Statement I: The max value of force F such that the block shown does not move is \$\dfrac{\mu mg}{cos\theta }\$ where \$\mu \$ is the coefficient of friction between block end surface.Statement II: Frictional Force = (Coefficient of friction) x (Normal reaction)

The max value of force F such that the block shown does not move is:\$Fcos\theta= \mu(mg+Fsin \theta)\$ \$F= \dfrac{\mu m g}{(cos\theta -\mu sin \theta)}\$So, the statement 1 is false.The frictional force is proportional to the normal reaction. So, Frictional Force = (Coefficient of friction) x (Normal reaction)

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>>Class 11
>>Physics
>>Laws of Motion
>> $Medium Questions$

Laws Of Motion

Q: A body of mass m is rotated at uniform speed along a vertical circle with the help of light string. If \$T_{1} and \ T_{2}\$ are tensions in the string when the body is crossing highest and lowest point of the vertical circle respectively, then which of the following expressions is correct?

Centripetal Force is: \$\dfrac { m{ v }^{ 2 } }{ r } \$At the highest point: \${ T }_{ 1 }=\dfrac { m{ v }^{ 2 } }{ r } -mg\$At the lowest point: \${ T }_{ 2 }=\dfrac { m{ v }^{ 2 } }{ r } +mg\$\$\therefore \quad { T }_{ 2 }-{ T }_{ 1 }=\dfrac { m{ v }^{ 2 } }{ r } +mg-(\dfrac { m{ v }^{ 2 } }{ r } -mg)=2mg\$

Q: A body is moving in a vertical circle such that the velocities of body at different points are critical. The ratio of velocities of body at angular displacements \$60^{0}\$ and \$120^{0}\$ from the lowest point is:

Given that velocities at different points are criticalso velocity at top point (\$T\$)\$\Rightarrow\$ \${ V }_{ T }\$\$=\sqrt { gR } ...(1) \$From the diagramin right triangle \$\triangle OMA\$\$\Rightarrow\$\$MA=OP=OP\cos { \theta } \$At point \$A\$, \$\theta =60°\$\${ \Rightarrow H }_{ A }=OG-OP\\ { \Rightarrow H }_{ A }=R-R\cos { 60° } =\frac { R }{ 2 } ...(2) \$in right triangle \$\triangle OMB\$\$\Rightarrow\$\$MB=QP=QP\cos { (180-\theta) } \$At point \$B\$, \$\theta =120°\$\${ \Rightarrow H }_{ B }=OG+OQ\\ { \Rightarrow H }_{ B }=R+R\cos { 60° } =\frac { 3R }{ 2 } ...(4) \$Using energy conservation equation at point \$T\$ and \$A\$ \$\Rightarrow \frac { 1 }{ 2 } m{ { V }_{ T } }^{ 2 }+mg(2R)=\frac { 1 }{ 2 } m{ { V }_{ A } }^{ 2 }+mg{ H }_{ 1 }...(2)\$putting the value of eqn (1), eqn (2) in eqn (4)\$\Rightarrow \frac { 1 }{ 2 } m{ (\sqrt { gR } ) }^{ 2 }+mg(2R)=\frac { 1 }{ 2 } m{ { V }_{ A } }^{ 2 }+mg\frac { R }{ 2 } \$\$\Rightarrow V_{ A }=2\sqrt { gR } \$Using energy conservation equation at point \$T\$ and \$B\$ \$\Rightarrow \frac { 1 }{ 2 } m{ { V }_{ T } }^{ 2 }+mg(2R)=\frac { 1 }{ 2 } m{ { V }_{ B } }^{ 2 }+mg{ H }_{2}...(2)\$putting the value of eqn (1), eqn (3) in eqn (4)\$\Rightarrow \frac { 1 }{ 2 } m{ (\sqrt { gR } ) }^{ 2 }+mg(2R)=\frac { 1 }{ 2 } m{ { V }_{ B} }^{ 2 }+mg\frac { 3R }{ 2 } \$\$\Rightarrow V_{ B }=\sqrt { 2gR } \$Ratio of \${ V }_{ A }\$ and \${ V }_{ B }\$\$=\sqrt { 2 } :1\$So the correct option is (\$D\$)

