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Physics MCQs for NEET — Practice Questions with Answers

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Work of 6.0 x 10 $^{- 4}$ Joule is required to be done in increasing the size of a soap film from 10cm x 6cm to 10cm x 11cm. The surface tension of the film is :

5 x 10 $^{- 2}$ N/m

6 x 10 $^{- 2}$ N/m

1.5 x 10 $^{- 2}$ N/m

1.2 x 10 $^{- 1}$ N/m

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Explanation

$S u r f a c e T e n s i o n = \frac{S u r f a c e E n e r g y}{A r e a} S u r f a c e E n e r g y = S u r f a c e T e n s i o n \times A r e a W o r k d o n e = c h a n g e i n s u r f a c e e n e r g y = S u r f a c e T e n s i o n \times C h a n g e i n a r e a S u r f a c e T e n s i o n = \frac{6 \times 10^{- 4}}{2 \times 10 \times 5 \times 10^{- 4}} = \frac{6}{100} = 6 \times 10^{- 2} N / m$

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When a cylindrical tube is dipped vertically into a liquid the angle of contact is 140^o. When the tube is dipped with an inclination of 40^o, the angle of contact is-

$100 °$

$140 °$

$180 °$

$60 °$

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Explanation

The angle of contact is independent of the tilting angle.

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Two liquid drops have their diameters as 1 mm and 2 mm. The ratio of excess pressure in them is :

1:2

2:1

4:1

1:4

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Explanation

$(2) ∆ P = \frac{2 T}{r} \frac{∆ P_{1}}{∆ P_{2}} = \frac{r_{2}}{r_{1}} = \frac{2}{1}$

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If a soap bubble of radius 3 cm coalesce with another soap bubble of radius 4 cm under isothermal conditions, the radius of the resultant bubble formed is in cm-

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Explanation

$P_{1} = \frac{4 T}{r_{1}} & V_{1} = \frac{4}{3} {πr}_{1}^{3} P_{2} = \frac{4 T}{r_{2}} & V_{2} = \frac{4}{3} {πr}_{2}^{3} Two bubbles coelesce isothermally : PV = P_{1} V_{1} + P_{2} V_{2} \frac{4 T}{r} \times \frac{4}{3} {πr}^{3} = \frac{4 T}{r_{1}} \times \frac{4}{3} {πr}_{1}^{3} + \frac{4 T}{r_{2}} \times \frac{4}{3} {πr}_{2}^{3} r^{2} = {r_{1}}^{2} + {r_{2}}^{2} r = \sqrt{{r_{1}}^{2} + {r_{2}}^{2}} = 5 c m$

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$Water flows through a non - uniform tube of area of cross section s A, B, and C whose value s are 25, 15, and 35 {cm}^{2} respectively . The ratio of the velocities of water at the sections A, B, and C is -$

5:3:7

7:3:5

21:35:15

1:1:1

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Explanation

$By equation of continuity Av = const . A_{1} v_{1} = A_{2} v_{2} = A_{3} v_{3} Since A_{1} : A_{2} : A_{3} = 25 : 15 : 35 v_{1} : v_{2} : v_{3} = 21 : 35 : 15$

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W is the work done in forming a bubble of radius r, the work is done in forming a bubble of radius 2r will be:-

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Explanation

$(1) W = T \times A = T \times 4 {πr}^{2} W \infty r^{2}$

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Equation of continuity based on :

Conservation of mass

Conservation of energy

Conservation of angular momentum

None of these

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Explanation

It is based on the conservation of mass

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Bernoulli's theorem is based on :

conservation of energy

conservation of mass

conservation of momentum

conservation of angular momentum

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Explanation

Conceptual

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Two spherical soap bubbles of radii r₁ and r₂ in vaccum collapse under isothermal condition. The resulting bubble has radius equal to :

$\frac{r_{1} + r_{2}}{2}$

$\frac{r_{1} r_{2}}{r_{1} - r_{2}}$

$\sqrt{r_{1} r_{1}}$

$\sqrt{r_{1}^{2} + r_{2}^{2}}$

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Explanation

$(4) for isothermal condition PV = P_{1} V_{1} + P_{2} V_{2} \frac{4 T}{r} \frac{4}{3} {πr}^{3} = \frac{4 T}{r_{1}} \frac{4}{3} {πr}_{1}^{3} + \frac{4 T}{r_{2}} . \frac{4}{3} {πr}_{2}^{3} r^{2} = r_{1}^{2} + r_{2}^{2}$

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The energy needed in breaking of a drop of liquid of radius R into n drops of radius r is given by (T is surface tension and P is atmospheric pressure) :

$(4 {πr}^{2} n - 4 {πR}^{2}) T$

$(4 / 3 {πr}^{2} n - 4 / 3 {πR}^{3}) T$

$(4 {πR}^{2} n - 4 {πr}^{2}) T$

$(8 {πr}^{2} n - 8 {πR}^{2}) T$

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Explanation

$S u r f a c e a r e a o f l a r g e s p h e r e = 4 {πR}^{2} S u r f a c e a r e a o f n s m a l l s p h e r e s = n 4 π r^{2} Work done = T \times ∆ A = T \times (n 4 {πr}^{2} - 4 {πR}^{2})$

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