Correct Answer is Option (C)

Let mass per unit length of wires are $\mu$₁ and $\mu$₂ respectively

$\because$ Materials are same, so density $\rho$ will be same.

$\therefore$ ${\mu }_1=\frac{\rho \pi r^{2}L}{L}=\mu$ and ${\mu }_2=\frac{\rho 4\pi r^{2}L}{L}=4\mu$

Tension in both are same = T, let speed of wave in wires are V₁ and V₂

$V_1=\frac{V_1}{2L}=\frac{V}{2L}\mathrm{\&}{V}_2=\frac{V_2}{2L}=\frac{V}{4L}$

Frequency at which both resonate is L.C.M. of both frequencies (i.e : $\frac{V}{2L}$ )

Hence number of loops in wires are 1 and 2 respectively

So, ratio of number of antinodes is 1 : 2.

Correct Answer is Option (D)

y = 0.03 sin(450 t $-$ 9x)

v = $\frac{\omega }{k}=\frac{450}{9}$ = 50m/s

v = $\sqrt{\frac{T}{\mu }}\Rightarrow \frac{T}{\mu }$ = 2500

$\Rightarrow$  T = 2500 $\times$ 5 $\times$ 10^$-$3

= 12.5 N

Correct Answer is Option (C)

Velocity of wave on string

$V=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{8}{5}\times 1000}=40m/s$

Now, wavelength of wave

$\lambda =\frac{v}{n}=\frac{40}{100}m$

Separation b/w successive nodes,

$\frac{\lambda }{2}=\frac{20}{100}m$ $=$ 20 cm

Correct Answer is Option (B)

Resultant force on the ball of mass M when car is moving with a acceleration a is ,

F_net = $\sqrt{{\left(Mg\right)}^2+{\left(Ma\right)}^2}$

   = $M\sqrt{g^2+a^2}$

$\therefore$   T = M$\sqrt{g^2+a^2}$

We know,

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

When Car is at rest then,

   60 = $\sqrt{\frac{Mg}{\mu }}$   . . . . (1)

and when is moving then

   60.5 = $\sqrt{\frac{M\sqrt{g^2+a^2}}{\mu }}$    . . . . (2)

By dividing (2) by (1) we get,

$\frac{60.5}{60}=\sqrt{\frac{\sqrt{g^2+a^2}}{g}}$

$\Rightarrow$   $\left(1+\frac{0.5}{60}\right)$ = ${\left(\frac{g^2+a^2}{g^2}\right)}^{\frac{1}{4}}$

$\Rightarrow$   $\frac{g^2+a^2}{g^2}$ = ${\left(1+\frac{0.5}{60}\right)}^4$

$\Rightarrow$   $\frac{g^2+a^2}{g^2}$ = 1 + 4 $\times$ $\frac{0.5}{60}$ [Using Binomial approximation]

$\Rightarrow$   $\frac{g^2+a^2}{g^2}$ = 1 + $\frac{1}{30}$

$\Rightarrow$   1 + $\frac{a^2}{g^2}$ = 1 + $\frac{1}{30}$

$\Rightarrow$   $\frac{a}{g}$ = $\frac{1}{\sqrt{30}}$

$\Rightarrow$   a = $\frac{g}{\sqrt{30}}$

$\therefore$   Closest answer, a = $\frac{g}{5}$

Correct Answer is Option (B)

As   rod length = 60 cm

$\therefore$ $\frac{\lambda }{2}$ = 60

$\Rightarrow$$$ $\lambda$ = 120 cm = 1.2 m

In solid, velocity of wave,

V = $\sqrt{\frac{Y}{\rho }}$

= $\sqrt{\frac{9.27\times {10}^{10}}{2.7\times {10}^3}}$

= 5.85 $\times$ 10³ m/sec.

As  we know,

v = f $\lambda$

$\therefore$ f = $\frac{v}{\lambda }$

= $\frac{5.85\times {10}^3}{1.2}$

= 4.88 $\times$ 10³ Hz

$\simeq$  5 kHz