QBManager

Motion in Plane

144044 What is the least radius of curve on a horizontal road, at which a vehicle can travel with a speed of $36 km / hr$ at an angle of inclination $45^{\circ}$ ?
[g $= 10 m / s^{2}, \tan 45^{\circ} = 1]$

1

15 m

2

20 m

3

10 m

4

25 m

Explanation:

C Given,
Vehicle speed $(v) = 36 km / hr \times \frac{5}{18} = 10 m / s$
Angle, $(θ) = 45^{\circ}$
We know,
Centripetal force $=$ Frictional force
$\frac{{mv}^{2}}{r} = μ mg (μ = \tan θ)$
$\frac{v^{2}}{r \times g} = \tan θ$
$r = \frac{v^{2}}{g \times \tan θ}$
$r = \frac{(10)^{2}}{10 \times \tan 45^{\circ}} = \frac{100}{10 \times 1} = 10 m$

MHT-CET 2020

Motion in Plane

144045 A particle is performing vertical circular motion. The difference in tension at lowest and highest point is

1

8 mg

2

2 mg

3

6 mg

4

4 mg

Explanation:

C According to the question
original image
Let, $T_{1} =$ Tension at lowest point $T_{2} =$ Tension at highest point $v =$ Velocity at highest point $u =$ Velocity at lowest point
At the lowest point-
$T_{1} = \frac{{mu}^{2}}{R} + mg$
At the highest point-
Substracting equation (i) from (ii)
$(T_{1} - T_{2}) = (\frac{{mu}^{2}}{R} + mg) - (\frac{{mv}^{2}}{R} - mg)$
$(T_{1} - T_{2}) = \frac{m}{R} (u^{2} - v^{2}) + 2 mg$
From conservation of energy,
$(P . E + K . E)_{Lowest} = (P . E + K . E)_{Highest}$
$0 + \frac{1}{2} {mu}^{2} = mg 2 R + \frac{1}{2} {mv}^{2}$
$\frac{1}{2} m (u^{2} - v^{2}) = mg 2 R$
$u^{2} - v^{2} = 4 gR$
Putting these value in equation (iii)
$T_{1} - T_{2} = \frac{m}{R} \times 4 gR + 2 mg$
$= 4 mg + 2 mg = 6 mg$

MHT-CET 2020

Motion in Plane

144046 A particle is moving along the circular path with constant speed and centripetal acceleration ' $a$ '. If the speed is doubled, the ratio of its acceleration after and before the change is

1

4 : 1

2

2 : 1

3

3 : 1

4

1 : 4

Explanation:

A We know,
Centripetal acceleration $(a) = \frac{v^{2}}{r}$
Acceleration before change, $a = \frac{v^{2}}{r}$
and acceleration after change, $a_{2} = \frac{(2 v)^{2}}{r} = \frac{4 v^{2}}{r}$
Ratio of both acceleration,
$\frac{a_{2}}{a} = \frac{4 v^{2} / r}{v^{2} / r} \Rightarrow \frac{a_{2}}{a} = \frac{4}{1}$

MHT-CET 2019

Motion in Plane

144047 A rod of length ' $L$ ' is hung from its one end and a mass ' $m$ ' is attached to its free end. What tangential velocity must be imparted to ' $m$ '. So that it reaches the top of the vertical circle? (g = acceleration due to gravity)

1

4 \sqrt{gL}

2

2 \sqrt{gL}

3

5 \sqrt{gL}

4

3 \sqrt{gL}

Explanation:

B According to the question
original image
From conservation of energy,
$(K \cdot E)_{A} + (P . E)_{A} = (K \cdot E)_{B} + (P . E)_{B}$
$\frac{1}{2} {mv}^{2} + 0 = 0 + mg \times (2 L)$
$\frac{1}{2} mv v^{2} = mg \times 2 L$
$\frac{1}{2} v^{2} = 2 gL$
$v^{2} = 4 gL$
$v = 2 \sqrt{gL}$

