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Rotational Motion question

2021 · 17 Mar · Shift 1 · Q48
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  5. /2021 · 17 Mar · Shift 1 · Q48

Rotational Motion question

2021 · 17 Mar · Shift 1 · Q48

JEE MainPhysicsRotational MotionMCQ+4 / −1
A triangular plate is shown. A force F→\overrightarrow FF = 4 i^−\widehat i-i− 3 j^\widehat jj​ is applied at point P. The torque at point P with respect to point 'O' and 'Q' are : JEE Main 2021 (Online) 17th March Morning Shift Physics - Rotational Motion Question 118 English
  1. A
    −-− 15 + 20 3\sqrt 33​, 15 + 20 3\sqrt 33​
  2. B
    15 −-− 20 3\sqrt 33​, 15 + 20 3\sqrt 33​
  3. C
    15 + 20 3\sqrt 33​, 15 −-− 20 3\sqrt 33​
  4. D
    −-− 15 −-− 20 3\sqrt 33​, 15 −-− 20 3\sqrt 33​
View written solutionFree

Correct answer: D

To compute torque, we need the position vector from the reference point to the point of application PPP.

1. Torque formula

For a force F⃗=4i^−3j^\vec F = 4\hat i - 3\hat jF=4i^−3j^​ applied at point PPP,

the torque about a point is

τ⃗=r⃗×F⃗\vec \tau = \vec r \times \vec Fτ=r×F

where r⃗\vec rr is the position vector from the reference point to PPP.

In 2D,

τz=xFy−yFx\tau_z = xF_y - yF_xτz​=xFy​−yFx​

if r⃗=xi^+yj^\vec r = x\hat i + y\hat jr=xi^+yj^​ and F⃗=Fxi^+Fyj^\vec F = F_x\hat i + F_y\hat jF=Fx​i^+Fy​j^​.

Here,

Fx=4,Fy=−3F_x = 4, \qquad F_y = -3Fx​=4,Fy​=−3

so

τ=x(−3)−y(4)=−3x−4y\tau = x(-3) - y(4) = -3x - 4yτ=x(−3)−y(4)=−3x−4y

Thus, to find the torques about OOO and QQQ, we must know the coordinates of PPP relative to OOO and relative to QQQ.

2. Missing geometric figure

The problem states “A triangular plate is shown”, but the actual diagram is not provided here. Without the figure, the required position vectors OP→\overrightarrow{OP}OP and QP→\overrightarrow{QP}QP​ cannot be determined uniquely.

So the torques cannot be derived rigorously from the text alone.

3. Consistency check using options

Let the torque about OOO be τO\tau_OτO​ and about QQQ be τQ\tau_QτQ​.

The options are:

  • A: −15+203,  15+203-15+20\sqrt3,\; 15+20\sqrt3−15+203​,15+203​
  • B: 15−203,  15+20315-20\sqrt3,\; 15+20\sqrt315−203​,15+203​
  • C: 15+203,  15−20315+20\sqrt3,\; 15-20\sqrt315+203​,15−203​
  • D: −15−203,  15−203-15-20\sqrt3,\; 15-20\sqrt3−15−203​,15−203​

Since the stored correct answer is D, the claimed torques are

τO=−15−203,τQ=15−203\tau_O = -15 - 20\sqrt3, \qquad \tau_Q = 15 - 20\sqrt3τO​=−15−203​,τQ​=15−203​

These values are plausible for some triangle geometry, but without the diagram they cannot be independently verified.

4. Final conclusion

Because the essential figure is missing, a strict derivation is not possible from the provided information alone.

However, comparing with the stored answer, the marked option is:

D\boxed{\text{D}}D​

which corresponds to

τO=−15−203,τQ=15−203\boxed{\tau_O = -15-20\sqrt3, \qquad \tau_Q = 15-20\sqrt3}τO​=−15−203​,τQ​=15−203​​
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