Correct Answer is Option (B)

$Y=A\sin (kx-\omega t)$

$A=\frac{6}{2}$ = 3cm = 0.03 m

$\omega =2\pi f=2\pi \times 245$

$\omega =1.5\times {10}^3$

$k=\frac{\omega }{v}=\frac{1.5\times {10}^3}{300}$

$k=5.1$

$y=0.03\sin (5.1x-(1.5\times {10}^3)t)$

Correct Answer is Option (C)

Y = F(x, t)

For travelling wave y should be linear function of x and t and they must exist as (x $\pm$ vt)

Y = A sin (15x $-$ 2t) $\to$ linear function in x and t.

Correct Answer is 2

As per question,

at t = 0, y = $\frac{1}{{(1+x)}^2}$

and at t = 1 s, y = $\frac{1}{1+{(x-2)}^2}$ .... (i)

As we know,

At t = t s, y = $\frac{1}{1+{(x-vt)}^2}$

So, at t = 1 s, y = $\frac{1}{1+{(x-v)}^2}$ .... (ii)

On comparing Eqs. (i) and (ii), we get

v = 2 ms^$-$1

Hence, the velocity of the wave will be 2 m/s.

Correct Answer is Option (B)

$1.5=\left(2n_1+1\right)\frac{\lambda }{2}$ ......(1)

5 = n₂$\lambda$ .....(2)

(1) $÷$ (2)

$\frac{1.5}{5}=\frac{\left(2n_1+1\right)}{2n_2}$

$\Rightarrow$ 3n₂ = 10n₁ + 5

n₁ = 1 ; n₂ = 5 $\Rightarrow$ $\lambda$ = 1

n₁ = 4 ; n₂ = 15 $\Rightarrow$ $\lambda$ = 1/3

n₁ = 7 ; n₂ = 25 $\Rightarrow$ $\lambda$ = 1/5

Correct Answer is Option (B)

y = A sin (kx – wt + $\phi$)

at x = 0, t = 0 and slope is negative

$\Rightarrow \phi =\pi$