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Name: Inertia and force mcq question answers
Subject: Physics
Topic: Inertia and force
Questions: 10
Time Allowed: 10 minutes
Important for: 11th & 12th school students, Engineering & Medical entrance exams etc.
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 Question 1 of 10
1. Question
1 pointsIf a person sitting in an open jeep, moving with a constant velocity along a straight line, throws a ball vertically upwards, then the ball will
CorrectWe know that the horizontal velocity of the ball, after it is thrown vertically upwards, will remain equal to the velocity of the jeep due to its inertia.
Therefore when a ball is thrown vertically upwards, then the ball will return to the thrower’s hand.IncorrectWe know that the horizontal velocity of the ball, after it is thrown vertically upwards, will remain equal to the velocity of the jeep due to its inertia.
Therefore when a ball is thrown vertically upwards, then the ball will return to the thrower’s hand.UnattemptedWe know that the horizontal velocity of the ball, after it is thrown vertically upwards, will remain equal to the velocity of the jeep due to its inertia.
Therefore when a ball is thrown vertically upwards, then the ball will return to the thrower’s hand.  Question 2 of 10
2. Question
1 pointsWhen a bus suddenly takes a bend, the passengers are thrown in the outward direction, due to
CorrectWe know that when a bus suddenly takes a bend, the passengers tend to maintain their direction of motion due to inertia of motion.
As a result of this, the passengers are thrown in the outward direction when the bus suddenly takes a bend.IncorrectWe know that when a bus suddenly takes a bend, the passengers tend to maintain their direction of motion due to inertia of motion.
As a result of this, the passengers are thrown in the outward direction when the bus suddenly takes a bend.UnattemptedWe know that when a bus suddenly takes a bend, the passengers tend to maintain their direction of motion due to inertia of motion.
As a result of this, the passengers are thrown in the outward direction when the bus suddenly takes a bend.  Question 3 of 10
3. Question
1 pointsA body is moving with a constant velocity along a straight line path. A force is not required to
CorrectWe know that according to the Newton’s first law of motion that no force is required to keep a body in motion with uniform velocity along a straight line path.
IncorrectWe know that according to the Newton’s first law of motion that no force is required to keep a body in motion with uniform velocity along a straight line path.
UnattemptedWe know that according to the Newton’s first law of motion that no force is required to keep a body in motion with uniform velocity along a straight line path.
 Question 4 of 10
4. Question
1 pointsA ball is dropped from a spacecraft revolving around the earth at a height of 120 km. What will happen to the ball at that instant?
CorrectWe know that when a ball is dropped from the spacecraft, no external force is applied on it.
Therefore according to Newton’s first law of motion, the ball will continue to move with the same speed along the original orbit of the spacecraft.IncorrectWe know that when a ball is dropped from the spacecraft, no external force is applied on it.
Therefore according to Newton’s first law of motion, the ball will continue to move with the same speed along the original orbit of the spacecraft.UnattemptedWe know that when a ball is dropped from the spacecraft, no external force is applied on it.
Therefore according to Newton’s first law of motion, the ball will continue to move with the same speed along the original orbit of the spacecraft.  Question 5 of 10
5. Question
1 pointsMaximum and minimum values of the resultant of two forces acting at a point are 7N and 3N respectively. Then smaller force is equal to
CorrectGiven maximum resultant force = 7N and
Minimum resultant force = 3N
We know that maximum resultant force = F_{1} + F_{2} = 7N —— (1)
And minimum resultant force = F_{1} – F_{2} = 3N —— (2)
Solving equation (1) and (2), we get F_{1}= 5N and F_{2} = 2N (where F_{1} and F_{2} = Bigger and smaller forces respectively.)IncorrectGiven maximum resultant force = 7N and
Minimum resultant force = 3N
We know that maximum resultant force = F_{1} + F_{2} = 7N —— (1)
And minimum resultant force = F_{1} – F_{2} = 3N —— (2)
Solving equation (1) and (2), we get F_{1}= 5N and F_{2} = 2N (where F_{1} and F_{2} = Bigger and smaller forces respectively.)UnattemptedGiven maximum resultant force = 7N and
Minimum resultant force = 3N
We know that maximum resultant force = F_{1} + F_{2} = 7N —— (1)
And minimum resultant force = F_{1} – F_{2} = 3N —— (2)
Solving equation (1) and (2), we get F_{1}= 5N and F_{2} = 2N (where F_{1} and F_{2} = Bigger and smaller forces respectively.)  Question 6 of 10
6. Question
1 pointsThe vector sum of two forces is perpendicular to their vector differences. In that case, the forces
CorrectGiven: Angle between vector sum and vector difference of two forces = 90°
We know that vector sum of forces (S) = F_{1} + F_{2} and their vector difference (D) = F_{1} – F_{2}.
