A uniform rope of mass 6 kg and length 6 m hangs vertically from a rigid support. A block of mass 2 kg is attached to the free end of the rope. A transverse pulse of wavelength 0.075 m is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is K cm where K is

Question

A uniform rope of mass 6 kg and length 6 m hangs vertically from a rigid support. A block of mass 2 kg is attached to the free end of the rope. A transverse pulse of wavelength 0.075 m is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is K cm where K is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

T1= 2 × 10 = 20 N T2 = (.2 + 6.) 10 = 80 N (V2/V1)=√(80/20)=2 Frequency remains unchanged So (λ 2/λ 1)=2 λ 2=2× 0.075 = 0.15 m K=15 cm

T1​=2×10=20N T2​=(2+6)10=80N V1​V2​​=2080​​=2 Frequency remains unchanged So λ1​λ2​​=2 λ2​=2×0.075 =0.15m K=15cm

$T_{1} = 2 \times 10 = 20 N$
$T_{2} = (2 + 6) 10 = 80 N$
$\frac{V _{2}}{V _{1}} = \frac{80}{20} = 2$
Frequency remains unchanged
So $\frac{λ _{2}}{λ _{1}} = 2$
$λ_{2} = 2 \times 0.075$
$= 0.15 m$
$K = 15 c m$