Euler's Equation MCQ Quiz in हिन्दी - Objective Question with Answer for Euler's Equation - मुफ्त [PDF] डाउनलोड करें
Last updated on Mar 28, 2025
Latest Euler's Equation MCQ Objective Questions
Euler's Equation Question 1:
यदि \(u = \sin^{-1} \left(\dfrac{x}{y}\right) + \tan^{-1} \left(\dfrac{y}{x}\right)\) है, तब \(x \dfrac{\partial u}{\partial x} + y \dfrac{\partial u}{\partial y}\)का मान क्या है?
Answer (Detailed Solution Below)
Euler's Equation Question 1 Detailed Solution
व्याख्या:
यदि \(u = {\sin ^{ - 1}}\left( {\frac{x}{y}} \right) + {\tan ^{ - 1}}\left( {\frac{y}{x}} \right)\)
\(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = ?\)
\(\frac{{\partial u}}{{\partial x}} = \frac{1}{{\sqrt {1 - {{\left( {\frac{x}{y}} \right)}^2}} }} \times \frac{1}{y} + \frac{1}{{1 + {{\left( {\frac{y}{x}} \right)}^2}}} \times y\left( {\frac{{ - 1}}{{{x^2}}}} \right)\)
\(\frac{{\partial u}}{{\partial y}} = \frac{{ - 1}}{{\sqrt {1 - {{\left( {\frac{x}{y}} \right)}^2}} }} \times \left( {\frac{x}{{{y^2}}}} \right) + \frac{1}{{1 + {{\left( {\frac{1}{x}} \right)}^2}}} \times \frac{1}{x}\)
\(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = \frac{x}{y} \times \frac{1}{{\sqrt {1 - {{\left( {\frac{x}{y}} \right)}^2}} }} - x \times \frac{y}{{{x^2}}} \times \frac{1}{{\left[ {1 + {{\left( {\frac{y}{x}} \right)}^2}} \right]}}\)
\( - y \times \left( {\frac{x}{{{y^2}}}} \right) \times \frac{1}{{\sqrt {1 - {{\left( {\frac{x}{y}} \right)}^2}} }} + y \times \frac{1}{x}\frac{1}{{\left( {1 + {{\left( {\frac{y}{x}} \right)}^2}} \right)}}\)
∴ \(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = 0\)
Top Euler's Equation MCQ Objective Questions
यदि \(u = \sin^{-1} \left(\dfrac{x}{y}\right) + \tan^{-1} \left(\dfrac{y}{x}\right)\) है, तब \(x \dfrac{\partial u}{\partial x} + y \dfrac{\partial u}{\partial y}\)का मान क्या है?
Answer (Detailed Solution Below)
Euler's Equation Question 2 Detailed Solution
Download Solution PDFव्याख्या:
यदि \(u = {\sin ^{ - 1}}\left( {\frac{x}{y}} \right) + {\tan ^{ - 1}}\left( {\frac{y}{x}} \right)\)
\(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = ?\)
\(\frac{{\partial u}}{{\partial x}} = \frac{1}{{\sqrt {1 - {{\left( {\frac{x}{y}} \right)}^2}} }} \times \frac{1}{y} + \frac{1}{{1 + {{\left( {\frac{y}{x}} \right)}^2}}} \times y\left( {\frac{{ - 1}}{{{x^2}}}} \right)\)
\(\frac{{\partial u}}{{\partial y}} = \frac{{ - 1}}{{\sqrt {1 - {{\left( {\frac{x}{y}} \right)}^2}} }} \times \left( {\frac{x}{{{y^2}}}} \right) + \frac{1}{{1 + {{\left( {\frac{1}{x}} \right)}^2}}} \times \frac{1}{x}\)
\(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = \frac{x}{y} \times \frac{1}{{\sqrt {1 - {{\left( {\frac{x}{y}} \right)}^2}} }} - x \times \frac{y}{{{x^2}}} \times \frac{1}{{\left[ {1 + {{\left( {\frac{y}{x}} \right)}^2}} \right]}}\)
\( - y \times \left( {\frac{x}{{{y^2}}}} \right) \times \frac{1}{{\sqrt {1 - {{\left( {\frac{x}{y}} \right)}^2}} }} + y \times \frac{1}{x}\frac{1}{{\left( {1 + {{\left( {\frac{y}{x}} \right)}^2}} \right)}}\)
∴ \(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = 0\)
Euler's Equation Question 3:
यदि \(u = {x^{n - 1}}yf\left( {\frac{y}{x}} \right)\) तो \(\frac{{x{\partial ^2}u}}{{\partial {x^2}}} + \frac{{y{\partial ^2}u}}{{\partial y\partial x}}\) किसके बराबर है?