Q: A ball of mass 0.6kg attached to a light inextensible string rotates in a vertical circle of radius 0.75m such that it has speed of 5 m/s when the string is horizontal. Tension in string when it is horizontal on other side is:\$(g=10ms^{-2})\$

Given: \$m=0.6 kg; R=0.75 m\$Force equilibrium\$T=\dfrac{mv^{2}}{R}\$ \$=\dfrac{(0.6)(5)^{2}}{(0.75)}\$ \$=20 N\$

Q: A simple pendulum is oscillating with an angular amplitude \$60^{0}\$ . If mass of bob is \$50\ g\$, the tension in the string at mean position is:Consider: \$g=10ms^{-2}\$, length of the string, \$L = 1\ m\$.

Apply work energy theorem between A and B\$mg L(1-cos\theta)=\dfrac{1}{2}mv^{2}\$\$10\times 1\times (1-\dfrac{1}{2})=\dfrac{1}{2}mv^{2}\$\$v^{2}=10\$Then, the tension in, \$T=mg+\dfrac{mv^{2}}{R}\$ \$=\dfrac{50\times 10}{1000}+\dfrac{50}{1000}\times \dfrac{10}{1}\$ \$=1 N\$

Q: Regarding linear momentum of a bodya. It is a measure of quantity of motion contained by the bodyb. Change in momentum is the measure of impulsec. Impulse and acceleration act in opposite direction to the change in momentumd. In the case of uniform circular motion the linear momentum is conserved.

Linear momentum (\$\vec{P}\$) of a body of mass m and velocity \$\vec{v}\$ is expressed as:\$\vec{P}=m \times \vec{v}\$. SO the linear momentum is the measure of the quantity of motion contained by the body. And impulse of a force is the change in momentum occurring due to the force.So, Impulse acts in the same direction to the change in momentum.Acceleration, \$\vec{a}= \dfrac{\vec{F}}{m}=\dfrac{1}{m} \dfrac{d \vec{P}}{dt}\$. So, acceleration also acts in the same direction as impulse.Uniform circular motion corresponds to zero angular acceleration leading to constant angular momentum. Linear momentum, in this case, is not conserved when an external force is present.

Q: A ball of mass 'm' is projected from the ground with a speed 'u' at an angle ' \$\alpha \$' with the horizontal. The magnitude of the change in momentum of the ball over a time interval from begining till it strikes the ground again is

Over the course of the journey, the velocity of the ball in the x-direction is constant. So the momentum in that direction remains the same.Initial momentum in y-direction \$p_{yi}=mu sin\alpha\$Final momentum in y-direction \$p_{yf}=-musin\alpha\$\$\therefore\$ change in momentum \$= -2musin\alpha\$Magnitude of change \$= 2musin\alpha\$

Q: A person carries a hammer on his shoulder and holds the other end of its light handle in his hand. Let \$y\$ be the distance between his hand and the point of support. If the person changes \$y\$, the pressure on his hand will be proportional to:

As it is clear from the Free body diagram, for the hammer to be in equilibrium condition following should be followed :\$ F \times y = w \times (l-y) \$which gives;\$ F \times y = w \times l - w \times y \$\$ y = \dfrac {w \times l}{F+w} \$\$ \dfrac {F+w}{w \times l} = \dfrac {1}{y} \$Let \$ w \times l =d \$ and \$ \dfrac {1}{l} = c \$Hence the new equation becomes;\$ \dfrac {F}{d} + c = \dfrac {1}{y} \$which gives \$ F = \dfrac {d}{y} - c\times d \$Let \$ c \times d =a \$ , which gives\$ F+ a = \dfrac {d}{y} \$As "\$a\$" being a constant does not effect proportionality we have \$F\$ inversely proportional to \$y\$.

