Practice Test 9 Archives

Which of the following temperatures is the highest?

100K
–13°F
–20°C
–23°C

Solution

–13°F is (13 + 32)° belowice point on F scale.

540 g of ice at 0°C is mixed with 540 g of water at 80°C. The final temperature of mixture is

0°C
40°C
80°C
less than 0°C

Solution

540 × 80 + 540 θ = 540 (80 – θ)

⇒ 80 + θ = 80 - θ ⇒ θ = 2 0 ⇒ θ = 0°

When a body is heated, which colour corresponds to high temperature

red
yellow
white
orange

Solution

When a body is heated then relation between colours and temperature is according to Prevost’s theory of radiation which states that everybody emitting radiant energy in all directions at a rate depending only on the nature of its surface and its temperature e.g., when a body is placed in an enclosure (furnace) it would acquire the temperature of furnace and seem white means radiate white light. So it becomes first dark and then white.

The two ends of a rod of length L and a uniform cross-sectional area A are kept at two temperatures T₁ and T₂ (T₁> T₂). The rate of heat transfer, \(\frac{dQ}{dt}\) through the rod in a steady state is given by

\(\frac{dQ}{dt}=\frac{k(T_{1}-T_{2})}{LA}\)
\(\frac{dQ}{dt}=kLA(T_{1}-T_{2})\)
\(\frac{dQ}{dt}=kLA(T_{1}-T_{2})\)
\(\frac{dQ}{dt}=\frac{kL(T_{1}-T_{2})}{L}\)

Solution

\(\frac{dQ}{dt}=kLA(T_{1}-T_{2})\)

[(T₁–T₂) is the temperature difference]

A black body at 227°C radiates heat at the rate of 7 cals/cm²s. At a temperature of 727°C, the rate of heat radiated in the same units will be

Solution

According to Stefan’s law E=σT⁴,
T₁= 500 K
T₂= 1000 K
\(\frac{E_{2}}{E_{1}}=\left ( \frac{T_{2}}{T_{1}} \right )^{4}\)
∴E2= 16 × 7 =112 cal/cm²s

The resistance of a resistance thermometer has values 2.71 and 3.70 ohms at 10°C and 100°C respectively. The temperature at which the resistance is 3.26 ohm is

40ºC
50ºC
60ºC
70ºC

Solution

R_t = R₀(1 + αt)

2. 71 = R₀(1 + α × 10).......(1)

3. 70 = R₀(1 + α × 100).......(2)

3. 26 = R₀(1 + αt).......(3)

Solve these equations to obtain the value of t.

A metallic rod l cm long, A square cm in cross-section is heated through t°C. If Young’s modulus of elasticity of the metal is E and the mean coefficient of linear expansion is α perdegree celsius, then the compressional force required to prevent the rod from expanding along its length is

E A α t
E A α t/(1 + αt)
EA α t/(1 – α t)
E l α t

Solution

\(E=\frac{F/A}{\Delta l/l}=\frac{stress}{strain}where \Delta l=(l'–l) =l\alpha t so F= EA\alpha t\)

A pendulum clock is 5 seconds fast at temperature of 15° C and 10 seconds slow at a temperature of 30°C. At what temperature does it give the correct time? (take time interval =24 hours)

18ºC
20ºC
22ºC
25ºC

Solution

\(\Delta t=\frac{1}{2}\alpha \Delta T\times t\)

\(∴5=\frac{1}{2}\alpha (T-15)\times 86400\)

and 10=\(\frac{1}{2}\alpha (30-T)\times 86400\)

If a bar is made of copper whose coefficient of linear expansion is one and a half times that of iron, the ratio of force developed in the copper bar to the iron bar of identical lengths and cross-sections, when heated through the same temperature range (Young’s modulus of copper may betaken to be equal to that of iron) is

Solution

F=Y α t A or F∝a
(∵Y t A is same for both copper and iron)
or F_C ∝ α_C and F₁∝ α₁
\(∴\frac{F_{C}}{F_{1}}=\frac{3/2}{1}=\frac{3}{2}\)

A metallic bar is heated from 0°C to 100°C. The coeficient of linear expansion is 10^-5 K^-1. What will be the percentage increase in length?

0.01%
0.1%
1%
10%

Solution

\(\frac{\Delta \iota }{\iota }=\alpha \Delta T=10^{-5}\times 100=10^{-3}\)

\(\frac{\Delta \iota }{\iota }\times 100\)%=10^-3×100=10^-1=0.1%