System of Particles and Rotational Motion - Result Question 45
48. A constant torque of $1000 N-m$ turns a wheel of moment of inertia $200 kg-m^{2}$ about an axis through its centre. Its angular velocity after 3 seconds is
(a) $1 rad / s$
(b) $5 rad / s$
(c) $10 rad / s$
(d) $15 rad / s$
[2001]
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Answer:
Correct Answer: 48. (d)
Solution:
- (d) $\tau=1000 N-m, I=200 kg-m^{2}$
$\tau=I . \alpha$ and $\alpha=(\frac{\omega_f-\omega_0}{t})$
$\Rightarrow \alpha=\frac{1000}{200}=5 rad / sec^{2}$
$\omega_1=\omega_0+\alpha t=0+3 \times 5=15 rad / s$