MHT-CET 2020

Motion in Plane

144044 What is the least radius of curve on a horizontal road, at which a vehicle can travel with a speed of $36 km / hr$ at an angle of inclination $45^{\circ}$ ?
[g $= 10 m / s^{2}, \tan 45^{\circ} = 1]$

1

15 m

2

20 m

3

10 m

4

25 m

Explanation:

C Given,
Vehicle speed $(v) = 36 km / hr \times \frac{5}{18} = 10 m / s$
Angle, $(θ) = 45^{\circ}$
We know,
Centripetal force $=$ Frictional force
$\frac{{mv}^{2}}{r} = μ mg (μ = \tan θ)$
$\frac{v^{2}}{r \times g} = \tan θ$
$r = \frac{v^{2}}{g \times \tan θ}$
$r = \frac{(10)^{2}}{10 \times \tan 45^{\circ}} = \frac{100}{10 \times 1} = 10 m$

MHT-CET 2020

Motion in Plane

144045 A particle is performing vertical circular motion. The difference in tension at lowest and highest point is

1

8 mg

2

2 mg

3

6 mg

4

4 mg

Explanation:

C According to the question
original image
Let, $T_{1} =$ Tension at lowest point $T_{2} =$ Tension at highest point $v =$ Velocity at highest point $u =$ Velocity at lowest point
At the lowest point-
$T_{1} = \frac{{mu}^{2}}{R} + mg$
At the highest point-
Substracting equation (i) from (ii)
$(T_{1} - T_{2}) = (\frac{{mu}^{2}}{R} + mg) - (\frac{{mv}^{2}}{R} - mg)$
$(T_{1} - T_{2}) = \frac{m}{R} (u^{2} - v^{2}) + 2 mg$
From conservation of energy,
$(P . E + K . E)_{Lowest} = (P . E + K . E)_{Highest}$
$0 + \frac{1}{2} {mu}^{2} = mg 2 R + \frac{1}{2} {mv}^{2}$
$\frac{1}{2} m (u^{2} - v^{2}) = mg 2 R$
$u^{2} - v^{2} = 4 gR$
Putting these value in equation (iii)
$T_{1} - T_{2} = \frac{m}{R} \times 4 gR + 2 mg$
$= 4 mg + 2 mg = 6 mg$

MHT-CET 2020

Motion in Plane

144046 A particle is moving along the circular path with constant speed and centripetal acceleration ' $a$ '. If the speed is doubled, the ratio of its acceleration after and before the change is

1

4 : 1

2

2 : 1

3

3 : 1

4

1 : 4

Explanation:

A We know,
Centripetal acceleration $(a) = \frac{v^{2}}{r}$
Acceleration before change, $a = \frac{v^{2}}{r}$
and acceleration after change, $a_{2} = \frac{(2 v)^{2}}{r} = \frac{4 v^{2}}{r}$
Ratio of both acceleration,
$\frac{a_{2}}{a} = \frac{4 v^{2} / r}{v^{2} / r} \Rightarrow \frac{a_{2}}{a} = \frac{4}{1}$

MHT-CET 2019

Motion in Plane

144047 A rod of length ' $L$ ' is hung from its one end and a mass ' $m$ ' is attached to its free end. What tangential velocity must be imparted to ' $m$ '. So that it reaches the top of the vertical circle? (g = acceleration due to gravity)

1

4 \sqrt{gL}

2

2 \sqrt{gL}

3

5 \sqrt{gL}

4

3 \sqrt{gL}

Explanation:

B According to the question
original image
From conservation of energy,
$(K \cdot E)_{A} + (P . E)_{A} = (K \cdot E)_{B} + (P . E)_{B}$
$\frac{1}{2} {mv}^{2} + 0 = 0 + mg \times (2 L)$
$\frac{1}{2} mv v^{2} = mg \times 2 L$
$\frac{1}{2} v^{2} = 2 gL$
$v^{2} = 4 gL$
$v = 2 \sqrt{gL}$

