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Motion in Plane

144031 A heavy mass is attached at one end of a thin wire and whirled in a vertical circle. The chances of breaking the wire are maximum when

1 The wire makes an angle of

60^{\circ}

with the horizontal

2 The mass is at the highest point of the circle

3 The mass is at the lowest point of the circle

4 The wire is horizontal

Explanation:

C original image
The body is rotating in a vertical circle.
So, at the bottom, $T_{2} = mg + F_{c}$
At the top, $T_{1} + mg = F_{c}$
$T_{1} = F_{c} - mg$
Comparing equation (i) and (ii), we get -
$T_{2} > T_{1}$
So, tension is maximum at the lowest position of mass Hence, the chance of breaking is maximum at lowest point.

MHT-CET 2020

Motion in Plane

144032 A particle of mass ' $m$ ' is rotating in a circle of radius ' $r$ ' having angular momentum ' $L$ '. Then the centripetal force will be

1

\frac{L^{2}}{{mr}^{2}}

2

\frac{L^{2}}{{mr}^{2}}

3

\frac{L^{2}}{mr}

4

\frac{L^{2}}{{mr}^{3}}

Explanation:

D Given, mass $= m$ , circle radius $= r$ , angular momentum $= L$
We know,
Centripetal force $(F_{c}) = \frac{{mv}^{2}}{r}$
Angular momentum $(L) = mvr$
$v = \frac{L}{mr}$
Substituting the value of ' $v$ ' from equation (ii) to equation (i),
$F_{c} = \frac{m}{r} {(\frac{L}{mr})}^{2}$
$F_{c} = \frac{m}{r} \times \frac{L^{2}}{m^{2} r^{2}} \Rightarrow F_{c} = \frac{L^{2}}{{mr}^{3}}$

MHT-CET 2020

Motion in Plane

144034 A particle is moving in a circle of radius ' $R$ ' with constant speed ' $V$ '. The magnitude of average acceleration after half revolution is

1

\frac{2 V^{2}}{π R}

2

\frac{2 π}{R V^{2}}

3

\frac{2 V}{π R^{2}}

4

\frac{2 R}{π V}

Explanation:

A In a half revolution, change in momentum, $Δ P = mV - m (- V) = 2 mV$
The amount of time it takes for a particle to complete a half revolution,
$t = \frac{π R}{V}$
Average force $(F) =$ Rate of change in momentum
$= \frac{Δ P}{t}$
$= \frac{2 mV}{\frac{π R}{V}} [From equation (i) and (ii)]$
$F = \frac{2 m V^{2}}{π R} {∵ F = ma}$
$ma = \frac{2 mV V^{2}}{π R}$
$a = \frac{2 V^{2}}{π R}$

MHT-CET 2020

Motion in Plane

144035 A particle of mass ' $m$ ' is performing U.C.M. along a circle of radius ' $r$ '. The relation between centripetal acceleration ' $a$ ' and kinetic energy ' $E$ ' is given by

1

a = {(\frac{2 E}{mr})}^{2}

2

a = \frac{E}{m r}

3

a = \frac{2 E}{mr}

4

a = 2 Em

Explanation:

C We know that
$K . E = \frac{1}{2} I ω^{2}$
and $ω^{2} = \frac{a}{r}$
from eqn (i) and (ii)
$∴ E = \frac{I}{2} \times \frac{a}{r}$
$E = \frac{1}{2} {mr}^{2} \times \frac{a}{r} [I = {mr}^{2}]$
$E = \frac{1}{2} mr \cdot a$
$a = \frac{2 E}{mr}$

MHT-CET 2020

Motion in Plane

144036 A train has to negotiate a curve of radius ' $r$ ' $m$ , the distance between the rails is ' $ℓ$ ' $m$ and outer rail is raised above inner rail by distance of ' $h$ ' $m$ . If the angle of banking is small, the safety speed limit on this banked road is

