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Physics MCQs for NEET — Practice Questions with Answers

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A spring of spring constant 5 × 10³ N/m is stretched initially by 5cm from the unstretched position. Then the work required to stretch it further by another 5 cm is

6.25 N-m

12.50 N-m

18.75 N-m

25.00 N-m

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Explanation

$W = \frac{1}{2} k (x_{2}^{2} - x_{1}^{2}) = \frac{1}{2} \times 5 \times 10^{3} (10^{2} - 5^{2}) \times 10^{- 4}$

$= 18 .75 J$

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A mass of 0.5kg moving with a speed of 1.5 m/s on a horizontal smooth surface, collides with a nearly weightless spring of force constant k = 50 N/m. The maximum compression of the spring would be

0.15 m

0.12 m

1.5 m

0.5 m

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Explanation

The kinetic energy of mass is converted into potential energy of a spring

$\frac{1}{2} m v^{2} = \frac{1}{2} k x^{2}$ ⇒ $x = \sqrt{\frac{m v^{2}}{k}} = \sqrt{\frac{0 .5 \times {(1 .5)}^{2}}{50}} = 0 .15 m$

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A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to

x²

e^x

log_e x

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Explanation

$According to the question : a = - k x \int_{u}^{v} v d v = - \int_{0}^{x} k x d x \frac{v^{2} - u^{2}}{2} = - \frac{k x^{2}}{2} \frac{m (v^{2} - u^{2})}{2} = - m \frac{k x^{2}}{2}$

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If a long spring is stretched by 0.02 m, its potential energy is U. If the spring is stretched by 0.1 m, then its potential energy will be

$\frac{U}{5}$

25U

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Explanation

$U \propto x^{2}$

⇒ $\frac{U_{2}}{U_{1}} = {(\frac{x_{2}}{x_{1}})}^{2} = {(\frac{0 .1}{0 .02})}^{2} = 25$

∴ $U_{2} = 25 U$

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The spring extends by x on loading, then energy stored by the spring is : (if T is the tension in spring and k is spring constant)

$\frac{T^{2}}{2 k}$

$\frac{T^{2}}{2 k^{2}}$

$\frac{2 k}{T^{2}}$

$\frac{2 T^{2}}{k}$

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Explanation

$x = \frac{T}{k} U = \frac{1}{2} k x^{2} = \frac{T^{2}}{2 k}$

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The potential energy of a body is given by, U = A – Bx² (Where x is the displacement). The magnitude of force acting on the particle is

Constant

Proportional to x

Proportional to x²

Inversely proportional to x

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Explanation

$U = A - B x^{2}$

⇒ $F = - \frac{d U}{d x} = 2 B x$

⇒ $F \propto x$

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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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