Laws of Motion — Page 4 | NEET Physics

Q31

A bullet from a gun is fired on a rectangular wooden block with velocity

u

. When bullet travels

24 \mathrm{~cm}

through the block along its length horizontally, velocity of bullet becomes

\dfrac{u}{3}

. Then it further penetrates into the block in the same direction before coming to rest exactly at the other end of the block. The total length of the block is :

A

24 \mathrm{~cm}

B

28 \mathrm{~cm}

C

30 \mathrm{~cm}

D

27 \mathrm{~cm}

Correct Answer

Option D

Solution

B y v^{2}=u^{2}+2 a s

\left(\frac{u}{3}\right)^{2}=u^{2}-2 a x

2 a x=u^{2}-\frac{u^{2}}{9}

2 \mathrm{ax}=\frac{8 \mathrm{u}^{2}}{9}

..... (1) Similarly from starting

v^{2}=u^{2}+2 a x

0=u^{2}-2 a x_{2}

2 \mathrm{ax}_{2}=\mathrm{u}^{2}

..... (2)

\mathrm{By}~(1) /(2)

\frac{x}{x_{2}}=\frac{8}{9}

\frac{24}{x_{2}}=\frac{8}{9}

\mathrm{x}_{2}=27 \mathrm{~cm}

Q32

If

\overrightarrow F = 2\widehat i + \widehat j - \widehat k

and

\overrightarrow r = 3\widehat i + 2\widehat j - 2\widehat k

, then the scalar and vector products of

\overrightarrow F

and

\overrightarrow r

have the magnitudes respectively as

A 10, 2

B 5,

\sqrt 3

C 4,

\sqrt 5

D 10,

\sqrt 2

Correct Answer

Option D

Solution

Given,

\overrightarrow F = 2\widehat i + \widehat j - \widehat k

and

\overrightarrow r = 3\widehat i + 2\widehat j - 2\widehat k

\overrightarrow F \,.\,\overrightarrow r = 2(3) + 2(1) + ( - 2)( - 1)

= 6 + 2 + 2 = 10

\overrightarrow F \times \overrightarrow r = \left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 2 & 1 & { - 1} \\ 3 & 2 & { - 2} \end{array} \right|

= \widehat i( - 2 + 2) - \widehat j( - 4 + 3) + \widehat k(4 - 3)

\Rightarrow \overrightarrow F \times \overrightarrow r = \widehat j + \widehat k

|\overrightarrow F \times \overrightarrow r | = \sqrt {{1^2} + {1^2}} = \sqrt 2

Q33

In the diagram shown, the normal reaction force between 2 kg and 1 kg is (Consider the surface, to be smooth) : (Given g = 10 ms

-

2 )

A 10 N

B 25 N

C 39 N

D 6 N

Correct Answer

Option B

Solution

Assuming all three blocks as system Hence, acceleration of system

a = {{60 - 30 - 18} \over 6} = {{12} \over 6} = 2

m/s 2

a = 2

m/s 2 , upward along the incline. Now, for 1 kg block

\Rightarrow {N_{12}} - 18 - 5 = 1 \times 2

\Rightarrow {N_{12}} - 23 = 2

\Rightarrow {N_{12}} = 25

N

Q34

Two bodies of mass 4 kg and 6 kg are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity (g) is :

A g/2

B g/5

C g/10

D g

Correct Answer

Option B

Solution

Acceleration of the masses is given to be :

a = {{\left( {{m_1} - {m_2}} \right)g} \over {\left( {{m_1} + {m_{}}} \right)}}

where,

{m_1} > {m_2}

a = {{\left( {6 - 4} \right)g} \over {6 + 4}}

a = {g \over 5}

Q35

A mass m is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break when :

A the wire is horizontal

B inclined at an angle of 60 o from vertical

C the mass is at the lowest point

D the mass is at the highest point

Correct Answer

Option C

Solution

Apply Newtons's 2nd law for equation of motion T - mg cos $\theta$ =

{{m{v^2}} \over R}

T will be maximum when $\theta$ = 0°, When mass is at lowest point the chance of breaking is maximum.

Q36

A block of mass 10 kg is in contact against the inner wall of a hollow cylindrical drum of radius 1 m. The coefficient of friction between the block and the inner wall of the cylinder is 0.1. The minimum angular velocity needed for the cylinder to keep the block stationary when the cylinder is vertical and rotating about its axis, will be : (g = 10 m/s 2 )

A

{{10} \over {2\pi }}

rad/s

B 10 rad/s

C 10

\pi

rad/s

D

\sqrt {10}

rad/s

Correct Answer

Option B

Solution

f L = $\mu$ N = $\mu$ mr

{\omega ^2}

f s = mg As, f s $\le$ f L $\Rightarrow$ mg $\le$ $\mu$ mr

{\omega ^2}

$\Rightarrow$

\omega \ge \sqrt {{g \over {\mu r}}}

$\Rightarrow$

{\omega _{\min }} \ge \sqrt {{{10} \over {0.1 \times 1}}}

= 10

Q37

Which one of the following statements is incorrect?

A Rolling friction is smaller than sliding friction.

B Limiting value of static friction is directly proportional to normal reaction.

C Frictional force opposes the relative motion.

D Coefficient of sliding friction has dimensions of length.

Correct Answer

Option D

Solution

We know, F = $\mu$ N $\Rightarrow$ MLT –2 = $\mu$ MLT –2 $\Rightarrow$ $\mu$ = M 0 L 0 T 0 $\therefore$ Coefficient of sliding friction has no dimension.

Q38

A block of mass m is placed on a smooth inclined wedge ABC of inclination q as shown in the figure. The wedge is given an acceleration ‘a’ towards the right. The relation between a and q for the block to remain stationary on the wedge is

A

a = {g \over {\cos ec\theta }}

B

a = {g \over {\cos ec\theta }}

C

a = g\cos \theta

D

a = g\tan \theta

Correct Answer

Option D

Solution

At equilibrium ma cos $\theta$ = mg sin $\theta$ $\Rightarrow$ a = g tan $\theta$

Q39

One end of string of length

l

is connected to a particle of mass 'm' and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed '

\upsilon

', the net force on the particle (directed towards centre) will be (T represents the tension in the string)

A

T + {{m{v^2}} \over l}

B

T - {{m{v^2}} \over l}

C zero

D

T

Correct Answer

Option D

Solution

Centripetal force

\left( {{{m{v^2}} \over l}} \right)

is provided by tension so net force on the particle will be equal to tension T.

Q40

Two blocks A and B of masses 3m and m respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of A and B immediately after the string is cut, are respectively

A

{g \over 3},g

B

g, g

C

{g \over 3},{g \over 3}

D

g,{g \over 3}

Correct Answer

Option A

Solution

Before cutting the string kx = T + 3 mg ...(i) T = mg ...(ii) $\Rightarrow$ kx = 4mg After cutting the string T = 0

{a_A} = {{4mg - 3mg} \over {3m}}

{a_A} = {g \over 3} \uparrow

and

{a_B} = {{mg} \over m} = g \downarrow