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Wondering what kind of questions that can be asked in next exam? Here you go

Question.

Which one of the following statement is false?

A vector has only magnitude, whereas a scalar has both magnitude and direction

Distance is a scalar quantity but displacement is a vector quantity

Momentum, force, torque are vector quantities

Mass, speed and energy are scalar quantities

Answer

Correct option is

A

vector

has only

magnitude,

whereas a

scalar has

both magnitude

and direction

Vector defined with a magnitude

as well as direction.
Exp:- Displacement, Velocity,
Momentum, Force, Torque.
Scalar have only magnitude
but not direction.
Exp:- Mass, Distance, speed
and energy
Option A

Question.

Of the vectors given below, the parallel vectors are

A = 6 \hat{i} + 8 \hat{j}

B = 210 \hat{i} + 280 \hat{k}

C = 5.1 \hat{i} + 6.8 \hat{j}

D = 3.6 \hat{i} + 8 \hat{j} + 48 \hat{k}

A

and

B

A

and

C

A

and

D

C

and

D

Answer

Correct option is

$A$ and

$C$

Vectors A and C are parallel to

each other due to the ratio of

both's i and j components are

same.
Also their unit vector is same i.e.
unit vector of
A is $\frac{6 i ^ + 8 j ^}{6 ^{2} + 8 ^{2}}$ $= \frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}$
Unit vector of
C is $\frac{5 . 1 i ^ + 6 . 8 j ^}{5 . 1 ^{2} + 6 . 8 ^{2}}$ $= \frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}$

Question.

The resultant of two forces P and Q is of magnitude P. If P be doubled, the resultant will be Inclined to Q at an angle

0^{o}

3 0^{o}

6 0^{o}

9 0^{o}

Answer

Correct option is

$9 0^{o}$

$R = P = [P^{2} + Q^{2} + 2 P Q c o s θ]^{1 / 2}$
$Q^{2} + 2 P Q c o s θ = 0$
$Q + 2 P c o s θ = 0$ ......{1}
let the angle between resultant of 2P
and Q be $α$
$t a n α = \frac{2 P s i n θ}{Q + 2 P c o s θ}$
from equation 1
$t a n α = \infty \to α = 90$

Question.

A large number of particles are moving towards each other with velocity

V

having directions of motion randomly distributed. What is the average relative velocity between any two particles averaged over all the pairs?

4 V / π

4 π V

V

π V / 4

Answer

Correct option is

$4 V / π$

$\overset{ˉ}{V}_{r e l} = \overset{ˉ}{V_{1}} - \overset{ˉ}{V_{2}}$
$∣ \overset{ˉ}{V}_{r e l} ∣ = V_{1}^{2} + V_{2}^{2} - 2 V_{1} V_{2} c o s θ$
$= 2 V^{2} - 2 V^{2} c o s θ$
$= 2 V 1 - c o s θ$
$= 2 V (2 s i n \frac{θ}{2}) = 2 V s i n \frac{θ}{2}$
$< V_{r e l} > = \frac{\int _{0}^{2 π} V _{r e l} d θ}{\int _{0}^{2 π} d θ}$
$= \frac{\int _{0}^{2 π} 2 V s i n \frac{θ}{2} d θ}{\int _{0}^{2 π} d θ}$
$= \frac{4 V [ - c o s \frac{θ}{2} ] _{0}^{2 π}}{2 π}$
$= \frac{2 V}{π} [- c o s (π) + c o s 0]$
$= \frac{4 V}{π}$

Video solution link below-

Consider a collection of a large number of particles each with speed v in a plane.The direction of velocity is randomly distributed in the collection.The magnitude of the average relative velocity of a particle with velocities of all other particles is (doubtnut.com)

Question.

A body is projected with a velocity

u

at an angle of

6 0^{o}

to the horizontal. The time after which it will be moving in a direction of

3 0^{o}

to the horizontal is

\frac{1}{3} \frac{u}{g}

\frac{3 u}{g}

\frac{3 u}{2 g}

\frac{2 u}{3 g}

Answer

Correct option is

$\frac{1}{3} \frac{u}{g}$

$u (t) = u_{x} \hat{i} + u_{y} \hat{j}$
$u (t) = (u cos θ) \hat{i} + (u sin θ - g t) \hat{j} (θ = 6 0^{0})$
Now, for this to make $3 0^{0}$ with horizontal.
$tan 3 0^{0} = \frac{u _{y}}{u _{x}} = \frac{u sin 6 0 ^{0} - g t}{u cos 6 0 ^{0}}$

$\Rightarrow \frac{1}{3} = \frac{\frac{3}{2} u - g t}{\frac{4}{2}} \Rightarrow u (\frac{3}{2} - \frac{1}{2 3}) = g t$

$\Rightarrow ∣ t ∣ = ∣ ∣ ∣ ∣ ∣ \frac{u}{g} (+ \frac{1}{3}) ∣ ∣ ∣ ∣ ∣$

$\Rightarrow t = \frac{u}{g} \frac{1}{3}$

Question.

A golfer standing on level ground hits a ball with a velocity of

u = 52 m / s

at an angle

α

above the horizontal. If

tan α = 5 / 12

, then the time for which the ball is at least 15 m above the ground (i.e. between A and B) will be (take

g = 10 m / s^{2}

)

1 s e c

2 s e c

3 s e c

4 s e c

Answer

Correct option is

$2 s e c$

$u = 52 m / s$
$tan x = \frac{5}{1 2}$

Initial vertical
Velocity $u_{y} = u sin α$
$n = 15$ $n = u_{y} t = \frac{1}{2} 9 t^{2}$

we get $15 = (52 sin α) t - 5 t^{2}$
$5 t^{2} - (52 sin α) t + 15 = 0$ ...(1)

It will give two values of $t$ i.e. $t_{1} & t_{2}$
$t_{1} < t_{2}, t_{1} =$ first time when it goes to $n = 15 m$ i.e. Point $A$
$t_{2}$ again $(a t B)$
So we need $t_{2} - t_{1} = (t_{2} + t_{1})^{2} - 4 t_{1} t_{2}$

Sum of roots $= t_{2} + t_{2} = \frac{5 2 sin α}{5}$

$t_{1} t_{2} = \frac{+ 1 5}{5} = 3$

So we get $t_{2} - t_{2} = \frac{( 5 2 ) ^{2} sin ^{2} α}{2 5} - 12$

as $tan α = \frac{5}{1 2} ⟹ sin α = \frac{5}{1 3}$

$⟹ t_{2} - t_{1} = \frac{( 5 2 ) ^{2}}{1 3 ^{2}} - 12 = 4 = 2$

Question.

For angles of projection of a projectile at angles

(4 5^{\circ} - θ)

and

(4 5^{\circ} + θ)

, the horizontal ranges described by the projectile are in the ratio of:

1 : 1

2 : 3

1 : 2

2 : 1

Answer

Question.

A large number of bullets are fired in all directions with the same speed v. What is the maximum area on the ground on which these bullets will cover:

π \frac{v ^{2}}{g}

π \frac{v ^{4}}{g ^{2}}

π^{2} \frac{v ^{4}}{g ^{2}}

π^{2} \frac{v ^{2}}{g ^{2}}

Answer

Question.

A point size body is moving along a circle at an angular velocity of

2.8 r a d s^{- 1}

. If the centripetal acceleration of the body is

7 m s^{- 2}

, then its speed is:

1.25 m s^{- 1}

2.5 m s^{- 1}

3.5 m s^{- 1}

7 m s^{- 1}

Answer

Question.

