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Motion in a Plane — Physics STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 SciencePhysicsMotion in a Plane1 Mark
MCQ
Consider a simple pendulum of length 1 m. Its bob performs a circular motion in horizontal plane with its string making an angle \[60^{\circ}\] with the vertical. The period of rotation of the bob is (Take $g =10 m / s ^2$ )
  • A
    2 s
  • B
    1.4 s
  • C
    1.98 s
  • D
    2.4 s
✓

Answer

(B)
Using,
$T \sin \theta=m \omega^2 r=m \omega^2 l \sin \theta$ ...(i)
$T \cos \theta= mg$ ... (ii)
From (i) and (ii), $\omega^2=\frac{g}{l \cos \theta}$
$\therefore \quad \omega=\sqrt{\frac{ g }{l \cos \theta}}$
$\therefore \quad$ Time period, $T =\frac{2 \pi}{\omega}$
$=2 \pi \sqrt{\frac{l \cos \theta}{g}}$
$=2 \times 3.14 \times \sqrt{\frac{1 \times \cos 60^{\circ}}{10}}$
= 1.4 s
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