Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Rotational Motion question

2022 · 25 Jun · Shift 1 · Q47
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Rotational Motion
  5. /2022 · 25 Jun · Shift 1 · Q47

Rotational Motion question

2022 · 25 Jun · Shift 1 · Q47

JEE MainPhysicsRotational MotionMCQ+4 / −1
If force F→=3i^+4j^−2k^\overrightarrow F = 3\widehat i + 4\widehat j - 2\widehat kF=3i+4j​−2k acts on a particle position vector 2i^+j^+2k^2\widehat i + \widehat j + 2\widehat k2i+j​+2k then, the torque about the origin will be :
  1. A
    3i^+4j^−2k^3\widehat i + 4\widehat j - 2\widehat k3i+4j​−2k
  2. B
    −10i^+10j^+5k^- 10\widehat i + 10\widehat j + 5\widehat k−10i+10j​+5k
  3. C
    10i^+5j^−10k^10\widehat i + 5\widehat j - 10\widehat k10i+5j​−10k
  4. D
    10i^+j^−5k^10\widehat i + \widehat j - 5\widehat k10i+j​−5k
View written solutionFree

Correct answer: B

  1. The torque about the origin is given by
τ⃗=r⃗×F⃗\vec\tau = \vec r \times \vec Fτ=r×F

where

r⃗=2i^+j^+2k^,F⃗=3i^+4j^−2k^\vec r = 2\hat i + \hat j + 2\hat k, \qquad \vec F = 3\hat i + 4\hat j - 2\hat kr=2i^+j^​+2k^,F=3i^+4j^​−2k^
  1. Compute the cross product using the determinant:
τ⃗=∣i^j^k^21234−2∣\vec\tau= \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 1 & 2 \\ 3 & 4 & -2 \end{vmatrix}τ=​i^23​j^​14​k^2−2​​
  1. Expand along the first row:
τ⃗=i^∣124−2∣−j^∣223−2∣+k^∣2134∣\vec\tau = \hat i \begin{vmatrix} 1 & 2 \\ 4 & -2 \end{vmatrix} - \hat j \begin{vmatrix} 2 & 2 \\ 3 & -2 \end{vmatrix} + \hat k \begin{vmatrix} 2 & 1 \\ 3 & 4 \end{vmatrix}τ=i^​14​2−2​​−j^​​23​2−2​​+k^​23​14​​
  1. Evaluate each minor:
  • For i^\hat ii^:
(1)(−2)−(2)(4)=−2−8=−10(1)(-2) - (2)(4) = -2 - 8 = -10(1)(−2)−(2)(4)=−2−8=−10
  • For j^\hat jj^​:
(2)(−2)−(2)(3)=−4−6=−10(2)(-2) - (2)(3) = -4 - 6 = -10(2)(−2)−(2)(3)=−4−6=−10

So,

−j^(−10)=+10j^-\hat j(-10) = +10\hat j−j^​(−10)=+10j^​
  • For k^\hat kk^:
(2)(4)−(1)(3)=8−3=5(2)(4) - (1)(3) = 8 - 3 = 5(2)(4)−(1)(3)=8−3=5
  1. Therefore,
τ⃗=−10i^+10j^+5k^\vec\tau = -10\hat i + 10\hat j + 5\hat kτ=−10i^+10j^​+5k^
  1. Comparing with the options, this matches:
B: −10i^+10j^+5k^\boxed{\text{B: } -10\hat i + 10\hat j + 5\hat k}B: −10i^+10j^​+5k^​
  1. Comparison with stored correct answer:

Stored correct answer is B, which matches the derived answer.

PreviousNext

More from Rotational Motion

  • Moment of Inertia (M.I.) of four bodies having same mass 'M' and radius '2R' are as follows: I1 = M.I. of solid sphere about its diameter I2 = M.I. of solid cylinder about its axis I3 = M.I. of solid circular disc about its diameter I4 =…2022 · Numerical
  • A disc of mass 1 kg and radius R is free to rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A body of same mass as that of disc is fixed at the highest point of the…2022 · Numerical
  • The radius of gyration of a cylindrical rod about an axis of rotation perpendicular to its length and passing through the center will be ​m. Given, the length of the rod is 103​ m.2022 · Numerical
  • A thin circular ring of mass M and radius R is rotating with a constant angular velocity 2 rads − 1 in a horizontal plane about an axis vertical to its plane and passing through the center of the ring. If two objects each of mass m be…2022 · MCQ
  • A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is2022 · MCQ
  • A pulley of radius 1.5 m is rotated about its axis by a force F=(12t−3t2)N applied tangentially (while t is measured in seconds). If moment of inertia of the pulley about its axis of…2022 · Numerical
  • A solid cylinder length is suspended symmetrically through two massless strings, as shown in the figure. The distance from the initial rest position, the cylinder should be unbinding the strings to achieve a speed of 4 ms−1,… Includes diagram2022 · Numerical
  • One end of a massless spring of spring constant k and natural length l0 is fixed while the other end is connected to a small object of mass m lying on a frictionless table. The spring remains horizontal on the table. If the object is made…2022 · MCQ