QBManager

Motion in Plane

144052 A mass attached to one end of a string crosses top-most point on a vertical circle with critical speed. Its centripetal acceleration when string becomes horizontal will be ( $g =$ gravitational acceleration)

1

g

2

3 g

3

4 g

4

6 g

Explanation:

B We know, velocity of particle at top most point on vertical circle,
$v_{T} = \sqrt{3 rg}$
original image
But centripetal acceleration $(a_{c}) = \frac{v_{T}^{2}}{r}$
$∴ a_{c} = \frac{(\sqrt{3 rg})^{2}}{r} [From equation (i)]$
$a_{c} = \frac{3 rg}{r} \Rightarrow a_{c} = 3 g$

MHT-CET 2018

Motion in Plane

144053 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:

A Given, initial speed of the particle $(u) = 0$
Final speed of the particle $= v$
Distance covering in 2 revolution $(s) = 2 \times 2 π r$
Applying equation of motion,
$= 4 π r$
$v^{2} = u^{2} + 2 as$
$v^{2} = 0 + 2 \times a \times 4 π r$
$a = \frac{v^{2}}{8 π r}$

MHT-CET 2016

Motion in Plane

144054 A particle of mass ' $m$ ' is moving in circular path of constant radius ' $r$ ' such that centripetal acceleration is varying with time ' $t$ ' as $K^{2} r t^{2}$ where $K$ is a constant. The power delivered to the particle by the force acting on it is

1

m^{2} K^{2} r^{2} t^{2}

2

m K^{2} r^{2} t

3

{mK}^{2} {rt}^{2}

4

{mKr}^{2} t

Explanation:

B Given,
Centripetal acceleration $(a_{c}) = K^{2} {rt}^{2}$
And we know,
Centripetal accelerate $(a_{c}) = \frac{v^{2}}{r}$
Comparing equation (i) and (ii),
$\frac{v^{2}}{r} = K^{2} r t^{2}$
$v^{2} = K^{2} r^{2} t^{2}$
$v = K r t$
Tangential acceleration $(a) = \frac{dv}{dt} = Kr$
Then, Force $(F) = ma = mKr$
So, Power $(P) = F . v$
$P = mKr \times Krt$
$P = {mK}^{2} r^{2} t$

MHT-CET 2015

Motion in Plane

144056 An aeroplane executes a horizontal loop at a speed of $720 km / h$ with its wings banked at $45^{\circ}$ . What is the radius of the loop? Take $g = 10 {ms}^{- 2}$ .

1

4 km

2

4.5 km

3

7.2 km

4

2 km

Explanation:

A Given, angle of banking $(θ) = 45^{\circ}$ , speed of the aeroplane $(v) = 720 km / h = 720 \times \frac{5}{18} = 200 m / s$
For radius $r$ of the loop, We have the relation,
$\tan θ = \frac{v^{2}}{rg}$
$r = \frac{v^{2}}{g \tan θ}$
$r = \frac{(200)^{2}}{10 \times \tan 45^{\circ}} (∵ θ = 45^{\circ})$
$r = \frac{40000}{10} = 4000 m$
$r = 4 km$

Karnataka CET-2014

Motion in Plane

144058 A cyclist starts from the centre $O$ of a circular park of radius one kilometre, reaches the edge $P$ of the park, then cycles along the circumference and returns to the centre along QO as shown in the figure. If the round trip takes ten minutes, the net displacement and average speed of the cyclist (in metre and kilometre per hour) is :
original image

1 0,1

2

\frac{π + 4}{2}, 0

3

21.4, \frac{π + 4}{2}

4

0, 21.4

Explanation:

D Since the net displacement of the cyclist is zero then the average velocity will also be zero.
by diagram, $OP =$ Radius $= 1 km, PQ = R θ = 1 \times π / 2$
Average speed of the cyclist $= \frac{Total distance travelled}{Total time}$
$= \frac{OP + PQ + QO}{10} km / \min$
$= \frac{1 + \frac{π}{2} \times 1 + 1}{10} km / \min$
$= \frac{4 + π}{20} \times 60 km / hr$
$= 7.14 \times 3 km / hr$
$= 21.4 km / hr$

