Question

An automobile engine develops $100H.P.$which is rotating at a speed of$1800\dfrac{rad}{\min }$. The torque it delivers is

\left( a \right)\;\;3.33{\rm{ }}W - s\\\ \left( b \right)\;\;200{\rm{ }}W - s\\\ \left( c \right)\;\;\;248.7{\rm{ }}W - s\\\ \left( d \right)\;\;2487{\rm{ }}W - s \end{array}$$

Answer 1

Hint : In this question we have been given Power$P$and Angular frequency${\overrightarrow \omega }$. We know$1H.P. \approx 746 Watt$. For easier calculation and the requirements of the question, converting the angular frequency from $\dfrac{Rad}{\min }$to $\dfrac{Rad}{\sec }$will be suggested. Apply the relation for power and we get the results.

Complete step-by-step solution:
Power is the amount of energy transferred or converted per unit time. SI units of power are taken to be$Watt$, but there are other units of power that are also used like Horse Power $HP$and Pascal.
Assuming the power given to be$P$, and
$P = 100H.P$
$1H.P. \approx 746Watt$
P$$$$inWatt = 746 \times 100$$$$Watt
$\approx$ 74600 $Watt$
Net torque is defined as the rate of change of angular momentum of an object. It is a measure of how much force is acting on an object causing it to rotate.
To understand this question better, let us derive the required equation to get Torque$\overrightarrow \tau$:
Keeping $\overrightarrow \tau$ constant we know,
$\dfrac{{dw}}{{dt}}$= $\overrightarrow \tau$. $\dfrac{{d\theta }}{{dt}}$ …… (1)
$\dfrac{{dw}}{{dt}}$Can also be written as $P$ …… (2)
$\dfrac{{d\theta }}{{dt}}$Can also be written as ${\overrightarrow \omega }$ …… (3)
Putting values of (2) and (3) in (1) we get
$P$ = $\overrightarrow \tau$.${\overrightarrow \omega }$
This can also be written as,
$\overrightarrow \tau$= $\dfrac{P}{{\overrightarrow \omega }}$ ……(4)
As calculated earlier,
P$$$$\approx 74600 $Watt$ ……(5)
${\overrightarrow \omega }$= 1800 $\dfrac{Rad}{\min }$
To convert $\dfrac{Rad}{\min }$to $\dfrac{Rad}{\sec }$ we divide the given amount by 60 because$1\min =60\sec$.
${\overrightarrow \omega }$=\dfrac{{1800}}{{60}}$$$$\dfrac{Rad}{\sec }
= 30 $\dfrac{Rad}{\sec }$
${\overrightarrow \omega }$ = 30 $\dfrac{Rad}{\sec }$ ……(6)
Putting the values of (5) and (6) in equation (4) we get
$\overrightarrow \tau$= \dfrac{{74600}}{{30}}$$$$W - s
= $\dfrac{{7460}}{3}$ $W - s$
= $2,486.666666666667{\rm{ }}W - s$
$\approx$ $2487{\rm{ }}W - s$
We can also find the torque using a different method.
We know that,
$P$= $\overrightarrow F \times \overrightarrow v$ …… (7)
And $\overrightarrow F = \dfrac{\tau }{r}$ ……(8)
$P$ = $\dfrac{\tau }{r} \times v$ ……(9)
= $\overrightarrow \tau \times \dfrac{{\overrightarrow v }}{r}$ ……(10)
We also know that,
$\dfrac{{\overrightarrow v }}{r} = \overrightarrow \omega$ ……(11)
Combining (9),(10), and (11) we get
$P$ = $\overrightarrow \tau$.${\overrightarrow \omega }$
This can also be written as,
$\overrightarrow \tau$= $\dfrac{P}{{\overrightarrow \omega }}$
Again, putting the values we had calculated earlier in this equation, we can get our answer.
To conclude this question,
**The torque delivered by the automobile will be 2487$W - s$ which means the Correct
Option is (D) **

Note: The earth's rotation is a prominent example of rotational kinetic energy. Torque is the measure of the force that can cause an object to rotate about an axis. Force is what causes an object to accelerate in linear kinematics. Similarly, torque is what causes an angular acceleration. Hence, torque can be defined as the rotational equivalent of linear force.