Answer (Detailed Solution Below)
Euler's Equation Question 3 Detailed Solution
अवधारणा :
चरों के रूप में 'x' और 'y' के साथ डिग्री 'n' के सजातीय समीकरण के लिए यूलर की प्रमेय में कहा गया है कि:
\(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = nu\)
विश्लेषण :
हमारे पास है:
\(u = {x^{n - 1}} \cdot yf\left( {\frac{y}{x}} \right)\)
\(u = {x^n} \cdot \left( {\frac{y}{x}} \right)f\left( {\frac{y}{x}} \right)\)
u डिग्री n का एक सजातीय फलन है।
\(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = nu\)
अब आंशिक रूप से x के संबंध में फिर से अवकलन करके हम प्राप्त करते हैं:
\(x\frac{{{\partial ^2}u}}{{\partial x}} + \frac{{\partial u}}{{\partial x}} + y\frac{{{\partial ^2}u}}{{\partial x\partial y}} = n\left( {\frac{{\partial u}}{{\partial x}}} \right)\)
\(\therefore x\frac{{{\partial ^2}u}}{{\partial x}} + y\frac{{{\partial ^2}u}}{{\partial x\partial y}} = \left( {n - 1} \right)\frac{{\partial u}}{{\partial x}}\)
Euler's Equation Question 4:
यदि \(u = \sin^{-1} \left(\dfrac{x}{y}\right) + \tan^{-1} \left(\dfrac{y}{x}\right)\) है, तब \(x \dfrac{\partial u}{\partial x} + y \dfrac{\partial u}{\partial y}\)का मान क्या है?
Answer (Detailed Solution Below)
Euler's Equation Question 4 Detailed Solution
व्याख्या:
यदि \(u = {\sin ^{ - 1}}\left( {\frac{x}{y}} \right) + {\tan ^{ - 1}}\left( {\frac{y}{x}} \right)\)
\(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = ?\)
\(\frac{{\partial u}}{{\partial x}} = \frac{1}{{\sqrt {1 - {{\left( {\frac{x}{y}} \right)}^2}} }} \times \frac{1}{y} + \frac{1}{{1 + {{\left( {\frac{y}{x}} \right)}^2}}} \times y\left( {\frac{{ - 1}}{{{x^2}}}} \right)\)
\(\frac{{\partial u}}{{\partial y}} = \frac{{ - 1}}{{\sqrt {1 - {{\left( {\frac{x}{y}} \right)}^2}} }} \times \left( {\frac{x}{{{y^2}}}} \right) + \frac{1}{{1 + {{\left( {\frac{1}{x}} \right)}^2}}} \times \frac{1}{x}\)
\(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = \frac{x}{y} \times \frac{1}{{\sqrt {1 - {{\left( {\frac{x}{y}} \right)}^2}} }} - x \times \frac{y}{{{x^2}}} \times \frac{1}{{\left[ {1 + {{\left( {\frac{y}{x}} \right)}^2}} \right]}}\)
\( - y \times \left( {\frac{x}{{{y^2}}}} \right) \times \frac{1}{{\sqrt {1 - {{\left( {\frac{x}{y}} \right)}^2}} }} + y \times \frac{1}{x}\frac{1}{{\left( {1 + {{\left( {\frac{y}{x}} \right)}^2}} \right)}}\)
∴ \(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = 0\)