Karnataka CET-2006

Motion in Plane

144052 A mass attached to one end of a string crosses top-most point on a vertical circle with critical speed. Its centripetal acceleration when string becomes horizontal will be ( $g =$ gravitational acceleration)

1

g

2

3 g

3

4 g

4

6 g

Explanation:

B We know, velocity of particle at top most point on vertical circle,
$v_{T} = \sqrt{3 rg}$
original image
But centripetal acceleration $(a_{c}) = \frac{v_{T}^{2}}{r}$
$∴ a_{c} = \frac{(\sqrt{3 rg})^{2}}{r} [From equation (i)]$
$a_{c} = \frac{3 rg}{r} \Rightarrow a_{c} = 3 g$

MHT-CET 2018

Motion in Plane

144053 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:

A Given, initial speed of the particle $(u) = 0$
Final speed of the particle $= v$
Distance covering in 2 revolution $(s) = 2 \times 2 π r$
Applying equation of motion,
$= 4 π r$
$v^{2} = u^{2} + 2 as$
$v^{2} = 0 + 2 \times a \times 4 π r$
$a = \frac{v^{2}}{8 π r}$

MHT-CET 2016

Motion in Plane

144054 A particle of mass ' $m$ ' is moving in circular path of constant radius ' $r$ ' such that centripetal acceleration is varying with time ' $t$ ' as $K^{2} r t^{2}$ where $K$ is a constant. The power delivered to the particle by the force acting on it is

1

m^{2} K^{2} r^{2} t^{2}

2

m K^{2} r^{2} t

3

{mK}^{2} {rt}^{2}

4

{mKr}^{2} t

Explanation:

B Given,
Centripetal acceleration $(a_{c}) = K^{2} {rt}^{2}$
And we know,
Centripetal accelerate $(a_{c}) = \frac{v^{2}}{r}$
Comparing equation (i) and (ii),
$\frac{v^{2}}{r} = K^{2} r t^{2}$
$v^{2} = K^{2} r^{2} t^{2}$
$v = K r t$
Tangential acceleration $(a) = \frac{dv}{dt} = Kr$
Then, Force $(F) = ma = mKr$
So, Power $(P) = F . v$
$P = mKr \times Krt$
$P = {mK}^{2} r^{2} t$

MHT-CET 2015

Motion in Plane

144056 An aeroplane executes a horizontal loop at a speed of $720 km / h$ with its wings banked at $45^{\circ}$ . What is the radius of the loop? Take $g = 10 {ms}^{- 2}$ .

1

4 km

2

4.5 km

3

7.2 km

4

2 km

Explanation:

A Given, angle of banking $(θ) = 45^{\circ}$ , speed of the aeroplane $(v) = 720 km / h = 720 \times \frac{5}{18} = 200 m / s$
For radius $r$ of the loop, We have the relation,
$\tan θ = \frac{v^{2}}{rg}$
$r = \frac{v^{2}}{g \tan θ}$
$r = \frac{(200)^{2}}{10 \times \tan 45^{\circ}} (∵ θ = 45^{\circ})$
$r = \frac{40000}{10} = 4000 m$
$r = 4 km$

Karnataka CET-2014

Motion in Plane

144058 A cyclist starts from the centre $O$ of a circular park of radius one kilometre, reaches the edge $P$ of the park, then cycles along the circumference and returns to the centre along QO as shown in the figure. If the round trip takes ten minutes, the net displacement and average speed of the cyclist (in metre and kilometre per hour) is :
original image

1 0,1

2

\frac{π + 4}{2}, 0

3

21.4, \frac{π + 4}{2}

4

0, 21.4

Explanation:

D Since the net displacement of the cyclist is zero then the average velocity will also be zero.
by diagram, $OP =$ Radius $= 1 km, PQ = R θ = 1 \times π / 2$
Average speed of the cyclist $= \frac{Total distance travelled}{Total time}$
$= \frac{OP + PQ + QO}{10} km / \min$
$= \frac{1 + \frac{π}{2} \times 1 + 1}{10} km / \min$
$= \frac{4 + π}{20} \times 60 km / hr$
$= 7.14 \times 3 km / hr$
$= 21.4 km / hr$

