SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

Question  4. Explain why

(a) To keep a piece of paper horizontal, you should blow over, not under, it.

(b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers.

(c) The size of a needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection.

(d) A fluid flowing out of a small hole in a vessel results in a backward thurst on the vessel.

(e) A spinning cricket ball in air does not follow a parabolic trajectory.

Answer:  (a) When we blow over the piece of paper, the velocity of air increases. As a result, the pressure on it decreases in accordance with the Bernoulli’s theorem whereas the pressure below remains the same (atmospheric pressure). Thus, the paper remains horizontal.

(b) By doing so the area of outlet of water jet is reduced, so velocity of water increases according to equation of continuity av = constant.

(c) For a constant height, Bernoulli’s theorem is expressed as P +1/2 ρ v2 = Constant

In this equation, the pressure P occurs with a single power whereas the velocity occurs with a square power. Therefore, the velocity has more effect compared to the pressure. It is for this reason that needle of the syringe controls flow rate better than the thumb pressure exerted by the doctor.

(d) This is because of principle of conservation of momentum. While the flowing fluid carries forward momentum, the vessel gets a backward momentum.

(e) A spinning cricket ball would have followed a parabolic trajectory has there been no air. But because of air the Magnus effect takes place. Due to the Magnus effect the spinning cricket ball deviates from its parabolic trajectory.

মেডিকেল (NEET) ইঞ্জিনিয়ারিংয়ে (JEE) এন্ট্রান্স ভর্তি চলছে : এক ক্লাসরুমে সব সাবজেক্ট 

ডক্টরেট উপাধিপ্রাপ্ত, গোল্ড মেডেলিস্ট, সাবজেক্ট এক্সপার্ট থেকে শিখতে হলে এখুনি ভর্তি হন।   ত্রিপুরাতে সব থেকে কম ফী ! সময় এবং টাকা দুটোই বাঁচবে 

৯৮ চ্যাপ্টার টেস্ট সিরিজ , সপ্তাহে ৬ দিন ক্লাস 

Question 5. A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?

Answer:

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

পদার্থের তাপীয় বৈশিষ্ট্য:Thermal Properties of matter

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

Question 11. 1. The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.

Answer: The relation between kelvin scale and Celsius scale is TK – 273.15 =TC => TC=TK– 273.15

Question 11. 2.  Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between TA and TB ?

Answer:  As we know, triple point of water on absolute scale = 273.16 K, Size of one degree of kelvin scale on absolute scale A

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

Question 11. 3.  The-electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: R = R0 [1 + α (T – T0)].

The resistances is 101.6 Ωat the triple-point of water 273.16 K, and 165.5 Ωat the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω ?

Answer: Here, R0 = 101.6 Ω; T0 = 273.16 K Case (i) R1= 165.5 Ω; T1 = 600.5 K, Case (ii) R2 = 123.4 , T2 = ?

Using the relation R = R0[1 + α (T – T0)]

Case (i) 165.5 = 101.6 [1 + α (600.5 – 273.16)]

Question 11. 4. Answer the following:

(a) The triple-point of water is a standard fixed point in modem thermometry. Why ? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale) ?SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0 °C and 100 °C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) Scale ?

(c) The absolute temperature (Kelvin scale) T is related to the temperature tc on the Celsius scale tc = T – 273.15

Why do we have 273.15 in this relation, and not 273.16 ?

(d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale ?

Answer: (a) Triple point of water has a unique value i.e., 273.16 K. The melting point and boiling points of ice and water respectively do not have unique values and change with the change in pressure.

(b) On Kelvin’s absolute scale, there is only one fixed point, namely, the triple-point of water and there is no other fixed point.

(c) On Celsius scale 0 °C corresponds to the melting point of ice at normal pressure and the value of absolute temperature is 273.15 K. The temperature 273.16 K corresponds to the triple point of water.

(d)The Fahrenheit scale and Absolute scale are related as

মেডিকেল (NEET) ইঞ্জিনিয়ারিংয়ে (JEE) এন্ট্রান্স ভর্তি চলছে : এক ক্লাসরুমে সব সাবজেক্ট 

ডক্টরেট উপাধিপ্রাপ্ত, গোল্ড মেডেলিস্ট, সাবজেক্ট এক্সপার্ট থেকে শিখতে হলে এখুনি ভর্তি হন।   ত্রিপুরাতে সব থেকে কম ফী ! সময় এবং টাকা দুটোই বাঁচবে 

৯৮ চ্যাপ্টার টেস্ট সিরিজ , সপ্তাহে ৬ দিন ক্লাস 

Question 11. 5. Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:

(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B ?

(b) What do you think is the reason behind the slight difference in answers of thermometers A and B ? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings ?

Answer: 

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

(b) The value of the melting point of sulphur found from the two thermometers differ slightly due to the reason that in practice, the gases do not behave strictly as perfect gases i.e., gases are not perfectly ideal.

To reduce the discrepency, readings should be taken for lower and lower pressures and the plot between temperature measured versus absolute pressure of the gas at triple point should be extrapolated to obtain the temperature in the limit pressure tends to zero (if P —> 0), when the gases approach ideal gas behaviour.

Question 11. 6.  A steel tape 1 m long is correctly calibrated for a temperature of 27.0 °C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 °C. What is the actual length of the steel rod on that day ? What is the length of the same steel rod on a day when the temperature is 27.0 °C ? Coefficient of linear expansion of steel = 1.20 x 10-1K-1.

