NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI11:THERMAL PROPERTIES OF MATTER
366624
For measuring temperature near absolute zero, the thermometer used is
1 Thermoelectric thermometer
2 Radiation thermometer
3 Magnetic thermometer
4 Resistance thermometer
Explanation:
Conceptual Question
PHXI11:THERMAL PROPERTIES OF MATTER
366625
Two absolute scales \(A\) and \(B\) have triple points of water defined to be \(200\;A\) and \(350\;B\). What is the relation between \(T_{A}\) and \(T_{B}\) ?
1 \(\dfrac{T_{A}}{T_{B}}=\dfrac{4}{7}\)
2 \(\dfrac{T_{A}}{T_{B}}=\dfrac{3}{7}\)
3 \(\dfrac{T_{A}}{T_{B}}=\dfrac{7}{3}\)
4 \(\dfrac{T_{A}}{T_{B}}=\dfrac{7}{4}\)
Explanation:
Given triple point of water on scale \(A=200 A\) Triple point of water on scale \(B=350 B\) We know that triple point of water on absolute scale \(\begin{aligned}& =273.16 K \\& 200 A=350 B=273.16 K \\& 1 A=\dfrac{273.16}{200} K \text { and } 1 B=\dfrac{273.16}{350} K\end{aligned}\) If \(T_{A}\) and \(T_{B}\) are the triple point of water on two scales \(A\) and \(B\) then \(\begin{aligned}& \dfrac{273.16}{200} T_{A}=\dfrac{273.16}{350} T_{B} \\& \dfrac{T_{A}}{T_{B}}=\dfrac{200}{350}=\dfrac{4}{7} \text { or } T_{A}=\dfrac{4}{7} T_{B} .\end{aligned}\)
NCERT Exempler
PHXI11:THERMAL PROPERTIES OF MATTER
366626
In a mercury thermometer the ice point (lower fixed point) is marked as \(10^{\circ}\) and steam point is marked as \(130^{\circ}\). At \(40^\circ C\) temperature, what will this thermometer read
366627
Oxygen boils at \( - 183^\circ C\). This temperature is approximately.
1 \(215^\circ F\)
2 \(-297^\circ F\)
3 \(329^\circ F\)
4 \(361^\circ F\)
Explanation:
According to the formula: \(\begin{aligned}& \dfrac{C}{5}=\dfrac{F-32}{9} \Rightarrow \dfrac{-183}{5}=\dfrac{F-32}{9} \\& F=-297^{\circ} F\end{aligned}\)
366624
For measuring temperature near absolute zero, the thermometer used is
1 Thermoelectric thermometer
2 Radiation thermometer
3 Magnetic thermometer
4 Resistance thermometer
Explanation:
Conceptual Question
PHXI11:THERMAL PROPERTIES OF MATTER
366625
Two absolute scales \(A\) and \(B\) have triple points of water defined to be \(200\;A\) and \(350\;B\). What is the relation between \(T_{A}\) and \(T_{B}\) ?
1 \(\dfrac{T_{A}}{T_{B}}=\dfrac{4}{7}\)
2 \(\dfrac{T_{A}}{T_{B}}=\dfrac{3}{7}\)
3 \(\dfrac{T_{A}}{T_{B}}=\dfrac{7}{3}\)
4 \(\dfrac{T_{A}}{T_{B}}=\dfrac{7}{4}\)
Explanation:
Given triple point of water on scale \(A=200 A\) Triple point of water on scale \(B=350 B\) We know that triple point of water on absolute scale \(\begin{aligned}& =273.16 K \\& 200 A=350 B=273.16 K \\& 1 A=\dfrac{273.16}{200} K \text { and } 1 B=\dfrac{273.16}{350} K\end{aligned}\) If \(T_{A}\) and \(T_{B}\) are the triple point of water on two scales \(A\) and \(B\) then \(\begin{aligned}& \dfrac{273.16}{200} T_{A}=\dfrac{273.16}{350} T_{B} \\& \dfrac{T_{A}}{T_{B}}=\dfrac{200}{350}=\dfrac{4}{7} \text { or } T_{A}=\dfrac{4}{7} T_{B} .\end{aligned}\)
NCERT Exempler
PHXI11:THERMAL PROPERTIES OF MATTER
366626
In a mercury thermometer the ice point (lower fixed point) is marked as \(10^{\circ}\) and steam point is marked as \(130^{\circ}\). At \(40^\circ C\) temperature, what will this thermometer read
366627
Oxygen boils at \( - 183^\circ C\). This temperature is approximately.
