Properties of Matter — Page 7 | NEET Physics

Q61

A slab of stone of area 0.36 m 2 and thickness 0.1 m is exposed on the lower surface to steam at 100 o C. A block of ice at 0 o C rests on the upper surface of the slab. In one hour 4.8 kg of ice is melted. The thermal conductivity of slab is (Given latent heat of fusion of ice = 3.36

\times

10 5 J kg

-

1 )

A 1.24 J/m/s/ o C

B 1.29 J/m/s/ o C

C 2.05 J/m/s/ o C

D 1.02 J/m/s/ o C

Correct Answer

Option A

Solution

Rate of heat given by steam = Rate of heat taken by ice where K = Thermal conductivity of the slab m = Mass of the ice L = Latent heat of melting/fusion A = Area of the slab

{{dQ} \over {dt}} = {{KA\left( {100 - 0} \right)} \over 1} = m{{dL} \over {dt}}

,

{{K \times 100 \times 0.36} \over {0.1}} = {{4.8 \times 3.36 \times {{10}^5}} \over {60 \times 60}}

K =1.24 J/m/s/°C

Q62

If the radius of a star is R and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is Q?

A

{Q \over {4\pi {R^2}\sigma }}

B

{\left( {{Q \over {4\pi {R^2}\sigma }}} \right)^{ - 1/2}}

C

{\left( {{{4\pi {R^2}Q} \over \sigma }} \right)^{1/4}}

D

{\left( {{Q \over {4\pi {R^2}\sigma }}} \right)^{1/4}}

Correct Answer

Option D

Solution

Stefan’s law for black body radiation

Q = \sigma eA{T^4}

\therefore T = {\left[ {{Q \over {\sigma \left( {4\pi {R^2}} \right)}}} \right]^{1/4}}

Q63

A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat Q in time t. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod, when placed in thermal contact with the two reservoirs in time t?

A

{Q \over 4}

B

{Q \over {16}}

C 2Q

D

{Q \over 2}

Correct Answer

Option B

Solution

The amount of heat flows in time t through a cylindrical metallic rod of length L and uniform area of cross-section

A\left( { = \pi {R^2}} \right)

with its ends maintained at temperatures T 1 and T 2 (T 1 > T 2 ) is given by

Q = {{KA\left( {{T_1} - {T_2}} \right)t} \over L}

...(i) where K is the thermal conductivity of the material of the rod. Area of cross-section of new rod

A' = \pi {\left( {{R \over 2}} \right)^2} = {{\pi {R^2}} \over 4} = {A \over 4}

...(ii) As the volume of the rod remains unchanged $\therefore$ AL = A'L' where L' is the length the new rod

\Rightarrow L' = L{A \over {A'}}

...(iii) = 4L (Using (ii)) Now, the amount of heat flows in same time t in the new rod with its ends maintained at the same temperatures T 1 and T 2 is given by

Q' = {{KA'\left( {{T_1} - {T_2}} \right)t} \over {L'}}

...(iv) Substituting the values of A' and L' from equations (ii) and (iii) in the above equation, we get

Q' = {{K(A/4)\left( {{T_1} - {T_2}} \right)t} \over {4L}} = {1 \over {16}}{{KA\left( {{T_1} - {T_2}} \right)t} \over L} = {1 \over {16}}Q

(Using (i))

Q64

The total radiant energy per unit area, normal to the direction of incidence, received at a distance R from the centre of a star of radius r, whose outer surface radiates as a black body at a temperature TK is given by

A

{{\sigma {r^2}{T^4}} \over {{R^2}}}

B

{{\sigma {r^2}{T^4}} \over {4\pi {R^2}}}

C

{{\sigma {r^2}{T^4}} \over {{R^4}}}

D

{{4\pi \sigma {r^2}{T^4}} \over {{R^2}}}

Correct Answer

Option A

Solution

According to the Stefan Boltzmann law, the power radiated by the star whose outer surface radiates as a black body at temperature T K is given by

P = \sigma 4\pi {r^2}{T^4}

where, r = radius of the star $\sigma$ = Stefan’s constant The radiant power per unit area received at a distance R from the centre of a star is

S = {P \over {4\pi {R^2}}} = {{\sigma 4\pi {r^2}{T^4}} \over {4\pi {R^2}}} = {{\sigma {r^2}{T^4}} \over {{R^2}}}

Q65

Assuming the sun to have a spherical outer surface of radius r, radiating like a black body at temperature t o C, the power received by a unit surface, (normal to the incident rays) at a distance R from the centre of the sun is where

\sigma

is the Stefan's constant.

