The temperature of two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively are T2 and T1 (T2 > T1). The rate of heat transfer through the slab, in a steady state is with f, equal to
1
1/2
2/3
2/3
A light ray is incident perpendicular to one face of a 90° prism and is totally internally reflected at the glass-air interface. If the angle of reflection is 45°, we conclude that the refractive index n
n<2
Time taken by a 836 W heater to heat one litre of water from 10°C to 40°C is
50 s
100 s
150 s
150 s
The thermo emf of a thermocouple varies with the temperature θ of the hot junction as E = a θ + bθ2 in volts where the ratio a/b is 700°C. If the cold junction is kept at 0°C, then the neutral temperature is
700°C
350°C
1400°C
1400°C
Initially a beaker has 100 g of water at temperature 90°C. Later another 600 g of water at temperature 20°C was poured into the beaker. The temperature, T of the water after mixing is
20°C
30°C
45°C
55°C
A metallic bar of coefficient of linear expansion 10-5 K-1 is heated from 0°C to 100°C. The percentage increase in its length is
0.1 %
1 %
10 %
0.01 %
Two perfectly black spheres A and B having radii 8 cm and 2 cm are maintained at temperatures 127°C and 527°C, respectively. The ratio of the energy radiated by A to that by B is
1 : 2
1 : 1
2 : 1
1 : 4
Hot water in a vessel kept in a room cools from 70°C to 65°C in t1 minutes, from 65°C to 60°C in t2 minutes and from 60°C to 55°C in t3 minutes then,
t1 < t2 > t3
t1 = t2 = t3
t1 < t2 < t3
t1 < t2 = t3
If the time taken by a hot body to cool from 50°C to 40°C is 10 min when the surrounding temperature is 25°C, then the time taken for it to cool from 40°C to 30°C when the surrounding temperature is 15°C, is
40 min
10 min
5 min
15 min
1 cc of water becomes 1681 cc of steam when boiled at a pressure of105 Nm-2. The increasing internal energy of the system is (Latent heat of steam is 540 cal g-1 , 1 cal = 4.2 J)
300 cal
500 cal
225 cal
600 cal
B.
500 cal
Initial volume of water = V1 = 1 cm3
Volume of steam = V2 = 1681 cm3
Change in volume dV = V2 − V1 = 1681
= 1680 cm3
= 1680 × 10-6 m3
Pressure, p = 105 N/m2
dQ = mL = 1 × 540 × 4.2
= 22.68 J
dW = pdV
= 105 × 1680 × 10-6
= 168
Internal energy dU = 2268 − 168 = 2100 J
= 500 cal