1 \(215^\circ F\)
2 \(-297^\circ F\)
3 \(329^\circ F\)
4 \(361^\circ F\)
Explanation:
According to the formula: \(\begin{aligned}& \dfrac{C}{5}=\dfrac{F-32}{9} \Rightarrow \dfrac{-183}{5}=\dfrac{F-32}{9} \\& F=-297^{\circ} F\end{aligned}\)
366624
For measuring temperature near absolute zero, the thermometer used is
1 Thermoelectric thermometer
2 Radiation thermometer
3 Magnetic thermometer
4 Resistance thermometer
Explanation:
Conceptual Question
PHXI11:THERMAL PROPERTIES OF MATTER
366625
Two absolute scales \(A\) and \(B\) have triple points of water defined to be \(200\;A\) and \(350\;B\). What is the relation between \(T_{A}\) and \(T_{B}\) ?
1 \(\dfrac{T_{A}}{T_{B}}=\dfrac{4}{7}\)
2 \(\dfrac{T_{A}}{T_{B}}=\dfrac{3}{7}\)
3 \(\dfrac{T_{A}}{T_{B}}=\dfrac{7}{3}\)
4 \(\dfrac{T_{A}}{T_{B}}=\dfrac{7}{4}\)
Explanation:
Given triple point of water on scale \(A=200 A\) Triple point of water on scale \(B=350 B\) We know that triple point of water on absolute scale \(\begin{aligned}& =273.16 K \\& 200 A=350 B=273.16 K \\& 1 A=\dfrac{273.16}{200} K \text { and } 1 B=\dfrac{273.16}{350} K\end{aligned}\) If \(T_{A}\) and \(T_{B}\) are the triple point of water on two scales \(A\) and \(B\) then \(\begin{aligned}& \dfrac{273.16}{200} T_{A}=\dfrac{273.16}{350} T_{B} \\& \dfrac{T_{A}}{T_{B}}=\dfrac{200}{350}=\dfrac{4}{7} \text { or } T_{A}=\dfrac{4}{7} T_{B} .\end{aligned}\)
NCERT Exempler
PHXI11:THERMAL PROPERTIES OF MATTER
366626
In a mercury thermometer the ice point (lower fixed point) is marked as \(10^{\circ}\) and steam point is marked as \(130^{\circ}\). At \(40^\circ C\) temperature, what will this thermometer read
366627
Oxygen boils at \( - 183^\circ C\). This temperature is approximately.
1 \(215^\circ F\)
2 \(-297^\circ F\)
3 \(329^\circ F\)
4 \(361^\circ F\)
Explanation:
According to the formula: \(\begin{aligned}& \dfrac{C}{5}=\dfrac{F-32}{9} \Rightarrow \dfrac{-183}{5}=\dfrac{F-32}{9} \\& F=-297^{\circ} F\end{aligned}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXI11:THERMAL PROPERTIES OF MATTER
366624
For measuring temperature near absolute zero, the thermometer used is
1 Thermoelectric thermometer
2 Radiation thermometer
3 Magnetic thermometer
4 Resistance thermometer
Explanation:
Conceptual Question
PHXI11:THERMAL PROPERTIES OF MATTER
366625
Two absolute scales \(A\) and \(B\) have triple points of water defined to be \(200\;A\) and \(350\;B\). What is the relation between \(T_{A}\) and \(T_{B}\) ?
1 \(\dfrac{T_{A}}{T_{B}}=\dfrac{4}{7}\)
2 \(\dfrac{T_{A}}{T_{B}}=\dfrac{3}{7}\)
3 \(\dfrac{T_{A}}{T_{B}}=\dfrac{7}{3}\)
4 \(\dfrac{T_{A}}{T_{B}}=\dfrac{7}{4}\)
Explanation:
Given triple point of water on scale \(A=200 A\) Triple point of water on scale \(B=350 B\) We know that triple point of water on absolute scale \(\begin{aligned}& =273.16 K \\& 200 A=350 B=273.16 K \\& 1 A=\dfrac{273.16}{200} K \text { and } 1 B=\dfrac{273.16}{350} K\end{aligned}\) If \(T_{A}\) and \(T_{B}\) are the triple point of water on two scales \(A\) and \(B\) then \(\begin{aligned}& \dfrac{273.16}{200} T_{A}=\dfrac{273.16}{350} T_{B} \\& \dfrac{T_{A}}{T_{B}}=\dfrac{200}{350}=\dfrac{4}{7} \text { or } T_{A}=\dfrac{4}{7} T_{B} .\end{aligned}\)
NCERT Exempler
PHXI11:THERMAL PROPERTIES OF MATTER
366626
In a mercury thermometer the ice point (lower fixed point) is marked as \(10^{\circ}\) and steam point is marked as \(130^{\circ}\). At \(40^\circ C\) temperature, what will this thermometer read
366627
Oxygen boils at \( - 183^\circ C\). This temperature is approximately.
1 \(215^\circ F\)
2 \(-297^\circ F\)
3 \(329^\circ F\)
4 \(361^\circ F\)
Explanation:
According to the formula: \(\begin{aligned}& \dfrac{C}{5}=\dfrac{F-32}{9} \Rightarrow \dfrac{-183}{5}=\dfrac{F-32}{9} \\& F=-297^{\circ} F\end{aligned}\)
