Question

the relation between force and density is f-x/√d the dimension of x is

Answer 1

f = a√x + bt2, where [f] = [M L T −2], [x] = [L] and [t] = [T]. LHS is force. So both the terms on the RHS have the dimensions of force. [f] = [a√x] = [bt2] [f] = [a√x] [M L T −2] = [a L½] ⇒ [a] = [M L½ T −2] [f] = [bt2] [M L T −2] = [b T2] ⇒ [b] = [M L T−4] [a]/[b] = [M L½ T −2]/[M L T−4] = [M0 L−½ T2] Therefore [a/b] = [M0 L−½ T2]

Answer 2

$f = \frac{x}{ \sqrt{d} } \\ x = f \sqrt{d} \\ x = {m}^{1} {l}^{1} {t}^{ - 2} \times {( {m}^{1} {l}^{ - 3} )}^{ \frac{1}{2} } \\ x = {m}^{1} {l}^{1} {t}^{ - 2} {m}^{ \frac{1}{2} } {l}^{ \frac{ - 3}{2} } \\ x = {m}^{ \frac{3}{2} } {l}^{ \frac{ - 1}{2} } {t}^{ - 2}$

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