Karnataka CET-2006

Motion in Plane

144052 A mass attached to one end of a string crosses top-most point on a vertical circle with critical speed. Its centripetal acceleration when string becomes horizontal will be ( $g =$ gravitational acceleration)

1

g

2

3 g

3

4 g

4

6 g

Explanation:

B We know, velocity of particle at top most point on vertical circle,
$v_{T} = \sqrt{3 rg}$
original image
But centripetal acceleration $(a_{c}) = \frac{v_{T}^{2}}{r}$
$∴ a_{c} = \frac{(\sqrt{3 rg})^{2}}{r} [From equation (i)]$
$a_{c} = \frac{3 rg}{r} \Rightarrow a_{c} = 3 g$

MHT-CET 2018

Motion in Plane

144053 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:

A Given, initial speed of the particle $(u) = 0$
Final speed of the particle $= v$
Distance covering in 2 revolution $(s) = 2 \times 2 π r$
Applying equation of motion,
$= 4 π r$
$v^{2} = u^{2} + 2 as$
$v^{2} = 0 + 2 \times a \times 4 π r$
$a = \frac{v^{2}}{8 π r}$

MHT-CET 2016

Motion in Plane

144054 A particle of mass ' $m$ ' is moving in circular path of constant radius ' $r$ ' such that centripetal acceleration is varying with time ' $t$ ' as $K^{2} r t^{2}$ where $K$ is a constant. The power delivered to the particle by the force acting on it is

1

m^{2} K^{2} r^{2} t^{2}

2

m K^{2} r^{2} t

3

{mK}^{2} {rt}^{2}

4

{mKr}^{2} t

Explanation:

B Given,
Centripetal acceleration $(a_{c}) = K^{2} {rt}^{2}$
And we know,
Centripetal accelerate $(a_{c}) = \frac{v^{2}}{r}$
Comparing equation (i) and (ii),
$\frac{v^{2}}{r} = K^{2} r t^{2}$
$v^{2} = K^{2} r^{2} t^{2}$
$v = K r t$
Tangential acceleration $(a) = \frac{dv}{dt} = Kr$
Then, Force $(F) = ma = mKr$
So, Power $(P) = F . v$
$P = mKr \times Krt$
$P = {mK}^{2} r^{2} t$

MHT-CET 2015

Motion in Plane

144056 An aeroplane executes a horizontal loop at a speed of $720 km / h$ with its wings banked at $45^{\circ}$ . What is the radius of the loop? Take $g = 10 {ms}^{- 2}$ .

1

4 km

2

4.5 km

3

7.2 km

4

2 km

Explanation:

A Given, angle of banking $(θ) = 45^{\circ}$ , speed of the aeroplane $(v) = 720 km / h = 720 \times \frac{5}{18} = 200 m / s$
For radius $r$ of the loop, We have the relation,
$\tan θ = \frac{v^{2}}{rg}$
$r = \frac{v^{2}}{g \tan θ}$
$r = \frac{(200)^{2}}{10 \times \tan 45^{\circ}} (∵ θ = 45^{\circ})$
$r = \frac{40000}{10} = 4000 m$
$r = 4 km$

Karnataka CET-2014

Motion in Plane

144058 A cyclist starts from the centre $O$ of a circular park of radius one kilometre, reaches the edge $P$ of the park, then cycles along the circumference and returns to the centre along QO as shown in the figure. If the round trip takes ten minutes, the net displacement and average speed of the cyclist (in metre and kilometre per hour) is :
original image

1 0,1

2

\frac{π + 4}{2}, 0

3

21.4, \frac{π + 4}{2}

4

0, 21.4

Explanation:

D Since the net displacement of the cyclist is zero then the average velocity will also be zero.
by diagram, $OP =$ Radius $= 1 km, PQ = R θ = 1 \times π / 2$
Average speed of the cyclist $= \frac{Total distance travelled}{Total time}$
$= \frac{OP + PQ + QO}{10} km / \min$
$= \frac{1 + \frac{π}{2} \times 1 + 1}{10} km / \min$
$= \frac{4 + π}{20} \times 60 km / hr$
$= 7.14 \times 3 km / hr$
$= 21.4 km / hr$