MHT-CET 2020

Motion in Plane

144044 What is the least radius of curve on a horizontal road, at which a vehicle can travel with a speed of $36 km / hr$ at an angle of inclination $45^{\circ}$ ?
[g $= 10 m / s^{2}, \tan 45^{\circ} = 1]$

1

15 m

2

20 m

3

10 m

4

25 m

Explanation:

C Given,
Vehicle speed $(v) = 36 km / hr \times \frac{5}{18} = 10 m / s$
Angle, $(θ) = 45^{\circ}$
We know,
Centripetal force $=$ Frictional force
$\frac{{mv}^{2}}{r} = μ mg (μ = \tan θ)$
$\frac{v^{2}}{r \times g} = \tan θ$
$r = \frac{v^{2}}{g \times \tan θ}$
$r = \frac{(10)^{2}}{10 \times \tan 45^{\circ}} = \frac{100}{10 \times 1} = 10 m$

MHT-CET 2020

Motion in Plane

144045 A particle is performing vertical circular motion. The difference in tension at lowest and highest point is

1

8 mg

2

2 mg

3

6 mg

4

4 mg

Explanation:

C According to the question
original image
Let, $T_{1} =$ Tension at lowest point $T_{2} =$ Tension at highest point $v =$ Velocity at highest point $u =$ Velocity at lowest point
At the lowest point-
$T_{1} = \frac{{mu}^{2}}{R} + mg$
At the highest point-
Substracting equation (i) from (ii)
$(T_{1} - T_{2}) = (\frac{{mu}^{2}}{R} + mg) - (\frac{{mv}^{2}}{R} - mg)$
$(T_{1} - T_{2}) = \frac{m}{R} (u^{2} - v^{2}) + 2 mg$
From conservation of energy,
$(P . E + K . E)_{Lowest} = (P . E + K . E)_{Highest}$
$0 + \frac{1}{2} {mu}^{2} = mg 2 R + \frac{1}{2} {mv}^{2}$
$\frac{1}{2} m (u^{2} - v^{2}) = mg 2 R$
$u^{2} - v^{2} = 4 gR$
Putting these value in equation (iii)
$T_{1} - T_{2} = \frac{m}{R} \times 4 gR + 2 mg$
$= 4 mg + 2 mg = 6 mg$

MHT-CET 2020

Motion in Plane

144046 A particle is moving along the circular path with constant speed and centripetal acceleration ' $a$ '. If the speed is doubled, the ratio of its acceleration after and before the change is

1

4 : 1

2

2 : 1

3

3 : 1

4

1 : 4

Explanation:

A We know,
Centripetal acceleration $(a) = \frac{v^{2}}{r}$
Acceleration before change, $a = \frac{v^{2}}{r}$
and acceleration after change, $a_{2} = \frac{(2 v)^{2}}{r} = \frac{4 v^{2}}{r}$
Ratio of both acceleration,
$\frac{a_{2}}{a} = \frac{4 v^{2} / r}{v^{2} / r} \Rightarrow \frac{a_{2}}{a} = \frac{4}{1}$

MHT-CET 2019

Motion in Plane

144047 A rod of length ' $L$ ' is hung from its one end and a mass ' $m$ ' is attached to its free end. What tangential velocity must be imparted to ' $m$ '. So that it reaches the top of the vertical circle? (g = acceleration due to gravity)

1

4 \sqrt{gL}

2

2 \sqrt{gL}

3

5 \sqrt{gL}

4

3 \sqrt{gL}

Explanation:

B According to the question
original image
From conservation of energy,
$(K \cdot E)_{A} + (P . E)_{A} = (K \cdot E)_{B} + (P . E)_{B}$
$\frac{1}{2} {mv}^{2} + 0 = 0 + mg \times (2 L)$
$\frac{1}{2} mv v^{2} = mg \times 2 L$
$\frac{1}{2} v^{2} = 2 gL$
$v^{2} = 4 gL$
$v = 2 \sqrt{gL}$

