krit.club logo

Where are we now and where might we be going - Least count and use of equipments

Grade 7IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Least Count (LCLC) is defined as the smallest value that can be measured accurately using a specific measuring instrument.

•

A smaller Least Count indicates that the instrument is more precise and can provide more detailed measurements.

•

Parallax Error is a common observational error where the measurement appears different because the eye is positioned at an angle to the scale. To avoid this, always keep the line of sight perpendicular to the scale.

•

When measuring liquids in a cylinder, always read the scale at the bottom of the concave meniscus (for water) at eye level.

•

Accuracy refers to how close a measurement is to the true value, while Precision refers to how consistent multiple measurements are with each other.

•

Zero Error occurs when an instrument does not read 00 when the actual measurement is zero. This must be added or subtracted from the final reading to get the corrected value.

📐Formulae

LC=Value of one main scale divisionTotal number of small divisions between two main marksLC = \frac{\text{Value of one main scale division}}{\text{Total number of small divisions between two main marks}}

LC=Value of higher mark−Value of lower markNumber of divisions between themLC = \frac{\text{Value of higher mark} - \text{Value of lower mark}}{\text{Number of divisions between them}}

Total Reading=Main Scale Reading+(Number of divisions×LC)\text{Total Reading} = \text{Main Scale Reading} + (\text{Number of divisions} \times LC)

💡Examples

Problem 1:

On a laboratory thermometer, there are 1010 small divisions between the 20∘C20^{\circ}\text{C} and 30∘C30^{\circ}\text{C} marks. Calculate the Least Count of the thermometer.

Solution:

LC=30∘C−20∘C10LC = \frac{30^{\circ}\text{C} - 20^{\circ}\text{C}}{10} LC=10∘C10=1∘CLC = \frac{10^{\circ}\text{C}}{10} = 1^{\circ}\text{C}

Explanation:

The difference between two consecutive major markings is 1010 units. Since there are 1010 small steps to cover this distance, each step represents 1∘C1^{\circ}\text{C}.

Problem 2:

A measuring cylinder has markings for 50 mL50 \text{ mL} and 100 mL100 \text{ mL}. There are 55 divisions between these two marks. What is the smallest volume this cylinder can measure accurately?

Solution:

LC=100 mL−50 mL5LC = \frac{100 \text{ mL} - 50 \text{ mL}}{5} LC=50 mL5=10 mLLC = \frac{50 \text{ mL}}{5} = 10 \text{ mL}

Explanation:

The smallest volume it can measure is its Least Count. By dividing the total volume between marks (50 mL50 \text{ mL}) by the number of divisions (55), we find that each mark represents 10 mL10 \text{ mL}.

Problem 3:

A ruler shows 1010 divisions between 00 and 1 cm1 \text{ cm}. If an object's edge falls on the 7th7^{th} small division after the 3 cm3 \text{ cm} mark, what is the length?

Solution:

LC=1 cm10=0.1 cmLC = \frac{1 \text{ cm}}{10} = 0.1 \text{ cm} Length=3 cm+(7×0.1 cm)\text{Length} = 3 \text{ cm} + (7 \times 0.1 \text{ cm}) Length=3 cm+0.7 cm=3.7 cm\text{Length} = 3 \text{ cm} + 0.7 \text{ cm} = 3.7 \text{ cm}

Explanation:

First, find the value of one small division (0.1 cm0.1 \text{ cm}). Then, add the value of the 77 small divisions to the base reading of 3 cm3 \text{ cm}.