A

{{{r^2}\sigma {{\left( {t + 273} \right)}^4}} \over {4\pi {R^2}}}

B

{{16{\pi ^2}{r^2}\sigma {t^4}} \over {{R^2}}}

C

{{{r^2}\sigma {{\left( {t + 273} \right)}^4}} \over {{R^2}}}

D

{{4\pi {r^2}\sigma {t^4}} \over {{R^2}}}

Correct Answer

Option C

Solution

Power radiated by the sun at t°C

= \sigma {\left( {t + 273} \right)^4}4\pi {r^2}

Power received by a unit surface

= {{\sigma {{\left( {t + 273} \right)}^4}4\pi {r^2}} \over {4\pi {R^2}}} = {{{r^2}\sigma {{\left( {t + 273} \right)}^4}} \over {{R^2}}}

Q66

The two ends of a rod of length L and a uniform cross-sectional area A are Kept at two temperatures T 1 and T 2 (T 1 > T 2 ). The rate of heat transfer,

{{dQ} \over {dt}}

through the rod in a steady state is given by :

A

{{dQ} \over {dt}} = {{k\left( {{T_1} - {T_2}} \right)} \over {LA}}

B

{{dQ} \over {dt}} = kLA({T_1} - {T_2})

C

{{dQ} \over {dt}} = {{kA\left( {{T_1} - {T_2}} \right)} \over L}

D

{{dQ} \over {dt}} = {{kL\left( {{T_1} - {T_2}} \right)} \over A}

Correct Answer

Option C

Solution

Similar to I = V/R

{{dQ} \over {dt}} = {{kA} \over L}\left( {{T_1} - {T_2}} \right)

k = conductivity of the rod.

Q67

A black body at 227 o C radiates heat at the rate of 7 cals/cm 2 s. At a temperature of 727 o C, the rate of heat radiated in the same units will be

A 50

B 112

C 80

D 60

Correct Answer

Option B

Solution

Rate of heat radiated at (227 + 273) K = 7 cals/(cm 2 s) Let rate of heat radiated at (727 + 273) K = x cals/(cm 2 s) By Stefan’s law, 7 $\propto$ (500) 4 and x $\propto$ (1000) 4 $\therefore$

{x \over 7} = {2^4} \Rightarrow x = 7 \times {2^4} = 112\,

cals/(cm 2 s)

Q68

On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are 39 o W and 239 o W respectively. What will be the temperature on the new scale, corresponding to a temperature of 39 o C on the Celsius scale ?

A 200 o W

B 139 o W

C 78 o W

D 117 o W

Correct Answer

Option D

Solution

W is the temperature on new scale corresponding to 39ºC on ºC scale.

So, (C – 0)/(100 – 0) = (W – 39)/(239 – 39) $\Rightarrow$ C/100 = (W – 39)/200 $\Rightarrow$ W = (C/100) × 200 + 39 = (39/100) × 200 + 39 = 78 + 39 = 117 So, temperature on new scale is 117°W corresponding to 39°C.

Q69

Assuming the sun to have a spherical outer surface of radius r, radiating like a black body at temperature t o C, the power received by a unit surface, (normal to the incident rays) at a distance R from the centre of the sun is where

\sigma

is the Stefan's constant.

A

{{{r^2}\sigma {{\left( {t + 273} \right)}^4}} \over {4\pi {R^2}}}

B

{{16{\pi ^2}{r^2}\sigma {t^4}} \over {{R^2}}}

C

{{{r^2}\sigma {{\left( {t + 273} \right)}^4}} \over {{R^2}}}

D

{{4\pi {r^2}\sigma {t^4}} \over {{R^2}}}

Correct Answer

Option C

Solution

Power radiated by the sun at t°C

= \sigma {\left( {t + 273} \right)^4}4\pi {r^2}

Power received by a unit surface

= {{\sigma {{\left( {t + 273} \right)}^4}4\pi {r^2}} \over {4\pi {R^2}}} = {{{r^2}\sigma {{\left( {t + 273} \right)}^4}} \over {{R^2}}}

Q70

A black body is at 727 o C. It emits energy at a rate which is proportional to

A (1000) 4

B (1000) 2

C (727) 4

D (727) 2

Correct Answer

Option A

Solution

According to Stefan’s law, rate of energy radiated

E \propto {T^4}

where T is the absolute temperature of a black body.

\therefore E \propto {\left( {727 + 273} \right)^4} \Rightarrow E \propto {\left[ {1000} \right]^4}