Laws of Motion — Page 7 | NEET Physics

Q61

A block B is pushed momentarily along a horizontaly surface with an initial velocity V. If

\mu

is the coefficient of sliding friction between B and the surface, block B will come to rest after a time

A g

\mu

/V

B g/V

C V/g

D V/(g

\mu

).

Correct Answer

Option D

Solution

Given u = V, final velocity = 0. Using v = u + at $\therefore$

0 = V - at\,

$\Rightarrow$

- a = {{0 - V} \over t} = - {V \over t}

f = $\mu$ R = $\mu$ mg (f is the force of friction) $\therefore$ Retardation, a =

\mu g

So,

t = {V \over a} = {V \over {\mu g}}

Q62

The coefficient of static friction,

\mu

s , between block A of mass 2 kg and the table as shown in the figure is 0.2. What would be the maximum mass value of block B so that the two blocks do not move? The string and the pulley are assumed to be smooth and massless. (g = 10 m/s 2 )

A 2.0 kg

B 4.0 kg

C 0.2 kg

D 0.4 kg

Correct Answer

Option D

Solution

Free body diagram of two masses is We get equations

T + ma = {f_\mu }

or

T = \mu {N_A}(for\,a = 0)

and T = ma + mg or T = m B g (for a = 0) $\therefore$

\mu {N_A} = {m_B}g

\Rightarrow {m_B} = \mu {m_A} = 0.2 \times 2 = 0.4\,kg

Q63

A block of mass m is placed on a smooth wedge of inclination

\theta

. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block will be (g is acceleration due to gravity)

A mg cos

\theta

B mg sin

\theta

C mg

D mg/cos

\theta

Correct Answer

Option D

Solution

The wedge is given an acceleration to the left.

$\therefore$ The block has a pseudo acceleration to the right, pressing against the wedge because of which the block is not moving. $\therefore$ mgsin $\theta$ = macos $\theta$ or

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

Total reaction of the wedge on the block is N = mgcos $\theta$ + masin $\theta$ .

\Rightarrow N = mg\cos \theta + {{mg\sin \theta \sin \theta } \over {\cos \theta }}

\Rightarrow N = {{mg\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)} \over {\cos \theta }} = {{mg} \over {\cos \theta }}

Q64

A monkey of mass 20 kg is holding a vertical rope. The rope will not break when a mass of 25 kg is suspended from it but will break if the mass exceeds 25 kg. What is the maximum acceleration with which the monkey can climb up along the rope ? (g = 10 m/s 2 )

A 5 m/s 2

B 10 m/s 2

C 25 m/s 2

D 2.5 m/s 2

Correct Answer

Option D

Solution

T = Tension caused in string by monkey = m (g + a)

T \le 25 \times 10 \Rightarrow 20\left( {10 + a} \right) \le 250

\Rightarrow 10 + a \le 12.5

\Rightarrow a \le 2.5

Q65

A man weight 80 kg. He stands on a weighting scale in a lift which is moving upwards with a uniform acceleration of 5 m/s 2 . What would be the reading on the scale ? (g = 10 m/s 2 )

A zero

B 400 N

C 800 N

D 1200 N

Correct Answer

Option D

Solution

Reading of the scale = Apparent wt. of the man = m(g + a) = 80 (10 + 5) = 1200 N

Q66

A block of mass 10 kg placed on rough horizontal surface having coefficient of friction m = 0.5, if a horizontal force of 100 N acting on it then acceleration of the block will be

A 10 m/s 2

B 5 m/s 2

C 15 m/s 2

D 0.5 m/s 2 .

Correct Answer

Option B

Solution

m = 10 kg, R = mg $\therefore$ Frictional force = f k =

{\mu _k}R = {\mu _k}mg

= 0.5 × 10 × 10 = 50 N [g = 10 m/sec 2 ] $\therefore$ Net force acting on the body = F = P – f k = 100 – 50 = 50 N.

$\therefore$ Acceleration of the block = a = F/m = 50/10 = 5 m/sec 2 .

Q67

An object of mass 3 kg is at rest. Now a force of

\overrightarrow F

= 6t 2

\widehat i

+ 4t

\widehat j

is applied on the object then velocity of object at t = 3 sec. is

A 18

\widehat i

+ 3

\widehat j

B 18

\widehat i

+ 6

\widehat j

C 3

\widehat i

+ 18

\widehat j

D 18

\widehat i

+ 4

\widehat j

Correct Answer

Option B

Solution

Mass (m) = 3 kg, force (F) =

6{t^2}\widehat i + 4t\widehat j

$\therefore$ acceleration

a = F/m = {{6{t^2}\widehat i + 4t\widehat j} \over 3} = 2{t^2}\widehat i + {4 \over 3}t\widehat j

Now,

a = {{dv} \over {dt}} = 2{t^2}\widehat i + {4 \over 3}t\widehat j

dv = \left( {2{t^2}\widehat i + {4 \over 3}t\widehat j} \right)dt

\therefore v = \int\limits_0^3 {\left( {2{t^2}\widehat i + {4 \over 3}t\widehat j} \right)} dt

=

\left. {{2 \over 3}{t^3}\widehat i + {4 \over 6}{t^2}\widehat j} \right|_0^3 = 18\widehat i + 6\widehat j

Q68

A lift of mass 1000 kg which is moving with acceleration of 1 m/s 2 in upward direction, then the tension developed in string which is connected to lift is

A 9800 N

B 10,800 N

C 11,000 N

D 10,000 N.

Correct Answer

Option B

Solution

For a lift which is moving in upward direction with an acceleration a, the tension T developed in the string connected to the lift is given by T = m (g + a).

Here m = 1000 kg, a = 1 m/s 2 , g = 9.8 m/s 2 $\therefore$ T = 1000(9.8 + 1) = 10,800 N.

Q69

A 1 kg stationary bomb is exploted in three parts having mass 1 : 1 : 3 respectively. Parts having same mass move in perpendicular direction with velocity 30 m/s, then the velocity of bigger part will be

A

10\sqrt 2

m/sec

B

{{10} \over {\sqrt 2 }}

m/sec

C

15\sqrt 2

m/sec

D

{{15} \over {\sqrt 2 }}

m/sec

Correct Answer

Option A

Solution

Apply conservation of linear momentum. Total momentum before explosion = total momentum after explosion

0 = {m \over 5}{v_1}\widehat i + {m \over 5}{v_2}\widehat j + {{3m} \over 5}{\overrightarrow v _3}

{{3m} \over 5}{\overrightarrow v _3} = - {m \over 5}\left[ {{v_1}\widehat i + {v_2}\widehat j} \right]

{\overrightarrow v _3} = {{ - {v_1}} \over 3}\widehat i - {{{v_2}} \over 3}\widehat j

$\therefore$

{v_1} = {v_2} = 30

m/sec

{\overrightarrow v _3} = - 10\widehat i - 10\widehat j;{v_3} = 10\sqrt 2

m/sec

Q70

A cricketer catches a ball of mass 150 gm in 0.1 sec moving with speed 20 m/s, then he experiences force of

A 300 N

B 30 N

C 3 N

D 0.3 N.

Correct Answer

Option B

Solution

Net force experienced =

{{Total\,{\mathop{\rm Impulse}\nolimits} } \over {Time\,taken}}

= {{m\Delta v} \over t} = 0.15 \times {{20} \over {0.1}} = 30N