JEE MAIN - Physics (2020 - 3rd September Morning Slot - No. 11)
A uniform thin rope of length 12 m and mass 6 kg hangs vertically from a rigid support and a block
of mass 2 kg is attached to its free end. A transverse short wavetrain of wavelength 6 cm is
produced at the lower end of the rope. What is the wavelength of the wavetrain (in cm) when it
reaches the top of the rope ?
12
3
9
6
Explanation
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T1 = 2g
T2 = 8g
V = $$\sqrt {{T \over \mu }} $$
$$ \therefore $$ V $$ \propto $$ $$\sqrt T $$
Also V = f$$\lambda $$
$$ \therefore $$ V1 = f1$$\lambda $$1
and V2 = f2$$\lambda $$2
We know frequency of sources are same.
$$ \therefore $$ f1 = f2
So $$\sqrt {{{{T_1}} \over {{T_2}}}} = {{{\lambda _1}} \over {{\lambda _2}}}$$
$$ \Rightarrow $$ $$\sqrt {{{2g} \over {8g}}} = {6 \over {{\lambda _2}}}$$
$$ \Rightarrow $$ $$\lambda $$2 = 12 cm
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