A student measures thickness of a human hair by looking at it through a microscope of magnification

English

Gujarati

Hindi

1.Units, Dimensions and Measurement

easy

A student measures the thickness of a human hair by looking at it through a microscope of magnification $100$. He makes $20$ observations and finds that the average width of the hair in the field of view of the microscope is $3.5 \;mm$. What is the estimate on the thickness(in $mm$) of hair ?

A

$0.35$

B

$3.5$

C

$0.0035$

D

$0.035$

Solution

Magnification of the microscope $=100$

Average width of the hair in the field of view of the microscope $=3.5 mm$

$\therefore$ Actual thickness of the hair is $\frac{3.5}{100}=0.035 mm$

Standard 11

Physics

Share

Similar Questions

If the screw on a screw-gauge is given six rotations, it moves by $3\; \mathrm{mm}$ on the main scale. If there are $50$ divisions on the circular scale the least count of the screw gauge is

medium

(JEE MAIN-2020)

The main scale of vernier callipers reads in millimetre and its vernier is divided into $8$ divisions, which coincides with $5$ divisions of main scale. When two jaws of instrument touch each other, the zero of the vernier coincides with the zero of main scale. A rod is tightly placed along its length between both jaws. It is observed that the zero of vernier scale lies just left to $36^{th}$ division of main scale and fourth division of vernier scale coincides with the main scale Then the measured value is ………. $cm$

hard

A spectrometer gives the following reading when used to measure the angle of a prism.

Main scale reading : $58.5$ degree

Vernier scale reading : $09$ divisions

Given that $1$ division on main scale corresponds to $0 .5$ degree. Total divisions on the Vernier scale is $30$ and match with $29$ divisions of the main scale. The angle of the prism from the above data ……. $degree$

medium

(AIEEE-2012)

A vernier callipers has $20$ divisions on the vernier scale, which coincides with $19^{\text {th }}$ division on the main scale. The least count of the instrument is $0.1 \mathrm{~mm}$. One main scale division is equal to $. . . . . ..$ $\mathrm{mm}$

medium

(JEE MAIN-2024)

There are two Vernier calipers both of which have $1 \mathrm{~cm}$ divided into $10$ equal divisions on the main scale. The Vernier scale of one of the calipers $\left(C_1\right)$ has $10$ squal divisions that correspond to $9$ main scale divisions. The Vernier scale of the other caliper $\left(C_2\right)$ has $10$ equal divisions that correspond to $11$ main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in $\mathrm{cm}$ ) by calipers $C_1$ and $C_2$, respectively, are

normal

(IIT-2016)