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The potential energy between two atoms in a molecule is given by $U (x) = \frac{a}{x^{12}} - \frac{b}{x^{6}}$ ; where a and b are positive constants and x is the distance between the atoms. The atoms are in stable equilibrium when:

$x = \sqrt[6]{\frac{11 a}{5 b}}$

$x = \sqrt[6]{\frac{a}{2 b}}$

x = 0

$x = \sqrt[6]{\frac{2 a}{b}}$

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Explanation

Condition for stable equilibrium $F = - \frac{d U}{d x} = 0$

⇒ $- \frac{d}{d x} [\frac{a}{x^{12}} - \frac{b}{x^{6}}] = 0$

⇒ $- 12 a x^{- 13} + 6 b x^{- 7} = 0$

⇒ $\frac{12 a}{x^{13}} = \frac{6 b}{x^{7}}$

⇒ $\frac{2 a}{b} = x^{6}$

⇒ $x = \sqrt[6]{\frac{2 a}{b}}$

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Which one of the following is not a conservative force

Gravitational force

Electrostatic force between two charges

Magnetic force between two magnetic dipoles

Frictional force

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Explanation

Friction is a non-conservative force.

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Work done in raising a box depends on

How fast it is raised

The strength of the man

The height by which it is raised

None of the above

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Explanation

Work in raising a box

= (weight of the box) × (height by which it is raised)

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If the water falls from a dam into a turbine wheel 19.6 m below, then the velocity of water at the turbine is (g = 9.8 m/s²)

9.8 m/s

19.6 m/s

39.2 m/s

98.0 m/s

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Explanation

Potential energy of water = kinetic energy at turbine

$m g h = \frac{1}{2} m v^{2}$

⇒ $v = \sqrt{2 g h} = \sqrt{2 \times 9 .8 \times 19 .6} = 19 .6 m / s$

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Two identical cylindrical vessels with their bases at the same level each contain a liquid of density; $ρ$ . The height of the liquid in one vessel is h₁ and that in the other vessel is h₂ $(h_{1} > h_{2})$ .The area of either base is A. The work done by gravity in equalizing the levels when the two vessels are connected is :

$(h_{1} - h_{2}) g ρ$

$(h_{1} - h_{2}) g A ρ$

$\frac{1}{2} {(h_{1} - h_{2})}^{2} g A ρ$

$\frac{1}{4} {(h_{1} - h_{2})}^{2} g A ρ$

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If the increase in the kinetic energy of a body is 22%, then the increase in the momentum will be

22%

44%

10%

300%

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Explanation

$P = \sqrt{2 m E} .$ If m is constant then

$\frac{P_{2}}{P_{1}} = \sqrt{\frac{E_{2}}{E_{1}}} = \sqrt{\frac{1 .22 E}{E}}$ ⇒ $\frac{P_{2}}{P_{1}} = \sqrt{1 .22} = 1 .1$

⇒ $P_{2} = 1 .1 P_{1}$ ⇒ $P_{2} = P_{1} + 0 .1 P_{1} = P_{1} + 10 % of P_{1}$

So the momentum will increase by 10%

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If a body of mass 200 g falls from a height 200 m and its total P.E. is converted into K.E. at the point of contact of the body with earth surface, then what is the decrease in P.E. of the body at the contact (g = 10 m/s²)

200 J

400 J

600 J

900 J

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Explanation

$Δ U = m g h = 0 .2 \times 10 \times 200$ = 400 J

∴ Gain in K.E. = decrease in P.E. = 400 J.

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If momentum is increased by 20%, then K.E. increases by

44%

55%

66%

77%

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Explanation

$E = \frac{P^{2}}{2 m} .$ If m is constant then $E \propto P^{2}$

⇒ $\frac{E_{2}}{E_{1}} = {(\frac{P_{2}}{P_{1}})}^{2} = {(\frac{1 .2 P}{P})}^{2} = 1 .44$

⇒ $E_{2} = 1 .44 E_{1} = E_{1} + 0 .44 E_{1}$

$E_{2} = E_{1} + 44 %$ of $E_{1}$

i.e. the kinetic energy will increase by 44%

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The kinetic energy of a body of mass 2 kg and momentum of 2 Ns is

1 J

2 J

3 J

4 J

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Explanation

$E = \frac{P^{2}}{2 m} = \frac{{(2)}^{2}}{2 \times 2} = 1 J$

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The decrease in the potential energy of a ball of mass 20 kg which falls from a height of 50 cm is

968 J

98 J

1980 J

None of these

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Explanation

$Δ U = m g h = 20 \times 9 .8 \times 0 .5 = 98 J$

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