Download App

Physics 3 Marks Question

PART - 2 CH - 14 Waves · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 5 questions.

Questions

3 Marks Question

Take a timed test

5 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks
A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20°C = 343 ms–1?
Answer
Length of steel wire = 12.0 meter
Mass $(M)=2.10 kg$, Tension in wire $T =$ ?
Unit length mass of wire
$m=\frac{M}{l}=\frac{2.10}{12.0}=0.175 kg / m$
Speed of sound in dry air $=343 m / s$
We know that $\quad v =\sqrt{\frac{ T }{m}}$
or $\quad$$v^2=\frac{T}{m}$
or$\quad$$T=v^2 m$
Put the value $\quad=(343)^2 \times 0.175$
$=343 \times 343 \times 0.175 $
$=2.06 \times 10^4 N$
View full question & answer→
Question 23 Marks
A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be 2.53 kHz. What is the speed of sound in steel?
Answer
According to question, we can represent the vibration of rod like as
Image
Given : $\quad n=2.53 kHz =2.53 \times 10^5 Hz$
or $\quad \frac{\lambda}{4}+\frac{\lambda}{4}=l$
$\Rightarrow \quad \frac{\lambda}{2}=1$ meter $\Rightarrow \lambda=2$ meter
Therefore, speed $v =n \lambda=2.53 \times 10^3 \times 2 $
$=5.06 \times 10^3 \text { meter } / \text { second }$
$=5.06$ $km / sec$.
View full question & answer→
Question 33 Marks
Two open tubes, whose length are 50 cm and 50.05 cm respectively. 10 beats are produced in 3 seconds. Find their fundamental frequencies.
Answer
Frequency of open tube
$\begin{aligned} n & =\frac{ v }{2 l} \\ v & \rightarrow \text { speed of sound } \\ l & \rightarrow \text { length of tube } \\ n_1 & =\frac{ v }{2 l_1}\ldots\ldots (1) \\ n_2 & =\frac{ v }{2 l_2}\ldots\ldots (2)\end{aligned}$
Given:
$\begin{aligned}n_1-n_2 & =\frac{10}{3} \\l_1 & =50 cm \\l_2 & =50.5 cm \\n_1-n_2 & =\frac{v}{2 l_1} \times \frac{v}{2 l_2} \\& =\frac{v}{2}\left[\frac{l_2-l_1}{l_1 l_2}\right] \\ \frac{10}{3} & =\frac{v}{2}\left[\frac{50.5-50}{50 \times 50.5}\right] \\\frac{10}{3} & =\frac{v}{2}\left[\frac{0.5}{2525}\right] \\\text { or } \quad v & =\frac{50500}{1.5}\end{aligned}$
$\begin{array}{l}=336.66 cm / s \\=336.66 m / \text { second }
\end{array}$
Put value in eqn. (1) and (2)
$\begin{array}{l}n_1=\frac{v}{2 l_1}=\frac{336.66}{2 \times 0.50} \\n_1=336.66 \cong 337 \text { vibration/sec. }\end{array}$
Similarly,
$\begin{array}{l}n_2=\frac{v}{2 l_2}=\frac{336.66}{2 \times 50.5 \times 10^{-2}} \\n_2=333 \text { vibration } / \text { second. }\end{array}$
View full question & answer→
Question 43 Marks
Define longitudinal and transverse waves and also draw diagram of both.
Answer
View full question & answer→
Question 53 Marks
Explain the difference between beats and interfrence.
Answer
S.No.BeatsInterference
1In this both the sound wave have almost the same frequencies.In this, the frequencies of both sound waves are same.
2At every point of medium amplitude value of time keeps changing with.at every point of the medium value of amplitude remain constant.
3The phase diference between waves at any point in the medium changes with time.at each point of medium phase difference is constant
4In beats the value of sound intensity at each point becomes heighest and lowest in alternating order with time in livesinterference the value of sound intensity at a particular point remain constant
View full question & answer→
Generate Paper FreeAll PART - 2 CH - 14 Waves topics
3 Marks Question - Physics STD 11 Science Questions - Vidyadip