MHT-CET 2020

Motion in Plane

144044 What is the least radius of curve on a horizontal road, at which a vehicle can travel with a speed of $36 km / hr$ at an angle of inclination $45^{\circ}$ ?
[g $= 10 m / s^{2}, \tan 45^{\circ} = 1]$

1

15 m

2

20 m

3

10 m

4

25 m

Explanation:

C Given,
Vehicle speed $(v) = 36 km / hr \times \frac{5}{18} = 10 m / s$
Angle, $(θ) = 45^{\circ}$
We know,
Centripetal force $=$ Frictional force
$\frac{{mv}^{2}}{r} = μ mg (μ = \tan θ)$
$\frac{v^{2}}{r \times g} = \tan θ$
$r = \frac{v^{2}}{g \times \tan θ}$
$r = \frac{(10)^{2}}{10 \times \tan 45^{\circ}} = \frac{100}{10 \times 1} = 10 m$

MHT-CET 2020

Motion in Plane

144045 A particle is performing vertical circular motion. The difference in tension at lowest and highest point is

1

8 mg

2

2 mg

3

6 mg

4

4 mg

Explanation:

C According to the question
original image
Let, $T_{1} =$ Tension at lowest point $T_{2} =$ Tension at highest point $v =$ Velocity at highest point $u =$ Velocity at lowest point
At the lowest point-
$T_{1} = \frac{{mu}^{2}}{R} + mg$
At the highest point-
Substracting equation (i) from (ii)
$(T_{1} - T_{2}) = (\frac{{mu}^{2}}{R} + mg) - (\frac{{mv}^{2}}{R} - mg)$
$(T_{1} - T_{2}) = \frac{m}{R} (u^{2} - v^{2}) + 2 mg$
From conservation of energy,
$(P . E + K . E)_{Lowest} = (P . E + K . E)_{Highest}$
$0 + \frac{1}{2} {mu}^{2} = mg 2 R + \frac{1}{2} {mv}^{2}$
$\frac{1}{2} m (u^{2} - v^{2}) = mg 2 R$
$u^{2} - v^{2} = 4 gR$
Putting these value in equation (iii)
$T_{1} - T_{2} = \frac{m}{R} \times 4 gR + 2 mg$
$= 4 mg + 2 mg = 6 mg$

MHT-CET 2020

Motion in Plane

144046 A particle is moving along the circular path with constant speed and centripetal acceleration ' $a$ '. If the speed is doubled, the ratio of its acceleration after and before the change is

1

4 : 1

2

2 : 1

3

3 : 1

4

1 : 4

Explanation:

A We know,
Centripetal acceleration $(a) = \frac{v^{2}}{r}$
Acceleration before change, $a = \frac{v^{2}}{r}$
and acceleration after change, $a_{2} = \frac{(2 v)^{2}}{r} = \frac{4 v^{2}}{r}$
Ratio of both acceleration,
$\frac{a_{2}}{a} = \frac{4 v^{2} / r}{v^{2} / r} \Rightarrow \frac{a_{2}}{a} = \frac{4}{1}$

MHT-CET 2019

Motion in Plane

144047 A rod of length ' $L$ ' is hung from its one end and a mass ' $m$ ' is attached to its free end. What tangential velocity must be imparted to ' $m$ '. So that it reaches the top of the vertical circle? (g = acceleration due to gravity)

1

4 \sqrt{gL}

2

2 \sqrt{gL}

3

5 \sqrt{gL}

4

3 \sqrt{gL}

Explanation:

B According to the question
original image
From conservation of energy,
$(K \cdot E)_{A} + (P . E)_{A} = (K \cdot E)_{B} + (P . E)_{B}$
$\frac{1}{2} {mv}^{2} + 0 = 0 + mg \times (2 L)$
$\frac{1}{2} mv v^{2} = mg \times 2 L$
$\frac{1}{2} v^{2} = 2 gL$
$v^{2} = 4 gL$
$v = 2 \sqrt{gL}$

MHT-CET 2020