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QJEEM21Pheb03: JEE Main 2021 Physics - Class 11 JEEM21Pheb03 Solution: A triangular plate is shown. A force = 4 3 is applied at point…

A triangular plate is shown. A force F→\overrightarrow F = 4 i^\widehat i −- 3 j^\widehat j 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 104 English

Multiple Choice Options

  • A −- 15 + 20 3\sqrt 3 , 15 + 20 3\sqrt 3
  • B 15 −- 20 3\sqrt 3 , 15 + 20 3\sqrt 3
  • C 15 + 20 3\sqrt 3 , 15 −- 20 3\sqrt 3
  • D −- 15 −- 20 3\sqrt 3 , 15 −- 20 3\sqrt 3

Step-by-Step Verified Solution

F→=4i^−3j^\overrightarrow F = 4\widehat i - 3\widehat j

r→1=5i^+53j^{\overrightarrow r _1} = 5\widehat i + 5\sqrt 3 \widehat j & r→2=−5i^+53j^{\overrightarrow r _2} = - 5\widehat i + 5\sqrt 3 \widehat j

Torque about 'O'

τ→O=r→1×F→=(−15−203)k^=(15+203)(−k^){\overrightarrow \tau _O} = {\overrightarrow r _1} \times \overrightarrow F = \left( { - 15 - 20\sqrt 3 } \right)\widehat k = \left( {15 + 20\sqrt 3 } \right)\left( { - \widehat k} \right)

Torque about 'Q'

τ→Q=r→2×F→=(−15+203)k^=(15−203)(−k^){\overrightarrow \tau _Q} = {\overrightarrow r _2} \times \overrightarrow F = \left( { - 15 + 20\sqrt 3 } \right)\widehat k = \left( {15 - 20\sqrt 3 } \right)\left( { - \widehat k} \right)

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