We also know that when these two vectors are perpendicular to each other, then
S.D = 0 or (F_{1} + F_{2}).( F_{1} – F_{2}) = 0 or (F_{1})^{2} + (F_{2})^{2} = 0 or F_{1}^{2} – F_{2}^{2} = 0
or F_{1}^{2} = F_{2}^{2} or F_{1} = F_{2}
Therefore these two forces are equal to each other in magnitude.IncorrectGiven: Angle between vector sum and vector difference of two forces = 90°
We know that vector sum of forces (S) = F_{1} + F_{2} and their vector difference (D) = F_{1} – F_{2}.
We also know that when these two vectors are perpendicular to each other, then
S.D = 0 or (F_{1} + F_{2}).( F_{1} – F_{2}) = 0 or (F_{1})^{2} + (F_{2})^{2} = 0 or F_{1}^{2} – F_{2}^{2} = 0
or F_{1}^{2} = F_{2}^{2} or F_{1} = F_{2}
Therefore these two forces are equal to each other in magnitude.UnattemptedGiven: Angle between vector sum and vector difference of two forces = 90°
We know that vector sum of forces (S) = F_{1} + F_{2} and their vector difference (D) = F_{1} – F_{2}.
We also know that when these two vectors are perpendicular to each other, then
S.D = 0 or (F_{1} + F_{2}).( F_{1} – F_{2}) = 0 or (F_{1})^{2} + (F_{2})^{2} = 0 or F_{1}^{2} – F_{2}^{2} = 0
or F_{1}^{2} = F_{2}^{2} or F_{1} = F_{2}
Therefore these two forces are equal to each other in magnitude.  Question 7 of 10
7. Question
1 pointsTwo concurrent equal forces of magnitude 5N each act at an angle 150°. The magnitude of their resultant is
CorrectGiven: First force (F_{1}) = 5N; Second force (F_{2}) = 5N and angle between them
(θ) = 150°.
We know that magnitude of the resultant (F) = √(F_{1}^{2} + F_{2}^{2} + 2 F_{1}F_{2} cosθ)
= √((5)^{2} + (5)^{2} + 2 × 5× 5 cos150°)
= √(25 + 25 + 50 × (0.866)
= √6.7
= 2.6 NIncorrectGiven: First force (F_{1}) = 5N; Second force (F_{2}) = 5N and angle between them
(θ) = 150°.
We know that magnitude of the resultant (F) = √(F_{1}^{2} + F_{2}^{2} + 2 F_{1}F_{2} cosθ)
= √((5)^{2} + (5)^{2} + 2 × 5× 5 cos150°)
= √(25 + 25 + 50 × (0.866)
= √6.7
= 2.6 NUnattemptedGiven: First force (F_{1}) = 5N; Second force (F_{2}) = 5N and angle between them
(θ) = 150°.
We know that magnitude of the resultant (F) = √(F_{1}^{2} + F_{2}^{2} + 2 F_{1}F_{2} cosθ)
= √((5)^{2} + (5)^{2} + 2 × 5× 5 cos150°)
= √(25 + 25 + 50 × (0.866)
= √6.7
= 2.6 N  Question 8 of 10
8. Question
1 pointsTwo equal forces of magnitude P each act at a point inclined to each other at an angle of 120°. The magnitude of their resultant is
CorrectGiven magnitude of first force (F_{1}) = P; Magnitude of second force (F_{2}) = P; and angle between them (θ) = 120°.