Answer:  On a day when the temperature is 27 °C, the length of 1 cm division on the steel tape is exactly 1 cm, because the tape has been calibrated for 27 °C.When the temperature rises to 45 °C (that is, ΔT = 45 – 27 = 18 °C), the increase in the length of 1 cm division is Δl = αlΔT = (1.2 x 10-5C-1) x 1 cm x 18 °C = 0.000216 cm Therefore, the length of 1 cm division on the tape becomes 1.000216 cm at 45 °C. As the length of the steel rod is read to be 63.0 cm on the steel tape at 45 °C, the actual length of the rod at 45 °C is 63.0 x 1.000216 cm = 63.0136 cm The length of the same rod at 27 °C is 63.0 cm, because 1 cm mark on the steel tape is exactly 1 cm at 27 °C.

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

Question 11. 7.  A large steel wheel is to befitted on to a shaft of the same material. At 27 °C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is, 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft ? Assume coefficient of linear expansion of the steel to be constant over the required temperature range αsteel= 1-20 x 10-5K-1.

Answer: 

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

Question 11. 8. A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C ? Coefficient of linear expansion of copper = 1.70 x 10-5K-1.

Answer:

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

Question 11. 9. A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of – 39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm ? Co-efficient of linear expansion of brass = 2.0 x 10-5K-1; Young’s modulus of brass = 0.91 x 1011 Pa

Ans.

মেডিকেল (NEET) ইঞ্জিনিয়ারিংয়ে (JEE) এন্ট্রান্স ভর্তি চলছে : এক ক্লাসরুমে সব সাবজেক্ট 

ডক্টরেট উপাধিপ্রাপ্ত, গোল্ড মেডেলিস্ট, সাবজেক্ট এক্সপার্ট থেকে শিখতে হলে এখুনি ভর্তি হন।   ত্রিপুরাতে সব থেকে কম ফী ! সময় এবং টাকা দুটোই বাঁচবে 

৯৮ চ্যাপ্টার টেস্ট সিরিজ , সপ্তাহে ৬ দিন ক্লাস 

Question 11. 10. A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C ? Is there a ‘thermal stress’ developed at the junction ? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 x 10-5 °C-1, steel = 1.2 x 10-5 °C-3.

Ans.

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

Question 11. 11. The coefficient of volume expansion of glycerine is 49 x 10-5K-1. What is the fractional change in its density for a 30 °C rise in temperature ?

Ans.

Question 11. 12. A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings? Specific heat of aluminium = 0.91 J g-1 K-1 .

Answer: Power = 10 kW = 104 W

Mass, m=8.0 kg = 8 x 103 g

Question 11. 13. A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500°C and then placed on a large ice block. What is the maximum amount of ice that can melt? Specific heat of copper is 0.39 Jg-1°C-1. Heat of fusion of water = 335 Jg-1.

Answer: 

মেডিকেল (NEET) ইঞ্জিনিয়ারিংয়ে (JEE) এন্ট্রান্স ভর্তি চলছে : এক ক্লাসরুমে সব সাবজেক্ট 

ডক্টরেট উপাধিপ্রাপ্ত, গোল্ড মেডেলিস্ট, সাবজেক্ট এক্সপার্ট থেকে শিখতে হলে এখুনি ভর্তি হন।   ত্রিপুরাতে সব থেকে কম ফী ! সময় এবং টাকা দুটোই বাঁচবে 

৯৮ চ্যাপ্টার টেস্ট সিরিজ , সপ্তাহে ৬ দিন ক্লাস 

Question 11. 14. In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27 °C. The final temperature is 40° C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?

Answer: Mass of metal block, m = 0.20 kg = 200 g

Fall in the temperature of metal block,

ΔT = (150 – 40) °C = 110 °C

If C be the specific heat of metal, then heat lost by the metal block = 200 x C x 110 cal Volume of water = 150 cm3

mass of water = 150 g

Increase in temperature of water = (40 – 27) °C = 13°C

Heat gained by water = 150 x 13 cal Water equivalent of calorimeter, w = 0.025 kg = 25g

Heat gained by calorimeter,

Question 11. 15. Given below are observations on molar specific heats at room temperature of some common gases.

The measured molar specific heats of these gases are markedly different from those for mono atomic gases. Typically, molar specific heat of a mono atomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine ?

Answer: The gases which are listed in the above table are diatomic gases and not mono atomic gases. For diatomic gases, molar specific heat =5/2 R = 5/2 x 1.98 = 4.95, which agrees fairly well with all observations listed in the table except for chlorine. A mono atomic gas molecule has only the translational motion. A diatomic gas molecule, apart from translational motion, the vibrational as well as rotational motion is also possible. Therefore, to raise the temperature of 1 mole of a diatomic gas through 1°C, heat is to be supplied to increase not only translational energy but also rotational and vibrational energies. Hence, molar specific heat of a diatomic gas is greater than that for mono atomic gas. The higher value of molar specific heat of chlorine as compared to hydrogen, nitrogen, oxygen etc. shows that for chlorine molecule, at room temperature vibrational motion also occurs along with translational and rotational motions, whereas other diatomic molecules at room temperature usually have rotational motion apart from their translational motion. This is the reason that chlorine has somewhat larger value of molar specific heat.

