Laws of Motion — Page 10 | NEET Physics

Q91

A block of mass

M

is pulled along a horizontal frictionless surface by a rope of mass

m.

If a force

P

is applied at the free end of the rope, the force exerted by the rope on the block is

A

{{Pm} \over {M + m}}

B

{{Pm} \over {M - m}}

C

P

D

{{PM} \over {M + m}}

Correct Answer

Option D

Solution

Taking the rope and the block as a system Acceleration of block(

a

) =

{{Force\,applied} \over {total\,mass}}

$\therefore$

a = {P \over {m + M}}

Force on block(T) = Mass of block $\times$

a

T=Ma

$\therefore$

T = {{MP} \over {m + M}}

Q92

A body of mass

m

is suspended by two strings making angles

\theta_1

and

\theta_2

with the horizontal ceiling with tensions

T_1

and

T_2

simultaneously.

T_1

and

T_2

are related by

T_1=\sqrt{3} T_2

, the angles

\theta_1

and

\theta_2

are

A

\theta_1=30^{\circ} \theta_2=60^{\circ}

with

T_2=\dfrac{3 \mathrm{mg}}{4}

B

\theta_1=45^{\circ} \theta_2=45^{\circ}

with

T_2=\dfrac{3 m g}{4}

C

\theta_1=30^{\circ} \theta_2=60^{\circ}

with

T_2=\dfrac{4 m g}{5}

D

\theta_1=60^{\circ} \theta_2=30^{\circ}

with

T_2=\dfrac{m g}{2}

Correct Answer

Option D

Solution

\begin{aligned} & \mathrm{T}_1 \sin \theta_1+\mathrm{T}_2 \sin \theta_2=\mathrm{mg} \& \mathrm{~T}_1=\sqrt{3} \mathrm{~T}_2 \\ & \Rightarrow \mathrm{~T}_2\left[\sqrt{3} \sin \theta_1+\sin \theta_2\right]=\mathrm{mg} \\ & \text{ for } \theta_1=60^{\circ} \& \theta_2=30^{\circ} \\ & \mathrm{T}_2=\frac{\mathrm{mg}}{2} \end{aligned}

Q93

Statement : 1 If three forces

{\overrightarrow F _1},{\overrightarrow F _2}

and

{\overrightarrow F _3}

are represented by three sides of a triangle and

{\overrightarrow F _1} + {\overrightarrow F _2} = - {\overrightarrow F _3}

, then these three forces are concurrent forces and satisfy the condition for equilibrium. Statement : 2 A triangle made up of three forces

{\overrightarrow F _1}

,

{\overrightarrow F _2}

and

{\overrightarrow F _3}

as its sides taken in the same order, satisfy the condition for translatory equilibrium. In the light of the above statements, choose the most appropriate answer from the options given below :

A Statement - I is false but Statement - II is true

B Statement - I is true but Statement - II is false

C Both Statement-I and Statement-II are false

D Both Statement-I and Statement-II are true

Correct Answer

Option D

Solution

Here,

{\overrightarrow F _1} + {\overrightarrow F _2} + {\overrightarrow F _3} = 0

{\overrightarrow F _1} + {\overrightarrow F _2} = - {\overrightarrow F _3}

Since

{\overrightarrow F _{net}} = 0

(equilibrium) Both statements correct.

Q94

A block of mass 1 kg , moving along

x

with speed

v_i=10 \mathrm{~m} / \mathrm{s}

enters a rough region ranging from

x=0.1 \mathrm{~m}

to

x=1.9 \mathrm{~m}

. The retarding force acting on the block in this range is

\mathrm{F}_{\mathrm{r}}=-\mathrm{kr} \mathrm{N}

, with k

=10 \mathrm{~N} / \mathrm{m}

. Then the final speed of the block as it crosses rough region is.

A

4 \mathrm{~m} / \mathrm{s}

B

10 \mathrm{~m} / \mathrm{s}

C

8 \mathrm{~m} / \mathrm{s}

D

6 \mathrm{~m} / \mathrm{s}

Correct Answer

Option C

Solution

Work Done = Final Kinetic Energy - Initial Kinetic Energy

W = \Delta K = K_f - K_i

Calculate the Work Done by the Retarding Force The problem gives a retarding force

F_r = -kr

. It seems 'r' is a typo for 'x', the position, since the constant

k

has units of N/m. So, the force is a function of position:

F_r = -kx

.

Because the force isn't constant, we need to find the work done by integrating the force over the distance the block travels in the rough region (from

x = 0.1 \text{ m}

to

x = 1.9 \text{ m}

).

