Motion in a Plane — Page 5 | NEET Physics

Q41

If

\left| {\overrightarrow A \times \overrightarrow B } \right| = \sqrt 3 \overrightarrow A .\overrightarrow B

then the value of

\left| {\overrightarrow A + \overrightarrow B } \right|

is

A (A 2 + b 2 + AB) 1/2

B

{\left( {{A^2} + {B^2} + {{AB} \over {\sqrt 3 }}} \right)^{1/2}}

C A + B

D (A 2 + B 2 +

{\sqrt 3 }

AB) 1/2 .

Correct Answer

Option A

Solution

\left| {\overrightarrow A \times \overrightarrow B } \right| = AB\sin \theta

\overrightarrow A \overrightarrow B = AB\cos \theta

\left| {\overrightarrow A \times \overrightarrow B } \right| = \sqrt 3 \overrightarrow A \overrightarrow B

\Rightarrow AB\sin \theta = \sqrt 3 AB\cos \theta

\Rightarrow \tan \theta = \sqrt 3

\therefore \theta

= 60 o $\therefore$

\left| {\overrightarrow A \times \overrightarrow B } \right| = \sqrt {{A^2} + {B^2} + 2AB\cos 60^\circ }

= \sqrt {{A^2} + {B^2} + AB}

= (A 2 + b 2 + AB) 1/2

Q42

The vector sum of two forces is perpendicular to their vector differences. In that case, the forces

A are equal to each other

B are equal to each other in magnitude

C are not equal to each other in magnitude

D cannot be predicted.

Correct Answer

Option B

Solution

Given :

\left( {\overrightarrow {{F_1}} + \overrightarrow {{F_2}} } \right) \bot \left( {\overrightarrow {{F_1}} - \overrightarrow {{F_2}} } \right)

\left( {\overrightarrow {{F_1}} + \overrightarrow {{F_2}} } \right).\left( {\overrightarrow {{F_1}} - \overrightarrow {{F_2}} } \right) = 0

F_1^2 - F_2^2 - \overrightarrow {{F_1}} .\overrightarrow {{F_2}} + \overrightarrow {{F_2}} .\overrightarrow {{F_1}} = 0

\Rightarrow F_1^2 = F_2^2

(i.e. F 1 , F 2 are equal to each other in magnitude.)

Q43

A particle A is dropped from a height and another particle B is projected in horizontal direction with speed of 5/sec from the same height then correct statement is

A particle A will reach at ground first with respect to particle B

B particle B will reach at ground first with respect to particle A

C both particles will reach at ground simultaneously

D both particles will reach at ground with same speed.

Correct Answer

Option C

Solution

Time required to reach the ground is dependent on the vertical motion of the particle.

Vertical motion of both the particles A and B are exactly same.

Although particle B has an initial velocity, but that is in horizontal direction and it has no component in vertical (component of a vector at a direction of 90 o = 0) direction.

Hence they will reach the ground simultaneously.

Q44

If

\left| {\overrightarrow A + \overrightarrow B } \right| = \left| {\overrightarrow A } \right| + \left| {\overrightarrow B } \right|

then angle between A and B will be

A 90 o

B 120 o

C 0 o

D 60 o .

Correct Answer

Option C

Solution

\left| {\overrightarrow A + \overrightarrow B } \right| = \left| {\overrightarrow A } \right| + \left| {\overrightarrow B } \right|

if

\overrightarrow A \parallel \overrightarrow B

. $\therefore$ $\theta$ = 0 o

Q45

Two particles having mass M and m are moving in a circular path having radius R and r. If their time period are same then the ratio of angular velocity will be

A

{r \over R}

B

{R \over r}

C 1

D

\sqrt {{R \over r}}

Correct Answer

Option C

Solution

\omega = {{2\pi } \over t}

. As t is same $\therefore$

{{{\omega _1}} \over {{\omega _2}}} = 1

Q46

The width of river is 1 km. The velocity of boat is 5 km/hr. The boat covered the width of river in shortest time 15 min. Then the velocity of river stream is

A 3 km/hr

B 4 km/hr

C

\sqrt {29}

km/hr

D

\sqrt {41}

km/hr.

Correct Answer

Option A

Solution

{v_{{\mathop{\rm Resultant}\nolimits} }} = {{1\,km} \over {1/4\,hr}} = 4\,km/hr

$\therefore$

{v_{River}} = \sqrt {{5^2} - {4^2}} = 3\,km/hr

Q47

A man is slipping on a frictionless inclined plane and a bag falls down from the same height. Then the velocity of both is related as

A

{v_B} > {v_m}

B

{v_B} < {v_m}

C

{v_B} = {v_m}

D

{v_B}

and

{v_m}

can't be related.

Correct Answer

Option C

Solution

Vertical acceleration in both the cases is g, whereas horizontal velocity is constant.

Q48

Two projectiles of same mass and with same velocity are thrown at an angle 60 o and 30 o with the horizontal, then which will remain same

A time of flight

B range of projectile

C maximum height acquired

D all of them.

Correct Answer

Option B

Solution

Given, u 1 = u 2 = u,

\theta _1

= 60º,

\theta _2

= 30º In 1 st case, we know that range

{R_1} = {{{u^2}\sin 2\left( {60^\circ } \right)} \over g} = {{{u^2}\sin 120^\circ } \over g}

= {{{u^2}\sin \left( {90^\circ + 30^\circ } \right)} \over g}

= {{{u^2}\left( {\cos 30^\circ } \right)} \over g} = {{ {u^2}\sqrt 3} \over {2g}}

In 2 nd case, when

\theta _2

= 30° , then

{R_2} = {{{u^2}\sin 60^\circ } \over g} = {{{u^2}\sqrt 3 } \over {2g}}

\Rightarrow {R_1} = {R_2}

Q49

A stone tied to the end of a string of 1 m long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revoluations in 44 seconds, what is the magnitude and direction of acceleration of the stone ?

A

{\pi ^2}\,

m s

-

2 and direction along the radius towards the centre

B

{\pi ^2}\,

m s

-

2 and direction along the radius away from the centre

C

{\pi ^2}\,

m s

-

2 and direction along the tangent to the circle

D

{\pi ^2}\,

/4 ms

-

2 and direction along the radius towards the centre.

Correct Answer

Option A

Solution

{a_r} = {\omega ^2}R\,

&

\,{a_t} = {{dv} \over {dt}} = 0

or, ar= (2 $\pi$ n)2R = 4

\pi ^2

n 2 R 2 =

4{\pi ^2}{\left( {{{22} \over {44}}} \right)^2}{\left( 1 \right)^2}

a net = a r =

{\pi ^2}

ms –2 and direction along the radius towards the centre.