Scalar and Vector Notation
Trending Questions
Q. A force →F=2^i+3^j−^k acts at a point (2, −3, 1). Then, magnitude of the torque of this force about point (0, 0, 2) will be
- 6 units
- 3√5 units
- 6√5 units
- None of these
Q. Angular momentum of a body is defined as the product of
- mass and angular velocity
- centripetal force and radius
- linear velocity and angular velocity
- moment of inertia and angular velocity
Q. Calculate the net torque on the system about the point O as shown in figure if F1=11 N, F2=9 N, F3=10 N, a=10 cm and b=20 cm (All the forces are along the tangents)
- 2.2 Nm out of the plane
- 3.0 Nm into the plane
- 2.2 Nm into the plane
- 3.0 Nm out of the plane
Q. The classroom door is of width 50 cm. The force of 5 N is applied on the handle which is situated in the middle of the smaller plate as shown below. Compute the torque about the hinge (lever arm) of the door.
- 3.5 Nm
- 1.5 Nm
- 2.5 Nm
- 2.0 Nm
Q. Two forces F1=(2^i–5^j−6^k) N and F2=(−^i+2^j−^k) N are acting on a body at the points (1, 1, 0) and (0, 1, 2) respectively. Find resultant torque acting on the body about point (−1, 0, 1) in (N-m).
- 14^i−9^j+10^k
- −14^i+10^j−9^k
- −10^i+14^j−9^k
- −14^i+5^j−9^k
Q. The 20 cm diameter disc in the figure can rotate on the axle through its center. What is the net torque about O (in Nm)?
- 2.94
- 5.06
- 0.06
- 0.94
Q. A force →F=4^i−5^j+3^k is acting at point →r1=^i+2^j+3^k. The torque acting about the point →r2=3^i−2^j−3^k is
- Zero
- 42^i−30^j+6^k
- 42^i+30^j+6^k
- 42^i+30^j−6^k
Q.
Two forces, each of magnitude have a resultant of the same magnitude . The angle between the two forces is
Q. A force →F=3^i+2^j−4^k acts at the point (1, −1, 2). Find its torque about the point (2, −1, 3).
- 6^i−7^j+^k
- 2^i+7^j−2^k
- 6^i+7^j+^k
- 2^i−7^j−2^k
Q. If a rigid body is subjected to two forces −→F1=2^i+3^j+4^k acting at (3, 3, 4) and −→F2=−2^i−3^j−4^k acting at (1, 0, 0), then which of the following is true ?
- The body is in equilibrium.
- The body is under the influence of a torque only.
- The body is under the influence of a force as well as a torque.
- The body is under the influence of a single force.
Q.
As the perpendicular distance of the application of force increases, the torque decreases.
- True
- False
Q. For which of the following forces, the torque about the origin is zero, if force is applied at →r=8^i m?
- 16^i N
- 16^j N
- 8^k N
- 8^i+16^j N
Q. Two second after projection, a projectile is travelling in a direction inclined at 30∘ to the horizontal. After 1 more
Second, it is travelling horizontally. Then (g=10m/s2)
Second, it is travelling horizontally. Then (g=10m/s2)
- the velocity of projection is 20√3m/s
- the angle of projection is 30∘ with horizontal
- Both (A) and (B) are correct
- Both (A) and (B) are wrong
Q. A particle having mass m is projected with a speed v at an angle α with the horizontal ground. Find the torque of the weight of the particle about the point of projection when the particle reaches the ground.
- 2mv2sin2α
- mv2sin2α2
- mv2sin2α
- 0
Q. In the pulley system shown in the figure, the radii of the bigger and smaller pulleys are 2 m and 1 m respectively. Find the resultant torque at point O. (Take g=10 m/s2)
- 200 N-m
- 100 N-m
- 0 N-m
- 50 N-m
Q. A particle of mass 10 kg is projected with a velocity 5 m/s from a point on horizontal ground, making an angle 37∘ with the horizontal. Find the total torque about the point of projection acting on the particle when it is at its maximum height.
