Physics Part Ii Chapter 11 Thermal Properties Of Matter

B.

We know that 
In steady state flow of heat

C.

(k₁

A.

C_p – C_v = R/28

C.

D.

C.

A.

doubled

D.

64

B.

2E/c

Initial momentum of surface
p_i = E/c
where c = velocity of light (constant)
Since, the surface is perfectly so, the same momentum will be reflected completely
Final momentum
p_f= E/c
therefore
Change in momentum
∆p = p_f − pi
= -E/c - E/c = -2E/c
Thus, momentum transferred to the surface is
∆p' = |∆p| = 2E/c

D.

1/3

B.

Angle of incidence i > C for total internal reflection.
Here i = 45° inside the medium. ∴ 45° > sin−1 (1/n)
⇒ n > √2.

C.

150 s

Let time taken in boiling the water by the heater is t sec. Then
Q = ms ∆ T

D.

11 RT

B.

de-Broglie wavelength

C.

1800

Rate of power loss

D.

256/81

Apply Wien's law

$$\begin{gathered} {\lambda }_{max} T = cons\tan t \\ i.e., {{\lambda }_{max}}_1 T_1 = {{\lambda }_{max}}_2 T_2 \\ \Rightarrow {\lambda }_{0}T = \frac{3{\lambda }_0}{4}T\text{'} \\ \Rightarrow T\text{'} = \frac{4}{3}T \\ Power radiated P\propto T^4 \\ So, \frac{P_2}{P_1} = n = {\left(\frac{T\text{'}}{T}\right)}^4 = {\left(\frac{4}{3}\right)}^4 = \frac{256}{81} \end{gathered}$$

B.

8.360 × 10⁴ K

Let at temperature T rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere

V_escape = 11200 m/s

Also, V_rms = V_escape = $\sqrt{\frac{3k_{B}T}{m_{O_2}}} = 11200 m/s$

Putting value of K_B and m_O2 we get,

T = 8.360 x 10⁴ K

C.

the blue will look brighter as compared to the red piece.

When the blue glass is heated at a high temperature. It absorbs all the radiation of higher wavelength except blue. If it is taken inside a dimly -lit room, it emits all the radiation of higher wavelength, hence it looks brighter as compared to the red piece.

A.

80.0096 cm

Using the relation 

$$\begin{gathered} {\mathrm{l}}_{\mathrm{t}} ={\mathrm{l}}_{\mathrm{o}}^{\text{'}} \left(1 + \mathrm{\alpha t}\right) \\ =1\times \left[1 + 11 \times {10}^{-6} \times \left({40}^{\mathrm{o}}-{20}^{\mathrm{o}}\right)\right] \\ =1.00022 \mathrm{cm} \\ \mathrm{Now} \mathrm{length} \mathrm{of} \mathrm{copper} \mathrm{rod} \mathrm{at} {40}^{\mathrm{o}}\mathrm{C} \\ {\mathrm{R}}_{\mathrm{t}}^{\text{'}} ={\mathrm{l}}_{\mathrm{o}}^{\text{'}} \left(1+\mathrm{\alpha }\text{'} \mathrm{t}\right) \\ =80 \left[1 + 17 \times {10}^{-6} \times \left({40}^{\mathrm{o}}-{20}^{\mathrm{o}}\right)\right] \\ =80.02727 \mathrm{cm} \\ \mathrm{Now} \mathrm{number} \mathrm{of} \mathrm{cms} \mathrm{observed} \mathrm{on} \mathrm{the} \mathrm{scale} \\ =\frac{80.0272}{1.00022} \\ =80.0096 \end{gathered}$$

C.

16

The total electromagnetic energy radiated by a body at absolute temperature T is proportional to its size, its ability to radiate (called emissivity) and most importantly to its temperature.