We know that magnitude of their resultant (F) = √(F_{1}^{2} + F_{2}^{2} + 2 F_{1}F_{2} cosθ)
= √((P)^{2} + (P)^{2} + 2 × P × P cos120°)
= √(P^{2} + P^{2} + 2 P^{2} × (0.5)
= √P^{2} = PIncorrectGiven magnitude of first force (F_{1}) = P; Magnitude of second force (F_{2}) = P; and angle between them (θ) = 120°.
We know that magnitude of their resultant (F) = √(F_{1}^{2} + F_{2}^{2} + 2 F_{1}F_{2} cosθ)
= √((P)^{2} + (P)^{2} + 2 × P × P cos120°)
= √(P^{2} + P^{2} + 2 P^{2} × (0.5)
= √P^{2} = PUnattemptedGiven magnitude of first force (F_{1}) = P; Magnitude of second force (F_{2}) = P; and angle between them (θ) = 120°.
We know that magnitude of their resultant (F) = √(F_{1}^{2} + F_{2}^{2} + 2 F_{1}F_{2} cosθ)
= √((P)^{2} + (P)^{2} + 2 × P × P cos120°)
= √(P^{2} + P^{2} + 2 P^{2} × (0.5)
= √P^{2} = P  Question 9 of 10
9. Question
1 pointsThe resultant of two forces, one double the other in magnitude, is perpendicular to the smaller of the two forces. The angle between the two forces is
CorrectGiven: First force (F_{1}) = P; Second force (F_{2}) = 2 and angle which the resultant marks with the smaller force (α) = 90°
We know that relation for the angle between the forces F_{1} and F_{2} (θ) is:
tan α = (F_{2} sinθ) / (F_{1} + F_{2} cos θ) or
tan 90° = (2P sinθ) / (P + 2P cos θ) or
∞ = (2P sinθ) / (P + 2P cos θ) or
(P + 2P cos θ) = (2P sinθ) / ∞ = 0
2P cos θ = – P or
cos θ = P / 2P = 1/2
Hence; θ = 120° (where θ = Angle between the two forces)IncorrectGiven: First force (F_{1}) = P; Second force (F_{2}) = 2 and angle which the resultant marks with the smaller force (α) = 90°
We know that relation for the angle between the forces F_{1} and F_{2} (θ) is:
tan α = (F_{2} sinθ) / (F_{1} + F_{2} cos θ) or
tan 90° = (2P sinθ) / (P + 2P cos θ) or
∞ = (2P sinθ) / (P + 2P cos θ) or
(P + 2P cos θ) = (2P sinθ) / ∞ = 0
2P cos θ = – P or
cos θ = P / 2P = 1/2
Hence; θ = 120° (where θ = Angle between the two forces)UnattemptedGiven: First force (F_{1}) = P; Second force (F_{2}) = 2 and angle which the resultant marks with the smaller force (α) = 90°
We know that relation for the angle between the forces F_{1} and F_{2} (θ) is:
tan α = (F_{2} sinθ) / (F_{1} + F_{2} cos θ) or
tan 90° = (2P sinθ) / (P + 2P cos θ) or
∞ = (2P sinθ) / (P + 2P cos θ) or
(P + 2P cos θ) = (2P sinθ) / ∞ = 0
2P cos θ = – P or
cos θ = P / 2P = 1/2
Hence; θ = 120° (where θ = Angle between the two forces)  Question 10 of 10
10. Question
1 pointsWhich of the following sets of concurrent forces may be in equilibrium?
CorrectWe know that three concurrent forces are in equilibrium, if they can be represented by the three sides of a triangle in magnitude and direction.
And for this condition, sum of magnitudes of any two smaller forces must be greater than that of the third force. This condition is satisfied by the set of forces F_{1} = 3N; F_{2} = 5N; F_{3} = 6N.IncorrectWe know that three concurrent forces are in equilibrium, if they can be represented by the three sides of a triangle in magnitude and direction.
And for this condition, sum of magnitudes of any two smaller forces must be greater than that of the third force. This condition is satisfied by the set of forces F_{1} = 3N; F_{2} = 5N; F_{3} = 6N.UnattemptedWe know that three concurrent forces are in equilibrium, if they can be represented by the three sides of a triangle in magnitude and direction.
And for this condition, sum of magnitudes of any two smaller forces must be greater than that of the third force. This condition is satisfied by the set of forces F_{1} = 3N; F_{2} = 5N; F_{3} = 6N.