Laws of Motion — Page 5 | NEET Physics

Q41

A car is negotiating a curved road of radius R. The road is banked at an angle

\theta

. The coefficient of friction between the tyres of the car and the road is

\mu

s . The maximum safe velocity on this road is

A

\sqrt {{g \over R}{{{\mu _s} + \tan \theta } \over {1 - {\mu _s}\tan \theta }}}

B

\sqrt {{g \over {{R^2}}}{{{\mu _s} + \tan \theta } \over {1 - {\mu _s}\tan \theta }}}

C

\sqrt {g{R^2}{{{\mu _s} + \tan \theta } \over {1 - {\mu _s}\tan \theta }}}

D

\sqrt {gR{{{\mu _s} + \tan \theta } \over {1 - {\mu _s}\tan \theta }}}

Correct Answer

Option D

Solution

On a banked road,

{{V_{\max }^2} \over {Rg}} = \left( {{{{\mu _s} + \tan \theta } \over {1 - {\mu _s}\tan \theta }}} \right)

Maximum safe velocity of a car on the banked road

{V_{\max }} = \sqrt {Rg\left[ {{{{\mu _s} + \tan \theta } \over {1 - {\mu _s}\tan \theta }}} \right]}

Q42

Two stones of masses m and 2m are whirled in horizontal circles, the heavier one in a radius

{r \over 2}

and the lighter one in radius r. The tangential speed of lighter stone is n times that of the value of heavier stone when they exprience same centripetal forces. The value of n is

A 4

B 1

C 2

D 3

Correct Answer

Option C

Solution

According to question, two stones experience same centripetal force i.e. F C 1 = F C 2 or

{{mv_1^2} \over r} = {{2mv_2^2} \over {\left( {r/2} \right)}} \Rightarrow V_1^2 = 4V_2^2

So, V 1 = 2V 2 i.e., n = 2

Q43

A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches 30 o , the box starts to slip and slides 4.0 m down the plank in 4.0 s. The coefficients of static and kinetic friction between the box and the plank will be, respectively

A 0.5 and 0.6

B 0.4 and 0.3

C 0.6 and 0.6

D 0.6 and 0.5

Correct Answer

Option D

Solution

Coefficient of static friction,

{\mu _s} = \tan 30^\circ = {1 \over {\sqrt 3 }} = 0.577 \cong 0.6

S = ut + {1 \over 2}a{t^2}

4 = {1 \over 2}a{\left( 4 \right)^2} \Rightarrow a = {1 \over 2} = 0.5

[ $\because$ s = 4m and t = 4s given] a = gsin $\theta$ –

{\mu _k}(g)\cos \theta

\Rightarrow {\mu _k} = {{0.9} \over {\sqrt 3 }} = 0.5

Q44

Three blocks A, B and C, of masses 4 kg, 2 kg and 1 kg respectively, are in contact on a frictionless surface, as shown. If a force of 14 N is applied on the 4 kg block, then the contact force between A and B is

A 8 N

B 18 N

C 2 N

D 6 N

Correct Answer

Option D

Solution

Here, M A = 4 kg, M B = 2 kg, M C = 1 kg, F = 14 N Net mass, M = M A + M B + M C = 4 + 2 + 1 = 7 kg Let a be the acceleration of the system.

Using Newton’s second law of motion, F = Ma 14 = 7a $\therefore$ a = 2 ms –2 Let F' be the force applied on block A by block B i.e. the contact force between A and B.

Free body diagram for block A Again using Newton’s second law of motion, F – F' = 4a 14 – F' = 4 × 2 $\Rightarrow$ 14 – 8 = F' $\therefore$ F' = 6 N

Q45

A block A of mass m 1 rests on a horizontal table. A light string connected to it passes over a frictionless pully at the edge of table and from its other end another block B of mass m 2 is suspended. The coefficient of kinetic friction between the block and the table is

\mu

k . When the block A is sliding on the table, the tension in the string is

A

{{{m_1}{m_2}(1 + {\mu _k})g} \over {({m_1} + {m_2})}}

B

{{{m_1}{m_2}(1 - {\mu _k})g} \over {({m_1} + {m_2})}}

C

{{\left( {{m_2} + {\mu _k}{m_1}} \right)g} \over {\left( {{m_1} + {m_2}} \right)}}

D

{{\left( {{m_2} - {\mu _k}{m_1}} \right)g} \over {\left( {{m_1} + {m_2}} \right)}}