For a body, which is a perfect radiator, the energy emitted per unit time (H) is given by,

H = A σ T⁴

And this relation first obtained experimentally but Stefan and later proved theoretically by Boltzmann law and later known as  Stefan- Boltzmann law  

According to Stefan's law for radiated energy

E ∝ T⁴

$$\begin{gathered} \Rightarrow \frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}=\frac{{\mathrm{T}}_1^4}{{\mathrm{T}}_2^4} \\ ={\left(\frac{\mathrm{T}}{2\mathrm{T}}\right)}^4 \\ =\frac{1}{16} \\ \therefore {\mathbf{E}}_{\mathbf{2}}\mathbf{ }\mathbf{=}\mathbf{16}\mathbf{ }{\mathbf{E}}_{\mathbf{1}} \end{gathered}$$

B.

3 min

When rods are joined in series then heat flow 

$$\begin{gathered} ∆\mathrm{Q} = \frac{\mathrm{A} \left({\mathrm{T}}_1 - {\mathrm{T}}_2\right) {\mathrm{t}}_1}{{\displaystyle \frac{{\mathrm{l}}_1}{{\mathrm{K}}_1} + \frac{{\mathrm{l}}_1}{{\mathrm{K}}_2}}} \\ = \frac{\mathrm{A} \left({\mathrm{T}}_{1 }+ {\mathrm{T}}_2\right) {\mathrm{t}}_1}{{\displaystyle \frac{1}{{\mathrm{K}}_1} + \frac{1}{{\mathrm{K}}_2}}} \\ ∆\mathrm{Q} = \frac{\mathrm{A} \left({\mathrm{T}}_1 -{\mathrm{T}}_2\right) {\mathrm{t}}_1}{{\mathrm{l}}_1} \frac{\mathrm{K}}{2} \end{gathered}$$

When rods are joined in parallel

$$\begin{gathered} ∆{\mathrm{Q}}_2 =\left({\mathrm{K}}_1 \mathrm{A} - {\mathrm{K}}_2 \mathrm{A}\right)\frac{\left({\mathrm{T}}_1 - {\mathrm{T}}_2\right) {\mathrm{t}}_2}{\mathrm{l}} \\ ∆{\mathrm{Q}}_2 = \frac{2 \mathrm{K} \mathrm{A} \left({\mathrm{T}}_1 - {\mathrm{T}}_2\right) {\mathrm{t}}_2}{\mathrm{l}} \\ \because ∆{\mathrm{Q}}_1= ∆{\mathrm{Q}}_2 \left(\mathrm{given}\right) \end{gathered}$$

$\therefore {\mathrm{t}}_2 = \frac{{\mathrm{t}}_1}{4} = \frac{12}{4} = 3 \min$

B.

r = 2r_o, l = l_o

Heat conduction through a rod is rate of change of heat $\left(\frac{∆\mathrm{Q}}{∆\mathrm{t}}\right)$

$$\begin{gathered} \therefore \mathrm{H} = \left(\frac{∆\mathrm{Q}}{∆\mathrm{t}}\right) \\ = \mathrm{KA} \left(\frac{{\mathrm{T}}_1 - {\mathrm{T}}_2}{\mathrm{l}}\right) \end{gathered}$$

The rate of heat H is proportional to the temperature difference (T₁ - T₂) and the area of cross section A and inversely proportional to length L. Here K is called thermal conductivity of the material.

