A) \[20.3\text{ }Nm\]
B) \[37.8\text{ }Nm\]
C) \[31.48\text{ }Nm\]
D) \[2.0045\text{ }Nm\]
Correct Answer: D
Solution :
[d] Thermal power \[P=\frac{dQ}{dt}=mc\left( \frac{dT}{dt} \right)\] \[=(180\times {{10}^{-3}})(0.10)(4200)(0.5)=37.8\,Watt\] Mechanical power \[\rho =r\omega \] \[\tau =\frac{37.8}{\left( 180\times \frac{2\pi }{60} \right)}=2.0045\,Nm\]
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