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1.

The angle through which a cyclist bends when he covers a circular path of 34.3 m circumference in $\sqrt{22}$ sec is (g=9.8 m/s)

15

^{0}30

^{0}60

^{0}45

^{0}

D.

45^{0}

.Given:-Speed of particle V=$\frac{\mathrm{S}}{\mathrm{t}}$=$\frac{34.3}{\sqrt{22}}$

radius r=$\frac{\mathrm{S}}{2\mathrm{\pi}}$=$\frac{34.3}{2\mathrm{\pi}}$

tan$\mathrm{\theta}$=$\frac{{\mathrm{V}}^{2}}{\mathrm{rg}}=\frac{{\displaystyle \frac{(34.3{)}^{2}}{({\sqrt{22)}}^{2}}}}{{\displaystyle \frac{34.3}{2\mathrm{\pi}}\times 9.8}}$=$\frac{34.3\times 2\times {\displaystyle \frac{22}{7}}}{22\times 9.8}$=$\frac{68.6}{68.6}=1$

tan$\mathrm{\theta}$ = 1 .....(tan45^{0}=1)

2.

**Assertion:** In an elastic collision of two billiard balls, the total kinetic energy is conserved during the short time of oscillation of the balls (i.e. when they are in contact).

**Reason:** Energy spent against friction does not follow the law of conservation of energy.

If both the assertion and reason are true and reason is a correct explanation of the assertion.

If both assertion and reason are true but assertion is not a correct explanation of the assertion.

If the assertion is true but the reason is false.

If both assertion and reason are false.

D.

If both assertion and reason are false.

A billiard ball is a small, hard ball used in cue sports such as carom billiards, pool and snooker.

The billiard balls in an elastic collision are in a deformed state. And their total energy is in the form of P.E. and K.E. Thus K.E. is less than the total energy. The energy spent against friction is dissipated in the form of heat which is not available for doing work.

3.

**Assertion:** Work done in uniform circular motion is zero.

**Reason:** Force is always directed along displacement.

If both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true but reason is not the correct explanation of assertion.

If assertion is true but reason is false.

If assertion is false but reason is true.

C.

If assertion is true but reason is false.

Uniform motion is the object which is moving with a constant angular velocity. So the angular acceleration of the object is zero.

Displacement use only the force componeent along the object's displacement. The force component perpendicular to the displacement direction does zero work.

In uniform circular motion, the force is always directed perpendicular to the displacement.

4.

A car of mass 1000 kg moves on a circular track of radius 20 m. If the coefficient of friction of 0.64. The maximum velocity with which the car can be moved is

11.2m/s

112m/s

$\frac{0.64\times 20}{1000\times 100}\mathrm{m}/\mathrm{s}$

$\frac{1000}{64\times 20}\mathrm{m}/\mathrm{s}$

A.

11.2m/s

${\mathrm{V}}_{\mathrm{max}}=\sqrt{\mathrm{\mu rg}}=\sqrt{0.64\times 20\times 9.8}\phantom{\rule{0ex}{0ex}}\mathrm{where},\mathrm{\mu}-\mathrm{coefficient}\mathrm{of}\mathrm{friction}\phantom{\rule{0ex}{0ex}}{\mathrm{V}}_{\mathrm{max}}=11.2\mathrm{m}/\mathrm{s}$

5.

A particle is revolving in a circle of radius R. If the force acting on it is inversely proportional to R, then the time period is proportional to

R

R

^{2}1/R

1/R

^{2}

A.

R

Given that the force is inversly proportional to R.

F ∝ $\frac{1}{\mathrm{R}}$

⇒ F = $\frac{\mathrm{k}}{\mathrm{R}}$ ......(i)

According to centripetal force

F = mRω^{2}

⇒ mRω^{2} = $\frac{\mathrm{k}}{\mathrm{R}}$

⇒ ω^{2} = $\frac{\mathrm{k}}{\mathrm{m}{\mathrm{R}}^{2}}$

∴ ω^{2} ∝ $\frac{1}{{\mathrm{R}}^{2}}$

⇒ ω ∝ $\frac{1}{\mathrm{R}}$

⇒ $\frac{2\mathrm{\pi}}{\mathrm{T}}\propto \frac{1}{\mathrm{R}}$

where ω = $\frac{2\mathrm{\pi}}{\mathrm{T}}$

where ω is the angular frequency or angular speed

T is the time period

⇒ T ∝ R

6.