$$\begin{gathered} \Rightarrow \mathrm{H} \propto \frac{{\mathrm{r}}^{ 2}}{\mathrm{l}} \\ \left(\mathrm{a}\right) \mathrm{When} \mathrm{r} = 2{\mathrm{r}}_{\mathrm{o}}, \mathrm{l} = 2 {\mathrm{l}}_{\mathrm{o}} \\ \mathrm{H}\propto \frac{{\left(2{\mathrm{r}}_{\mathrm{o}}\right)}^2}{{\mathrm{l}}_{\mathrm{o}}} \Rightarrow \mathrm{H} \propto \frac{2 {\mathrm{r}}_{\mathrm{o}}^2}{{\mathrm{I}}_{\mathrm{o}}} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{b}\right) \mathrm{When} \mathrm{r} = 2{\mathrm{r}}_{\mathrm{o}}, \mathrm{l} = {\mathrm{l}}_{\mathrm{o}} \\ \mathrm{H} \propto \frac{{\left(2{\mathrm{r}}_{\mathrm{o}}\right)}^2}{2{\mathrm{l}}_{\mathrm{o}}}\Rightarrow \mathrm{H} \propto \frac{4 {\mathrm{r}}_{\mathrm{o}}^2}{{\mathrm{l}}_{\mathrm{o}}} \\ \left(\mathrm{c}\right) \mathrm{When} \mathrm{r} = {\mathrm{r}}_{\mathrm{o}}, \mathrm{l} = {\mathrm{l}}_{\mathrm{o}} \\ \mathrm{H} \propto \frac{{\mathrm{r}}_{\mathrm{o}}^2}{{\mathrm{l}}_{\mathrm{o}}} \\ \left(\mathrm{d}\right) \mathrm{When} \mathrm{r} = {\mathrm{r}}_{\mathrm{o}}, \mathrm{l} = 2{\mathrm{l}}_{\mathrm{o}} \\ \mathrm{H} \propto \frac{{\mathrm{r}}_{\mathrm{o}}^2}{2{\mathrm{l}}_{\mathrm{o}}} \end{gathered}$$

it is obvious that heat conduction will be more in case (b).

C.

-20^o C

The temperature of inversion is higher than the neutral temperature by the same amount as the neutral temperature is higher than the temperature of cold junction.

As per the above statement

t_n - t₀ = t_i - t_n

⇒ t₀ = 2t_n - t_i

Given, t_i = 620^o C, t_n =300^o C

Hence,  

t_o = temperature of cold junction

   = 2 × 300 - 620

t_o = -20^o C

Neutral temperature does not change for a metal whether the temperature of cold junction and inversion temperature change by any value.

B.

filling nitrogen gas at high pressure above the mercury column

If we fill nitrogen gas at high pressure above mercury level, the boiling point of mercury is increased which can extend the range upto 500°C.

B.

r = 2r_o ; l = l_o

Heat conduction through a rod is given by

$$\begin{gathered} \mathrm{H} = \frac{∆\mathrm{Q}}{∆\mathrm{t}} \\ =\mathrm{KA} \left(\frac{{\mathrm{T}}_1 - {\mathrm{T}}_2}{\mathrm{l}}\right) \\ \Rightarrow \mathrm{H} \propto \frac{{\mathrm{r}}^2}{\mathrm{l}} \end{gathered}$$

(a) When r = 2r_o ; l = 2l_o

 $$\begin{gathered} \mathrm{H} \propto \frac{{\left(2 {\mathrm{r}}_{\mathrm{o}}\right)}^2}{2{\mathrm{l}}_{\mathrm{o}}} \\ \Rightarrow \mathrm{H} \propto \frac{2{\mathrm{r}}_{\mathrm{o}}^2}{{\mathrm{l}}_{\mathrm{o}}} \end{gathered}$$

(b) When r = 2r_o; l = l_o

 $$\begin{gathered} \mathrm{H} \propto \frac{{\left(2{\mathrm{r}}_{\mathrm{o}}\right)}^2}{{\mathrm{l}}_{\mathrm{o}}} \\ \Rightarrow \mathrm{H} \propto \frac{4{\mathrm{r}}_{\mathrm{o}}^2}{{\mathrm{l}}_{\mathrm{o}}} \end{gathered}$$

(c) When r = r_o ; l = l_o

$\mathrm{H} \propto \frac{{\mathrm{r}}_{\mathrm{o}}^2}{{\mathrm{l}}_{\mathrm{o}}}$

(d) When r = r_o ; l = 2l_o

$\mathrm{H} \propto \frac{{\mathrm{r}}_{\mathrm{o}}^2}{2{\mathrm{l}}_{\mathrm{o}}}$

It is obvious that heat conduction will be more in case (b).

C.

