Moment of inertia of a fan is 0.06kg-m2 . If its angular speed is increased from 0 to 5 revolution/sec. Then its maximum angular momentum in M.K.S. unit is:

Question

Moment of inertia of a fan is 0.06kg-m2 . If its angular speed is increased from 0 to 5 revolution/sec. Then its maximum angular momentum in M.K.S. unit is: (A)  0.3 π  (B)  0.6 π  (C)  (0.03/ π )  (D)  (0.6/ π )

Tardigrade Admin · Accepted Answer

Angular momentum J=Iω J=I. 2π n=0.6× 2π × 5=0.6π

Q. Moment of inertia of a fan is $0.06 k g - m^{2}$ . If its angular speed is increased from 0 to 5 revolution/sec. Then its maximum angular momentum in M.K.S. unit is:

$0.3 π$

$0.6 π$

$\frac{0.03}{π}$

$\frac{0.6}{π}$

Angular momentum $J = I ω$ $J = I . 2 πn = 0.6 \times 2 π \times 5 = 0.6 π$

Q. Moment of inertia of a fan is 0.06kg−m2 . If its angular speed is increased from 0 to 5 revolution/sec. Then its maximum angular momentum in M.K.S. unit is:

0.3π

0.6π

π0.03​

π0.6​