One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM

DETAILS

MRP

DISCOUNT

FINAL PRICE

Total Price: Rs.

There are no items in this cart.

Continue Shopping

Continue Shopping

If two soap bubbles of different radii are connected by a tube then, what happen..?

8 years ago

The excess pressure inside the soap bubble is inversely proportional to radius of soap bubble i.e.P ∝1/ r, r being the radius of bubble. It follows that pressure inside a smaller bubble is greater than that inside a bigger bubble. Thus, if these two bubbles are connected by a tube, air will flow from smaller bubble to bigger bubble and the bigger bubble grows at the expense of the smaller one.

8 years ago

*Air flows from ** the bigger bubble to the smaller bubble till the sizes become equal*

8 years ago

air flows from smaller bubble to bigger bubble...hence,smaller bubble gets smaller and bigger bubble gets even bigger

8 years ago

The excess pressure inside the soap bubble is inversely proportional to radius of soap bubble i.e.P ∝1/ r, r being the radius of bubble. It follows that pressure inside a smaller bubble is greater than that inside a bigger bubble. Thus, if these two bubbles are connected by a tube, air will flow from smaller bubble to bigger bubble and the bigger bubble grows at the expense of the smaller one

8 years ago

When these bubbles are joined through tube..the smaller bubble(with more pressure) gets its size decresed as the time passes by..and large bubble(with less pressure) gets its size increased..The smaller bubble size decreases upto that extent when the radius of bubble becomes infinity..(i.e...it becomes flat with respect to tube)

2 years ago

Excess pressure in a soap bubble ( in concave part ) is 4S/r .

Pressure inside a soap bubble will be = P + 4S/r.

S is the surface tension of the soap liquid and r is the radius of the bubble and P is the atmospheric pressure.

When two soap bubble will come in contact then contact area will be bulge towards the bigger sphere. air pressure inside the smaller soap bubble will be greater as compared to the bigger sphere.

If pressure inside bigger bubble is P1 and inside smaller bubble is P2

P1= P +4S/r1,

P2= P+ 4S/r2

P2>P1 as r1> r2

To find the radius of common surface (r3 say) , we can write P2-P1= 4S/r3.

If they are coalescing then air will flow from smaller bubble to the bigger bubble as the pressure inside the smaller bubble is greater compare to the bigger bubble.

10 months ago

Please find below the solution to your problem.

Pressure inside a bubble =Po + 4S/R

When they are connected using a tube then, as smaller bubbles {due to small radius} has higher pressure, the air flows from smaller bubble to bigger bubble until the smaller bubble vanishes completely.

Thanks and Regards

copyright © 2006-2021 askIITians.com

info@askiitians.com