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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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A pulley fixed to the ceilling carries a string with blocks of mass m and 3 m attached to its ends. The masses of string and pulley are negligible. When the system is released, its centre of mass moves with what acceleration

g/4

g/2

–g/2

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Explanation

Let acceleration of each mass be a

$a = (\frac{m_{2} - m_{1}}{m_{2} + m_{1}}) g U \sin g f o r m u l a f o r c e n t e r o f m a s s, a_{c m} = \frac{m_{1} a_{1} + m_{2} a_{2}}{m_{1} + m_{2}} a_{2} = - a a_{1} = a P u t t i n g v a l u e o f a i n a b o v e e q u a t i o n, a_{c m} = {(\frac{m_{2} - m_{1}}{m_{2} + m_{1}})}^{2} g$

$a_{c m} = {(\frac{m_{1} - m_{2}}{m_{1} + m_{2}})}^{2} g = {(\frac{3 m - m}{3 m + m})}^{2} g = \frac{g}{4}$

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A uniform rope of length l lies on a table. If the coefficient of friction is μ, then the maximum length l₁ of the part of this rope which can overhang from the edge of the table without sliding down is

$\frac{l}{μ}$

$\frac{l}{μ + l}$

$\frac{μ l}{1 + μ}$

$\frac{μ l}{μ - 1}$

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Explanation

For a uniform rope of length l lying on a table with coefficient of friction Î¼, the maximum length lâ‚ that can overhang from the edge without sliding down is given by lâ‚ = (Î¼l)/(1+Î¼). This is because the frictional force acts on the part of the rope on the table, and the condition for no sliding is that this force must balance the weight of the overhanging part.

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A heavy uniform chain lies on a horizontal table-top. If the coefficient of friction between the chain and table surface is 0.25, then the maximum fraction of length of the chain, that can hang over one edge of the table is

20%

25%

35%

15%

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A uniform chain of length L hangs partly from a table which is kept in equilibrium by friction. The maximum length that can withstand without slipping is l, then coefficient of friction between the table and the chain is

$\frac{l}{L}$

$\frac{l}{L + l}$

$\frac{l}{L - l}$

$\frac{L}{L + l}$

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Explanation

$μ = \frac{Lenght of chain hanging from the table}{Lenght of chain lying on the table} = \frac{l}{L - l}$

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When two surfaces are coated with a lubricant, then they

Stick to each other

Slide upon each other

Roll upon each other

None of these

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Explanation

Surfaces always slide over each other.

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A 20 kg block is initially at rest on a rough horizontal surface. A horizontal force of 75 N is required to set the block in motion. After it is in motion, a horizontal force of 60 N is required to keep the block moving with constant speed. The coefficient of static friction is

0.38

0.44

0.52

0.60

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Explanation

Coefficient of friction $μ_{s} = \frac{F_{l}}{R} = \frac{75}{m g} = \frac{75}{20 \times 9 .8} = 0 .38$

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