Laws of Motion — Page 8 | NEET Physics

Q71

On the horizontal surface of a truck a block of mass 1 kg is placed (m = 0.6) and truck is moving with acceleration 5 m/sec 2 then the frictional force on the block will be

A 5 N

B 6 N

C 5.88 N

D 8 N

Correct Answer

Option A

Solution

Maximum friction force = $\mu$ mg = .6 × 1 × 9.8 = 5.88 N But here required friction force = ma = 1 × 5 = 5 N

Q72

250 N force is required to raise 75 kg mass from a pulley. If rope is pulled 12 m then the load is lifted to 3 m. the efficiency of pulley system will be

A 25%

B 33.3%

C 75%

D 90%

Correct Answer

Option C

Solution

Load W = Mg = 75 × 10 = 750 N Effort (P) = 250 N $\therefore$ Mechanical advantage

= {{load} \over {effort}} = {W \over P} = {{750} \over {250}} = 3

Velocity ratio

= {{distance\,travelled\,by\,effort} \over {distance \,travelled\,by\,load}} = 4

Efficiency

(\eta) = {{Mechanical\,advantage} \over {Velocity\,ratio}}

= \left( {3/4} \right) \times 100 = 75\%

Q73

Two masses as shown in the figure are suspended from a massless pulley. The acceleration of the system when masses are left free is

A

{{2g} \over 3}

B

{g \over 3}

C

{g \over 9}

D

{g \over 7}

.

Correct Answer

Option B

Solution

Let T be the tension in the string. $\therefore$ 10g – T = 10a ....(i) T – 5g = 5a ....(ii) Adding (i) and (ii),

5g = 15a \Rightarrow a = {g \over 3}m/{s^2}

Q74

A body of mass 3 kg hits a wall at an angle of 60 o and returns at the same angle. The impact time was 0.2 sec. The force exerted on the wall

A 150

{\sqrt 3 }

N

B 50

{\sqrt 3 }

N

C 100 N

D 75

{\sqrt 3 }

N.

Correct Answer

Option A

Solution

Change in momentum = 2 × 3 × 10 × sin60° =

60 \times {{\sqrt 3 } \over 2}

Force = Change in momentum/Impact time

= {{30\sqrt 3 } \over {0.2}} = 150\sqrt 3 N

Q75

A particle moves in

x

-

y

plane under the influence of a force

\vec{F}

such that its linear momentum is

\overrightarrow{\mathrm{p}}(\mathrm{t})=\hat{i} \cos (\mathrm{kt})-\hat{j} \sin (\mathrm{kt})

. If

\mathrm{k}

is constant, the angle between

\overrightarrow{\mathrm{F}}

and

\overrightarrow{\mathrm{p}}

will be :

A

\dfrac{\pi}{2}

B

\dfrac{\pi}{3}

C

\dfrac{\pi}{4}

D

\dfrac{\pi}{6}

Correct Answer

Option A

Solution

To find the angle between

\vec{F}

and

\overrightarrow{\mathrm{p}}

, we first need to understand the relationship between force and momentum. The force

\vec{F}

acting on a particle is related to the rate of change of its linear momentum

\overrightarrow{\mathrm{p}}

with respect to time, as described by Newton's second law of motion:

\vec{F} = \frac{d\overrightarrow{\mathrm{p}}}{dt}

Given the expression for the momentum

\overrightarrow{\mathrm{p}}(t) = \hat{i} \cos (kt) - \hat{j} \sin (kt)

, we can find

\vec{F}

by differentiating

\overrightarrow{\mathrm{p}}

with respect to

t

:

\frac{d\overrightarrow{\mathrm{p}}}{dt} = -k\hat{i} \sin (kt) - k\hat{j} \cos (kt)

So,

\vec{F} = -k\hat{i} \sin (kt) - k\hat{j} \cos (kt)

. Now, to find the angle between

\vec{F}

and

\overrightarrow{\mathrm{p}}

, we use the dot product formula:

\vec{F} \cdot \overrightarrow{\mathrm{p}} = |\vec{F}| |\overrightarrow{\mathrm{p}}| \cos(\theta)

, where $\theta$ is the angle between

\vec{F}

and

\overrightarrow{\mathrm{p}}

. However, in this case, it's more insightful to see if

\vec{F}

and

\overrightarrow{\mathrm{p}}

are orthogonal (at a

\frac{\pi}{2}

angle to each other), because the dot product of two perpendicular vectors is zero. The dot product of

\vec{F}

and

\overrightarrow{\mathrm{p}}

is:

(-k\hat{i} \sin (kt) - k\hat{j} \cos (kt)) \cdot (\hat{i} \cos (kt) - \hat{j} \sin (kt)) =

- k \sin (kt) \cos (kt) + k \cos (kt) \sin (kt) = 0

The result is zero, indicating that the angle between

\vec{F}

and

\overrightarrow{\mathrm{p}}

is indeed

\frac{\pi}{2}

. Therefore, the correct option is: Option A

\frac{\pi}{2}

.