$\frac{4}{3} \mathrm{K}$

Thermal conductivity equatio is given by 

         H = KA $\frac{{\mathrm{T}}_1- {\mathrm{T}}_2}{\mathrm{L}}$

Where, ( T₁ - T₂ ) is the temperature difference two points, A is the area of cross-section and L is the length. As in question different materials having equal thickness and at constant temperature A, L and temperature difference remains constant.

Equivalent thermal conductivity of the compound slab,

        K_eq = $\frac{{\mathrm{l}}_1 + {\mathrm{l}}_2}{{\displaystyle \frac{{\mathrm{l}}_1}{{\mathrm{K}}_1} + \frac{{\mathrm{l}}_2}{{\mathrm{K}}_2}}}$

         K_eq = $\frac{\mathrm{l} + \mathrm{l}}{{\displaystyle \frac{\mathrm{l}}{\mathrm{K}} + \frac{\mathrm{l}}{2\mathrm{k}}}}$

                = $\frac{2 \mathrm{l}}{{\displaystyle \frac{3 \mathrm{l}}{2 \mathrm{K}}}}$

        K_eq = $\frac{4}{3} \mathrm{K}$

A.

$\frac{2\mathrm{\lambda }}{3}$

Below figure gives the experimental curves for radiation energy per unit area per unit wavelength emitted by blackbody versus wavelength for different temperatures.

The wavelength λ_m for which energy is maximum decreases with increasing temperature. The relation between λ_m and T is given by Wein's Displacement Law

According to Wein's displacement law,

             λ_m T = Constant

 ⇒           $\frac{{\mathrm{\lambda }}_2}{{\mathrm{\lambda }}_1} = \frac{2000}{3000}$

                $\frac{{\mathrm{\lambda }}_2}{\mathrm{\lambda }} = \frac{{\displaystyle 2}}{3}$

                 ${\mathrm{\lambda }}_2 = \frac{2}{3} \mathrm{\lambda }$

C.

Stefan's Boltzmann law

(a) Wein's law is useful for estimating the surface temperatures of celestial bodies like the moon, sun and other stars.

(b)Kepeler's law described the orbits of planets around the sun or stars around the galaxy in classical mechanics.

(c) The temperature of the surface of the sun was found out to be 5761 k approx by using Stefan's law i.e

      E =  σ T⁴ 

Where E = amount of heat energy radiated / sec area

σ = Stefan's Boltzmann constant

(d)Planck's law is to explain the spectral -energy distribution of radiation emitted by black body absorbs all radiant energy falling upon it, reaches some equilibrium temperature, and then reemits.

So Stefan's Boltzmann law is used to find the temperature of the sun.

D.

no neutral temperature is possible for this thermocouple

When two wires composed of dissimilar metals are joined at both ends and one of the ends is heated, there is a continuous current which flows in the thermoelectric circuit.

The emf set in the thermocouple, when its two junctions are kept at different temperatures is called thermo e.m.f

Given:-

E = aθ + bθ²

  $\mathrm{P} = \frac{\mathrm{dE}}{\mathrm{dt}}$

     = $\frac{\mathrm{d}}{\mathrm{d\theta }} \left(\mathrm{a\theta } + {\mathrm{b\theta }}^2\right)$

  P = a + 2 bθ

At θ = T_n  ,P = 0

⇒    0 = a + 2 b T

⇒   T_n = $- \frac{\mathrm{a}}{2 \mathrm{b}}$

⇒ T_n = -$\frac{{700}^{\mathrm{o}}}{2}$

 T_n  = - 350^o C

The temperature of the hot junction of a thermocouple at which the elctromotive force on the thermocouple attains its maximum value, when the cold junction is maintained at constant temperature of 0^o C , called neutral temperature.

But  neutral temperature can never be negative ( less than zero) 

A.

