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Cell - How to Study Cells?

Grade 9CBSE

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

🔑Concepts

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The cell is the fundamental structural and functional unit of all living organisms. It was first discovered by Robert Hooke in 1665 using a primitive microscope to observe cork cells.

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Microscopes are essential tools for studying cells. A Compound Microscope uses visible light and glass lenses. The total magnification is the product of the ocular lens and the objective lens magnification.

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Cell Theory, proposed by Schleiden and Schwann and later expanded by Virchow, states: 1. All living things are composed of cells. 2. The cell is the basic unit of life. 3. All cells arise from pre-existing cells (Omnis cellula e cellulaOmnis \ cellula \ e \ cellula).

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To observe cells clearly under a microscope, specimens are stained with dyes. Common stains include Safranin (for plant cells/lignin) and Methylene Blue (for animal cells/nuclei).

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Mounting is the process of placing the specimen on a slide. Glycerin is often used as a mounting medium because it prevents the specimen from drying out (desiccation).

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Cells are measured in very small units: the micrometre (μm\mu m) and the nanometre (nmnm). 1μm=10−6m1 \mu m = 10^{-6} m and 1nm=10−9m1 nm = 10^{-9} m.

📐Formulae

Total Magnification=Magnification of Ocular Lens×Magnification of Objective Lens\text{Total Magnification} = \text{Magnification of Ocular Lens} \times \text{Magnification of Objective Lens}

1μm=10−3mm=10−6m1 \mu m = 10^{-3} mm = 10^{-6} m

1nm=10−6mm=10−9m1 nm = 10^{-6} mm = 10^{-9} m

1μm=1000nm1 \mu m = 1000 nm

💡Examples

Problem 1:

A student is using a compound microscope. If the eyepiece (ocular lens) has a magnification of 15x15x and the high-power objective lens has a magnification of 40x40x, calculate the total magnification of the cell being observed.

Solution:

Using the formula: Total Magnification=15×40=600x\text{Total Magnification} = 15 \times 40 = 600x

Explanation:

The total magnification of a microscope is determined by multiplying the magnifying power of the ocular lens by the magnifying power of the objective lens currently in use.

Problem 2:

The diameter of a typical animal cell is approximately 20μm20 \mu m. Convert this value into millimetres (mmmm).

Solution:

Since 1μm=10−3mm1 \mu m = 10^{-3} mm, then: 20μm=20×10−3mm=0.02mm20 \mu m = 20 \times 10^{-3} mm = 0.02 mm

Explanation:

To convert micrometres to millimetres, we divide the value by 10001000 because there are 10001000 micrometres in one millimetre.

Problem 3:

A bacterial cell is 2μm2 \mu m long. Express its length in nanometres (nmnm).

Solution:

We know 1μm=1000nm1 \mu m = 1000 nm. Therefore: 2μm=2×1000nm=2000nm2 \mu m = 2 \times 1000 nm = 2000 nm

Explanation:

Nanometres are a smaller unit than micrometres. To convert from μm\mu m to nmnm, we multiply by 10310^3.