Laws of Motion Test

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Laws of Motion Test - 46

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Question 1
1 / -0

The one which does not represent a force in any context is:

A
friction

B
impulse

C
tension

D
weight

E
viscous drag

Solution

Impulse is equal to the charge in momentum whereas all the other given terms are referred to as force.
So, option (B) is the correct answer.
Question 2
1 / -0

If a body of mass $$5\ g$$ initially at rest is acted upon by a force of $$50$$ dynes for $$3\ s$$. Then the impulse will be

A
$$0.98\times 10^{-3} Ns$$

B
$$1.5\times 10^{-3} Ns$$

C
$$2.0\times 10^{-3} Ns$$

D
$$2.5\times 10^{-3} Ns$$

Solution

Impulse, $$I = F\times \triangle t$$
$$= 50\times 10^{-5} \times 3 = 1.5\times 10^{-3}Ns$$.
Question 3
1 / -0

A motor car is travelling at speed of $$30m/s$$ on a circular road of radius $$500m$$. If the speed of the car is increasing at the rate of $$2.0m/{ s }^{ 2 }$$, then the total acceleration will be

A
$$1.8m/{ s }^{ 2 }$$

B
$$2m/{ s }^{ 2 }$$

C
$$2.7m/{ s }^{ 2 }$$

D
$$3.8m/{ s }^{ 2 }$$

Solution

Radial acceleration, $${ a }_{ r }=\cfrac { { v }^{ 2 } }{ r } =\cfrac { { (30) }^{ 2 } }{ 500 } =1.8m/{ s }^{ 2 }\quad $$
Tangential acceleration, $${ a }_{ r }=2m/{ s }^{ 2 }\quad $$
$$\therefore$$ Resultant acceleration, $$a=\sqrt { { a }_{ r }^{ 2 }+{ a }_{ r }^{ 2 }2{ a }_{ n }{ a }_{ r }\cos { \theta } } $$
$$\quad \therefore a=\sqrt { { (1.8) }^{ 2 }+{ (2) }^{ 2 } } =\sqrt { 7.24 } $$
$$\Rightarrow a=2.7m/{ s }^{ 2 }$$
Question 4
1 / -0

A block of mass $$M$$ at the end of the string is whirled round a vertical circle of radius $$R$$. The critical speed of the block at the top of the swing is

A
$${ \left( R/g \right) }^{ 1/2 }$$

B
$$g/R$$

C
$$M/Rg$$

D
$${ \left( Rg \right) }^{ 1/2 }$$

Solution

Using conservation of energy
Total mechanical energy at lowest point = Total mechanical energy at top
$$\dfrac{1}{2}m{v_{lowest}^2} = \dfrac{1}{2}m{v_{High}^2} + 2mgR$$
we know that $$ {v_{lowest}} = \sqrt{5Rg}$$
$$\therefore \dfrac{1}{2}(5Rg) = \dfrac{1}{2}{v_{high}^2} + 2gR$$
$${v_{high}} = \sqrt{Rg}$$
Question 5
1 / -0

A railway carriage has its centre of gravity at a height of $$ 1 m $$ above the rails , which are $$ 1.5 m $$ apart.The maximum safe speed at which it could travel round an unbanked curve of radius $$ 100 m $$ is

A
$$ 12 m/s$$

B
$$ 18 m/s$$

C
$$ 22 m/s$$

D
$$ 27 m/s$$

Solution

Here, $$ h=1 m , r=100 m , and 2x=1.5 m$$
For no skidding
$$\dfrac{mv^{2}}{r}\times h=mgx$$
$$v=\sqrt{\dfrac{grx}{h}}$$
$$=\sqrt{\dfrac{9.8 \times 100 \times 0.75}{1}}$$
$$=v=27.1 m/s$$
Question 6
1 / -0

A particle tied to a string describes a vertical circular motion of radius r continually. If it has a velocity $$\sqrt{3gr}$$ at the highest point, then the ratio of the respective tensions in the string holding it at the highest and lowest points is

A
4 : 3

B
5 : 4

C
1 : 4

D
3 : 2

E
1 : 2

Solution

Tension at highest point
$$T_H=\dfrac{mv^2}{r}-mg=3mg-mg=2mg$$
and tension at lowest point
$$T_L=\dfrac{mv^2}{r}+mg$$
Here, $$V^2_L=3gr+2g.2r=7gr$$
So, $$T_L=7mg+mg=8mg$$
Hence, $$\dfrac{T_H}{T_L}=\dfrac{2mg}{8mg}=\dfrac{1}{4}$$
Question 7
1 / -0

A batsman deflects a ball by an angle of $$45^o$$ without changing its initial speed which is equal to $$54$$km/hr. Mass of the ball is $$0.15$$kg. What is the impulse imparted to the ball?