The length of a seconds hand in a watch is

1

cm. The change in its velocity in

15

s is

0 c m / s

\frac{π}{3 0 2} c m / s

\frac{π}{3 0} c m / s

\frac{π}{3 0} 2 c m / s

Answer

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CLASS 11 | CHAPTER 4 | Motion In A Plane-Types Of Questions | PHYSICS | NEET

Wondering what kind of questions that can be asked in next exam? Here you go

A

vector

has only

magnitude,

whereas a

scalar has

both magnitude

and direction

Vector defined with a magnitude

as well as direction.
Exp:- Displacement, Velocity,
Momentum, Force, Torque.
Scalar have only magnitude
but not direction.
Exp:- Mass, Distance, speed
and energy
Option A

$A$ and

$C$

Vectors A and C are parallel to

each other due to the ratio of

both's i and j components are

same.
Also their unit vector is same i.e.
unit vector of
A is $\frac{6 i ^ + 8 j ^}{6 ^{2} + 8 ^{2}}$ $= \frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}$
Unit vector of
C is $\frac{5 . 1 i ^ + 6 . 8 j ^}{5 . 1 ^{2} + 6 . 8 ^{2}}$ $= \frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}$

$9 0^{o}$

$R = P = [P^{2} + Q^{2} + 2 P Q c o s θ]^{1 / 2}$
$Q^{2} + 2 P Q c o s θ = 0$
$Q + 2 P c o s θ = 0$ ......{1}
let the angle between resultant of 2P
and Q be $α$
$t a n α = \frac{2 P s i n θ}{Q + 2 P c o s θ}$
from equation 1
$t a n α = \infty \to α = 90$

$4 V / π$

$\frac{1}{3} \frac{u}{g}$

$2 s e c$

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CLASS 11 | CHAPTER 4 | Motion In A Plane-Types Of Questions | PHYSICS | NEET

Wondering what kind of questions that can be asked in next exam? Here you go

A

vector

has only

magnitude,

whereas a

scalar has

both magnitude

and direction

Vector defined with a magnitude

as well as direction. Exp:- Displacement, Velocity, Momentum, Force, Torque.Scalar have only magnitude but not direction.Exp:- Mass, Distance, speed and energyOption A

A and

C

Vectors A and C are parallel to

each other due to the ratio of

both's i and j components are

same. Also their unit vector is same i.e. unit vector of A is 62+82​6i^+8j^​​=53​i^+54​j^​ Unit vector of C is 5.12+6.82​5.1i^+6.8j^​​=53​i^+54​j^​

90o

R=P=[P2+Q2+2PQcosθ]1/2Q2+2PQcosθ=0Q+2Pcosθ=0......{1}let the angle between resultant of 2P and Q be αtanα=Q+2Pcosθ2Psinθ​from equation 1tanα=∞→α=90

4V/π

3​1​gu​

u(t)=ux​i^+uy​j^​u(t)=(ucosθ)i^+(usinθ−gt)j^​ (θ=600)Now, for this to make 300 with horizontal.tan300=ux​uy​​=ucos600usin600−gt​⇒3​1​=24​23​​u−gt​⇒u(23​​−23​1​)=gt⇒∣t∣=∣∣∣∣∣​gu​(+3​1​)∣∣∣∣∣​⇒t=gu​3​1​

2sec

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as well as direction.
Exp:- Displacement, Velocity,
Momentum, Force, Torque.
Scalar have only magnitude
but not direction.
Exp:- Mass, Distance, speed
and energy
Option A

$A$ and

$C$

same.
Also their unit vector is same i.e.
unit vector of
A is $\frac{6 i ^ + 8 j ^}{6 ^{2} + 8 ^{2}}$ $= \frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}$
Unit vector of
C is $\frac{5 . 1 i ^ + 6 . 8 j ^}{5 . 1 ^{2} + 6 . 8 ^{2}}$ $= \frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}$

$9 0^{o}$

$R = P = [P^{2} + Q^{2} + 2 P Q c o s θ]^{1 / 2}$
$Q^{2} + 2 P Q c o s θ = 0$
$Q + 2 P c o s θ = 0$ ......{1}
let the angle between resultant of 2P
and Q be $α$
$t a n α = \frac{2 P s i n θ}{Q + 2 P c o s θ}$
from equation 1
$t a n α = \infty \to α = 90$

$4 V / π$

$\frac{1}{3} \frac{u}{g}$

$2 s e c$