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Physics MCQs for NEET — Practice Questions with Answers

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The radill of two soap bubbles are r1 and r2. In isothermal conditions two meet together is vacum Then the radius of the resultant bubble is given by

A. $ R = ( r_1 + r_2 ) / 2 $ B. $ R = r_1 ( r_1 + r_2 + r_3) $ C. $ R^2 = r_1^2 + r_2^2 $ D. $ R = r_1 + r_2 $

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Explanation

When two soap bubbles combine, the volume is conserved. The relationship between the radii of the initial bubbles and the resultant bubble is given by $R^3 = r_1^3 + r_2^3$. Simplifying for radius in terms of the areas, we get $R^2 = r_1^2 + r_2^2$.

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A spherical drop of coater has radius 1 mm if surface tension of contex is $70 \times 10^ {-3} N / m $ difference of pressures between inside and outside of the spherical drop is

A. $ 35 N/m^2 $ B. $ 70 N/m^2 $ C. $ 140 N/m^2 $ D. Zero

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Explanation

$ \triangle p = { 2T \over R } $

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In capilley pressure below the curved surface at water will be

A. Equal to atomospheric B. Equal to upper side pressure C. More than upper side pressure D. Lesser than upper side pressure

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Two bubbles A and B (A>B) are joined through a narrow tube than

A. The size of a will increase B. The size of B will increase C. The size of B will increase until then pressure equals D. None of these

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Explanation

$ r_A \gt r_B and p \alpha {1 \over r } $ $ So P_A \lt P_B $ So air will flow from B to A i.e. size ofA will increase

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A capillary tube at radius R is immersed in water and water rises in it to a height H. Mass of water in the capillary tube is M. If the radius of the tube is doubled. Mass of water that will rise in the capillary tube will now be

A. M B. 2M D. 4M

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Explanation

Mass of liquid in capillary tube $ M = \pi R^2 - \rho $ $ M \alpha R^2 \times (1 / R } $ $ M \alpha R $ If radius become double then mass will become twice.

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A vessel whose bottom has round holes with diametre of 0.1 mm is filled with water. The maximum height to which the water can be filled without leakage is $ ( S.T of water = 75 dyne /cm , g = 1000 m/s^2 ) $

A. 100c m B. 75cm C. 50 cm D. 30 cm

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The correct relation is

A. $ r ={ 2T cos \theta \over hdg } $ B. $ r ={ hdg \over 2T cos \theta } $ C. $ r ={ 2T dgh \over cos \theta } $ D. $ r ={ T cos \theta \over 2 hdg } $

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Explanation

The correct relation for capillary rise is $r = rac{2T ext{ cos } heta}{hdg}$. Here, $r$ is the radius, $T$ is the surface tension, $ heta$ is the contact angle, $h$ is the height of the liquid column, $d$ is the density of the liquid, and $g$ is the acceleration due to gravity. Hence, option o1 is correct.

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In a capillary tube water rises by 1.2 mm. The height of water that will rise in another capillary tube having half the radius of the first is

A. 1.2 mm B. 2.4 mm C. 0.6 mm D. 0.4 mm

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Explanation

$ h = { 2T \over rdg} $ $ \therefore h \alpha { 1 \over r }$ $ \therefore r_1 h_1 = r_2 h_2 $

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Water rises in a vertical capillary tube upto a beight of 2.0 cm. If tube is inclined at an angle of $ 60^ \circ $ with the verticalthen the what length the water will rise in the tube.

A. 2.0 cm B. 4.0 cm C. $ { 4 \over \sqrt 3 } cm $ D. $ 2 \sqrt 2 cm $

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Explanation

$l = { h \over cos \theta } = { \alpha \over cos 60 } = 4.0 cm $

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The lower end of a glass capillary tube is dipped in water rises to a height of 8 cm the tube is then broken at a height of 6 cm. The height of water column and engled as contact will be

A. $ 6cm .sin ^ {-1} {3 \over 4} $ B. $ 6cm .sin ^ {-1} {4 \over 5 } $ C. $ 6cm .sin ^ {-1} {3 \over 4} $ D. $ 6cm .sin ^ {-1} {1 \over 2 } $

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Explanation

When a capillary tube is broken at a height of 6 cm the height of water column will be 6 cm. $ As h = { 25 cos \theta \over \rho rg } or { h \over cos \theta } = constant $ $ { 8 \over cos 0 ^\circ } = { 6 \over cos \theta } or cos \theta = { 6 cos \theta \over 8 } = {3 \over 4} $ $ \therefore \theta = cos ^ {-1} \left( {3 \over 4} \right) $

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