JEE Main 2023 · 25 January · evening · Single correct · +4 / −1
The graph between two temperature scales P and Q is shown in the figure. Between upper fixed point and lower fixed point there are 150 equal divisions of scale P and 100 divisions on scale Q. The relationship for conversion between the two scales is given by :-
- (a) t_P/100 = (t_Q - 180)/150
- (b) t_P/180 - (t_Q - 40)/100
- (c) t_Q/150 = (t_P - 180)/100
- (d) t_Q/100 = (t_P - 30)/150
Answer and solution
Answer (d) t_Q/100 = (t_P - 30)/150
Solution
Physics sir · 4 steps
Read both fixed points off the graph, then match fractions of the way up on each scale.
- Read the graph. The line starts on the P axis, where t_Q = 0. The upper fixed point is at t_P = 180, and P has 150 divisions between the fixed points. So the lower fixed point of P is 180 - 150 = 30.
- Fixed points. Scale P runs from 30 to 180 (150 divisions). Scale Q runs from 0 to 100 (100 divisions).
- Same fraction. Any temperature sits the same fraction of the way from the lower to the upper fixed point on every scale. That fraction is (reading minus lower point) divided by the number of divisions.
(t_Q - 0)/100 = (t_P - 30)/150
- So. The conversion is t_Q/100 = (t_P - 30)/150, which is option (d).
The idea to keep. For any two linear scales: (reading minus lower fixed point) divided by (upper minus lower) is the same on both. Always find both fixed points first.
Answer from the NTA final answer key. Solution by PrepWiser.