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Laws of Motion Test - 29

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Laws of Motion Test - 29
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  • Question 1
    1 / -0
    A simple pendulum is oscillating with an angular amplitude $$60^o$$. If $$m$$ is mass of bob and $$T_1$$, $$T_2$$ are tensions in the string, when the bob is at extreme position, mean position respectively then is:

    A) $$T_1 = \dfrac{mg}{2}$$
    B) $$T_2 = 2\ mg$$
    C) $$T_1 = 0$$
    D) $$T_2 = 3\ mg$$
    Solution

    Angular Magnitude is given  $$\theta=60^o$$

    As we see in figure, At extreme position: 
    $$T_A=mg\cos\theta$$
    $$T_A=mg\cos 60^o$$

    $$T_A=T_1=\dfrac{mg}{2}$$

    As we see in figure, At middle position: 

    $$T_B=mg+F_c$$   where $$F_c$$ is centrifugal force

    $$T_B=mg+\dfrac{mv^2}{R}$$    where $$v$$ is velocity of pendulum.

    Using energy conservation for extreme position and middle position

    $$v=\sqrt{gR}$$

    putting values :

    $$T_B=T_2=mg+\dfrac{mgR}{R}=2mg$$

    Hence option A is correct.

  • Question 2
    1 / -0
    The length of a simple pendulum is  '$$L$$'. Its bob from rest position is projected horizontally with a velocity $$\sqrt{\dfrac{7gL}{2}}$$. The maximum angular displacement of bob, such that the string does not slack, is:
    Solution
    At point where tension become just zero, the centripetal force  must be balanced by gravitational force
    $$ mg cos\theta = \dfrac{mv_{1}^{2}}{L}$$  ------ (1)
    BY conservation of energy 
    $$\dfrac{1}{2}m   \dfrac{7gL}{2} = \dfrac{mv_{1}^{2}}{2}+mgL(1+cos\theta)$$
    $$\sqrt{2gL} \sqrt{\dfrac{3}{4}-cos\theta} = v_{1}$$
    Put $$v_{1}$$ in (1)
    $$mg$$$$ cos\theta = $$$$\dfrac{m\times2gL}{L}\left ( \dfrac{3}{4}-cos\theta \right ).$$
    $$cos\theta =\dfrac{3}{2}- 2cos\theta$$
    $$cos\theta=\dfrac{1}{2}$$
    $$\theta=60^{0}$$
    $$\therefore $$Total maximum angular displacement
    $$= 180^0-60^0$$
    $$=120^{0}$$
  • Question 3
    1 / -0
    The length of a ballistic pendulum is $$1 m$$ and mass of its block is $$1.9 kg$$. A bullet of mass $$0.1 kg$$ strikes the block of ballistic pendulum in horizontal direction with a velocity $$100ms^{–1}$$ and got embedded in the block. After collision the combined mass (block & bullet) swings away from lowest point. The tension in the string when it makes an angle $$60°$$ with vertical is  $$(g=10ms^{-2})$$:
    Solution
    Momentum Conservation
    $$0.1\times 100= \left ( 1.9+0.1 \right )V_{o}$$
    $$V_o = 5m s^{-1}.$$
    $$ H= L(1-cos60^{\circ}). =\dfrac{1}{2}= \dfrac{1}{2}m.$$

    Energy Conservation
    $$\dfrac{1}{2}(2)(5)^{2} = \dfrac{1}{2} \times 2 \times V^{2}+2\times10 \times \dfrac{1}{2}$$
    $$25-10= V^{2}.$$
    $$ V=\sqrt{15}mvs $$

    By Force Balance 
    $$ T-mg\ cos\ 60^{\circ}= m\dfrac{V^{2}}{L}.$$
    $$ T= 2\times10\times\dfrac{1}{2}+2\times\dfrac{15}{1}$$
    $$= 40N.$$
  • Question 4
    1 / -0
    A simple pendulum is oscillating with an angular amplitude $$60^{0}$$. If mass of bob is $$50\ gram$$, the tension in the string at mean position is:
    $$(g=10\ m/s^{2})$$
    Solution
    $$H = R(1 - cos\ 60^0)$$

    $$H = \dfrac {R}{2}$$

    Apply Work Energy Theorem,

    $$mg\dfrac {R}{2} = \dfrac {1}{2}mV^2$$

    $$V = \sqrt {Rg}$$

    At mean position,
    $$T - mg = \dfrac {mV^2}{R}$$

    $$T = 2\ mg$$

    $$= 2 \times 0.05 \times 10$$
    $$= 1\ N$$
  • Question 5
    1 / -0
    The bob of a simple pendulum of mass $$'m'$$ is performing oscillations such that the tension in the string is equal to twice the weight of the bob while it is crossing the mean position. The tension in the string when the bob reaches extreme position is :
    Solution
    At bottom $$T = 2mg$$
    $$T - mg = \dfrac{mV^2}{R}$$
    $$2mg - mg = \dfrac{mV^2}{R}$$
    $$\sqrt {Rg} = V$$
    Applying Work Energy Theorem between pt. (1) and (2)
    $$mg R(1-cos\theta) = \dfrac{1}{2}mV^2$$
    $$mg R(1-cos\theta) = \dfrac{mRg}{2}$$