Question 11. 16. (a) At what temperature and pressure can the solid, liquid and vapour phases of CO2 co-exist in equilibrium ?

(b) What is the effect of decrease of pressure on the fusion and boiling point of CO2 ?

(c) What are the critical temperature and pressure for CO2 ? What is their significance 1

(d) Is CO2 solid, liquid or gas at (a) – 70 °C under 1 atm (b) – 60 °C under 10 atm (c) 15°C under 56 atm?

Answer:  (a) At the triple point, temperature = – 56.6 °C and pressure = 5.11 atm.

(b) Both the boiling point and freezing point of CO2 decrease if pressure decreases.

(c) The critical temperature and pressure of CO2 are 31.1°C and 73.0 atm respectively. Above this temperature, CO2 will not liquefy/even if compressed to high pressures.

(d) (i) The point (- 70 °C, 1.0 atm) lies in the vapour region. Hence, CO2 is vapour at this point.

(ii) The point (- 60 °C, 10 atm) lies in the solid region. Hence, CO2 is solid at this point.

(iii) The point (15 °C, 56 atm) lies in the liquid region. Hence, CO2 is liquid at this point.

Question 11. 17. Answer the following questions based on the P – T phase diagram of CO2 (Fig. of question 17 given above)

(a) COSCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

2 at 1 atm pressure and temperature – 60 °C is compressed isothermally. Does it go through a liquid phase ?

(b) What happens when CO2 at 4 atm pressure is cooled from room temperature at constant pressure ?

(c) Describe qualitatively the changes in a given mass of solid CO2 at 10 atm pressure and temperature – 65 °C as it is heated up at room temperature at constant pressure.

(d) CO2 is heated to a temperature 70 °C and compressed isothermally. What changes in its properties do you expect to observe ?

Answer:  (a) No, the CO2 does not go through the liquid phase. The point (1.00 atm, – 60°C) is to the lift of the triple-point O and below the sublimation curve OA. Therefore, when CO2 is compressed at this point at constant temperature, the point moves perpendicular to the temperature-axis and enters the solid phase region. Hence, the CO2 vapour condenses to solid directly without going through the liquid phase.

(b) CO2 at 4.0 atm pressure and room temperature (say, 27 °C) is in vapour phase. This point (4.0 atm, 27°C) lies below the vaporation curve OC and to the right of the triple point O. Therefore, when CO2 is cooled at this point at constant pressure, the point moves perpendicular to the pressure-axis and enters the solid phase region. Hence, the CO2 vapour condenses directly to solid phase without going through the liquid phase.

(c) When the solid CO2 at – 65 °C is heated at 10 atm pressure, it is first converted into liquid. A further increase in its temperature brings it into the vapour phase. If a horizontal line at P = 10 atm is drawn parallel to the T-axis, then the points of intersection of line with the fusion and vaporization curve give the fusion and boiling points at 10 atm.

(d) Above 31.1°C, the gas cannot be liquefied. Therefore, on being compressed isothermally at 70°C, there will be no transition to the liquid region. However, the gas will depart, more and more from its perfect gas behaviour with the increase in pressure.

Question 11. 18. A child running a temperature of 101°F is given an antipyrin (i.e., a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98° F in 20 minutes, what is the average rate of extra evaporation caused by the drug ? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g-1.

Answer:SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

Question 11. 19. A ‘thermacole’ icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45°C, and coefficient of thermal conductivity of thermacole is 0.01 Js-1 m-1 °C-1 [Heat of fusion of water = 335 x 103 J kg-1 ].

Answer:  Each side of the cubical box (having 6 faces) is 30 cm = 0.30 m. Therefore, the total surface area’ of the icebox exposed to outside air is A = 6 x (0.30 m)2 = 0.54 m2. The thickness of the icebox is d = 5.0 cm = 0.05 m, time of exposure t = 6h = 6 x 3600 s and temperature difference T1 – T2 = 45°C – 0°C = 45°C.

.•. Total heat entering the icebox in 6 h is given by

Question 11. 20. A brass boiler has a base area 0.15 m2SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

 and thickness 1.0 cm. It boils water at the rate of 6.0 kg/ min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 Js-1 m-1 K-1.(Heat of vaporization of water = 2256 x 103 J kg-1 )

Answer:

Question 11. 21. Explain why:

(a) a body with large reflectivity is a poor emitter.

(b) a brass tumbler feels much colder than a wooden tray on a chilly day.

(c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace.

(d) the earth without its atmosphere would be inhospitably cold.

(e) heat systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water.

Answer: (a) According to Kirchh off’s law of black body radiations, good emitters are good absorbers and bad emitters are bad absorbers. A body with large reflectivity is a poor absorber of heat and consequently, it is also a poor emitter.

(b) Brass is a good conductor of heat, while wood is a bad conductor. When we touch the brass tumbler on a chilly day, heat starts flowing from our body to the tumbler and we feel it cold. However, when the wooden tray is touched, heat does not flow from our hands to the tray and we do not feel cold.

(c) An optical pyrometer is based on the principle that the brightness of a glowing surface of a body depends upon its temperature. Therefore, if the temperature of the body is less than 600°C, the image formed by the optical pyrometer is not brilliant and we do not get the reliable result. It is for this reason that the pyrometer gives a very low value for the temperature of red hot iron in the open.

