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 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: Consider the quantities, pressure, power, energy, impulse, gravitational potential, electrical charge, temperature, area. Out of these, the only vector quantities are

We know that impulse J = F. \Delta{t} = \Delta{p} , where F is force, At is time duration and Ap is change in momentum. As \Delta p is a vector quantity, hence impulse is also a vector quantity. Sometimes area can also be treated as vector direction of area vector is perpendicular to its plane.

Q: A water bucket of mass ' m ' is revolved in a vertical circle with the help of a rope of length ' r '. If the velocity of the bucket at the lowest point is \sqrt{7gr} . Then the velocity and tension in the rope at the highest point are:

V_{b} = \sqrt{7gr} Work Energy Theorem applied between top and bottom. mg(2R) = \dfrac{1}{2}m (V_{b} )^{2} - \dfrac{1}{2}m (V_{T} )^{2} 2mgr = \dfrac{7}{2}mgr - \dfrac{1}{2}m (V_{T} )^{2} V_{T} = \sqrt{3gr} At top point: T = \dfrac{mV_{T}^{2}}{r} - mg =m \dfrac{3gr}{r} -mg = 2mg

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

Q: A bog of mass m = 50\ gm is oscillating with the help of a string, the maximum value of angle from the vertical attained by the bog is 60^{o} .The tension in the string at extreme position is

In the given problem at the extreme position, the tension in the string and the weight mg balance each other. But weight mg is resolved into mgcos\theta and hence T = mgcos\theta = \left(50 \times 10^{-3}\right) \times (10) \times \frac{1}{2}=0.25 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: 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

Physics
617 Views

JEE Mains JEE Advanced NEET AIIMS BITSAT Easy Qs Med Qs Hard Qs

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

View solution>

Consider the quantities, pressure, power, energy, impulse, gravitational potential, electrical charge, temperature, area. Out of these, the only vector quantities are

Impulse, pressure and area

Impulse and area

Area and gravitational potential

Impulse and pressure

Medium

View solution>

A water bucket of mass ' $m$ ' is revolved in a vertical circle with the help of a rope of length ' $r$ '. If the velocity of the bucket at the lowest point is $7 g r$ . Then the velocity and tension in the rope at the highest point are:

3 g r, 2 m g

2 g r, m g

g r, m g

Z e r o

Z e r o

Medium

View solution>

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

View solution>

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

View solution>

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>

A bog of mass $m = 50 g m$ is oscillating with the help of a string, the maximum value of angle from the vertical attained by the bog is $6 0^{o}$ .The tension in the string at extreme position is

0.25 N

1.5 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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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

View solution>

Laws Of Motion

Physics 617 Views

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:

Consider the quantities, pressure, power, energy, impulse, gravitational potential, electrical charge, temperature, area. Out of these, the only vector quantities are

A water bucket of mass 'm' is revolved in a vertical circle with the help of a rope of length 'r'. If the velocity of the bucket at the lowest point is 7gr​ . Then the velocity and tension in the rope at the highest point are:

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:

A bog of mass m=50 gm is oscillating with the help of a string, the maximum value of angle from the vertical attained by the bog is 60o .The tension in the string at extreme position is

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 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?

Physics
617 Views

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 water bucket of mass ' $m$ ' is revolved in a vertical circle with the help of a rope of length ' $r$ '. If the velocity of the bucket at the lowest point is $7 g r$ . Then the velocity and tension in the rope at the highest point are:

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:

A bog of mass $m = 50 g m$ is oscillating with the help of a string, the maximum value of angle from the vertical attained by the bog is $6 0^{o}$ .The tension in the string at extreme position is

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 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?