Q76

A given object takes n times more time to slide down a

{45^ \circ }

rough inclined plane as it takes to slide down a perfectly smooth

{45^ \circ }

incline. The coefficient of kinetic friction between the object and the incline is :

A

{1 \over {2 - {n^2}}}

B

1 - {1 \over {{n^2}}}

C

\sqrt {1 - {1 \over {{n^2}}}}

D

\sqrt {{1 \over {1 - {n^2}}}}

Correct Answer

Option B

Solution

Let, t1 and t2 are time taken to move on the smooth and rough surface for smooth surface, S =

{1 \over 2}

g sin45o

t_1^2

$\Rightarrow$

\,\,\,\,

t1 =

\sqrt {{{2\sqrt 2 S} \over g}}

For rough surface, S =

{1 \over 2}

g (sin45o $-$ $\mu$ k cos45o)

t_2^2

$\Rightarrow$

\,\,\,\,

t2 =

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

Here

{\mu _k}

= Kinetic friction. According to question, t2 = n t1 $\Rightarrow$

\,\,\,\,

{{2\sqrt 2 \,S} \over {g\left( {1 - {\mu _k}} \right)}}

= n2 $\times$

{{2\sqrt 2 \,S} \over g}

$\Rightarrow$

\,\,\,\,

1 $-$ $\mu$ k =

{1 \over {{n^2}}}

$\Rightarrow$

\,\,\,\,

{\mu _k}

= 1 $-$

{1 \over {{n^2}}}

Q77

A block of mass

5 \mathrm{~kg}

is placed at rest on a table of rough surface. Now, if a force of

30 \mathrm{~N}

is applied in the direction parallel to surface of the table, the block slides through a distance of

50 \mathrm{~m}

in an interval of time

10 \mathrm{~s}

. Coefficient of kinetic friction is (given,

g=10 \mathrm{~ms}^{-2}

):

A 0.25

B 0.75

C 0.60

D 0.50

Correct Answer

Option D

Solution

\begin{aligned} & S=u t+\frac{1}{2} a t^2 \\\\ & 50=0+\frac{1}{2} \times a \times 100 \\\\ & a=1 \mathrm{~m} / \mathrm{s}^2 \\\\ & F-\mu m g=m a \\\\ & 30-\mu \times 50=5 \times 1 \\\\ & 50 \mu=25 \\\\ & \mu=\frac{1}{2} \end{aligned}

Q78

A block of mass M slides down on a rough inclined plane with constant velocity. The angle made by the incline plane with horizontal is

\theta

. The magnitude of the contact force will be :

A Mg

B

\mathrm{Mg} \cos \theta

C

\sqrt{\mathrm{Mg} \sin \theta+\mathrm{Mg} \cos \theta}

D

\operatorname{Mg} \sin \theta \sqrt{1+\mu}

Correct Answer

Option A

Solution

As the body is moving with constant velocity so forces acting on the body must be balanced.

$\Rightarrow$ Contact force from incline should balance weight of the body.

$\Rightarrow$ | Fcontact | = Mg

Q79

A balloon and its content having mass M is moving up with an acceleration ‘a’. The mass that must be released from the content so that the balloon starts moving up with an acceleration ‘3a’ will be(Take ‘g’ as acceleration due to gravity)

A

\dfrac{3Ma}{2a + g}

B

\dfrac{2Ma}{3a + g}

C

\dfrac{3Ma}{2a - g}

D

\dfrac{2Ma}{3a - g}

Correct Answer

Option B

Solution

\begin{aligned} &\begin{aligned} & F-m g=m a \\ & F=m a+m g \\ & F-(m-x) g=(m-x) 3 a \end{aligned}\\ &\text{ Put F }\\ &\begin{aligned} & \mathrm{Ma}+\mathrm{mg}-\mathrm{mg}+\mathrm{xg}=3 \mathrm{ma}-3 \mathrm{xa} \\ & \mathrm{x}=\frac{2 \mathrm{ma}}{\mathrm{~g}+3 \mathrm{a}} \end{aligned} \end{aligned}

Q80

A smooth block is released at rest on a

{45^ \circ }

incline and then slides a distance

'd'

. The time taken to slide is

'n'

times as much to slide on rough incline than on a smooth incline. The coefficient of friction is

A

{\mu _k} = \sqrt {1 - {1 \over {{n^2}}}}

B

{\mu _k} = 1 - {1 \over {{n^2}}}

C

{\mu _k} = \sqrt {1 - {1 \over {{n^2}}}}

D

{\mu _s} = 1 - {1 \over {{n^2}}}

Correct Answer

Option B

Solution

For smooth surface,

d = {1 \over 2}\left( {g\,\sin \,\theta } \right)t_1^2,

{t_1} = \sqrt {{{2d} \over {g\,\sin \,\theta }}} ,

When surface is rough

d = {1 \over 2}\left( {g\,\sin \,\theta - \mu g\,\cos \theta } \right)t_2^2

{t_2} = \sqrt {{{2d} \over {g\,\sin \,\theta - \mu g\,\cos \theta }}}

According to question,

{t_2} = n{t_1}

n\sqrt {{{2d} \over {g\,\sin \,\theta }}} = \sqrt {{{2d} \over {g\,\sin \,\theta - \mu g\,\cos \theta }}}

n = {1 \over {\sqrt {1 - {\mu _k}} }}

( as

\cos \,{45^ \circ } = \sin \,{45^ \circ } = {1 \over {\sqrt 2 }}

)

{n^2} = {1 \over {1 - {\mu _k}}}

or

1 - {\mu _k} = {1 \over {{n^2}}}

or

{\mu _k} = 1 - {1 \over {{n^2}}}