Question

the difference between two marked values is 5 ml and there are 5 spaces between the marked values. what is the calibration of this graduated cylinder?
2 ml
10 ml
5 ml
1 ml
(image of a graduated cylinder with markings 60, 65, 70 ml)

Explanation:

Step1: Understand calibration concept

Calibration (interval between marks) = Total difference / Number of spaces.

Step2: Plug in values

Total difference is 5 mL, number of spaces is 5. So calibration = $\frac{5}{5}$ = 1 mL? Wait, no, wait. Wait, looking at the diagram, between 60 and 65: let's check. Wait the problem says "the difference between two marked values is 5 mL and there are 5 spaces between the marked values". Wait, maybe I misread. Wait, let's re-express. Wait, suppose two marked values (like 60 and 65) have a difference of 5 mL, and between them, how many spaces? Wait the diagram shows between 60 and 65: let's count the lines. Wait the problem statement says "there are 5 spaces between the marked values". So difference is 5 mL, spaces is 5. So each space is 5/5 = 1 mL? Wait but let's check the options. Wait no, wait maybe the marked values are, say, 65 and 70? Wait no, the problem says "the difference between two marked values is 5 mL and there are 5 spaces between the marked values". So calibration (the value per space) is 5 mL / 5 spaces = 1 mL? Wait but let's check the diagram. Wait the diagram has 60, 65, 70. Between 60 and 65: how many spaces? Let's count the lines. From 60 to 65: the lines are 60, then a line, then another, up to 65. Wait maybe the number of spaces between 60 and 65: let's see, 60 to 65 is 5 mL. If there are 5 spaces, then each space is 1 mL. Wait but let's do the calculation again.

Wait the formula for calibration (the value of each small division) is: Calibration = (Difference between two major marks) / (Number of divisions between them). So here, difference is 5 mL, number of spaces (divisions) between is 5. So 5 / 5 = 1 mL. Wait but let's check the options. The options are 2,10,5,1. So 1 mL is an option. Wait maybe I made a mistake earlier. Wait let's re-express. Suppose the two marked values are, for example, 65 and 70: difference is 5 mL. Between them, how many spaces? Looking at the diagram, between 65 and 70: let's count the lines. From 65 to 70: the lines are 65, then a line, then another, up to 70. Wait maybe the number of spaces between 65 and 70 is 5? Then 5 mL /5 spaces =1 mL. So the calibration is 1 mL.

Wait but let's check the problem statement again: "the difference between two marked values is 5 mL and there are 5 spaces between the marked values". So yes, 5 mL difference, 5 spaces. So 5/5=1 mL. So the answer should be 1 mL.

Answer:

1 mL (Option: 1 mL)