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Laws of Motion Test - 72

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Laws of Motion Test - 72
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  • Question 1
    1 / -0
    A stone of mass 1 kg is tied to the end of a string of $$1m$$ length. It is whired in a vertical circel. If the velocity of the stone at the top be $$4m/s$$. What is the tension in the string (at that instant)?
    Solution

  • Question 2
    1 / -0
    A stone of mass $$1kg$$ is tied to one end of a string of length $$0.5\ m$$. It is whirled in a veretical circular. If the maximum tension in the string is $$58.8N$$, the velocity at the top is
    Solution

  • Question 3
    1 / -0
    A stone tied to the end of a string of 1 m long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in 44 s, what is the magnitude & direction of acceleration of the stone: 
    Solution

  • Question 4
    1 / -0
    Mass $$2m$$ is kept on a smooth circular track of mass $$m$$ , which is kept on a smooth horizontal surface. The circular track is given a horizontal velocity $$\sqrt{2gR}$$ towards left and released. The maximum height reached by $$2m$$ will be ?

    Solution

  • Question 5
    1 / -0
    A stone is tied to a rope is rotated in a vertical circle with unifrom speed. if the difference between maximum and minimum tension in the rope is $$20N$$, mass of the stone in $$Kg$$ is $$(g=10m/s^{2})$$
    Solution

  • Question 6
    1 / -0
    A ball of mass m strikes the fixed inclined plane inclined plane after falling through a height h. If it rebounds elastically, the impulse on the ball is:

    Solution
    Refer image 1
    Velocity by which it strikes the ground (v) is:
    $$v^2-u^2=2gh$$
    $$\Rightarrow v^2-0=2gh$$
    $$\therefore v=\sqrt{2gh}$$
    Refer image, 2
    [As the collision is elastic ; the velocity of approval and  velocity of separation along line of impact will be same]
    As velocity of approach along Line of impact is $$\sqrt{2gh}\cos\theta$$ so velocity of separation will be $$-\sqrt{2gh}\cos\theta$$ [as it rebounds in the opposite direction]

    $$\therefore P_i=m\times v_i=m\times \sqrt{2gh}\cos\theta$$

    $$\therefore P_f=m\times v_f=-m\times \sqrt{2gh}\cos\theta$$

    Now according to impulse - momentum theorem:
    $$J=\Delta P$$
    $$\Rightarrow J=m\times \sqrt{2gh}\cos\theta-(-m\times \sqrt{2gh}\cos\theta)$$
    $$\Rightarrow J=2m\cos\theta\sqrt{2gh}$$

    So the impulse on the ball is $$2m\cos\theta\sqrt{2gh}$$

  • Question 7
    1 / -0
    Two balls of the same mass are dropped from the same height h, on to the floor. The first ball bounces to a height $$h/4$$, after the collision & the second ball to a height $$h/16$$. The impulse applied by the first & second ball on the floor are $$I_1$$ and $$I_2$$ respectively. Then
    Solution

  • Question 8
    1 / -0
    A system of two blocks $$A$$ and $$B$$ are connected by an inextensible massless strings as shown. The pulley is massless and frictionless. Initially the system is at rest when, a bullet of mass $$m$$ moving with a velocity $$u$$ as shown hits the block $$B$$ and gets embedded into it.  The impulse imparted by tension force to the block of mass $$3m$$ is:

    Solution

  • Question 9
    1 / -0
    Select the incorrect option:-
    ($$\tau = torque, \,v = velocity, \,\omega = angular \,velocity, \, a_c = centripetal \, acceleration$$)
    Solution

  • Question 10
    1 / -0
    A road is banked at an angle of $$30^{o}$$ to the horizontal for negotiating a circular road of radius $$10 m$$. At what velocity a car will experience no friction while negotiating the curve?
    Solution

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