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0Factorise: x2 – y 2 + 4yz – 4z2​

Answer» <html><body><p>Given :The <a href="https://interviewquestions.tuteehub.com/tag/period-1151023" style="font-weight:bold;" target="_blank" title="Click to know more about PERIOD">PERIOD</a> of revolution (t) of a planet around the sun depends upon <a href="https://interviewquestions.tuteehub.com/tag/radius-1176229" style="font-weight:bold;" target="_blank" title="Click to know more about RADIUS">RADIUS</a> (r) of the orbit, mass (m) of the sun and <a href="https://interviewquestions.tuteehub.com/tag/gravitational-476409" style="font-weight:bold;" target="_blank" title="Click to know more about GRAVITATIONAL">GRAVITATIONAL</a> <a href="https://interviewquestions.tuteehub.com/tag/constant-930172" style="font-weight:bold;" target="_blank" title="Click to know more about CONSTANT">CONSTANT</a> (G), mathematically :\implies \sf [T] \propto [r]^a [m]^b [G]^c⟹[T]∝[r] a [m] b [G] c Adding k as dimensionaless constant :\begin{gathered}\implies \sf [T] = k [r]^a [m]^b [G]^c \: \: \: \: \:... (i)\\ \end{gathered} ⟹[T]=k[r] a [m] b [G] c ...(i) • DIMENSIONS:\begin{gathered}\<a href="https://interviewquestions.tuteehub.com/tag/dag-430679" style="font-weight:bold;" target="_blank" title="Click to know more about DAG">DAG</a> \bf \: Dimensions \: of \: period\:of\: revolution \: (t) = [M^0L^0T^1] \\ \end{gathered} †Dimensionsofperiodofrevolution(t)=[M 0 L 0 T 1 ] \begin{gathered}\dag \bf \: Dimensions \: of \: radius \: (r) = [M^0L^1T^0] \\ \end{gathered} †Dimensionsofradius(r)=[M 0 L 1 T 0 ] \begin{gathered}\dag \bf \: Dimensions \: of \: mass \: (m) = [M^1L^0T^0] \\\end{gathered} †Dimensionsofmass(m)=[M 1 L 0 T 0 ] \begin{gathered}\dag \bf \: Dimensions \: of \: gravitational \:constant \: (G) = [M^{ - 1} L^3T^{ - 2} ] \\\end{gathered} †Dimensionsofgravitationalconstant(G)=[M −1 L 3 T −2 ] \begin{gathered}\implies \sf [M^0L^0T^1] =[M^0L^1T^0] ^a [M^1L^0T^0]^b [M^{ - 1} L^3T^{ - 2} ]^c \\ \end{gathered} ⟹[M 0 L 0 T 1 ]=[M 0 L 1 T 0 ] a [M 1 L 0 T 0 ] b [M −1 L 3 T −2 ] c \begin{gathered}\implies \sf [M^0L^0T^1] =[M]^{b - c} [L]^{a + 3c} [T]^{ - 2c} \\ \end{gathered} ⟹[M 0 L 0 T 1 ]=[M] b−c [L] a+3c [T] −2c • On comparing the powers we have:\implies b - c = 0⟹b−c=0\implies b = c \: \: \: ....(ii)⟹b=c....(ii)\implies a + 3c = 0 \: \: \: ....(iii)⟹a+3c=0....(iii)\implies - 2c = 1⟹−2c=1\implies \bf c = \dfrac{ -1 }{2} \: \: \: ....(iv)⟹c= 2−1 ....(iv)Substituting value of c = -1/2 from equation (iv) to equation (ii) :\begin{gathered}\implies b = c \\ \end{gathered} ⟹b=c \implies \bf b = \dfrac{ - 1}{2}⟹b= 2−1 Also substitue the value of c from equation (iv) to equation (ii)\implies a + 3c = 0⟹a+3c=0\implies a + 3 \times \dfrac{ - 1}{2} = 0⟹a+3× 2−1 =0\implies a - \dfrac{ 3}{2} = 0⟹a− 23 =0\implies \bf a = \dfrac{ 3}{2}⟹a= 23 Now, Substituting the value of a, b and c in the equation (I):\begin{gathered}\implies \sf [T] = k [r]^{ \frac{3}{2} }[m]^{ \frac{ - 1}{2} } [G]^{ \frac{ - 1}{2} } \\ \end{gathered} ⟹[T]=k[r] 23 [m] 2−1 [G] 2−1 \begin{gathered}\implies \sf [T]^{2} = k [r]^{3 }[m]^{ - 1 } [G]^{ - 1 } \\ \end{gathered} ⟹[T] 2 =k[r] 3 [m] −1 [G] −1 \implies \sf [T]^{2} = k [r]^{3 }⟹[T] 2 =k[r] 3 \implies \sf [T]^{2} \propto [r]^{3 }⟹[T] 2 ∝[r] 3 \implies \underline{ \boxed{ \orange{\bf T^{2} \propto r^{3 }}}}⟹ T 2 ∝r 3 Hence, Proved!</p></body></html>


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