Motion In Two Dimension(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

MOTION IN TWO DIMENSION MCQs

Total Questions : 31 | Page 2 of 4 pages

Question 11. A metal sphere is hung by a string fixed to a wall. The sphere is pushed away from the wall by a stick. The forces acting on the sphere are shown in the second diagram. Which of the following statements is wrong

P = W tan θ
T= P+W
T2 = P2 + W2
none of these

Discuss Question

Answer: Option B. -> T= P+W
:
B

As the metal sphere is in equilibrium under the effect of three forces therefore

\to T

\to P

\to W

= 0
From the figure T cos

θ

=W ...(i)
T sin

θ

= P ..(ii)
From equation (i) and (ii) we get P= Wtan

θ

and

T^{2}

P^{2}

W^{2}

Question 12. The resultant of two vectors A and B is perpendicular to the vector A and its magnitude is equal to half the magnitude of vector B. The angle between A and B is

120∘
150∘
135∘
None of these

Discuss Question

Answer: Option B. -> 150∘
:
B

\frac{B}{2}

\sqrt{A^{2} + B^{2} + 2 A B c o s θ}

..(i)

∴

tan

90^{\circ}

\frac{B s i n θ}{A + B c o s θ}

\Rightarrow

A + Bcos

θ

= 0

∴

cos

θ

= -

\frac{A}{B}

Hence, from (i)

\frac{B^{2}}{4}

A^{2} + B^{2} - 2 A^{2}

\Rightarrow

A =

\sqrt{3} \frac{B}{2}

\Rightarrow

cos

θ

= -

\frac{A}{B}

= -

\frac{\sqrt{3}}{2}

∴

θ

150^{\circ}

Question 13. The maximum and minimum magnitudes of the resultant of two given vectors are 17 units and 7 units respectively. If these two vectors are at right angles to each other, the magnitude of their resultant is

Discuss Question

Answer: Option D. -> 13
:
D

R_{m a x}

= A + B = 17 when

θ

0^{\circ}

R_{m i n}

= A - B = 7 when

θ

180^{\circ}

by solving we get A = 12 and B = 5
Now when

θ

90^{\circ}

then R =

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

\Rightarrow

\sqrt{{(12)}^{2} + {(5)}^{2}}

\sqrt{169}

= 13

Question 14. The sum of the magnitudes of two forces acting at point is 18 and the magnitude of their resultant is 12. If the resultant is at

90^{\circ}

with the force of smaller magnitude, what are the, magnitudes of forces

12, 5
14, 4
5, 13
10, 8

Discuss Question

Answer: Option C. -> 5, 13
:
C
Let P be the smaller force and Q be the greater force then according to problem-
P + Q = 18 ..(i)
R =

\sqrt{P^{2} + Q^{2} + 2 P Q c o s θ}

= 12 ..(ii)
tan

ϕ

\frac{Q s i n θ}{P + Q c o s θ}

= tan 90 =

\infty

∴

P + Q cos

θ

= 0 ..(iii)
By solving (i), (ii) and (iii) we will get P = 5 , and Q = 13

Question 15. A scooter going due east at 10 m

s^{- 1}

turns right through an angle of

90^{\circ}

. If the speed of the scooter remains unchanged in taking turn, the change is the velocity of the scooter is

20.0 ms−1 south eastern direction
Zero
10.0 ms−1 in southern direction
14.14 ms−1 in south-west direction

Discuss Question

Answer: Option D. -> 14.14 ms−1 in south-west direction
:
D
If the magnitude of vector remains same, only direction change by

θ

then

⃗ △ v

{⃗ v}_{2}

{⃗ v}_{1}

⃗ △ v

{⃗ v}_{2}

+ (-

{⃗ v}_{1}

)
Magnitude of change in vector |

⃗ △ v

| = 2 v sin (

\frac{θ}{2}

)
|

⃗ △ v

| = 2

\times

\times

sin(

\frac{90^{\circ}}{2}

) = 10

\sqrt{2}

= 14.14 m/s
Direction is south - west as shown in figure.

