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Physics MCQs for NEET — Practice Questions with Answers

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If two forces of 5 N each are acting along X and Y axes, then the magnitude and direction of resultant is

$5 \sqrt{2}, π / 3$

$5 \sqrt{2}, π / 4$

$- 5 \sqrt{2}, π / 3$

$- 5 \sqrt{2}, π / 4$

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Explanation

$| \vec{F} | = \sqrt{5^{2} + 5^{2}} = 5 \sqrt{2} N .$

and $\tan θ = \frac{5}{5}$ =1

$\Rightarrow θ = π / 4 .$

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A block is kept on a frictionless inclined surface with angle of inclination 'α'. The incline is given an acceleration 'a' to keep the block stationary. Then a is equal to

g tan α

g / tan α

g cosec α

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Explanation

For a block to remain stationary on a frictionless inclined surface with angle Î±, the acceleration a given to the incline must be equal to g tan Î±, where g is the acceleration due to gravity. This acceleration provides just enough force component along the incline to balance the weight component.

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Two masses 2 kg and 3 kg are attached to the end of the string passed over a pulley fixed at the top. The tension and acceleration are

$\frac{7 g}{8}; \frac{g}{8}$

$\frac{21 g}{8}; \frac{g}{8}$

$\frac{21 g}{8}; \frac{g}{5}$

$\frac{12 g}{5}; \frac{g}{5}$

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Explanation

$T = \frac{2 m_{1} m_{2}}{m_{1} + m_{2}} g = \frac{2 \times 2 \times 3}{2 + 3} g = \frac{12}{5} g$

$a = (\frac{m_{2} - m_{1}}{m_{1} + m_{2}}) g = (\frac{3 - 2}{3 + 2}) g = \frac{g}{5}$

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Two bodies of mass 3 kg and 4 kg are suspended at the ends of massless string passing over a frictionless pulley. The acceleration of the system is (g = 9.8 m/s²)

4.9 m/s²

2.45 m/s²

1.4 m/s²

9.5 m/s²

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Explanation

Acceleration $= \frac{(m_{2} - m_{1})}{(m_{2} + m_{1})} g$

$= \frac{4 - 3}{4 + 3} \times 9 .8 = \frac{9 .8}{7} = 1 .4 m / \sec^{2}$

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Three solids of masses m₁, m₂ and m₃ are connected with weightless string in succession and are placed on a frictionless table. If the mass m₃ is dragged with a force T, the tension in the string between m₂ and m₃ is

$\frac{m_{2}}{m_{1} + m_{2} + m_{3}} T$

$\frac{m_{3}}{m_{1} + m_{2} + m_{3}} T$

$\frac{m_{1} + m_{2}}{m_{1} + m_{2} + m_{3}} T$

$\frac{m_{2} + m_{3}}{m_{1} + m_{2} + m_{3}} T$

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The mass of a body measured by a physical balance in a lift at rest is found to be m. If the lift is going up with an acceleration a, its mass will be measured as

$m (1 - \frac{a}{g})$

$m (1 + \frac{a}{g})$

Zero

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Explanation

Mass measured by physical balance remains unaffected due to variation in acceleration due to gravity.

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Three weights W, 2W and 3W are connected to identical springs suspended from a rigid horizontal rod. The assembly of the rod and the weights fall freely. The positions of the weights from the rod are such that

3W will be farthest

W will be farthest

All will be at the same distance

2W will be farthest

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Explanation

For W, 2W, 3W apparent weight will be zero because the system is falling freely. So the distances of the weight from the rod will be same.

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When forces F₁, F₂, F₃ are acting on a particle of mass m such that F₂ and F₃ are mutually perpendicular, then the particle remains stationary. If the force F₁ is now removed then the acceleration of the particle is

$F_{1} / m$

$F_{2} F_{3} / m F_{1}$

$(F_{2} - F_{3}) / m$

$F_{2} / m$

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Explanation

For equilibrium of system, $F_{1} = \sqrt{F_{2}^{2} + F_{3}^{2}}$ As $θ = 90 °$

In the absence of force F₁, Acceleration $= \frac{Net force}{Mass}$

$= \frac{\sqrt{F_{2}^{2} + F_{3}^{2}}}{m} = \frac{F_{1}}{m}$

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Consider the following statement: When jumping from some height, you should bend your knees as you come to rest, instead of keeping your legs stiff. Which of the following relations can be useful in explaining the statement?

$Δ \vec{P_{1}} = - Δ \vec{P_{2}}$

$Δ E = - Δ (P E + K E) = 0$

$\vec{F} Δ t = m Δ \vec{v}$

$Δ \vec{x} \propto Δ \vec{F}$

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Explanation

$\vec{F} Δ t = m Δ \vec{v}$ ⇒ $F = \frac{m Δ \vec{v}}{t}$

By doing so, time of change in momentum increases and impulsive force on knees decreases.

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A false balance has equal arms. An object weigh X when placed in one pan and Y when placed in other pan, then the weight W of the object is equal to

$\sqrt{X Y}$

$\frac{X + Y}{2}$

$\frac{X^{2} + Y^{2}}{2}$

$\frac{2}{\sqrt{X^{2} + Y^{2}}}$

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Explanation

When false balance has equal arms then, $W = \frac{X + Y}{2}$

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