3 × 10²⁰

Given:- Current - 1.6 A

     t = 1 × 60s = 60 s

Charge flowing in one minute

     q = I × t 

         = 1.6 × 60

Charge required to liberate one copper ion 2C

Number of Cu²⁺ liberated 

            =$\frac{1.6 \times 60}{2 \times 1.6 \times {10}^{-19}}$

            = 3 × 10²⁰

C.

7500 k

Wein's displacement law:- Relationship between the temperature of a black body ( an ideal substance that emits and absorbs all frequencies of light) and the wavelength at which it emits the most light.

            ${\mathrm{\lambda }}_{\mathrm{m}} \mathrm{T} = \mathrm{Constant}$

By using Wein's Displacement law,

         ${\left({\mathrm{\lambda }}_{\mathrm{m}} \right)}_1 {\mathrm{T}}_1 = {\left({\mathrm{\lambda }}_{\mathrm{m}} \right)}_2 {\mathrm{T}}_2$

⇒     500 × 6000 = 400 × T₂

⇒     T₂ = $\frac{500 \times 6000}{400}$

⇒      T₂ = 7500 K 

A.

65^o C

Let the final temperature of mixture be t.

According to Newton's law of cooling

Heat lost by water at 80°C

               = m s Δt

Where m is the mass of body and s is the specific heat capacity

                = 0.1 × 10³ ×  s_water ×  ( 80^o - t )

                     (since m = v ×  d = 0.1 ×  10³ kg )

Heat gained by water at 60^o C

               = 0.3 ×  10³ ×  s_water ×  ( t - 60^o )

According to principle of calorimetry

            Heat lost = Heat gained

∴  0.1 × 10³ × s_water × (80^o - t) = 0.3 × 10³ ×s_water × ( t - 60^o )

⇒           ( 80^o - t) = 3 × ( t - 60^o )

⇒             4t = 260^o

⇒               t = 65^o C

A.

$\frac{2 \mathrm{\lambda }}{3}$

According to Wien's displacement law wavelength λ_m for which energy is the maximum decrease with increasing temperature.

This relation between λ_m and T is given by 

            λ_m T = constant

 ⇒           $\frac{{\mathrm{\lambda }}_2}{{\mathrm{\lambda }}_1} = \frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}$

⇒          $\frac{{\lambda }_2}{\lambda } = \frac{2000}{3000}$

⇒           $\frac{{\mathrm{\lambda }}_2}{\mathrm{\lambda }} = \frac{2}{3}$

⇒            λ₂ = $\frac{2}{3}$ λ 

A.

10^oC

D.

above 20^oC

According to Newton's law of cooling,

   Heat lost = Heat gained

⇒   m₁ s₁ ( 100 - t ) = mL + m₂ s₂ ( t - 0)

⇒    50 × 1( 100 - t) = 50 × 80 + 50 × 1 (t - 0)

⇒                        t = 10^oC

A.

5800 K

Let diameter of the sun is d while radius of the earth is r then

          $\frac{\mathrm{D}}{2} \simeq 0.53 \times \frac{\mathrm{\pi }}{180}$

               = 0.00925

  ⇒     $\frac{\mathrm{D}}{2}$  =   9.25 × 10^-3

The radiation emitted per unit area of the sun is

   $4\mathrm{\pi } {\left(\frac{\mathrm{d}}{2}\right)}^2{\mathrm{\sigma T}}^4 = {\mathrm{\pi D}}^2\mathrm{\sigma } {\mathrm{T}}^4$

The radiation received at the earth's surface per unit time per unit area is

      $\frac{\mathrm{\pi \sigma d} {\mathrm{T}}^4}{4{\mathrm{\pi r}}^2} = \frac{\mathrm{\sigma } {\mathrm{T}}^4}{4}{\left(\frac{\mathrm{d}}{\mathrm{R}}\right)}^2$

⇒       $\frac{{\mathrm{\sigma T}}^4}{4} {\left(\frac{\mathrm{D}}{\mathrm{R}}\right)}^2$ = 8.2 Jcm^-2 min^-1

⇒    $\frac{1}{4} \times 5.67 \times {10}^{-8} \times {\mathrm{T}}^4 \times {\left(9.25 \times {10}^{-3} \right)}^2$ = $\frac{8.2}{{10}^{-4} \times 60}{\mathrm{Wm}}^{-2}$

⇒                           T = 5794 K

                              T ≅ 5800 K

C.