1

rg \frac{h}{ℓ}

2

\frac{{(\frac{h}{ℓ})}^{2}}{rg}

3

\sqrt{rg (\frac{h}{ℓ})}

4

{(rg \frac{h}{ℓ})}^{2}

Explanation:

C Given, radius of curve $= r$ , distance between the rails $= l$ , raised distance $= h$
original image
At optimum speed,
Centrifugal force $(F_{c}) =$ Friction force $(F_{s})$
$\frac{{mv}^{2}}{r} = μ N = μ mg$
$v^{2} = μ gr \Rightarrow v = \sqrt{μ gr}$
$v = \sqrt{\tan θ gr} (∴ \tan θ = \frac{h}{l})$
$v = \sqrt{gr \frac{h}{l}}$

MHT-CET 2020

Motion in Plane

144031 A heavy mass is attached at one end of a thin wire and whirled in a vertical circle. The chances of breaking the wire are maximum when

1 The wire makes an angle of

60^{\circ}

with the horizontal

2 The mass is at the highest point of the circle

3 The mass is at the lowest point of the circle

4 The wire is horizontal

Explanation:

C original image
The body is rotating in a vertical circle.
So, at the bottom, $T_{2} = mg + F_{c}$
At the top, $T_{1} + mg = F_{c}$
$T_{1} = F_{c} - mg$
Comparing equation (i) and (ii), we get -
$T_{2} > T_{1}$
So, tension is maximum at the lowest position of mass Hence, the chance of breaking is maximum at lowest point.

MHT-CET 2020

Motion in Plane

144032 A particle of mass ' $m$ ' is rotating in a circle of radius ' $r$ ' having angular momentum ' $L$ '. Then the centripetal force will be

1

\frac{L^{2}}{{mr}^{2}}

2

\frac{L^{2}}{{mr}^{2}}

3

\frac{L^{2}}{mr}

4

\frac{L^{2}}{{mr}^{3}}

Explanation:

D Given, mass $= m$ , circle radius $= r$ , angular momentum $= L$
We know,
Centripetal force $(F_{c}) = \frac{{mv}^{2}}{r}$
Angular momentum $(L) = mvr$
$v = \frac{L}{mr}$
Substituting the value of ' $v$ ' from equation (ii) to equation (i),
$F_{c} = \frac{m}{r} {(\frac{L}{mr})}^{2}$
$F_{c} = \frac{m}{r} \times \frac{L^{2}}{m^{2} r^{2}} \Rightarrow F_{c} = \frac{L^{2}}{{mr}^{3}}$

MHT-CET 2020

Motion in Plane

144034 A particle is moving in a circle of radius ' $R$ ' with constant speed ' $V$ '. The magnitude of average acceleration after half revolution is

1

\frac{2 V^{2}}{π R}

2

\frac{2 π}{R V^{2}}

3

\frac{2 V}{π R^{2}}

4

\frac{2 R}{π V}

Explanation:

A In a half revolution, change in momentum, $Δ P = mV - m (- V) = 2 mV$
The amount of time it takes for a particle to complete a half revolution,
$t = \frac{π R}{V}$
Average force $(F) =$ Rate of change in momentum
$= \frac{Δ P}{t}$
$= \frac{2 mV}{\frac{π R}{V}} [From equation (i) and (ii)]$
$F = \frac{2 m V^{2}}{π R} {∵ F = ma}$
$ma = \frac{2 mV V^{2}}{π R}$
$a = \frac{2 V^{2}}{π R}$

MHT-CET 2020

Motion in Plane

144035 A particle of mass ' $m$ ' is performing U.C.M. along a circle of radius ' $r$ '. The relation between centripetal acceleration ' $a$ ' and kinetic energy ' $E$ ' is given by

1

a = {(\frac{2 E}{mr})}^{2}

2

a = \frac{E}{m r}

3

a = \frac{2 E}{mr}

4

a = 2 Em

Explanation:

C We know that
$K . E = \frac{1}{2} I ω^{2}$
and $ω^{2} = \frac{a}{r}$
from eqn (i) and (ii)
$∴ E = \frac{I}{2} \times \frac{a}{r}$
$E = \frac{1}{2} {mr}^{2} \times \frac{a}{r} [I = {mr}^{2}]$
$E = \frac{1}{2} mr \cdot a$
$a = \frac{2 E}{mr}$

MHT-CET 2020

Motion in Plane

144036 A train has to negotiate a curve of radius ' $r$ ' $m$ , the distance between the rails is ' $ℓ$ ' $m$ and outer rail is raised above inner rail by distance of ' $h$ ' $m$ . If the angle of banking is small, the safety speed limit on this banked road is

1

rg \frac{h}{ℓ}

2

\frac{{(\frac{h}{ℓ})}^{2}}{rg}

3

\sqrt{rg (\frac{h}{ℓ})}

4

{(rg \frac{h}{ℓ})}^{2}

Explanation:

C Given, radius of curve $= r$ , distance between the rails $= l$ , raised distance $= h$
original image
At optimum speed,
Centrifugal force $(F_{c}) =$ Friction force $(F_{s})$
$\frac{{mv}^{2}}{r} = μ N = μ mg$
$v^{2} = μ gr \Rightarrow v = \sqrt{μ gr}$
$v = \sqrt{\tan θ gr} (∴ \tan θ = \frac{h}{l})$
$v = \sqrt{gr \frac{h}{l}}$

MHT-CET 2020

Motion in Plane

144031 A heavy mass is attached at one end of a thin wire and whirled in a vertical circle. The chances of breaking the wire are maximum when

1 The wire makes an angle of

60^{\circ}

with the horizontal

2 The mass is at the highest point of the circle

3 The mass is at the lowest point of the circle

4 The wire is horizontal

Explanation:

C original image
The body is rotating in a vertical circle.
So, at the bottom, $T_{2} = mg + F_{c}$
At the top, $T_{1} + mg = F_{c}$
$T_{1} = F_{c} - mg$
Comparing equation (i) and (ii), we get -
$T_{2} > T_{1}$
So, tension is maximum at the lowest position of mass Hence, the chance of breaking is maximum at lowest point.

MHT-CET 2020

Motion in Plane

144032 A particle of mass ' $m$ ' is rotating in a circle of radius ' $r$ ' having angular momentum ' $L$ '. Then the centripetal force will be

1

\frac{L^{2}}{{mr}^{2}}

2

\frac{L^{2}}{{mr}^{2}}

3

\frac{L^{2}}{mr}

4

\frac{L^{2}}{{mr}^{3}}

Explanation:

D Given, mass $= m$ , circle radius $= r$ , angular momentum $= L$
We know,
Centripetal force $(F_{c}) = \frac{{mv}^{2}}{r}$
Angular momentum $(L) = mvr$
$v = \frac{L}{mr}$
Substituting the value of ' $v$ ' from equation (ii) to equation (i),
$F_{c} = \frac{m}{r} {(\frac{L}{mr})}^{2}$
$F_{c} = \frac{m}{r} \times \frac{L^{2}}{m^{2} r^{2}} \Rightarrow F_{c} = \frac{L^{2}}{{mr}^{3}}$

MHT-CET 2020

Motion in Plane

144034 A particle is moving in a circle of radius ' $R$ ' with constant speed ' $V$ '. The magnitude of average acceleration after half revolution is

1

\frac{2 V^{2}}{π R}

2

\frac{2 π}{R V^{2}}

3

\frac{2 V}{π R^{2}}

4

\frac{2 R}{π V}

Explanation:

A In a half revolution, change in momentum, $Δ P = mV - m (- V) = 2 mV$
The amount of time it takes for a particle to complete a half revolution,
$t = \frac{π R}{V}$
Average force $(F) =$ Rate of change in momentum
$= \frac{Δ P}{t}$
$= \frac{2 mV}{\frac{π R}{V}} [From equation (i) and (ii)]$
$F = \frac{2 m V^{2}}{π R} {∵ F = ma}$
$ma = \frac{2 mV V^{2}}{π R}$
$a = \frac{2 V^{2}}{π R}$

MHT-CET 2020

Motion in Plane

144035 A particle of mass ' $m$ ' is performing U.C.M. along a circle of radius ' $r$ '. The relation between centripetal acceleration ' $a$ ' and kinetic energy ' $E$ ' is given by

1

a = {(\frac{2 E}{mr})}^{2}

2

a = \frac{E}{m r}

3

a = \frac{2 E}{mr}

4

a = 2 Em

Explanation:

C We know that
$K . E = \frac{1}{2} I ω^{2}$
and $ω^{2} = \frac{a}{r}$
from eqn (i) and (ii)
$∴ E = \frac{I}{2} \times \frac{a}{r}$
$E = \frac{1}{2} {mr}^{2} \times \frac{a}{r} [I = {mr}^{2}]$
$E = \frac{1}{2} mr \cdot a$
$a = \frac{2 E}{mr}$

MHT-CET 2020

Motion in Plane

144036 A train has to negotiate a curve of radius ' $r$ ' $m$ , the distance between the rails is ' $ℓ$ ' $m$ and outer rail is raised above inner rail by distance of ' $h$ ' $m$ . If the angle of banking is small, the safety speed limit on this banked road is

1

rg \frac{h}{ℓ}

2

\frac{{(\frac{h}{ℓ})}^{2}}{rg}

3

\sqrt{rg (\frac{h}{ℓ})}

4

{(rg \frac{h}{ℓ})}^{2}

Explanation:

C Given, radius of curve $= r$ , distance between the rails $= l$ , raised distance $= h$
original image
At optimum speed,
Centrifugal force $(F_{c}) =$ Friction force $(F_{s})$
$\frac{{mv}^{2}}{r} = μ N = μ mg$
$v^{2} = μ gr \Rightarrow v = \sqrt{μ gr}$
$v = \sqrt{\tan θ gr} (∴ \tan θ = \frac{h}{l})$
$v = \sqrt{gr \frac{h}{l}}$

MHT-CET 2020

Motion in Plane

144031 A heavy mass is attached at one end of a thin wire and whirled in a vertical circle. The chances of breaking the wire are maximum when

1 The wire makes an angle of

60^{\circ}

with the horizontal

2 The mass is at the highest point of the circle

3 The mass is at the lowest point of the circle

4 The wire is horizontal

Explanation:

C original image
The body is rotating in a vertical circle.
So, at the bottom, $T_{2} = mg + F_{c}$
At the top, $T_{1} + mg = F_{c}$
$T_{1} = F_{c} - mg$
Comparing equation (i) and (ii), we get -
$T_{2} > T_{1}$
So, tension is maximum at the lowest position of mass Hence, the chance of breaking is maximum at lowest point.

MHT-CET 2020

Motion in Plane

144032 A particle of mass ' $m$ ' is rotating in a circle of radius ' $r$ ' having angular momentum ' $L$ '. Then the centripetal force will be

1

\frac{L^{2}}{{mr}^{2}}

2

\frac{L^{2}}{{mr}^{2}}

3

\frac{L^{2}}{mr}

4

\frac{L^{2}}{{mr}^{3}}

Explanation:

D Given, mass $= m$ , circle radius $= r$ , angular momentum $= L$
We know,
Centripetal force $(F_{c}) = \frac{{mv}^{2}}{r}$
Angular momentum $(L) = mvr$
$v = \frac{L}{mr}$
Substituting the value of ' $v$ ' from equation (ii) to equation (i),
$F_{c} = \frac{m}{r} {(\frac{L}{mr})}^{2}$
$F_{c} = \frac{m}{r} \times \frac{L^{2}}{m^{2} r^{2}} \Rightarrow F_{c} = \frac{L^{2}}{{mr}^{3}}$