W = \int_{0.1}^{1.9} F_r \, dx = \int_{0.1}^{1.9} (-kx) \, dx

Plugging in

k = 10 \text{ N/m}

:

W = -10 \int_{0.1}^{1.9} x \, dx

W = -10 \left[ \frac{x^2}{2} \right]_{0.1}^{1.9}

W = -5 \left[ (1.9)^2 - (0.1)^2 \right]

W = -5 [3.61 - 0.01] = -5(3.6)

W = -18 \text{ J}

The negative sign just means the force did negative work, slowing the block down. Calculate the Initial Kinetic Energy (

K_i

) The initial kinetic energy is found using the formula

K_i = \frac{1}{2}mv_i^2

. Given: Mass

m = 1 \text{ kg}

Initial speed

v_i = 10 \text{ m/s}

K_i = \frac{1}{2}(1 \text{ kg})(10 \text{ m/s})^2 = \frac{1}{2}(100) = 50 \text{ J}

Use the Work-Energy Theorem to find the Final Speed (

v_f

) Now we can put everything together into the Work-Energy Theorem equation:

W = K_f - K_i

. We know

W = -18 \text{ J}

and

K_i = 50 \text{ J}

. The final kinetic energy is

K_f = \frac{1}{2}mv_f^2

.

-18 = \left(\frac{1}{2}(1)v_f^2\right) - 50

Now, let's solve for

v_f

:

-18 + 50 = \frac{1}{2}v_f^2

32 = \frac{1}{2}v_f^2

v_f^2 = 64

v_f = \sqrt{64}

v_f = 8 \text{ m/s}

The final speed of the block as it leaves the rough region is 8 m/s. Therefore, the correct option is C.

Q95

A particle of mass m is acted upon by a force F given by the empirical law F =

{R \over {{t^2}}}\,v\left( t \right).

If this law is to be tested experimentally by observing the motion starting from rest, the best way is to plot :

A

\upsilon

(t) against t2

B log

\upsilon

(t) against

{1 \over {{t^2}}}

C log

\upsilon

(t) against t

D log

\upsilon

(t) against

{1 \over {{t}}}

Correct Answer

Option D

Solution

Given, F =

{R \over {{t^2}}}

v(t) $\Rightarrow$ m

{{dv} \over {dt}}

=

{R \over {{t^2}}}

(v) $\Rightarrow$

{{dv} \over v}

=

{R \over m}

{{dt} \over {{t^2}}}

Intergrating both sides,

\int {{{dv} \over v} = {R \over m}\int {{{dt} \over {{t^2}}}} }

$\Rightarrow$ lnv =

{{R \over m}}

$\times$

\left( { - {1 \over t}} \right)

+ C $\Rightarrow$ lnv = $-$

{{R \over m}}

\left( {{1 \over t}} \right)

+ C Graph between lnv and

{{1 \over t}}

will be straight line curve.

Q96

A force

\overrightarrow F = (40\widehat i + 10\widehat j)N

acts on a body of mass 5 kg. If the body starts from rest, its position vector

\overrightarrow r

at time t = 10 s, will be :

A

(100\widehat i + 400\widehat j)m

B

(100\widehat i + 100\widehat j)m

C

(400\widehat i + 100\widehat j)m

D

(400\widehat i + 400\widehat j)m

Correct Answer

Option C

Solution

{{d\overrightarrow v } \over {dt}} = \overrightarrow a = {{\overrightarrow F } \over m} = (8\widehat i + 2\widehat j)m/{s^2}

{{d\overrightarrow r } \over {dt}} = \overrightarrow v = (8t\widehat i + 2t\widehat j)m/s

\overrightarrow r = (8\widehat i + 2\widehat j){{{t^2}} \over 2}m

At t = 10 sec

\overrightarrow r = \left[ {(8\widehat i + 2\widehat j)50} \right]m

\Rightarrow \overrightarrow r = (400\widehat i + 100\widehat j)m

Q97

A light string passing over a smooth light pulley connects two blocks of masses

m_1

and

m_2\left(\right.

where

\left.m_2>m_1\right)

. If the acceleration of the system is

\dfrac{g}{\sqrt{2}}

, then the ratio of the masses

\dfrac{m_1}{m_2}

is:

A

\dfrac{1+\sqrt{5}}{\sqrt{5}-1}

B

\dfrac{\sqrt{3}+1}{\sqrt{2}-1}

C

\dfrac{\sqrt{2}-1}{\sqrt{2}+1}

D

\dfrac{1+\sqrt{5}}{\sqrt{2}-1}

Correct Answer

Option C

Solution

Given that the acceleration of the system is

\frac{g}{\sqrt{2}}

, let's analyze the forces acting on both masses to find the ratio

\frac{m_1}{m_2}

.

The tension force in the string is equal for both masses since the pulley and string are light and smooth.

Assuming the pulley only changes the direction of the tension force without affecting its magnitude, we can analyze the forces in the vertical direction for both masses.

