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Movements of Mimosa pudica, prayer plant, venus flytrap, etc.

(1) Mean (or average) speed of molecules of a gas : The mean speed of gas molecules is defined as the arithmetic mean of the speeds of all molecules of the gas at a given temperature.

(2) Mean square speed of molecules of a gas : The mean square speed of gas molecules is defined as the arithmetic mean of the squares of the speeds of all molecules of the gas at a given temperature.

(3) Root-mean-square speed of molecules of a gas : The root-mean-square (rms) speed of gas molecules is defined as the square root of the arithmetic mean of the squares of the speeds of all molecules of the gas at a given temperature.

If there are N molecules in an enclosed pure gas and v₁ , v₂ , v₃ , …, v_N are the speeds of different molecules,

1. the mean speed, \(\bar v\) = \(\frac{v_1+v_2+,,,+v_N}N\)

2. the mean square speed,

\(\overline {v^2}\) = \(\frac{v_1^2+v_2^2+,,,+v_N^2}N\)

3. the rms speed, v_rms = \(\sqrt{\overline{v^2}}\)

[Note : The mean square velocity is numerically equal to the mean square speed. Similarly, the rms velocity is numerically equal to the rms speed. But in random motion, the mean velocity would be statistically zero, but the mean speed cannot be zero. ]

Data : T = 273 K, M₀ = 28 × 10^-3 kg/mol, R = 8.314 J/mol.K, k_B = 1.38 × 10^-23 J/K

(i) The KE per mole = \(\frac32RT\)

 = \(\frac32\)(8.314)(273) = 3.404 × 10³ J/mol

(ii) The KE per unit mass = \(\frac32\)\(\frac{RT}{M_0}\)

= \(\frac32\). \(\frac{(8.314)(273)}{28\times10^{-3}}\) = 1.216 × 10⁵ J/kg

(iii) The KE per molecule

= \(\frac32\)k_BT

= \(\frac32\). (1.38 x 10^-23)(273) = 5.651 x 10^-21)

Data : ρ = 1.44 kg/m³ , v_rms = 456.4 m/s

The pressure of oxygen,

P = \(\frac13\)ρv²_rms

= \(\frac12\)(1.44)(456.4)² = 9.98 × 10⁴ Pa

(1) Emissive power or radiant power of a body (symbol, R) : The emissive power or radiant power of a body at a given temperature is defined as the quantity of radiant energy emitted by the body per unit time per unit surface area of the body at that temperature.

(2) Coefficient of emission (or emissivity) of a body (symbol, e) : The coefficient of emission (or emissivity) of a body is defined as the ratio of the emissive power of the body (R) to the emissive power of a perfect blackbody (R ) at the same temperature as that of the body. e = \(\frac{R}{R_b}\)

[Note : The SI unit and dimensions of emissive power are the watt per square metre (W/m² Or Js^-1 m^-2) and [M¹L°T^-3]. The coefficient of emission is a dimensionless and unitless quantity. For a perfect blackbody, e = 1 and for a perfect reflector, e = 0.]

Characteristics of blackbody radiation spectrum :

1. The emissive power R_λ for every wavelength λ increases with increasing temperature.

2. Each curve has a characteristic form with a maximum for R_λ at a certain wavelength λ_m .

3. λ_m depends on the absolute temperature of the body and, with increasing temperature, shifts towards shorter wavelengths, i.e., towards the ultraviolet end of the spectrum.

4. λ_mT = a constant.

5. The area under each curve gives the total radiant power per unit area of a blackbody at that temperature and is proportional to T⁴ , (Stefan-Boltzmann law)

Notes :

1. Experimental work on the distribution of energy in blackbody radiation, was carried out by German physicists Otto Lummer (1860-1925), Wilhelm Wien (1864-1928) and Ernst Pringsheim (1859-1917).

2. Explanation of the radiation spectrum, given by Wien on the basis of thermodynamics could account only for the short wavelength region. The formula obtained by Rayleigh and Jeans, on the basis of the equipartition of energy could account only for long wavelength region. Planck’s empirical formula, put forward in 1900, could account for the entire spectrum.

