

361879 A particle is moving along a circular path of radius $6 m$ with uniform speed of $8 m s^{- 1}$ . The average acceleration when the particle completes one half of the revolution is-

1

\frac{16}{3 π} m / s^{2}

2

\frac{32}{3 π} m / s^{2}

3

\frac{64}{3 π} m / s^{2}

4 None of these

Explanation:

Average acceleration $= \frac{change in velocity}{change in time}$
supporting img
$= \frac{\vec{v} - (\vec{v})}{Δ t} = \frac{2 \vec{v}}{Δ t}$
$= \frac{64}{3 π} m / s^{2}$

PHXI04:MOTION IN A PLANE

361880 A small ball describes a horizontal circle on the smooth inner surface of a cone. Speed of the ball at a height of 0.1 $m$ above the vertex is
supporting img

1

4 m s^{- 1}

2

3 m s^{- 1}

3

2 m s^{- 1}

4

1 m s^{- 1}

Explanation:

Forces acting on the ball are shown in the figure.
Using $\frac{m v^{2}}{R} = N \cos θ a n d m g = N \sin θ$
we get $\tan θ = \frac{g R}{v^{2}}$
or $\frac{R}{h} = \frac{g R}{v^{2}} i . e ., v = \sqrt{g h}$
$\Rightarrow v = \sqrt{10 \times 0.1} = 1 m s^{- 1}$
supporting img

PHXI04:MOTION IN A PLANE

361881 A cylindrical vessel partially filled with water is rotated about its vertical central axis. Its surface will

1 Rise equally

2 Rise from the sides

3 Rise from the middle

4 Lowered equally

PHXI04:MOTION IN A PLANE

361882 A particle describes a horizontal circle in a conical funnel whose inner surface is smooth with speed of 0.5 $m$ / $s$ . What is height of the plane of circle from vertex of the funnel?

1

0.25 m

2

2 c m

3

4 c m

4

2.5 c m

Explanation:

The particle is moving in circular path
From the figure $m g = R \sin θ$
$\frac{m v^{2}}{r} = R \cos θ$
From Eqs. (1) and (2), we get
$\tan θ = \frac{r g}{v^{2}}$ and $\tan θ = \frac{r}{h}$
$h = \frac{v^{2}}{g} = \frac{{(0.5)}^{2}}{10} = 0.25 m = 2.5 c m$
supporting img

PHXI04:MOTION IN A PLANE

361878 A particle moves in a circle of radius 25 $c m$ at two revolutions per sec. The acceleration of the particle in $m / s^{2}$ is :

1

π^{2}

2

8 π^{2}

3

4 π^{2}

4

2 π^{2}

Explanation:

Since, $n = 2, ω = 2 π \times 2 = 4 π r a d / s^{2}$
So acceleration $= ω^{2} r = (4 π)^{2} \times \frac{25}{100} m / s^{2} = 4 π^{2}$

PHXI04:MOTION IN A PLANE

361879 A particle is moving along a circular path of radius $6 m$ with uniform speed of $8 m s^{- 1}$ . The average acceleration when the particle completes one half of the revolution is-

1

\frac{16}{3 π} m / s^{2}

2

\frac{32}{3 π} m / s^{2}

3

\frac{64}{3 π} m / s^{2}

4 None of these

Explanation:

Average acceleration $= \frac{change in velocity}{change in time}$
supporting img
$= \frac{\vec{v} - (\vec{v})}{Δ t} = \frac{2 \vec{v}}{Δ t}$
$= \frac{64}{3 π} m / s^{2}$

PHXI04:MOTION IN A PLANE

361880 A small ball describes a horizontal circle on the smooth inner surface of a cone. Speed of the ball at a height of 0.1 $m$ above the vertex is
supporting img

1

4 m s^{- 1}

2

3 m s^{- 1}

3

2 m s^{- 1}

4

1 m s^{- 1}

Explanation:

Forces acting on the ball are shown in the figure.
Using $\frac{m v^{2}}{R} = N \cos θ a n d m g = N \sin θ$
we get $\tan θ = \frac{g R}{v^{2}}$
or $\frac{R}{h} = \frac{g R}{v^{2}} i . e ., v = \sqrt{g h}$
$\Rightarrow v = \sqrt{10 \times 0.1} = 1 m s^{- 1}$
supporting img

PHXI04:MOTION IN A PLANE

361881 A cylindrical vessel partially filled with water is rotated about its vertical central axis. Its surface will

1 Rise equally

2 Rise from the sides

3 Rise from the middle

4 Lowered equally

PHXI04:MOTION IN A PLANE

361882 A particle describes a horizontal circle in a conical funnel whose inner surface is smooth with speed of 0.5 $m$ / $s$ . What is height of the plane of circle from vertex of the funnel?

1

0.25 m

2

2 c m

3

4 c m

4

2.5 c m

Explanation:

The particle is moving in circular path
From the figure $m g = R \sin θ$
$\frac{m v^{2}}{r} = R \cos θ$
From Eqs. (1) and (2), we get
$\tan θ = \frac{r g}{v^{2}}$ and $\tan θ = \frac{r}{h}$
$h = \frac{v^{2}}{g} = \frac{{(0.5)}^{2}}{10} = 0.25 m = 2.5 c m$
supporting img