A
$$4.2$$kg m/s in the direction of the final velocity

B
$$4.2$$kg m/s in the direction of the initial velocity

C
$$4.2$$kg m/s in the direction opposite to the initial velocity

D
$$4.2$$kg m/s directed along the bisector of the initial and the final directions of the velocity

Solution

Initial momentum of ball $$ = mucos\theta$$ along NO
Final momentum of ball $$ = mucos\theta$$ along ON
Impulse = Change in momentum
=$$2 mu Cos\theta$$
On subsituitng values we get,
$$\therefore Impulse = 4.2 kg/s$$
Question 8
1 / -0

A pendulum string of length $$l$$ is moved upto a horizontal position and released as shown in figure. If the mass of pendulum is $$m$$, then what is the minimum strength of the string that can withstand the tension as the pendulum passes through the position of equilibrium?
(neglect mass of the string and air resistance)

A
$$mg$$

B
$$2mg$$

C
$$3mg$$

D
$$5mg$$

Solution

We have, $$\quad T-mg=\cfrac { m{ v }^{ 2 } }{ I } $$
$$\quad { v }^{ 2 }=2gI\Rightarrow T=mg+\cfrac { m2gI }{ I } =3mg$$
Question 9
1 / -0

Two blocks of masses $${m}_{1}$$ and $${m}_{2}$$ are connected by spring constant $$K$$. Initially the spring is at its natural length. The coefficient of friction between the bars and the surface is $$\mu$$. What minimum constant force has to be applied in the horizontal direction on the block of mass $${m}_{1}$$, in order to shift the other block?

A
$$F=\mu \left( 2{ m }_{ 1 }+\cfrac { { m }_{ 2 } }{ 2 } \right) g$$

B
$$F=\mu \left( 2{ m }_{ 1 }+{ m }_{ 2 } \right) g$$

C
$$F=\mu \left( { m }_{ 2 }+\cfrac { { m }_{ 1 } }{ 2 } \right) g\quad $$

D
$$F=\mu \left( { m }_{ 1 }+\cfrac { { m }_{ 2 } }{ 2 } \right) g$$

Solution

$$f_1 = \mu m_1 g$$

$$f_2 = \mu m_2 g$$

where, $$m_2$$ is fixed and $$F - f_1 = F - \mu m_1 g >0$$

Under the influence of the spring $$m_1$$ moves like $$a$$ damped oscillation.

Let at point A,

$$ F - f_1 = kx$$ ..............(1)

Net force is zero then $$m_1$$ have $$E$$. So that $$m_1$$ move upto length $$2x$$ from their normal position where A point is mean position.

Then when m_1 at point B. End of first half ;

$$f_2 = k(2x) = \mu m_2 g$$

$$ kx = \cfrac{\mu m_2}{2} g$$

Hence from eq. (1);

$$ F - f_1 = kx = \mu m_1 g + \mu \cfrac{m_2}{2} g = \mu (m_1 + \cfrac{m_2}{2})g$$
Question 10
1 / -0

Two people push a car for 3 sec, with a combined net force of 200 N.The impulse provided to the car............

A
400 N-sec.

B
500 N-sec.

C
600 N-sec.

D
300 N-sec.

Solution

Time of action $$t = 3 \ s$$
Force applied $$F = 200 \ N$$
Impulse $$I = F\times t = 200\times 3 = 600 \ N \ s$$

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Re-Attempt Weekly Quiz Competition

Laws of Motion Test - 46

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Laws of Motion Test - 46

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SHARING IS CARING

Question 1

friction

impulse

tension

weight

viscous drag

Solution

Question 2

$$0.98\times 10^{-3} Ns$$

$$1.5\times 10^{-3} Ns$$

$$2.0\times 10^{-3} Ns$$

$$2.5\times 10^{-3} Ns$$

Solution

Question 3

$$1.8m/{ s }^{ 2 }$$

$$2m/{ s }^{ 2 }$$

$$2.7m/{ s }^{ 2 }$$

$$3.8m/{ s }^{ 2 }$$

Solution

Question 4

$${ \left( R/g \right) }^{ 1/2 }$$

$$g/R$$

$$M/Rg$$

$${ \left( Rg \right) }^{ 1/2 }$$

Solution

Question 5

$$ 12 m/s$$

$$ 18 m/s$$

$$ 22 m/s$$

$$ 27 m/s$$

Solution

Question 6

4 : 3

5 : 4

1 : 4

3 : 2

1 : 2

Solution

Question 7

$$4.2$$kg m/s in the direction of the final velocity

$$4.2$$kg m/s in the direction of the initial velocity

$$4.2$$kg m/s in the direction opposite to the initial velocity

$$4.2$$kg m/s directed along the bisector of the initial and the final directions of the velocity

Solution

Question 8

$$mg$$

$$2mg$$

$$3mg$$

$$5mg$$

Solution

Question 9

$$F=\mu \left( 2{ m }_{ 1 }+\cfrac { { m }_{ 2 } }{ 2 } \right) g$$

$$F=\mu \left( 2{ m }_{ 1 }+{ m }_{ 2 } \right) g$$

$$F=\mu \left( { m }_{ 2 }+\cfrac { { m }_{ 1 } }{ 2 } \right) g\quad $$

$$F=\mu \left( { m }_{ 1 }+\cfrac { { m }_{ 2 } }{ 2 } \right) g$$

Solution

Question 10

400 N-sec.

500 N-sec.

600 N-sec.

300 N-sec.

Solution

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