(d) The lower layers of earth’s atmosphere reflect infrared radiations from earth back to the surface of earth. Thus the heat radiation received by the earth from the sun during the day are kept trapped by the atmosphere. If atmosphere of earth were not there, its surface would become too cold to live.

(e) Steam at 100°C possesses more heat than the same mass of water at 100°C. One gram of steam at 100°C possesses 540 calories of heat more than that possessed by 1 gm of water at 100°C. That is why heating systems based on circulation of steam are more efficient than those based on circulation of hot water.

Question 11. 22. A body cools from 80 °C to 50°C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30°C. The temperature of the surroundings is 20 °C.

Answer:

Class 11 Physics Chapter 12 Thermodynamics:

Section Name Topic Name

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

12 Thermodynamics

12.1 Introduction

12.2 Thermal equilibrium

12.3 Zeroth law of thermodynamics

12.4 Heat, internal energy and work

12.5 First law of thermodynamics

12.6 Specific heat capacity

12.7 Thermodynamic state  variables and equation of state

12.8 Thermodynamic processes

12.9 Heat engines

12.10 Refrigerators and heat pumps

12.11 Second law of thermodynamics

12.12 Reversible and irreversible processes

12.13 Carnot engine

ক্লাস 11 পদার্থবিদ্যা অধ্যায় 12 তাপগতিবিদ্যা:

বিভাগের নাম বিষয়ের নাম

12 তাপগতিবিদ্যা

12.1 ভূমিকা

12.2 তাপীয় ভারসাম্য

12.3 তাপগতিবিদ্যার জিরোথ সূত্র

12.4 তাপ, অভ্যন্তরীণ শক্তি এবং কাজ

12.5 তাপগতিবিদ্যার প্রথম সূত্র

12.6 নির্দিষ্ট তাপ ক্ষমতা

12.7 থার্মোডাইনামিক স্টেট ভেরিয়েবল এবং স্টেটের সমীকরণ

12.8 থার্মোডাইনামিক প্রক্রিয়া

12.9 তাপ ইঞ্জিন

12.10 রেফ্রিজারেটর এবং তাপ পাম্প

12.11 তাপগতিবিদ্যার দ্বিতীয় সূত্র

12.12 বিপরীত এবং অপরিবর্তনীয় প্রক্রিয়া

12.13 কার্নট ইঞ্জিন

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

Class 11 Physics Chapter 14 Oscillations

Section Name Topic Name

14 Oscillations

14.1 Introduction

14.2 Periodic and oscilatory motions

14.3 Simple harmonic motionSCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

14.4 Simple harmonic motion and uniform circular motion

14.5 Velocity and acceleration in simple harmonic motion

14.6 Force law for simple harmonic motion

14.7 Energy in simple harmonic motion

14.8 Some systems executing Simple Harmonic Motion

14.9 Damped simple harmonic motion

14.10 Forced oscillations and resonance

14.1 ভূমিকা

14.2 পর্যায়ক্রমিক এবং দোলনীয় গতি

14.3 সরল সুরেলা গতি

14.4 সরল সুরেলা গতি এবং অভিন্ন বৃত্তাকার গতি

14.5 সরল সুরেলা গতিতে বেগ এবং ত্বরণ

14.6 সরল সুরেলা গতির জন্য বল আইন

14.7 সরল সুরেলা গতিতে শক্তি

14.8 কিছু সিস্টেম সরল হারমোনিক মোশন নির্বাহ করে

14.9 স্যাঁতসেঁতে সরল সুরেলা গতি

14.10 জোরপূর্বক দোলন এবং অনুরণন

Question 14. 1. Which of the following examples represent periodic motion?

(a) A swimmer completing one (return) trip from one bank of a river to the other and back.

(b) A freely suspended bar magnet displaced from its N-S direction and released.

(c) A hydrogen molecule rotating about its centre of mass.

(d) An arrow released from a bow.

Answer:  (a) It is not a periodic motion. Though the motion of a swimmer is to and fro but will not have a definite period.

(b) Since a freely suspended magnet if once displaced from N-S direction and released, it oscillates about this position, it is a periodic motion.

(c) The rotating motion of a hydrogen molecule about its centre of mass is periodic.

(d) Motion of an arrow released from a bow is non-periodic.

Question 14. 2. Which of the following examples represent (nearly) simple harmonic motion and which represent

periodic but not simple harmonic motion?

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

(a) the rotations of earth about its axis.

(b) motion of an oscillating mercury column in a U-tube.

(c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lowermost point.

(d) general vibrations of a polyatomic molecule about its equilibrium position.

Answer: (a) Since the rotation of earth is not to and fro motion about a fixed point, thus it is periodic but not S.H.M.

(b) It is S.H.M.

(c) It is S.H.M.

(d) General vibrations of a polyatomic molecule about its equilibrium position is periodic but non SHM. In fact, it is a result of superposition of SHMs executed by individual vibrations of atoms of the molecule.

Question 14. 3. Fig. depicts four x-t plots for linear motion of a particle. Which of ike plots represent periodic motion? What is the period of motion (in case of periodic motion)?

Answer: Figure (b) and (d) represent periodic motions and the time period of each of these is 2 seconds, (a) and (c) are non-periodic motions.