Karnataka CET-2006

Motion in Plane

144052 A mass attached to one end of a string crosses top-most point on a vertical circle with critical speed. Its centripetal acceleration when string becomes horizontal will be ( $g =$ gravitational acceleration)

1

g

2

3 g

3

4 g

4

6 g

Explanation:

B We know, velocity of particle at top most point on vertical circle,
$v_{T} = \sqrt{3 rg}$
original image
But centripetal acceleration $(a_{c}) = \frac{v_{T}^{2}}{r}$
$∴ a_{c} = \frac{(\sqrt{3 rg})^{2}}{r} [From equation (i)]$
$a_{c} = \frac{3 rg}{r} \Rightarrow a_{c} = 3 g$

MHT-CET 2018

Motion in Plane

144053 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:

A Given, initial speed of the particle $(u) = 0$
Final speed of the particle $= v$
Distance covering in 2 revolution $(s) = 2 \times 2 π r$
Applying equation of motion,
$= 4 π r$
$v^{2} = u^{2} + 2 as$
$v^{2} = 0 + 2 \times a \times 4 π r$
$a = \frac{v^{2}}{8 π r}$

MHT-CET 2016

Motion in Plane

144054 A particle of mass ' $m$ ' is moving in circular path of constant radius ' $r$ ' such that centripetal acceleration is varying with time ' $t$ ' as $K^{2} r t^{2}$ where $K$ is a constant. The power delivered to the particle by the force acting on it is

1

m^{2} K^{2} r^{2} t^{2}

2

m K^{2} r^{2} t

3

{mK}^{2} {rt}^{2}

4

{mKr}^{2} t

Explanation:

B Given,
Centripetal acceleration $(a_{c}) = K^{2} {rt}^{2}$
And we know,
Centripetal accelerate $(a_{c}) = \frac{v^{2}}{r}$
Comparing equation (i) and (ii),
$\frac{v^{2}}{r} = K^{2} r t^{2}$
$v^{2} = K^{2} r^{2} t^{2}$
$v = K r t$
Tangential acceleration $(a) = \frac{dv}{dt} = Kr$
Then, Force $(F) = ma = mKr$
So, Power $(P) = F . v$
$P = mKr \times Krt$
$P = {mK}^{2} r^{2} t$

MHT-CET 2015

Motion in Plane

144056 An aeroplane executes a horizontal loop at a speed of $720 km / h$ with its wings banked at $45^{\circ}$ . What is the radius of the loop? Take $g = 10 {ms}^{- 2}$ .

1

4 km

2

4.5 km

3

7.2 km

4

2 km

Explanation:

A Given, angle of banking $(θ) = 45^{\circ}$ , speed of the aeroplane $(v) = 720 km / h = 720 \times \frac{5}{18} = 200 m / s$
For radius $r$ of the loop, We have the relation,
$\tan θ = \frac{v^{2}}{rg}$
$r = \frac{v^{2}}{g \tan θ}$
$r = \frac{(200)^{2}}{10 \times \tan 45^{\circ}} (∵ θ = 45^{\circ})$
$r = \frac{40000}{10} = 4000 m$
$r = 4 km$

Karnataka CET-2014

Motion in Plane

144058 A cyclist starts from the centre $O$ of a circular park of radius one kilometre, reaches the edge $P$ of the park, then cycles along the circumference and returns to the centre along QO as shown in the figure. If the round trip takes ten minutes, the net displacement and average speed of the cyclist (in metre and kilometre per hour) is :
original image

1 0,1

2

\frac{π + 4}{2}, 0

3

21.4, \frac{π + 4}{2}

4

0, 21.4

Explanation:

D Since the net displacement of the cyclist is zero then the average velocity will also be zero.
by diagram, $OP =$ Radius $= 1 km, PQ = R θ = 1 \times π / 2$
Average speed of the cyclist $= \frac{Total distance travelled}{Total time}$
$= \frac{OP + PQ + QO}{10} km / \min$
$= \frac{1 + \frac{π}{2} \times 1 + 1}{10} km / \min$
$= \frac{4 + π}{20} \times 60 km / hr$
$= 7.14 \times 3 km / hr$
$= 21.4 km / hr$

