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Physics MCQs for NEET — Practice Questions with Answers

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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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The maximum speed that can be achieved without skidding by a car on a circular unbanked road of radius R and coefficient of static friction μ, is

$μ R g$

$R g \sqrt{μ}$

$μ \sqrt{R g}$

$\sqrt{μ R g}$

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Explanation

In the given condition the required centripetal force is provided by frictional force between the road and tyre.

$\frac{m v^{2}}{R} = μ m g$

∴ $v = \sqrt{μ R g}$

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A car is moving along a straight horizontal road with a speed v₀. If the coefficient of friction between the tyres and the road is μ, the shortest distance in which the car can be stopped is

$\frac{v_{0}^{2}}{2 μ g}$

$\frac{v_{0}}{μ g}$

${(\frac{v_{0}}{μ g})}^{2}$

$\frac{v_{0}}{μ}$

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Explanation

Retarding force $F = m a = μ R = μ m g$

∴ $a = μ g$

Now from equation of motion $v^{2} = u^{2} - 2 a s$

$\Rightarrow 0 = u^{2} - 2 a s$

⇒ $s = \frac{u^{2}}{2 a} = \frac{u^{2}}{2 μ g}$

∴ $= \frac{v_{0}^{2}}{2 μ g}$

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A block of mass 50 kg can slide on a rough horizontal surface. The coefficient of friction between the block and the surface is 0.6. The least force of pull acting at an angle of 30° to the upward drawn vertical which causes the block to just slide is

29.43 N

219.6 N

21.96 N

294.3 N

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Explanation

For a block of mass 50 kg on a rough horizontal surface with coefficient of friction Î¼ = 0.6, the least force F required to cause sliding at an angle of 30Â° is given by F = Î¼N/sin(30Â°), where N is the normal force (equal to mg). Substituting the values, we get F = 0.6 Ã— 50 Ã— 9.8/0.5 = 294.3 N. This force must overcome the maximum static friction to initiate sliding.

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Assuming the coefficient of friction between the road and tyres of a car to be 0.5, the maximum speed with which the car can move round a curve of 40.0 m radius without slipping, if the road is unbanked, should be

25 m/s

19 m/s

14 m/s

11 m/s

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Explanation

$v = \sqrt{μ g r} = \sqrt{0 .5 \times 9 .8 \times 40} = \sqrt{196} = 14 m / s$

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