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51.

The degrees of freedom of a molecul of a non-linear triatomic gas is (ignore vibraional motion)A. 2B. 4C. 6D. 8

Answer» Correct Answer - C
f= 3N -K
For non-linear triatomc gas,f = `3xx3-3=6`
Discussion
52.

How many degrees of freedom have the gas molecules, if under standard conditions the gas density is `1.3 kg m^(-3)` and the velocity of sound propagation in it is `C=330 m s^(-1)`.A. 5B. 6C. 7D. 8

Answer» Correct Answer - A
We know that, `v=sqrt((gammap)/(p))`
`gamma=(v^(2)p)/(p)=((330)^(2)xx1.3)/(1.015xx10^(5))=1.4`
`rArr" "1+1/n=1.4rArrn=5`
Discussion
53.

The readings of a bath on Celsius and Fahrenheit thermometers are in the ratio 2 :5. The temperature of the bath isA. `-26.66^(@)C`B. `40^(@)C`C. `45.71^(@)C`D. `26.66^(@)C`

Answer» Correct Answer - C
Given, `(T_(C))/(T_(F))=2/5rArrT_(F)=5/2T_(C)`
`therefore" " (T_(C))/100=((5//2)T_(C)-32)/180rArr9T_(C)=25/2-160`
`rArr" " T_(C)=45.71^(@)C`
Discussion
54.

The graph AB shown in figure is a plot of temperature of a body in degree celsius and degree Fahrenheit. Then A. Slope of line AB is 9/5B. Slope of line AB is 5/9C. Slope of line AB is 1/9D. Slope of line AB is 3/9

Answer» Correct Answer - B
Relation between celsius and Fahrenheit scale of temperature is `C/5=(F-32)/9`
`rArr" "C=5/9F(160)/9`
Equating about with standard equation of the line y = mx c we slope of the line. AB is m = `5/9`
Discussion
55.

On which of the following scales of temperature, the temperature is never negativeA. CelsiusB. FahrenheitC. ReaumurD. Kelvin

Answer» Correct Answer - D
Zero kelvin =- `273^(@)C` (absolute temperature). As no matter can attain this temperature. Hence temperature can never be negative on kelvin scale.
Discussion
56.

The folliwing observation were recorded on a platinum resistance thermometer. Resistance at melting point of pressure `=4.71 Omega,` and resistance at `t^(@)C=2.29Omega.` Calculate Temperature coefficient of resistants of platinum. Value of temperature t.

Answer» Given, resistance at melting point of ice, `R_(o)=3.70Omega` Resistance at boiling point of water at normal pressure,
`R_(100)=4.71Omega`
Resitance at `t^(@)C,R_(t) = 5.29 Omega`
According of the formuls, temperature coefficient of resistance is given by
`alpha=(R_(100)-R_(o))/(R_(o)xx100)=(4.71-3.70)/(3.70xx100)=(1.01)/(370)=2.73xx10^(3)//^(@)C`
According to the formula, for temperature t, we have `t=100^(@)Cxx(R_(t)-R_(o))/(R_(100)-R_(o))`
`100^(@)Cxx(5.29-370)/(4.71-3.70)=100^(@)Cxx(1.59)/(1.01)=157.4^(@)C`
Discussion
57.

The absolute zero temperature in Fahrenheit scale isA. `-273^(@)F`B. `-32^(@)F`C. `-460^(@)F`D. `-132^(@)F`

Answer» Correct Answer - C
Since, `(F-32)/9=(K-23)/5rArr(F-32)/9=(0-273)/5`
`rArr" " F=-459.4^(@)F~~-460^(@)F`
Discussion
58.

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

Answer» In general, whenever we are to go from any known scale to any unknow scale, then we follow the equation
`=(("Temperature on one scale")-("LFP for known scale"))/((UFP-LFP)_("known"))`
`=(("Temperature on other scale")-("LFP for unknown scale"))/((UFP-LFP)_("unknown"))`
`rArr " "(39-0)/(100-0)=(t-39)/(239-39)rArrt=117^(@)W`
Discussion
59.

Coefficient of volume expansion of mercury is `0.18xx10^(-3)//.^(@)C`. If the density of mercury at `0^(@)C` is 13.6g/cc, then its density at `200^(@)C` isA. 13.11 g/ccB. 52.11 g/ccC. 16.11 g/ccD. 26.11 g/cc

Answer» Correct Answer - A
Discussion
60.

The density of water is maximum at:A. `4^(@)C`B. `gt 4^(@)C`C. ` lt 4^(@)C`D. `10^(@)C`

Answer» Correct Answer - A
When cooled from room temperature liquid water becames dense, as compared with other substances, but at approximately `4^(@)C(39^(@)F),` pure water reaches its maximum density. If it is cooled further, it expands to become less dense.
Discussion
61.

At constant pressure, the ratio of increases in volume of an ideal gas per degree rise in kelbin temperature to its volume isA. 1/TB. `1//T^(2)`C. TD. `T^(2)`

Answer» Correct Answer - A
According to the ideal gas law,
`rArrpVRT=V=[(R)/(p)]T`
`rArrVpropT rArr(V_(1))/(V_(2))=(T_(1))/(T_(2))or (V_(2))/(V_(1))-1=(T_(2))/(T_(1))-1`
`rArr(V_(2)-V_(1))/(V_(1))=(T_(2)-T_(1))/(T_(1))rArr(V_(2)-V_(1))/(V_(1))=(1)/(T_(1))=1/T`
Discussion
62.

A cyclic process 1-2-3-4-1 is depicted on V-T diagram. The p-T and p-V diagrams for this cyclic process are given below. Select the correct choices (more than one options is/are correct) A. B. C. D. None of these

Answer» Correct Answer - C
Process 1-2 V = constant p `prop` T
T is increasing. Therefore pressure will be increase.
process 2-3 p = constant, V `prop` T
V and T both are increasing.
Process 3-4 T = constant `pprop1/V`
V is increasing. Therefore p will decrease.
Process 4-1 Inverses of process 2-3
Discussion