Dimensional Formula of Torque – Rankers Physics

Dimensions: Practice Problem & Solution

The dimensional formula of torque is: [1989]
\(ML^2T^{-2}\)
\(MLT^{-2}\)
\(ML^{-1}T^{-2}\)
\(ML^{-2}T^{-2}\)

Solution Explained:

To solve this problem, we apply the core principles of Dimensions. Understanding the underlying formula is key to arriving at the correct answer below:

Torque is calculated as Force \(\times\) perpendicular distance. So its dimensions are \([MLT^{-2}] \times [L] = [ML^2T^{-2}]\).

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. Following observations were taken with a vernier calliper while measuring the length of a cylinder: \(3.29 \text{cm}\), \(3.28 \text{cm}\), \(3.29 \text{cm}\), \(3.31 \text{cm}\), \(3.28 \text{cm}\). The most accurate length of the cylinder will be
2. The unit of thermal conductivity is [2019]
3. The angle \(1'\) (minute of arc) in radian is nearly equal to, [2020- Covid]
4. Dashrath measures the length of a wire using a meter scale with a least count of 1 mm and finds it to be L = 75.0 cm. He also measures diameter of thin wire using a screw gauge with a least count of 0.01 mm and finds it to be d = 0.500 cm. He uses these measurements to calculate the volume of wire. The maximum percentage error in volume of wire is nearly
5. Assertion (A): In SHM let \(x\) be the maximum speed, \(y\) the frequency of oscillation and \(z\) the maximum acceleration, then \(\frac{xy}{z}\) is a constant quantity. Reason (R): This is because \(\frac{xy}{z}\) becomes a dimensionless quantity