Question 16. The torque of the force

\to F

= (2

^i

- 3

^j

+ 4

^k

) N acting at the point

\to r

= (3

^i

+ 2

^j

+ 3

^k

) m about the origin be

6^i - 6^j + 12^k
17^i - 6^j - 13^k
-6^i + 6^j - 12^k
-17^i + 6^j + 13^k

Discuss Question

Answer: Option B. -> 17^i - 6^j - 13^k
:
B

\to τ

\to τ

\times

\to F

⎡ ⎢⎢⎢ ⎢ \begin{matrix} ^i & ^j & ^k 3 & 2 & 3 2 & - 3 & 4 \end{matrix} ⎤ ⎥⎥⎥ ⎥

= [(2

\times

4)-(3

\times

-3)]

^i

+ (2

\times

3)-(3

\times

4)]

^j

+ (3

\times

-3)-(2

\times

2)]

^k

= 17

^i

- 6

^j

- 13

^k

Question 17. A stone of mass 1 kg tied to a light inextensible string of length is whirling in a circular path of radius L in a vertical plane. If the ratio of the maximum tension in the string to the minimum tension in the string is 4 and if g is taken to be 10 m/

s^{2}

, the speed of the stone at the highest point of the circle is

A Stone Of Mass 1 Kg Tied To A Light Inextensible String Of ...

20 m/sec
10√3 m/sec
5√2 m/sec
10 m/sec

Discuss Question

Answer: Option D. -> 10 m/sec
:
D
Since the maximum tension

T_{B}

in the string moving in the vertical circle is at the bottom and minimum tension

T_{T}

is at the top.

T_{B}

\frac{m {v_{B}}^{2}}{L}

= mg and

T_{T}

\frac{m {v_{T}}^{2}}{L}

- mg

∴

\frac{T_{B}}{T_{T}}

\frac{\frac{m {v_{B}}^{2}}{L} + m g}{\frac{m {v_{T}}^{2}}{L} - m g}

\frac{{v_{B}}^{2} + g L}{{v_{T}}^{2} + g L}

\frac{4}{1}

{v_{B}}^{2}

+ gL = 4

{v_{T}}^{2}

- 4gL but

{v_{B}}^{2}

{v_{T}}^{2}

+ 4gL

∴

{v_{T}}^{2}

+ 4gL +gL

\Rightarrow

{v_{T}}^{2}

= 3gL

∴

{v_{T}}^{2}

= 3

\times

\times

L = 3

\times

\times

\frac{10}{3}

v_{T}

= 10 m/sec

Question 18. If the sum of two unit vectors is a unit vector, then magnitude of difference is

√2
√3
1√2
√5

Discuss Question

Answer: Option B. -> √3
:
B
Let

{^n}_{1}

and

{^n}_{2}

are the two unit vectors, then the sum is

{⃗ n}_{s}

{^n}_{1}

{^n}_{2}

{n_{s}}^{2}

{n_{1}}^{2}

{n_{2}}^{2}

+ 2

n_{1} n_{2} c o s θ

= 1 + 1 + 2cos

θ

Since it is given that

n_{s}

is also a unit vector, therefore 1 = 1+ 1 + 2 cos

θ

\Rightarrow

cos

θ

= -

\frac{1}{2}

∴

θ

120^{\circ}

Now the difference vector is

{^n}_{d}

{^n}_{1}

{^n}_{2}

{n_{1}}^{2}

{n_{2}}^{2}

- 2

n_{1} n_{2} c o s θ

= 1 + 1 - 2 cos(

120^{\circ}

)

∴

{n_{d}}^{2}

= 2 - 2(-

\frac{1}{2}

) = 2 + 1 = 3

\Rightarrow

n_{d}

\sqrt{3}

Question 19. A man standing on a road holds his umbrella at

30^{\circ}

with the vertical to keep the rain away. He throws the umbrella and starts running at 10 km/hr. He finds that raindrops are hitting his head vertically, the speed of raindrops with respect to the road will be

10 km/hr
20 km/hr
30 km/hr
40 km/hr

Discuss Question

Answer: Option B. -> 20 km/hr
:
B
When the man is at rest w.r.t the the ground, the rain comes to him at an angle

30^{\circ}

with the vertical. This is the direction of the velocity
of raindrops with respect to the ground.
Here

{⃗ v}_{r g}

= velocity of rain with respect to the ground

{⃗ v}_{m g}

= velocity of the main with respect to the ground
and

{⃗ v}_{r m}

= velocity of the rain with respect to the man.
We have

{⃗ v}_{r g}

{⃗ v}_{r m}

{⃗ v}_{m g}

... (i)

{⃗ v}_{r g}

sin

30^{\circ}

{⃗ v}_{m g}

= 10 km/hr
or

{⃗ v}_{r g}

\frac{10}{s i n 30^{\circ}}

= 20 km/hr

Question 20. A bomber plane moves horizontally with a speed of 500 m/s and a bomb released from it, strikes the ground in 10 sec. Angle with the horizontal at which it strikes the ground will be (g=10 m/