The bob of a simple pendulum is of mass $10 \, g$. It is suspended with a thread of $1 \, m$. If we hold the bob so as to stretch the string horizontally and release it,what will be the tension at the lowest position? (Take $g = 10 \, m/s^2$)

  • A
    $1.0$
  • B
    $0.3$
  • C
    $0.1$
  • D
    $0$

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$A$ small sphere is moving at a constant speed in a vertical circle. Below is a list of quantities that could be used to describe some aspect of the motion of the sphere.
$I$ - Kinetic energy
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In vertical circular motion of a bob,match the entries of List-$I$ with entries of List-$II$. Here,$v_0$ is the velocity of the bob at the lowest point.
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$(Q) v_0 = \sqrt{g \ell}$ $(2)$ String will slack for a finite time
$(R) v_0 = 2 \sqrt{g \ell}$ $(3)$ Bob will oscillate
$(S) v_0 = 3 \sqrt{g \ell}$ $(4)$ Tension at highest point $= 4 mg$

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$A$ mass $m$ is released from point $A$ as shown in the figure. The tension in the string at point $B$ will be:

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In order to just complete the vertical circular motion,the ratio of kinetic energy of a particle at the highest point to that at the lowest point is

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