R_b = σT⁴= 5.67 × 10^-8 × (10³)⁴

= 5.67 × 10⁴ \(\frac{W}{m^2}\) is the required emissive power.

Correct Option is (C) twice the rms speed at 200 K

We know that ,

\(V_{rms} = \sqrt \frac{3RT}{M}\)

\(V_1 \propto \sqrt{T_1}\)

\(\frac{V_2}{V_1} = \sqrt \frac{800}{200}\)

\(V_2 = 2V_1\)

(C) twice the rms speed at 200 K

Data : Mass of hydrogen = 4 grams,

V = 16 litres = 16 × 10^-3 m³,

T = 273 + 10 = 283 K, M₀ = 2 g/mol,

R = 8.314 J/mol.K

Number of moles (n) = \(\frac{mass}{M_0}\) = \(\frac42\) = 2

PV = nRT 

∴ P = \(\frac{nRT }V\) = \(\frac{2\times8.314\times283}{16\times10^{-3}}\)

P = 2.941 × 10⁵ N/m²

This is the pressure exerted by the gas.

Correct Option is (C) \(\frac{400}{\sqrt3} \ m/s\)

\(V_{rms} = \sqrt \frac{3RT}{M}\)

\(V_{rms} \times \sqrt T\)

\(\frac{V_2}{V_1} = \sqrt \frac{T_2}{T_1}\)

\(V_2 = V_1 \times \sqrt \frac{T_2}{T_1}\)

= \(200 \sqrt \frac{400}{300}\)

\(V_2 = 400 \sqrt \frac{1}{3} \ m/s\)

(C) \(\frac{400}{\sqrt3}\)m/s

The coefficient of emission of the body,

e = \(\frac{R}{R_b}\) = \(\frac{2000}{1000}\) = 0.2

Data : M = 10 grams = 10 × 10^-3 kg,

S_v = 0.16 kcal/kg.K, J = 4186 J/kcal

Rise in the temperature of the gas, 

∆T = 305 – 300 = 5K

∆E = J S_v M∆T = (4186)(0.16)(10 × 10^-3) × (5) J 

= 33.49 J 

The increase in the internal energy of the gas, ∆E = 33.49 J

Correct Option is (D) \(400^\circ C\)

We know that ,

\(P \propto T\)

\(\frac{P_1}{P_2} = \frac{T_1}{T_2}\)

\(\frac{T_1}{T_2} = \frac{P_2}{P_1}\)

\(\frac{P_2}{P_1} = \frac{T_2}{T_1}\)

\(\because P_2 = P_1 + P_1 \times \frac{0.5}{100}\)

\(P_2 = 1.005 P_1\)

\(\frac{1.005 \ P_1}{P_1} = \frac{T_1 + 2}{T_1}\)

\(1.005 T_1 = T_1 + 2\)

\(0.005T_1 = 2\)

\(T_1 = \frac{2000}{0.005}\)

\(T_1 = 400^\circ C\)

Correct option is (D) 400 K.

Data : C_v = 12.5 J/mol.K, 

R = 8.31 J/mol.K The molar specific heat of helium at constant press-ure,

C_p = C_v + R = (12.5 + 8.31) J/mol.K 

= 20.81 J/mol.K.

(i) It is responsible for the climate changes in different parts of the world and it has forced people to migrate from one place to the other. 

(ii) It has affected blooming season of the different plants.

The rising temperature will lower water quality in lakes and oceans through a fall in oxygen concentrations, release of phosphorus from sediments and increased thermal stability due to which many marine species will either die or they will become extinct.

The anomalous behaviour of lithium is due to the exceptionally small size of the atom and high polarizing power, which is a ratio of charge to radius and hydration energy.

• Both react with nitrogen to form nitrides. 
• Both react with O₂ to form monoxides.
• Both the elements have the tendency to form covalent compounds. 
• Both can form complex compounds.

