A solid cylinder is released from rest from the top of an inclined plane of inclination $30^{\circ}$ and length $60 \mathrm{~cm}$. If the cylinder rolls without slipping, then the speed when it reaches the bottom is

• A

  $1.5 \mathrm{~m} / \mathrm{s}$
• Correct answer
  B

  $2.0 \mathrm{~m} / \mathrm{s}$
• C

  $3.0 \mathrm{~m} / \mathrm{s}$
• D

  $6.0 \mathrm{~m} / \mathrm{s}$

Solution

According to the question, given situation is shown below
In $\triangle A B C$, $\sin \theta=\frac{B C}{A C} \Rightarrow B C=A C \sin \theta=l \sin \theta$ From law of conservation of mechanical energy, $m g(l \sin \theta)=\frac{1}{2} I \omega^2+\frac{1}{2} m v^2$ $m g(l \sin \theta)=\frac{1}{2}\left(\frac{1}{2} m r^2\right)\left(\frac{v}{r}\right)^2+\frac{1}{2} m v^2$ $\left[\because I=\frac{1}{2} m r^2\right.$ and $\left.\omega=\frac{v}{r}\right]$ $m g l \sin \theta=\left(\frac{1}{4}+\frac{l}{2}\right) m v^2$ $g l \sin \theta=\frac{3}{4} v^2 \Rightarrow v^2=\frac{4 g l \sin \theta}{3}$ $\Rightarrow \quad v=\sqrt{\frac{4 g l \sin \theta}{3}}=\sqrt{\frac{4 \times 10 \times 0.6 \times \sin 30^{\circ}}{3}}$ $[\because l=60 \mathrm{~cm}=0.6 \mathrm{~m}]$ $=\sqrt{\frac{12}{3}}=\sqrt{4}=2 \mathrm{~m} / \mathrm{s}$

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