MHT-CET 2020

Motion in Plane

144034 A particle is moving in a circle of radius ' $R$ ' with constant speed ' $V$ '. The magnitude of average acceleration after half revolution is

1

\frac{2 V^{2}}{π R}

2

\frac{2 π}{R V^{2}}

3

\frac{2 V}{π R^{2}}

4

\frac{2 R}{π V}

Explanation:

A In a half revolution, change in momentum, $Δ P = mV - m (- V) = 2 mV$
The amount of time it takes for a particle to complete a half revolution,
$t = \frac{π R}{V}$
Average force $(F) =$ Rate of change in momentum
$= \frac{Δ P}{t}$
$= \frac{2 mV}{\frac{π R}{V}} [From equation (i) and (ii)]$
$F = \frac{2 m V^{2}}{π R} {∵ F = ma}$
$ma = \frac{2 mV V^{2}}{π R}$
$a = \frac{2 V^{2}}{π R}$

MHT-CET 2020

Motion in Plane

144035 A particle of mass ' $m$ ' is performing U.C.M. along a circle of radius ' $r$ '. The relation between centripetal acceleration ' $a$ ' and kinetic energy ' $E$ ' is given by

1

a = {(\frac{2 E}{mr})}^{2}

2

a = \frac{E}{m r}

3

a = \frac{2 E}{mr}

4

a = 2 Em

Explanation:

C We know that
$K . E = \frac{1}{2} I ω^{2}$
and $ω^{2} = \frac{a}{r}$
from eqn (i) and (ii)
$∴ E = \frac{I}{2} \times \frac{a}{r}$
$E = \frac{1}{2} {mr}^{2} \times \frac{a}{r} [I = {mr}^{2}]$
$E = \frac{1}{2} mr \cdot a$
$a = \frac{2 E}{mr}$

MHT-CET 2020

Motion in Plane

144036 A train has to negotiate a curve of radius ' $r$ ' $m$ , the distance between the rails is ' $ℓ$ ' $m$ and outer rail is raised above inner rail by distance of ' $h$ ' $m$ . If the angle of banking is small, the safety speed limit on this banked road is

1

rg \frac{h}{ℓ}

2

\frac{{(\frac{h}{ℓ})}^{2}}{rg}

3

\sqrt{rg (\frac{h}{ℓ})}

4

{(rg \frac{h}{ℓ})}^{2}

Explanation:

C Given, radius of curve $= r$ , distance between the rails $= l$ , raised distance $= h$
original image
At optimum speed,
Centrifugal force $(F_{c}) =$ Friction force $(F_{s})$
$\frac{{mv}^{2}}{r} = μ N = μ mg$
$v^{2} = μ gr \Rightarrow v = \sqrt{μ gr}$
$v = \sqrt{\tan θ gr} (∴ \tan θ = \frac{h}{l})$
$v = \sqrt{gr \frac{h}{l}}$

MHT-CET 2020

Motion in Plane

144031 A heavy mass is attached at one end of a thin wire and whirled in a vertical circle. The chances of breaking the wire are maximum when

1 The wire makes an angle of

60^{\circ}

with the horizontal

2 The mass is at the highest point of the circle

3 The mass is at the lowest point of the circle

4 The wire is horizontal

Explanation:

C original image
The body is rotating in a vertical circle.
So, at the bottom, $T_{2} = mg + F_{c}$
At the top, $T_{1} + mg = F_{c}$
$T_{1} = F_{c} - mg$
Comparing equation (i) and (ii), we get -
$T_{2} > T_{1}$
So, tension is maximum at the lowest position of mass Hence, the chance of breaking is maximum at lowest point.