Question 14. 4. Which of the following function of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion (to is any positive constant).

(a) sin wt – cos wt (b) sin2 wt (c) 3 cos -2 cos (π/4-2 wt) (d) cos wt + cos 3 wt + cos 5 wt

(e) exp (- w2t2) (f) 1 + wt + w2t2.

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

Answer:  The function will represent a periodic motion, if it is identically repeated after a fixed interval of time and will represent S.H.M if it can be written uniquely in the form of a cos

(e) e-w2t2 . It is an exponential function which never repeats itself. Therefore it represents non-periodic motion.

(f) 1 + wt + w2t2 also represents non periodic motion.

Question 14. 5. A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is

(a) at the end A,

(b) at the end B,

(c) at the mid-point of AB going towards A,

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

(d) at 2 cm away from B going towards A,

(e) at 3 cm away from A going towards B, and (f)  at 4 cm away from B going towards A.

Answer:  In the fig. (given below), the points A and B, 10 cm apart, are the extreme positions of the particle in SHM, and the point O is the mean position. The direction from A to B is positive, as indicated.

(a) At the end A, i.e., extreme position, velocity is zero, acceleration and force are directed towards O and are positive.

(b) At the end B, i.e., second extreme position, velocity is zero whereas the acceleration and force are directed towards the point O and are negative.

(c) At the mid point O, while going towards A, velocity is negative and maximum. The acceleration and force both are zero.

(d) At 2 cm away from B, that is, at C and going towards A: v is negative; acceleration and F, being directed towards O, are also negative.

(e) At 3 cm away from A, that is, at D and going towards B: v is positive; acceleration and F, being directed towards O, are also positive.

(f)  At a distance of 4 cm away from A and going towards A, velocity is directed along BA, therefore, it is positive. Since acceleration and force are directed towards OB, both of them are positive.

Question 14. 6. Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?

(a) a = 0.7x (b) a = – 200 x2

(c) a = – 10x (d) a = 100 x3

Answer:  Only (c) i.e., a = – 10x represents SHM. This is because acceleration is proportional and opposite to displacement (x).

Question 14. 7. The motion of a particle executing simple harmonic motion is described by the displacement function. x(t) = A cos (wt +Ф ).If the initial (t = 0) position of the particle is 1 cm and its initial velocity is w cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is π s-1. If instead of the cosine function, we choose the sine function to describe the SHM: x = B sin (wt + α), what are the amplitude and initial phase of the particle with the above initial conditions?

Answer: 

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

Question 14. 8. A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?

Answer:

Question 14. 9. A spring having with a spring constant 1200 Nm-1 is mounted on a horizontal table as shown in Fig. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released. Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.

Answer:

Question 14. 10. In Exercise 9, let us take the position of mass when the spring is unstreched as x = 0, and the direction from left to right as the positive direction of x – axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is (a)at the mean position,(b)at the maximum stretched position, and (c)at the maximum compressed position.In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

Answer: 

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

Note: The functions neither differ in amplitude nor in frequency . They differ only in initial phase.

Question 14. 11. The following figures correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e., clockwise or anticlockwise) are indicated on each figure.

Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P in each case.

Answer:

Question 14. 12. Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anti-clockwise in every case:

(x is in cm and t is in s)

(a) x = – 2 sin (3t + π /3)

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

(b) x = cos (π /6 – t)

(c) x = 3 sin (2πt + π /4)

(d) x = 2 cos π t.

Answer:

Question 14. 13. Figure (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Figure – (b) is stretched by the same force F.

(a) What is the maximum extension of the spring in the two cases?

(b) If the mass in Fig. (a) and the two masses in Fig. (b) are released free, what is the period of oscillation in each case?

Answer:

Question 14. 14. The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rev/min, what its maximum speed?

Answer:

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

Question 14. 15. The acceleration due to gravity on the surface of moon is 1.7 ms-2. What is the time period of a simple pendulum on the surface of moon if its time-period on the surface of Earth is 3.5 s? (g on the surface of Earth is 9.8 ms-2 .)

Answer:

Question 14. 16.

Answer:

(c) The wristwatch uses an electronic system or spring system to give the time, which does not change with acceleration due to gravity. Therefore, watch gives the correct time.

(d) During free fall of the cabin, the acceleration due to gravity is zero. Therefore, the frequency of oscillations is also zero i.e., the pendulum will not vibrate at all.

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

Question 14. 17. A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?

Answer: In this case, the bob of the pendulum is under the action of two accelerations.

Question 14. 18.  A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρ1 . The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period

where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).

Answer:

Question 14. 19. One end of a U-tube containing mercury is connected to a suction pump and the other end to the atmosphere.

A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.

Answer: The suction pump creates the pressure difference, thus mercury rises in one limb of the U-tube. When it is removed, a net force acts on the liquid column due to the difference in levels of mercury in the two limbs and hence the liquid column executes S.H.M. which can be expressed as:

Consider the mercury contained in a vertical U-tube upto the level P and Q in its two limbs.

Let P = density of the mercury.

L = Total length of the mercury column in both the limbs.

A = internal cross-sectional area of U-tube. m = mass of mercury in U-tube = LAP.

Assume, the mercury be depressed in left limb to F by a small distance y, then it rises by the same amount in the right limb to position Q’.

.’. Difference in levels in the two limbs = P’ Q’ = 2y.

