A conical pendulum of length $l$ makes an angle $\theta =45°$ with respect to $Z-$axis and moves in a circle in the $XY$ plane. The radius of the circle is $0.4 \mathrm{m}$ and its center is vertically below $O$. The speed of the pendulum, in its circular path, will be - $\left(\mathrm{T}\mathrm{a}\mathrm{k}\mathrm{e} g=10 \mathrm{m} {\mathrm{s}}^{-2}\right)$

• A

  $0.2 \mathrm{m} {\mathrm{s}}^{-1}$
• B

  $0.4 \mathrm{m} {\mathrm{s}}^{-1}$
• Correct answer
  C

  $2 \mathrm{m} {\mathrm{s}}^{-1}$
• D

  $4 \mathrm{m} {\mathrm{s}}^{-1}$

Solution

$T\sin \theta =\frac{m{v}^2}{r}$

$T\cos \theta =mg$

$\tan \theta =\frac{v^2}{rg}$

$\theta ={45}^o$

$v^2=rg$

$v=\sqrt{rg}=\sqrt{0.4\times 10}$

$v=2 \mathrm{m} {\mathrm{s}}^{-1}$
 

Hence, the correct answer is option: 'C'.