Properties of Matter — Page 8 | NEET Physics

Q71

A black body at 1227 o C emits radiations with maximum intensity at a wavelength of 5000

\mathop A\limits^ \circ

. If the temperature of the body is increased by 1000 o C, the maximum intensity will be observed at

A 3000

\mathop A\limits^ \circ

B 4000

\mathop A\limits^ \circ

C 5000

\mathop A\limits^ \circ

D 6000

\mathop A\limits^ \circ

Correct Answer

Option A

Solution

Applying Wein's displacement law,

{\lambda _m}T = contant

5000{\bf{{Å}}} \times \left( {1227 + 273} \right) = \left( {2227 + 273} \right) \times {\lambda _m}

{\lambda _m} = {{5000 \times 1500} \over {1000}} = 3000Å

Q72

Which of the following rods, (given radius r and length

l

) each made of the same material and whose ends are maintained at the same temperature will conduct most heat ?

A r = r 0 ,

l

=

l

0

B r = 2r 0 ,

l

=

l

0

C r = r 0 ,

l

= 2

l

0

D r = 2r 0 ,

l

= 2

l

0

Correct Answer

Option B

Solution

Heat conducted

= {{KA\left( {{T_1} - {T_2}} \right)t} \over l} = {{K\pi {r^2}\left( {{T_1} - {T_2}} \right)t} \over l}

The rod with the maximum ratio of A/

l

will conduct most. Here the rod with r = 2r 0 and

l = l_0

will conduct most.

Q73

If

\lambda

m denotes the wavelength at which the radioactive amission from a black body at a temperature TK is maximum, then

A

{\lambda _m}\, \propto \,{T^4}

B

{\lambda _m}

is independent of T

C

{\lambda _m}

\propto

T

D

{\lambda _m}\, \propto \,{T^{ - 1}}

Correct Answer

Option D

Solution

From Wein’s displacement law

{\lambda _m}T

= constant

\Rightarrow {\lambda _m} \propto {T^{ - 1}}

Q74

Consider a compound slab consisting of two different materials having equal thicknesses and thermal conductivities K and 2K, respectively. The equivalent thermal conductivity of the slab is

A

{2 \over 3}K

B

\sqrt 2 K

C 3 K

D

{4 \over 3}K

Correct Answer

Option D

Solution

In series, equivalent thermal conductivity

{K_{eq}} = {{2{K_1}{K_2}} \over {{K_1} + {K_2}}}

\Rightarrow {K_{eq}} = {{2 \times K \times 2K} \over {K + 2K}} = {4 \over 3}K

Q75

Unit of Stefan's constant is

A watt m 2 K 4

B watt m 2 /K 4

C watt/m 2 K

D watt/m 2 K 4

Correct Answer

Option D

Solution

Unit of Stefan’s constant is watt/m 2 K 4 .

Q76

Which of the following is best close to an ideal black body?

A black lamp

B cavity maintained at constant temperature

C platinium black

D a lump of charcoal heated to high temperature.

Correct Answer

Option B

Solution

An ideal black body is one which absorbs all the incident radiation without reflecting or transmitting any part of it.

Black lamp absorbs approximately 96% of incident radiation.

An ideal black body can be realized in practice by a small hole in the wall of a hollow body (as shown in figure) which is at uniform temperature.

Any radiation entering the hollow body through the holes suffers a number of reflections and ultimately gets completely absorbed.

This can be facilitated by coating the interior surface with black so that about 96% of the radiation is absorbed at each reflection.

The portion of the interior surface opposite to the hole is made conical to avoid the escape of the reflected ray after one reflection.

Q77

Consider two rods of same length and different specific heats (S 1 , S 2 ), conductivities (K 1 , K 2 ) and area of cross-sections (A 1 , A 2 ) and both having temperatures T 1 and T 2 at their ends. If rate of loss of heat due to conduction is equal, then

A K 1 A 1 = K 2 A 2

B

{{{K_1}{A_1}} \over {{S_1}}} = {{{K_2}{A_2}} \over {{S_2}}}

C K 2 A 1 = K 1 A 2

D

{{{K_2}{A_1}} \over {{S_2}}} = {{{K_1}{A_2}} \over {{S_1}}}

Correct Answer

Option A

Solution

Rate of heat loss in rod 1 = Q 1

= {{{K_1}{A_1}\left( {{T_1} - {T_2}} \right)} \over {{l_1}}}

Rate of heat loss in rod 2 = Q 2

= {{{K_2}{A_2}\left( {{T_1} - {T_2}} \right)} \over {{l_2}}}

By problem, Q1 = Q2.

\therefore {{{K_1}{A_1}\left( {{T_1} - {T_2}} \right)} \over {{l_1}}} = {{{K_2}{A_2}\left( {{T_1} - {T_2}} \right)} \over {{l_2}}}

$\therefore$ K 1 A 1 = K 2 A 2 .

\left[ \because{{l_1} = {l_2}} \right]

Q78

For a black body at temperature 727 o C, its radiating power is 60 watt and temperature of surrounding is 227 o C. If temperature of black body is changed to 1227 o C then its radiating power will be

A 304 W

B 320 W

C 240 W

D 120 W

Correct Answer

Option B

Solution

Radiating power of a black body

= {E_0} = \sigma \left( {{T^4} - T_0^4} \right)A

where $\sigma$ is known as the Stefan-Boltzmann constant, A is the surface area of a black body, T is the temperature of the black body and T 0 is the temperature of the surrounding.

$\therefore$ 60 = $\sigma$ (1000 4 – 500 4 ) ...(i) [T = 727 o C = 727 + 273 = 1000 K, T0 = 227 o C = 500 K].

In the second case, T = 1227 o C = 1500 K and let E' be the radiating power.

$\therefore$ E' = $\sigma$ (1500 4 – 500 4 ) ...(ii) From (i) and (ii) we have

{{E'} \over {60}} = {{{{1500}^4} - {{500}^4}} \over {{{1000}^4} - {{500}^4}}} = {{{{15}^4} - {5^4}} \over {{{10}^4} - {5^4}}} = {{50000} \over {9375}}

$\therefore$

E' = {{50000} \over {9375}} \times 60 = 320\,W

Q79

The Wien's displacement law express relation between

A wavelength corresponding to maximum energy and temperature

B radiation energy and wavelength

C temperature and wavelength

D colour of light and temperature.

Correct Answer

Option A

Solution

Wien’s displacement law states that the product of absolute temperature and the wavelength at which the emissive power is maximum is constant i.e.

\lambda _{max}

T = constant. Therefore it expresses relation between wavelength corresponding to maximum energy and temperature.

Q80

A cylindrical rod having temperature T 1 and T 2 at its end. The rate of flow of heat Q 1 cal/sec. If all the linear dimension are doubled keeping temperature constant, then rate of flow of heat Q 2 will be

A 4Q 1

B 2Q 1

C

{{{Q_1}} \over 4}

D

{{{Q_1}} \over 2}

Correct Answer

Option B

Solution

Heat flow rate

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

When linear dimensions are double.

{A_1} \propto r_1^2,{L_1} = L

{A_2} \propto 4r_1^2,{L_2} = 2L \Rightarrow {Q_2} = 2{Q_1}