Q: An impulse is supplied to a moving object with the force at an angle of \$120^{0}\$ with the velocity vector. The angle between the impulse vector and the change in momentum vector is

Impulse of a force which is defined as force times its duration of action is equal to the change in the momentum of the body. Hence the direction of impulse will be along the direction of change in momentum.

Q: A pilot of mass \$m\$ can bear a maximum apparent weight \$7\$ times of \$mg\$. The aeroplane is moving in a vertical circle. If the velocity of aeroplane is \$210\ m/s\$ while driving up from the lowest point of vertical circle, the minimum radius of vertical circle should be:

Given maximum apparent weight is \$7mg\$Apparent weight at the lowest point is (\$mg + m \dfrac{v^2}{r} \$)So, \$7mg = mg + m \dfrac{v^2}{r} \$or, \$6mg = m \dfrac{v^2}{r} \$or, \$r = \dfrac{v^2}{6g} \$or, \$r = \dfrac{210^2}{6g} \$or, \$ r = 750\ m\$So , minimum radius of circle is \$750\ m \$

Physics
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Easy Qs Med Qs Hard Qs

A body of mass m is rotated at uniform speed along a vertical circle with the help of light string. If $T_{1} a n d T_{2}$ are tensions in the string when the body is crossing highest and lowest point of the vertical circle respectively, then which of the following expressions is correct?

T_{2} - T_{1} = 6 m g

T_{2} - T_{1} = 4 m g

T_{2} - T_{1} = 2 m g

T_{2} - T_{1} = m g

Medium

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Statement I: The max value of force F such that the block shown does not move is $\frac{μ m g}{c o s θ}$ where $μ$ is the coefficient of friction between block end surface.
Statement II: Frictional Force = (Coefficient of friction) x (Normal reaction)

Statement 1 is True, Statement 2 is True; Statement 2 is a correct explanation for Statement 1

Statement 1 is True, Statement 2 is True; Statement 2 is NOT a correct explanation for Statement 1

Statement 1 is True, Statement 2 is False

Statement 1 is False, Statement 2 is True

Medium

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A body is moving in a vertical circle such that the velocities of body at different points are critical. The ratio of velocities of body at angular displacements $6 0^{0}$ and $12 0^{0}$ from the lowest point is:

5 : 2

3 : 2

3 : 1

2 : 1

Medium

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A ball of mass 0.6kg attached to a light inextensible string rotates in a vertical circle of radius 0.75m such that it has speed of 5 m/s when the string is horizontal. Tension in string when it is horizontal on other side is:
$(g = 10 m s^{- 2})$

30 N

26 N

20 N

6 N

Medium

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A simple pendulum is oscillating with an angular amplitude $6 0^{0}$ . If mass of bob is $50 g$ , the tension in the string at mean position is:
Consider: $g = 10 m s^{- 2}$ , length of the string, $L = 1 m$ .

0.5 N

1 N

1.5 N

2 N

Medium

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Regarding linear momentum of a body
a. It is a measure of quantity of motion contained by the body
b. Change in momentum is the measure of impulse
c. Impulse and acceleration act in opposite direction to the change in momentum
d. In the case of uniform circular motion the linear momentum is conserved.

a & b are true

b & c are true

c & d are true

a, b & c are ture

Medium

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A ball of mass 'm' is projected from the ground with a speed 'u' at an angle ' $α$ ' with the horizontal. The magnitude of the change in momentum of the ball over a time interval from begining till it strikes the ground again is

\frac{m u sin α}{2}

2 m u cos α

\frac{m u cos α}{2}

2 m u sin α

Medium

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A person carries a hammer on his shoulder and holds the other end of its light handle in his hand. Let $y$ be the distance between his hand and the point of support. If the person changes $y$ , the pressure on his hand will be proportional to:

y

y^{2}

\frac{1}{y}

\frac{1}{y ^{2}}

Medium

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An impulse is supplied to a moving object with the force at an angle of $12 0^{0}$ with the velocity vector. The angle between the impulse vector and the change in momentum vector is

12 0^{0}

0^{0}

6 0^{0}

24 0^{0}

Medium

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A pilot of mass $m$ can bear a maximum apparent weight $7$ times of $m g$ . The aeroplane is moving in a vertical circle. If the velocity of aeroplane is $210 m / s$ while driving up from the lowest point of vertical circle, the minimum radius of vertical circle should be:

375 m

420 m

750 m

840 m

Medium

View solution>

Laws Of Motion

Physics 7806 Views

A body of mass m is rotated at uniform speed along a vertical circle with the help of light string. If T1​and T2​ are tensions in the string when the body is crossing highest and lowest point of the vertical circle respectively, then which of the following expressions is correct?

