ટોર્ક \(\tau = 6.9 \times 10^2 Nm,\)

પ્રારંભિક કોણીય ઝડપ = \(\omega_0­ = 4.6 rad s^{-1}\) ,

અંતિમ કોણીય ઝડપ = \(\omega =0rad/s\)

હવે \(\omega \,\, = \,\,{\omega _0}\, + \,\alpha t\,\,\)

\(\Rightarrow \,\,\,\alpha \,\, = \,\,\,\frac{{\omega \, - \,{\omega _0}}}{t}\,\,\,\)'

\(= \,\,\frac{{0 - 4.6}}{t}\,\)

\( = \,\,\frac{{ - 4.6}}{t}\,rad\,{s^{ - 2}}\,\)

ઋણ નિશાની પ્રતિપ્રવેગ તોર્ક દર્શાવેછે 

\(\,\tau \,\, = \,\,I\alpha \,\, \Rightarrow \,\,\,\,6.9\, \times \,{10^2}\,\, = \,\,3\,\, \times \,{10^2}\,\, \times \,\frac{{4.6}}{t}\,\,\,t\,\)

\( = \,\,\frac{{3\, \times \,{{10}^2}\, \times \,4.6}}{{6.9\, \times \,{{10}^2}}}\,\, = \,\,2\,s\)