Answer: (b) Lithium

(b) magnesium (diagram pending)

(d) aircraft parts

Answer: (d) Mg

Metals lies on the extreme left, metalloids lie in the middle and non-metals lie on the right side.

Boron, silicon, germanium and arsenic.

Metalloids: Boron (B) and Silicon (Si).

Merlin’s class teacher arranged a trip to Ooty. Merlin was so excited, when she stepped into the toy train. The train moved with a jerk. As it moved, they saw monkeys racing along the tracks. A monkey tried to snatch a banana from a girl. She let out a scream and moved back. The train stopped suddenly because there was a baby elephant on the track. It was a memorable trip for everyone. Muthu, one of the boys, acted wisely by getting the baby elephant off the track. All were excited about this trip to Ooty.

Merlin writes a letter to Malli about her trip to Ooty with her friends. Their class teacher Mrs. Geetha arranged this trip. Merlin was so excited when she stepped into the toy train. The train moved slowly allowing the children to have a glimpse of,the beautiful landscape. One of the boys, Muthu wanted to walk along the side of the train. But his teacher scolded him to get back to his seat. The scene outside was beautiful with the purple blue mountains forming a lovely backdrop to the green fields and tea estates. There were monkeys racing along the tracks.

One of the monkeys tried to snatch a banana from a girl. She let out a scream and moved back Suddenly the train stopped. Everyone got down to see what happened. There was a baby elephant sitting on the track. Everyone tried to persuade the baby off the track with the bunch of bananas. But it was no use. Muthu acted wisely with a presence of mind. He took a bunch of bananas near the baby elephant. As it moved forward to eat the bananas, Muthu moved backwards. He kept doing this, till the baby was on the side of the tracks. Everyone got into the train and the train started to move.

Merlin ended the letter saying that it was a fantastic trip. It was enchanting to hear the sound around them. The whole group was so excited about this trip.

Muthu and his friends were the first to step out to see what had happened.

(b) wanted to see what had happened

(a) they wanted to move the baby elephant out of the track

(b) train never passes through such places

The maximum summer temperature of udhagamandalam is 25°c.

Correct answer is (b) 2240 m

1. False. 

Correct: The kingdoms of (ihazni and Ghor were established by the Turks.

2. True.

3. False.

Correct: Iltutmish crush the power of the governor of Bengal.

4. False. 

Correct: Razia Sultan was brave, intelligent and just woman.

5.False.

Correct : alban made the monarchy absolute and all powerful.

6. True.

7. False. 

Correct: Muhammad Ghori was the founder of the Turkish rule in India.

8.False.

Corrent: Muhammad Ghori was not as great a military leader as Mahmud of Ghazni.

Correct answer is (c) D.P.D. is zero

Protoplasm of the plant cell shrink is Plasmolysis

Correct answer is (c) deplasmolysis

When cell is plasmolysed, protoplast of cell shrinks and recedes from the cell wall thus gap is observed between cell wall and protoplast.

When water absorbed by root hair passes across the roots through cell wall and intercellular spaces of root cortex it is called apoplast pathway.

(a) induces partial stomatal closure for two weeks.

When a stem of potted plant is cut above the soil, xylem sap is seen oozing through cut end, that proves presence of root pressure.

(b) decreases the pH level

Analysis of plant ash contents is indication of absorbed minerals.

(c) Abscisic acid

When stomata closes during the day and open during the night, such type of stomata is called scotoactive type of stomata. When stomata open in the day and close during the night, this type of opening of stomata is called photoactive type. 

xerophytes based on stomata is hypostomatic

The epistomatic

Ions of phosphorus, sulphur and nitrogen are remobilized from older plants, parts (leaves) to younger parts.

(b) hydrolyses reaction

The salts present in the soil dissolve in the irrigated water and form hypertonic solution outside the root hairs of the plant and the root hairs cannot absorb water from hypertonic solution, since water molecules cannot move from hypertonic solution to hypotonic solution in the cells of root hair. Hence the plants become wilt even the field is irrigated.