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A force of 10 N is applied to a lever arm of length 5 m. If the torque produced by the force is 50 Nm, what is the value of the dimensionless constant in the torque formula T = k F r^a ?

Option: 1

2


Option: 2

5


Option: 3

10


Option: 4

25


Answers (1)

best_answer

Using the formula T = k F r^a , we can solve for the dimensionless constant k as:

k = \frac{T}{F r^a}

Substituting the given values of F, r, and T, we get:

k = \frac{50\text{ Nm}}{10\text{ N} \times (5\text{ m})^a}

Simplifying, we get :

k = \frac{10\text{ m}}{(5\text{ m})^a}

Using dimensional analysis, we can determine the value of the exponent 'a' to be 1. Therefore, we have:

k = \frac{10\text{ m}}{5\text{ m}} = 2\text{ m/m} = 2

So, the value of the dimensionless constant k is 2.

Posted by

Suraj Bhandari

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