If momentum (P ), area (A) and time (T ) are taken to be fundamental quantities, then energy has the dimensional formula

(a) $\mathit{\rho }$ (A^-1 T¹)

(b) ${\mathit{\rho }}^2$ (A T)

(c) $\mathit{\rho }$ ( A^-1/2 T)

(d) $\mathit{\rho }$ ( A^1/2 T^-1)

<p><span data-teams="true">This is a multiple choice answer as classified in NCERT Exemplar</span></p><p> (d) E=&nbsp;<span contenteditable="false"> <math> <mo>∝</mo> <msup> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>a</mi> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> <mrow> <mrow> <mi>b</mi> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> </msup> </math> </span></p><p>E= <span contenteditable="false"> <math> <mo>=</mo> <mi>k</mi> <msup> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>a</mi> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> <mrow> <mrow> <mi>b</mi> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> </msup> </math> </span></p><p>E= [ML²T^-2] and <span contenteditable="false"> <math> <mi>ρ</mi> <mo>=</mo> <mfenced open=" [" close="]" separators="|"> <mrow> <mrow> <mi>M</mi> <mi>L</mi> <msup> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfenced> </math> </span></p><p>A= [L²] and T= [T]</p><p>E= [k] [ <span contenteditable="false"> <math> <mi>ρ</mi> </math> </span>]^a [A]^b [T]^c</p><p>ML²T^-2= [MLT^-1]^a [L²]^b [T]^c</p><p>= M^aL^2b+aT^-a+c</p><p>by principle of homogeneity we get, </p><p>ML²T^-2= M^aL^2b+aT^-a+c</p><p>After solving</p><p>A=1, 2b+a=2</p><p>So 2b+1=2</p><p>2b=1, b= ½</p><p>C=-2+a, c=-2+1, c=-1</p><p>So formula become</p><p>E=<span contenteditable="false"> <math> <mi>ρ</mi> </math> </span> A^1/2T^-1</p>

0 Upvotes 0 Upvotes 0 Downvotes

• • Report Abuse
• Comment