MHT-CET 2020

Motion in Plane

144032 A particle of mass ' $m$ ' is rotating in a circle of radius ' $r$ ' having angular momentum ' $L$ '. Then the centripetal force will be

1

\frac{L^{2}}{{mr}^{2}}

2

\frac{L^{2}}{{mr}^{2}}

3

\frac{L^{2}}{mr}

4

\frac{L^{2}}{{mr}^{3}}

Explanation:

D Given, mass $= m$ , circle radius $= r$ , angular momentum $= L$
We know,
Centripetal force $(F_{c}) = \frac{{mv}^{2}}{r}$
Angular momentum $(L) = mvr$
$v = \frac{L}{mr}$
Substituting the value of ' $v$ ' from equation (ii) to equation (i),
$F_{c} = \frac{m}{r} {(\frac{L}{mr})}^{2}$
$F_{c} = \frac{m}{r} \times \frac{L^{2}}{m^{2} r^{2}} \Rightarrow F_{c} = \frac{L^{2}}{{mr}^{3}}$

MHT-CET 2020

Motion in Plane

144034 A particle is moving in a circle of radius ' $R$ ' with constant speed ' $V$ '. The magnitude of average acceleration after half revolution is

1

\frac{2 V^{2}}{π R}

2

\frac{2 π}{R V^{2}}

3

\frac{2 V}{π R^{2}}

4

\frac{2 R}{π V}

Explanation:

A In a half revolution, change in momentum, $Δ P = mV - m (- V) = 2 mV$
The amount of time it takes for a particle to complete a half revolution,
$t = \frac{π R}{V}$
Average force $(F) =$ Rate of change in momentum
$= \frac{Δ P}{t}$
$= \frac{2 mV}{\frac{π R}{V}} [From equation (i) and (ii)]$
$F = \frac{2 m V^{2}}{π R} {∵ F = ma}$
$ma = \frac{2 mV V^{2}}{π R}$
$a = \frac{2 V^{2}}{π R}$

MHT-CET 2020

Motion in Plane

144035 A particle of mass ' $m$ ' is performing U.C.M. along a circle of radius ' $r$ '. The relation between centripetal acceleration ' $a$ ' and kinetic energy ' $E$ ' is given by

1

a = {(\frac{2 E}{mr})}^{2}

2

a = \frac{E}{m r}

3

a = \frac{2 E}{mr}

4

a = 2 Em

Explanation:

C We know that
$K . E = \frac{1}{2} I ω^{2}$
and $ω^{2} = \frac{a}{r}$
from eqn (i) and (ii)
$∴ E = \frac{I}{2} \times \frac{a}{r}$
$E = \frac{1}{2} {mr}^{2} \times \frac{a}{r} [I = {mr}^{2}]$
$E = \frac{1}{2} mr \cdot a$
$a = \frac{2 E}{mr}$

MHT-CET 2020

Motion in Plane

144036 A train has to negotiate a curve of radius ' $r$ ' $m$ , the distance between the rails is ' $ℓ$ ' $m$ and outer rail is raised above inner rail by distance of ' $h$ ' $m$ . If the angle of banking is small, the safety speed limit on this banked road is

1

rg \frac{h}{ℓ}

2

\frac{{(\frac{h}{ℓ})}^{2}}{rg}

3

\sqrt{rg (\frac{h}{ℓ})}

4

{(rg \frac{h}{ℓ})}^{2}

Explanation:

C Given, radius of curve $= r$ , distance between the rails $= l$ , raised distance $= h$
original image
At optimum speed,
Centrifugal force $(F_{c}) =$ Friction force $(F_{s})$
$\frac{{mv}^{2}}{r} = μ N = μ mg$
$v^{2} = μ gr \Rightarrow v = \sqrt{μ gr}$
$v = \sqrt{\tan θ gr} (∴ \tan θ = \frac{h}{l})$
$v = \sqrt{gr \frac{h}{l}}$

MHT-CET 2020