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Laws of Motion Test - 51

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Laws of Motion Test - 51
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  • Question 1
    1 / -0
    A road is $$8\ m$$ wide. Its radius of curvature is $$40\ m$$. The outer edge is above the lower edge by  distance of $$1.2\ m$$. The most suited velocity on the road is nearly:
    Solution
    We know,
    $$\frac{v^2}{rg}=\frac{h}{l}$$
    $$v=\sqrt{\frac{40\times1.2\times10}{8}}$$
    $$v=7.74\approx8 m/s$$
  • Question 2
    1 / -0
    A vehicle is at rest on a banked road with angle of banking $$\theta$$. The normal reaction of the vehicle is $${N}_{1}$$. When the vehicle takes a turn on the same road the normal reaction is $${N}_{2}$$. Then $${N}_{1}/{N}_{2}$$ is equal to:
    Solution
    $${N_1} = mg\cos \theta $$                           $$------(1)$$
    When it takes a turn on the same road $$\left( {\dfrac{{m{v^2}}}{r}} \right)$$ is the centripetal force acting on it $$.$$
    Balancing the force 
    $$mg\sin{\theta}  = \dfrac{{m{v^2}}}{r}\cos \theta $$ $$;$$  $$\dfrac{{{v^2}}}{r} = g\tan \theta $$
    $${N_2} = mg\cos \theta  + \dfrac{{m{v^2}}}{r}\sin \theta \, = mg\cos \theta  + mg\tan \theta \,\sin \theta $$                     $$-----------(2)$$
    $$\dfrac{{{N_1}}}{{{N_2}}} = \dfrac{{mg\cos \theta }}{{mg\cos \theta  + mg\,\sin \theta \,\tan \theta \,}}\, = \dfrac{{\cos \theta }}{{\cos \theta \dfrac{{{{\sin }^2}\theta }}{{\cos \theta }} = \dfrac{{\cos \theta }}{{\dfrac{{{{\sin }^2}\theta  + {{\cos }^2}\theta }}{{\cos \theta }}}}}} = {\cos ^2}\theta $$
    Hence$$,$$
                  Option $$(c)$$ is correct

  • Question 3
    1 / -0
    A boy is swinging in a swing. If he stands the time period will
    Solution
    As the child stand up the effective length of pendulum decreases due to the reason that center of gravity rises up.
    According to $$T=2\pi\sqrt{\left(\dfrac{l}{g}\right)}$$
    Thus, the time peroid decreases
  • Question 4
    1 / -0
    A particle is moving in a vertical circle the tension in the string when passing through two position at angle $${30}^{o}$$ and $${60}^{o}$$ from vertical lowest position are $${T}_{1}$$ and$${T}_{2}$$ respectively then-
    Solution
    $$T-mgcos\theta =\dfrac { m{ v }^{ 2 } }{ r } \\ T=mgcos\theta +\dfrac { m{ v }^{ 2 } }{ r } \\ for\quad \theta =30\\ { T }_{ 1 }=mgcos30+\dfrac { m{ v }^{ 2 } }{ r } \\ for\quad \theta =60\\ { T }_{ 2 }=mgcos60+\dfrac { m{ v }^{ 2 } }{ r } \\ { T }_{ 1 }>{ T }_{ 2 }$$
  • Question 5
    1 / -0
    A weightless thread can support tension upto $$30N$$. A particle of mass $$0.5kg$$ is tied to it and is revolved in a circle of radius $$2m$$ in a vertical plane. If $$g=10m/{s}^{2}$$, then the maximum angular velocity of the stone will be
    Solution
    $$T-mg=m\omega ^{ 2 }r\\ { T }_{ max }=mg+m\omega ^{ 2 }r\\ { \omega  }^{ 2 }=\left( \frac { T-mg }{ mr }  \right) \\ =\left( \frac { 30-5 }{ 0.5(2) }  \right) \\ =5rad/sec$$

  • Question 6
    1 / -0
    A hammer of mass 1 kg moving with a speed of 6 m/s strikers a wall and comes to rest in 0.1 s, calculate Retarding force that stops the hammer.
    Solution
    $$m=1kg$$
    $$u = 6m/s            ,v=0       ,t=0.15$$
    impulse = $$f\times t = m(v-u) = 1(0-6) = -6Ns$$
    retarding force stops the hammer 
    $$F = \dfrac{I}{T} = \dfrac{6}{0.1} = 60N$$
  • Question 7
    1 / -0
    A boy is swinging on a swing such that his lowest and highest positions are at heights of $$2m$$ and $$4.5m$$ respectively. His velocity at the lowest position is:
    Solution
    The height of the boy falls with the swing,
    $$h=4.5-2=2.5m$$
    Therefore,
    The velocity at the lowest position $$=\sqrt{2gh}$$
    $$v=\sqrt{2\times9.8\times2.5}=\sqrt{49}$$
    $$v=7 m/s$$
  • Question 8
    1 / -0
    A stone of mass $$0.2kg$$ is tied to one end of a thread of length $$0.1m$$ whirled in a vertical circle. When the stone is at the lowest point of circle, tension in thread is $$52N$$, then velocity of the stone will be:
    Solution
    $$T-mg=\dfrac { { mv }^{ 2 } }{ r } \\ T=\dfrac { { mv }^{ 2 } }{ r } +mg\\ 52=0.2(10)\quad +\dfrac { 0.2({ v }^{ 2 }) }{ 0.1 } \\ 2{ v }^{ 2 }=52-2=50\\ \qquad v=5m/sec$$
  • Question 9
    1 / -0
    A stone of mass $$1kg$$ is tied to the end of a string of $$1m$$ length. It is whirled in a vertical circle. If the velocity of the stone at the top be $$4m/s$$. What is the tension in the string (at that instant)?
    Solution
    $$T+mg=\dfrac { { mv }^{ 2 } }{ r } \\ T=\dfrac { { mv }^{ 2 } }{ r } -mg\\ \dfrac { 1\times { 4 }^{ 2 } }{ 1 } -1(10)\\ =16-10\\ =6N$$

  • Question 10
    1 / -0
    A stone at one end of a light string is whirled around a vertical circle. If the difference between the maximum and minimum tension experience by the string were $$2\ Kg$$, the mass of the stone must be:
    Solution
    Given
    $$T_{max}-T_{min}=2kg wt...............(i)$$
    WE have the relation 
    $$T_{max}-T_{min}=6mg................(ii)$$
    Now putting the value from (i) to (ii), we get
    $$2g=6mg$$
    $$m=\frac{1}{3}kg$$

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