Laws of Motion Test

Result

Laws of Motion Test - 80

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Question 1
1 / -0

Two particles A and B having mass m each and charge $$q_1$$ and $$-q_2$$ respectively, are connected at the ends of a non conducting flexible and inextensible string of the length l. The particle A is fixed and B is whirled along a vertical circle with centre at A. If a vertically upward electric field of strength E exists in the space, then for minimum velocity of particle B:

A
The tension force in string at lowest point is zero

B
The tension force at the lowest point is non-zero

C
The tension force at the highest point is zero

D
The work done by interaction force between particles A and B is non-zero

Solution
Question 2
1 / -0

Rocket propulsion theory is analogues to-

A
Newton`s third law

B
Relative theory

C
First law of thermodynamics

D
none

Solution
Question 3
1 / -0

Metal wire in the form of a ring of radius $$R$$ and mass m placed on a smooth horizontal table is set rotating about its own axis in such a way that each part of the ring moves with speed $$V$$.The tension in the ring Will be;-

A
$$ \dfrac {2\pi m v^2 }{R} $$

B
$$ \dfrac {mv^2}{4\pi R} $$

C
$$ \dfrac {mv^2}{R} $$

D
$$ 2 \pi m v^2 $$
Question 4
1 / -0

In which example is it not possible for the underlined body to be in equilibrium?

A
An aeroplane climbs at a steady rate.

B
An aeroplane tows a glider at a constant altitude.

C
A speedboat changes direction at a constant speed.

D
Two boats tow a ship into harbour.
Question 5
1 / -0

The angle of banking $$\theta$$ for a cyclist taking curve is given by $$tan\theta=\dfrac { { v }^{ n } }{ rg }$$, where symbols( v= speed of the cyclist, r= radius of the curved path, g= acceleration due to gravity) have their usual meanings.Then the value of $$n$$ is equal to

A
$$1$$

B
$$3$$

C
$$2$$

D
$$4$$

Solution

Mass of cyclist + cycle $$= M$$
radius of curvature of the curved path $$= R$$
ANgle of inclination from vertical$$ = \theta$$
acceleration due to gravity$$ = g$$
velocity of cyclist $$= v$$

Let the normal force : N from the ground on to cyclist. Balance forces in horizontal and vertical directions.

$$N sin \theta = \dfrac{M v^2}{ R}$$ = centripetal force for the cyclist to move in a curve of radius R

$$N cos \theta = M g $$ = weight balanced by normal reaction from ground

SO divide equ 1 by equ 2
$$tan \theta =\dfrac{ v^2}{ Rg} $$

Compairing with the equation
$$ tan \theta= \dfrac{ v^n}{ Rg} $$ as given in question
Required answer is $$n= 2$$
Question 6
1 / -0

A uniform bar of mass $$m$$ is supported by a pivot at its top about which the bar can swing like a pendulum. If a force $$F$$ is applied perpendicular to the lower end of the bar as shown in figure, what is the value of $$F$$ in order to hold the bar in equilibrium at an angle $$(\theta )$$ from the vertical

A
$$2mg\sin\theta$$

B
$$mg\sin\theta$$

C
$$mg\cos\theta$$

D
$$\dfrac{mg}{2}\sin \theta$$

E
$$\dfrac{mg}{2}\cos \theta$$

Solution

Let the length of the bar is, then according to the problem,
Refer image,
For equilibrium of the bar
$$F^{net}_{ext}=0$$ and $$\tau^{net}_{ext}=0$$

taking torque about pivot $$0$$,
$$OB\times Mg+OA\times F=0$$

$$\Rightarrow -OB\,Mg\sin\theta+OA.F\sin 90^0=0$$
(clockwise) (anti-clockwise)

$$\therefore 1\times F=\dfrac{1}{2}Mg\sin\theta$$

$$\Rightarrow F=\dfrac{Mg\sin\theta}{2}$$
Question 7
1 / -0

A heavy stone hanging from a massless string of length $$15$$m is projected horizontally with speed $$147$$ m/s. The speed of the particle at the point where the tension in the string equals the weight of the particle is?

A
$$10$$ m/s

B
$$7$$ m/s

C
$$12$$ m/s

D
None of these

Solution

Let at angle $$\theta$$ from vertical

speed be $$u$$ m/s and tension equal weight of particle i.e

$$\boxed{T=mg}$$.........(1) [given]

at this location, $$T-mg\cos\theta=\dfrac{mu^2}{l}$$

$$mg-mg\cos\theta=\dfrac{mu^2}{l}$$ [From (1)]

$$\boxed{lg(1-\cos\theta)=u^2}$$

by work energy theorem
$$W_{gravity}=\Delta K.E$$

$$-mgl(1-\cos\theta)=\dfrac{1}{2}mu^2-\dfrac{1}{2}mv^2$$

$$-2gl(1-\cos\theta)=u^2-v^2$$

$$-2u^2=u^2-v^2$$ [From (2)]

$$3u^2=v^2$$

$$u=\dfrac{v}{\sqrt{3}}=\dfrac{147}{\sqrt{3}}=85m/s$$

So answer is $$85m/s$$
Question 8
1 / -0

A particle originally at rest at the hoghest point of a smooth vertical circle is slightly displaced. it will leave the circle at a vertical distance h below the highest point, such that h is equal to

A
R

B
R/4

C
R/2

D
R/3

Solution
Question 9
1 / -0

A stone of mass $$0.3$$ kg attached to a $$1.5$$m long string is whirled around in a horizontal circle at a speed of $$6$$ m/s. The tension in the string is?

A
$$10$$N

B
$$20$$N

C
$$7.2$$N

D
None of these

Solution
Question 10
1 / -0

A block of mass 10 kg is suspended through two light spring balances as shown below

A
Both the scales will read 5 kg

B
The upper scale will read 10 kg & the lower zero

C
Both the scale will read 10 kg

D
The readings may be anything but their sum will be 10 kg

Solution