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NEET PHYSICSThermal Properties of MatterMedium

Question

On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are 39W39^{\circ}\text{W} and 239W239^{\circ}\text{W} respectively. What will be the temperature on the new scale, corresponding to a temperature of 39C39^{\circ}\text{C} on the Celsius scale?

A

78W78^{\circ}\text{W}

B

117W117^{\circ}\text{W}

C

200W200^{\circ}\text{W}

D

139W139^{\circ}\text{W}

Step-by-Step Solution

Let TWT_W be the temperature on the W scale and TCT_C be the temperature on the Celsius scale. For any two linear temperature scales, the ratio of the difference between any temperature and the lower fixed point (freezing point) to the fundamental interval (difference between boiling and freezing points) is constant. TWLFPWUFPWLFPW=TCLFPCUFPCLFPC\frac{T_W - \text{LFP}_W}{\text{UFP}_W - \text{LFP}_W} = \frac{T_C - \text{LFP}_C}{\text{UFP}_C - \text{LFP}_C} Given: LFPW=39W\text{LFP}_W = 39^{\circ}\text{W}, UFPW=239W\text{UFP}_W = 239^{\circ}\text{W} LFPC=0C\text{LFP}_C = 0^{\circ}\text{C}, UFPC=100C\text{UFP}_C = 100^{\circ}\text{C} TC=39CT_C = 39^{\circ}\text{C} Substituting the values: TW3923939=3901000\frac{T_W - 39}{239 - 39} = \frac{39 - 0}{100 - 0} TW39200=39100\frac{T_W - 39}{200} = \frac{39}{100} TW39=39×200100T_W - 39 = 39 \times \frac{200}{100} TW39=78T_W - 39 = 78 TW=78+39=117WT_W = 78 + 39 = 117^{\circ}\text{W}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Thermal Properties of Matter. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSThermal Properties of Mattertemperaturelinearcalledfreezingboiling

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