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If speed (v), acceleration (a), and force (f) are considered as fundamental units, the dimension of Young's modulus will be:

a) [M¹ L⁰ T⁻2]

b) [M¹ L¹ T⁻2]

c) [M¹ L¹ T⁻1]

d) [M¹ L¹ T⁻2]

Answer :

Final answer:

The dimension of Young's modulus, considering speed, acceleration, and force as fundamental units, is [M¹ L¹ T¹²], matching option d).

Explanation:

The dimension of Young's modulus in terms of speed (v), acceleration (a), and force (f) as fundamental units is option d) [M¹ L¹ T¹²]. Young's modulus is defined as stress over strain. Stress has the same dimensions as pressure, which is force per area; hence, it is equivalent to force divided by length squared ([F][L⁻²]). Strain is dimensionless as it's a ratio of lengths.

Therefore, Young's modulus in fundamental units would use the dimension of force ([F]), which is essentially mass times acceleration ([M][a]) and thus it equals mass ([M¹]) times length ([L¹]) times time to the negative second power ([T²]). Using the provided information, where [a] is given as LT⁻² and the force ([F]) is the product of mass and acceleration (ma), the dimension of Young's modulus becomes [M][a][L⁻²], simplifying to [M¹ L¹ T¹²].

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