Two mating spur gears have $40$ and $120$ teeth respectively. The pinion rotates at $1200 r p m$ and transmits a torque of $20 N . m .$ . The torque transmitted by gear is

60 N m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

6.6 N m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

20 N m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

40 N m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $60 N m$
Power transmitted,
$P = \frac{2 π N_{G} T_{G}}{60} = \frac{2 π N_{P} T_{P}}{60}$

$\Rightarrow N_{G} T_{G} = N_{P} T_{P}$ (where $T$ is torque)
and, $\frac{N_{P}}{N_{G}} = \frac{Z_{G}}{Z_{P}}$
(where $Z$ is number of teeths)
$\Rightarrow \frac{1200}{N_{G}} = \frac{120}{40}$
$N_{G} = 400 r . p . m$
Hence, $400 \times T_{G} = 1200 \times 20$
$T_{G} = 60 N . m$

Suggest Corrections

Similar questions

A spur pinion of pitch diameter 50 mm rotates at 200 rad/s and transmits 3 kW power. The pressure angle of the tooth of the pinion is 20^{\circ} . Assuming that only one pair of the teeth is in contact,the total force (in Newtom) exerted by a tooth of the pinion on the tooth on a mating gear is

20^{o}

20^{\circ}

Join BYJU'S Learning Program