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• A slight change in temperature causes considerable change in pressure as well as volume of the gas. 
• Therefore, two principal specific heats are defined for a gas viz., specific heat capacity at constant volume (S_V) and specific heat capacity at constant pressure (S_p).

A solid or a liquid when heated, does not undergo any change in the volume or pressure. But in case of a gas, both the pressure and volume change on heating. Therefore, specific heat of a gas is defined either at constant volume or at constant pressure and hence a gas has two specific heats.

For crystalline solids, the specific heat capacity is 6 cal mole^-1 K^-1 or 3 R at room temperature. Specific heat capacity decreases rapidly at low temperature

No, water and ice do not have the same values of specific heats : 

For water c = 1 cal g^-1 °C^-1 

For ice c = 0.5 al g^-1 °C^-1 .

1. Liquid-in-glass thermometer depends on the change in volume of the liquid with temperature.

2. When the bulb is heated, the liquid in a glass bulb expands upward in a capillary tube.

3. The liquid is such that it is easily seen and expands (or contracts) rapidly and by a large amount over a wide range of temperature.

4. Most commonly used liquids are mercury and alcohol as they remain in liquid state over a wide range. Mercury freezes at -39 °C and boils at 357 °C; alcohol freezes at -115 °C and boils at 78 °C.

At new temperature T °C,

change in length of rod is,

∆L_S = L₀(α_S∆T) = 50.00(12 × 10^-6 × 100)

= 0.06 cm.

Hence, the actual length of rod at T °C,

L_S = 50.06 cm

Due to change in temperature, along with rod, the scale will also increase in length.

For aluminium, at T °C,

∆L_a = L₀(α_a∆t) = 100.00 × 24 × 10^-6 × 100

= 0.24 cm

∴ Length of the scale will be,

L_a = 100.24 cm

As, the expansion in scale is more than that in rod, the reading recorded by the scale at t°C will be less than 50 cm.

Due to weakened bonds liquids do not possess definite shape but have definite volume.

A gas can be heated under two definite conditions, hence it has two principal specific heats which are : 

(i) Specific heat at constant volume. 

(ii) Specific heat at constant pressure.

(b) A system X is in thermal equilibrium with Y but not with Z. Systems Y and Z are not in thermal equilibrium with each other.

(d) A system X is neither in thermal equilibrium with Y nor with Z. The system Y and Z may be in thermal equilibrium with each other

1. Resistance thermometer uses the change of electrical resistance of a metal wire with temperature.

2. It measures temperature accurately in the range -2000 °C to 1200 °C is best for steady temperatures.

3. It is bulky and hence not easily portable.

Correct option is: (B) 120.18 cm

Correct option is: (D) Invar steel

The solid particles possess potential energy due to the interatomic forces that hold the particles together at some mean fixed positions.

When two or more systems/bodies are in thermal equilibrium, their temperatures are same. This principle is used to measure the temperature of a system by using a thermometer.

Sodium carbonate.

1. False. 

Coefficient of linear expansion is different for different substances. 

2. True.

1. Matter consists of particles which are in continuous vibrational motion and thereby possess kinetic energy.

2. When external energy is provided to these particles, internal energy of particles increases.

3. This increase in internal energy is in the form of increased kinetic energy of atoms/molecules and raises temperature of body (except at melting or boiling point of the body).

4. Greater the kinetic energy, faster the atoms/ molecules move and temperature of body becomes higher.

5. Thus, temperature of body is directly proportional to its kinetic energy. Hence, temperature is said to be a measure of average kinetic energy of the atoms and molecules of the body.

(a) kinetic energy of random motion of molecules.

1. A change in the temperature of a body causes change in its dimensions.

2. The increase in the dimensions of a body due to an increase in its temperature is called thermal expansion.

3. There are three types of thermal

expansion:

• Linear expansion
• Areal expansion
• Volume expansion

(i) Water is used as an effective coolant. 

(ii) How water is used in rubber bottles for fomentation. It has more thermal capacity

We know that PV = nRT   …(i) 

Due to friction between tyre and rod the temperature of the tyre and the air inside it increases, since volume of the tyre does not change (practically) 

∴ by equation (i), 

∴ pressure increase

Correct option is: (A) 2.4 cm²

D) both (a) and (b) are true

Correct option is: (B) 1.9 litre

1. The atoms in a solid vibrate about their mean positions.

2. When heated, they vibrate faster and force each other to move a little farther apart. This results into expansion.

3. The molecules in a liquid or gas move with certain speed.

4. When heated, they move faster and force each other to move a little farther apart. This results in expansion of liquids and gases on heating.

5. The expansion is more in liquids than in solids; gases expand even more.

Correct option is  C) 1.5

Correct option is  A) 30°

The speedometer measures the instantaneous speed of the vehicle at a given instant of time.

Yes,it can be so in case of uniform circular motion of an object.

Gas is cooled when expanded due to decrease in internal energy.

(a) Critical angle r = 90° 

(b) Total internal reflection.

As light goes from a denser to rarer medium, if the value of the angle of incidence increases, then the value of the angle of refraction also increases. But after a specific angle of incidence called the critical angle, the light gets reflected back into the denser medium.

The ray of light incident at the boundary separating the two media bends away from the normal on refraction. Here, the angle of refraction r, is greater than the angle of incidence i.

Now _an_w = \(\frac{sin\,i}{sin\,r}\) < 1. Here, _an_w is the refractive index of sin r air with respect to water. As _an_w is constant, r increases as i increases. For r = 90°, the ray travels long the boundary. If i is increased further, as r cannot be greater than 90°, light does not enter air. There is no refraction of light and all the light enters water on reflection. This is called total internal reflection.

For r = 90°, _an_w = \(\frac{sin\,i}{sin\,90^\circ}\) = sin i. This angle i is sin 90° called the critical angle.

Here,

r = 49 x 10^-5°C^-1;

ΔT = 30°C

∴ Fractional change, Δρ/ρ = γΔT

= 49 x 10^-5 x 30 = 1.47 x 10^-2

D) total internal reflection

It is so because steam has extra heat of amount 540 cal per gram as its latent heat. It causes more severe burns.

The area of the hole will increase. The hole will expand in same manner as the brass if it was present in place of the hole

The heat contents of the tea gets shared with sugar on adding sugar to it. Due to which, the temperature of the tea decreases

Mass, m = 1 kg

Acceleration, 

a =1m/s²

Force F = m x a = 1

1 x 1= 1N

Correct option is  A) increases

Correct option is  A) 24.4°

A) total internal reflection

Given: l_brass = 50cm, l_steel = 50cm

T₁ = 40°C = 313K

T₂ = 250°C = 523 K

α_brass = 2.0 × 10^-5K^-1

α_steel = 1.2 × 10^-5K^-1

∆l_brass = l_brass × α_brass × ∆T

= 50 × 2.0 × 10^-5 × (523 – 313)

∆l_brass = 0.21 cm

∆l_steel = l_steel × α_steel × ∆T

= 50 × 1.2 × 10^-5 × (523 – 313)

= 0.126 cm

∆l_steel = 0.13cm

(round off to two significant figures) Since the rods are free to expand, no thermal stress is developed at the junction.

Here, l_brass = l_steel = 50 cm

d_brass = d_steel = 3 mm = 0.3 cm

Δl_brass = ?

Δl_steel = ?

ΔT = 250 - 40 = 210°C

α_brass = 2 x 10^-5°C^-1

and α_steel = 1.2 x 10^-5°C^-1

Now, Δl_brass = α_brass x l_brass x ΔT

= 2 x 10^-5 x 50 x 210 = 0.21 cm

Again, Δl_steel = α_steel x l_steel x ΔT

= 1.2 x 10^-5 x 50 x 210 = 0.126 cm

≈ 0.13 cm

∴ Total change in length, Δl = Δl_brass + Δl_steel

= 0.21 + 0.13 = 0.34 cm

Since, the rod is not clamped at its ends no thermal stress is developed at the junction.

The system must be cooled which causes more contraction in brass disc and less contraction in the hole of steel plate. Hence, the disc becomes loose.

(B) sin^-1(1/2) 

Correct option is  A) very low

Yes, it is possible below :

For difference of length to remains constant, the change in length of each rod must be equal at all temperatures. Let L_b and L_s be the length of the brass and the steel rod and

\(\alpha_{b}L_{b}ΔT=\alpha_{s}L_{s}ΔT\)

\(\alpha_{b}L_{b}=\alpha_{s}L_{s}\)

or L_b/L_s =\(\alpha_{s}/\alpha_{b}\).

Hence, the lengths of the rods must be in the inverse ratio of the coefficient of linear expansion of their materials.

Given,

Distance (D) = 3.84 × 10⁸ m

Subtended angle (α)

= 1° 54′ = (60’+ 54′) = 114′

= 114 × 2.91 × 10^-4 rad

= 3.317 × 10^-2 rad

To find: Diameter of Earth (d)

Formula: d = αD

Calculation: From formula,

d = 3.317 × 10^-2 × 3.84 × 10⁸

= 1.274 × 10⁷ m

Diameter of Earth is 1.274 × 10⁷ m.

Given: Diameter (D) = 3474 km

∴ Radius of moon (R) = 1737 km

= 1.737 × 10⁶ m

Distance from Earth r = 3.84 × 10⁸ m

To find: Solid angle (dΩ)

Formula: dΩ = \(\frac{dA}{r^2}\)

Calculation:

From formula,

dΩ = \(\frac{\pi R^2}{r^2}\) ……..( cross-sectional area of disc of moon = πR²)

dΩ = \(\frac{\pi\times(1.737\times10^5)^2}{(3.84\times10^8)^2}\)

= \(\frac{3.412\times(1.737)^2\times10^{10}}{(3.84)^2\times10^{16}}\)

= antilog{log(3.142) + 2log(1.737) – 2log(3.84)} × 10^-6

= antilog {0.4972 + 2(0.2397) – 2(0.5843)} × 10^-6

= antilog{0.4972 + 0.4794 – 1.1686} × 10^-6

= antilog{\(\overline{1}\) .8080} × 10^-6

= 6.428 × 10^-1 × 10^-6

= 6.43 × 10^-5 sr

Solid angle subtended by moon at Earth is 6.43 × 10^-5 sr