QJEEAdv09Pha50d: JEE Advanced 2009 Physics - Class 11 JEEAdv09Pha50d Solution: A 20 cm long string, having a mass of 1.0 g, is fixed at both…

The distance between the successive nodes is $\lambda/2$ . Therefore,

${\lambda \over 2} = {V \over {2v}} = {1 \over {2 \times 100}}\sqrt {{{Tl} \over m}} = {1 \over {200}}\sqrt {{{0.5 \times 0.2} \over {{{10}^{ - 3}}}}} = {1 \over {20}}$ m = 5 cm