The net force acting on

m_1

(towards the pulley) would be the tension

T

minus its weight component in the direction of motion, which we will assume is

m_1g

, and for

m_2

, it would be its weight component in the direction of motion, which we can assume is

m_2g

minus the tension

T

. Given that

m_2

is moving downwards and

m_1

is moving upwards, and considering that

m_2>m_1

, the acceleration

a

of both masses will be the same and can be described as: For

m_1

:

T - m_1g = m_1a

For

m_2

:

m_2g - T = m_2a

Adding these equations to eliminate

T

gives:

m_2g - m_1g = m_1a + m_2a

Given that the acceleration

a

of the system is

\frac{g}{\sqrt{2}}

, we can substitute

a

with

\frac{g}{\sqrt{2}}

in the equation:

m_2g - m_1g = m_1\left(\frac{g}{\sqrt{2}}\right) + m_2\left(\frac{g}{\sqrt{2}}\right)

Simplifying and factoring out

g

, we get:

m_2 - m_1 = \frac{m_1}{\sqrt{2}} + \frac{m_2}{\sqrt{2}}

Multiplying every term by

\sqrt{2}

to clear the denominator, we get:

\sqrt{2}(m_2 - m_1) = m_1 + m_2

Rearranging the terms to isolate the masses on one side, we obtain:

\sqrt{2}m_2 - m_2 = m_1(\sqrt{2} + 1)

This simplifies to:

m_2(\sqrt{2} - 1) = m_1(\sqrt{2} + 1)

Hence, the ratio

\frac{m_1}{m_2}

is:

\frac{m_1}{m_2} = \frac{\sqrt{2} - 1}{\sqrt{2} + 1}

This corresponds to Option C

\frac{\sqrt{2}-1}{\sqrt{2}+1}

.

Q98

A block of mass

m

is placed on a surface with a vertical cross section given by

y = {{{x^3}} \over 6}.

If the coefficient of friction is

0.5,

the maximum height above the ground at which the block can be placed without slipping is:

A

{1 \over 6}m

B

{2 \over 3}m

C

{1 \over 3}m

D

{1 \over 2}m

Correct Answer

Option A

Solution

At limiting equilibrium,

\mu = \tan \theta

Equation of the surface,

y = {{{x^3}} \over 6}

Slope,

\tan \theta = \mu = {{dy} \over {dx}} = {{{x^2}} \over 2}

Given that, Coefficient of friction

\mu = 0.5

$\therefore$

\,\,\,\,0.5 = {{{x^2}} \over 2}

\Rightarrow \,\,\,x = \pm \,1

Now,

y = {{{x^3}} \over 6} = {1 \over 6}m

Q99

A small ball of mass m is thrown upward with velocity u from the ground. The ball experiences a resistive force mkv2 where v is its speed. The maximum height attained by the ball is :

A

{1 \over k}{\tan ^{ - 1}}{{k{u^2}} \over {2g}}

B

{1 \over {2k}}{\tan ^{ - 1}}{{k{u^2}} \over g}

C

{1 \over {2k}}\ln \left( {1 + {{k{u^2}} \over g}} \right)

D

{1 \over k}\ln \left( {1 + {{k{u^2}} \over {2g}}} \right)

Correct Answer

Option C

Solution

Fnet = ma $\Rightarrow$ -mg - mkv2 =

mv{{dv} \over {ds}}

$\Rightarrow$

ds = {{ - vdv} \over {g + k{v^2}}}

$\Rightarrow$

\int\limits_{s = 0}^{{H_{\max }}} {ds} = \int\limits_{v = u}^{v = 0} {{{ - vdv} \over {g + k{v^2}}}}

$\Rightarrow$ Hmax =

{1 \over {2k}}\ln \left( {{{g + k{u^2}} \over g}} \right)

=

{1 \over {2k}}\ln \left( {1 + {{k{u^2}} \over g}} \right)

Q100

At any instant the velocity of a particle of mass

500 \mathrm{~g}

is

\left(2 t \hat{i}+3 t^{2} \hat{j}\right) \mathrm{ms}^{-1}

. If the force acting on the particle at

t=1 \mathrm{~s}

is

(\hat{i}+x \hat{j}) \mathrm{N}

. Then the value of

x

will be:

A 2

B 4

C 6

D 3

Correct Answer

Option D

Solution

Given the velocity vector of a particle $v = (2t \hat{i}+3 t^{2} \hat{j}) \, \text{ms}^{-1}$ , the acceleration $a$ is the derivative of the velocity vector with respect to time.

So, we have: $a = \dfrac{dv}{dt} = (2 \hat{i} + 6t \hat{j}) \, \text{ms}^{-2}$ .

At $t=1 \, \text{s}$ , the acceleration $a$ is $(2 \hat{i} + 6 \hat{j}) \, \text{ms}^{-2}$ .

According to Newton's second law, the force $F$ is equal to the mass $m$ times acceleration $a$ .

The mass $m$ is given as $500 \, \text{g}$ , or equivalently, $0.5 \, \text{kg}$ .

Therefore, the force $F$ on the particle at $t=1 \, \text{s}$ is: $F = m \cdot a = 0.5 \cdot (2 \hat{i} + 6 \hat{j}) = (1 \hat{i} + 3 \hat{j}) \, \text{N}$ .

So, the force acting on the particle at $t=1 \, \text{s}$ is $(\hat{i} + x \hat{j}) \, \text{N}$ , where $x=3$ .

Therefore, the answer is $x=3$ .