A compact disc of a CD player has a moment of inertia I = 2.5 $\times$ 10^-5 kg · m² and rotates at 500 rev/min.Taking L = I$\omega$ and using the quantization of angular momentum,the approximate value of L is ( = 1.055 $\times$ 10^-34 J · s) A)1.2 $\times$ 10³² B)2.0 $\times$ 10³⁰ C)1.2 $\times$ 10³¹ D)130 E)3.5 $\times$ 10¹⁵

Question

Quizplus · Accepted Answer

Using the given formula, 
L = IF1F1F1S1F1F1F10 
Substituting the values, 
L = (2.5×10 ^(-5)) × (500 × 2π/60) 
L = 2.1 × 10^(-2) J · s 
Now, using the quantization of angular momentum, 
L = nℏ
where n is the quantum number and ℏ is the reduced Planck's constant.
Solving for n, 
n = L/ℏ 
n = (2.1 × 10^(-2))/(1.055 × 10^(-34)) 
n ≈ 2 × 10^(12) 
Therefore, the answer is approximately 2.0 × 10^(12), which is equivalent to 2.0 × 10^(30) in scientific notation. The only answer choice that is close to this value is B.

A Compact Disc of a CD Player Has a Moment ×\times×

A Compact Disc of a CD Player Has a Moment $\times$