12th Grade > Physics
VECTORS MCQs
Total Questions : 26
| Page 3 of 3 pages
Answer: Option A. -> The magnitude of the vector is doubled but its direction remains the same
:
A
Multiplication of a vector by a real positive number n makes its magnitude n times but does not change thedirection of the vector.
Hence the correct choice is (a).
:
A
Multiplication of a vector by a real positive number n makes its magnitude n times but does not change thedirection of the vector.
Hence the correct choice is (a).
Answer: Option B. -> 0 vector
:
B
0 vector. Watch the next video to refine this concept.
:
B
0 vector. Watch the next video to refine this concept.
Answer: Option B. -> Yes, Running slow might help sometimes.
:
B
Running slow can also work sometimes. Watch the next video to understand why.
:
B
Running slow can also work sometimes. Watch the next video to understand why.
Answer: Option A. -> 60∘
:
A
From the triangle law of vector addition, it is clear that the vectors R, A and B must be representedby the three sides of an equilateral triangle because the magnitudes of the three vectors are equal. Hencethe angle α between vectors R and A is 60∘. Alternatively, the value of αis given by
tanα=BsinθA+Bcosθ=sinθ1+cosθ(∵A=B)
=sin120∘1+cos120∘=1.732
Which gives α=60∘.
Hence the correct choice is (a)
:
A
From the triangle law of vector addition, it is clear that the vectors R, A and B must be representedby the three sides of an equilateral triangle because the magnitudes of the three vectors are equal. Hencethe angle α between vectors R and A is 60∘. Alternatively, the value of αis given by
tanα=BsinθA+Bcosθ=sinθ1+cosθ(∵A=B)
=sin120∘1+cos120∘=1.732
Which gives α=60∘.
Hence the correct choice is (a)
Answer: Option B. -> π2
:
B
The magnitude R of the resultant vector R of two vectors A and B inclined at an angle θ is given by
R2=A2+B2+2ABcosθ
cosθ=R2−A2−B22AB=(5)2−(3)2−(4)22×3×4=0
∴ θ=π2.
Hence the correct choice is (b)
:
B
The magnitude R of the resultant vector R of two vectors A and B inclined at an angle θ is given by
R2=A2+B2+2ABcosθ
cosθ=R2−A2−B22AB=(5)2−(3)2−(4)22×3×4=0
∴ θ=π2.
Hence the correct choice is (b)
Answer: Option C. -> There is no such thing as Negative Vector
:
C
We can assign any random direction to be positive and then all the vectors directing towards that direction will be represented with a positive sign and the vectors directing opposite to it will be accompanied with a negative sign. We define such signs for our convenience and there is no positive or negative vector in absolute terms.
:
C
We can assign any random direction to be positive and then all the vectors directing towards that direction will be represented with a positive sign and the vectors directing opposite to it will be accompanied with a negative sign. We define such signs for our convenience and there is no positive or negative vector in absolute terms.
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