Three different objects m_{1 }, m_{2} and m_{3} to fall from rest and from the same point O along three different frictionless paths. The speeds of the three objects, on reaching the ground, will be in the ratio of

m

_{1}: m_{2}: m_{3}1 : 1 : 1

m

_{1}: 2 m_{2 }: 3m_{3}$\frac{1}{{\mathrm{m}}_{1}}:\frac{1}{{\mathrm{m}}_{2}}:\frac{1}{{\mathrm{m}}_{3}}$

B.

1 : 1 : 1

As the speed of an object, falling freely under gravity, depends only upon its height from which it is allowed to fall and not upon its mass. Since the paths are frictionless and all objects are falling through the same vertical height, therefore their speeds on reaching the ground must be same or ratio of their speeds = 1 : 1 : 1.

7.

**Assertion:** It is difficult to move a cycle along the road with its brakes on.

**Reason:** Sliding friction is greater than rolling friction.

If both the assertion and reason are true and reason is a correct explanation of the assertion.

If both assertion and reason are true but assertion is not a correct explanation of the assertion.

If the assertion is true but the reason is false.

If both assertion and reason are false.

A.

If both the assertion and reason are true and reason is a correct explanation of the assertion.

When brakes are on, there is no rolling of the wheels and the wheels slide. The sliding friction is greater than the rolling friction. Thus it is difficult to move a cycle along the road with its breaks on.

Rolling friction is the resistance to motion experienced by a body when it rolls upon another. It is much less than sliding friction for same pair of bodies. When one body rolls upon another, there is theoretically no sliding or slip between them. And if both are perfectly rigid, there is no surface of contact.

Above figure of rolling friction.

8.

A body is moving in a circular path with acceleration 'a'. If its velocity gets doubled, find the ratio of acceleration after and before the change

1:4

$\frac{1}{4}:1$

2:1

4:1

A.

1:4

D.

4:1

In a circular motion $\mathrm{\alpha}=\frac{{\mathrm{\nu}}^{2}}{\mathrm{r}}$

$\frac{{\mathrm{\alpha}}_{2}}{{\mathrm{\alpha}}_{1}}=\left(\frac{{\mathrm{\nu}}_{2}}{{\mathrm{\nu}}_{1}}\right)={\left(\frac{2{\mathrm{\nu}}_{1}}{{\mathrm{\nu}}_{1}}\right)}^{2}=4\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{\alpha}}_{2}:{\mathrm{\alpha}}_{1}=4:1$

9.

A body of mass 0.1 kg attains a velocity of 10 ms' in 0.1 sec. The force acting on the body is

10N

0.01N

0.1N

100N

A.

10N

$\mathrm{F}=\mathrm{ma}=\mathrm{m}\times \frac{\u2206\mathrm{V}}{\u2206\mathrm{t}}\phantom{\rule{0ex}{0ex}}=0.1\times \frac{10}{0.1}=10\mathrm{N}$

10.

If mass of an atom M moving with speed v, what will be its speed after the emission of an $\mathrm{\alpha}$-particle is zero?

$\frac{\mathrm{Mv}}{\mathrm{M}+2}$

$\frac{\mathrm{Mv}}{\mathrm{M}-4}$

$\frac{\mathrm{M}\mathrm{v}}{\mathrm{M}+4}$

$\frac{\mathrm{M}-4}{\mathrm{Mv}}$

B.

$\frac{\mathrm{Mv}}{\mathrm{M}-4}$

Applying the principle of momentum conservation,

Mv = ( M $-$ 4 ) v'

[ since the speed of the $\mathrm{\alpha}$-particle is zero ]

⇒ v' = $\frac{\mathrm{M}\mathrm{v}}{\mathrm{M}-4}$