Karnataka CET-2006

Motion in Plane

144052 A mass attached to one end of a string crosses top-most point on a vertical circle with critical speed. Its centripetal acceleration when string becomes horizontal will be ( $g =$ gravitational acceleration)

1

g

2

3 g

3

4 g

4

6 g

Explanation:

B We know, velocity of particle at top most point on vertical circle,
$v_{T} = \sqrt{3 rg}$
original image
But centripetal acceleration $(a_{c}) = \frac{v_{T}^{2}}{r}$
$∴ a_{c} = \frac{(\sqrt{3 rg})^{2}}{r} [From equation (i)]$
$a_{c} = \frac{3 rg}{r} \Rightarrow a_{c} = 3 g$

MHT-CET 2018

Motion in Plane

144053 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:

A Given, initial speed of the particle $(u) = 0$
Final speed of the particle $= v$
Distance covering in 2 revolution $(s) = 2 \times 2 π r$
Applying equation of motion,
$= 4 π r$
$v^{2} = u^{2} + 2 as$
$v^{2} = 0 + 2 \times a \times 4 π r$
$a = \frac{v^{2}}{8 π r}$

MHT-CET 2016

Motion in Plane

144054 A particle of mass ' $m$ ' is moving in circular path of constant radius ' $r$ ' such that centripetal acceleration is varying with time ' $t$ ' as $K^{2} r t^{2}$ where $K$ is a constant. The power delivered to the particle by the force acting on it is

1

m^{2} K^{2} r^{2} t^{2}

2

m K^{2} r^{2} t

3

{mK}^{2} {rt}^{2}

4

{mKr}^{2} t

Explanation:

B Given,
Centripetal acceleration $(a_{c}) = K^{2} {rt}^{2}$
And we know,
Centripetal accelerate $(a_{c}) = \frac{v^{2}}{r}$
Comparing equation (i) and (ii),
$\frac{v^{2}}{r} = K^{2} r t^{2}$
$v^{2} = K^{2} r^{2} t^{2}$
$v = K r t$
Tangential acceleration $(a) = \frac{dv}{dt} = Kr$
Then, Force $(F) = ma = mKr$
So, Power $(P) = F . v$
$P = mKr \times Krt$
$P = {mK}^{2} r^{2} t$

MHT-CET 2015

Motion in Plane

144056 An aeroplane executes a horizontal loop at a speed of $720 km / h$ with its wings banked at $45^{\circ}$ . What is the radius of the loop? Take $g = 10 {ms}^{- 2}$ .

1

4 km

2

4.5 km

3

7.2 km

4

2 km

Explanation:

A Given, angle of banking $(θ) = 45^{\circ}$ , speed of the aeroplane $(v) = 720 km / h = 720 \times \frac{5}{18} = 200 m / s$
For radius $r$ of the loop, We have the relation,
$\tan θ = \frac{v^{2}}{rg}$
$r = \frac{v^{2}}{g \tan θ}$
$r = \frac{(200)^{2}}{10 \times \tan 45^{\circ}} (∵ θ = 45^{\circ})$
$r = \frac{40000}{10} = 4000 m$
$r = 4 km$

Karnataka CET-2014

Motion in Plane

144058 A cyclist starts from the centre $O$ of a circular park of radius one kilometre, reaches the edge $P$ of the park, then cycles along the circumference and returns to the centre along QO as shown in the figure. If the round trip takes ten minutes, the net displacement and average speed of the cyclist (in metre and kilometre per hour) is :
original image

1 0,1

2

\frac{π + 4}{2}, 0

3

21.4, \frac{π + 4}{2}

4

0, 21.4

Explanation:

D Since the net displacement of the cyclist is zero then the average velocity will also be zero.
by diagram, $OP =$ Radius $= 1 km, PQ = R θ = 1 \times π / 2$
Average speed of the cyclist $= \frac{Total distance travelled}{Total time}$
$= \frac{OP + PQ + QO}{10} km / \min$
$= \frac{1 + \frac{π}{2} \times 1 + 1}{10} km / \min$
$= \frac{4 + π}{20} \times 60 km / hr$
$= 7.14 \times 3 km / hr$
$= 21.4 km / hr$

Karnataka CET-2006

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