Correct Answer is 10

$\mu =9.0\times {10}^{-4}\frac{kg}{m}$

T = 900 N

$V=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{900}{9\times {10}^{-4}}}=1000$ m/s

f₁ = 500 Hz

f = 550

$\frac{nV}{2l}=500$ .... (i)

$\frac{(n+1)V}{2l}=500$ .... (ii)

(ii) (i) $\frac{V}{2l}=50$

$l=\frac{1000}{2\times 50}=10$

Correct Answer is 1215

$\mu =0.135$ gm/cm

$\mu =0.135\times \frac{{10}^{-3}}{{10}^{-2}}\frac{kg}{m}$

y = $-$0.21 sin (x + 30t)

$v=\frac{\omega }{K}=\frac{30}{1}$ = 30 m/s

v = $\sqrt{\frac{T}{\mu }}$

T = v² $\times$ $\mu$

T = (30)² $\times$ 0.135 $\times$ 10^$-$1

T = 900 $\times$ 0.135 $\times$ 10^$-$1

T = 12.15 N

T = 1215 $\times$ 10^$-$2 N

x = 1215

Correct Answer is 2

Speed of transverse wave is

$V=\sqrt{\frac{T}{\mu }}$

$\ln V=\frac{1}{2}\ln T-\frac{1}{2}\ln \mu$

$\frac{\mathrm{\Delta }V}{V}=\frac{1}{2}\frac{\mathrm{\Delta }T}{T}$

$=\frac{1}{2}\times 4$

$\Rightarrow$ $\frac{\mathrm{\Delta }V}{V}=2\mathrm{\%}$

Correct Answer is Option (A)

T₁ = 2g

T₂ = 8g

V = $\sqrt{\frac{T}{\mu }}$

$\therefore$ 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 $\sqrt{\frac{T_1}{T_2}}=\frac{{\lambda }_1}{{\lambda }_2}$

$\Rightarrow$ $\sqrt{\frac{2g}{8g}}=\frac{6}{{\lambda }_2}$

$\Rightarrow$ $\lambda$₂ = 12 cm

Correct Answer is Option (A)

f = $\frac{1}{2l}\sqrt{\frac{T}{\mu }}$

For identical string $l$ and $\mu$ will be same

f $\propto$ $\sqrt{T}$

$\therefore$ $\frac{450}{300}=\sqrt{\frac{T_x}{T_y}}$

$\Rightarrow$ $\frac{T_x}{T_y}=\frac{9}{4}$ = 2.25