$\frac{2 {\mathrm{K}}_1 {\mathrm{K}}_2}{{\mathrm{K}}_1 + {\mathrm{K}}_2}$

As both slabs are of same thickness made of different materials i.e d₁ = d₂ = d

As two portions of the slab are connected in series, 

∴    K = $\frac{\mathrm{\Sigma } {\mathrm{x}}_{\mathrm{i}}}{\mathrm{\Sigma }{\displaystyle \raisebox{1ex}{$ {\mathrm{x}}_{\mathrm{i} }$}\!\left/ \!\raisebox{-1ex}{$ {\mathrm{K}}_{\mathrm{i}}$}\right.}}$

      K = $\frac{\mathrm{d} + \mathrm{d}}{{\displaystyle \frac{\mathrm{d}}{{\mathrm{K}}_1} + \frac{\mathrm{d}}{{\mathrm{K}}_2}}}$

         = $\frac{2 \mathrm{d}}{\mathrm{d} \left({\displaystyle \frac{1}{{\mathrm{K}}_1}}+ {\displaystyle \frac{1}{{\mathrm{K}}_2}}\right)}$

     K = $\frac{2 {\mathrm{K}}_1 {\mathrm{K}}_2}{{\mathrm{K}}_1+ {\mathrm{K}}_2}$

A.

49.6 W

Given, initial temperature of metal rod (T)= 145° C = 418 K

Rate of radiated energy (E₁) = 17W

Final temperature ( T₂) = 273^o C

                                   = 273 + 273

                                   = 546 k

We know, from the Stefan's law,

             E ∝ T⁴

            $\frac{{\mathrm{E}}_1}{{\mathrm{E}}_2} = {\left(\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}\right)}^4$

                  = ${\left(\frac{418}{546}\right)}^4$

           $\frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}$= 0.343

Therefore, final radiated energy

      E₂ = $\frac{{\mathrm{E}}_1}{0.343}$

           = $\frac{17}{0.343}$

       E₂ = 49.6 W

C.

$\frac{7}{4}$K

∴              R₁ = $\frac{\mathrm{L}}{{\mathrm{K}}_1 . {\mathrm{\pi R}}^2}$      

   and         R₂ = $\frac{\mathrm{L}}{{\mathrm{K}}_2 \left( 4{\mathrm{\pi R}}^2 - {\mathrm{\pi R}}^2 \right)}$

                       = $\frac{\mathrm{L}}{{\mathrm{K}}_{\mathrm{eq}}. 4{\mathrm{\pi R}}^2}$

R₁ and R₂ are in parallel.

∴                  $\frac{1}{{\mathrm{R}}_{\mathrm{eq}}} = \frac{1}{{\mathrm{R}}_1} + \frac{1}{{\mathrm{R}}_2}$

⇒         $\frac{4 {\mathrm{\pi R}}^2 {\mathrm{K}}_{\mathrm{eq}}}{\mathrm{L}} = \frac{{\mathrm{\pi R}}^2 {\mathrm{K}}_1}{\mathrm{L}} + \frac{3{\mathrm{\pi R}}^2 {\mathrm{K}}_2}{\mathrm{L}}$

⇒               4K_eq = K₁ + 3K₂

On putting values of K₁ and K₂, we get

                  4K_eq = K₁ + 3 ( 2K)                 [ $\because$ K₁ = K, K₂ = 2K ]

                  K_eq = $\frac{\mathrm{K} + 6\mathrm{K}}{4}$

                  K_eq = $\frac{7 \mathrm{K}}{4}$

B.

