\(v=f \lambda\)
\(=\lambda \text { is constant for an organ pipe }\)
\(v=\sqrt{\frac{\gamma R T}{M}}\)
\(v=\frac{1}{\sqrt{M}}\)
\(v\) is maximum when \(M\) is least. Thus, velocity of sound in air is maximum when \(M\) is least.
Since \(f \propto v\) so \(f \propto \frac{1}{\sqrt{M}}\)
Hence, frequency is maximum when \(M\) is least.
Hence, among options answer is \(H _2\).
${y}=1.0\, {mm} \cos \left(1.57 \,{cm}^{-1}\right) {x} \sin \left(78.5\, {s}^{-1}\right) {t}$
${x}>0$ ના ક્ષેત્રમાં ઉગમબિંદુથી નજીકનું નિસ્પંદ બિંદુ ${x}=\ldots \ldots \ldots\, {cm}$ અંતરે હશે.