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# Tag Archives: density

## Dimensions and conversion of units

The height of mercury column in a barometer in a Calcutta laboratory was recorded to be 75 cm. Calculate this pressure in SI and CGS units using the following data : specific gravity is 13.6, density of water is 1000 kg/m3, g is 9.8 m/s2 at Calcutta. Pressure = hρg in usual symbols

## Current electricity – Numerical

A 10m length of aluminium wire has a diameter of 1.5mm. It carries a current of 12A. Find (i) the current density. (ii) the drift velocity (iii) the electric field in the wire. aluminium has approximately 10^29 free electrons per m^3

## A numerical problem on Drift velocity

Two conducting wires X and Y of same diameter but different materials are joined in series across a battery . if the number density of electrons in X is twice that in Y . Find the ratio of drift velocity of electrons in the two wires.

## Dimensional Analysis – A solved example

Deduce the dimensional formula for workdone,pressure and density. [posted by boineelo ]

workdone

We know that workdone = force x distance

Therefore [work done] =  [force] [distance]

but force = ma

[force]=[m][a]

[a]=[velocity]/[t]

[velocity]=[displacement]/[t]

So, putting all these together

[work done]=[m][distance][distance]/[time][time]

[work done]=ML2T-2

Similarly, you can work out the dimensional formula for pressure and density as

[pressure]=[force]/[area]=ML-1T-2

[density]=[mass]/[volume]=ML-3

## Explanation For Empirical Horopter: Density Of Photoreceptors

Karolis  asks: ” External SiteLink -In this website (and in many more) in page 5, it’s written that photoreceptors are more densely packed in nasal areas of retina than temporal. That’s one of the explanations why empirical horopter (that’s the unit of all points in visual field, that are seen in the same position monocularly) deviates from theoretical. In page 6, in graphs we see that the same segment in visual field has longer arc in nasal retina than temporal (binocular vision). I don’t see how these two things agree: density of photoreceptors in nasal and temporal retinas and the magnitude of arcs. If compensation would be the case, then I think it should be the reverse situation. Given there are less amount of photoreceptors in temporal area, arc should be longer than that of nasal – so that the quantity of photoreceptors in both areas would be the same.
What’s wrong with my thinking?”

## why sound travel faster in moist air than in dry air?

“Why sound travel faster in moist air than in dry
air?”

The density of dry air is more than that of moist air (Wonder why? Just answer me – which is denser – skimmed milk or fresh milk. The cream is lighter and when removed from milk, we get skimmed milk and therefore skimmed milk will be denser than the fresh one with cream content. Just like that the water vapor is lighter than dry air. When moisture is removed from air, its density increases). The speed of sound in a medium is inversely proportional to the square root of its density. Therefore, the speed of sound in moist air is more than that in dry air.

## A problem from Hydraulic

“a stone of [wiki]density [/wiki] 4gm/cm3 is dropped freely in a liquid of density 0.8 gm/cm3 .what will be the acceleration of the sinking stone?”

Ans:

the force acting are

weight =mg= $\frac{4}{3}\pi r^{3}\rho g$ downwards (where $\rho$ is the density of the body)

[wiki]Buoyant force[/wiki] =$\frac{4}{3}\pi r^{3}\sigma g$

SO, THE NET DOWNWARD [wiki]FORCE [/wiki]IS
$\frac{4}{3}\pi r^{3}(\rho -\sigma ) g$

Therefore, acceleration = net downward force / mass

=$\frac{\frac{4}{3}\pi r^{3}(\rho -\sigma ) g}{\frac{4}{3}\pi r^{3}\rho }$
=$\left (1-\frac{\sigma }{\rho } \right )g$

Now substitute the values and calculate

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