:. Volume of mercury contained in the column of length 2y = A X 2y

.•. m – A x 2y x ρ.

If W = weight of liquid contained in the column of length 2y.

Then W = mg = A x 2y x ρ x g

This weight produces the restoring force (F) which tends to bring back the mercury to its equilibrium position.

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

Question 14. 20. An air chamber of volume V has a neck area of cross-section into which a ball of mass m just fits and can move up and down without any friction (Fig.). Shaw that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal.

Answer:  Consider an air chamber of volume V with a long neck of uniform area of cross-section A, and a frictionless ball of mass m fitted smoothly in the neck at position C, Fig. The pressure of air below the ball inside the chamber is equal to the atmospheric pressure.

Increase the pressure on the ball by a little amount p, so that the ball is depressed to position D, where CD = y.

There will be decrease in volume and hence increase in pressure of air inside the chamber. The decrease in volume of the air inside the chamber, ΔV = Ay

Note. If the ball oscillates in the neck of chamber under isothermal conditions, thru E = P = picture of air inside the chamber, when ball is at equilibrium position. If the ball oscillate in the neck of chamber under adiabatic conditions, then E = gP. where g = Cp/Cv.

Question 14. 21. You are riding an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of (a) the spring constant k and (b) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg. g = 10 m/s2.

Answer:

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

Question 14. 22.  Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.

Answer:  Let the particle executing SHM starts oscillating from its mean position. Then displacement equation is

Question 14. 23. A circular disc, of mass 10 kg, is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations of found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant) a is defined by the relation J = – αθ, where J is the restoring couple and θ the angle of twist).

Answer:

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

Question 14. 24. A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm (b) 3 cm (c) 0 cm

Answer:

Question 14. 25. A mass attached to a spring is free to oscillate, with angular velocity w, in a horizontal plane without friction or damping. It is pulled to a distance x0 and pushed towards the centre with a velocity v0 at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters w, x0 and v0.

Answer:

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

Class 11 Physics Chapter 15 Waves:

Section Name Topic Name

15 Waves

15.1 Introduction

15.2 Transverse and longitudinal waves

15.3 Displacement relation in a progressive wave

15.4 The speed of a travelling wave

15.5 The principle of superposition of waves

15.6 Reflection of waves

15.7 Beats

15.8 Doppler effect

ক্লাস 11 পদার্থবিদ্যা অধ্যায় 15 তরঙ্গ:

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

বিভাগের নাম বিষয়ের নাম

15 তরঙ্গ

15.1 ভূমিকা

15.2 অনুপ্রস্থ এবং অনুদৈর্ঘ্য তরঙ্গ

15.3 একটি প্রগতিশীল তরঙ্গে স্থানচ্যুতি সম্পর্ক

15.4 একটি ভ্রমণ তরঙ্গের গতি

15.5 তরঙ্গের সুপারপজিশনের নীতি

15.6 তরঙ্গের প্রতিফলন

15.7 বিট

15.8 ডপলার প্রভাব

Question 15. 1.  A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?

Answer:

Question 15. 2. A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 ms-1 ? (g = 9.8 ms-2 )

Answer:  Here, h = 300 m, g = 9.8 ms-2  and velocity of sound, v = 340 ms-1 Let t1 be the time taken by the stone to reach at the surface of pond.

Question 15. 3.  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 2 0°C = 340 ms-1.

Answer:

Question 15. 4.

Answer:

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

Question 15. 5. You have learnt, that a travelling wave in one dimension is represented by a function y = f(x, t), where x and t must appear in the combination x-vtorx + vt i.e., y =f (x ± vt). Is the converse true? That is, does every function of (x – vt) or (x + vt) represent a travelling wave? Examine, if the following functions for y can possibly represent a travelling wave?

Answer: No, the converse is not true. The basic requirement for a wave function to represent a travelling wave is that for all values of x and t, wave function must have a finite value. Out of the given functions for y none satisfies this condition. Therefore, none can represent a travelling wave.

Question 15. 6.  A bat emits ultrasonic sound of frequency 1000 kHz in air. If this sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air = 340 ms-1 and in water = 1486 ms-1.

Answer:

Question 15. 7. A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in a tissue in which the speed of sound is 1.7 km s-1? The operating frequency of the scanner is 4.2 MHz.

Answer: Here speed of sound => υ = 1.7 km s-1 = 1700 ms-1 and

frequency υ= 4.2 MHz = 4.2 x 106 Hz

.’. Wavelength, A = υ/V = 1700/(4.2 x 106) =4.1 x 10-4 m.

Question 15. 8. A transverse harmonic wave on a string is described by

y(x, t) = 3.0 sin (36 t + 0.018 x + π/4)

where x and y are in cm and t in s. The positive direction of x is from left to right.

(a) Is this a travelling wave or a stationary wave? If it is travelling, what are the speed and direction of its propagation ?

(b) What are its amplitude and frequency?

(c) What is the initial phase at the origin?

(d) What is the least distance between two successive crests in the wave?

Answer:

Question 15. 9. For the wave described in Exercise 8, plot the displacement (y) versus (t) graphs for x = 0, 2 and 4 cm. What are the shape of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another : amplitude, frequency or phase?SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

Answer: The transverse harmonic wave is

Using the values of t and y (as in the table), a graph is plotted as under The graph obtained is sinusoidal.