16^oC

According to Newton's law of cooling

          $\frac{{\mathrm{\theta }}_1 - {\mathrm{\theta }}_2}{\mathrm{t}} = \mathrm{K} \left[\frac{{\mathrm{\theta }}_1 + {\mathrm{\theta }}_2}{2} - {\mathrm{\theta }}_0\right]$

In the first case,

            $\frac{80 - 64}{5}$ = K $\left[\frac{80 + 64}{2} - {\mathrm{\theta }}_0\right]$

⇒        3.2 = K [ 72 $-$ θ₀ ]                  ....(i)

In the second case,

          $\frac{64 - 52}{5} = \mathrm{K} \left[\frac{64 + 52}{2} - {\mathrm{\theta }}_0\right]$

⇒        2.4 = K [ 58 $-$ θ₀ ]                 ......(ii)

Dividing (i) by (ii),

       $\frac{3.2}{2.4} = \frac{72 - {\mathrm{\theta }}_0}{58 - {\mathrm{\theta }}_0}$

⇒     185.6 $-$ 3.2 θ₀ = 172.8 $-$ 2.4 θ₀

⇒      185.6 $-$ 172.8 = ( 3.2 $-$ 2.4 ) θ₀

⇒                   θ₀  = $\frac{185.6 - 172.8}{3.2 - 2.4}$

⇒                        θ₀ = 16⁰C

A.

$\frac{{\mathrm{P}}_1{\mathrm{P}}_2 \left( {\mathrm{m}}_1 + {\mathrm{m}}_2 \right)}{\left( {\mathrm{P}}_2 {\mathrm{m}}_1 + {\mathrm{P}}_1 {\mathrm{m}}_2 \right)}$

According to Boyle's law, PV== k (a constant )

          p$\frac{\mathrm{m}}{\mathrm{\rho }}$= k

         ρ = $\frac{\mathrm{P} \mathrm{m}}{\mathrm{k}}$                           $\left(\because \mathrm{V} = \frac{\mathrm{m}}{\mathrm{\rho }}\right)$

       ${\mathrm{\rho }}_1 = \frac{{\mathrm{P}}_1}{\mathrm{K}}$ 

and V₁ = $\frac{{\mathrm{m}}_1}{{\mathrm{\rho }}_1}$

          = $\frac{{\mathrm{m}}_1}{{\displaystyle \raisebox{1ex}{${\mathrm{P}}_1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.}}$

      V₁ = $\frac{\mathrm{K} {\mathrm{m}}_1}{{\mathrm{P}}_1}$

Simillarly

      V₂ = $\frac{\mathrm{K} {\mathrm{m}}_2}{{\mathrm{P}}_1}$

∴ Total volume = V₁ + V₂

                       = K $\left( \frac{{\mathrm{m}}_1}{{\mathrm{P}}_1} + \frac{{\mathrm{m}}_2}{{\mathrm{P}}_2} \right)$

Let P be the common pressure and ρ be the common density of mixture. Then 

             ρ = $\frac{{\mathrm{m}}_1 + {\mathrm{m}}_2}{{\mathrm{V}}_1 + {\mathrm{V}}_2}$

                 = $\frac{{\mathrm{m}}_1 + {\mathrm{m}}_2}{\mathrm{K} \left({\displaystyle \frac{{\mathrm{m}}_1}{{\mathrm{P}}_1}} + {\displaystyle \frac{{\mathrm{m}}_2}{{\mathrm{P}}_2}}\right)}$

∴        P = K ρ

              = $\frac{{\mathrm{m}}_1+ {\mathrm{m}}_2}{{\displaystyle \frac{{\mathrm{m}}_1}{{\mathrm{P}}_1} + \frac{{\mathrm{m}}_2}{{\mathrm{P}}_2}}}$

⇒      P = $\frac{{\mathrm{P}}_1 {\mathrm{P}}_2 \left({\mathrm{m}}_1 + {\mathrm{m}}_2\right)}{\left({\mathrm{m}}_1 {\mathrm{P}}_2 + {\mathrm{m}}_2 {\mathrm{P}}_2\right)}$

A.