    here, $$\theta$$ is the angle made by the particle with the axis at its extreme point, 

    $$1-cos\theta = \dfrac{1}{2}$$
    $$\theta = 60^0$$
    at, $$\theta = 60^0$$
    $$ T = mgcos\theta$$
    $$\therefore T = mg\dfrac{1}{2}$$
  • Question 6
    1 / -0
    Mass of the bob of a simple pendulum of length $$L$$ is $$m$$. If the bob is projected horizontally from its mean position with velocity $$\sqrt{4gL}$$ , then the tension in the string becomes zero after a vertical displacement of :
    Solution
    Let the angle $$\theta$$ be when tension becomes zero.
    Under this condition $$mg sin\theta = \dfrac{mv^{2}}{L}$$ ....... eq(1)

    Also, by conservation of energy:
    $$\dfrac { m (\sqrt {4gL})^{2}}{2} = mg(L+L sin \theta) + \dfrac{mv^{2}}{2}$$

    => $$\dfrac{mv^{2}}{L} = 2gmL - 2gmLsin \theta$$ ------- eq(2)

    From eq (1) and eq(2):
    $$mg sin\theta = 2mg - 2mg sin\theta$$
    $$sin \theta = 2/3$$

    So, the tension in the string becomes zero after a vertical displacement of:  $$L+Lsin\theta = 5L/3$$

  • Question 7
    1 / -0
    Two bodies A, B of masses $$m_1, m_2$$ are knotted to a mass-less string at different points rotated along concentric circles in horizontal plane. The distances of A, B from common centre are 50cm, 1m. If the tensions in the string between centre to A and A to B are in the ratio 5:4, then the ratio of $$m_{1}$$ to $$m_{2}$$ is:
    Solution
    The tension will be more between center to A as it has to bear the centripetal force of 2 bodies.

    applying force balance on the masses, 
    $$T_{1}-T_2=m_1 \omega^{2}r_1$$.........(1)
    $$T_2=m_2 \omega^{2}r_2$$.........(1)

    $$\Rightarrow \dfrac { { T }_{ 2 } }{ { T }_{ 1 } } =\dfrac { 4 }{ 5 } \quad $$ 

    Now: $${ r }_{ 1 } = 0.5$$, $${ r }_{ 2 } =1 $$ 

    => $$\dfrac { 5 }{ 4 } = \dfrac { { m }_{ 1 }(0.5) }{ { m }_{ 2 }+1 } $$ 

    => $$\dfrac{m_1}{m_2} = \dfrac{2\times 1}{4} = \dfrac{1}{2}$$
  • Question 8
    1 / -0
    A simple pendulum consists of a light string from which a spherical bob of mass, $$M$$, is suspended. The distance between the point of suspension and the center of bob is $$L$$. At the lowest position, the bob is given tangential velocity of $$\sqrt{5gL}$$. The K.E of the bob when the string becomes horizontal is:
    Solution
    From Work Energy Theorem,

    $$(WD)_G = K_f - K_i$$ ,  where $$(WD)_G$$ is work done by gravity 

    $$-MgL = \dfrac{-1}{2}M\times 5gL + K_i$$

    $$K_i = \dfrac {3}{2}MgL$$
  • Question 9
    1 / -0
    A curved road is $$7.5\ m$$ wide and its outer edge is raised by $$1.5\ m$$ over the inner edge. The radius of curvature is $$50\ m$$. For what speed of the car is this road suited?$$(g=10m/s^{2})$$:
    Solution
    $$N  sin  \theta =\dfrac{mv^2}{R}$$

    $$N  cos  \theta =mg$$

    $$tan  \theta =\dfrac{v^2}{rg}$$

    $$tan  \theta =\dfrac{1.5}{7.5}$$

    $$\dfrac{1}{5}=\dfrac{v^2}{50\times 10}$$

    $$U=10  \ m/s$$
  • Question 10
    1 / -0
    The bob of a simple pendulum is given a velocity in horizontal direction when the bob is at lowest position ,so that the bob describes vertical circle of radius equal to length of pendulum and tension in the string is $$10 \ N$$ when the bob is at an angle $$60^0$$ from lowest position of vertical circle. The tension in the string when the bob reaches highest position is (The mass of bob is $$ 100$$ gram. $$g = 10\ ms^{–2}$$):
    Solution
    For a body of mass $$'m'$$  moving in a vertical circle, Tension in the string at highest point and Tension at any angle $$\theta$$ with vertical is related as:-
    $$T-{ T }_{ H }=3mg(1+cos\theta )$$
    Substituting the values 
    $$10-{ T }_{ H }=3(0.1)(10)(1+cos60^0)=4.5\\ \Rightarrow { T }_{ H }=5.5\ N$$
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