- 120 Nm along ^k direction
- 60 Nm along ^k direction
- 120 Nm along −^k direction
- 60 Nm along −^k direction
Q. A rod of length 20 cm is kept along x axis. Two forces 5 N and 10 N are applied at distances 10 cm and 15 cm from point A as shown in figure. Find the point of application of net force from point A.
- 20 cm
- 12.5 cm
- 15 cm
- 17.5 cm
Q. A force →F=(^i–^j+^k) N acts at point P. Find the torque due to the force →F about O.
- ^i+6^j+5^k Nm
- 5^i−6^j+^k Nm
- 5^i+6^j+^k Nm
- −^i−6^j−5^k Nm
Q. Derive an expression for maximum height and range of an object in projectile motion.
Q. In the figure shown below, find the torque due to the force of 5 N, about point O and A.
- 6 Nm along +z direction, 6 Nm along +z direction.
- 6 Nm along +z direction, 6 Nm along −z direction
- 6 Nm along −z direction, 6 Nm along −z direction
- 6 Nm along −z direction, 6 Nm along +z direction
Q. Force −→F1=(2^i–3^j) N and −→F2=(−5^i+2^j) N act through the points whose position vectors are ^k and 3^k respectively. The vector sum of the torque due to these forces is
- 3^i+13^j Nm
- −3^i−13^j Nm
- −13^i−3^j Nm
- −3^i+13^j Nm
Q. Two uniform rods of equal lengths but different masses are rigidly joined to form an L- shaped body, which is then pivoted about O as shown in the figure. Find the net torque about point O. Given M=m√3
- 0
- mgL(√34)
- mgl(2√3)
- mgL(√32)
Q. What is the torque of a force F=(2ˆi−3ˆj+4k)N acting at a point r=(3ˆi+2ˆj+3ˆk)m about the origin in N-m ? (Given τ=r×F )
- 6ˆi−6ˆj+12ˆk
- 17ˆi−6ˆj−13ˆk
- −6ˆi+6ˆj−2ˆk
- −17ˆi+6ˆj+13ˆk
Q. A projectile thrown with an initial speed u and the angle of projection 15o to the horizontal has a range R. If the same projectile is thrown at an angle of 45o to the horizontal with speed 2u, what will be its range?
- 7R1
- 10R1
- 8R1
- 9R1
Q. When air resistance can be neglected, the maximum range of a projectile is achieved for what launch angle? (in degrees)
- 90
- 45
- 30
- 60
Q. There is some change in length when a 33000N tensile force is applied on a steel rod of area of cross-section 10−3m2 . The change of temperature required to produce the same elongation of the steel rod when heated is
(Y=3×1011N/m2, α=1.1×10−5/0C)
(Y=3×1011N/m2, α=1.1×10−5/0C)
- 20∘C
- 15∘C
- 10∘C
- 0∘C
Q. In the figure shown below, find the torque due to the force of 5 N, about point O and A.
- 6 Nm along +z direction, 6 Nm along +z direction.
- 6 Nm along +z direction, 6 Nm along −z direction
- 6 Nm along −z direction, 6 Nm along −z direction
- 6 Nm along −z direction, 6 Nm along +z direction
Q. Two forces −→F1=2^i−5^j−6^k and −→F2=−^i+2^j−^k are acting on a body at a point (1, 1, 0) and (0, 1, 2) respectively. Find the torque acting on the body about the point (−1, 0, 1).
- −14^i+10^j−9^k
- −11^i+10^j−12^k
- −3^i+3^k
- −14^i−10^j−9^k
Q. An electric dipole placed with its axis in the direction of a uniform electric field experiences:
- a force but not torque
- a torque but no force
- a force as well as a torque
- neither a force nor a torque
Q. Assertion :If a particle moves with a constant velocity, the angular momentum of this particle about any point remains constant. Reason: Angular momentum has the units of Plank's constant.
- Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
- Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
- Assertion is correct but Reason is incorrect
- Both Assertion and Reason are incorrect