If both assertion and reason are true and reason is the correct explanation of assertion.

The water level remains the same when ice cube melts. A floating object displaces an amount of water equal to its own weight.

D.

81

Emissive power is the energy of thermal radiation emitted in all directions per unit time per area of a surface given temperature.

Radiation from a body = R

          R = e σ T⁴ × time × area

where e = emissive power

∴        eT⁴ = constant

           e  = $\left( \frac{\mathrm{constant}}{{\mathrm{T}}^4} \right)$

⇒       $\frac{{\mathrm{e}}_1}{{\mathrm{e}}_2} = {\left(\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\right)}^4$

⇒       $\frac{{\mathrm{e}}_1}{{\mathrm{e}}_2} = {\left(\frac{273 + 327}{273 - 73}\right)}^4$

⇒        $\frac{{\mathrm{e}}_1}{{\mathrm{e}}_2} = \frac{{\displaystyle {\left(3\right)}^4}}{1}$

⇒              = $\frac{81}{1}$

⇒        $\frac{{\mathrm{e}}_1}{{\mathrm{e}}_2} = \frac{81}{1}$

C.

45 ^oC

Joule heating in a wire

          H = I² Rt

For given R and t

          H ∝ t²

 ∴        $\frac{{\mathrm{H}}_1}{{\mathrm{H}}_{2\mathrm{I}}} = \frac{{\mathrm{I}}_1^{ 2}}{{\mathrm{I}}_2^{ 2}}$                   ....(i)

Also   H = m s Δ

For given m and s 

∴        $\frac{{\mathrm{H}}_1}{{\mathrm{H}}_2} = \frac{∆{\mathrm{T}}_1}{∆{\mathrm{T}}_2}$                   ....(ii)

From equation (i) and (ii)

         $\frac{{\mathrm{I}}_1^{ 2}}{{\mathrm{I}}_2^{ 2}} = \frac{∆{\mathrm{T}}_1}{∆{\mathrm{T}}_2}$    

Here    I₁ = I 

          ΔT₁ = 5 ^oC

          I₂ = 3I

           ΔT₂ = ?

∴       ${\left(\frac{\mathrm{I}}{3\mathrm{I}}\right)}^2 = \frac{5}{∆{\mathrm{T}}_2}$

⇒     ΔT₂ = 45 ^oC

D.

0 ^oC

Heat lost by water to come from 15 ^oC to 0 ^oC is

          H₁ = m s ΔT

           H₁ = $\frac{40}{1000}$ × (4.2 × 10³ ) × ( 15 $-$ 0 )

           H₁   = 2520 J

Heat required to convert 10 g ice into 10 g water at 0°C is

           H₂ = mL

                 = $\frac{10}{1000}$ × ( 3.36 × 10⁵ )

           H₂ = 3360 J

Since H₂ > H₁, so the whole ice will not be converted into water, whereas the temperature of the whole water will be 0^oC.

Therefore the temperature of the mixture is 0^oC.

C.

11.0 W

The surface area of the ball 

               A = 4 $\mathrm{\pi } {\mathrm{r}}^{ 2}$

                    = (4 $\mathrm{\pi }$) ( 0.02 m)²

               A = 0.05 m²

and its absolute temperature is

               T = 327 + 273

                T = 600 K

       P = e σ AT⁴

           = [ 0.3 × 5.67 × 10^-8 × 0.05 × (600)⁴ ]  

       P = 11.0 W

B.

higher than 0 K

All objects with a temperature greater than absolute 0 emit electromagnetic radiation, which is commonly called the object's thermal emission. The temperature of the radiating body determines the intensity and characteristics of the radiation it emits.

C.

If assertion is true but reason is false.

 A black body is an idealized object which absorbs and emits all frequencies. Blackbody radiation as a function of frequency for a fixed temperature.

The hollow metallic closed container maintained at a uniform temperature can act as a source of the black body. It is also well known that all metals cannot ac as a black body because if we take highly metallic polished surface. It will not behave as a perfect black body.