
An automobile engine develops 100kW when rotating at a speed of \[1800rev/min\]. The torque it delivers is
\[\begin{align}
& \text{A}\text{. }3.33Nm \\
& \text{B}\text{. 200}Nm \\
& \text{C}\text{. 530}\text{.5}Nm \\
& \text{D}\text{. 2487}Nm \\
\end{align}\]
Answer
570.9k+ views
Hint: Torque is produced due to the force applied on the object which results in the rotational motion. Here we have given the power produced by an engine and power can be defined in terms of torque. By using the formula of power we can find the torque which is produced by the automobile engine.
Formula used:
\[\tau =\dfrac{P}{\omega }\]
Complete answer:
Power is the amount of work done per unit time and in terms of torque it can be given as a product of torque and angular velocity. The mathematical expression for it, is given as
\[P=\tau \omega \]
Here we have given the power of the engine and the angular velocity and we have to calculate the torque. Above equation can be rewritten as
\[\tau =\dfrac{P}{\omega }\]
Before substituting the values we have to change the units into SI units. The SI unit of angular velocity is rad/s. We can change it as shown below
\[\begin{align}
& \omega =1800rev/min=\dfrac{1800\times 2\pi rad}{60s} \\
& \omega =\dfrac{3600\times 3.14}{60}rad/s \\
& \omega =188.4rad/s \\
\end{align}\]
And the SI unit of power is watt so we can write
\[\begin{align}
& P=100kW \\
& P=100\times {{10}^{3}}W \\
& P={{10}^{5}}W \\
\end{align}\]
Now substituting these values in the formula for torque we get,
\[\begin{align}
& \tau =\dfrac{{{10}^{5}}}{188.4}Nm \\
& \tau =530.5Nm \\
\end{align}\]
So, the correct answer is “Option C”.
Note:
The answer may vary considering the decimal points. Also it is essential to change the unit to avoid error. 1 revolution is equal to 2π in radians and 360 in degrees.
The formula for power in terms of torque can be verified as the power is given by the product of force and the velocity, condition is the angle of rotation should be perpendicular to the applied force.
Formula used:
\[\tau =\dfrac{P}{\omega }\]
Complete answer:
Power is the amount of work done per unit time and in terms of torque it can be given as a product of torque and angular velocity. The mathematical expression for it, is given as
\[P=\tau \omega \]
Here we have given the power of the engine and the angular velocity and we have to calculate the torque. Above equation can be rewritten as
\[\tau =\dfrac{P}{\omega }\]
Before substituting the values we have to change the units into SI units. The SI unit of angular velocity is rad/s. We can change it as shown below
\[\begin{align}
& \omega =1800rev/min=\dfrac{1800\times 2\pi rad}{60s} \\
& \omega =\dfrac{3600\times 3.14}{60}rad/s \\
& \omega =188.4rad/s \\
\end{align}\]
And the SI unit of power is watt so we can write
\[\begin{align}
& P=100kW \\
& P=100\times {{10}^{3}}W \\
& P={{10}^{5}}W \\
\end{align}\]
Now substituting these values in the formula for torque we get,
\[\begin{align}
& \tau =\dfrac{{{10}^{5}}}{188.4}Nm \\
& \tau =530.5Nm \\
\end{align}\]
So, the correct answer is “Option C”.
Note:
The answer may vary considering the decimal points. Also it is essential to change the unit to avoid error. 1 revolution is equal to 2π in radians and 360 in degrees.
The formula for power in terms of torque can be verified as the power is given by the product of force and the velocity, condition is the angle of rotation should be perpendicular to the applied force.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

