a) State one principle of thermometric measurements. (1mark) b) (II). A circular ring at 30.0^circ C has a hole with an area of 21.3 cm^2 . What minimum temperature must it have so that it can be slippe onto a steel metal rod having a cross-sectional area of 22.00cm^2 (a copper is 17times 10^-6 and a steel is 11times 10^-6 (5marks) Suppose the ring and the rod are heated simultaneously. Wh minimum change in temperature of both will allow the ring to be slippe onto the end of the rod. (5marks)

a) One principle of thermometric measurements is the principle of expansion. This principle states that the length or volume of a substance will change in direct proportion to the change in temperature, provided the change in temperature is within the range of linearity for that substance.b) To find the minimum temperature at which the circular ring can be slipped onto the steel metal rod, we need to consider the thermal expansion of both the ring and the rod. The thermal expansion of a material is given by the formula: where:- is the change in length,- is the coefficient of linear expansion,- is the original length,- is the change in temperature.Given:- The area of the hole in the ring is ,- The cross-sectional area of the steel rod is ,- The coefficient of linear expansion for copper is ,- The coefficient of linear expansion for steel is ,- The initial temperature of the ring is .First, we need to find the initial lengths of the ring and the rod. Since the area of the hole in the ring is given, we can assume the ring's outer diameter is such that the area of the hole represents a significant portion of the ring's area. Let's denote the outer radius of the ring as and the inner radius (hole) as . The area of the hole is given by: Solving for : Assuming the ring is a perfect circle, the outer radius is: For simplicity, let's assume the thickness is negligible compared to the radius, so .Next, we need to find the initial length of the steel rod. Since the cross-sectional area is given, we can assume the rod is a perfect cylinder. The length of the rod can be found if we know the volume or diameter, but since we don't have that information, we'll assume the length is such that the cross-sectional area is .Now, we need to find the temperature change that will cause the ring to slip onto the rod. The expansion of the ring and the rod must be equal. The expansion of the ring is given by: The expansion of the rod is given by: For the ring to slip onto the rod, the expansion of the ring must be greater than the expansion of the rod. Therefore, we set up the inequality: Simplifying, we get: Solving for : Substituting the given values: \[\Delta T >