physics ncert solutions class 11th

Speed of sound in tissue, v = 1.7 km/s = 1.7 $*{10}^3$ m/s

Operating frequency of the scanner,  $\nu$ = 4.2 MHz = 4.2 $*{10}^6$ Hz

The wavelength of sound in the tissue is given as:

$\lambda =\frac{v}{\nu }$ = $\frac{1.7\mathrm{}*{10}^3}{4.2\mathrm{}*{10}^6}$ = 4.05 $*{10}^{-4}$ m

Frequency of ultrasound,  $\nu$ = 1000 kHz = ${10}^6$ Hz

Speed of sound in air,  $v_a$ = 340 m/s

Speed of sound in water,  $v_w$ = 1486 m/s

The wavelength of the reflected sound is given by the relation

${\lambda }_r=\frac{v_a}{\nu }$ = $\frac{340}{{10}^6}$ = 3.4 $*{10}^{-4}$ m

The wavelength of the transmitted sound wave is given by

${\lambda }_t=\frac{v_w}{\nu }$ = $\frac{1486}{{10}^6}$ = 1.486 $*{10}^{-3}$ m

(a) For x =0 and t=0, the function (x – vt )² becomes 0

Hence for x=0 and t=0, the function represents a point and not a wave.

(b) For x =0 and t=0, the function $\log ?\frac{x+vt}{x_0}$ = log 0 = $\propto$

Since the function does not converge to a finite value for x =0 and t = 0, it represents a travelling wave.

(c) For x = 0 and t = 0, the function $\frac{1}{x+vt}$ = $\frac{1}{0}$ = $\propto$

Since the function does not converge to a finite value for x = 0 and t = 0, it does not represent a travelling wave.

Length of the steel wire, l = 12 m

Mass of the steel wire, m = 2.1 kg

Velocity of the transverse wave, v = 343 m/s

Mass per unit length,  $\mu$ = $\frac{m}{l}$ = $\frac{2.1}{12}$ = 0.175 kg/m

The velocity (v) of the transverse wave in the string is given by the relation:

$v=\sqrt{\frac{T}{\mu }}$ , where T is the tension

T = $v^2*\mu$ = ${343}^2*0.175$ = 20588.575 N = 2.06 $*{10}^4$ N

Height of the tower, h = 300 m

Initial velocity of the stone, u = 0

Acceleration, a = g = 9.8 m/ $s^2$

Speed of sound in air, V = 340 m/s

The time taken by the stone (t), to strike the water can be calculated from the relation

s =us + $\frac{1}{2}$ a $t^2$ as

300 = 0 + $\frac{1}{2}*9.8*t^2$ or t = 7.82 s

Time taken by the sound to reach the top of the tower,  $t_1$ = $\frac{h}{V}$ = $\frac{300}{340}$ = 0.88 s

Therefore, the time when the splash can be heard = 7.82 + 0.88 = 8.7 s

Mass of the string, M = 2.5 kg

Tension in the string, T = 200 N

Length of the string, l = 20 m

Mass per unit length,  $\mu$ = $\frac{M}{l}$ = $\frac{2.5}{20}$ = 0.125 kg/m

The velocity (v) of the transverse wave in the string is given by the relation:

$v=\sqrt{\frac{T}{\mu }}$ = $\sqrt{\frac{200}{0.125}}$ = 40 m/s

Therefore, time taken by the disturbance to reach the other end, t = $\frac{l}{v}$ = $\frac{20}{40}$ = 0.5 s

Water pressure at the bottom, p = 1.1 $*{10}^8$ Pa

Initial volume of the steel ball, V = 0.32 $m^3$

Bulk modulus of the steel, B = 1.6 $*{10}^{11}$ N/ $m^2$

The ball falls at the bottom of the Pacific ocean which is 11 km beneath the surface

Let the change in volume of the ball on reaching the bottom of the trench be ΔV

We know, bulk modulus, B = $\frac{p}{\frac{\mathrm{\Delta }\mathrm{V}}{V}}$ or ΔV = $\frac{pV}{B}$

ΔV = $\frac{1.1*{10}^8*0.32}{1.6*{10}^{11}}$ = 2.2 $*{10}^{-4}{m}^3$

Diameter of the metal strip, d = 6.0 mm = 6 $*{10}^{-3}$ m

Radius, r = d/2 = 3 $*{10}^{-3}$ m

Maximum shearing stress = 6.9 $*{10}^7$ Pa = $\frac{Maximumforce}{Area}$

Maximum force = Maximum stress $*Area$

= 6.9 $*{10}^7*\pi *r^2$ = 6.9$*{10}^7*\pi *{(3*{10}^{-3})}^2$ = 1950.93 N

Since each rivet carries 1/4th of load,

Maximum tension on each rivet = 4 $*1950.93$ N = 7803.72 N