Statement I: The max value of force F such that the block shown does not move is cosθμmg​ where μ is the coefficient of friction between block end surface. Statement II: Frictional Force = (Coefficient of friction) x (Normal reaction)

A body is moving in a vertical circle such that the velocities of body at different points are critical. The ratio of velocities of body at angular displacements 600 and 1200 from the lowest point is:

A ball of mass 0.6kg attached to a light inextensible string rotates in a vertical circle of radius 0.75m such that it has speed of 5 m/s when the string is horizontal. Tension in string when it is horizontal on other side is:(g=10ms−2)

A simple pendulum is oscillating with an angular amplitude 600 . If mass of bob is 50 g, the tension in the string at mean position is:Consider: g=10ms−2, length of the string, L=1 m.

A ball of mass 'm' is projected from the ground with a speed 'u' at an angle ' α' with the horizontal. The magnitude of the change in momentum of the ball over a time interval from begining till it strikes the ground again is

A person carries a hammer on his shoulder and holds the other end of its light handle in his hand. Let y be the distance between his hand and the point of support. If the person changes y, the pressure on his hand will be proportional to:

An impulse is supplied to a moving object with the force at an angle of 1200 with the velocity vector. The angle between the impulse vector and the change in momentum vector is

A pilot of mass m can bear a maximum apparent weight 7 times of mg. The aeroplane is moving in a vertical circle. If the velocity of aeroplane is 210 m/s while driving up from the lowest point of vertical circle, the minimum radius of vertical circle should be:

Physics
7806 Views

A body of mass m is rotated at uniform speed along a vertical circle with the help of light string. If $T_{1} a n d T_{2}$ are tensions in the string when the body is crossing highest and lowest point of the vertical circle respectively, then which of the following expressions is correct?

Statement I: The max value of force F such that the block shown does not move is $\frac{μ m g}{c o s θ}$ where $μ$ is the coefficient of friction between block end surface.
Statement II: Frictional Force = (Coefficient of friction) x (Normal reaction)

A body is moving in a vertical circle such that the velocities of body at different points are critical. The ratio of velocities of body at angular displacements $6 0^{0}$ and $12 0^{0}$ from the lowest point is:

A ball of mass 0.6kg attached to a light inextensible string rotates in a vertical circle of radius 0.75m such that it has speed of 5 m/s when the string is horizontal. Tension in string when it is horizontal on other side is:
$(g = 10 m s^{- 2})$

A simple pendulum is oscillating with an angular amplitude $6 0^{0}$ . If mass of bob is $50 g$ , the tension in the string at mean position is:
Consider: $g = 10 m s^{- 2}$ , length of the string, $L = 1 m$ .

A ball of mass 'm' is projected from the ground with a speed 'u' at an angle ' $α$ ' with the horizontal. The magnitude of the change in momentum of the ball over a time interval from begining till it strikes the ground again is

A person carries a hammer on his shoulder and holds the other end of its light handle in his hand. Let $y$ be the distance between his hand and the point of support. If the person changes $y$ , the pressure on his hand will be proportional to:

An impulse is supplied to a moving object with the force at an angle of $12 0^{0}$ with the velocity vector. The angle between the impulse vector and the change in momentum vector is

A pilot of mass $m$ can bear a maximum apparent weight $7$ times of $m g$ . The aeroplane is moving in a vertical circle. If the velocity of aeroplane is $210 m / s$ while driving up from the lowest point of vertical circle, the minimum radius of vertical circle should be: