An automobile engine develops 100 kW power when rotating at a speed of 1800 rpm. The torque delivered by the engine is

Question

An automobile engine develops 100 kW power when rotating at a speed of 1800 rpm. The torque delivered by the engine is (A) (102/6π) N m (B) (104/6π) N m (C) (106/6π) N m (D) (108/6π) N m

Tardigrade Admin · Accepted Answer

Here, P = 100 kW = 100×103 W = 105W upsilon = 1800 rpm = (1800/60) rps = 30 rps ∴ ω = 2 π upsilon =2π (30) = 60 π rad s-1 As P= τω ∴ τ = (p/ω) = (105 W/60 π rad s-1) = (104/6π) N m

Q. An automobile engine develops $100 kW$ power when rotating at a speed of $1800 r p m$ . The torque delivered by the engine is

$\frac{1 0 ^{2}}{6 π} N m$

$\frac{1 0 ^{4}}{6 π} N m$

$\frac{1 0 ^{6}}{6 π} N m$

$\frac{1 0 ^{8}}{6 π} N m$

Here, $P = 100 kW$
$= 100 \times 1 0^{3} W = 1 0^{5} W$
$υ = 1800 r p m = \frac{1800}{60} r p s = 30 r p s$
$∴ ω = 2 π υ = 2 π (30) = 60 π r a d s^{- 1}$
As $P = τ ω$
$∴ τ = \frac{p}{ω} = \frac{1 0 ^{5} W}{60 π r a d s ^{- 1}}$
$= \frac{1 0 ^{4}}{6 π} N m$

Q. An automobile engine develops 100kW power when rotating at a speed of 1800rpm. The torque delivered by the engine is

6π102​Nm

6π104​Nm

6π106​Nm

6π108​Nm