Similar graphs are obtained for y x = 2 cm and x = 4 cm. The (incm) oscillatory motion in the travelling wave only differs in respect of phase. Amplitude and frequency of oscillatory motion remains the same in all the cases.

Question 15. 10. For the travelling harmonic wave

y(x, t) = 2.0 cos 2π (10t – 0.0080x + 0.35)

where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of

(a) 4 m (b) 0.5 m

(c) λ/2 (d) 3λ/4.

Answer:

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

Question 15. 11. The transverse displacement of a string (clamped at its two ends) is given by y(x, t) = 0.06 sin 2π /3 x cos (120π t)

where x, y are in m and t in s. The length of the string is 1.5 m and its mass is 3 x 10-2 kg. Answer the following:

(i) Does the function represent a travelling or a stationary wave?

(ii) Interpret the wave as a superimposition of two waves travelling in opposite directions. What are the wavelength, frequency and speed of propagation of each wave?

(iii) Determine the tension in the string.

Answer:  The given equation is

y(x, t) = 0.06 sin 2π/3 x cos 120 πt …(1)

(i)As the equation involves harmonic functions of x and t separately, it represents a stationary wave.

Question 15. 12. (i) For the wave on a string described in Question 11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers, (ii) What is the amplitude of a point 0.375 m away from one end?

Answer: (i) For the wave on the string described in questions we have seen that l = 1.5 m and λ = 3 m. So, it is clear that λ = λ /2 and for a string clamped at both ends, it is possible only when both ends behave as nodes and there is only one antinode in between i.e., whole string is vibrating in one segment only.

(a) Yes, all the sring particles, except nodes, vibrate with the same frequency v = 60 Hz.

(b) As all string particles lie in one segment, all of them are in same phase.

(c) Amplitude varies from particle to particle. At antinode, amplitude = 2A = 0.06 m. It gradually falls on going towards nodes and at nodes, and at nodes, it is zero.

(ii) Amplitude at a point x = 0.375 m will be obtained by putting cos (120 πt) as + 1 in the wave equation.

Question 15. 13. Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave or (iii) none at all

Answer: (a) It represents a stationary wave.

(b) It does not represent either a travelling wave or a stationary wave.

(c) It is a representation for the travelling wave.

(d) It is a superposition of two stationary wave.

Question 15. 14. A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 x 10-2 kg and its linear mass density is 4.0 x 10-2 kg m-3SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?

Answer:

Question 15. 15. A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a turning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effect may be neglected.

Answer:

[Approximation has been used because edge effect is being ignored. Moreover, we know that in the case of a closed organ pipe, the second resonance length is three times the first resonance length.]

On simplification, n = 1

Now, (2n-1)υ/4l1 = 340. Substituting values

(2 x 1 -1) υ x 100/4 x 25.5 = 340 or υ =346.8 ms-1

Question 15. 16. A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod is given to be 2.53 k Hz. What is the speed of sound in steel?

Answer:  Here, L = 100 cm = 1m, v = 2.53 k Hz = 2.53 x 103 Hz

When the rod is clamped at the middle, then in the fundamental mode of vibration of the rod, a node is formed at the middle and ant mode is formed at each end.

Question 15. 17. A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 ms-1).

Answer:  Here length of pipe, 1 = 20 cm = 0.20 m, frequency v = 430 Hz and speed of sound in air υ = 340 ms-1

Question 15. 18. Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3Hz. If the original frequency of A is 324 Hz, what is the frequency of B?

Answer: Let υ1 and υ2 be the frequencies of strings A and B respectively.

Then, υ1 = 324 Hz, υ2 = ?

Number of beats, b = 6

υ2 = υ1 ± b = 324 ± 6 !.e., υ2 = 330 Hz or 318 Hz

Since the frequency is directly proportional to square root of tension, on decreasing the tension in the string A, its frequency υ1 will be reduced i.e., number of beats will increase if υ2 = 330 Hz. This is not so because number of beats become 3.

Therefore, it is concluded that the frequency υ2 = 318 Hz. because on reducing the tension in the string A, its frequency may be reduced to 321 Hz, thereby giving 3 beats with υ2 = 318 Hz.

Question 15. 19. Explain why (or how):

(a) in a sound wave, a displacement node is a pressure antinode and vice versa.

(b) bats can ascertain distances, directions, nature and sizes of the obstacles without any “eyes”.

(c) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes.

(d) solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and

(e) the shape of a pulse gets distorted during propagation in a dispersive medium.

Answer:  (a) In a sound wave, a decrease in displacement i.e., displacement node causes an increase in the pressure there i.e., a pressure antinode is formed. Also, an increase in displacement is due to the decrease in pressure.

(b) Bats emit ultrasonic waves of high frequency from their mouths. These waves after being reflected back from the obstacles on their path are observed by the bats. These waves give them an idea of distance, direction, nature and size of the obstacles.

(c) The quality of a violin note is different from the quality of sitar. Therefore, they emit different harmonics which can be observed by human ear and used to differentiate between the two notes.

(d) This is due to the fact that gases have only the bulk modulus of elasticity whereas solids have both, the shear modulus as well as the bulk modulus of elasticity.

(e) A pulse of sound consists of a combination of waves of different wavelength. In a dispersive medium, these waves travel with different velocities giving rise to the distortion in the wave.