Correct Answer

Option A

Solution

For the motion of both the blocks m 1 a = T –

{\mu _k}

m 1 g m 2 g – T = m 2 a

a = {{{m_2}g - {\mu _k}{m_1}g} \over {{m_1} + {m_2}}}

{m_2}g - T = \left( {{m_2}} \right)\left( {{{{m_2}g - {\mu _k}{m_1}g} \over {{m_1} + {m_2}}}} \right)

solving we get tension in the string

T = {{{m_1}{m_2}\left( {1 + {\mu _k}} \right)g} \over {{m_1} + {m_2}}}

Q46

A system consists of three masses m 1 , m 2 and m 3 connected by a string passing over a pulley P. The mass m 1 hangs freely and m 2 and m 3 are on a rough horizontal table (the coefficient of friction =

\mu

). The pulley is frictionless and of negligible mass. The downward acceleration of mass m 1 is (Assume m 1 = m 2 = m 3 = m)

A

{{g\left( {1 - g\mu } \right)} \over 9}

B

{{2g\mu } \over 3}

C

{{g\left( {1 - 2\mu } \right)} \over 3}

D

{{g\left( {1 - 2\mu } \right)} \over 2}

Correct Answer

Option C

Solution

Force of friction on mass m 2 = $\mu$ m 2 g Force of friction on mass m 3 = $\mu$ m 3 g Let a be common acceleration of the system.

$\therefore$

a = {{{m_1}g - \mu {m_2}g - \mu {m_3}g} \over {{m_1} + {m_2} + {m_3}}}

Here, m 1 = m 2 = m 3 = m $\therefore$

a = {{mg - \mu mg - \mu mg} \over {m + m + m}} = {{mg - 2\mu mg} \over {3m}} = {{g\left( {1 - 2\mu } \right)} \over 3}

Hence, the downward acceleration of mass m 1 is

{{g\left( {1 - 2\mu } \right)} \over 3}

Q47

A balloon with mass m is descending down with an acceleration a (where a < g). How much mass should be removed from it so that it starts moving up with an acceleration a ?

A

{{2ma} \over {g + a}}

B

{{2ma} \over {g - a}}

C

{{ma} \over {g + a}}

D

{{ma} \over {g - a}}

Correct Answer

Option A

Solution

Let upthrust of air be F a then For downward motion of balloon F a = mg – ma mg – Fa = ma For upward motion

{F_a} - \left( {m - \Delta m} \right)g = \left( {m - \Delta m} \right)a

$\Rightarrow$

\Delta m = {{2ma} \over {g + a}}

Q48

A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 m/s. A bob is suspended from the roof of the car by a light wire of length 1.0 m. The angle made by the wire with the vertical is

A

{\pi \over 3}

B

{\pi \over 6}

C

{\pi \over 4}

D 0 o

Correct Answer

Option C

Solution

Let $\theta$ is the angle made by the wire with the vertical. $\therefore$

\tan \theta = {{{v^2}} \over {rg}}

Here, v = 10 m/s, r = 10 m, g = 10 m/s 2 $\therefore$

\tan \theta = {{{{\left( {10\,\,m/s} \right)}^2}} \over {10\,m\left( {10\,\,m/{s^2}} \right)}} = 1

\theta = {\tan ^{ - 1}}\left( 1 \right) = {\pi \over 4}

Q49

The upper half of an inclined plane of inclination

\theta

is perfectly smooth while lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by

A

\mu

= 2 tan

\theta

B

\mu

= tan

\theta

C

\mu

=

{1 \over {\tan \theta }}

D

\mu = {2 \over {\tan \theta }}

Correct Answer

Option A

Solution

For upper half of inclined plane v 2 = u 2 + 2a S/2 = 2 (g sin $\theta$ ) S/2 = gS sin $\theta$ For lower half of inclined plane 0 = u 2 + 2 g (sin $\theta$ – $\mu$ cos $\theta$ ) S/2

\Rightarrow - gS\sin \theta = gS\left( {\sin \theta - \mu \cos \theta } \right)

\Rightarrow 2\sin \theta = \mu \cos \theta

\Rightarrow \mu = {{2\sin \theta } \over {\cos \theta }} = 2\tan \theta

Q50

Three blocks with masses m, 2m and 3m are connected by strings, as shown in the figure. After an upward force F is applied on block m, the masses move upwards at constant speed v. What is the net force on the block of mass 2m ? (g is the acceleration due to gravity)

A 3mg

B 6mg

C zero

D 2mg

Correct Answer

Option C

Solution

From figure F = 6 mg, As speed is constant, acceleration a = 0 $\therefore$ 6 mg = 6ma = 0, F = 6 mg $\therefore$ T = 5 mg , T' = 3 mg T'' = 0 F net on block of mass 2 m = T – T' – 2 mg = 0 ALTERNATE : $\because$ v = constant so, a = 0, Hence, F net = ma = 0