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NEET
Exam
NEET
Subject
Physics
Chapter
Mechanical Properties of Fluids

NEET · Physics · Mechanical Properties of Fluids

A soap bubble, blown by a mechanical pump at the mouth of a tube...

A soap bubble, blown by a mechanical pump at the mouth of a tube increases in volume with time at a constant rate. The graph that correctly depicts the time dependence of pressure inside the bubble is given by:A soap bubble, blown by a mechanical pump at the mouth of a tube increases in volume with time at a constant rate. TheA soap bubble, blown by a mechanical pump at the mouth of a tube increases in volume with time at a constant rate. TheA soap bubble, blown by a mechanical pump at the mouth of a tube increases in volume with time at a constant rate. TheA soap bubble, blown by a mechanical pump at the mouth of a tube increases in volume with time at a constant rate. The

Options

  1. AA mass M is tied to the top of two identical poles of height H using massless strings of equal length. The mass is at
  2. BA mass M is tied to the top of two identical poles of height H using massless strings of equal length. The mass is at
  3. CA mass M is tied to the top of two identical poles of height H using massless strings of equal length. The mass is at
  4. DA mass M is tied to the top of two identical poles of height H using massless strings of equal length. The mass is at

Correct answer

D. A mass M is tied to the top of two identical poles of height H using massless strings of equal length. The mass is at

Step-by-step solution

dVdt=k\dfrac{dV}{dt} = k

⇒V=kt⇒ V = kt

⇒43πr3=kt⇒ \dfrac{4}{3}\pi{r}^{3} = kt

⇒r=(3k4π)13t13⇒ r = {\left(\dfrac{3k}{4\pi}\right)}^{\dfrac{1}{3}}{t}^{\dfrac{1}{3}}

P−P0=4TrP - {P}_{0} = \dfrac{4T}{r}

⇒P−P0=4T.(3k4π)13.1t13⇒ P - {P}_{0} = 4T . \left(\dfrac{3k}{4\pi}\right)\dfrac{1}{3} . \dfrac{1}{t\dfrac{1}{3}}⇒P=4T.(3k4π)13.1t13+P0⇒ P = 4T . \left(\dfrac{3k}{4\pi}\right)\dfrac{1}{3} . \dfrac{1}{t\dfrac{1}{3}} + {P}_{0}

 

 

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