Question 15. 20. A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air. (i) What is the frequency of the whistle for a platform observer when the train (a) approaches the platform with a speed of 10 ms~1. (b) recedes from the platform with a speed of 10 ms-1(ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 ms-1   .

Answer: Frequency of whistle, v = 400 Hz; speed of sound, υ= 340 ms-1 speed of train, υs= 10 ms1

(i) (a) When the train approaches the platform (i.e., the observer at rest),

(ii) The speed of sound in each case does not change.It is 340 ms-1 in each case.

Question 15. 21. A train, standing in a station-yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10 ms-1. What are the frequency, wavelength, and speed of sound for an observer standing on the station’s platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 ms-1? The speed of sound in still air can be taken as 340 ms-1?

Answer:

Question 15. 22. A travelling harmonic wave on a string is described by y (x, t) = 7.5 sin (0.0050x + 12t + π /4)

(a) what are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1s? Is this velocity equal to the velocity of wave propagation?

(b) Locate the points of the string which have the same transverse displacement and velocity as the x = 1 cm point at t = 2s, 5s and 11s.

Answer:

Question 15. 23. A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium, (a) Does the pulse have a definite (i) frequency, (ii) wavelength, (iii) speed of propagation? (b) If the pulse rate is 1 after every 20 s, (that is the whistle is blown for a split of second after every 20 s), is the frequency of the note produced by the whistle equal to 1/20 or 0.05 Hz?

Answer:  (a) In a non dispersive medium, the wave propagates with definite speed but its wavelength of frequency is not definite.

(b) No, the frequency of the note is not 1/20 or 0.50 Hz. 0.005 Hz is only the frequency ‘ of repetition of the pip of the whistle.

Question 15. 24. One end of a long string of linear mass density 8.0 x 10-3 kg m-1 is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At t = 0, the left end (fork end) of the string x = 0 has zero transverse displacement (y = 0) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as function of x and t that describes the wave on the string.

Answer: Here, mass per unit length, g = linear mass density = 8 x 10-3 kg m-1;

Tension in the string, T = 90 kg = 90 x 9.8 N= 882 N;

Question 15. 25. A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h-1. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 ms-1.

Answer:

Question 15. 26. Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically the speed of S wave is about 4.0 km s-1 . A seismograph records P and S waves from an earthquake. The first P wave arrives 4 min before the first S wave. Assuming the waves travel in straight line, at what distance does the earthquake occur?

Answer: Here speed of S wave, υs = 4.0 km s-1 and speed of P wave, υp = 8.0 km s-1. Time gap between P and S waves reaching the resimograph, t = 40 min = 240 s.

Let distance of earthquake centre = sKm

Question 15. 27. A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall ?

Answer:  Here, the frequency of sound emitted by the bat, υ = 40 kHz. Velocity of bat, υs = 0.03 υ, where υ is velocity of sound. Apparent frequency of sound striking the wall

মেডিকেল (NEET) ইঞ্জিনিয়ারিংয়ে (JEE) এন্ট্রান্স ভর্তি চলছে : এক ক্লাসরুমে সব সাবজেক্ট 

ডক্টরেট উপাধিপ্রাপ্ত, গোল্ড মেডেলিস্ট, সাবজেক্ট এক্সপার্ট থেকে শিখতে হলে এখুনি ভর্তি হন।   ত্রিপুরাতে সব থেকে কম ফী ! সময় এবং টাকা দুটোই বাঁচবে 

৯৮ চ্যাপ্টার টেস্ট সিরিজ , সপ্তাহে ৬ দিন ক্লাস 

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

CLASS 12 TBSE CBSE NEET JEE ADMISSION 2024

LLB BBA CUET TPSC COACHING

ALL SUBJECTS IN ONE CLASSROOM

CLASS STARTS 1st April 2023

Learn with Gold Medalist, Doctorate, NITian, Asst Professor, AIR-1, Toppers Exp. Guide

What We Guide: School Of Learning Coaching

TBSE CBSE ICSE Board from Class VI-XII, College, University Students (Arts, Science and Commerce), Drawing, Music also available here.

SCHOOL OF LEARNING COACHING AGARTALA
CRACK NEET JEE TBJEE WBJEE IIT NIT ENTRANCE
WITH SUBJECT EXPERTS & DOCTORATE, GOLD MEDALIST
LOWEST FEE IN INDIA

We also guide Common University Entrance Test (CUET)- All India Level Entrance for the admission in Colleges and Universities.

*BA BSC LLB BBA BA-BEd BSC-BEd, IMD

Medical (NEET) and Engineering(JEE) Entrance,TPSC-Civil Services, Competitive Exams guidance also done by the experts.

Guidance for Teaching Job: Tripura TET & CTET, UGC NET/SLET

Competitive Govt Job Coaching

Vocational Training Course

• Medical & Engineering Entrance

Board + NEET + TBJEE

JEE (Main & Advanced)

• CUET (Common University Entrance Test)

Degree College Entrance

Eligibility: Class 12 Final Students

Master Degree Entrance

Eligibility: Degree 5-6th Semester Final Students

• College Tuition

B.A B.Sc B.Com (Pass & Honours) all subjects

• University Tuition

M.A. M.SC, M.COM

• TPSC-COMPETITIVE EXAMS
• COMPETITIVE EXAMS