DIRECTIONS:These are Assertion-Reason type questions. Each of these question contains two statements:Statement-1 (Assertion) and Statement-2 (Reason). Answer these questions from the following four options.

Statement 1 : While measuring the thermal conductivity of liquid experimentally, the upper layer is kept hot and the lower layer is kept cold.

Statement 2 : This avoids heating of liquid by convection.

(a)Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement -1
(b)Statement-1 is True, Statement -2 is True; Statement-2 is NOT a correct explanation for Statement - 1
(c)Statement-1 is True, Statement- 2 is False
(d)Statement-1 is False, Statement -2 is True

Solution

(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement -1

We know that to measure thermal conductivity of liquids experimentally, they must be heated from the topi.e., upper layer is kept hot and lower layer is kept cold, so as to prevent convection in liquids.

Statement 1 : It is hotter over the top of a fire than at the same distance on the sides.

Statement 2 : Air surrounding the fire conducts more heat upwards.

(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement -1
(b) Statement-1 is True, Statement -2 is True; Statement-2 is NOT a correct explanation for Statement - 1
(c)Statement-1 is True, Statement- 2 is False
(d)Statement-1 is False, Statement -2 is True

Solution

(c)Statement-1 is True, Statement- 2 is False

Heat is carried away from a fire sideways mainly by radiations. Above the fire, heat is carried by both radiation and by convection of air. The latter process carries much more heat.

Statement 1 :As the temperature of the black body increases, the wavelength at which the spectral intensity(E_λ) is maximum decreases.

Statement 2 :The waveleng that which the spectral intensity will be maximum for a black body is proportional to the fourth power of its absolute temperature.

(a)Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement -1
(b)Statement-1 is True, Statement -2 is True; Statement-2 is NOT a correct explanation for Statement - 1
(c)Statement-1 is True, Statement- 2 is False
(d)Statement-1 is False, Statement -2 is True

Solution

(c) Statement-1 is True, Statement- 2 is False

From Wein's law λ_mT = constant i.e. peak emission

wavelength λ_m α ¹⁄_T.

Hence as T increases λ_m decreases.

Statement 1 :The equivalent thermal conductivity of two plates of same thickness in series is less than the smaller value of thermal conductivity.

Statement 2 :For two plates of equal thickness in series the equivalent thermal conductivity is given by

¹⁄_K = ¹⁄_K₁ + ¹⁄_K₂

(a)Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement -1
(b)Statement-1 is True, Statement -2 is True; Statement-2 is NOT a correct explanation for Statement - 1
(c)Statement-1 is True, Statement- 2 is False
(d)Statement-1 is False, Statement -2 is True

Solution

(a)Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement -1

The reading of Centigrade thermometer coincides with that of Fahrenheit thermometer in a liquid. The temperature of the liquid is

–40ºC
313°C
0°C
100°C

Solution

The ratio of thermal conductivity of two rods of different materials is 5 : 4. The two rods having same area of crosssection and same thermal resistance have the lengths in the ratio

5 : 4
1 : 9
9 : 1
4 : 5

Solution

With the increases of temperature of a black body the peak of the spectral energy distribution of the radiation shift towards lower wavelength. This is given by :

Kirchoff’s law
Wein’s displacement law
Stefan’s law
Rayleigh-Jeans law

Solution

According to Wein’s displacement law, λ T = const.

A slab of stone of area 0.36 m2and thickness0.1 m is exposed on the lower surface to steam at 100°C. A block of ice at 0°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 × 10⁵ Jkg^–1.) :

1.24 J/m/s/°C
1.29 J/m/s/°C
2.05 J/m/s/°C
1.02 J/m/s/°C

Solution

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

Q/4πR² σ
(Q/4πR²σ)^–1/2
(4πR²Q/σ)^1/4
(Q/4πR²σ)^1/4

Solution

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, who seouter surface radiates as a black body at a temperature T K is given by

(where σ is Stefan’s constant)

\(\frac{\sigma r^{2}T^{4}}{R^{2}}\)
\(\frac{\sigma r^{2}T^{4}}{4\pi r^{2}}\)
\(\frac{\sigma r^{4}T^{4}}{r^{4}}\)
\(\frac{4\pi \sigma r^{4}T^{4}}{R^{2}}\)

Solution

A cylindrical metallic rod in therrnal contact with two reservoirs of heat at its two ends conductsan 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?

^Q⁄₄
^Q⁄₄
2Q
^Q⁄₂

Solution

Six identical conducting rods are joined as shown in figure.Points A and D are maintained at 200°C and 20° C respectively. The temperature of junction B will be

120°C
100°C
140°C
80°C

Solution

The equivalent electrical circuit of the arrangement is shown in figure.

Temperature difference between the end points A and D = 200 – 20 = 180°C

As the resistances for the three parts are equal, the temperature difference must be distribuited equally in the three parts (= 180/3 = 60°C)

∴ Temperature of B =200°C – 60° = 140°C.

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